at.algebraic topology - Why does one think to Steenrod squares and powers?

Steenrod operations are an example of what's known as a power operation. Power operations result from the fact that cup product is "commutative, but not too commutative". The operations come from a "refinement" of the operation of taking $p$th powers (squares if $p=2$), whose construction rests on this funny version of commutativity.

A cohomology class on $X$ amounts to a map $a: Xto R$, where $R = prod_{ngeq0} K(F_2,n)$. So the cup product of $a$ and $b$ is given by
$$Xtimes X to Rtimes R xrightarrow{mu} R.$$
In other words, the space $R$ carries a product, which encodes cup product. (There is another product on $R$ which encodes addition of cohomology classes.)

You might expect, since cup product is associative and commutative, that if you take the $n$th power of a cohomology class, you get a cohomology class on the quotient $X^n/Sigma_n$, where $Sigma_n$ is the symmetric group, i.e.,
$$X^n xrightarrow{a^n} R^n rightarrow R$$
should factor through the quotient $X^n/Sigma_n$. This isn't quite right, because cup product is really only commutative up to infinitely many homotopies (i.e., it is an "E-infinity structure" on $R$). This means there is a contractible space $E(n)$ with a free action of $Sigma_n$, and a product map:
$$mu_n' : E(n)times R^nto R$$
which is $Sigma_n$ invariant, so it factors through $(E(n)times R^n)/Sigma_n$. Thus, given $a: Xto R$, you get
$$P'(a): (E(n)times X^n)/Sigma_n to (E(n)times R^n)/Sigma_n to R.$$
If you restrict to the diagonal copy of $X$ in $X^n$, you get a map
$$P(a):E(n)/Sigma_n times Xto R.$$

If $n=2$, then $E(2)/Sigma_2$ is what Hatcher seems to call $L^infty$; it is the infinite real proj. space $RP^infty$. So $P(a)$ represents an element
in $H^* RP^infty times X approx H^*X[x]$; the coefficients of this polynomial in $x$ are the Steenrod operations on $a$.

Other cohomology theories have power operations (for K-theory, these are the Adams operations).

You can also describe the steenrod squares directly on the chain level: the account in the book by Steenrod and Epstein is the best place to find the chain level description.

mg.metric geometry - Is there a neat formula for the volume of a tetrahedron on the surface of $S^3$?

On the volume of a hyperbolic and spherical tetrahedron, by Murakami and Yano. The volume is obtained as a linear combination of dilogarithms and squares of logarithms. The origin of their formula is really interesting: Asymptotics of quantum $6j$ symbols. (These asymptotics have also been studied by many other people: D. Thurston, Roberts, Woodward, Frohman, Kania-Bartoszynska, etc.)

Note that the 3-dimensional formula has to be much more complicated. The 2-dimensional formula comes from Euler characteristic and Gauss-Bonnet, but the Euler characteristic of the 3-sphere, or any odd-dimensional manifold, vanishes. In fact every characteristic class of a 3-sphere vanishes, because the tangent bundle is trivial. There can't be a purely linear treatment of volumes in isotropic spaces in odd dimensions. In even dimensions, there is always a purely linear extension from lower dimensions using generalized Gauss-Bonnet.

stochastic calculus - Relation between regularities of the trajectory of a mean zero gaussian process and its covariance operator

Suppose we already know that the process is continuous with probability one, so that the process takes values in the Banach space $X = C([0,1])$ with distribution $mathbb P$. The covariance operator $C : X^* to X$ is then a map from the dual space $X^*$ to $X$. The support of the Gaussian measure $mathbb P$ is then the closure of the image of $X^*$ under $C$: $$operatorname{supp} mathbb P = overline{ CX^* }.$$ (This is the main theorem of [Vakhania 1975])

Now let's construct the Cameron-Martin space. The operator $C$ defines an inner product on the dual space $X^*$ by $$langle f, g rangle = f(Cg)$$ for $f, g in X^*$. The space $X^*$ isn't necessarily closed under the topology induced by the inner product, so let $H$ be the Hilbert space completion, and let $iota : X^* hookrightarrow H$ be the inclusion map. Define a map $iota^* : H hookrightarrow X$ first on the dense subspace $iota X^* subseteq H$ by $$iota^*(iota f) = Cf,$$ and extend continuously to all of $H$. Thus the covariance operator factors as $C = iota^* circ iota$.

The Hilbert space $iota^* H$ is a subspace of $X$, and is called the Cameron-Martin space of the process. Interpreting this in the context of the support of the Gaussian measure $mathbb P$, we have $$operatorname{supp} mathbb P = overline{iota^* H},$$ so that the closure of the Cameron-Martin space (with respect to the original norm of $X$) is exactly the support of $mathbb P$.

I go into these ideas in more detail in Section 2 of my preprint [LaGatta 2010].

Suppose your covariance operator is an integral operator with kernel $c(s,t)$, called the covariance function of the process. That is, if $mu$ is a Radon measure on $[0,1]$, then $$(Cmu)(s) = int_0^1 c(s,t) , dmu(t).$$

If we write $c_s(t) = c(s,t)$, then the support of $mathbb P$ is the closure of the span of the functions $c_s$ in $C([0,1])$. So to answer your question: if you know that the closure $overline{operatorname{span}{c_s}}$ is a space with regularity property $P$, then the process $xi_t$ satisfies property $P$ with probability one.

homotopy theory - Do h-coequalizers and coproducts give all h-colimits?

It is well known that if a category has all coequalizers and all (small) coproducts then in fact it has all (small) colimits. More important is the proof which shows that every colimit can be built by using coproducts and coequalizers. This implies that if a functor commutes with coproducts and coequalizers, then it must commute with all (small) colimits as well.

Is there homotopical analog of this? If I have a functor which commutes with all small (homotopy) coproducts and all homotopy coequalizers, does it necessarily commute with all homotopy colimits in general?

This question makes sense for general model categories, but I am particularly interested in the usual model structure on spaces.

foliations - Foliated bundles and suspensions

Hi,

Is there any criteria, except for the existence of a flat connection, for a foliated bundle $E$ to be a suspension ( a foliated flat bundle)? For example, the Kronecker foliation on the torus is a suspension $mathbb{R}times_{mathbb{Z}} mathbb{S}^1$ , i.e. its of the form $Mtimes_Gamma F$, where $Gamma$ acts freely and transitively on the manifold $M$ and there is a free action $
rho: Gamma rightarrow Diff(F)$.

Note: The Kronecker foliation is induced by the vector field , $a frac{partial}{partial x} + b frac{partial}{partial y}$ on $mathbb{R}^2$ with $a,b$ constants.
The action of $mathbb{Z}$ on $mathbb{R}times mathbb{S}^1$ is given by:

$(r, exp{iz}).m= (r+m, exp{(iz+malpha)})$ for some $alpha in mathbb{R}$.

Thanks

gn.general topology - Non-homeomorphic spaces that have continuous bijections between them

Recycling an old (ca. 1998) sci.math post:

" Anyone know an example of two topological spaces $X$ and $Y$
with continuous bijections $f:Xto Y$ and $g:Yto X$ such that
$f$ and $g$ are not homeomorphisms?

Let $X = Y = Z times {0,1}$ as sets, where $Z$ is the set of integers.
We declare that the following subsets of $X$ are open for each $n>0$.
$${(-n,0)}, {(-n,1)}, {(0,0)}, {(0,0),(0,1)}, {(n,0),(n,1)}$$
This is a basis for a topology on $X$.

We declare that the following subsets of $Y$ are open for each $n>0$.
$${(-n,0)}, {(-n,1)}, {(0,0),(0,1)}, {(n,0),(n,1)}$$
This is a basis for a toplogy on $Y$.

Define $f:Xto Y$ and $g:Yto X$ by $f((n,i))=(n,i)$ and $g((n,i))=(n+1,i).$
Then $f$ and $g$ are continuous bijections, but $X$ and $Y$ are not homeomorphic.

This example is due to G. Paseman.

David Radcliffe "

More generally, take a space X with three successively
finer topologies T, T' and T''. Form two spaces which have underlying
set ZxX, and "form the infinite sequences" .... T T T T' T'' T'' T'' ....
and ... T T T T T'' T'' T'' T'' .... The continuous maps will take a finer
topology in one sequence to a rougher topology in the other. You can
make them bijective, and show that they are obviously non-homeomorphic
for a judicious choice of X, T, T', and T''.

Gerhard "Ask Me About System Design" Paseman, 2010.07.05

ca.analysis and odes - Closed forms for Monotonic polynomial recurrences?

A lot depends of what you mean by "fair accuracy" and on what exactly you are going to do with your formula. If a 30% upside error in each $d_n$ is tolerable, you can do the following.

We look at the recursion $d_{n+1}=qd_n(1-d_n)$ with $0<q=2p<1$ starting with some $d_1in[0,1]$. It'll be convenient to do the first step separately, so we have $d_2=q d_1(1-d_1)$. The reason is that $q^{-1}d_2in[0,1/4]$. Now, denote $b_n=q^{-(n-1)}d_n$ and rewrite the recurrence as $b_{n+1}^{-1}=b_n^{-1}+q^{n-1}frac{b_n}{b_{n+1}}$. Note now that the ratio of each term to the next is between $1$ and $4/3$ and tends to $1$, so, replacing it by $1$, we get $b_n^{-1}approx b_2^{-1}+frac{q-q^{n-1}}{1-q}$ with accuracy 30% and being sure that it is an underestimate. Putting all this stuff together, we get
$$
d_napprox q^{n-1}left[frac 1{d_1(1-d_1)}+frac{q-q^{n-1}}{1-q}right]^{-1}
$$
for $nge 2$.

Note also that 30% is a very rough estimate for the accuracy. In practice, I've never seen it being worse that 6% (though I haven't done too many simulations).

P.S. The estimate
$$
d_napprox q^{n-1}left[frac 1{d_1(1-d_1)}+frac{q-q^{n-1}}{1-q}
+logleft(1+d_1(1-d_1)frac{q^2-q^{2(n-1)}}{1-q^2}right)right]^{-1}
$$
is even better (3% accuracy for small $n$ and about 0.5% accuracy for large $n$) but noticeably uglier. As several people have already mentioned, there is no exact formula, so the higher precision you want, the longer and uglier the approximation gets.

P.P.S. The main idea behind the second approximation is that if $B_n=b_{n}^{-1}$, then we make an error $q^{n-1}frac{B_{n+1}-B_n}{B_n}approx frac{q^{2(n-1)}}{B_n}$ during each step that led us to the first approximation. To compensate, we would like to take the partial sums of that series. We also know that $B_napprox frac 1{d(1-d)}+q+q^2+dots+q^{n-2}$. Unfortunately, we cannot sum the resulting series nicely. But if we double all the exponents in the approximate formula for $B_n$ (which will lead only to a small error when $qapprox 1$), then we'll get the Riemann sum type expression, which we can replace by the corresponding integral. Clearly, it may overshoot on the long run but to my own surprise, it not only corrects the first few terms nicely, but also corrects the asymptotics giving an error below 0.5% for large $n$ in the entire range of parameters (well, at least that is so for all values I looked at and I tried quite a few). Why it works this way remains a mystery, but it does. It can be shown rigorously that the total relative overshot coming from all steps after $n$ is at most $n^{-1}$ and I've never seen the overshots of more than 0.1% up to $n=300$.

intuition - What is a symplectic form intuitively?

Incidentally, I more or less disagree that symplectic geometry captures what I would consider "classical mechanics". The reason is that in all the examples that I think deserve to be called "classical mechanics", I actually have a configuration space $N$, and your symplectic manifold is $X = {rm T}^*N$ the cotangent bundle. Then, of course, the symplectic form is precisely (part of) the cotangent structure.

This is not to say that symplectic geometry isn't interesting — it's led to great mathematics, and certainly captures some of "classical mechanics". From the physics perspective, what I think makes it most interesting is that it shows that there are strange symmetries between mechanical systems, when you have a symplectomorphism ${rm T}^*N to {rm T}^*N'$ that does not arise from a diffeomorphism $N to N'$.

But physics is not invariant under all symplectomorphisms. Otherwise, how would I know which coordinates are "position" and which are "momentum"? And I do believe that I know this, although maybe I'm wrong. You and I should get together and compare if our Darboux coordinates differ only by a map $N to N'$, or by some more interesting symplectomorphism.

Do plane projections determine a convex polytope?

Suppose a compact convex body $P subset Bbb R^3$ has only polygonal orthogonal projections onto a plane. Does this imply that $P$ is a convex polytope?

This question occurred to me when I was making exercises for my book. I figured this is probably easy and well known, but the literature hasn't been any help. One remark: if the number of sides of all polygons is bounded by $n$, the problem might be easier. Furthermore, if $P$ is assumed to be a convex polytope, this elegant paper by Chazelle-Edelsbrunner-Guibas (1989) gives a (perhaps, unexpectedly large) sharp $exp O(n log n)$ upper bound on the number of vertices of $P$ (ht Csaba Toth who generalized this to higher dimensions).

soft question - Use of the word "data" not in the statistical sense

This use seems in line -- although perhaps not identical -- with the following dictionary definition:

---

Data:

1. Factual information, especially information organized for analysis or used to reason or make decisions.




2. Computer Science Numerical or other information represented in a form suitable for processing by computer.




3. Values derived from scientific experiments.

---

It is often used in mathematics in the way you have identified above. Namely, when defining a mathematical structure, it gives the reader a heads up as to the fact that that the structure is "multi-sorted" and involves more than one object. In more formal language, one might say "tuple", e.g.,

"A topological group is a triple $(G,m,tau)$, where G is a set, $m: G times G rightarrow G$ is a binary operation, and $tau$ is a family of subsets of $G$, such that...."

One could also have said "A topological group is given by the data G,m,$tau$..."

pr.probability - easy(?) probability/diff eq. question

I've been wondering about this ever since I was a little kid and I used to ride in the back of the car and my mom would speed like hell towards a green light, only to slam on the brakes when she realized she wasn't going to make it.

Here is my question, loosely phrased: Given that we want to make it across the intersection before the light is up, what should be our position function? To make things precise, we could specify some initial conditions, put a cap on the car's acceleration/deceleration (or its speed or its jerk or whatever), fix a probability distribution on when the light will change, and assign some point values to our happiness if we make the light vs. don't make the light vs. end up entering the intersection after the light has changed and get a ticket for it. And then of course we could throw in the extra curveball of a yellow light warning you that it's about to turn red...

A similar question arises if you're approaching a red light but you think it might turn green soon. Ideally you'd like to enter the intersection at the highest possible speed just after the light has turned green, but then again you don't want to enter when it's still red.

I'm sure a computer could solve such problems easily, but it seems like there should be some better way to think about this than just asking a machine to do it for me. For the first question, it seems like the answer will just be either "hit the gas and go for it" or "cruise to a stop and don't plan on going through", depending on the parameters. (Of course, there might be something in the "go just fast enough that you can slow down and not enter on a red" plan if the cost of a ticket is high enough.) On the other hand, the second question seems to admit much more interplay between probability and differential equations. The real issue here is that I know almost nothing about either of these two fields. Any ideas?

lo.logic - Is there a formal notion of what we do when we 'Let X be ...'?

This is likely an elementary question to logicians or theoretical computer scientists, but I'm less than adequately informed on either topic and don't know where to find the answer. Please excuse the various vague notions that appear here, without their appropriate formal setting. My question is exactly about what I should have said instead of what follows:

When we make a definition, of either a property, or the explicit value of a symbol, it seems that we are somehow changing the language. Prescribing meaning to a word might be viewed as a kind of transformation of the formal language akin to taking a quotient, where we impose further relations on a set of generators.

I don't know how to describe a 'semantic' object, but am assuming an ad-hoc definition could be as a class of words under an equivalence relation supplied by an underlying logic. If the complexity of such an object is the size of the smallest word in the language that describes it, then making a definition lowers the complexity of some objects (and doesn't change the rest.) The obvious example is that if I add the word group to my language, then saying G is a group is a lot shorter than listing its properties.

It seems that lowering complexity is a main point of making definitions. Further, that one reason mathematical theory-building works, is a compression effect through which I am able to use less resources to describe more complex objects, at the cost of the energy it takes to cram definitions from a textbook.

Likely there is some theory out there that describes this process, but I've not been able to google it. I would appreciate being pointed towards the right source, even if it's a wikipedia link. Specifically:

Where can I find a theory of formal logic or complexity theory that studies the process of adding definitions to a mathematical language, viewed as a transformation that changes complexity?

soft question - Alternative Undergraduate Analysis Texts

Nobody has mentioned Folland's "Real Analysis with Applications"?? This was the textbook for my undergraduate real analysis course (measure theory, Banach spaces, Hilbert spaces), and I still go back to it all the time. I am not yet all that experienced (I just finished my third year of graduate school), but overall I have gotten more use out of this book than any other that I own.

It has the most comprehensive swath of applications of analysis of any introductory text I have ever encountered: basic functional analysis, Fourier theory, probability theory, distributions, Hausdorff measures, Haar measure, smooth measures, and more. The early material is covered with all the appropriate detail, while the later material quickly provides the essential definitions and results needed to come to grips with an unfamiliar idea in the literature. Also, the exercises are abundant and uniformly fantastic. My only complaints are that some of the later proofs are hard to read, and there is sadly no discussion of the spectral theorem.

nt.number theory - English reference for a result of Kronecker?

I don't know a reference, but here is a quick proof: Let the roots of the polynomial be $alpha_1$, $alpha_2$, ..., $alpha_r$. Let

$$f_n(x) = prod_{i=1}^r (x- alpha_i^n).$$

All the coefficients of $f_n$ are rational, because they are symmetric functions of the $alpha$'s, and are algebraic integers, because the $alpha$'s are, so they are integers.
Also, since $|alpha_i| leq 1$, the coefficient of $x^k$ in $f_n$ is at most $binom{r}{k}$.

Combining the above observations, the coefficients of the $f_n$ are integers in a range which is bounded independent of $n$. So, in the infinite sequence $f_i$, only finitely many polynomials occur. In particular, there is some $k$ and $ell$, with $ell>0$, such that $f_{2^k} = f_{2^{k + ell}}$. So raising to the $2^{ell}$ power permutes the list $(alpha_1^{2^{k}}, ldots, alpha_r^{2^k})$. For some positive $m$, raising to the $2^{ell}$ power $m$ times will be the trivial permutation. In other words,

$$alpha_i^{2^k} = alpha_i^{2^{k+ell m}}$$.

Every root of the above equation is $0$ or a root of unity.

nt.number theory - Repeated digits of squares in different bases

Hello, I am Mahima. I would like to ask the following clarifications. If any one answered, I am so thankful to you.

In which bases is 1111 a square?
b^3 + b^2 + b + 1 = n^2.
(b + 1)(b^2 + 1) = n^2.
We look at the gcd(b+1, b^2 +1) using the Euclidean algorithm.
And find that gcd(b+1, b^2 +1) = 2 if b is odd, but 1 if b is even.
If b is even, we have both (b + 1) and (b^2 + 1) a square.
But that is not possible as no positive squares differ by 1.
So b is odd, and both b + 1 and b^2 + 1 are even, so they are both twice a
square.
So we have:
b + 1 = 2a^2 and b^2 + 1 = 2c^2.

These are simultaneous diophantine equations.
We solve the one with least solutions and test these with the easier one.
The second is a Pellian equation.
The smallest solution is b = 7, c = 5. This also satisfies the first.
So we have one solution.
base 7 1111 = 1 + 7 + 7^2 + 7^3 = 20^2.
Using the method for solving the Pellian, I can't find another solution
for both equations. I may be able to produce a proof by induction that
the solution is unique.
I have had a look at base 12, and think it might be a limited base.
Please see what you can do there.

I want generalizations also.
Thanks in advance.
with LOVE,
Mahima.

dg.differential geometry - Can minimal surfaces be characterized by some universal property?

As objects which are minimal (in some respect), this seems entirely plausible, but I'm not sure what category we should be working in, and what restrictions we would need, to actually have a situation where minimal surfaces would be characterized by a universal property, if they ever can be. An uneducated guess on one possible setup where minimal surfaces would be universal: the objects are surfaces whose boundary is a given simple closed curve, and the morphisms are the area-decreasing isometries - it seems like a minimal surface should be a final object, though we would probably need to introduce an equivalence relation on the morphisms to get the maps to be unique?

I'm also curious about the same question, but for geodesics. Perhaps for them, we would use the collection of paths from point $x$ to point $y$ on a given surface, and use the length-decreasing homotopies?

Being a final object isn't the only option - maybe, for any surface, some kind of map will factor through a minimal surface associated to it?

EDIT: I'm worried this is perhaps too soft a question for MathOverflow - I'm not sure there's really a "right" answer.

ag.algebraic geometry - What cohomology theories would be interesting for nilpotent cones/nullcones?

As I understand, when we have a nilpotent cone, or a nullcone of a Lie group representation, what seems to be done in a lot of the literature (e.g. Achar&Henderson-"Orbit closures in the enhanced nilpotent cone") is to compute the intersection cohomology sheaves and find polynomials that determine the dimensions of various stalks.

But what other cohomology theories (that are different to intersection cohomology, I understand sometimes two different cohomology/homology can coincide under special circumstances), would be interesting in nilpotent cones?

Here's a bit about the problem I'm working on, and some theories that I hope (?) might be interesting, can anyone tell me some more that might be interesting? I am very far from being knowledgeable about cohomology, so if some-one could tell me if the following questions are stupid/trivial/ill-defined or not, please tell me.

• I have the orbits, which themselves are usually quasi-affine or quasi-projective varieties, which I could compute cohomology of? (perhaps Cech cohomology?)




• The set of orbits inherits a Zariski topology structure from the Zariski topology structure (that coincides with that inherited from the classical topology), perhaps I can compute some (co)homology of this topological space? In my case the set of orbits is uncountably infinite, but I am not completely sure if it has a triangulation - any theory that doesn't involve triangulations?




• As standard, one computes the orbit closures, and instead of doing intersection cohomology of these singular varieties, compute perhaps some of the lower K-groups?



• perhaps Hochschild cohomology of the coordinate rings of some of the affine coordinate rings of these varieties could be interesting?

soft question - Number theory textbook with an algebraic perspective

Most of the number theory textbooks I've dealt with take a very classical approach to the subject. I'm looking for a textbook that's something like a first course in number theory for people who have a decent command of modern algebra (at the level of something like Lang's Algebra). Does such a book exist, and if it does, what is it called?

Edit: As I posted in a comment below:

In the introduction to Ireland and Rosen, they note something that was bugging me for a while, "Nevertheless it is remarkable how a modicum of group and ring theory introduces unexpected order into the subject."

This is precisely the perspective I was looking for, so if anyone passes by this topic looking for a book that approaches number theory in this way, I feel like this quote should point him (her?) in the right direction.

alexandrov geometry - Example of non-closed convex hull in a CAT(0) space

There are such examples already in Riemannian world!
In fact in any generic Riemannian manifold of dimension $ge3$ convex hull of 3 points in general position is not closed.
BUT it is hard to make explicit and generic at the same time :)

If it is closed then there are a lot of geodesics lying in its boundary --- that is rare!
To see it do the following exercise first: Show that in generic 3-dimensional manifold, arbitrary smooth convex surface contains no geodesic. (Here geodesic = geodesic in ambient space.)

To make word "generic" more clear: show that any metric admits $C^infty$-perturbation such that above property holds.

Semisolution:
Assume that a geodesic $gamma$ lies in the boundary of a convex set $K$ with smooth boundary. Let $N(t)$ be the outer normal vector to $K$ at $gamma(t)$. Note that $N(t)$ is parallel.
Further note that from convexity of $K$ we get that for any Jacoby field $J(t)$ such that
$$langle N(t_0),J(t_0)ranglele 0 text{and} langle N(t_1),J(t_1)ranglele 0,$$
we have
$$langle N(t),J(t)ranglele 0 text{if} t_0<t<t_1.$$
Note that this condition does not hold if the curvature tensor on $gamma$ is generic.

P.S. Roughly it means that convex hulls in Riemannian world are too complicated. But I know one example where it is used, see Kleiner's An isoperimetric comparison theorem.
But he is only using that Gauss curvature of non-extremal points on the boundary of convex hulls is zero...

Appendix. (A construction of convex hull.) To construct convex hull you can do the following: start with some set $K_0$ and construct a sequence of sets $K_n$ so that $K_{n+1}$ is a union of all geodesics with ends in $K_n$. The union $W$ of all $K_n$ is convex hull. Now assume it coincides with its closure $bar w$. In particular if $xinpartialbar W$ then $xin K_n$ for some $n$. I.e. there is a geodesic in $bar W$ passing through $x$ (if $xnotin K_0$). From convexity, it is clear that such geodesic lies in $partial bar W$...

ct.category theory - What are the 'standard' applications of the duals of the adjoint functor theorems?

There are some 'standard' applications of the adjoint functor
theorem (AFT) and the special adjoint functor theorem (SAFT), for
example, the existence of a free $tau$-algebra (where
$tau=$(operations,identities)) on a small set by the AFT,
Stone-Cech compactification by the SAFT, and, if I am not mistaken, the proof that the category of $tau$-algebras is cocomplete (by
using the AFT to establish a left adjoint to the appropriate diagonal
functor).

However, I was not able to find any applications of the duals of
the AFT and the SAFT, neither in MacLane, nor in the Joy of
Cats.

The Joy of Cats contains the following intriguing
remark on p. 311:

Since many familiar categories have separators but
fail to have coseparators, the dual of the Special Adjoint Functor
Theorem is applicable even more often than the theorem itself.

But what are the mentioned application of the dual of the SAFT?

So my questions is: What are the 'standard' applications of the duals
of the AFT and the SAFT?

Googling for combinations of phrases like ''adjoint functor theorem''
and ''dual'' is not very useful, so I have tried ''dual of the adjoint
functor theorem'' and ''dual of the special adjoint functor theorem.''
This resulted in a total of 7 papers/books, from which I was not able
to get a clear answer to the current question. I have also tried in
The Wikipedia article on adjoint
functors, in nLab's
article on the adjoint functor
theorems, and
in some
MO
questions, but without success.

ct.category theory - Lax and Colax Monads

This isn't technically an answer, but depending on your examples, you might want to think about lax/colax monads on (pseudo) double categories instead. Part of the problem with lax monads on bicategories is that there is no tricategory of bicategories and lax functors, whereas there is a 2-category of pseudo double categories and lax functors, so that all of the "formal theory of monads" can be applied directly to lax monads on double categories. Most of the lax functors and lax monads that I've seen on bicategories have "actually" lived on double categories, except that people tend to forget about the extra direction of arrows and think only about the bicategory.

at.algebraic topology - blowing up the graphs

Here the graph product is defined as in the Hatcher.

Anyway I am posting a response by my professor which has answered my question:

In algebraic geometry, blowing up is a process of replacing a point inan algebraic variety by a projective space. In graph products, it is the process of replacing each vertex v of the graph by K(G_v,1), and each edge by the appropriate mapping cylinder (as described in Hatcher).

See section 1B on K(G,1) spaces for further detail.

Adeel

gr.group theory - Which finitely presented groups can be distinguished by decidable properties?

This question continues the line of inquiry
of these
three
questions.

Question. Which finitely presented groups can be
distinguished by decidable properties?

To be precise, let us say that φ is a decidable property of finitely
presented groups, if there is a class A of finitely
presented groups, closed under group isomorphisms, such that
{ p | 〈p〉 ∈ A } is decidable, where 〈p〉 denotes the group presented by p. That is, we insist that the decision procedure give the same answer for presentations leading to the same group up to isomorphism.

One extreme case, perhaps unlikely, would be that any two non-isomorphic
finitely presented groups can be distinguished by decidable
properties, so that for any two finitely presented
non-isomorphic groups 〈p〉 and 〈q〉,
there is a decidable property φ where φ(p) holds
and φ(q) fails. That would be quite remarkable.

If this is not the case, then there would be two finite group
presentations p and q, such that the groups presented
〈p〉 and 〈q〉 are not isomorphic, but
they have all the same decidable properties. This also
would be remarkable.

Which is the case?

Another way to describe the question is in terms of the
equivalence relation ≡, which I introduced in my
previous question, where p
≡ q if φ(p) and φ(q) have the same answer for
any decidable property φ of finitely presented groups. This
is precisely the equivalence relation of "having all the
same decidable properties". Of course, this includes the
group-isomorphism relation, and the current question is
asking: What is this relation ≡? In particular, is
≡ the same as the group isomorphism relation?
If it is, then any two non-isomorphic finitely presented
groups can be distinguished by decidable properties; if
not, then there are two finitely presented non-isomorphic
groups 〈p〉, 〈q〉 having all the same
decidable properties.

Henry Wilton has emphasized several times that there are
relatively few truly interesting decidable properties of
finitely presented groups. This may very well be true.
Nevertheless, the answers to the previous MO questions on
this topic have provided at least some decidable
properties, and my question here is asking the extent to
which these properties are able to distinguish any two
finitely presented groups.

In particular, in these previous MO questions, Chad Groft
inquired whether there were
any nontrivial decidable properties of finitely presented
groups. John Stillwell's
answer was
that one could decide many questions about the
abelianization of the group. In a subsequent question, I
inquired whether all decidable properties were really about
the abelianization, and David Speyer's
answer was
that no, there were questions about other quotients, such
as whether the group had a nontrivial homomorphism into a
particular finite group, such as A₅. In a third question, David generalized further and inquired
whether all decidable properties depended on the
profinitization, and the answer again was no (provided by David and Henry). So at least
in these cases we have been increasingly able to separate
groups by decidable properties.

A generalization of the question would move beyond the
decidable properties. For example, if we consider the
computably enumerable (c.e.) properties, then we have quite
a lot more ability to distinguish groups. A property is
c.e. if there is a computable algorithm to determine the
positive instances of φ(p), but without requiring the
negative instances to ever converge on an answer. For
example, the word problem for any finitely presented group,
or indeed, for any computably presented group, is
computably enumerable, since if a word is indeed trivial, we will eventually discover this. Using the same idea as David's answer to my question, it follows that the question of
whether a finitely presented group 〈p〉 admits a
nontrivial homomorphism into the integers Z, say, or many other
groups, is computably enumerable. One may simply try out all possible maps of the generators. A generalization of
this establishes:

Theorem. The question of whether one finitely
presented group 〈p〉 maps homomorphically onto (or into) another 〈q〉 is computably enumerable.

The proof is that given p and q, one can look for a map of
the generators of p to the words of q, such that all
relations of p are obeyed by the image in q, and such that
all the generators of q are in the range of the resulting
map. This is a c.e.
property, since one can look for all possible candidates
for the map of the generators of p into words of q, and
check whether the relations are obeyed and the generators
of q are in the range of the map and so on. If they are,
eventually this will be observed, and at the point one can
be confident that 〈p〉 maps onto 〈q〉. More generally, is the isomorphism relation itself c.e.? It is surely computable from the halting problem 0', since we could ask 0' whether the kernel of the proposed map was trivial or not, and it would know the answer.

• Where does the isomorphism relation on finitely
  presented groups fit into the hierarchy of Turing
  degrees? Is it c.e.? Is it Turing equivalent to the Halting problem?

Once one moves to the c.e. properties, it is similarly
natural to move beyond the finitely presented groups to the
computably presented groups (those having a computable
presentation, not necessarily finitely generated). In this
context, the proof above no longer works, and the natural
generalization of the question asks:

• Which computably presented groups are distinguished by
  c.e. properties?

The isomorphism relation on finitely generated computably
presented groups (given the presentations) seems to be
computable from the halting problem for the same reason as
in the proof above, but now one doesn't know at a finite
stage that the proposed map of the generators will
definitely work, since one must still check all the
relations-yet-to-be-enumerated. But 0' knows the answer, so
we get it computably in 0'. In the infinitely generated case,
however, things are more complicated.

ag.algebraic geometry - Existence of hyperelliptic curve with specific number of points in a family

I think that this is equivalent to a known open question. Here are
the details. For $K:=mathbb{F}_{2^n}$, the function $f:ymapsto
y+y^2:Kto K$ is $mathbb{F}_2$-linear, and its kernel ${0,1}$ has
dimension 1. The image is therefore of dimension $n-1$, and for $z$ in
the image, the fiber $f^{-1}(z)$ has exactly 2 elements.

Hence, to prove that $y^2+y=x^k+ax$ has exactly $2^n$ solutions
for some fixed $ain K$, we have to show that $|{xin K|x^k+axin
mathrm{Im}(f)}|=2^{n-1}$.

Since $sigma:ymapsto y^2$ is a generator of the Galois group of
$K/mathbb{F}_2$, Hilbert's Theorem 90 (in additive form) says that
$zin mathrm{Im}(f)$ if and only if $mathrm{Tr}(z)=0$, where
$mathrm{Tr}$ stands for the trace map from $K$ to $mathbb{F}_2$.

So the problem is equivalent to showing that there exists an $aneq 0$ in $K$
such that $|{xin K|mathrm{Tr}(x^k+ax)=0}|=2^{n-1}$. In other
words, we would like to show that there exists a nonzero $ain K$ such that
$$
S_k(a):=sum_{xin K}(-1)^{mathrm{Tr}(x^k+ax)}=0.
$$

Apparently, this question was addressed in the coding community. In
detail, in [1, p. 258], the following conjecture (of Helleseth) is
mentioned:

``Conjecture 3. For any $m$ and $k$ such that
$mathrm{gcd}(2^m-1,k)=1$, the sum
$sum_{xinmathbb{F}_{2^m}}(-1)^{mathrm{Tr}(x^k+ax)}$ is null for at
least one nonzero $a$.'' (Note that $n$ in the current question is
$m$ in 1).

It seems that in [1, Corollary 1, p. 253], Conjecture 3 is proved for
even $m$ and for certain values of $k$ (the ``Niho exponents,''
defined on p. 252 of 1).

Interestingly, at least at a first glance it seems that 1 has
nothing to say on $kin{1,ldots,2^{n-1}}$, but to me it seems that
this case is trivial (am I missing something?): Consider a normal
basis for $K/mathbb{F}_2$, that is, a basis $B$ consisting of an orbit of
an element $gammain K$ under the Galois group of $K/mathbb{F}_2$
(the $i$th element of $B$ is $b_i:=gamma^{2^i}$ for
$iin{0,ldots,n-1}$).

From the linearity of the trace and the fact that the trace is onto,
we must have $mathrm{Tr}(b)=1$ for at least one element $bin B$, and
from $mathrm{Tr}(b^2)=mathrm{Tr}(b)$ we then have
$mathrm{Tr}(b)=1$ for all $bin B$. So the trace of an element in $K$
is just the modulo-2 sum of the coefficients in its decomposition
according to the basis $B$.

Let $a$ be any element in the
trace-dual basis of $B$, say $mathrm{Tr}(ab_i)=delta_{i,0}$. Then
for $k=2^j$, if we write $x=sum_i alpha_i b_i$, we
get: $mathrm{Tr}(ax)=alpha_0$,
$mathrm{Tr}(x^k)=mathrm{Tr}(x)=sum alpha_i$ (sum in
$mathbb{F}_2$). These agree for half of the $xin K$, as required.

That's about it. I hope at least some of this makes sense :) I also
hope that the original person asking this question didn't actually want
to solve the above open question by converting it to a question about
curves, for then this answer is useless.

1 P. Charpin, ``Cyclic codes with few weights and Niho exponents,''
Journal of Combinatorial Theory, Series A 108 (2004)
247--259.

riemannian geometry - Special Holonomy Groups for Lorentzian Manifolds

I think it would be more accurate to say that the real reason why
Calabi-Yau, hyperkähler, $G_2$ and $mathrm{Spin}(7)$ manifolds are of
interest in string theory is not their Ricci-flatness, but the fact
that they admit parallel spinor fields. Of course, in
positive-definite signature, existence of parallel spinor fields
implies Ricci-flatness, but the converse is still open for compact
riemannian manifolds, as discussed in this
question, and known to fail for noncompact manifolds as pointed
out in an answer to that question.

The similar question for lorentzian manifolds has a bit of history.
First of all, the holonomy principle states that a spin manifold
admits parallel spinor fields if and only if (the spin lift of) its
holonomy group is contained in the stabilizer subgroup of a nonzero
spinor. Some low-dimensional (i.e., $leq 11$, the cases relevant to
string and M-theories) investigations (by Robert Bryant and myself,
independently) suggested that these subgroups are either of two types:
subgroups $G < mathrm{Spin}(n) < mathrm{Spin}(1,n)$, whence $G$ is
the ones corresponding to the cases 5-8 in the question, or else $G =
H ltimes mathbb{R}^{n-1}$, where $H < mathrm{Spin}(n-1)$ is one of
the groups in cases 5-8 in the question. Thomas Leistner showed that
this persisted in the general case and, as Igor pointed out in his
answer, arrived at a classification of possible lorentzian holonomy
groups. Anton Galaev then constructed metrics with all the possible
holonomy groups, showing that they all arise. Their work is reviewed
in their
paper (MR2436228).

The basic difficulty in the indefinite-signature case is that the de
Rham decomposition theorem is modified. Recall that the de Rham
decomposition theorem states that if $(M,g)$ is a complete, connected
and simply connected positive-definite riemannian manifold and if the
holonomy group acts reducibly, then the manifold is a riemannian
product, whence it is enough to restrict to irreducible holonomy
representations. This is by no means a trivial problem, but is
tractable.

In contrast, in the indefinite signature situation, there is a
modification of this theorem due to Wu, which says that it is not
enough for the holonomy representation to be reducible, it has to be
nondegenerately reducible. This means that it is fully reducible
and the direct sums in the decomposition are orthogonal with respect
to the metric. This means that it is therefore not enough to restrict
oneself to irreducible holonomy representations. For example,
Bérard-Bergery and Ikemakhen proved that the only lorentzian holonomy
group acting irreducibly is $mathrm{SO}_0(1,n)$ itself: namely, the
generic holonomy group.

It should be pointed out that in indefinite signature, the
integrability condition for the existence of parallel spinor fields is
not Ricci-flatness. Instead, it's that the image of the Ricci
operator $S: TM to TM$, defined by $g(S(X),Y) = r(X,Y)$, with $r$ the
Ricci curvature, be isotropic. Hence if one is interested in
supersymmetric solutions of supergravity theories (without fluxes) one
is interested in Ricci-flat lorentzian manifolds (of the relevant
dimension) admitting parallel spinor fields. It is now not enough to
reduce the holonomy to the isotropy of a spinor, but the
Ricci-flatness equation must be imposed additionally.

cv.complex variables - Conformal maps in higher dimensions

I think you're looking for Liouville's theorem. This theorem states that for $n >2$, if $V_1,V_2 subset mathbb{R}^n$ are open subsets and $f : V_1 rightarrow V_2$ is a smooth conformal map, then $f$ is the restriction of a higher-dimensional analogue of a Mobius transformation.

By the way, observe that there are no assumptions on the topology of the $V_i$ -- they don't have to be simply-connected, etc.

---

EDIT : I'm updating this ancient answer to
link to a blog post by Danny Calegari which contains a sketch of a beautifully geometric argument for Liouville's theorem.

dg.differential geometry - Is there a complex structure on the 6-sphere?

Here is a shot in the dark (Disclosure: I really know nothing about this problem).

Let $G:=mathsf{SU}(2)$ act on $G^3$ by simultaneous conjugation; namely, $$gcdot(a,b,c)=(gag^{-1},gbg^{-1},gcg^{-1}).$$ Then the quotient space is homeomorphic to $S^6$ (see Bratholdt-Cooper).

The evaluation map shows that the character variety $mathfrak{X}:=mathrm{Hom}(pi_1(Sigma),G)/G$ is homeomorphic to $G^3/G,$ where $Sigma$ is an elliptic curve with two punctures.

Fixing generic conjugation classes around the punctures, by results of Mehta and Seshadri (Math. Ann. 248, 1980), gives the moduli space of fixed determinant rank 2 degree 0 parabolic vector bundles over $Sigma$ (where we now think of the punctures are marked points with parabolic structure). In particular, these subspaces are projective varieties.

Letting the boundary data vary over all possibilities gives a foliation of $mathfrak{X}cong G^3/Gcong S^6$. Therefore, we have a foliation of $S^6$ where generic leaves are projective varieties; in particular, complex.

Moreover, the leaves are symplectic given by Goldman's 2-form; making them Kähler (generically). The symplectic structures on the leaves globalize to a Poisson structure on all of $mathfrak{X}$.

Is it possible that the complex structures on the generic leaves also globalize?

Here are some issues:

1. As far as I know, the existence of complex structures on the leaves is generic. It is known to exist exactly when there is a correspondence to a moduli space of parabolic bundles. This happens for most, but perhaps not all, conjugation classes around the punctures (or marked points). So I would first want to show that all the leaves of this foliation do in fact admit a complex structure. Given how explicit this construction is, if it is true, it may be possible to establish it by brute force.


2. Assuming item 1., then one needs to show that the structures on the leaves globalize to a complex structure on all of $mathfrak{X}$. Given that in this setting, the foliation is given by the fibers of the map: $mathfrak{X}to [-2,2]times [-2,2]$ by $[rho]mapsto (mathrm{Tr}(rho(c_1)),mathrm{Tr}(rho(c_2)))$ with respect to a presentation $pi_1(Sigma)=langle a,b,c_1,c_2 | aba^{-1}b^{-1}c_1c_2=1rangle$, it seems conceivable that the structures on the leaves might be compatible.


3. Moreover, $mathfrak{X}$ is not a smooth manifold. It is singular despite being homeomorphic to $S^6$. So lastly, one would have to argue that everything in play (leaves, total space and complex structure) can by "smoothed out" in a compatible fashion. This to me seems like the hardest part, if 1. and 2. are even true.

Anyway, it is a shot in the dark, probably this is not possible...just the first thing I thought of when I read the question.

Relative version of sheaf cohomology?

To elaborate on Kevin's comment: If $f: X to S$, and $mathcal F$ is a sheaf on
$X$, then $f_*mathcal F$ is the sheaf on $S$ defined by $H^0(U,f_*mathcal F)
:= H^0(f^{-1}(U),mathcal F).$

Taking the derived functors of $f_*$ gives functors $R^if_*$, and it turns out
(fairly easily) that $R^if_*(mathcal F)(U)$ is the sheaf associated to the presheaf
$U mapsto H^i(f^{-1}(U),mathcal F)$. If $i > 0,$ then this presheaf may not be
a sheaf (unlike the $i = 0$ case), and this is related to the fact that it can be a little subtle to compute the stalks of $R^if_*mathcal F$ in general; for example, it need not
be the case in general that the stalk $(R^if_*mathcal F)_s$ is equal to
$H^i(f^{-1}(s),mathcal F)$. (E.g. think about the case when $f$ is the inclusion of
a punctured disk into a disk, $mathcal F$ is the constant sheaf ${mathbb Z}$,
and $s$ is the centre of the disk (so that $f^{-1}(s)$ is empty).)

In other words, $R^if_*mathcal F$ does not always literally interpolate the cohomology
of the fibres.

There is one case where one knows that $R^if_*mathcal F$ does interpolate
the cohomology of the fibres: if the map $f$ is proper, than the proper base-change theorem
says that the stalk of $R^if_*mathcal F$ at $s$ is the cohomology of $mathcal F$ along
the fibre of $s$. (One good place for these kinds of facts is the beginning of Borel's book on Intersection Cohomology.)

Also, in the context of maps of varieties, if $f$ is proper and $mathcal F$ is a coherent
sheaf, then the completed stalk of $R^if_* mathcal F$ at $s$ coicides with the cohomology
of the pull-back of $mathcal F$ to the formal completion of $f^{-1}(s)$ in $X$. (This is Grothendieck's proper base-change theorem, proved in some form in Hartshorne, Ch. III.)

real analysis - Embedding of BV and L^P spaces

An $L^1_{loc}$ function on $mathbb{R}^n$ is in $BV_{loc}$ iff its distributional derivatives $partial_i finmathcal{M}^1_{loc}$, i.e. they are all locally finite (Radon) measures. If $n=1$, the situation is well-known, and $BV_{loc}subset L^infty_{loc}$. So assume $ngeq 2$. Since $W^{s,p}_{c}(mathbb{R}^n)subset C^0_{c}(mathbb{R}^n)$ if $s>n/p$, you have that $$BV_{loc}(mathbb{R}^n)subset W^{1-s,p'}_{loc}(mathbb{R}^n)subset L^{p'}_{loc}(mathbb{R}^n)$$ if $sleq 1$ and $1/p+1/p'=1$, that is if $p'<n/(n-1)$. On the other hand, when $n>1$, $1/r^alpha$ is in $BV_{loc}(mathbb{R}^n)$ if $alpha<n-1$, since partial derivatives are in $L^1_{loc}$, but it is in $L^q$ only for $q<n/alpha$, so that $BV_{loc}(mathbb{R}^n)subset L^q_{loc}(mathbb{R}^n)$ fails for any $q>n/(n-1)$. I wouldn't bet on the limiting case.

pr.probability - A Local CLT with large variance

For n an even integer, $0 leq i leq$ ${n} choose{j}$, $1 leq j leq n$ let $X_{i,j}$ be a
random variable taking values $frac{n}{2}-j,0,j - frac{n}{2}$ with equal probability. Let $S_{n}$ be the sum of these $2^{n}$
random variables.

My question is, what is the 'correct' local limit theorem for this sum as n goes to infinity?
That is, what is a local limit theorem that is in some sense sharp?

To those who have not heard of the term: A local limit theorem is one that describes probabilities
of the sum being a specific number rather than being in a region of size roughly proportional
to the square root of the variance.

To those who might think this is a little specialized: You're right, of course. I wanted to
post a question with a concrete answer, and this is the simplest one that seems to be 'on the
edge' of where local CLTs hold. The versions you see in textbooks fail for this (those that
I'm familiar with fail in a few places), but 'just barely'. Also, a local CLT does at least hold
here. I'm interested in other borderline cases as well.

To those who might think this is trivial: It is true that a CLT for this sum follows from
standard textbook theorems (e.g. the Lindeberg CLT). It is even true that the martingale
local CLT can be used to get rates here - unfortunately, they seem to be wrong (that is,
quite far from sharp).