mg.metric geometry - Malfatti Circles - Limiting point

"Three circles packed inside a triangle such that each is tangent to the other two and to two sides of the triangle are known as Malfatti circles" (for a brief historical account on this topic, see here and here on MathWorld).

Consider the triangle formed by the centers of these circles, one can draw a new set of smaller Malfatti circles in this triangle. What is the limiting point of this process?

One thing sort of discouraging is that I tried on an isosceles triangle, unfortunately did not find the limiting point matching any of the known relevant points (e.g., incenter or the first Ajima-Malfatti point).

arithmetic geometry - Torsion of an abelian variety under reduction.

This is Hartshorne, Exercise IV.4.19: Let $mathcal{A}/mathbb{Z} setminus S =: T$ be an Abelian scheme. The multiplication by $n$ morphism is flat [I don't know how to show this, but I think it can be found in Katz-Mazur.], so the $n$-Torsion $mathcal{A}[n] to mathcal{A}$ is also flat as it is a base change and $mathcal{A} to T$ also since it is flat. It is also proper and quasi-finite, and therefore finite. So we have a finite flat group scheme. Because of $(n,p) = 1$ $mathcal{A}[n]$ is étale over $mathbb{Z}_{(p)}$ (how to show this?). We have for $X/S$ finite étale

Consider the reduction map $mathcal{A}[n]_eta(mathbf{Q}) = mathcal{A}[n](T) to mathcal{A}[n]_p(mathbf{F}_p)$ for $(n,p) = 1$, confer Liu, Chapter 10.1.3. Liu, Proposition 10.1.40(b) gives us one-point fibres of the reduction map.

oc.optimization control - Term to describe how much harder an optimization problem can become after constraining a small part of the domain?

This is a follow up to this question.

I'm interested in discrete optimization problems formulated as 0-1 integer programs; essentially, anything of the form

$$Phi = max_{mathbf{x} in left{0,1right} ^N} f(mathbf{x})$$

I'm looking for terminology or references that describe a property of these problems related to how much easier or more difficult they are to optimize as we constrain or expand the domain. This is motivated by looking at search procedures that successively assign values to a subset of variables, and I'd like to understand problem characteristics that make it so assigning more variables in the search tree will make the constrained combinatorial problem easier or harder.

It's not always the case that assigning variables makes the problem easier.

Example: suppose we have an optimization version of the subset-sum problem: find a subset of integers $S = {x_1, ldots, x_n}$ whose sum is as large as possible but not larger than $t$. If one of the elements of $S$ is $t$, then the problem is trivial. As a less extreme example, if one of the elements of $S$, say $x_i$, is $t - k$ and another element, say $x_j$, is $k$, the problem is also trivial. However, if we branch in the search tree on $x_i$ not being used (or any other $x_{i'}$ being used where $i' not = i, i' not = j$), then we get a potentially hard optimization problem.

Conversely, Jonah raised the point in the previous question comments that if you think about this in the other direction--of trying to expand the domain to make a hard problem become easy, then

it's pretty unlikely that you can
trivialize the problem of finding a
perfect matching in a graph by adding
just one edge. I haven't read all that
much about this, but I'd even go so
far as to guess that you need O(n)
edges in general to make this easy (where n is the number of vertices).

Is there a term to describe this property--how much easier or more difficult can you make a class of problem by constraining $k$ variable values? Are there other cases where constraining variable values makes a problem harder (other than more examples of constraining a variable eliminating a trivial solution)?

I realize that "harder" and "trivial" need to be defined better. I'd also be curious about ideas on that.

Analytic density of the set of primes starting with 1

I think instead of posting my own explanation (which will only lose something in the translation) I'll instead refer you to two very interesting papers (thanks for posting this question, I haven't thought about this stuff in a couple years, and these papers were interesting reads to solve your problem.)

The first (among other things) proves that the density of primes with leading coefficient $k$ is $log_{10}left(frac{k +1}{k}right).$

Prime numbers and the first digit phenomenon
by
Daniel I. A. Cohen* and Talbot M. Katz
in Journal of number theory 18, 261-268 (1984)

The second is a more general statement about first digits. It is

The first digit problem
by
Ralph Raimi
in American Math Monthly vol 83 No 7

Hope this all helps.

ca.analysis and odes - Inverse gamma function?

This is an analysis question I remember thinking about in high school. Reading some of the other topics here reminded me of this, and I'd like to hear other people's solutions to this.

We have the gamma function, which has a fairly elementary form as we all know,

$Gamma(z) = int_0^infty e^{-t} t^{z-1} dt = int_0^1 left[ ln(t^{-1}) right]^{z-1}$

Which satisfies of course, $Gamma(n) = (n-1)!$, $nin mathbb{N}$, and the various recurrence relations and other identities that we can all look up on wikipedia or mathwolrd or wherever. We note that the gamma function is increasing on the interval $[a,infty]$ where $aapprox 1.46163$.

The question is--can we come up with an explicit inverse function to the gamma function on this interval which looks similarly simple?

My techniques at the time were to write down a differential equation that the inverse would satisfy, and solve it, which I could do in terms of a power series expansion (being in high school, ignoring the issues of convergence) to get an approximate solution. But I was never able to get a very nice looking or exact solution. I have a few more sophisticated tricks now to do this, but I would be interested to see how people with more experience with these kinds of questions would go about answering this.

The gamma function also satisfies a reasonable number of somewhat interesting looking functional relations like $Gamma(z)Gamma(1-z)=pi/sin(pi z)$. Does the inverse function satisfy any similar relations?

linear algebra - On permutation of elements of two bases of a vector space (Greub´s book)

Let {a₁,a₂,...,a_n} and {b₁,b₂,...,b_n} be two bases for a vector space E. Fix p, 1 ≤ p ≤n. Is there a permutation σ such that
{a₁,a₂,...,a_p,b_σ(p+1),...,b_σ(n)} and {b_σ(1),b_σ(2),...,,b_σ(p),a_p+1,...,a_n} are both bases of E?

This question is the last exercise of the first chapter in the book Linear Algebra by Greub. I can prove the case p=n-1.

oa.operator algebras - Does equality of the operator norm and the cb norm for every bimodule map over a C*-subalgebra imply that the subalgebra is matricially norming?

Dear Jonas,
My colleague Bill Johnson has drawn this to my attention.
Your question has a positive answer after noting that a $C$-bimodule map lifts to a $Cotimes M_n$-bimodule map on $Aotimes M_n$ If $X$ is an $ntimes n$ matrix over $A$ of norm 1 then for any row and column contractions $R$, $C$ over $C$ we have

$$|phi_n(X)|=sup{|Rphi_n(X)C|: |R|,|C|leq 1} =sup{|phi(RXC)|: |R|,|C|leq 1} leq |phi|$$
showing that $|phi_n|leq |phi|. Of course, the reverse inequality holds and we have proved that the norm and cb-norm coincide.
Best of luck with your studies,
Roger Smith

Sorry, I (mis)read your question rather quickly so the answer above is not an answer at all. Will think about it.

st.statistics - Distribution of the sum of the $m$ smallest values in a sample of size $n$

This is following on from Douglas Zare's answer. While not an answer in its own right,it was getting too long for a comment. Briefly, we can hit the question with brute force and look for a generating function for the desired expected values.

So, put

$$f_j(y) = sum_{k=0}^{j-1} {n choose k} y^k (1-y)^{n-k}$$
so that using the notation of Douglas' answer, $g_j(x)=f_j(F(x))$, and the expectation of the $j$th order statistic is
$$E_j := int_0^infty f_j(F(x)) dx$$

Put $G(y,z) = sum_{j=1}^n f_j(y) z^j$ where $z$ is a formal variable. By linearity we have

$$sum_{j=1}^n E_j z^j = int_0^infty G(F(x),z) dx$$

We can try to write $G$ as a rational function in $y$ and $z$.
Expanding out and interchanging the order of summation gives

$$eqalign{ G(y,z) & = sum_{j=1}^nsum_{k=0}^{j-1} {nchoose k} y^k (1-y)^{n-k} z^j \\ & = sum_{k=0}^{n-1} sum_{j=k+1}^n z^j {nchoose k} y^k (1-y)^{n-k} \\ & = sum_{k=0}^{n-1} left( sum_{j=1}^{n-k} z^j right) cdot z^k{nchoose k} y^k (1-y)^{n-k} \\ }$$

Now

$$eqalign{ sum_{j=1}^{M-1}jz^j = z frac{d}{dz}sum_{j=0}^{M-1} z^j & = z frac{d}{dz}left[frac{1-z^M}{1-z}right] \\ & = frac{z-z^{M+1}}{(1-z)^2} - frac{Mz^M}{1-z} & = frac{z-z^M}{(1-z)^2} - frac{(M-1)z^M}{1-z} }$$
so substituting this back in we get
$$eqalign{ G(y,z) & = sum_{k=0}^{n-1} left[ frac{z-z^{n-k+1}}{(1-z)^2} - frac{(n-k)z^{n-k+1}}{1-z} right] cdot z^k{nchoose k} y^k (1-y)^{n-k} \\ & = sum_{k=0}^{n-1} frac{z}{(1-z)^2}{nchoose k} (yz)^k (1-y)^{n-k} \\ & - sum_{k=0}^{n-1} frac{z^{n+1}}{(1-z)^2} {nchoose k} y^k (1-y)^{n-k} \\ & - sum_{k=0}^{n-1} frac{nz^{n+1}}{1-z} { {n-1} choose k } y^k(1-y)^{n-k} \\ & = frac{z}{(1-z)^2} left[ (1-y+yz)^n - (yz)^n right] - frac{z^{n+1}}{(1-z)^2} (1-y^n) - frac{nz^{n+1}}{1-z} }$$

I guess that in theory one could plug this back in to obtain a "formula" for the generating function $sum_{j=1}^n E_j z^j$, but I can't see how that formula might then simplify to something calculable, unless $F$ has a rather special form.

ac.commutative algebra - Is every integral epimorphism of commutative rings surjective?

If I'm not mistaken, there is a counter-example. Have a look at Lazard's second counter-example in:
"Deux mechants contre-exemples" in Séminaire Samuel, Algèbre commutative, 2, 1967-1968.

For any field $k$, Lazard provides a non-surjective epimorphism of local $k$-algebras $Cto D$, both of Krull dimension zero, and both of residue field equal to $k$. It is then easy to show that $D$ is also integral over $C$, which is what we need here. Indeed, every $din D$ can be written as $d=a+b$ with $ain k$ and $b$ in the maximal ideal (and unique prime) of $D$, which is therefore nilpotent $b^n=0$, hence trivially integral. Since $ain k$ is also in $C$, our $d$ is the sum of two integral elements. (Or simply, $D$ is a $k$-algebra, hence a $C$-algebra, generated by nilpotent, hence integral, elements.)

In cash, for those who don't want to click, the rings are constructed as follows:
Consider the local ring in countably many pairs of variables $S=(k[X_i,Y_i]_{igeq 0})_M$ localized at $M=langle X_i,Y_irangle_{igeq0}$.
For every $igeq 0$ choose an integer $p(i) > 2^{i-1}$. Define $J=langle Y_i-X_{i+1} Y_{i+1}^2 , X_i^{p(i)}rangle_{igeq0}subset S$ and define $D=S/J$. Note immediately that $D$ is a local $k$-algebra, say with maximal ideal $m$ and with residue field $D/mcong S/Mcong k$. Finally, he defines $C$ to be the localization (at $C_0cap m$) of the subalgebra $C_0:=k[x_i,x_iy_i]_{igeq 0}subset D$ where the $x_i$ are the classes of the $X_i$ in $D$ and I let you guess what the $y_i$ are. By construction, the residue field of $C$ is an extension of $k$ which is also a subfield of $D/m=k$, so the residue field of $C$ must be $k$ and we are in the announced situation.

pr.probability - Existence of probability measure defined on all subsets

Here is some elaboration on Joel David Hamkins's answer.

A (two-valued) measurable cardinal is an uncountable cardinal $kappa$ such that there is a continuous $({<kappa})$-additive ${0,1}$-valued probability measure $mu$ defined on all subsets of $kappa$. To say that $mu$ is $({<kappa})$-additive means that if $(X_i)_{i in I}$ is a family of pairwise disjoint subsets of $kappa$ with $|I|<kappa$, then $muleft(bigcup_{i in I} X_iright) = sum_{iin I} mu(X_i)$. Since $mu$ can only take values $0$ and $1$, this is equivalent to saying that (a) at most one of the $X_i$ can have measure $1$, and (b) if they all have measure $0$ then so does $bigcup_{i in I} X_i$. (Some people say $kappa$-additive instead of $({<kappa})$-additive, but I prefer to use $kappa$-additive to mean the above for families with index set of size equal to $kappa$.)

The existence of a continuous countably additive ${0,1}$-valued probability measure $mu$ defined on all subsets of a set $S$ implies the existence of a measurable cardinal. Indeed, I claim that if $kappa geq aleph_1$ is the smallest cardinal such that $mu$ is not $kappa$-additive, then $kappa$ is a measurable cardinal. To see this, let $(X_i)_{i<kappa}$ be a family of pairwise disjoint measure $0$ sets such that $bigcup_{i<kappa} X_i$ has measure $1$ (i.e. the family contradicts $kappa$-additivity in the only possible way). Defining $bar{mu}(I) = muleft(bigcup_{i in I} X_iright)$ for every $I subseteq kappa$, we obtain a $({<kappa})$-additive ${0,1}$-valued probability measure $barmu$ defined on all subsets of $kappa$.

So the existence of a continuous countably additive ${0,1}$-valued measure defined on all subsets of a set $S$ is exactly equivalent to the existence of a measurable cardinal. However, since a probability measure is also allowed to take values strictly between $0$ and $1$, this is not quite equivalent to the statement you asked about.

By analogy with the above, a real-valued measurable cardinal is an uncountable cardinal $kappa$ such that there is a continuous $({<kappa})$-additive probability measure defined on all subsets of $kappa$. The existence of a real-valued measurable cardinal is equivalent to your statement by a variation of the trick used above.

In 1930, Ulam (Zur Masstheorie in der allgemeinen Mengenlehre, Fund. Math. 16) showed that if $kappa$ is real-valued measurable then $kappa geq 2^{aleph_0}$, and that if $kappa > 2^{aleph_0}$ is real-valued measurable then $kappa$ is in fact measurable (with a possibly different measure). Ulam also showed that successor cardinals like $aleph_1$ cannot be real-valued measurable.

In the 1960's, Solovay (MR290961) finally resolved the boundary case. He showed that if $kappa = 2^{aleph_0}$ is real-valued measurable then there is an inner model (namely $L[I]$ where $I$ is the ideal of null sets) wherein $kappa$ is still real-valued measurable and GCH holds, therefore $kappa$ is measurable in that inner model by Ulam's earlier results. While this doesn't mean that the existence of a real-valued measurable cardinal and the existence of a measurable cardinal are equivalent, it shows that the two statements are equiconsistent over ZFC.

Using forcing (and another result of Ulam), Solovay also showed that if there is a model with a measurable cardinal then there is a model in which the Lebesgue measure on $[0,1]$ can be extended to a probability measure defined on all subsets of $[0,1]$.

co.combinatorics - Checking if two graphs have the same universal cover

It's possible I just haven't thought hard enough about this, but I've been working at it off and on for a day or two and getting nowhere.

You can define a notion of "covering graph" in graph theory, analogous to covering spaces. (Actually I think there's some sense -- maybe in differential topology -- in which the notions agree exactly, but that's not the question.) Anyway, it behaves like a covering space -- it lifts paths and so on.

There's also a "universal cover," which I think satisfies the same universal property as topological universal covers but I'm not sure. Universal covers are acyclic (simply connected) in graph theory, so they're trees, usually infinite. The universal cover doesn't determine the graph; for instance, any two k-regular graphs (k > 1) have the same universal cover. You can construct plenty of other pairs, too.

I'm interested in necessary and sufficient conditions for two graphs $G, H$ to have the same universal cover. One such condition (I'm pretty sure!) is whether you can give a 1-1 correspondence between trails in $G$ and trails in $H$ that preserves degree sequences. Unfortunately this doesn't help me much, since this is still basically an infinite condition. Is there some less-obvious but more easily checkable condition? In particular is it possible to determine if two (finite) graphs have the same universal cover in polynomial time?

at.algebraic topology - How should I visualise RP^n?

I find it easiest to get $mathbb RP^n$ from $S^n$ with two cells in every dimension. That is to say, you start with two points for $S^0$, attach two one-cells oriented opposite ways for $S^1$, attach one 2-cell on top and one on the bottom for $S^2$, and so on until you have $S^n$. Then the antipodal map actually switches the two cells in each dimension, so it defines a free $mathbb Z/2$ action on $S^n$. If you quotient out by that action, then you get a cell structure on $mathbb RP^n$ with one cell in each dimension, and you have a universal double cover ($n$>$1$) for it.

I like this construction for a couple of reasons. First, it displays the (co)boundary maps in cellular (co)homology: the two $k$-cells share the same boundary (up to sign $(-1)^k$) which is the two $k-1$-cells. Because it's a covering space you can sort of see how after quotienting you get one $k$-cell whose boundary folds twice onto the $k-1$-cell. So that also shows you why the maps alternate between the 0 map and the multiplication by 2, or are all the 0 map in $mathbb Z/2$ coefficients.

Also, this generalizes to a construction of $S^infty$ and thus $mathbb RP^infty$, with $mathbb RP^n$ inside it just by taking the $n$-skeleton. It looks like you could modify Qiaochu's definition to account for the infinite case by taking a $k$-cell to be the points that vanish after the first $k+1$ dimensions and are scaled to 1 in the $k+1$st.

human biology - Why do sneezes come in pairs or more?

The very simple answer is that the causal factor or trigger hasn't disappeared after the first sneeze. Sneezing is a reflex, partially autonomous, to clear the nasal cavity of particles that don't belong there. As long as the trigger is not removed, the reflex is repeated.

This is also one of the reasons that allergies that cause sneezing, such as hay fever, continue as long as the factor causing the allergy is present.

oc.optimization control - Minimizing a function containing an integral

It looks like a mixed-integer dynamic optimization problem. Your problem can be rewritten as follows: (notice the transformation of the integral into a differential equation? It's a standard trick. Also, note that you need an initial condition for $Y$)

$min_{x(t)} L(T)$

s.t.
$frac{dL(t)}{dt} = AR(t)-x(t)$, with $L(0) = 0$

$frac{dR(t)}{dt} = ax(t)R(t)Y(t) - bR(t)$, with $R(0) = R_{0}$

$frac{dY(t)}{dt}=−x(t)R(t)Y(t)$, with $Y(0) = Y_{0}$

$x(t) = delta(t) x_{min} + (1 - delta(t)) x_{max}$ where $delta(t) in {0,1}$

To solve this problem numerically, simply discretize the differential equations using backward Euler (easy), or implicit Runge Kutta (harder, but more accurate). Pose this as a Mixed Integer Nonlinear Program (MINLP) and use one of these solvers to find the solution.

Bonmin
https://projects.coin-or.org/Bonmin

Couenne
https://projects.coin-or.org/Couenne

These solvers will traverse the branch-and-bound tree more intelligently and efficiently than your method of enumerating every single case, which will grow with the no. of discretization grid points you have. (e.g. let's say you discretize over 20 points; the no. of cases you have to search is $2^{20} = 1048576$. Not nice.)

With a branch-and-bound|cut|reduce MINLP solver (and a bit of luck), on average you are unlikely to hit the worst case scenario where every single case is enumerated.

There are other ways of solving this problem -- multiple-shooting, sequential dynamic optimization, etc. In my opinion, optimal control methods (Pontryagin's maximum principle) are typically intractable on problems like this.

mg.metric geometry - Can different bicycles leave the same tracks?

The answer, I believe, is no in the case of finite length tracks and yes in the case of infinite length tracks. I will show this in the (probably not very) simplified case of assuming that we can idenitify a back wheel track and a front wheel track from the tracks left.

Finite length:
We will show that if they have the same back wheel path then the lengths of their front wheel paths are different.
Let $gamma$ be the arclength parameterization of the back wheel path. Then, for a bicycle of length $l$, the front wheel path is given by $gamma + lcdotgamma'$ and so we can calculate its length by:

$int |gamma' + lcdotgamma''|$

Since we chose the arclength parameterization, $gamma'$ and $gamma''$ are orthogonal so this becomes

$int sqrt{|gamma'| + l^2cdot|gamma''|^2}$

And clearly this quantity is increasing with $l$ unless $gamma''$ is identically zero, which only occurs for a straight line. Thus, two different lengths of bicycle will leave front wheel tracks of different lengths, which therefore cannot be the same (note - I just realized this only says that the distance traveled by the two front wheels is different, but due to overlap perhaps the tracks could still have the same length. I think we should be able to discount this possibility, but it is not immediately obvious to me exactly how).

Infinite length:
We will provide a general construction. Start by fixing two bikes of different lengths, call them bike 1 and bike 2 of lengths $l_1$ and $l_2$. Pick some finite length curve, call it $gamma_1$ as the first back wheel curve, and let $Gamma_1$ be the corresponding front wheel curve for bike 1. Then we will take $gamma_2$ to be $gamma_1$ followed by a backwheel curve starting at the end of $gamma_1$ that if ridden by bike 2 will end up with the front wheel of bike 2 tracing over all of $Gamma_1$ (of course it will also trace over much more). Then $Gamma_2$ will be the front wheel curve associated to bike 2 with back wheel curve $gamma_2$. Then we will let $gamma_3$ be $gamma_2$ followed by a backwheel curve starting at the end of $gamma_2$ that if ridden by bike 1 will end up with the front wheel of bike 1 tracing over all of $Gamma_2$ and $Gamma_3$ will be the front wheel curve associated to bike 1 with back wheel curve $gamma_3$. We repeat this process, alternating back and forth between bikes each time adding to the backwheel path what we need to cover what the previous bike covered with its front wheel tracks. This algorithm can be used to produce an infinite length backwheel path that will give the same frontwheel paths when ridden by either bike (we can consider the front wheel path abstractly as just the extension along the tangent lines of the appropriate bicycle length)

For an interesting account of bicycle curves (including some exposition on the model I am implicitly using), I recommend the paper by Levi and Tabachnikov, "On bicycle tire tracks geometry, hatchet planimeter, Menzin's conjecture and oscillation of unicycle tracks" available at http://arxiv.org/abs/0801.4396

Noether-Lefschetz locus in enumerative geometry.

It is well known that if you have a smooth quartic surface $Xsubset mathbb{P}^3$, it may or may not have lines in it. Indeed, $X$ has the following options, 64 (the maximal number), 32, 16, or none.

Between the space of all such quartics in $mathbb{P}^3$ which is $|mathcal{O}_{mathbb{P}^3}(4)|=mathbb{P}^{34}$, those with at least one line in it form a divisor $D$. Therefore, intersections of such a divisor with curves in $mathbb{P}^{34}$ may give enumerative information about points(quartics) in the intersection.

Is all the enumerative information about such quartics encoded in $D$?

I'd like to know more about such a divisor $D$. So, Could someone recommend free-references of this online?

oc.optimization control - Recommendations for a large scale bounded variable least squares (BVLS) solver for sparse matrices

This is such a well-solved problem that there are many software packages that have built in functions for this.

Here are a selection of built-in functions in different software packages that can be used:

In Matlab: lsqlin (type help lsqlin into Matlab and it tells you exactly what to type. I have just (approximately) solved your problem with random sparse matrices and it works great.)

KNITRO for Mathematica this package also solves this exact problem but I don't have this software so I can't tell you which exact function.

For a free solver I have found this: http://sourceforge.net/projects/quadprog/
However it assumes that $A$ has full column rank. This is just because this algorithm uses the dual problem which exists when the Hessian $A^TA$ is positive definite.

exactness in triangulated categories is reflected by hom-functor

let $T$ be a triangulated category and $A to B to C to A[1]$ a triangle in $T$ such that for every $A_0 in T$ the induced long sequence

$... to hom(A_0,A) to hom(A_0,B) to hom(A_0,C) to hom(A_0,A[1]) to hom(A_0,B[1]) to ...$

is exact. is then $A to B to C to A[1]$ exact? I'm a beginner, so this could be rather trivial. I've checked it in special cases, but I can't translate the proof into $T$. I would like the result because it would imply that the homological algebra you know for abelian groups takes over to triangulated categories. comparable to the fact that you can work with group objects in arbitrary categories as with ordinary groups. you can do it directly with the axioms, but this is a big mess.

in many answers I've seen here so far, even standard facts are enriched with wonderful, ample insights. so you are also invited to dish durt about these basics of triangulated categories because I'm just beginning to learn them.

immunology - Do we have enough diversity of antibodies to overcome almost any antigen?

As my answer is quite big I've highlighted the different regions I answer your question.

Intro and basic info

I take it you know how antibodies are made, but as a quick recap they're made by VDJ recombination of a limited but still quite numerous pool of genes which are recombined randomly, and then further recombination occurs by hypermutation of certain segments particularly in the regions that identify antigens.

What is Hypermutation?

Hypermutation occurs after the original antibody is made in order to perfect it. The same mechanism I talk about in moulding antigens below. It is exactly what it says: a high rate of mutation compared to that of the low rate in cells normally. When B cells are activated because they've met their antigen to which they respond (cognate antigen), they undergo somatic hypermutation. While the B cells undergo somatic hypermutation, the genes which code for the antibody are mutated so that the antibody binds to its cognate antigen with a different specificity. If it binds better the B cell gets to live, if it doesn't bind or it is worse then the B cell dies.

Moulding antigens

What we cannot do is the other way around. We can't recognise an antigen and then make an antibody to it, because recognising the antigen requires an antibody. What we can do is once we've recognised the antigen, then mould the antibody so it is a better fit. How we do this is by B cells mutating their antibodies and competing which binds better. However, B cells just don't have the knowledge to see an antigen and think a tyrosine residue at this position will be perfect to change my shape to this configuration - that's really complex stuff!

Do antibodies recognise every molecule?

Antibodies are made to essentially every antigen which is very costly in terms of energy as you mentioned. In fact, the number of genes that make antibodies are thought to be the perfect amount as to protect us without being too numerous so that it is not only too costly, but we have so many different antibodies the chance of ever finding the right B cell is so low we die before we manage. This is because of the way antibodies are mass produced, in the typical pathway of new antigens, firstly a T cell must recognise the antigen then B cells that produce an antibody that binds to the antigen (rarely in the same region, and may not even be to a protein antigen).

So these two paragraphs highlight the limitations and answer your first question. Antibodies are so diverse that we can expect quite a few to recognise every pathogen, but not every single possible antigen will have a cognate antibody. Diversity without so much diversity that it becomes redundant. The second limitation is that antibodies will ideally not form if they recognise self (our own proteins etc.) Which leads on to your next question.

What if a pathogen makes antigens that look like our own proteins?

Autoantibodies are made but are typically non-pathogenic (i.e., they don't cause harm). Certain HLA/MHC polymorphisms that make us likely to form autoreactive T cells, also make us make autoreactive antibodies. Interestingly, this increases our risk of autoimmune diabetes but ALSO makes us more resistant to infection by HIV. This can be practically seen in the DQ8 polymorphism. This is a perfect example to illustrate the answer to your second question. Ideally, pathogens which have antigens similar to our own would not be detected. HIV wraps itself with an envelope of our own lipid membrane for this purpose, and the best neutralising antibodies are often self-reactive and therefore are those that failed elimination.

However even these pathogens are going to do something or make something that isn't something our cells usually do. This is how the body finds and targets cancer cells (in a way), they do things our own cells won't typically do like make the wrong proteins or wrong amounts etc. Thus cancer cells and pathogens can be targeted by our cells that find something that isn't self (CD8 cytotoxic/killer T cells). As for pathogens, we all make MHC/HLA which is unique to us and pathogens could never manage to invade and make the specific HLA/MHC that we do, so our body also targets missing HLA/MHC.

zoology - How do sharks and other fish conserve and gain water in marine environments?

The mechanisms of osmoregulation is different for sharks (and other elasmobranch fishes) and teleost fishes.

In Elasmobranchs the body osmolarity is maintained equal to the seawater by Na⁺ Cl⁻ and urea. Toxicity because of high concentrations of urea (strong chaotrope) is counteracted by high levels of trimethylamine oxide (TMAO). So, the elasmobranchs do not need to drink water to prevent dehydration. But sometimes they do drink water for adapting to different salinity levels [1].

Teleost fishes, however, need to drink water in order to prevent dehydration. Initially Na⁺ and Cl⁻ are absorbed in the esophagus via cotransporters (Na-Cl/Na-K-Cl), thereby reducing the osmolarity of water. The cells that absorb Na⁺ maintain its low concentration by pumping extra Na⁺ in the Extracellular fluid, using Na⁺/K⁺-ATPase. Cl⁻ is also exchanged with Bicarbonate (HCO3⁻) via anion exchangers [2].

.

ct.category theory - What categorical mathematical structure(s) best describe the space of "localized events" in "relational quantum mechanics"?

In a recent (and to me, very beautiful) paper, entitled "Relational EPR",

Smerlak and Rovelli present a way of thinking about EPR which relies upon Rovelli's previously published work on relational quantum mechanics (see http://arxiv.org/abs/quant-ph/9609002 ). In relational quantum mechanics, there is no non-locality, but the definition of when an event occurs is weakened from Einstein's strict definition and instead is localized to each observer-measurement apparatus, including subsequent observers. There are (informal) coherence assumptions to ensure the consistency of reports from different subsequent observers (all possible friends of Wigner).

All of this seems very similar to various results in modern categorical mathematics. Is there a standard mathematical structure which well describes the structure of the space of localized measurements which Rovelli has envisioned? I know of Isham's work on topos theory and quantum mechanics, but I think he is aiming at something a little different.

PS I first asked this on mathunderflow, but was advised to repost here.

soft question - Which mathematical ideas have done most to change history?

I'm planning a course for the general public with the general theme of "Mathematical ideas that have changed history" and I would welcome people's opinions on this topic. What do you think have been the most influential mathematical ideas in terms of what has influenced science/history or changed the way humans think, and why?

I won't expect my audience to have any mathematical background other than high-school.

My thoughts so far are: non-Euclidean geometry, Cantor's ideas on uncountability, undecidability, chaos theory and fractals, the invention of new number systems (i.e. negative numbers, zero, irrational, imaginary numbers), calculus, graphs and networks, probability theory, Bayesian statistics.

My apologies if this has already been discussed in another post.

molecular biology - Does bleach destroy RNAse activity, and if so, how does it do it?

I am working with RNA samples, and I'm trying to be very careful about RNAse contamination.

I have some questions about bleach, though. Some people say that a solution of bleach is enough to destroy RNAse activity, but is it really? How does it do it?

I'm not sure if wiping the surface using a solution of bleach would be enough to kill RNAse activity, or if it is necessary to use products such as RNAseZap or RNAseAway.

human biology - Adipocyte Density

Is the volume of an adipocyte proportional to the mass of fat that it stores?

The answer is: yes.

Since fat is stored in a lipid droplet within the adipocyte, it is already in the densest form possible. In other words there is no way to make the stored fat more compact, so that a given mass of stored fat takes up less volume.

Response to comment from Chris:

I think you are perhaps being a little pedantic here. Anyway the graph below conveys my view, together with your refinement. So yes, adipocyte volume changes in proportion to mass of fat stored. (The line on my graph should be at 45 degrees, but I drew it by hand so it may be a little out.) There is no way to escape the increase in volume when fat is stored.

evolution - Genetic Diversity and Adaptation

Great question! A lot of things affect how quickly a population or species can adapt to a new environment, including population size, mutation rate, generation time, standing genetic diversity, and selective pressure.

The diversity of life encompasses practically all combinations of those variables. A bacterial population might very well contain enough diversity to allow a portion of the population to overcome a rapid change. In fact, applying antibiotics to a bacterial population and counting the survivors is a common measure of mutation rates.

On the other hand, organisms with small populations and long generation times will be much less likely to overcome a rapid environmental change. This is why there is so much concern over anthropogenic changes to the environment, including climate change.

It's hard to provide a definitive answer, since "rapid change" is a relative term, and the difficulty of adapting isn't known. Some striking adaptations can be caused by a single base pair mutation, such as in beetles and Monarch butterflies that are insensitive to toxic plant compounds.

It's important to point out that natural selection is based on relative fitness. Therefore, an adaptive mutation will spread because it's likely that carriers will be more fit than all of their neighbors. This does not necessarily mean that individuals without the mutation will die or fail to reproduce, only that those with the mutation will do it better.

Likewise, adaptation is not necessarily caused by the environment changing and wiping out all but a few lucky mutants like in the bacteria example. Instead, the environmental change might make things more difficult, but as long as the population can persist, then mutations will continue to enter the population which could confer a selective advantage against the change. So, no, a population does not (cannot) carry all possible adaptations. A population cannot adapt to an environment it hasn't encountered.

Finally, it's worth pointing out that a species can expand its range through migration. A new, unsuitable environment can act as a migration sink (that is, migrants make it there but fail to establish) for a nearby population. If this happens for long enough, some of the migrants may have a mutation allowing them to establish in the new environment.

botany - Do plant-plant interactions inhibit growth?

The phenomenon you are asking about is allelopathy.

Remarkably, it has its own journal - Allelopathy Journal (ISSN:
0971-4693) published since 2009 and currently at volume 30, issue 1.

Here are the titles of the papers from the most recent issue:

In-vitro assessment of allelopathic effects of wheat on potato.

Allelopathic effects of sunflower on seed germination and seedling growth of Trianthema portulacastrum.

Effects of simulated acid rain on the allelopathic potential of invasive weed Wedelia trilobata.

Allelopathic effects of Brachiaria ruziziensis and aconitic acid on Ipomoea triloba weed.

Effects of Butia capitata pyrenes extracts on the germination of lettuce seeds.

Effects of root exudates from crop plants on the growth of Fusarium oxysporum f.sp. niveum.

So, if you have specific questions, this might be the place to look.

Stability of medians in Median graphs

Here is an attempt to prove that the answer is yes.

Claim 1: Median graphs are bipartite.

This surely appears in the literature and is easy to verify. (Consider for a contradiction the shortest odd cycle and a median of 3 vertices on it: a pair of adjacent ones and a third one "opposite" of this pair.)

Claim 2: If $z neq m(x,y,z)$ then there exists a vertex $z'$ adjacent to $z$ such that $d(x,z')=d(x,z)-1$ and $d(y,z')=d(y,z)-1$. Further, for each such vertex $z'$ we have $m(x,y,z')=m(x,y,z)$.

Let $m=m(x,y,z)$ and let $P(z,m)$ be as in Tony's comment. Then the neighbor of $z$ on $P(z,m)$ satisfies the claim. The second part of the claim holds as one can extend to $z$ the shortest paths between $z'$ and $x$ and $y$.

Main argument: By induction on $d(x,y)+d(x,z)+d(y,z)$. By Claim 1 $d(x,y)=d(x',y)pm 1$ and $d(x,z)=d(x',z)pm 1$. If the signs in both of these identities are the same then $m(x,y,z) = m(x',y,z)$ by Claim 2. Thus, wlog, $d(x,y)=d(x',y)+1$ and $d(x,z)=d(x',z)-1$.

If $z neq m(x,y,z)$ then let $z'$ be as in Claim 2. We have $m(x,y,z')=m(x,y,z)$. As $d(x',z') leq d(x,z')+1 = d(x',z)-1$, by the second part of the claim we have $m(x',y,z')=m(x',y,z)$. We can now replace $z$ by $z'$ and apply induction hypothesis.

We assume therefore that $z = m(x,y,z)$. Symmetrically, $y=m(x',y,z)$. We have

$(d(x,z) + d(z,y)) + (d(x',y)+d(y,z))=d(x,y)+d(x',z) leq (d(x',y)+1)+(d(x,z)+1)$.

Thus $d(y,z) leq 1$, as desired.

physiology - Would muscle fatigue still occur if aerobic conditions for a working muscle is maintained?

Put another way if the muscle is given everything it needs to contract and do work will it ever get tired or have a reduction in energy efficiency?

As far as I understand muscles depend upon a blood supply delivering oxygen and nutrients (e.g. glucose and calcium) to effectively contract at its best level of performance. With the ability to work under anaerobic conditions if need be but producing lactic acid as a by-product which reduces the muscles ability to contract and therefore producing fatigue.

I also know that muscles are dependent upon temperature to work efficiently like the rest of the body. So, if the muscles temperature was able to be regulated well enough to maintain efficiency and aerobic conditions are met could fatigue be negated?

ag.algebraic geometry - Colimits of schemes

This is not an answer to the question, but is intended to show that Martins's concern about the possible distinction between the colimit in the category of schemes vs. in the category of locally ringed spaces is a valid one. Indeed, if I haven't blundered below, then it seems that in some circumstances at least the direct limit (colimit) of schemes exists, but does not coincide with the direct limit in the category of locally ringed spaces.

For example, suppose that $X_n$ is the Spec of $k[x]/(x^n),$ for some field $k$ (and the transition maps are the obvious ones). Then the direct limit of the $X_n$ in the category of locally ringed spaces is a formal scheme which is not a scheme, whose underlying topological space is a point, and whose structure sheaf (which is in this context simply a ring, namely the stalk at the unique point) is $k[[x]]$.

On the other hand, suppose given compatible maps from the $X_n$ to a scheme $S$. These must all map the common point underlying the $X_n$ to some point $s in S$, which lies in some affine open Spec $A$. Thus the maps from the $X_n$ factor through Spec $A$, and correspond to compatible maps $A rightarrow k[x]/(x^n),$ i.e. to a map $A rightarrow k[[x]].$ This in turn gives a map Spec $k[[x]] rightarrow$ Spec $Asubset X,$ and so we see that the natural compatible maps from the $X_n$ to Spec $k[[x]]$ identify Spec $k[[x]]$ with the direct limit of the $X_n$ in the category of schemes.

EDIT: As is noted in a comment of David Brown's attached to his answer, this example
generalizes, e.g. if $I$ is an ideal in a ring $A$, then the direct limit in the category of schemes of Spec $A/I^n$ coincides with Spec $hat A$, where $hat A$ is the $I$-adic completion of $A$.

FURTHER EDIT: I am no longer certain about the claim of the previous paragraph. If $A/I$ (and hence each $A/I^n$) is local then for any scheme $S$ the maps Spec $A/I^n to S$ factor through an affine open subscheme, so one reduces to computations in the category of rings, and hence finds that indeed the direct limit of the Spec $A/I^n$ equals Spec $hat{A}$. More generally, I'm currently unsure ... .

ag.algebraic geometry - Functorial characterization of morphisms of schemes

Check out the paper of Kontsevich-Rosenberg Noncommutative spaces and Noncommutative Grassmannian and related constructions. You will get what you want. i.e. the definition of properness and separatedness of presheaves(as functors, taken as "space") and morphism between presheaves(natural transformations).

Notice that these definitions are general treatment for algebraic geometry in functorial point of view,nothing to do with noncommutative.

Definition for separated morphism and separated presheaves

Let $X$ and $Y$ be presheaves of sets on a category $A$(in particular, $CRings^{op}$). We call a morphism $Xrightarrow Y$ separated if the canonical morphism

$Xrightarrow Xtimes _{Y}X$ is closed immersion We say a presheaf $X$ on $A$ is separated if the diagonal morphism:
$Xrightarrow Xtimes X$ is closed immersion

Definition for strict monomorphism and closed immersion

For a morphism $f$: $Yrightarrow X$ of a category $A$, denote by $Lambda _{f}$ the class of all pairs of morphisms

$u_{1}$,$u_{2}$:$XRightarrow V$ equalizing $f$, then $f$ is called a strict monomorphism if any morphism $g$: $Zrightarrow X$ such that $Lambda_{f}subseteq Lambda_{g}$ has a unique decomposition: $g=fcdot g'$

Now we have come to the definition of closed immersion: Let $F,G$ be presheaves of sets on $A$. A morphism $Frightarrow G$ a closed immersion if it is representable by a strict monomorphism.

Example

Let $A$ be the category $CAff/k$ of commutative affine schemes over $Spec(k)$, then strict monomorphisms are exactly closed immersion(classcial sense)of affine schemes. Let $X,Y$ be arbitrary schemes identified with the correspondence sheaves of sets on the category $CAff/k$. Then a morphism $Xrightarrow Y$ is a closed immersion iff it is a closed immersion in classical sense(Hartshorne or EGA)

Definition for proper morphism just follows the classical definition: i.e. universal closed and separated. You can also find the definition of universal closed morphism in functorial flavor in the paper I mentioned.

ag.algebraic geometry - Algebraic geometric model for symplectic $T^* Sigma_g$?

The cotangent bundle to $T^2$ is $(mathbb{C}^ast)^2$. But a theorem of Totaro (Internat. J. Math. 2 (1991), 563-566; MR1124283) implies that, if $S$ is a closed orientable surface of negative Euler characteristic, there is no diffeomorphism, sending the symplectic orientation to the complex orientation, from $T^*S$ to a smooth affine complex surface.

co.combinatorics - Is there a group whose cardinality counts non-intersecting paths?

Hi Gjergji. First, when there is a Gessel-Viennot matrix to count non-intersecting lattice paths, there is also a Gessel-Viennot cokernel. This will happen when the graph is planar and when the sources are all "on the left" and the sinks are all "on the right". Otherwise, if you can't find an integer matrix whose determinant counts the families of lattice paths or whatever, then there is no particular reason to believe in a cokernel.

Second, the Gessel-Viennot determinant is not essentially different from the Kasteleyn-Percus determinant. Every planar non-intersecting lattice path problem can be expressed as a planar perfect matching problem for a modified graph, and the Kasteleyn-Percus matrix reduces to the Gessel-Viennot matrix in a predictable way. In particular, the cokernels are the same. This is explained in my paper that you cite.

Third, even the tree group of a planar graph is not all that different from either Gessel-Viennot cokernel or the Kasteleyn cokernel. Again, you can modify a planar, bipartite graph to turn the matchings into trees. The tree group is more general in the sense that it does not require planarity.

---

Actually, although the Gessel-Viennot method is a very nice and very important result, it is something of a social accident that it is (or has at times been) much more popular than the Kasteleyn method in enumerative combinatorics circles. Kasteleyn published in mathematical physics journals, and his work was known but not widely known among combinatorialists until the 1990s. Then, he published a more complicated Pfaffian expression that applies to non-bipartite graphs, and this was only simplified by Percus in the bipartite case. (Although this simplification is easy.) Then, although Kasteleyn-Percus determinants explain and generalize Gessel-Viennot determinants, the matrices are larger and the more condensed Gessel-Viennot form can look more convenient for explicit calculations. Then too, unless you are studying cokernels, you might not need to know that Gessel-Viennot matrices can come from Kasteleyn-Percus matrices.

On the other hand, there are some things that Gessel-Viennot matrices do not easily show you. Three examples: (1) The same Kasteleyn-Percus matrix may have more than one Gessel-Viennot reduction, which will then have the same cokernel. (2) The minors of a Kasteleyn-Percus matrix give you edge probabilities, a fact which has been used to great effect by Rick Kenyon. (3) You may have to discard symmetry to write down the Gessel-Viennot matrix, which can make it more difficult to analyze matchings that are invariant under a group action.

co.combinatorics - Collection of subsets closed under union and intersection

The answer is Yes. Furthermore, such a family can be found of size at most the cardinality of A, even when S is much larger.

The key to the solution is to realize that every such family S arises as the collection of downward-closed sets for a certain partial pre-order on A, which I shall define. (Conversely, every such order also leads to such a family.)

An interesting special case occurs when the family S is linearly ordered by inclusion. For example, one might consider the family of cuts in the rational line, that is, downward-closed subsets of Q. (I had thought briefly at first that this might be a counterexample, but after solving it, I realized a general solution was possible by moving to partial orders.)

Suppose that S is such a collection of subsets of A. Define the induced partial pre-order on A by

• a <= b if whenever B in S and b in B, then also a in B.

It is easy to see that this relation is transitive and reflexive.

I claim, first, that S consists of exactly the subsets of A that are downward closed in this order. It is clear that every set in S is downward closed in this order. Conversely, suppose that X is downward closed with respect to <=. For any b in X, consider the set X_b, which the intersection of all sets in S containing b as an element. This is in S. Also, X_b consists of precisely of the predecessors of b with respect to <=. So X_b subset X. Thus, X is the union of the X_b for b in X. So X is in S.

Next, define f_a(b) = a if a <= b, and otherwise f_a(b) = b. Let F be the family of all such functions f_a for a in A.

Clearly, every B in S is closed under every f_a, by the definition of <=. Conversely, suppose that X is closed under all f_a. Thus, whenever b is in X and a <= b, then a is in X also. So X is downward closed, and hence by the claim above, X is in S.

Incidently, the sets S are exactly the open sets in the topology on A induced by the lower cones of <=.

mg.metric geometry - Advanced view of the napkin ring problem?

The key in seeing this lies in breaking it down into a series of stacked washers or discs.

• Define the napkin ring of height $h$ as being hollowed out of a sphere of radius centered at the origin with radius $R=alpha h$, with $alpha>=1$




• Define the hole as being drilled along the $z$-axis, creating the napkin ring with an outer diameter $r_{out}$ and inner diameter $r_{in}$ defined as functions of $z$.




• $r_{out}=sqrt(R^2-z^2)$ and inner diameter $r_{in}=sqrt(R^2-(h/2)^2)$




• Define the volume as the integral over $z$ ranging over ($-h/2, +h/2$)

$$pi int_{-h/2}^{h/2} (r_{out}^2-r_{in}^2) dz$$

as the volume being height ($dz$) multiplied by the area ($pi r_{out}^2 - pi r_{in}^2)$ of the outer disk minus the inner disk. Note that the $R^2$ cancels out, showing that the radius of the sphere of material does not affect the AREA of the annulus at height $z$.

This is the intuitive step that may be hard to grasp. The thickness of the napkin ring ($r_{out}-r_{in}$) certainly does change with the radius $R$ of the carving sphere. As $R$ increases, the thickness of the napkin ring decreases inversely proportional to the square of $R$. This inverse relationship is what keeps the areas of the annuli at height $z$ at a value that is independent of $R$.

The intuitive mis-step is in thinking that even as $R$ increases, the thickness of the napkin ring stays the same. This is incorrect. A similar intuitive mis-step also occurs with this example.

---

You are wearing a belt of circumference 1 meter. You let out the belt ~31.416 centimeters (~ ten belt notches). How far over your body will the belt float? (also, how much larger can you now become?) The answer is 10 centimeter increase in radius.

The earth is wearing a belt at its equator of circumference $C$. How much do you have to let the belt out to create a an increase in radius of 10 centimeters for the earth? The answer is the same ~31.416 centimeters, since circumference is linearly proportional to radius. $C=2pi r$

In the napkin ring problem, the key is that the area of the annuslus as a function of distance along the longitudinal-axis remains constant despite any change in $R$ because of the inverse-square proportionality. Intuition fools us into thinking that the thickness must remain constant for such a thing.

gt.geometric topology - Classification problem for non-compact manifolds

This answer complement the answers of Henry and Algori. I think, it is worth to strees, that a classification of open manifolds does not follow from a classification of compact manifolds. Open surfaces were classified, but open 3-folds are not, their classification does not follow from the classification of compact ones at least at the present time. In particular, {it prime decomposition}, or Kneser's theorem (http://en.wikipedia.org/wiki/Prime_decomposition_(3-manifold)) does not hold for non-compact 3-manfiolds. There is a constructiion due to Scott (http://plms.oxfordjournals.org/cgi/pdf_extract/s3-34/2/303) of an example of a simply connected 3-fold that can not be a connected sum of finite or infinite number of prime manifolds.

Open 3-manifold are actively studdied now. Let me give two citations that confirm further that our knowlage of compact 3-manfiolds is not sufficient for understending of non-compact ones.

---

1) Ricci flow on open 3-manifolds and positive scalar curvature.
Laurent Bessi`eres, G´erard Besson and Sylvain Maillot.
http://arxiv.org/PS_cache/arxiv/pdf/1001/1001.1458v1.pdf

Thanks to G. Perelman’s proof of W. Thurston’s
Geometrisation Conjecture, the topological structure of compact 3-manifolds
is now well understood in terms of the canonical geometric decomposition.
The first step of this decomposition, which goes back to H. Kneser,
consists in splitting such a manifold as a connected sum of prime 3-manifolds,
i.e. 3-manifolds which are not nontrivial connected sums themselves.
It has been known since early work of J. H. C.Whitehead [Whi35] that the
topology of open 3-manifolds is much more complicated. Directly relevant
to the present paper are counterexamples of P. Scott [ST89] and the third
author [Mai08] which show that Kneser’s theorem fails to generalise to open
manifolds, even if one allows infinite connected sums.

---

The refference for the article of Maillot is the followning.

2) Some open 3-manifolds and 3-orbifolds
without locally finite canonical decompositions. http://arxiv.org/PS_cache/arxiv/pdf/0802/0802.1438v2.pdf

Here is the citation

---

Much of the theory of compact 3-manifolds relies on decompositions into
canonical pieces, in particular the Kneser-Milnor prime decomposition [12,
16], and the Jaco-Shalen-Johannson characteristic splitting [10, 11]. These
have led to important developments in group theory [22, 7, 9, 24], and form
the background of W. Thurston’s geometrization conjecture, which has recently
been proved by G. Perelman [19, 20, 21].

For open 3-manifolds, by contrast, there is not even a conjectural description
of a general 3-manifold in terms of geometric ones. Such a description
would be all the more useful that noncompact hyperbolic 3-manifolds are
now increasingly well-understood, thanks in particular to the recent proofs
of the ending lamination conjecture [17, 4] and the tameness conjecture [5, 1].
The goal of this paper is to present a series of examples which show
that naive generalizations to open 3-manifolds of the canonical decomposition
theorems of compact 3-manifold theory are false.

ergodic theory - Fundamental domains of measure preserving actions

No, you need some other condition, since you aren't guaranteed to have many measurable sets.

For example, a probability pace consisting of a single atom (which need not be a set with one element) with measure $1$ has no measurable subsets of probability $|G|^{-1}$. You can let the space be $(-1/2,0) cup (0,1/2)$, let $C_2$ act by reversing signs, and let only the countable and cocountable subsets be measurable, with measures $0$ and $1$, respectively. Then no set has measure $1/2$.

Ok, assume $(X,mu)$ has no atoms. You still aren't guaranteed that there are enough measurable sets. Let the space be $(-1/2,0) cup (0,1/2)$, let $C_2$ act by reversing sign, and let the measurable sets be $Acup-A$ where $Asubset(0,1/2)$ is Lebesgue measurable. Then all measurable sets are fixed by the action of $C_2$, so there can't be a fundamental domain. As far as measure theory is concerned, this is still the trivial action, even though nothing is fixed by the nontrivial element.

You need some condition like that any set of positive measure contains subsets of positive measure which are moved by the action, whose images are measure-disjoint. Then (at least using some level of choice, but perhaps this isn't necessary) you can build a fundamental domain as a maximal set which has measure $0$ intersection with its images by nontrivial elements of $G$.

ag.algebraic geometry - Elliptic Curves over F_1?

In most of the current schemes, it is very unlikely that elliptic curves are defined over F_1.
They are certainly not in Deitmar's or Toen-Vaquie since they restric to toric varieties. For Soule/Connes-Consani old notions, all the examples found so far seem to come from torified varieties (as defined by Oliver and myself in this paper. Also in CC new notion, up to the torsion part of the monoidal scheme their schemes appear to be generalized torified (see Theorem 2.2 in the reference you mention). But all torified schemes are rational, so elliptic curves are not in there.

On the other hand, Manin was very confident that the set of torsion points of an elliptic curve would define a convenient model for it. But if that was the case I don't think it would fit CC models, but rather Manin/Marcolli analytic approach, conjecturally related to Borger's, but no there is clear explanation on that relation just yet.

pr.probability - On operator ranges in Hilbert & Banach spaces

Lemma 1 from Anderson & Trapp's Shorted Operators, II is

Let $A$ and $B$ be bounded operators on the Hilbert space $mathcal H$. The following statements are equivalent:

(1) ran($A$) $subset$ ran($B$).

(2) $AA^* le lambda^2 BB^*$ for some $lambda ge 0$.

(3) There exists a bounded operator $C$ such that $A = BC$.

Moreover, if (1), (2) and (3) are satisfied, there exists a unique operator $C$ so that ker($A$) = ker($C)$ and ran($C$) $subset$ closure(ran($B^*$)).

They follow this with the statement, "the lemmas and the original proofs remain valid for operators between two Hilbert spaces."

Question:I would like to know if there is a similar statement for more general Banach spaces, and if so, where I might find it.

My context: I am considering the Banach space $Omega = C(U_1) times C(U_2)$ of continuous functions over two domains. I have a covariance operator

$$K : Omega^* to Omega$$
which is decomposed as $$K = binom{K_{11} ~ K_{12}}{K_{21} ~ K_{22}}.$$ I want to apply the above lemma to $A = K_{21}$ and $B = K_{22}^{1/2}$.

Edit: If we have a probability measure $mathbb P$ on $Omega$, then continuous linear functionals $Omega^*$ are random variables. Thus the expectation $mathbb Efg$ for $f, g in Omega^*$ is well-defined. The covariance operator is the bilinear form defined by $f(Kg) = mathbb Efg$.

nt.number theory - Do six consecutive numbers always contain an abundant or perfect number?

Any multiple of 6 (more generally, of any perfect number) is abundant.

To answer a question asked in the comments, the asymptotic density of the abundant numbers is between 0.2474 and 0.2480; see Marc Deleglise, Bounds for the density of abundant integers, Experiment. Math. Volume 7, Issue 2 (1998), 137-143. That is, if $A(n)$ is the number of abundant integers less than $n$, $lim_{n to infty} A(n)/n$ exists and is in that interval. Perfect numbers have density zero, so deficient numbers have density just over 0.75. Unfortunately this doesn't give a proof of the result asked for in the original question.

Edited to add: Any six consecutive integers contain an abundant number. So it's natural to ask: Is there any $k$ such that $k$ consecutive integers always contain a deficient number? Erdos showed in 1935 that this is not so. More specifically, there are constants $c_1, c_2$ such that for all large enough $n$, there exist $c_1 log log log n$ consecutive abundant integers less than $n$, but not $c_2 log log log n$ consecutive abundant integers less than $n$. This Google preview of a book by Pickover states that the smallest pair of consecutive abundant numbers is 5775, 5776 and the smallest triplet is 171078830 + {0, 1, 2}. See the sequence A094268 in Sloane. The smallest known quadruplet of abundant numbers begins with $N_4 = 141363708067871564084949719820472453374$. These numbers suggest $c_1, c_2$ are somewhere around $0.37$. In particular, if you conjectured that any six consecutive integers contain a deficient number, the first counterexample would be around $exp exp exp (6 times 0.37)$ which has thousands of digits.

This is somewhat surprising, I think -- that the deficients are more common but the abundants come in longer runs. But in a way it makes sense. Abundant numbers tend to have lots of small factors, which means that they come in "nice" families; deficient numbers tend to not have lots of small factors, and so are essentially the holes left when the abundant numbers are left out. So perhaps it's reasonable that the holes have less structure than the thing that they're holes in.

evolution - Examples of virus originated from a living system

Prion diseases arise spontaneously fairly often. There are only ~170 total cases of human infection by mad cow disease, but there are ~50 cases/year of Creutzfeldt-Jakob disease, which is a spontaneous strain of prion diseases in humans. There is a strong case that Alzheimer's and Parkinson's diseases (among others) are similar in nature, in which case it's a much more frequent occurrence.

However, they are the consequence of a misfolding and aggregation of a naturally occurring protein produced in our brains. In this sense, they are nothing like the complexity required to produce a de novo virus. The closest viral analogy would be retroviruses, which could insert their genome into your DNA, die off, and then possibly spontaneously revive at a later date.

mathematics education - Teaching Methods and Evaluating them

Hey,

As a lowly graduate student, I'm on a committee (I'm not sure how important my role really is) trying to evaluate how effective different approaches teaching undergraduates. We are looking at all sorts of methods like the standard lecturing style seen in most schools, the Moore method, a Dewey/Montessori approach, Polya, etc. Now, my question is this: how do you go about developing some sort of effective criteria evaluating these different approaches?

One of the reasons I'm asking is that I'm sitting on the committee with another graduate student, and we were discussing what we thought about our measure theory class. It was taught in a slightly different manner; basically the students gave the lectures to one another. Our prof sat in the back and would occasionally ask a question directing us to a subtle point or an important topic to be covered. Now, I loved the class. It was challenging and forced me to work very hard to understand the material. Further, I enjoyed reconstructing the entire theory with my classmates. And being asked pointed questions by my prof was traumatizing but also very good preparation for my quals. Now, in contrast, my classmate didn't like the method at all, felt it went too slow, and he didn't think that many of the standard tricks, techniques or approaches that one ought to get out the class is what he actually got out of it. I'm not interested in debating the merits of our respective conclusions, but what I am interested in is developing some sort of analysis that would help us objectively evaluate various approaches/methods to teaching students. Our dept is looking at implementing some of these methods next academic year. We'd like to come up with a way of objectively evaluating the a particular method based off of firmer ground than anecdotal evidence and end of semester surveys. In sum, I was wondering if anybody here had experience in this area of curriculum analysis/development? If so I'd love to hear about it.

I hope this question isn't too far off from the mainstream questions of MO.

Regards,
Ben

mg.metric geometry - Is the ratio Perimeter/Area for a finite union of unit squares at most 4?

I cannot give a full proof, but I can reduce the problem to another one that I think some people might know the answer to.

Here is the problem:

PROBLEM 1.
Let $U$ be an open subset of the unit square with rectifiable boundary.
Then

$$Pge 4A,$$ where $P$ is the length of the boundary of $U$ and $A$ is the area of $U$.

"PROOF:" I can prove this if $Ale pi/4$ and I have a precise idea how to prove it in general.

First assume $Alepi/4$, say $A=pi r^2$ for some $rlefrac12$.
Since the circle minimizes the boundary length for a given area, we get $Pge 2pi r$ which gives the claim.
(Note that this also works if you assume that the diameter of $U$ is $le 1$ and so this gives a second proof of Proposition 7.1 in the mentioned thesis.)

If $A>pi/4$ this proof doesn't work. We have to replace the circle with the corresponding minimizer inside the square.
I do not know what this minimizer looks like, somewhat like a balloon you blow up inside a box. But minimal surface people might know and once you know the shape of that set, you can compute the boundary length to solve the problem as above.
Q.E.D.

Next I show how to solve the original problem once you have Problem 1.
I use induction on the number of squares.
For $n=1$ there is nothing to show.
We do $nto n+1$.
So let a set $F$ in the plane be given which is the union of $n$ unit squares.
Let $A(F)$ be its area and $P(F)$ its boundary length. By induction hyothesis we have $P(F)le 4A(F)$.
Add another unit square $S$, then the boundary length of the new set will be $P(F)+4-P$, where $P$ is the boundary lenght of some open set $U$ inside $S$, more precisely, $U$ is the intersection of the interior of $S$ and the interior of $F$.
Anyway, $U$ is the interior of a polygon, hence has rectifiable boundary.
The area of $Fcup S$ is $A(F)+1-A$, where $A$ is the area of $U$.
Problem 1 now tells us $Pge 4A$.
Together with the induction hypothesis this gives the claim.
Q.E.D.

ag.algebraic geometry - Nature of Invertible Sheaves in which there are no global sections.

They can still give you (non-canonical) rational maps to $mathbb{P}^n$:

Even when an invertible sheaf $L$ on $X$ has no global sections, one can still find open subsets $U$ of $X$ such that $L|_U$ is globally generated, for example when $U$ is affine. This isn't so interesting, but if you can find such a $U$ that is not contained in an affine, then $L|_U$ might not be trivial, and then you might get morphisms from $U$ to $mathbb{P}^n$ (by choosing generators) not coming from $mathcal{O}_U$. If $X$ was integral, then $U$ will be automatically dense, so you get a rational map from $X$ to $mathbb{P}^n$.

One way you could look for such a $U$ is to pick $n$ elements of the stalk $L_x$ at a point $x$, then intersect $n$ neighborhoods on which these elements extend, remove their common zero locus, and let the result be
$U$. If $U$ isn't contained in an affine, maybe you've found something cool. If you really wanted you could try working out a nice description for the rational map you've defined (though I've never done this). Even if $U$ was affine, maybe you've found a more interesting description of a less interesting map.

From a very different perspective, since $mathcal{O}(-d)$ is the inverse of $mathcal{O}(d)$, in some sense any "information" it contains is obtainable from inverting the transition functions of $mathcal{O}(d)$... so with this restrictive view of "information", perhaps it would be interesting to study what sort of morphisms/maps to $mathbb{P}^n$ one can get from an invertible sheaf $L$ on $X$ when neither $L$ nor $L^vee$ is ample/very ample.

biochemistry - What does it mean for a distribution to be "consistent with a two rate-limiting stochastic steps"?

WYSIWYG is almost there, but you need one more piece of information to make this explicit.

The distribution cited in the paper is $h(t)\propto te^{-t/\tau}$ (we're going to ignore normalizing constants today). We can recognize this as a particular case of the Gamma distribution, with PDF:

$$f_{\mathrm{Gamma}}(t\,\big|\,k,\theta)\propto t^{k-1}e^{-t/\theta}.$$

In particular, $h(t)$ looks like the PDF of a $\mathrm{Gamma}(2,\tau)$ random variable,

$$h(t)\propto f_{\mathrm{Gamma}}(t\,\big|\,2,\tau)\propto te^{-t/\tau}.$$

So the authors of the paper are claiming that the burst duration $\tau_{\mathrm{burst}}$ is a random variable with a $\mathrm{Gamma}(2,70\,\mathrm{ sec})$ distribution.

How do we get from that to "This distribution was consistent with two rate-limiting stochastic steps?"
If we recall from our stats class (or look up properties of the Gamma distribution on Wikipedia), a Gamma distribution with integer values of $k$ is the distribution of the waiting time for $k$ events to occur in a Poisson process. The setup of a Poisson process is that you are tracking the occurrence of unlikely/infrequent events in time, so the fact that the distribution $h(t)$ of burst times looks like a $\mathrm{Gamma}(2,\tau)$ suggests that the burst duration $\tau_{\mathrm{burst}}$ is determined by the occurrence of 2 stochastic events (with the same "frequency" $1/\tau$). In other words, this distribution is consistent with the following scenario: a nuclear localization occurs, and then will stop after two particular events, where the timing of the events is governed by a Poisson process (which, as WYSIWYG pointed out, is a reasonable model for the occurrence of chemical reactions with reasonably slow kinetics, i.e. you can count individual reactions in time). If this is true, the distribution of $\tau_{\mathrm{burst}}$ will look like $te^{-t/\tau}$.

The authors generalize that statement somewhat by saying that technically there could be more events/reactions required to terminate a localization burst, but all but 2 of those reactions happen extremely rapidly, i.e. there are 2 rate-limiting steps in the decision process (both of which happen to have the same value for $\tau$).

---

Edit: I realized that one more mathematical point might make this more clear: The statement about the Gamma distribution and Poisson processes above is equivalent to the statement that the sum of $k$ iid random variables with distribution $\mathrm{Exponential}(\tau)$ is a random variable with a $\mathrm{Gamma}(k,\tau)$ distribution. Thus $h(t)$ is literally (up to normalization) the distribution of a sum of 2 independent exponential random variables. If you read up on the connection between the exponential distribution and waiting times that might make the claim in the paper seem more transparent.

Source: https://en.wikipedia.org/wiki/Gamma_distribution#Special_cases

http://en.wikipedia.org/wiki/Exponential_distribution

Literature about putative epigenetic state changes in mammal sequences after cloning steps in Escherichia coli

I would like you to point me out some literature about putative epigenetic state changes in mouse/mammal sequences after cloning steps in Escherichia coli.

This are the last search details I used in NCBI PubMed:

"epigenomics"[MeSH Terms] AND "cloning, molecular"[MeSH Terms] OR
"molecular cloning"[All Fields] AND "sequence"[All Fields] AND
("escherichia coli"[All Fields] OR "e. coli"[All Fields])

life history - Is there a standard definition for plant "maturity"?

Is there a species-agnostic metric for identifying plant maturity?

There seems to be plenty of literature defining life-history stages for specific crops, but it is not clear if there is a generic definition.

In the context of crops, I would like to be able to identify the stage at which plants are no longer exhibiting juvenile traits.

nt.number theory - Definition of L-function attached to automorphic representation

As far as I understand, attaching an $L$-function to an automorphic representation attached to a general reductive group $G$ is conjectural and still open.

The way one attaches $L$-function depends on a representation $r$ of ${^L}G$ and partitioning the set of irreducible admissible representation of $G(k_v)$ in to $L$-packets (which is conjectural in general and known in very few cases). Assuming one can define local $L$-packets, Borel and Tate's article in Corvalis explains how to attach $L$-function to it. But still this $L$-function depends on the chosen representation $r$.

If $pi$ is an irreducible admissible representation of $G_A$ then $pi= otimes_v pi_v$, where $pi_v$ is an irreducible admissible representation of $G(k_v)$. So assuming we can partition the set of irreducible admissible representation of $G(k_v)$ in to $L$-packets, $pi_v$ belongs to $L$-packet $Pi_{phi_v}$ corresponding to some admissible homomorphism $phi_v$ of Weil-Deligne group to ${^L}G/k_v$ . The repesenation $r$ defines a representation $r_v$ of ${^L}G/k_v$.

Then the $L$-function attached to $pi$ and $r$ is defined as:
$L(s,pi,r) = prod_v L(s,pi_v,r_v)$,
$L(s,pi_v,r_v)=L(s, r_v circ phi_v)$

Now $r_v circ phi_v$ is a represenatation Wiel-Deilgne group, so by Tate's article in Corvalis, we know local $L$-factor.

gn.general topology - Minimal Hausdorff

A Hausdorff space $(X,tau)$ is said to be minimal Hausdorff if for each topology $tau' subseteq tau$ with $tau' neq tau$ the space $(X,tau')$ is not Hausdorff.

Every compact Hausdorff space is minimal Hausdorff.

I would like to know:

1) Is every minimal Hausdorff space compact?

2) Does every Hausdorff topology contain a minimal Hausdorff topology?

Many thanks!

tag removed - Hat Problem/Hamming Codes

Well, I wouldn't say there's just one case where they lose, but it depends on how you count cases. Remember, the idea is that the number of people is of the form $2^n - 1$. Accordingly, if we take a case to be an ordered assignment of colors, then there are $2^{2^n - 1}$ different cases. The fraction of these in which the prisoners is win is $(2^n - 1)/2^n$, but the denominator here isn't the number of cases, on this method of counting them; there's more than one case in which they lose.

Anyway, the way it works is that in a Hamming code, some cases (specifically, $2^{2^n - 1 - n}$ of them) are picked as "well-formed" such that for every case A, there is a unique well-formed case B such that the number of color changes between A and B is at most 1.

So, suppose the prisoners use the strategy "Consider both possible cases compatible with the information available to you. If one of them is well-formed (they won't both be), then guess in accordance with the other one. Otherwise, in the case where neither of them is well-formed, you should refrain from guessing". What will the outcome be?

What happens is that, in those cases which are well-formed, every prisoner guesses wrong, since every prisoner will take the first branch of the above strategy. On the other hand, in those cases which aren't well-formed, there is a unique prisoner who goes down the first branch and guesses correctly, while every other prisoner goes down the second branch and refrains.

Accordingly, the number of cases in which the prisoners lose is the number of well-formed cases; that is, $2^{2^n - 1 - n}$. So the fraction of cases in which the prisoners lose is $2^{2^n - 1 - n}/2^{2^n - 1} = 1/2^n$. Like I said, though, this isn't just one case (if we identify cases with ordered assignments of colors).

(Incidentally, if you'd like to know how to actually construct a Hamming code, you can take the well-formed cases to be those with the property that, for every $i$, there are an even number of black hats at positions whose $i$th bit is 1 (using position numbering starting at 1). It should be straightforward to verify that this has all the properties mentioned above)

rt.representation theory - Is the real Jacquet module of a Harish-Chandra module still a Harish-Chandra module?

As I understand it, the Jacquet module for $(mathfrak g, K)$-modules is defined so as to again be a $mathfrak g$-module, and in fact it is a Harish-Chandra module, not for $({mathfrak g},K)$, but rather for $(mathfrak g,N)$ (where $N$ is the unipotent radical of the parabolic with respect to which we compute the Jacquet module). (I am probably assuming that the original $(mathfrak g, K)$-module has an infinitesimal character here.)

I am using the definitions of this paper, in particular the discussion of section 2. This in turn refers to Ch. 4 of Wallach's book. So probably this latter reference will cover things in detail.

Added: I may have misunderstood the question (due in part to a confusion on my part about definitions; see the comments below), but perhaps the following remark is helpful:

If one takes the Jacquet module (say in the sense of the above referenced paper,
which is also the sense of Wallach), say for a Borel, then it is a category {mathcal O}-like
object: it is a direct sum of weight spaces for a maximal Cartan in ${mathfrak g},$
and any given weight appears only finitely many times. (See e.g. Lemma 2.3 and Prop. 2.4 in the above referenced paper; no doubt this is also in Wallach in some form; actually
these results are for the geometric Jacquet functor of that paper rather than for Wallach's
Jacquet module, but I think they should apply just as well to Wallach's.

Maybe they also apply with Casselman's definition; if so, doesn't this give the desired admissibility?

pr.probability - Kullback-Leibler divergence of scaled non-central Student's T distribution

What is the Kullback-Leibler divergence of two Student's T distributions that have been shifted and scaled? That is, $textrm{D}_{textrm{KL}}(k_aA + t_a; k_bB + t_b)$ where $A$ and $B$ are Student's T distributions.

If it makes things easier, $A$ could be a Gaussian. (That is, it could have infinite degrees of freedom.)

The motivation behind this question is that the scaled non-central Student's T distribution is the posterior predictive distribution of normally distributed data with unknown mean and variance. Thus, I would like to compare the true distribution $k_aA + t_a$ with the estimate $k_bB + t_b$.

botany - What kind of fruit is this?

Just spotted this fruit while walking to school. It's the size of a small coin.The taste is almost sour and tangy and somewhat sweet (I only tried one of them and very little of it). I admit, I've never seen it before. Does anyone know what this is?

Update: Location is two or three blocks north of UC Berkeley (Berkley, CA, USA).

ag.algebraic geometry - What classes am I missing in the Picard lattice of a Kummer K3 surface?

The lattice $L_{K3}=H^2(K3,mathbb Z)$ is $2E_8+3U$, with $E_8$ negative definite and $U$ the hyperbolic lattice for the bilinear form $xy$. It is unimodular and has signature $(3,19)$.

The 16 (-2)-curves $E_i$ form a sublattice $16A_1$ of determinant $2^{16}$. It is not primitive in $L_{K3}$. The primitive lattice $K$ containing it is computed as follows. Consider a linear combination $F=frac12sum a_i E_i$ with $a_i=0,1$. Recall that $E_i$ are labeled by the 2-torsion points of the torus $A$, i.e. the elements of the group $A[2]$.

Then $F$ is in $K$ $iff$ the function $a:A[2]to mathbb F_2$,
$imapsto a_i$, is affine-linear. You will find the proof of this statement in Barth-(Hulek-)Peters-van de Ven "Compact complex surfaces", VIII.5.
(The element $frac12sum E_i$ in your example corresponds to the constant function 1, which is affine linear).
Thus, $K$ has index $2^5$ in $16A_1$ and its determinant is $2^{16}/(2^5)^2=2^6$.

$K$ is called the Kummer lattice. By the above, it is a concrete negative-definite lattice of rank 16 with determinant $2^6$. Nikulin proved that a K3 surface is a Kummer surface iff $Pic(X)$ contains $K$.

The orthogonal complement $K^{perp}$ of $K$ in $L_{K3}$ is $H^2(A,mathbb Z)$ but with the intersection form multiplied by 2. As a lattice, it is isomorphic to $3U(2)$. It has determinant $2^6$, the same as $K$. The lattice $L_{K3}=H^2(K3,mathbb Z)$ is recovered from the primitive orthogonal summands $K$ and $K^{perp}$.

However, your question has "Picard lattice" in the title. The Picard group of $X$ is strictly smaller than $H^2(X,mathbb Z)$. To begin with, it has signature $(1,r-1)$, not $(3,19)$. For a Kummer surface, it contains Kummer lattice $K$ described above, and its intersection with $K^{perp}$ is the image of the Picard group of $A$. For a Kummer surface one has $r=17,18,19$ or 20.

For the Mori-Kleiman cone of effective curves, which you would need for Gromov-Witten theory, the description you put in a box is already the best possible.

co.combinatorics - How does this relationship between the Catalan numbers and SU(2) generalize?

This is a question, or really more like a cloud of questions, I wanted to ask awhile ago based on this SBS post and this post I wrote

As the SBS post describes, the Catalan numbers can be obtained as the moments of the trace of a random element of $SU(2)$ with respect to the Haar measure. This is equivalent to the integral identity

$$int_{0}^{1} (2 cos pi x)^{2k} (2 sin^2 pi x) , dx = C_k.$$

I can prove this identity "combinatorially" as follows: let $A_n$ denote the Dynkin diagram with $n$ vertices and $n-1$ undirected edges connecting those vertices in sequence. The adjacency matrix of $A_n$ has eigenvectors $mathbf{v}_i$ with entries $mathbf{v}_{i,j} = sin frac{pi ij}{n+1}$ with corresponding eigenvalues $2 cos frac{pi i}{n+1}$. If $k le n-1$, then a straightforward computation shows that the number of closed walks from one end of $A_n$ to itself of length $2k$ is

$$frac{1}{n+1} sum_{i=1}^{n} left( 2 cos frac{pi i}{n+1} right)^{2k} 2 sin^2 frac{pi}{n+1} = C_k$$

by the combinatorial definition of the Catalan numbers. Taking the limit as $n to infty$ gives the integral identity; in other words, the integral identity is in some sense equivalent to the combinatorial definition of the Catalan numbers in terms of closed walks on the "infinite path graph" $A_{infty}$. (Is this the correct notation? I mean the infinite path graph with one end.)

Now, closed walks of length $2k$ from one end of $A_n$ to itself can be put in bijection with ordered rooted trees of depth at most $n$ and $k$ non-root vertices. (Recall that the Catalan numbers also count ordered rooted trees of arbitrary depth.) The generating function $P_n$ of ordered rooted trees of depth at most $n$ satisfies the recursion

$$P_1(x) = 1, P_n(x) = frac{1}{1 - x P_{n-1}(x)}.$$

This is because an ordered rooted tree of depth $n$ is the same thing as a sequence of ordered rooted trees of depth $n-1$ together with a new root. (Recall that the generating function of the Catalan numbers satisfies $C(x) = frac{1}{1 - x C(x)}$. In other words, $C(x)$ has a continued fraction representation, and $P_n$ is its sequence of convergents.) On the other hand, since $P_n(x)$ counts walks on the graph $A_n$, it is possible to write down the generating function $P_n$ explicitly in terms of the characteristic polynomials of the corresponding adjacency matrices, and these polynomials have roots the eigenvalues $2 cos frac{pi i}{n+1}$. This implies that they must be the Chebyshev polynomials of the second kind, i.e. the ones satisfying

$$q_n(2 cos x) = frac{sin (n+1) x}{sin x}.$$

But the Chebyshev polynomials of the second kind are none other than the characters of the irreducible finite-dimensional representations of $SU(2)$! In particular, they're orthogonal with respect to the Haar measure.

The overarching question I have is: how does this sequence of computations generalize, and what conceptual framework ties it together? But I should probably ask more specific sub-questions.

Question 1: I remember hearing that the relationship between the Catalan numbers and the Chebyshev polynomials generalizes to some relation between moments, continued fractions, and orthogonal polynomials with respect to some measure. Where can I find a good reference for this?

Question 2: I believe that adding another edge and considering the family of cycle graphs gives the sequence ${2k choose k}$ and the Chebyshev polynomials of the first kind, both of which are related to $SO(2)$. According to the SBS post, this is a "type B" phenomenon, whereas the Catalan numbers are "type A." What exactly does this mean? What would happen if I repeated the above computations for other Dynkin diagrams? Do I get continued fractions for the other infinite families?

Question 3: Related to the above, in what sense is it natural to relate walks on the Dynkin diagram $A_n$ to representations of $SU(2)$? This seems to have something to do with question #16026. How do the eigenvectors of the adjacency matrices fit into the picture? I want to think of the eigenvectors as "discrete harmonics"; does this point of view make sense? Does it generalize?

As you can see, I'm very confused, so I would greatly appreciate any clarification.

ho.history overview - Good books on problem solving / math olympiad

From a review for Polya's book on Amazon, the books to be read in sequence:

• Mathematical Problem Solving by Alan Schoenfeld


• Thinking Mathematically by J. Mason et al.


• The Art and Craft of Problem Solving by Paul Zeitz


• Problem Solving Strategies by Arthur Engel


• Mathematical Olympiad Challenges by Titu Andreescu


• Problem Solving Through Problems by Loren Larson

Full text of the review below:

By Abhi:

Good aspects of this book have been said by most of the other
reviewers. The main problem with such books is that for slightly
experienced problem solvers, this book probably does not provide a
whole lot of information as to what needs to be done to get better.
For instance, for a kid who is in 10th grade struggling with math,
this is a very good book. For a kid who is in his 11th grade trying
for math Olympiad or for people looking at Putnam, this book won't
provide much help.

Most people simply say that "practice makes perfect". When it comes to
contest level problems, it is not as simple as that. There are
experienced trainers like Professor Titu Andreescu who spend a lot of
time training kids to get better. There is lot more to it than simply
trying out tough problems.

The most common situation occurs when you encounter extremely tough
questions like the Olympiad ones. Most people simply sit and stare at
the problem and don't go beyond that. Even the kids who are extremely
fast with 10th grade math miserably fail. Why?

The ONE book which explains this is titled "Mathematical Problem
Solving" written by Professor Alan Schoenfeld. It is simply amazing. A
must buy. In case you have ever wondered why, in spite of being
lightning fast in solving textbook exercises in the 10th and 11th
grade, you fail in being able to solve even a single problem from the
IMO, you have to read this book. I am surprised to see Polya's book
getting mentioned so very often bu nobody ever mentions Schoenfeld's
book. It is a must read book for ANY math enthusiast and the math
majors.

After reading this book, you will possibly get a picture as to what is
involved in solving higher level math problems especially the
psychology of it. You need to know that as psychology is one of the
greatest hurdles to over when it comes to solving contest problems.
Then you move on to "Thinking Mathematically" written by J. Mason et
al. It has problems which are only few times too hard but most of the
times, have just enough "toughness" for the author to make the point
ONLY IF THE STUDENT TRIES THEM OUT.

The next level would be Paul Zeitz's The Art and Craft of Problem
Solving. This book also explains the mindset needed for solving
problems of the Olympiad kind. At this point, you will probably
realize what ExACTLY it means when others say that "problem solving is
all about practice". All the while you would be thinking "practice
what? I simply cannot make the first move successfully and how can I
practice when I can't even solve one problem even when I tried for
like a month". It is problem solving and not research in math that you
are trying to do. You will probably get a better picture after going
through the above three books.

Finally, you can move on to Arthur Engel's Problem Solving Strategies
and Titu Andreescu's Mathematical Olympiad Challenges if you managed
to get to this point. There is also problem solving through problems
by Loren Larson. These are helpful only if you could solve Paul
Zeitz's book successfully.

To conclude, if you are looking for guidance at the level of math
Olympiad, look for other books. This book won't be of much assistance.
On the other hand, if you are simply trying to get better at grade
school math, this book will be very useful.

ag.algebraic geometry - K3 surface of genus 8

To be able calculate the degree it is worth to read a bit of Griffiths-Harris about Grassmanians (chapter 1 section 5). To prove that $S$ is $K3$ one needs to caluclate the canonical bundle of $G$, use simple facts about Plucker embedding, use adjunction formula and finally the fact that a simply connected surface with $Kcong O$ is $K3$.

I will make the second bit of calculation, that proves that $S$ is a $K3$ (so I don't calculate the degree $14$).

First we want to calculate the canonical bundle of $G$. Denote by $E$ the trivial $6$-dimensional bundle over $G$, and by $S$ the universal (tautological) rank $2$ sub-bundle. Then the tangent bundle to $G$ is $TG=S^* otimes (E/S)$ . It follows from the properties of $c_1$ that $c_1(TG)=6c_1(S^*)$. Similarly for the canonical bundle $K_G$ we have the expression $K_Gcong (detS^*)^{otimes -6}$.

Now we will use the (simple) statement from Griffiths-Harris that under the Plucker embedding we have the isomorphism of the line bundles $det S^*=O(1)$. Using the previous calculation we see $K_Gcong O(-6)$.
Finally, the surface $S$ is an iterated (6 times) hyperplane section of $G$. So by Lefshetz theorem it has same fundamental group as $G$, i.e., it is simply-connected. It suffices now to see that its canonical bundle is trivial. This is done using the adjunction formula $K_D=K_X+D|_D$. Every time we cut $G$ by a hyper-plane we tensor the canonical by $O(1)$, but $O(-6)otimes O(6)cong O$.

Added. The calculation of the degree is done by Andrea

Added. The number 19 is obtaied in the following way. The dimension of the grassmanian of $8$-planes in $CP^{14}$ is $6cdot 9=54$. At the same time the Grassmanian of $2$-planes in $mathbb C^6$ has symmetires, given by $SL(6,mathbb C)$, whose dimension is 35. We should quotient by these symmetries and get $54-35=19$

differential equations - When is Sobolev space a subset of the continuous functions?

If we let $Omegasubsetmathbb{R}^d$ with $d=1,2,3$ and define $mathcal{H}^1(Omega)=(win L_2(Omega): frac{partial w}{partial x_i}in L_2(Omega), i=1,...,d)$. My tutor has repeated several times:

1. If $d=1$ then $mathcal{H}^1(Omega)subsetmathcal{C}^0(Omega)$.


2. If $d=2$ then $mathcal{H}^2(Omega)subsetmathcal{C}^0(Omega)$ but $mathcal{H}^1(Omega)notsubsetmathcal{C}^0(Omega)$.


3. If $d=3$ then $mathcal{H}^3(Omega)subsetmathcal{C}^0(Omega)$ but $mathcal{H}^2(Omega)notsubsetmathcal{C}^0(Omega)$.

I was interested in trying to show these relationships. Does anyone know any references that would be useful.

Thanks in advance.