nt.number theory - Stark's conjecture and p-adic L-functions

Not long back I asked a question about the existence of p-adic L-functions for number fields that are not totally real; and I was told that when the number field concerned has a nontrivial totally real or CM subfield, then there is a construction due to various people including Coates-Sinnott and Katz.

But my favourite number field at the moment is K = $mathbb{Q}(sqrt[3]{2})$, and sadly K contains no totally real or CM subfield, so for trivial reasons $L(n, chi) = 0$ for every Groessencharacter $chi$ of $K$ and every $n le 0$. So in this case the above constructions just give zero. When I learnt this, I thought "that can't be the whole story, what about higher derivatives at 0"? Asking around, I was told about Stark's conjectures, which apparently predict that the leading term at $s = 0$ of the L-function of any GC of K should be the product of an explicit transcendental regulator and an algebraic number (which, if I've understood this right, should lie in the field $mathbb{Q}$(values of $chi$).)

My question is this: assuming Stark's conjecture, can we construct a distribution on the Galois group of the maximal unramified-outside-p abelian extension of K whose evaluation at any locally constant character of this group gives the algebraic part of the leading term at 0 of the L-series of the corresponding Groessencharacter?

pr.probability - most general way to generate pairwise independent random variables?

I'm sure that Gil's answer is wise and that it is a good idea to look at Alon and Spencer's book. Here also is a quick summary of what is going on.

Suppose that $X_1,ldots,X_n$ are random variables, and suppose for simplicity that they take finitely many values. Suppose that you prescribe the distribution of each $X_i$, and suppose also that you want the random variables to be pairwise independent or $k$-wise independent. Then the constraints on the joint distribution are a finite list of equalities and inequalities. The solution set is a polytope whose dimension is fairly predictable, and the fully independent distribution is always in the interior of this polytope. If you are interested in $k$-wise independent distributions that are far from $k+1$-wise independent, then it can be difficult to determine what is achievable because the polytope is complicated. (The vertices are a particularly interesting and non-trivial class: $k$-wise independent distributions with small support. These are called "weighted orthogonal arrays".) However, if you're just intersted in examples, it is much easier to write down a small deviation of the fully independent distribution. The deviation just satisfies linear equations.

For example, suppose that $X,Y,Z$ are three unbiased Bernoulli random variables (coin flips) that take values $0$ and $1$. Then there are 8 probabilities $p_{ijk}$, one for each outcome $(X,Y,Z) = (i,j,k)$. Then you can set
$$p_{ijk} = frac18 + (-1)^{ijk}epsilon. qquadqquadqquad text{(1)}$$
to get a pairwise independent but not independent distribution. In this simple example, there is a 1-dimensional space of deviations and it is easy to compute how far you can vary the independent solution. (Up to $|epsilon| = frac18$.) In larger cases, the variations can be multidimensional and the polytope of deviations can be more complicated.

Addendum: If I have not made a mistake, all deviations for any finite list of discrete random variables are linear combinations of those of the form (1). More precisely, given discrete random variables $X_1,ldots,X_n$, let $f_i$ be some function of the value of $X_i$ which is 1 for one value, $-1$ for another value, and $0$ otherwise. Then you can make deviations proportional to $prod f_{i_j}$ as long as there are at least $k+1$ factors. It looks like all deviations are a linear combination of those of this form.

pr.probability - Can you explain a step in an expectation maximization algorithm in a Nature article?

These numbers are the normalized likelihoods that the results given in the 10 toss vector
are obtained from the current distributions the coin A (or respectively B).

I'll work out the first two rows for illustration:

The guessed Bernoulli parameter for type A is 0.6 and for type B is 0.5.
According to the binomial distribution formula,
the unnormalized likelihood for obtaining 5H 5T are
From A:

L_A = C(10,5)(0.6)^5(0.4)^5

where C(10,5) is the binomial coefficient 10!/5!5!

Similarly from B we obtain:

L_B = C(10,5)(0.5)^5(0.5)^5

The normalized likelihoods are obtained as

For A: L_A/(L_A+L_B) = 0.4491

For B: L_B/(L_A+L_B) = 0.5509

For the second case 9H 1T

L_A = C(10,9)(0.6)^9(0.4)^1

L_B = C(10,9)(0.5)^9(0.5)^9

The normalized likelihoods:

For A: L_A/(L_A+L_B) = 0.8050

For B: L_B/(L_A+L_B) = 0.1950

fa.functional analysis - Is the category of Banach spaces with contractions an algebraic theory?

I think this is some kind of infinitary algebraic theory, but that it is not a monadic adjunction. That is, if you take the "closed unit ball functor" $B$ from ${bf Ban}_1$ to ${bf Set}$ and the "free Banach space functor" $L: {bf Set} to {bf Ban}_1$, then $L$ is left adjoint to $B$ but this adjunction is not monadic (IIRC, and I often don't).

See, for instance, the first few pages of this paper by Pelletier and Rosicky.

You say something about a closed structure on ${bf Ban}_1$, if I understand this right then this is symmetric monoidal with the tensor being the projective tensor product of Banach spaces. That seems to be well known but little-used, although IMHO having this kind of perspective takes some of the tedium/clutter out of certain computations/constructions in my corner of functional analysis.

I think the ball functor from (Hilbert spaces & contractions) to Set doesn't have a left adjoint, but that's more of a guess than an intuition. Certainly the `natural' attempt to build a left adjoint falls over.

As for putting a closed structure on Hilb .... well, the fact that the natural norm on B(H) is not Hilbertian suggests to me that this won't work. (Put another way, the natural Hilbertian tensor product would dualise to only considering Hilbert-Schmidt class maps between your Hilbert spaces, which in infinite dimensions rules out the identity morphism.)

pr.probability - measurable sets not depending on even coordinates

Let $Asubset{0,1}^omega$ be a measurable set (w.r.t. the usual borel sigma algebra) which does not depend on any even coordinate (that is, if $xin A$ and $x$ and $y$ agree except on a finite number of even coordinates, then $yin A$).

Is it true that $A$ belongs to the sigma-algebra generated by all the odd coordinates + the tail sigma algebra?

To clarify: the tail sigma algebra consists of all the events which do not depend on any coordinate.

It seems to me that this should be some easy/well known measure theory fact/counterexample, but perhaps I'm wrong? Suggestions on where to look for an answer would be welcome.

Note that it is well known the this statement is false if the ground space would be $[0,1]^omega$.

computability theory - {transcendental numbers} {computable transcendental numbers}

Note: Answer is pending update per attached comments.

The difference, stated informally, is that that the non-computable transcendentals in their k-base digit representation are entirely random and non compressible. A computable transcendental, such as e, can be represented by a finite algorithmic description, such as a series expansion, which is a form of compression. For the non-computable numbers no such shorter representation exist. Their shortest computational description is their own infinite digit sequence.
You can read more about computational complexity here:
http://en.wikipedia.org/wiki/Kolmogorov_complexity.

There is a wealth of similar numbers to the Ω class of numbers. In general it is "easy" to come up with new definitions for such numbers. These all belong to the countably infinite set of non-computable definable numbers.

To make matters worse, what is left are the non-definable (and therefore also non-computable) numbers. They are the numbers that cannot be described in any way what-so-ever, other than by just iterating through their infinite non-compressible digit sequence. The set of all non-definable numbers is uncountable.

How many dimensions I need to embed a graph?

As Charles points out, you can always embed a graph in three dimensions. The interesting question is how complicated a surface one needs to embed a graph into. The number of handles one has to attach to a spehere in order for a graph to become embeddable is called the genus of the graph, see graph embedding on Wikipedia, which offers other useful information.

gt.geometric topology - Lipschitz orthonormal frames on submanifolds of $mathbf{R}^n$ ?

You will probably need to require $C^2$ smoothness of your submanifold. Take a simple example of $d=1$ and the manifold the graph of the function $f(x) = x^alpha$ for $x > 0$ and $f(x) = 0$ if $x le 0$. Then for $1 < alpha < 2$, the normal bundle is only Hölder continuous, so no $L$ exists at $x = 0$. Or would you exclude this counterexample for the reason that the exponential map from the normal bundle is not a diffeomorphism onto any ball of radius $r > 0$ at $x = 0$?

Generally, you will probably need to have a bound on the extrinsic curvature, for which $C^2$ and compactness would suffice.

rt.representation theory - Acyclic quivers differing only in arrow directions: functorial isomorphism of representation categories?

The categories are not equivalent. In fact, an acyclic quiver is determined (up to non-unique isomorphism) by the equivalence class of its category of representations. The construction is simple: Find the isomorphism classes of simple objects. These are in bijection with the vertices. For any two simple objects $S$ and $S'$, the number of arrows from vertex $S$ to vertex $S'$ is the dimension of $mathrm{Ext}^1 (S, S')$. The nonuniqueness is because we need to choose a basis of $mathrm{Ext}^1 (S, S')$.

---

There is a lot more to say on this subject, but it doesn't go in the direction your questions are pointing. When the underlying graphs of $Q$ and $R$ are trees, then it is true that the bounded derived categories of $mathrm{Rep} Q$ and $mathrm{Rep} R$ are isomorphic. The basic point here is that, although the reflection functors are not equivalences of categories, the derived reflection functors are equivalences of derived categories. Of course, every Dynkin diagram is a tree, so in particular this is true for Dynkin diagrams.

There is a version of this for non-trees, but I don't know a reference for it nor the exact statement and I don't want to hunt one down until I know whether derived categories are something that you love or that you fear. The proof, of course, is completely different.

Here is a result of Kac you might be happier with. Let our ground field be a finite field $mathbb{F}_q$ and fix a dimension vector $d$. Then the number of isomorphism classes of representations of $Q$ and $R$, of dimension $d$, over $mathbb{F}_q$, is the same. (Infinite root systems, representations of graphs and invariant theory, Theorem 1.)

Morally, one wants to work over an arbitrary field. The statement then is a statement about the stacks of $Q$ and $R$ representations of dimension $d$. But, again, to formulate this you need to know about a lot of machinery: stacks, perverse sheaves, derived categories again.

---

In short, the categories are not equivalent. There are very close relations between them, but the best formulations of those relations use sophisticated category theory. You can often see shadows of these relations by counting points over $mathbb{F}_q$.

ag.algebraic geometry - Explicit equations for Schubert varieties

If one is learning about this, computing directly with matrices seems like the easiest way (though not as powerful as standard monomials and the toric degenerations that result). Alex Woo's references are a good source for this point of view; I'd also add the first couple chapters of the book Schubert varieties and degeneracy loci by Fulton and Pragacz. And a quick example, in that spirit: to find equations for $X_{2143}$ inside $SL_4/B$, do the following:

(1) Form the rank matrix $(r_{ij})$ for the permutation:
$$left[begin{array}{cccc} 0 & 1 & 0 & 0 \
1 & 0 & 0 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 end{array}right] to
left[begin{array}{cccc} 0 & 1 & 1 & 1 \
1 & 2 & 2 & 2 \
1 & 2 & 2 & 3 \
1 & 2 & 3 & 4 end{array}right].$$

(2) Write down the equations on the $4 times 4$ generic matrix that say "upper-left $itimes j$ submatrix has rank at most $r_{ij}$". These are your polynomials cutting out the matrix Schubert variety, e.g.,
$$x_{11}=0, \ det( x_{ij} )_{1leq i,jleq 3} = 0.$$

(3) If you want equations in an open affine patch, set appropriate $x_{ij}$'s to $0$ or $1$, e.g., the opposite cell would have free variables in the $*$ positions:
$$left[begin{array}{cccc} * & * & * & 1 \
* & * & 1 & 0 \
* & 1 & 0 & 0 \
1 & 0 & 0 & 0 end{array}right].$$

(4) If you want equations in the Plucker embedding, the determinants are built-in to the definition of Plucker coordinates, and you just intersect with appropriate linear subspaces.

In general, these equations are highly redundant, and a big part of the work of Lakshmibai et al, Fulton, and Woo-Yong (as I understand) is to find minimal sets of equations. For the matrix Schubert varietes for $SL_n/B$, the Fulton paper cited by Alex gives a simple answer.

nt.number theory - When is the extension defined by an Eisenstein polynomial galoisian or abelian or cyclic ?

Let $p$ be a prime number, $K$ a finite extension of $mathbb{Q}_p$, $mathfrak{o}$ its ring of integers, $mathfrak{p}$ the unique maximal ideal of $mathfrak{o}$, $k=mathfrak{o}/mathfrak{p}$ the residue field, and $q=operatorname{Card} k$.

Recall that a polynomial $varphi=T^n+c_{n-1}T^{n-1}+cdots+c_1T+c_0$ ($n>0$) in $K[T]$ is said to be Eisenstein if $c_iinmathfrak{p}$ for $iin[0,n[$ and if $c_0notinmathfrak{p}^2$.

Question. When is the extension $L_varphi$ defined by $varphi$ galoisian (resp. abelian, resp. cyclic) over $K$ ?

Background. Every Eisenstein polymonial $varphi$ is irreducible, the extension $L_varphi=K[T]/varphi K[T]$ is totally ramified over $K$, and every root of $varphi$ in $L_varphi$ is a uniformiser of $L_varphi$. There is a converse.

If the degree $n$ of $varphi$ is prime to $p$, then the extension $L_varphi|K$ is tamely ramified and can be defined by the polynomial $T^n-pi$ for some uniformiser $pi$ of $K$. Thus $L_varphi|K$ is galoisian if and only if $n|q-1$, and, when such is the case, $L_varphi|K$ is actually cyclic.

Real question. Is there a similar criterion, in case $n=p^m$ is a power of $p$, for deciding if $L_varphi|K$ is galoisian (resp. abelian, resp. cyclic) ?

Sets that can be mapped onto R^n by a polynomial

The answer to the precise version of the question (the one asked at the end) is no.

The map $(x,y) mapsto (x,xy)$ maps $mathbb{R}^2$ onto all of $mathbb{R}^2$ except the positive and negative parts of the $y$-axis. Following this with $z mapsto z^3$ (where $z=x+iy in mathbb{C}$) yields a polynomial surjection $(x,y) mapsto (f_1(x,y),f_2(x,y))$ such that the preimage of $(0,0)$ is the $y$-axis. Define $f(x,y,z) = (f_1(x,y),f_2(x,y),z)$, so $f$ defines a surjection $mathbb{R}^3 to mathbb{R}^3$. Let $U := mathbb{R}^3 - lbrace (0,y,z) : y le e^z rbrace$, so $U$ is a topological ball. The value of $f$ on each deleted point $(0,y,z)$ is the same as its value at $(0,e^z+1,z) in U$, so $f|_U$ is still surjective.

For $r>0$, if $u in U$ and $f(u) = (0,0,log r)$, then $u=(0,y,log r)$ with $y>r$, so $|u|>r$. So $f(U cap B(r))$ does not even contain $(0,0,log r)$, let alone $B(r^epsilon)$ for large $r$ (for fixed $epsilon>0$).

gr.group theory - subgroups of graph groups

This isn't a proper answer, but here are some remarks.

• Wise's work actually purports to tell you that most (all?) hyperbolic 3-manifold groups virtually embed in graph groups - ie, a finite-index subgroup embeds. So you might lose a lot of torsion when you pass to this finite-index subgroup.


• Haglund and Wise have a very nice criterion for injecting fundamental groups of cube complexes into graph groups. It would be completely reasonable to try to use this to construct interesting examples of torsion in the homology of subgroups of graph groups.


• Because the embeddings in Haglund--Wise are quasiconvex, it follows from work of Haglund that every such subgroup H is a virtual retract - that is, there is a finite-index subgroup K that contains H and the inclusion map has a left inverse K->H. In particular, any torsion you see in the homology of H will also show up in the homology of K. It would be easy to use a computer algebra package like GAP to look for torsion in finite-index subgroups of graph groups.

UPDATE:

In a comment below, Stefan points out a very nice way of constructing torsion in finite-index subgroups of graph groups, due to Bridson and Miller. As I'm familiar with the argument, I'll explain it. The graph group in question will always be a direct product of two free groups.

Let Q be any group you like and let q:F->Q be a surjection from a free group. Now take the fibre product K of q with itself; in other words, consider the subgroup

$K= {(g,h)in Ftimes Fmid q(g)=q(h)}~.$

There is a standard five-term exact sequence in homology that derives from the map $q:Fto Q$, which reduces to

$0to H_2(Q)to H_0(Q,H_1(ker f))to H_1(F)to H_1(Q)$

where, by definition, $H_0(Q,H_1(ker f))$ is the quotient of the abelianisation of $ker f$ by the natural action of $Q$ by conjugation. On the other hand, projecting onto a factor decomposes $K$ as $ker frtimes F$ (where the action is, again, by conjugation), from which it follows that

$H_1(K) = H_0(Q,H_1(ker f))oplus H_1(F)~.$

Putting these together, we see that $H_2(Q)$ embeds into $H_1(K)$.

The fibre product $K$ can also be characterised as the preimage of the diagonal subgroup of $Qtimes Q$ in $Ftimes F$, so if $Q$ is finite then $K$ is of finite index in $Ftimes F$. This proves the following.

Proposition. For any finite group $Q$ there is a finite-index subgroup $K$ of $Ftimes F$ with $H_2(Q)subseteq H_1(K)$.

This shows that all sorts of torsion can arise in $H_1$ of subgroups of graph groups.

Remark. We took $Q$ to be finite because otherwise $K$ isn't quasiconvex. But the rest of the argument goes through.

pr.probability - Exponential tails for a functional of a subcritical branching process.

Let $(m_i, i in mathbb{N})$ be positive weights with $sum_{i in mathbb{N}} m_i^2 < 0.1$.
Consider a subcritical branching process in discrete time and continuous space,
started from some initial mass $x>0$, and with branching mechanism as follows:
given mass $m$ in generation $n$, generation $n+1$ has total mass
$$
sum_{i in mathbb{N}} m_i N_i,
$$
where the $N_i$ are independent and $N_i$ has distribution $mathrm{Poisson}(mcdot m_i)$. (One can think of a total number $N_i$ of copies of mass $m_i$ in generation $n+1$.)

Let $Z_n$ be the total mass at level $n$. Does $sum_{n=0}^{infty} Z_n^{1/2}$ then have exponential tails?

computer science - Prove a function is primitive recursive

This is not an exact answer, but it helps to quickly determine in many cases that a given function is primitive recursive. The idea is to use a reasonable programming language in which your function can be expressed more easily than with "raw" arithmetic and primitive recursion. Of course, the programming language must be limited enough to prevent unbounded searches and other constructs that lead to general recursion. I mention here just one easy possibility.

Suppose a function $f : mathbb{N} to mathbb{N}$ is computed by a simple imperative program in polynomial running time (actually, as Grigory points out in his answer, any primitive recursive bound will do). More precisely, the program is allowed to use basic arithmetic operations, boolean values, variables, arrays, loops, and conditional statements. Then the function $f$ is primitive recursive.

The reason that such a program can only define a primitive recursive function is that the entire execution of the program may be encoded by a number $N$ whose bit size is bounded by a polynomial $p(n)$, where $n$ is the input. Because verifying that a number encodes the execution of a simple imperative program is primitive recursive, we can perform a bounded search up to $2^{p(n)}$ to find the number $N$ (such a bounded search is primitive recursive) and extract the answer from it.

Let us apply this to your question. The following Python program computes the function $pi(n)$ and uses just a couple of loops and basic arithmetic (we could replace the remainder function % with another loop). Its running time is quadratic in $n$, assuming all basic operations are constant time:

def pi(n):
   '''Computes the number of primes which are less than or equal n.'''
   p = 0 # the number of primes found
   k = 2 # for each k we test whether it is prime
   while k <= n:
       j = 1 # possible divisors of k
       d = 0 # number of divisors of k found
       while j <= n:
           if k % j == 0: d = d + 1
           j = j + 1
       if d == 2: p = p + 1
       k = k + 1
   return p

Therefore your function is primitive recursive.

nt.number theory - Legitimacy of reducing mod p a complex multiplication action of an elliptic curve?

I scoured Silverman's two books on arithmetic of elliptic curves to find an answer to the following question, and did not find an answer:

Given an elliptic curve E defined over H, a number field, with complex multiplication by R, and P is a prime ideal in the maximal order of H and E has good reduction at P. Is it legitimate to reduce an endomorphism of E mod P?

In the chapter "Complex Multiplication" of the advanced arithmetic topics book by Silverman, a few propositions and theorems mention reducing an endomorphism mod P.

A priori, this doesn't seem trivial to me. Sure, the endomorphism is comprised of two polynomials with coefficients in H. But I still don't see why if a point Q is in the kernel of reduction mod P, why is phi(Q) also there. When I put Q inside the two polynomials, how can I be sure that P is still in the "denominator" of phi(Q)?

(*) I looked at the curves with CM by sqrt(-1), sqrt(-2) and sqrt(-3), and it seems convincing that one can reduce the CM action mod every prime, except maybe in the case of sqrt(-2) at the ramified prime.

complex geometry - Mirror of Flop?

I assume the question regards the coherent sheaves on these two CY's.
These CY's should be regarded as the "same" complex manifold with two
different choices of complexified symplectic forms ("Kahler form," in
physics terminology).

The mirrors are a "single" symplectic manifold with two different
complex structures on it. There is a curve of complex structures
relating the two.

That's about it. The tricky part is to "parallel transport" the
category of coherent sheaves along this curve, using
a "flat family of categories" defined by stability conditions.
Doing so should provide a preferred isomorphism of the categories.
Examples have been studied, but general statements (like the ones
I have glibly been making) are not proven.

fields - transcendental Galois theory

Even in our day of sophisticated search engines, it still seems that the success of a search often turns on knowing exactly the right keyword.

I just followed up on Sylvain Bonnot's comment above. The property of a field extension $K/F$ that for all subextensions $L$ we have $K^{operatorname{Aut}(K/L)} = L$ is apparently most commonly called Dedekind. This terminology appears in Exercise V.9 of Bourbaki's Algebra II, where the reader is asked to show that if $L/K$ is a nonalgebraic Dedekind extension and $T$ is a transcendence basis, then $L/K(T)$ must have infinite degree. Ironically, this is exactly what I could show in my note. One can (in the general case, even...) immediately reduce to the case $T = {t}$ and then the exercise is saying that the function field $K(C)$ of an algebraic curve (again, it is no loss of generality to assume the function field is regular by enlarging $K$) is not Dedekind over $K$. This is kind of a strange coincidence! [However, the proof I give is openly geometric so is probably not the one that N.B. had in mind...]

It also appears in

MR0067098 (16,669f)
Barbilian, D.
Solution exhaustive du problème de Steinitz. (Romanian. Russian, French summary)
Acad. Repub. Pop. Române. Stud. Cerc. Mat. 2, (1951). 195–259 (misprinted 189–253).

In this paper, the author shows that $L/K$ is a Dedekind extension iff for all subextensions $M$, the algebraic closure $M^*$ of $M$ in $L$ is such that $M^*/M$ is Galois in the usual sense: i.e., normal and separable. (This is a nice fact, I suppose, and I didn't know it before, but it seems that the author regarded this as a solution of the problem of which extensions are Dedekind. I don't agree with that, since it doesn't answer my question!)

Apparently one is not supposed to read the above paper but rather this one:

MR0056588 (15,97b)
Krull, Wolfgang
Über eine Verallgemeinerung des Normalkörperbegriffs. (German)
J. Reine Angew. Math. 191, (1953). 54–63.

Here is the MathSciNet review by E.R. Kolchin (who knew something about transcendental
Galois extensions!):

The author reviews a definition and some results of D. Barbilian [Solutia exhaustiva a problemai lui Steinitz, Acad. Repub. Pop. Române. Stud. Cerc. Mat. 2, 189--253 (1950), unavailable in this country], providing proofs which are said to be simpler, and further results. Let L be an extension of a field K. Then L is called normal over K if for every intermediate field M the relative algebraic closure M∗ of M in L is normal (in the usual sense) over M. If L has the property that every M is uniquely determined by the automorphism group U(M) of L over M, then L is normal over K and, if the characteristic p=0, conversely; if p>0 the converse fails but a certain weaker conclusion is obtained. Various further results are found, and constructive aspects of normal extensions are explored. Some open questions are discussed, the most important one being: Do there exist transcendental normal extensions which are not algebraically closed?

So it seems that my question is a nearly 60 year-old problem which was considered but left unsolved by Krull. I am tempted to officially give up at this point, and perhaps write up an expository note informing (and warning?) contemporary readers about this circle of ideas. Comments, suggestions and/or advice would be most welcome...

P.S.: Thanks very much to M. Bonnot.

co.combinatorics - fibonacci identity using generating function

There are many nice ways of showing that $f_0^2+f_1^2+cdots+f_n^2=f_{n+1}f_n$. I was wondering if there is a way of showing this using the generating function $F(x)=frac{1}{1-x-x^2}=sum_{igeq0}f_ix^i$. In other words, is there any operation (perhaps the Hadamard product) that can be applied to $F(x)$ that would yield the identity above?

What about other identities that involve sums and squares, like $f_1f_2+cdots +f_nf_{n+1}=f_{n+1}^2$ for $n$ odd?

books - Errata for Emil Artin's 'The Gamma Function'?

In the English translation of The Gamma Function by Emil Artin (1964 - Holt, Rinehart and Winston) there appears to be a mistake in the formula given for the gamma function on page 24:

$$Gamma(x) = sqrt{2pi}x^{x-1/2}e^{-x+mu(x)}$$
$$mu(x)=sum_{n=0}^infty(x+n+frac{1}{2})text{log}(1+frac{1}{x+n})-1=frac{theta}{12x}, 0 < theta < 1$$

and on page 22 where this is derived, it is noted that '$theta$ is a number independent of $x$ between 0 and 1'.

This sounds incorrect, as $theta$ does depend on $x$, but since the wording is a little ambiguous it may just be an unclear translation. The original German might have meant that $0< theta(x) < 1$ for any $x$. That the variable $x$ is suppressed from $theta$ could be just confusing notation, or someone's misunderstanding (possibly mine.)

The preface does mention that a (different) formula had to be corrected for the English reprint.

I would like to know if there are mistakes in this book, and if so, whether they exist in the German edition. Is there an available list of errata?

set theory - When does collection imply replacement?

The answer is no, if you allow me to adopt some weak-but-equivalent forms of the other axioms. And the reason is interesting:

• A shocking number of the axioms of set theory are true in the non-negative real line R⁺, with the usual order < being used to interpret set membership. (!)

Let's just check. For example, Extensionality holds, because if two real numbers have the same predecessors, then they are equal. The emptyset axiom holds, since there are no non-negative reals below 0. The Union axiom holds, since for any real x, the reals less than x are precisely the reals that are less than something less than x. (Thus, every set is its own union.) A weakened version of the Power set axioms holds, the Proper Power set, which asserts that for every x, there is a set p whose elements are the strict substs of x. This is because for any real number x, the reals below x are precisely the reals y (other than x), all of whose predecessors are less than x. (Thus, every real is its own proper power set.) An alternative weakening of power set would say: for every x, there is p such that y subset x implies y in p. This is true in the reals by using any p > x. A weakened pairing axiom states similarly: for every x,y, there is z with x ε z and y ε z. This is true in the reals by using any z above both x and y. The Foundation axiom is no problem, since 0 is in every nonempty set. Also, similarly AC holds in the form about families of disjoint nonempty sets, since this never occurs in this model. The Weak Collection Axiom holds since if every y < x has phi(x,y,w), then in fact any y in the same interval will work (since this structure has many automorphisms), and so we may collect witness with any B above x and w. Note that Separation fails, since, for example, {0} does not exist in this model. Also, Replacement fails for the same reason.

Similar interesting models can be built by considering the structure (ORD,<) built using only ordinals, or the class { V_α | α in ORD }. These also satisfy all of the weakened forms of the ZFC axioms without Separation, using Collection in place of Replacement.

Thus, part of the answer to your question is that it depends on what you mean by the "other axioms of ZFC".

Apart from this, however, let me say that the term Weak Collection is usually used to refer to the axioms that restrict the complexity of the formulas in the usual Collection scheme, rather than the axiom that you state. For example, in Kripke Platek set theory KP, a weak fragment of ZFC, one has collection only for Sigma_1 formulas, and this is described as a Weak Collection axiom. (What you call Weak Collection is usually just called Collection.) And there is a correspondingly weakened version of Separation in KP.

But I am happy to adopt your terminology here. You stated that AC plus Replacement implies Weak Collection, but this is not quite right. You don't need any AC. Instead, as your argument shows, what you need is the cumulative V_α hierarchy, which is built on the Power set axiom, not AC. That is, If you have Replacement and Power set and enough else to build the V_α hierarchy, then you get Collection as you described, even if AC fails. For example, ZF can be axiomatized equivalently with either Collection or Replacement.

Your question is a bit unusual, since usually Separation is regarded as a more fundamental axiom than Replacement and Collection, and more in keeping with what we mean by set theory. After all, if one has a set A and a property phi(x), particularly when phi is very simple, it is one of the most basic set theoretic constructions to be able to form { x in A | phi(x) }, and any set theory violating this is not very set-like. We don't really want to consider models of set theory where many instances of Separation fail (for example, Separation for atomic formula is surely elemental).

Incidentally, there is another version of a weakening of Collection in the same vein as what you are considering. Namely, consider the scheme of assertions, whenever phi(x,y) is a property, that says for every set A, there is a set B such that for every a in A, if there is b with phi(a,b), then there is b in B such that phi(a,b).

---

OK, let me now give a positive answer, with what I think is a more sensible interpretation of your question. I take what I said above (and Dorais's comment) to show that we shouldn't consider set theories where the Separation Axiom fails utterly. Rather, what we want is some very weak set theory, such as the Kripke Platek axioms, and then ask the relationship between Weak Collection and Replacement over those axioms. And here, you get the postive result as desired.

Theorem. If KP holds, then Weak Collection implies Replacement.

Proof. Assume KP plus (Weak) Collection. First, I claim that this is enough to prove a version of the Reflection Principle, since that proof amounts to taking successive upward Skolem hulls, which is what Collection allows. That is, I claim that for every set x and any formula phi, there is a transitive set Y such that x in Y and phi(w) is absolute between Y and the universe V. This will in effect turns any formula into a Delta₀ formula using parameter Y.

Applying this, suppose we have a set A and every a in A has a unique b such that phi(a,b). By the Reflection Principle, let Y be a large set transitive containing A such that phi(a,b) and "exists b phi(a,b)" are absolute between Y and V. So Y has all the desired witnesses b for a in A. But also, now { b | exists a in A phi(a,b) } is a Delta₀ definable subset of Y, since we can bound the quantifier again by Y. So the set exists. So Replacement holds. QED

I think we can get away with much less than KP. Perhaps one way to do the argument is to just prove Separation by induction on the complexity of formulas. One can collect witnesses by (weak) Collection, and this turns the formulas into lower complexity, using the new bound as a bound on the quantifiers.

pr.probability - A random variable: is it a function or an equivalence class of functions?

I think this question may be slightly deeper than some people are giving it credit for being. I lectured a course in probability to first-year undergraduates at Cambridge recently, and a previous lecturer, who was a genuine probabilist, was very keen to impress on me the importance of talking "correctly" about random variables. It took me a while to understand what he meant, but basically his concern was that the notion of a sample space should be very much in the background. It's tempting to define a random variable as a function on a probability measure space (not that this particular course used measure theory -- but some more elementary substitute for the definition would have been needed), but his view was that this was absolutely not how probabilists think about random variables.

The practical point was not so much to come up with a better formal definition of random variables, but rather to try, whenever possible, to prove results about random variables without referring to the sample space. It's surprising how little you need to mention them (or not surprising if you're a real probabilist). I seem to remember that the one place where I found I really wanted sample spaces was when it came to proving linearity of expectations.

The compromise I reached in that particular course was to define random variables using sample spaces (which makes them seem fairly straightforward objects) but then to tell people to prove as much as they could just with reference to the distribution of the random variable itself. In other words, I gave the "wrong" definition and immediately admitted that it was wrong.

Added very slightly later: I see that I am interpreting the question differently from everyone else. I am not talking about two functions on a measure space being equivalent if they agree outside a set of measure zero -- which is indeed not a very interesting issue. I am talking about two functions on different measure spaces being the same random variable if you can find a nice map between the measure spaces such that one function is (up to a set of measure zero) the obvious transformation of the other. One of the big advantages of not specifying a sample space is when you start talking about several random variables. For instance, if you start by discussing the tossing of a coin, it's sort of clear that your sample space is ${0,1}$, but if another coin enters the picture does that mean you have to go back and prove everything for a more complicated function defined on ${0,1}^2$? Not if you didn't mention the sample space in the first place but just the random variable.

pr.probability - What's the use of a complete measure?

In light of the comments here, I'm going to show why completeness can be a pain. In exercise 9 of section 2.1 of Folland, he develops a function $g: [0,1] to [0,2]$ by $g(x) = f(x) + x$ where $f : [0,1] to [0,1]$ is the Cantor function. In that exercise it is established that $g$ is a (monotonic increasing) bijection, and that its inverse $h = g^{-1}$ is continuous from $[0,2]$ to $[0,1]$.

Since $h$ is continuous, it is Borel measurable. On the other hand, $h$ is not $(mathcal{L}, mathcal{L})$-measurable!! In particular, let $C$ be the Cantor set; $m(g(C)) = 1$, but this means there is a subset $A subseteq g(C)$ which is not Lebesgue measurable. On the other hand $B := g^{-1}(A) subseteq C$ whereas $m(C) = 0$; thus this preimage $B$ is Lebesgue measurable (with measure zero). But therefore $h^{-1}(B) = A$ is not Lebesgue measurable, meaning $h$ is not $(mathcal{L}, mathcal{L})$-measurable.

On one hand, this function is contrived. On the other hand, it shows that completing measures can mess things up. The typical definition of "measurable function" is a Borel
measurable function, and I suppose reasons like the above led to this convention. I do not know the material Bridge references above, and so can't say what breaks when completeness is dropped. Although it seems mathematically convenient to throw in completeness, I don't know any examples in basic probability theory where it helps. For instance, Fubini-Tonelli can be formulated just fine without completeness. Your statement of the theorem only need mention completeness if your measures happen to be complete!

EDIT I corrected the nonsense in the second paragraph; also I meant to talk about $(mathcal L, mathcal L)$-measurable functions, which I accidentally refered to as Lebesgue measurable (which means $(mathcal L, mathcal B)$-measurable). My whole point is that if you take completion in $sigma$-algebra of the range space, the extra sets you added could map back to basically anything. IE it is somewhat nonsensical to add in all sorts of null sets, but not all sorts of finite measure sets. Sometimes completion gives you something you want, but sometimes it does not, as I showed here--the function is better behaved wrt the non-completed measure.

nt.number theory - Fermat's Bachet-Mordell Equation

Fermat once claimed that the only integral solutions to $y^2 = x^3 - 2$ are $(3, pm 5)$.
Fermat knew Bachet's duplication formulas (more precisely, Bachet had a formula for computing what we call $-2P$), which for $y^2 = x^3 + ax + b$ says
$$ x_{2P} = frac{x^4-8bx}{4x^3+4b}
= frac x4 cdot frac{x^3 - 8b}{x^3+b}.$$

Using this formula it is easy to prove the following:

Consider the point $P = (3,5)$ on the elliptic curve $y^2 = x^3 - 2$.
The $x$-coordinate $x_n$ of $[-2]^nP$ has a denominator divisible by
$4^n$; in particular, $[-2]^nP$ has integral coordinates only if $n = 0$.

In fact, writing $x_n = p_n/q_n$ for coprime integers $p_n$, $q_n$, we find
$$ x_{n+1} = frac{x_n}4 cdot frac{x_n^3 + 16}{x_n^3 - 2}
= frac{p_n}{4q_n} cdot
frac{p_n^3 + 16q_n^3}{p_n^3 - 2q_n^3}. $$
Since $p_n$ is odd for $n ge 1$ and $q_n = 4^nu$ for some odd number $u$ (use induction), we deduce that the power of $2$ dividing $q_{n+1}$ is $4$ times that dividing $q_n$.

My question is whether the general result that $kP$ has integral affine coordinates if and only if $k = pm 1$ can be proved along similarly simple lines. The modern proofs based on the group law, if I recall it correctly, use Baker's theorem on linear forms in logarithms.

ag.algebraic geometry - Proof of Borel-Weil-Bott Theorem

The simplest proof of Borel-Weil-Bott that I know is due to Demazure: he has two papers in Inventiones (one in 1968 the other in 1976) on the theorem, and the second is two pages long -- it gives a simplification of his previous proof, and he uses only algebro-geometric techniques. Both papers are readable.

pr.probability - Equality in the union bound.

Your "lemma" is false for finite probability spaces, e.g.,
$Omega = {a,b}, mathbb P({a})=0,mathbb P({a,b})=1, mathbb P({a} cup {a,b})=1.$

After you fix it, a cannon to swat the fly is inclusion-exclusion, or more specifically, the Bonferroni inequalities.

I think people are confusing your question as stated with the natural and very elementary question of whether countably additive probabilities must be uncountably additive, and the example of Lebesgue measure on $[0,1]$ shows this this is not the case.

You should very rarely do anything with the sample space itself in any intrinsic probability question. See the answer gowers gave to this question and this Tao blog entry. It's ok to have a sample space when you apply probability to something like an analysis question (e.g., proving the Weierstrauss approximation theorem using probability) or use the probabilistic method in combinatorics.

reference request - Numerical algorithms on mixed-precision computational models.

I wonder whether this approach is worthwhile on modern hardware. There's not as much difference between integer, float, and double operations as there used to be.

One way it may matter, however, is saving space. With smaller data types, you can fit more numbers at a time into cache, and that may give you a speed up. The time savings doesn't come from arithmetic operations but from faster access to the data if you can avoid a cache miss.

References for theorem about unipotent algebraic groups in char=0?

There is a textbook theorem that the categories of unipotent algebraic groups and nilpotent finite-dimensional Lie algebras are equivalent in characteristic zero. Indeed, the exponential map is an algebraic isomorphism in this case and the group structure can be defined in terms of the Lie algebra structure and vice versa via the Campbell-Hausdorff series, which is finite due to nilpotency.

My problem is that I am unable to locate any textbook where this textbook theorem is stated. The books by Borel, Humphreys, Springer, Serre do not seem to mention this theorem.

The only reference I was able to locate is this original paper by Hochschild (which refers to his earlier papers), but he does it in a heavy Hopf-algebra language that is good, too, but still leaves one desiring to find also a simple textbook-style exposition. Later Hochschild wrote a book "Basic Theory of Algebraic Groups and Lie Algebras" on the subject, to which I have presently no access, but judging by Parshall's review, it is certainly not textbook-style.

Could anyone suggest a simple reference for this textbook theorem?

ag.algebraic geometry - Deligne-Simpson problem in the symmetric group

Question.
Let $C_1,dots,C_k$ be conjugacy classes in the symmetric group $S_n$. (More explicitly,
each $C_i$ is given by a partition of $n$; $C_i$ consists of permutations whose cycles
have the length prescribed by the partition.) Give a necessary and sufficient condition on
$C_i$ that would ensure that there are permutations $sigma_iin C_i$ with
$$prodsigma_i=1.$$

Variant.
Same question, but now $sigma_i$'s are required to be irreducible in the sense that
they have no common invariant proper subsets $Ssubsetlbrace 1,dots,nrbrace$.

I am not certain how hard this question is, and I would appreciate any comments or observations. (I was unable to find references, but perhaps I wasn't looking for the right things.)

This question is inspired by Jonah Sinick's question via the simple

Geometric interpretation.
Consider the Riemann sphere with $k$-punctures $X=mathbb{CP}^1-lbrace x_1,dots,x_krbrace$.
Its fundamental group $pi_1(X)$ is generated by loops $gamma_i$ ($i=1,dots,k$) subject to the relation
$$prodgamma_i=1.$$
Thus, homomorphisms $pi_1(X)to S_n$ describe degree $n$ covers of $X$, and the problem
can be stated as follows: Determine whether there exists a cover of $X$ with prescribed
ramification over each $x_i$. The variant requires in addition the cover to be irreducible.

Background.
The Deligne-Simpson Problem refers to the following question:

Fix conjugacy classes $C_1,dots,C_kinmathrm{GL}(n,mathbb{C})$ (given explicitly by $k$ Jordan forms). What is the necessary and sufficient condition for existence of matrices
$A_iin C_i$ with $$prod A_i=1$$
(variant: require that $A_i$'s have no common proper invariant subspaces)?

There are quite a few papers on the subject; my favorite is Simpson's paper, which has references to other relevant papers. The problem has a very non-trivial solution (even stating the answer is not easy): first there is a certain descent procedure (Katz's middle convolution algorithm) and then the answer is constructed directly (as far as I understand, there are two answers: Crawley-Boevey's argument with parabolic bundles, and Simpson's construction using non-abelian Hodge theory).

The same geometric interpretation shows that the usual Deligne-Simpson problem is
about finding local systems (variant: irreducible local systems) on $X$ with prescribed local monodromy.

So: any remarks on what happens if we go from $mathrm{GL}(n)$ to $S_n$?

mg.metric geometry - Stable Tables on Fluctuating Floors

If a four-legged, rectangular table is rickety, it can nearly always be stabilised just by turning it a little. This is very useful in everyday life! Of course it relies on the floor being the source of the ricketiness; if the table's legs are different lengths, it doesn't work (this is why I said 'nearly always').

Here is a quick 'proof': Label the feet A, B, C, D in order, and integrate the function F(A) + F(C) - F(B) - F(D) while the table is turned through 180° (here F is the height of the floor). The result is ∫ F - ∫ F = 0 (these integrals are over 360°), so the function must be zero somewhere; at this point, the table is stable.

This proof can easily be adapted for any cyclic quadrilateral ABCD.

The proof only works if the slope S of the floor is small enough that S² can be ignored. If not, and if the stable solution is not horizontal, then the feet will not lie on the projection of the circle that we integrated over. So a more complicated argument is required:

Suppose ABCD is a square, and suppose that |F(x)-F(y)| <= S|x-y| everywhere, where S = sin^-1(1/√3). Fix a vertical line V. Then ABCD can always be positioned with its centre on V, so that each corner touches the floor.

To see this, consider the two diagonals AC and BD separately. Choose a starting position such that

(i) their endpoints lie on the floor, 
(ii) their midpoints are on V,
(iii) they are perpendicular to each other.

We are not requiring that their centres coincide (so we will have to dismantle the table to achieve this). We may suppose that in the starting position, the centre of AB lies above the centre of BD. Now rotate them so that constraints (i)-(iii) remain satisfied, until AC occupies the starting position of BD and vice versa. (The constraint on the slope ensures that we can do this continuously.) Now the centre of AB lies below the centre of BD, so the two centres must have coincided at some time.

(This proof generalises to rectangles, but the maximum slope depends on the rectangle. It does not generalise to cyclic quadrilaterals.)

My first question is this: Is the condition on the slope (or a less stringent slope condition) necessary? Or can we always stabilise the table, whatever the slope of the floor?

My second question (if the answer to the first question is that we can always stabilise the table): Given any partition of $mathbb R$³ into {L,U} with L bounded above and U bounded below, and a vertical line V, can we always find a unit square whose corners lie in the closures of both L and U, and whose centre lies on V?

Update The paper referenced below by Q.Q.J. answers my first question: a rectangular table can always be stabilised, if the floor function F is continuous. But it can only be stabilised by a smooth rotation if the floor satisfies the slope condition. (I was wrong about different rectangles requiring different slope conditions.)

What is the difference between matrix theory and linear algebra?

Let me elaborate a little on what Steve Huntsman is talking about. A matrix is just a list of numbers, and you're allowed to add and multiply matrices by combining those numbers in a certain way. When you talk about matrices, you're allowed to talk about things like the entry in the 3rd row and 4th column, and so forth. In this setting, matrices are useful for representing things like transition probabilities in a Markov chain, where each entry indicates the probability of transitioning from one state to another. You can do lots of interesting numerical things with matrices, and these interesting numerical things are very important because matrices show up a lot in engineering and the sciences.

In linear algebra, however, you instead talk about linear transformations, which are not (I cannot emphasize this enough) a list of numbers, although sometimes it is convenient to use a particular matrix to write down a linear transformation. The difference between a linear transformation and a matrix is not easy to grasp the first time you see it, and most people would be fine with conflating the two points of view. However, when you're given a linear transformation, you're not allowed to ask for things like the entry in its 3rd row and 4th column because questions like these depend on a choice of basis. Instead, you're only allowed to ask for things that don't depend on the basis, such as the rank, the trace, the determinant, or the set of eigenvalues. This point of view may seem unnecessarily restrictive, but it is fundamental to a deeper understanding of pure mathematics.

at.algebraic topology - Contractible manifold with boundary - is it a disc?

Given a function $psi:mathbb Rto mathbb R$,
set
$$Psi=psicircmathrm{dist}_ {partial M}, f=Psicdot(R-mathrm{dist}_ p)$$
for some fixed $R>mathrm{diam}\, M$.

Further,
$$d\,f=
(R-mathrm{dist}_ p)cdot d\,Psi-Psicdot d\,mathrm{dist}_ p$$
Thus, we may choose smooth increasing $psi$,
such that $psi(0)=0$
and it is constant outside of little nbhd of $0$ so that
$Psi$ is smooth.
(It is possible since the function $mathrm{dist}_ {partial M}$ is smooth and has no critical points in a small neighborhood of $partial M$.)
Note that $d\,Psi$ is positive muliple of $d\,mathrm{dist}_ {partial M}$.
Thus $d_x\,f=0$ means that geodesic from $x$ to $p$ goes directly in the direction of minimizing geodesic from $x$ to $partial M$, which can not happen.

Now we can apply Morse theory for $f$...

rt.representation theory - Can you find linear recurrence relation for dimensions of invariant tensors?

Let $V$ be a finite dimensional highest weight representation of a (semi)-simple Lie algebra. For each $nge 0$ take $a_n$ to be the dimension of the space of invariant tensors in $otimes^n V$.

In certain cases there is a formula for $a_n$. For example, for $V$ the two dimensional representation of $sl(2)$ we get $a_n=0$ if $n$ is odd and for $n$ even we get the ubiquitous Catalan numbers. In general I don't expect a formula but the sequence does satisfy a linear recurrence relation with polynomial coefficients (known as D-finite).

For example, for the seven dimensional representation of $G_2$ this sequence starts:
1, 0, 1, 1, 4, 10, 35, 120, 455, 1792, 7413, 31780, 140833, 641928, 3000361, 14338702, 69902535, 346939792, 1750071307, 8958993507, 46484716684, 244187539270, 1297395375129, 6965930587924
for more background see http://www.oeis.org/A059710

This satisfies the recurrence
$(n+5)(n+6)a_n=2(n-1)(2n+5)a_{n-1}+(n-1)(19n+18)a_{n-2}+ 14(n-1)(n-2)a_{n-3}$

Question How does one find these recurrence relations?

Then I also have a more challenging follow-up question. The space of invariant tensors in $otimes^n V$ also has an action of the symmetric group $S_n$ and so a Frobenius character which is a symmetric function of degree $n$.

Question How does one calculate these symmetric functions?

I know these can be calculated using plethysms individually. I am hoping for something along the lines of the first question.

Further remarks David's answer solves the problem theoretically but I want to make some remarks about the practicalities. This is in case anyone wants to experiment and also because I believe there is a more efficient method.

The $sl(2)$ example can easily be extended. For the $n$-dimensional representation $a_k$ is the coefficient of $ut^k$ in
$$frac{u-u^{-1}}{1-tleft(frac{u^n-u^{-n}}{u-u^{-1}}right)}$$
For the case $n=3$ see http://www.oeis.org/A005043 and
http://www.oeis.org/A099323
I am not aware of any references for $nge 4$. I don't know if these are algebraic.

The limitation of this method is that there is a sum over the Weyl group. This means it is impractical to implement this method for $E_8$. For the adjoint representation of $E_8$ the start of the sequence is
1 0 1 1 5 16 79 421 2674 19244 156612 1423028 14320350
(found using LiE)

If K/k is a finite normal extension of fields, is there always an intermediate field F such that F/k is purely inseparable and K/F is separable?

I was feeling a bit rusty on my field theory, and I was reviewing out of McCarthy's excellent book, Algebraic Extensions of Fields. Out of Chapter 1, I was able to work out everything "left to the reader" or omitted except for one corollary, stated without proof (see here for the page in the book):

Let $K/k$ be a finite normal
extension. Then $K$ can be obtained by
a purely inseparable extension,
followed by a separable extension.

The text immediately preceding this implies that the intermediate field that's going to make this happen is $F={ain K:sigma(a)=a$ for all $sigmain Gal(K/k)}$, and I understand his argument as to why $F/k$ is purely inseparable (in fact, that's the theorem, Theorem 21, which this is a corollary to). What I don't understand is why $K/F$ is separable; I don't see how we've ruled out it being non-purely inseparable.

Note that I will be making a distinction between non-purely inseparable (inseparable, but not purely inseparable) and not purely inseparable (either separable or non-purely inseparable).

Here are some observations / my general approach:

• One big thing that seemed promising was Theorem 11 (at the bottom of this page), which is basically the reverse of the corollary I'm having trouble with:

Let $K$ be an arbitrary algebraic extension of $k$. Then $K$ can be obtained by separable extension followed by a purely inseparable extension.

(the separable extension referred to is of course the separable closure of $k$ in $K$). It seems like we want to use Theorem 11 on $K/F$, and argue that there can't be "any more" pure inseparability, but I couldn't figure out a way of doing this.

• Theorem 21 is actually an "if and only if" (that is, $ain K$ is purely inseparable over $k$ iff $sigma(a)=a$ for all $sigmain Gal(K/k)$). Because this implies that any $ain K$ with $anotin F$ is not purely inseparable over $k$, we have that $F$ is the maximum (not just maximal) purely inseparable extension of $k$ in $K$.




• If any $ain K$ were purely inseparable over $F$, by Theorem 8 (see here), there is some $e$ for which $a^{p^e}in F$. But by the same theorem, since $F/k$ is purely inseparable, there is some $b$ for which $(a^{p^e})^{p^b}=a^{p^{e+b}}in k$. Thus $a$ would be purely inseparable over $k$ by the converse (Corollary 1 to Theorem 9, see here), and hence be in $F$. Thus, $K$ (and any field between $K$ and $F$, besides $F$ itself) is not purely inseparable over $F$.

So, that's why I don't see how we've ruled out $K/F$ being non-purely inseparable. Sorry about making lots of references to the book - I'm just not sure what previously established results McCarthy intended to be used, and I wanted to point out what I saw as the important ones for people not familiar with the book. I'm sure I'm missing something obvious here. Does anyone see the last bit of the argument?

soft question - Simple definition of the Hausdorff measure using squared paper

I am giving a "non-technical" seminar in which I would like to give an elementary introduction to the Hausdorff dimension and measure.

For simplicity, I was hoping to give a more intuitive introduction using 'squared paper': suppose you draw your figure on squared paper with squares of size $2^{-n}$, and let $M(n)$ the number of squares that your figure touches; take the limit of $M(n)(2^{-n})^d$ for $ntoinfty$.

This seems to be ok for defining the dimension $d$ (the only issue is that you only allow coverings with very specific sets, the 'elementary squares'), but not for the measure. In fact, if you just consider line segments, then this definition converges to their 1-norm length ($|x_1-x_0|+|y_1-y_0|$) instead of the usual Euclidean length.

So, my questions are:

1- is there any hope to save this approach, or do I really have to introduce the measure using balls (this would be unpleasant as I would also have to deal with the proper constants for the volume of a ball in fractional dimension, ouch)

2- is there a simpler introduction that I am overlooking here?

ag.algebraic geometry - Why is a variety of general type hyperbolic?

I heard people mentioned this in one sentence, but don't see the reason.

Why a (smooth) variety of general type, i.e. an algebraic variety X with K_X big, is hyperbolic, i.e. has no non-constant map from the complex number into it?

I don't know what are the necessary assumption on the variety, do we need properness or smoothness?

Edit: according to David Lehavi's reply, we should certainly put some more condition on it. What's the correct statement of the fact?

rt.representation theory - Is every finite group a group of "symmetries"?

The permutohedron may have additional symmetries. For example, the order 3 permutohedron {(1,2,3),(1,3,2),(2,1,3),(3,1,2),(3,2,1)} is a regular hexagon contained in the plane x+y+z=6, which has more than 6 symmetries.

I think we can solve it as follows:

Let G be a group with finite order n thought via Cayley's representation as a subgroup of S_n.

Let S={A_1,...,A_n} be the set of vertices of a regular simplex centered at the origin in a (n-1)-dimensional real inner product space V. Let r be the distance between the origin and A_1. The set of vertices S is an affine basis for V.

First unproven claim: If a closed ball that has radius r contains S, then it is centered at the origin. Let B be this ball.

The group of isometries that fix S hence contains only isometries that fix the origin and permute the vertices, which can be identified with S_n in the obvious way. The same is true if we replace S by its convex hull.

Now G can be thought of as a group containing some of the symmetries of S.

Let C=k(A_1+2A_2+3A_3+...+nA_n)/(1+2+...+n), with k a positive real that makes the distance between C and the origin a number r' slightly smaller than r.

Let GC={g(C) / g in G}. It has n distinct points, as a consequence of being S an affine basis of V.

Let P be the convex hull of the points of S union GC.

Remark: A closed ball of radius r contains P iff it is B. The intersection of the border of B and P is S.

Second unproven claim: The extremal points of P are the elements of S union GC.

Claim: G is the group of symmetries of P.

If g is in G, g is a symmetry of GB and of S, and it is therefore a symmetry of P.
If T is a symmetry of P, then T(P)=P, and in particular, T(P) is contained in B, and hence T(0)=0 (i.e. T is also a symmetry of B). T must also fix the intersection of P and the border of B, so T permutes the points of S, and it can be thought of as an element s of S_n sending A_i to A_s(i). And since T fixes the set of extremal points of P, T also permutes GC. Let's see that s is in G.

Since T(C) must be an element of g(C) of GC, we have T(C)=g(C). But since T is linear, T(C/k)=g(C/k). Expanding,

(A_s(1)+2A_s(2)+...+nA_s(n)/(1+...+n)=(A_g(1)+2A_g(2)+...+nA_g(n))/(1+...+n).

For each i in {1,...,n} the coefficient that multiplyes A_i is s⁻1(i)/(1+...+n) in the left hand side and g⁻1(i) in the right hand side. It follows that s=g.

I think that, taking n into account, the ratio r'/r can be set to substantiate the second unproven claim. The first unproven claim may be a consequence of Jung's inequality.

EDIT: With the previous argument, we can represent a finite group of order n as the group of linear isometries of a certain polytope in an n-1 dimensional real inner product space.

Now, if a finite group G of linear isometries of an (n-1)-dimensional inner product space V is given, can we define a polytope that has G as its group of symmetries? Yes. I'll give a somehow informal proof.

Let G={g_1,...,g_m}. Let A={a_1,...,a_n} be the set of vertices of a regular n-simplex centered at the origin of V. Let S be the sphere centered at the origin that contains A, and let C be the closed ball. Notice that C is the only minimun closed ball contaiing A.

(Remark: The set A need not be a regular simplex. It may be any finite subset of S that intersects all the possible hemispheres of S. C will then still be only minimum closed ball
containing it.)

Remark: An isometry of V is linear iff it fixes the origin.

Before proceeding, we need to be sure that the m copies of A obtained by making G act on it are disjoint. If that is not the case, our set A is useless but we can find a linear isometry T such that TA does de job. We consider the set M of all linear isometries with the usual operator metric, and look into it for an isometry T such that for all (g,a) and (h,b) distinct elements of GxA the equation g(Ta)=h(Tb) does not hold. Because each of the n*m(n*m-1) equations spoils a closed subset of M with empty interior(*), most of the choices of T will do.

Let K={ga/g in G, a in A}. We know that it has n*m points, which are contained in the sphere S. Now let e be a distance that is smaller than a quarter of any of the distances between different points of K. Now, around each vertex a=a_i of A make a drawing D_i. The drawing consists of a finite set of points of the sphere S, located near a (at a distance smaller than e). One of the points must be a itself, and the others (if any) should be apart from a and very near each other, so that a can be easily distinguished. Furthermore, for i=1 the drawing D_i must have no symmetries, i.e, there must be no linear isometries fixing D_1 other than the identity. For other values of i, we set D_i={a_i}. The union F of all the drawings contains A, but has no symmetries. Notice that each drawing has diameter less than 2*e,

Now let G act on F and let Q be the union of the m copies obtained. Q is a union of n*m drawings. Points of different drawings are separated by a distance larger than 2*e. Hence the drawings can be identified as the maximal subsets of Q having diameter less than 2*e. Also, the ball C can be identified as the only sphere with radius r containing Q. S can be identified as the border of C.

Let's prove that the set of symmetries of Q is G. It is obvious that each element of G is a symmetry. Let T be an isometry that fixes Q. It must fix S, so it must be linear. Also, it must permute the drawings. It must therefore send D_1 to some gD_i with g in G and 1<=i<=n. But i must be 1, because for other values of i, gD_i is a singleton. So we have TD_1=gD_1. Since D_1 has no nontrivial symmetries, T=g.

We have constructed a finite set Q with group of symmetries G. Q is not a polytope, but its convex hull is a polytope, and Q is the set of its extremal points.

(*) To show that for any (g,a) and (h,b) distinct elements of GxA the set of isometries T satisfying equation g(Ta)=h(Tb) has empty interior, we notice that if an isometry T satisfies the equation, any isometry T' with T'a=Ta and T'b=/=Tb must do (since h is injective). Such T' may be found very near T, provided dimV>2. The proof doesn't work for n=1 or 2, but these are just the easy cases.

nt.number theory - Tate module of CM elliptic curves

Here is the standard argument: you can decide whether it is prettier than the one you had in mind.

Let $V_{ell}(E)$ be ${mathbb Q}_{ell}otimes_{{mathbb Z}_{ell}}T_{ell}(E)$; it is a two-dimensional ${mathbb Q}_{ell}$ vector space. When $E$ has CM by a quad. imag. field
$F$, it is free of rank one over $Fotimes_{mathbb Q}{mathbb Q}_{ell}$. Thus
the image of $Gal(bar{K}/K)$ acting on $V_{ell}(E)$ (or equivalently, $T_{ell}(E)$)
lies in $GL_1(Fotimes_{mathbb Q}{mathbb Q}_{ell})$, and so is abelian.

Note that this argument gets to the very heart of CM theory, and its relation to the class field theory (i.e. to the construction of abelian extensions): the elliptic curve $E$
(or, more precisely, its Tate modules) look 1-dim'l as modules over $F$, and so give
abelian Galois reps. (Just as the $ell$-adic Tate modules of the multiplicative group
${mathbb G}_m$ give 1-dim'l. reps. of $Gal(bar{mathbb Q}/{mathbb Q}).$)

You might also want to compare with Lubin--Tate theory, which is very similar: one uses
formal groups with an action of the ring of integers ${mathcal O}$ in an extension of
${mathbb Q}_p$, and again they are constructed so that the $p$-adic Tate module is free of rank one over ${mathcal O}$, and hence gives abelian Galois reps.

---

Added in response to basic's questions below: To say that $E$ has CM over $K$ is to say that it has an action by an order in a quad. imag. field $F$. By a standard theorem (in Silverman, say) the ring $F_{ell} := Fotimes_{mathbb Q}{mathbb Q}_{ell}$ acts faithfully on $V_{ell}(E)$. Counting dimensions over ${mathbb Q_{ell}},$ we find that $V_{ell}(E)$ is free of rank 1 over $F_{ell}$. The Galois action on $V_{ell}(E)$ is
being $F_{ell}$-linear (we have assumed that the action of $F$ is defined over $K$).

Thus we have a group, $Gal(bar{K}/K)$, acting on a free rank 1 module over a ring
(namely, the $F_{ell}$-module $V_{ell}(E)$). Such an action must be given by $1times 1$ invertible matrices (just choose a basis for $V_{ell}(E)$ as an $F_{ell}$-module),
i.e. is described by a homomorphism $Gal(bar{K}/K) to GL_1(F_{ell})$.

Since the group of invertible $1times 1$-matrices over any commutative ring is
itself commutative, we see that the $Gal({bar K}/K)$ action on $V_{ell}(E)$ is
through an abelian group, as required.

gr.group theory - When do the sizes of conjugacy classes and squares of degrees of irreps give the same partition for a finite group?

My standard rant about "what can we say about $G$": what we can say about $G$ is that the two partitions are the same. If the questioner doesn't find that a helpful answer then they might want to consider the possibility that they asked the wrong question ;-)

But as to the actual question: "is $G$ forced to be abelian?", the answer is no, and I discovered this by simply looping through magma's database of finite groups. Assuming I didn't make a computational slip, the smallest counterexample has order 64, is the 73rd group of order 64 in magma's database, which has 8 representations of degree 1, 14 representations of degree 2, 8 elements in the centre and 14 more conj classes each of order 4.

Letting the loop go further, I see counterexamples of size 64, 128, 192 (I guess these are just the counterexamples of size 64 multiplied by Z/3Z) and then ones of order 243 (a power of 3). So I guess all examples I know are nilpotent. Are they all nilpotent? That's a question I don't know the answer to.

teaching - Undergraduate Probability Topics

Kenneth Levasseur's paper, "How to Beat Your Kids at Their Own Game"
analyzes the simple game of guessing whether the next card in a deck is red or black.
He computes the expected score of correct guesses if you count carefully.
There is a nice geometric flavor to his analysis.
With the standard 52-card deck, the expected score is slightly over 30.

Mathematics Magazine Vol. 61, No. 5 (Dec., 1988), pp. 301-305.

nt.number theory - What's the analogue of the Hilbert class field in the following analogy?

This is a great question. Someone will come along with a better answer I'm sure, but here's a bit off the top of my head:

1) The Hilbert class field of a number field $K$ is the maximal everywhere unramified abelian extension of $K$. (Here when we say "$K$" we really mean "$mathbb{Z}_K$", the ring of integers. That's important in the language of etale maps, because any finite separable field extension is etale.)

In the case of a curve over $mathbb{C}$, the "problem" is that there are infinitely many unramified abelian extensions. Indeed, Galois group of such is the abelianization of the fundamental group, which is free abelian of rank $2g$ ($g$ = genus of the curve). Let me call this group G.

This implies that the covering space of C corresponding to G has infinite degree, so is a non-algebraic Riemann surface. In fact, I have never really thought about what it looks like. It's fundamental group is the commutator subgroup of the fundamental group of C, which I believe is a free group of infinite rank. I don't think the field of meromorphic functions on this guy is what you want.

2) On the other hand, the Hilbert class group $G$ of $K$ can be viewed as the Picard group of $mathbb{Z}_K$, which classifies line bundles on $mathbb{Z}_K$. This generalizes nicely: the Picard group of $C$ is an exension of $mathbb{Z}$ by a $g$-dimensional complex torus $J(C)$, which has exactly the same abelian fundamental group as $C$ does: indeed their first homology groups are canonically isomorphic. $J(C)$ is called the Jacobian of $C$.

3) It is known that every finite unramified abelian covering of $C$ arises by pulling back an isogeny from $J(C)$.

So there are reasonable claims for calling either $G cong mathbb{Z}^{2g}$ and $J(C)$ the Hilbert class group of $C$. These two groups are -- canonically, though I didn't explain why -- Pontrjagin dual to each other, whereas a finite abelian group is (non-canonically) self-Pontrjagin dual. [This suggests I may have done something slightly wrong above.]

As to what the Hilbert class field should be, the analogy doesn't seem so precise. Proceeding most literally you might take the direct limit of the function fields of all of the unramified abelian extensions of $C$, but that doesn't look like such a nice field.

Finally, let me note that things work out much more closely if you replace $mathbb{C}$ with a finite field $mathbb{F}_q$. Then the Hilbert class field of the function field of that curve is a finite abelian extension field whose Galois group is isomorphic to $J(C)(mathbb{F}_q)$, the (finite!) group of $mathbb{F}_q$-rational points on the Jacobian.

teaching - Blackboard rendering of math fonts

Suggestion number one:

Learn Calligraphy! It's a lot of fun and does mean that you can write the fonts in genuinely nice ways. Books on calligraphy tend to have detailed instructions on how to do at least the basic alphabets: explaining which stroke to do first, and how to hold the pen. Although not all of it transfers to the blackboard, it helps a lot. For example, once you seen how the different "g"s are written, you'll know how to write the Lie algebra symbol correctly. However, I do find that a script S ($mathcal{S}$ is not even close) can take me a couple of goes to make it look right - it shouldn't be pointy at the top but should sweap backwards.

(Note: to anyone reading the comments, I originally had this answer together with my other answer in the same post. Given the comments, I decided to split them.)

Following up on aleksander's comment to the original question, I had a go at doing a video of how to draw a fraktur g (𝔤) (well, actually it's gothic but if you know the difference you don't need this video, and the gothic g is probably more distinguishable from a normal g than a fraktur one on a blackboard). It's not very polished, but you can see what it looks like here. It was quite fun to do so if this would be helpful, I can easily do more.

Suggest effective heuristic (not precise) graph colouring algorithm

There are a number of heuristics that work fairly well. They all work by prescribing some kind of ordering on the vertices, and then coloring the vertices one by one, using the least unused color to color the next one.

• First Fit does precisely the above, with an arbitrary initial ordering. It's fast, but needless to say performs rather poorly.


• LDO orders the vertices in decreasing order of degree, the idea being that the large degree vertices can be colored more easily.


• SDO (saturation degree ordering) is a variant on LDO where the vertices are ordered in decreasing order by "saturation degree", defined as the number of distinct colors in the vertex neighborhood.


• IDO (incidence degree ordering) is a variant of SDO where the "degree" of a vertex is defined as the number of colored vertices in its neighborhood.

The latter two heuristics require the order to be rebuilt after each step, and so are more expensive, but there's empirical evidence suggesting that they do reasonably well, especially in parallel.

None of these algorithms come with any kind of formal guarantees, so be warned.

at.algebraic topology - Conventions for definitions of the cap product

This is just an expanded version of Tyler's comment, I think.

Let's use a, b, c for cochains, x, y, z for chains, [a,x] for the value of a cochain on a chain. I'll be lazy and write $ab$ for $acup b$ and $ax$ for $acap x$. Let $[aotimes b,yotimes z]=(-1)^{|b||y|}[a,y][b,z]$.

I like to define $bx$ in such a way that $[a,bx]$ is $[ab,x]$. Then associativity and unit of cup makes cap into a module structure: $(bc)x=b(cx)$ because $[a,(bc)x]=[a(bc),x]=[(ab)c,x]=[ab,cx]=[a,b(cx)]$, and $1x=x$ follows from $[a,1x]=[a1,x]=[a,x]$.

If you have a product $aotimes bmapsto ab$ of cochains and a coproduct of chains defined in such a way that $[ab,x]=[aotimes b,y_iotimes z_i]$ where the coproduct of $x$ is $Sigma y_iotimes z_i$, then that means I have to define $bx$ to be $Sigma (-1)^{|y_i||b|}[b,z_i]y_i$, so as to get

$[a,bx]=[ab,x]=[aotimes b,Sigma y_iotimes z_i]=Sigma (-1)^{|y_i||b|}[a,y_i][b,z_i]=[a,Sigma (-1)^{|y_i||b|}[b,z_i]y_i]$.

The sign is a bit unexpected, as is the jump you mention, but it's worth it for the neat formulas in the third paragraph above.

It's not a bad idea to define $xa$ as well. I'd do it by first declaring $[x,a]$ to be $(-1)^p[a,x]$ when $a$ is a $p$-cochain and $x$ is a $p$-chain, then defining $xa$ in such a way that $[xa,b]=[x,ab]$. This insures $x(ab)=(xa)b$ and $x1=x$. The chain-level formula is no better and no worse than the formula for $ax$.

Of course, $ax$ and $xa$ end up differing only by a sign when you get to homology, but the sign is hard to remember; in working it out you have to use the commutativity law for the cup product.

career - Where are mathematics jobs advertised if not on mathjobs (e.g. in Europe and elsewhere)?

My impression is that in the US, there is a canonical place for finding math jobs, namely mathjobs.org. For those of us who live and apply for jobs elsewhere, life is more complicated, and searching for advertised academic mathematics jobs for example in Europe can be a real hassle, with loads of different sites, different systems, and some jobs apparently advertised only on the web page of the hiring institution, or one some obscure mailing list.

So, where are academic math jobs advertised when they for some reason are not or cannot be on mathjobs.org? Of course I know of a few such places, but I am sure there must be many more.

All answers welcome, this would help me and probably many others.

qa.quantum algebra - Can "premodular" be relaxed as a condition for uniqueness of Bruguieres/Mueger modularization?

I think this is a counterexample to the result I was looking for. Let C be the idempotent completion of TL_{-1}, the Temperley-Lieb category with loop value -1. (Or equivalently, the category of finite dimensional representations of U_q(sl_2) where q is an third root of unity.)

Let D_1 be the quotient of C by negligibles. This category is just Rep(Z/2) but with the dimension of the nontrivial rep being -1.

Let B be the Fibonacci category and let B' be its Galois conjugate, let x and x' be the nontrivial objects. Let D_2 be the Deligne tensor product of B and B'. Let F_2 be the functor sending V_2 to $x boxtimes x'$, this exists by the universal property of Temperley-Lieb (i.e. $x boxtimes x'$ is symmetrically self-dual and has quantum dimension -1). This functor is dominant because $F_2(V_2 otimes V_2)$ has every object of D_2 as a summand.

As Victor Ostrik points out F_2 is not a ribbon functor, so this is not a counterexample.