ct.category theory - What is the motivation for maps of adjunctions?

In Mac Lane, there is a definition of an arrow between adjunctions
called a map of adjunctions. In detail, if a functor $F:Xto A$ is left
adjoint to $G:Ato X$ and similarly $F':X'to A'$ is left adjoint to
$G':A'to X'$, then a map from the first adjunction to the second is a
pair of functors $K:Ato A'$ and $L:Xto X'$ such that $KF=F'L$,
$LG=G'K$, and $Leta=eta'L$, where $eta$
and $eta'$ are the units of the first and second adjunction. (The
last condition makes sense because of the first two conditions; also,
there are equivalent conditions in terms of the co-units, or in terms
of the natural bijections of hom-sets).

As far as I can see, after the definition, maps of adjunctions do not
appear anywhere in Mac Lane. Googling, I found this definition also
in the unapologetic mathematician,
again with the motivation of being an arrow between adjunctions.

But what is the motivation for defining arrows between adjunctions
in the first place? I find it hard to believe that the only
motivation to define such arrows is, well, to define such arrows...

So my question is: What is the motivation for defining a map of
adjunctions? Where are such maps used?

Besides the unapologetic mathematician, the only places on the web
where I found the term ''map of adjunctions'' were sporadic papers,
from which I was not able to get an answer to my question (perhaps
''map of adjunctions'' is non-standard terminology and I should have
searched with a different name?).

I came to think about this when reading Emerton's first answer
to a question about completions of metric spaces.
In that question, $X$ is metric spaces with isometric embeddings, $A$
is complete metric spaces with isometric embeddings, $X'$ is metric
spaces with uniformly continuous maps, $A'$ is complete metric
spaces with uniformly continuous maps, and $G$ and $G'$ are the
inclusions. Now, if I understand the implications of Emerton's answer
correctly, then it
is possible to choose left adjoints $F$ and $F'$ to $G$ and $G'$ such
that the (non-full) inclusions $Ato A'$ and $Xto X'$ form a map of
adjunctions. This made me think whether the fact that we have a map
of adjunctions has any added value. Then I realized that I do not
even know what was the motivation for those maps in the first place.

[EDIT: Corrected a typo pointed out by Theo Johnson-Freyd (thanks!)]

gr.group theory - Does $mathrm{Aut}(mathrm{Aut}(...mathrm{Aut}(G)...))$ stabilize?

I don't know about non-stabilizing, but rigidity provides many examples that stabilize quickly.

1) Let π be the fundamental group of a finite volume hyperbolic manifold M of dimension ≥ 3 with no symmetries (that is, no nontrivial self-isometries). Negative curvature implies that π is centerless, so the map π -> Aut(π) is injective. Mostow-Prasad rigidity says that Out(π) = Isom(M), so the lack of isometries implies that Out(π) is trivial and Aut(π) = π. [This works verbatim for lattices in higher-rank semi-simple Lie groups subject to appropriate conditions.]

2) Let π=F_d be a free group of rank 2≤d<∞. Then Aut(F_n) is a much larger group; however, Dyer-Formanek showed that Out(Aut(F_n)) is trivial. Thus since Aut(F_n) is clearly centerless, we have Aut(Aut(F_n)) = Aut(F_n).

3) Interpolating between these two examples, if π=π₁(S_g) is the fundamental group of a surface of genus g≥2, then Aut(π) is the so-called "punctured mapping class group" Mod_g,*, which is much bigger than π. Ivanov proved that Out(Mod_g,*) is trivial, and since Mod_g,* is again centerless, we have Aut(Aut(π₁(S_g))) = Aut(π₁(S_g)).

In each of these cases, rigidity in fact gives stronger statements: Let H and H' be finite index subgroups of G = Aut(F_n) or Mod_g,*. (This class of groups can be widened enormously, these are just some examples.) Then any isomorphism from H to H' comes from conjugation by an element of G, by Farb-Handel and Ivanov respectively. In particular, Aut(H) is the normalizer of H in G. Rigidity gives the same conclusion for H = π₁(M) as in the first example and G = Isom(Hⁿ) [which is roughly SO(n,1)]. It seems that by carefully controlling the normalizers, you could use this to construct examples that stabilize only after n steps, for arbitrary large n.

---

Edit: I find the examples of D₈ and D_∞ unsatisfying because even though Inn(D) is a proper subgroup of Aut(D), we still have Aut(D) isomorphic to D. Here is a general recipe for building similarly liminal examples. Let G be an infinite group with no 2-torsion so that Aut(G) = G and H¹(G;Z/2Z) = Z/2Z. (Edited: For example, by rigidity, any hyperbolic knot complement with no isometries has these properties; by Thurston, most knot complements are hyperbolic.) The condition on the 2-torsion implies that for any automorphism G x Z/2Z -> G x Z/2Z, the composition

G -> G x Z/2Z -> G x Z/2Z -> G

is an isomorphism. From this we see that Aut(G x Z/2Z) / G = H¹(G;Z/2Z) = Z/2Z. By examination the extension is trivial, and thus Aut(G x Z/2Z) = G x Z/2Z. However, the image Inn(G x Z/2Z) is the proper subgroup G.

Comments: looking back, this feels very close to your original example of R x Z/2Z. Interesting that it's (seemingly) much harder to find group-theoretic conditions to force the behavior the way you want, while topologically it's easy.

Also, if you instead take G with H¹(G;Z/2Z) having larger dimension, say H¹(G;Z/2Z) = (Z/2Z)², this blows up quickly. You get Aut(G x Z/2Z) = G x (Z/2Z)², but then Aut(Aut(G x Z/2Z)) is the semidirect product of H¹(G;Z/2Z²) = (Z/2Z)⁴ with Aut(G) x Aut(Z/2Z²) = G x GL(2,2). Already the next step seems very hard to figure out. However, if you had enough control over the finite quotients of G, perhaps you could show that the linear parts of these groups don't get "entangled" with the rest, so that the automorphism groups would act like a product of G x (Z/2Z)ⁿ with something else, with n going to infinity. If so, this could yield an example where the isomorphism types of the groups never stabilize.

What is an example of a presheaf P where P^+ is not a sheaf, only a separated presheaf?

I think this works:

Consider a topological space consisting of 4 points $A$, $B$, $C$, $D$, where the topology is given by open sets $ABC$, $BCD$, $B$, $C$, $ABCD$, $emptyset$.

Then let the presheaf $mathcal{F}$ be given by:
$$mathcal{F}(ABC)=mathbb{Z}$$
$$mathcal{F}(BCD)=mathbb{Z}$$
$$mathcal{F}(BC)=mathbb{Z}$$
$$mathcal{F}(ABCD)=mathbb{Z}$$
$$mathcal{F}(B)=mathbb{Z}/2mathbb{Z}$$
$$mathcal{F}(C)=mathbb{Z}/2mathbb{Z}$$
$$mathcal{F}(emptyset)=0$$

where all restrictions are what you expect (identity in the case of $mathbb{Z} to mathbb{Z}$ and canonical surjection in the case $mathbb{Z} to mathbb{Z}/2 mathbb{Z}$).

Then if we we get $mathcal{F}^+$ is given by:

$$mathcal{F}^+(ABC)=mathbb{Z}$$
$$mathcal{F}^+ (BCD)=mathbb{Z}$$
$$mathcal{F}^+ (BC)= mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$$
$$mathcal{F}^+ (ABCD)=mathbb{Z}$$
$$mathcal{F}^+ (B)= mathbb{Z}/2mathbb{Z} $$
$$mathcal{F}^+ (C)=mathbb{Z}/2mathbb{Z}$$
$$mathcal{F}^+ (emptyset)=0$$

where the map from $mathcal{F}^+ (BCD)$ to $mathcal{F}^+ (BC)$ is given by taking the canonical surjection on both copies, and other restrictions are obvious. Then note that if we take 1 over $BCD$ and 3 over $ABC$, these two are compatible over $BC$ but they do not patch.

The key point is that being compatible over a refinement is not the same thing as being compatible. That is, the way the plus construction works is by taking $F^+$ of a space to be some direct limit over open covers of guys on the covers which are compatible on intersections. If we had said instead take direct limit over open covers of guys on the covers which compatible on some refinement of the intersection, then applying just once probably works.

So in our example, 1 and 3, over $ABC$ and $BCD$, in our original presheaf were compatible on a refinement of $BC$ but not on $BC$.

pr.probability - What is hidden in Hidden Markov Models?

The unobserved state.

Let's consider a hidden Markov model for my cat's behavior. Bella can be in five states: hungry, tired, playful, cuddly, bored. She can respond to these states with six behaviors: whining, scratching, cuddling, pouncing, sleeping and stalking.

A hidden Markov model would consist of two matrices, one 5x5 and the other 5x6. The 5x5 matrix gives the probabilities that, if she is hungry at time $t$, she will be tired at time $t+1$, and so forth. So we can compute the probability that she is in different emotional states by taking powers of this matrix.

However, we can't observe her emotions -- they are hidden. The 5x6 matrix gives the probability that, if she is hungry at time $t$, she will whine at time $t+1$. (Very close to $1$.) These are the behaviors we observe.

In an ordinary Markov model, there would just be a single 6x6 matrix, which directly described the probability of transitions like whining ---> clawing. As you can see, an ordinary Markov model is less able to reflect the complexity of my cat's inner life.

See the wikipedia article for much more information.

Question on the decimal expansion of algebraic numbers

I'm not an expert on such topics, but I would say "probably not". The wikipedia page on normal numbers says $sqrt{2}$ is "widely believed" to be normal, though this hasn't been proven. Normality basically means the decimal (and also base $n$ for all $n$) expansions look random, which is true of almost all real numbers. It's conceivable that such decimal expansions could still have some structure in the case of algebraic numbers, so it depends what you're looking for, but "probably not". Either no such structure is known, or the known structure is weak enough not to interfere with normality.

mp.mathematical physics - the Toda Lattice hierarchy of B and C type

In the K.Ueno and K.Takasaki's paper ``Toda Lattice hierarchy", advanced sdudies in pure mathematics 4(1984),pp1-95, they mentioned the Toda Lattice hierarchy of B and C type(we denote BTL and CTL hierarchies).In their paper, the bilinear relation for BTL and CTL hierarchies are almost the same as the one for Toda Lattice hierarchy except with the even time flows being zero( see eq.2.4.1 in reference).

My question is as follows:

1. Since the KP and TL hierarchies are almost the same (see Mark Adler et al's paper Comm.math.Phys.171,547-588(1995)), and considering the great difference between the bilinear relations for BKP and KP hierarchies(see E.Date,Jimbo,et al's paper,Transformation groups for soltion equations), why BTL and TL lattice hierarches are almost the same? Is there something wrong?




2. Can the BTL and CTL hierarchies have the fay like identities just as the KP and BKP hierarchies(see H.F.Shen and M.H.Tu's paper,arxiv:0811.1469)




3. why the literatures for the BTL and CTL hierarchies are so few? is there someting difficult in studying them?

   
   
   

   thanks

ag.algebraic geometry - Decomposition of k[G]

This is true for reductive groups, more or less by definition. An algebraic representation of an algebraic group is a comodule V over the algebra of functions O(G) of the group. Therefore, every representation V induces a map
V -> V ⊗ O(G), or equivalently V^* ⊗ V --> O(G) (call the source of this map C(V) for coefficient space of V). It is not hard to see that the latter is a map of G x G modules. If G is reductive, then its representation category is semi-simple, and thus so is the representation category of G x G. In this case the simples of G x G are external tensor product of simples of V, and Hom(A ⊗' B, C ⊗' D) = d(A,C) ⊗ d(B,D) where d(V,W)=0 if v cong W, C else. Here ⊗' means external tensor product. There doesn't appear to be a ⊠

For non-reductive groups, you can still form O(G) in an analogous way:

Let A = ⊕_V V^* ⊗' V, where here the sum is over ALL finite dimensional modules V (not just isoclass representatives, and not just simples), and again the tensor product is external, so this lives in a completion of Rep(G) ⊗' Rep(G), and ⊗' means Deligne tensor product of categories.

Well this A is way too big, but now let's quotient A by the images of f^* ⊗' id - id ⊗' f, for all f:V-->W. This cuts A back down, for instance it identifies C(V) and C(V') whenever V and V' are isomorphic. If the category Rep(G) is semi-simple, you can similarly use the projectors and inclusions of simple objects to reduce to a Peter-Weyl type decomposition.

One nice thing about this construction (even in the semi-simple case) is that it is basis free because you don't choose representatives of simple objects, and also it makes the multiplication structure completely trivial:
V^* ⊗' V ⊗₂ W^* ⊗' W = V^* ⊗ W^* ⊗' V ⊗ W --> W^* ⊗ V^* ⊗' V ⊗ W,
using the braiding (tensor swap). It also works in braided tensor categories and explains the multiplication structure on the "covariantized" quantum group.

ag.algebraic geometry - Reverse Langlands transform

My guess is that, traditionally, the Geometric Langlands program seems to be looking for a functor from D_{coh}(Loc,O)to D_{coh}(^L Bun,D), that is, from the derived category on the space of local systems to the derived category of D-modules on the space of bundles for the Langlands dual. So, by reverse Langlands transformation, he might be asking for a functor in the other direction, to try to establish the equivalence.

gr.group theory - normalizer of algebras and groups

This is true, and in fact more has been shown in the recent preprint http://arxiv.org/abs/1005.3049 of Fang, Gao, and Smith. One can also give the following alternative argument based on ideas of Popa:

If $LH subset LG$ is a MASA then it follows from the condition $ ( hgh^{-1} | h in H ) = infty$ for all $g in G setminus H$, that the normalizer of $H$ in $G$ is the same as the set of elements $g in G$ such that $[H: H cap gHg^{-1}] < infty$. (This set is not in general closed under inversion but in this case it is since it coincides with the normalizer.)

Suppose we fix $g in G$ such that $[H: H cap gHg^{-1}] = infty$ and let's show that $u_g$ is orthogonal to $mathcal N_{LG}(LH)''$. Since $mathcal N_{LG}(LH)''$ is spanned by $mathcal N_{LG}(LH)$ it is enought to show that $u_g$ is orthogonal to this set and so let's fix $v in mathcal N_{LG}(LH)$.

Before we show that $u_g$ and $v$ are orthogonal let's rewrite the condition $[H: H cap gHg^{-1}] = infty$ in a more von Neumann algebraic friendly context which states that there are always "large" subalgebras of $LH$ which are almost moved orthogonal to $LH$.

Lemma:
For all $n in mathbb N, delta > 0$ there exists a finite dimensional subalgebra $A_0 subset LH$ such that if $p$ is any minimal projection in $A_0$ then $tau(p) = 1/2^n$ and $| langle x, u_g^* p u_g - tau(p) rangle | < delta |x |_2$ for all $x in LH$.

Proof. This essentially follows from Popa's intertwining techniques since the condition $[H: H cap gHg^{-1}] = infty$ translates in this context to $LH notprec_{LH} L(H cap gHg^{-1})$ (See Popa's paper http://www.ams.org/mathscinet-getitem?mr=2231961).

Let's show this by induction on $n$. For the case when $n = 1$ consider the group $mathcal G = ( u in mathcal U(LH) | u = 1 - 2p, p in mathcal P(LH), tau(p) = 1/2 ) cup (1)$. Since $mathcal G$ generates $LH$ as a von Neumann algebra and since $LH notprec_{LH} L(H cap gHg^{-1})$ it follows from Popa's intertwining Theorem that there exists a sequence $p_k in mathcal P(LH)$ with $tau(p_k) = 1/2$ such that $lim_{k to infty} | E_{L(H cap gHg^{-1})}(1 - 2p_k ) |_2 = 0$ (see Popa, op. cit.). In particular, for some $k$ this is less than $2delta$ and so if $x in LH$, $| x |_2 < 1$ we have $| langle x, u_g^*p_ku_g - tau(p) rangle | leq | E_{LH}(u_g^* p u_g - tau(p) ) |$ $_2 = | E_{L(H cap gHg^{-1})} (p_k - 1/2) |_2 < delta$. The same inequality holds for the other minimal projection $1 - p_k$.

Once we have produced such an $A_0$ for $1/2^n$ then given any minimal projection $p in A_0$ we again have that $pLH notprec_{pLH} pL(H cap gHg^*)$ and so the argument above shows that there exists $p_1$ and $p_2$ in $mathcal P(LH)$ such that $p_1 + p_2 = p$, each has half the trace and $| langle x, u_g^* p_j u_g - tau(p_j) rangle | < delta$. This proves the induction step. QED

Now that we have established the above lemma, the fact that $u_g$ and $v$ are orthogonal follows from a lemma of Popa's in http://www.ams.org/mathscinet-getitem?mr=703810. Let's give the proof here.

Let $varepsilon > 0$ be given and take $n in mathbb N$ such that $1/2^n < varepsilon/2$. From the above lemma let's consider a finite dimensional subalgebra $A_0 subset LH$ such that if $p$ is any minimal projection in $A_0$ then $tau(p) = 1/2^n$ and $| langle x, u_g^*pu_g - tau(p) rangle | < | x |_2 varepsilon/2^{n + 1}$. Let's denote the minimal projections in $A_0$ by $p_k$ where $1 leq k leq 2^n$.
Denote by $B_0$ the commutant of $A_0$ in $LG$.

Since $v in mathcal N_{LG}(LH)$ we have that $vLHv^* = LH$, hence $v^* p_k v in LH$ for each $k$. Therefore
$| langle v, u_g rangle |^2 leq | E_{B_0} ( vu_g^*) |_2^2$ $=
| $ $Sigma_k$ $ p_k v u_g^* p_k |_2^2 = Sigma_k langle v^* p_k v, u_g^* p_k u_g rangle < (Sigma_k tau(p_k)^2 ) + Sigma_k varepsilon/2^{n + 1} < varepsilon$.

Since $varepsilon$ was arbitrary we conclude that $u_g$ and $v$ are orthogonal. Hence since $v$ was arbitrary we conclude that $mathcal N_{LG}(LH)'' = L(mathcal N_G(H))$.

fa.functional analysis - Is there a use for a Hilbert space that uses a different norm than the one induced by the inner product?

One place where this is used is with Hilbert bundles. Taking $L^2$-functions on a space generally works very badly, so one instead takes $L^{2,1}$-functions - functions which are differentiable (almost everywhere) with square-integrable first derivative. However, the transition functions aren't isometries with respect to this norm so one does tend to use the $L^2$-norm on this space, remembering that the fibres aren't complete with respect to the norm.

Another use is in Wiener integration. Depending on one's point of view, one either has a Hilbert space with a weaker norm, or a Banach space with a dense subspace equipped with a Hilbert norm. Essentially, having the two norms means that one can "tame" stuff with respect to one norm by using the stronger one.

This extends further to the notion of "rigged Hilbert spaces".

Generally, the idea is that you want to work in one space but you don't have enough control over convergence, so you introduce a stronger norm and then ordinary convergence with respect to the stronger norm implies fantastic convergence with respect to the weaker one.

(I apologise for using vague language, but the question doesn't give much indication of how deep an answer to give. For those who know a little about ideals of operators, one generally wants the inclusion from the strong norm to the weak to be at least trace class, and often Hilbert-Schmidt.)

mathematics education - Depressed graduate student.

You have to ask yourself some basic questions.

1) Perhaps you lost your enthusiasm for a good reason. Maybe your initial enthusiasm was naive. Maybe you liked mathematics for unsustainable reasons? You're likely to go through far larger "down" cycles in the future if you stay in mathematics (we all do, it's a chronic problem in the field) so if you're going to stick with mathematics you have to find some kind of joy you can hold on to, through all kinds of messy situations.

2) Maybe you really do love mathematics but there's aggravating factors causing problems. Maybe you're not working on enough easy problems. Solving easy problems is fun, and if they're the right kind of problems you build up new skills. This is one of the reasons why I frequent this webpage.

I had a bunch of issues like (2) as a grad student. IMO I'm mostly better off for them. I'm talking about issues like solving a problem (or making progress on a problem) and finding out perhaps way too late that the problem had been solved by someone else. I found it pretty tricky to balance focus with awareness of what other people are doing. The math arXiv and MathSciNet are excellent resources that help with that.

Being in a very active place where you can talk to lots of people about various areas of mathematics helps. Being surrounded by enthusiastic people helps. Going to small conferences where you get to know people can help. Talking to people about what you're interested in helps. Barring external impetus, "computing the daylights out of things" is an excellent fall-back procedure. I know quite a few very successful mathematicians for which this is one of the main approaches to things. You start piling up enough computations on things that interest you and you notice patterns -- maybe not what you were looking for, but sometimes of interest to people for reasons you never expected. Sometimes publishable. :)

edit: After reading your recent edit I can say I saw some similar things as a grad student. Sometimes the most talented/bright/whatever grad students have a hard time completing a Ph.D. Some students have too high expectations of themselves. They give up because they realize they're not going to prove the Riemann hypothesis -- they want that great big creative insight. In that regard it's good to ensure such grad students are working on both big hard problems and medium-sized publishable work, so that they can complete a Ph.D even if they never prove the Riemann hypothesis or whatever. Basically, always make sure you have a managable goal in sight. If your goals are only huge enormous things, you're setting yourself up for a potentially horrible failure. On the other hand, some people want that kind of situation, and if they're conscious of it, IMO you might as well let them be. It's their life. If they prove a major theorem, we're all the better off for it. If they don't, well at least they tried.

graph theory - Number of trees with n nodes and m leaves

The answer to this (very natural) question depends on your notion of "tree" (e.g. free, rooted) and the equivalence relation you employ (e.g. labelled, unlabelled). I haven't gone into the nitty-gritty details of all these results, but here's what I've found so far. There's likely published results I haven't found yet, but hopefully this helps to get you started.

We can compute $T_{m,n}$, the number of non-isomorphic free trees with $m$ leaves and $n$ vertices, for small $m$ and large $m$. For example, (a) $T_{3,n}$ is the number of partitions of $n-1$ into $3$ positive integer parts (Sloane's A001399), (b) $T_{n-2,n}=lfloor (n-2)/2 rfloor$ and (c) $T_{n-3,n}=sum_{j=0}^{n-5} lfloor (n-3-j)/2 rfloor$. The first result can be observed by deleting the vertex of degree 3 and the last two can be observed by colouring each non-leaf vertex by the number of adjacent leaves, then deleting the leaves.

Yu (8) seems to have given an algorithm for generating rooted trees with $m$ leaves. Wang (6) and Liu (3,4) considered the number of "structurally different" trees with $m$ leaves (according to MathSciNet). Bergeron, Labelle and Leroux (1) consider the expected number of leaves in trees that admit a certain automorphism. Lam (2) discusses embeddings of trees with $m$ leaves and discusses trees with $(d+1)d^{r+1}$ leaves for integers d and r.

Wilf (7. p. 163) gave a generating function for $sum_k T_{k,n}^{text{lab}}$ where $T_{k,n}^{text{lab}}$ is the number of labelled free trees with $m$ leaves and $n$ vertices. He also gives a formula for the average number of leaves in a labelled tree with $n$ vertices.

There is also this: K. Yamanaka, Y. Otachi, S.-I. Nakano Efficient Enumeration of Ordered Trees with k Leaves, which I haven't looked at yet.

(1) F. Bergeron, G. Labelle, and P. Leroux, Computation of the expected number of leaves in a tree having a given automorphism, and related topics, Discrete Appl. Math., 34 (1991), pp. 49-66.

(2) P. C. B. Lam, On number of leaves and bandwidth of trees, Acta Math. Appl. Sinica (English Ser.), 14 (1998), pp. 193-196.

(3) B. L. Liu, The enumeration of directed trees with a given number of leaves and the enumeration of free trees, Kexue Tongbao, 32 (1987), pp. 244-247. In Chinese.

(4) B. L. Liu, Enumeration of oriented trees and free trees with a given number of leaves, Kexue Tongbao (English Ed.), 33 (1988), pp. 1577-1581.

(5) Q. Q. Nong, The degree sequence and number of leaves in a tree, J. Yunnan Univ. Nat. Sci., 24 (2002), pp. 167-171. In Chinese.

(6) Z. Y. Wang, An enumeration problem on ordered trees, J. Math. (Wuhan), 6 (1986), pp. 201-208.

(7) H. C. Wilf, Generatingfunctionology, Academic Press, 1990.

(8) Q. L. Yu, An algorithm for lexicographically generating ordered rooted trees with constraints on the number of leaves, Chinese J. Oper. Res., 6 (1987), pp. 71-72

nt.number theory - What is it that makes some diophantine equations interesting, while others are less so

Your question is probably too general, I simply hope that you'd like to learn some personal experience of people who do their research in diophantine equations or who apply the equations to other areas.

Although my starting research (under- and postgraduate level) was the theory of transcendental numbers, quite tied to diophantine equations (especially the ones related to linear forms in logarithms), I did not try to work on these. (Maybe, because of my father's attempts to prove Fermat's Last Theorem.) But at one occasion I was "introduced" to the Erdős--Moser equation $1^k+2^k+dots+(m−1)^k=m^k$ ($m$ and $k$ positive integers), Applications of pattern-free continued fractions, and was impressed by the beauty and power of the method which Leo Moser used in 1953 to show that no solution (except $1+2=3$) exists with $mle 10^{10^6}$. (!) Moser's method could not give much more, and I was happy enough to collaborate on some new ideas and computational achievements of nowadays to significantly extend (after more than 50 years) Moser's bound.

Another favourite diophantine equation is Catalan's equation $x^p-y^q=1$ ($p,q>1$) which was rather recently solved by Preda Mihăilescu. He even managed to avoid linear forms in logs (which are far from beauty because of so many technicalities). There are many steps in the proof, treating some special cases, most of them using completely different methods of diophantine analysis and algebraic number theory. But they are just beautiful! For example, there are two different proofs of the nonsolvability of $x^2-y^q=1$ in integers $x>1$, $y>1$ for a prime $q>3$. The original one, due to Ko Chao ([On the diophantine equation $x^2=y^n+1$, $xyne0$, Sci. Sinica 14 (1965) 457--460]; also given in Mordell's "Diophantine equations"), uses the law of quadratic reciprocity in a very elegant way. Another one, due to E.Z. Chein [A note on the equation $x^2=y^q+1$, Proc. Amer. Math. Soc.
56 (1976) 83--84], is extremely short and elementary.

Summarizing, I would say that natural criteria for considering some diophantine equations (and ignoring other) are the simplicity of the equation (isn't $x^n+y^n=z^n$ simple?) and the beauty and novelty of methods to solve it. Probably, the usefullness of the equation has to be taken into account as well.

dg.differential geometry - Extension of strictly plurisubharmonic functions on a Kähler manifold

It is instructive to conisder the case of Kahler metrics invariant under torus action. In this case your question becomes a certain (nontivial) question on convex functions.

Recall first, that Kahler metrics on $(mathbb C^*)^n$ invariant under the action of $(S^1)^n$ have global potential that is given by a convex function $F$ on $mathbb R^n$. Here $mathbb R^n$ is identified with the quotient

$(mathbb C^*)^n/(S^1)^n$ and we take coordinates $log|z_i|$ on $mathbb R^n$.
So we can translate your original question as follows

QESTION. Suppose you have a smooth convex function $F$, defined on $mathbb R^n$ outside compact $Omega$. Is it possible to extend $F$ to a smooth convex function on the whole $mathbb R^n$?

It easy to construct an example of a non-convex $Omega$ on $mathbb R^2$, with convex $F$ defined on $mathbb R^2setminus Omega$, so that $F$ can not be extended. For the moment I don't see how to make such an example when $Omega$ is the unite disk, but it sounds plausible that such examples exist.

ca.analysis and odes - Has coarse continuity been known already?

Functions like this are called Lipschitz. The definition works for maps between any two metric spaces. There is also the notion of being coarse lipschitz:

If you have a function $f : X to Y$ between two metric spaces, and constants $K geq 1$ and $C geq 0$, then $f$ is $(K,C)$--coarse lipschitz if
$d_Y(f(x),f(y)) leq K d_X(x,y) + C$ for any $x$ and $y$ in $X$.

gr.group theory - Generation in a group versus generation in its abelianization.

Background

I have been spending a lot of time in my research with subsets of groups that are very close to being generating sets. To make this precise:

Let $G$ be a group. If a subset $S$ of $G$ projects onto a generating set of $G/[G,G]$, we say that $S$ weakly generates $G$. The following fact (see page 350 in this book for a proof) shows that weak generation in nilpotent groups is a strong condition.

Fact. Let $G$ be a nilpotent group. If $S$ weakly generates $G$, then $S$ generates $G$.

In light of this result, we ask the following question:

Does there exist a finitely presented but non-nilpotent group $G$ such that every weakly generating subset of $G$ generates $G$?

If we drop the condition "finitely presented" then the first Grigorchuk group suffices. I'd be pretty surprised if no finitely presented examples exist.

Surface groups and free groups

In response to Matt's question below: For the free group $F(a,b)$, the set ${a[[a,b],a],b }$ weakly generates but doesn't generate (you can show this directly using uniqueness of freely reduced word form in a free group). You can use this to show that any non-abelian closed surface group has subsets that weakly generate but don't generate. For instance, in the genus two case, suppose $G$ has the standard presentation with generators $a,b,c,d$ and relation $[a,b][c,d]$. Consider the set $S = ${$a,b[[b,c],b],c,d$}. If this set generates G, then it generates the image $G/N$, where $N$ is the normal subgroup generated by $a$ and $d$. This image is a free group generated by the images of $b$ and $c$. The set S projects to a set which does not generate.

ag.algebraic geometry - An optimization problem for points on the sphere (master's dissertation)

First, by means of a disclaimer, some background. I am entering the fourth and final year of an undergraduate master's degree in maths, and a quarter of the maximum credit for this year will be for a project on the subject of my choice, which I will describe below. In this project, I am allowed to cite and quote the work of others to any extent so long as it is correctly referenced. My current understanding is that final-year projects in pure maths tend to include little or none of the author's original research, but rather take the form of a summary with perhaps some relatively minor original investigation. I have checked that asking about my chosen subject on this website is within regulations, so long as I cite the thread. Assuming it's alright, please bear in mind that I might quote you unless you ask not to be quoted.

Moving on to the subject of the project, I am interested in "sphere-like" polyhedra. More specifically: given a polyhedron with a fixed number n of sides, I would like to minimize its surface area relative to its volume. Some preliminary research turned up Michael Goldberg's paper "The Isoperimetric Problem For Polyhedra", from Tohoku Mathematical Journal vol. 40 (1934), which can be found at http://staff.aist.go.jp/d.g.fedorov/Tohoku_Math_J_1934_40_226-236.pdf

This is the only paper in English that I was able to find directly addressing the problem. It cites a number of previous non-English papers, which it summarizes as not having established very much, the main result being that a solution polyhedron can be considered as a set of n points on the unit sphere, representing the centres of mass of the polyhedron's faces which are tangent to the sphere at this point. Then, if we define sphericity as inverse to the surface area of such a polyhedron circumscribed about a unit sphere, we are looking to maximize the sphericity given n.

Goldberg also makes some further progress on the problem, and conjectures that the solutions fall into a particular class of polyhedra, which he terms "medial". He published a later paper on the subject of medial polyhedra, which can be found at http://staff.aist.go.jp/d.g.fedorov/Tohoku_Math_J_1936_43_104-108.pdf

As an undergraduate making his first foray into mathematical research, I have little intuition for the directions I should be going in (and hence, apologies if any of the tags I've added are inappropriate); but I have considered looking at what happens if I add the additional constraint that the polyhedron should have a nontrivial symmetry group, as the few known results do (tetrahedron, triangular prism, cube, pentagonal prism, dodecahedron). To this end I thought using the stereographic projection, fixing one of the n points to be at infinity, might make it easier to study the symmetry and perhaps even view the points as roots of a polynomial. Then I may be able to provide new proofs for the known solutions, which might generalize so that I can investigate unknown solutions for small values of n, at least those which have nontrivial symmetry.

In the other direction, I am interested in the asymptotic behaviour of the solution set. Goldberg places some bounds on its behaviour in his first paper; I've been wondering about possible bounds on the minimum or maximum area of a single face in terms of n, or on the maximum number of edges it can have (I've got a gut feeling that most of the faces are going to be hexagons for large enough n). An important question is of course how sphericity behaves asymptotically with respect to n. While I expect the solution set will eventually become quite irregular, is it possible to find a well-behaved family of polyhedra for which sphericity behaves similarly, if not quite as well?

Finally, how if at all does this problem relate to other optimization problems for points on the sphere? The known solutions at least coincide, for example, with those for minimizing the energy of n point charges on a sphere (the Thomson problem), though I suspect the coincidence only occurs because n is small.

In summary, my question is this: are there any English-language publications, or translations, on this subject of which I am currently unaware? Has any further progress been made? If not, what avenues sound like they may be worth exploring? If anyone is interested in pursuing the problem themselves and would be willing to give me permission to cite their work I would of course be delighted.

Many thanks,

Robin

Sheaves and bundles in differential geometry

If $X$ is a manifold, and $E$ is a smooth vector bundle over $X$ (e.g. its tangent bundle),
then $E$ is again a manifold. Thus working with bundles means that one doesn't have to leave
the category of objects (manifolds) under study; one just considers manifold with certain extra structure (the bundle structure). This is a big advantage in the theory; it avoids introducing another class of objects (i.e. sheaves), and allows tools from the theory of manifolds to be applied directly to bundles too.

Here is a longer discussion, along somewhat different lines:

The historical impetus for using sheaves in algebraic geometry comes from the theory of several complex variables, and in that theory sheaves were introduced, along with cohomological techniques, because many important and non-trivial theorems can be stated as saying that certain sheaves are generated by their global sections, or have vanishing higher cohomology. (I am thinkin of Cartan's Theorem A and B, which have as consequences many
earlier theorems in complex analysis.)

If you read Zariski's fantastic report on sheaves in algebraic geometry, from the 50s, you will see a discussion by a master geometer of how sheaves, and especially their cohomology,
can be used as a tool to express, and generalize, earlier theorems in algebraic geometry.
Again, the questions being addressed (e.g. the completeness of the linear systems of hyperplane sections) are about the existence of global sections, and/or vanishing of higher cohomology. (And these two usually go hand in hand; often one establishes existence results about global sections of one sheaf by showing that the higher cohomology of some related sheaf vanishes, and using a long exact cohomology sequence.)

These kinds of questions typically don't arise in differential geometry. All the sheaves
that might be under consideration (i.e. sheaves of sections of smooth bundles) have global sections in abundance, due to the existence of partions of unity and related constructions.

There are difficult existence problems in differential geometry, to be sure: but these are very often problems in ODE or PDE, and cohomological methods are not what is required to solve them (or so it seems, based on current mathematical pratice). One place where a sheaf theoretic perspective can be useful is in the consideration of flat (i.e. curvature zero) Riemannian manifolds; the fact that the horizontal sections of a bundle with flat connection form a local system, which in turn determines the
bundle with connection, is a useful one, which is well-expressed in sheaf theoretic language. But there are also plenty of ways to discuss this result without sheaf-theoretic language, and in any case, it is a fairly small part of differential geometry, since typically the curvature of a metric doesn't vanish, so that sheaf-theoretic methods don't seem to have much to say.

If you like, sheaf-theoretic methods are potentially useful for dealing with problems,
especially linear ones, in which local existence is clear, but the objects are suffiently rigid that there can be global obstructions to patching local solutions.

In differential geomtery, it is often the local questions that are hard: they become difficult non-linear PDEs. The difficulties are not of the "patching local solutions"
kind. There are difficult global questions too, e.g. the one solved by the Nash embedding theorem, but again, these are typically global problems of a very different type to those that are typically solved by sheaf-theoretic methods.

ag.algebraic geometry - How to unify various reconstruction theorems (Gabriel-Rosenberg, Tannaka,Balmers)

What I am talking about are reconstruction theorems for commutative scheme and group from category. Let me elaborate a bit. (I am not an expert, if I made mistake, feel free to correct me)

Reconstruction of commutative schemes

Given a quasi compact and quasi separated commutative scheme $(X,O_{X})$ (actually, quasi compact is not necessary), we can reconstruct the scheme from $Qcoh(X)$ as category of quasi coherent sheaves on $(X,O_{X})$. This is Gabriel-Rosenberg theorem. Let me sketch the statement of this theorem which led to the question I want to ask:

The reconstruction can be taken as geometric realization of $Qcoh(X)$ (abelian category). Let $C_X$=$Qcoh(X)$. We define spectrum of $C_X$ and denote it by $Spec(X)$

(For example, if $C_X=R-mod$.where $R$ is commutative ring, then $Spec(X)$ coincides with prime spectrum $Spec(R)$). We can define Zariski topology on $Spec(X)$, we have open sets respect to Zariski topology. Then we have contravariant pseudo functor from category of Zariski open sets of the spectrum $Spec(X)$ to $Cat$,$Urightarrow C_{X}/S_{U}$,where $U$ is Zariski open set of $Spec(X)$ and $S_U=bigcap_{Qin U}^{ }hat{Q}$

(Note:$Spec(X)$ is a set of subcategories of $C_{X}$ satisfying some conditions, so here,$Q$ is subcategory belongs to open set $U$ and $hat{Q}$=union of all topologizing subcategories of $C_{X}$ which do not contain $Q$.)

For each embedding:$Vrightarrow U$, we have correspondence localization functor: $C_{X}/S_{U}rightarrow C_{X}/S_{V}$. Then we have fibered category over the Zariski topology of $Spec(X)$, so we have given a geometric realization of $Qcoh(X)$ as a stack of local category which means the fiber(stalk)at each point $Q$,

$colim_{Qin U} C_{X}/S_{U}$=$C_{X}/hat{Q}$ is a local category.

Zariski geometric center

Define a functor $O_{X}$:$Open(Spec(X))rightarrow CRings$

$U|rightarrow End(Id_{C_{X}/S_{U}})$

It is easy to show that $O_{X}$ is a presheaf of commutative rings on $Spec(X)$, then Zariski geometric center is defined as $(Spec(X),hat{O_{X}})$,where $hat{O_{X}}$ is associated sheaf of $O_{X}$.

Theorem:

Given X=$(mathfrak{X},O_{mathfrak{X}})$ a quasi compact and quasi separated commutative scheme. Then the scheme X is isomorphic to Zariski geometric center of $C_X=Qcoh(X)$. If we denote the fibered category mentioned above by $mathfrak{F}_{X}$.

then center of each fiber (stalk) of $mathfrak{F}_{X}$ recover the presheaf of commutative ring defined above and hence Zariski geometric center. Moreover, Catersion section of this fibered category is equivalent to $Qcoh(X)$ when $X$ is commutative scheme

Question

It is well known that for a compact group $G$, we can use Tanaka formalism to reconstruct this group from category of its representation $(Rep(G),bigotimes _{k},Id)$. Is this reconstruction Morally the same as Gabriel-Rosenberg pattern reconstruction theorem?

From my perspective, I think $Qcoh(X)$ and category of group representations are very similar because $Qcoh(X)$ can be taken as "category of representation of scheme". On the otherhand, group scheme is a scheme compatible with group operations. Therefore, I think it should have united formalism to reconstruct both of them. Then I have following questions:

1 Is there a Tanaka formalism for reconstruction of general scheme$(X,O_{X})$ from $Qcoh(X)$?

2.Is there a Gabriel-Rosenberg pattern reconstruction (geometric realization of category of category of representations of group) to recover a group scheme? I think this formalism is natural (because the geometric realization of a category is just a stack of local categories and the center(defined as endomorphism of identical functor)of the fiber of this stack can recover the original scheme) and has good generality (can be extend to more general settings)

Maybe one can argue that these two reconstruction theorems live in different nature because reconstruction of group schemes require one to reconstruct the group operations which force one to go to monoidal categories (reconstruct co-algebra structure) while reconstruction of
non-group scheme doesn't (just need to reconstruct algebra structure).

However, If we stick to the commutative case. $Qcoh(X)$ has natural symmetric monoidal structure. Then, in this case, I think this argue goes away.

More Concern

I just look at P.Balmer's paper on reconstruction from derived category, it seems that he also used Gabriel-Rosenberg pattern in triangulated categories: He defined spectrum of triangulated category as direct imitation of prime ideals of commutative ring. Then, he used triangulated version of geometric realization (geometric realization of triangulated category as a stack of local triangulated category) hence, a fibered category arose, then take the center of the fiber at open sets of spectrum of triangulated categories to recove the presheaf (hence sheaf) of commutative rings. This triangulated version of geometric realization is also mentioned in Rosenberg's lecture notes Topics in Noncommutative algebraic geometry,homologial algebra and K-theory page 43-44, it also discuss the relation with Balmer's construction. But in Balmer's consideration, there is tensor structure in his derived category

Therefore, the further question is:

Is there a triangulated version of Tannaka Formalism which can recover P.Balmer's reconstruction theorem. ? I heard from some experts in derived algebraic geometry that Jacob Lurie developed derived version of Tannaka formalism. I know there are many experts in DAG on this site. I wonder whether one can answer this question.

In fact, I still have some concern about Bondal-Orlov reconstruction theorem but I think it seems that I should not ask too many questions at one time.So, I stop here.All the related and unrelated comments are welcome

Thanks in advance!

Reconstruction theorem in nLab reconstruction theorem

EDIT: A nice answer provided by Ben-Zvi for related questions

lo.logic - Algebraic description of compact smooth manifolds?

There is a very cool answer to your question, and it goes by the name well-adapted models for synthetic differential geometry. Andrew Stacey already indicated it in his reply, but maybe I can expand a bit more on this.

Synthetic differential geometry is an axiom system that characterizes those categories whose objects may sensibly be regarded as spaces on which differential calculus makes sense. These categories are called smooth toposes.

A model for this is a particular such category with these properties. A well-adapted model is one which has a full and faithful embedding of the category of smooth manifolds. (This is "well adapted" from the point of view of ordinary differential geoemtry: ordinary differential geometry embeds into these more powerful theories of smooth structures).

The striking insight is that this perspective in particular usefully unifies the ideas of algebraic geometry with that of differential geoemtry to a grander whole.

Indeed, the category of presheaves on the opposite of (finitely generated) commutative rings is a model for the axioms, and of course this is the context in which algebraic geometry takes place.

But we are entitled to take probe categories considerably richer than just that of duals of commutative rings. In particular, we may consider a category of commutative rings that have a notion of being "smooth" the way a ring of smooth functions is "smooth". These are the C-infinity rings or generalized smooth algebras. Every ring of smooth functions on a smooth manifold is an example, but there are more.

The formal dual of these rings are spaces called smooth loci. This is a smooth analog of the notion of affine scheme. (Notice that the notion of "smooth" as used here is that of differential geometry, not quite that of algebraic geometry, which is more like "singularity free". But they are not unrelated).

The main theorem going in the direction of an answer to your question is that the category of manifolds embeds fully and faithfully into that of smooth loci. See at the link smooth loci for the details.

But inside the category SmoothLoci, manifolds are characterized as the formal dual to their smooth rings of functions, so that's one way to answer your question.

There is a grand story developing from this point on, but for the moment this much is maybe sufficient as a reply.

computer science - What is a universal function?

In that argument, he just means that g is defined on all the two-element subsets that may arise in the argument, that is, for a given family F of four-element sets, g(A) should be defined on any two-element set A that is a subset of a four-element set in F.

The reason he needs to assume that is that he cannot allow that we need to choose the choice function g itself to be used with each separate A. There are many choice functions that work for families of 2-element sets, and different choices of g will give rise to different functions on the families of four-element sets. The way the argument works is that you make one choice of the function g that works on all the two element sets that arise, and then you define the choice function on the given family of four-element sets by the clever construction in the article.

In particular, he is not using some technical meaning of universal.

at.algebraic topology - Simple question of topological cofibration

I'm not sure whether this will be helpful to mpdude, but I thought I'd just point out that given a smooth manifold M and a smooth submanifold N, you can see very easily that the inclusion $Nto M$ is a cofibration, without any cell structure.

Simply observe that a tubular neighborhood gives you a mapping cylinder neighborhood of N inside M (if you choose a metric, then the disk bundle is the mapping cylinder of the projection from the sphere bundle down to N). The homotopy extension property is then easy to verify, because in general for a map $f: Xto Y$, the pair $(M_f, Xcup Y)$ has the homotopy extension property ($M_f$ is the mapping cylinder) so given a map on M and a homotopy on the submanifold M, you can extend your homotopy trivially outside the disk bundle of the tubular neighborhood, and use the HEP to extend it on the disk bundle itself (leaving the homotopy constant on the sphere bundle).

I suppose this procedure would also work for infinite-dimensional manifolds (with appropriate hypotheses).

Closely related is the characterization of (closed) cofibrations in terms of neighborhood deformation retracts, which appears for example in Arne Strom's Note on Cofibrations II.

dimension theory - Why do modules with small support have high Exts?

To understand what "nice" is in your sense has been a very interesting question in commutative algebra.

In the following discussion I will assume, unless otherwise notice, that $(R,m,k)$ is Noetherian local, and $M$ is finitely generated. Let $(1)$ be the codimension of support of
$M$ and $(2)$ be the biggest non-vanishing index of $text{Ext}(M,-)$.

First, the number (2) is finite forces $M$ to have finite projective dimension by taking $N=k$ the residue field. So we will assume $text{pd} M <infty$. Then, as BCrd pointed out:

$$ (2) = text{pd} M = text{depth} R - text{depth} M$$

The first inequality is easy by computing Ext via a projective res. of M + Nakayama lemma. The second is the Auslander-Buchsbaum theorem.

On the other hand:

$$(1) = text{dim} R - text{dim} M $$

So

$$(1) - (2) = (text{dim} R - text{depth} R) - (text{dim} M-text{depth} M) $$

Thus, if both $R,M$ are Cohen-Macaulay (which by def. means dim=depth) and $text{pd} M <infty$ then $(1) = (2)$. If $R$ is "more Cohen-Macaulay" then $M$, we will have $(1)<(2)$.

The situation described in Emerton's answer is also very interesting. In general, the smallest index for which $text{Ext}^i(M,N) neq 0$ is the biggest length of an $N$-regular sequence inside the annihilator of $M$. When $N=R$ this number is called the grade of $M$, which I will call (3).

It is easy to see that $(3) leq (1)$ in general. One can prove that $(1) = (3)$ if $R$ is Gorenstein as follows: By Local Duality, $text{Ext}^i(M,R)$ is Matlis dual to the local cohomology module $text{H}_{m}^{d-i}(M)$, here $d= text{dim} R$. It is not hard to show that local cohomology vanish beyond $text{dim} M$, QED.

Amazingly, it has been an open conjecture for 50 years that $(1)=(3)$ whenever $M$ has finite projective dimension!

For "intuitive" understanding, I would offer the following: often when study modules of finite projective dimension one draw inspirations from those of the form $R$ modulo a regular sequence (so the resolution is a Koszul complex). In such case one can easily see that $(1) = (2) =(3)$.

EDIT: I got too caught up in the results and forgot your main question: why bigger codimension implies bigger projective resolution? A very low-tech way to see it is: bigger codimension means bigger annihilator of $M$. Now each element of the annihilator of $M$ gives a non-trivial relation on elements of $M$, namely $ax=0$, so it is not surprising that the modules with bigger annihilators have more complicated resolutions.

nt.number theory - Is there a nice proof of the fact that there are (p-1)/24 supersingular elliptic curves in characteristic p?

One argument (maybe not of the kind you want) is to use the fact that the wt. 2 Eisenstein series on $Gamma_0(p)$ has constant term (p-1)/24.

More precisely: if ${E_i}$ are the s.s. curves, then for each $i,j$,
the Hom space $L_{i,j} := Hom(E_i,E_j)$ is a lattice with
a quadratic form (the degree of an isogeny), and we can form the corresponding
theta series $$Theta_{i,j} := sum_{n = 0}^{infty} r_n(L_{i,j})q^n,$$
where as usual $r_n(L_{i,j})$ denotes the number of elements of degree $n$.
These are wt. 2 forms on $Gamma_0(p)$.

There is a pairing on the $mathbb Q$-span $X$ of the $E_i$ given by $langle E_i,E_jrangle
= $ # $Iso(E_i,E_j),$ i.e. $$langle E_i,E_jrangle = 0 text{ if } i neq jtext{ and equals # }Aut(E_i) text{ if }i = j,$$
and another formula for $Theta_{i,j}$ is
$$Theta_{i,j} := 1 + sum_{n = 1}^{infty} langle T_n E_i, E_jrangle q^n,$$
where $T_n$ is the $n$th Hecke correspondence.

Now write $x := sum_{j} frac{1}{text{#}Aut(E_j)} E_j in X$. It's easy to see
that for any fixed $i$, the value of the pairing $langle T_n E_i,xrangle$
is equal to $sum_{d |n , (p,d) = 1} d$. (This is just the number of $n$-isogenies
with source $E_i,$ where the target is counted up to isomorphism.)
Now
$$sum_{j}
frac{1}{text{#}Aut(E_j)} Theta_{i,j} =
bigg{(}sum_{j} frac{1}{text{#}Aut(E_j)}bigg{)} + sum_{n =1}^{infty} langle T_n E_i, xrangle
q^n
= bigg{(}sum_{j}frac{1}{text{#}Aut(E_j)}bigg{)} + sum_{n = 1}^{infty} bigg{(}sum_{d | n, (p,d) = 1} dbigg{)}q^n.$$

Now the LHS is modular of wt. 2 on $Gamma_0(p)$, thus so is the RHS. Since we know
all its Fourier coefficients besides the constant term, and they coincide with those of the Eisenstein series, it must be the Eisenstein series.
Thus we know its constant term as well, and that gives the mass formula.

(One can replace the geometric aspects of this argument, involving s.s. curves and Hecke
correspondences, with pure group theory/automorphic forms: namely the set ${E_i}$ is
precisely the idele class set of the multiplicative
group $D^{times}$, where $D$ is the quat. alg. over $mathbb Q$ ramified at $p$ and $infty$. This formula, writing the Eisenstein series as a sum of theta series, is then
a special case of the Seigel--Weil formula, I believe, which in general, when you pass to constant
terms, gives mass formulas of the type you asked about.)

computational complexity - Decision problem restricted to inputs that satisfy some necessary condition.

Consider the following decision problem:

Problem 1

INPUT: A graph G.

OUTPUT: YES if G is 3-colorable, NO if not.

This is a well-known NP-complete problem. Now suppose that we have a necessary (but not sufficient) condition for a graph to be 3-colorable, called NC. Consider the following problem:

Problem 2

INPUT: A graph G that satisfies NC.

OUTPUT: YES if G is 3-colorable, NO if not.

Now suppose that it is not known whether NC can be determined in polynomial time. Can we say that Problem 2 is in NP? It seems to me that it should be, seeing as there is a succinct certificate for a YES answer.

(However, I've been told by someone who I trust on other matters that this is a "promise problem" and not in NP, which is why I'm posting it here.)

---

Update

I'm not entirely happy with the answers below (although the subsequent discussions in the comments were useful), so I will attempt to answer the question myself.

Consider:

Problem 3

INPUT: A planar graph G.

OUTPUT: YES if G is 3-colorable, NO if not.

Now, by the usual definitions, an NP problem is one where, from the set of all binary strings, certain ones (those in the "language") must be recognised. Now, it is not easy to come up with a way to represent planar graphs such that every binary string corresponds to a planar graph. So by convention, Problem 3 means the following:

Given an input string, determine if it represents a planar graph that is 3-colorable.

So by the same convention, Problem 2 defines an NP problem: Given an input string, determine if it represents a graph that satisfies NC that is 3-colorable.

Now, the strings that should give "yes" answers are exactly the same as those that should give "yes" answers to Problem 1. So as NP problems, Problems 1 and 2 are exactly the same. (In other words, the languages they define are the same.)

So to make Problem 2 useful, we need to cast it as a Promise problem, where the input is not all binary strings, but is restricted in some way. Posed in this form, it is not an NP problem.

ag.algebraic geometry - Extending Functions on Closed Submanifolds of C^n

Yes, this is true. It follows from "Cartan's Theorem B" which says that H^1 of any coherent analytic sheaf on a closed submanifold of C^n is 0; the same result is also true for analytic subspaces. Look up any book on several complex variables for a proof. (It is quite possible that there is a more elementary proof.)

(One uses the theorem as follow: Let X be the submanifold or analytic space and consider the exact sequence of sheaves on C^n

0 --> I --> O_{C^n} --> O_X --> 0

where I is the ideal sheaf of X. The vanishing of H^1(C^n,I) implies that the map H^0(C^,O_{C^n}) to H^0(X,O_X) is surjective, which is what you want.)

graph theory - Complete tree invariants?

If we take a graph invariant to be "a property that depends only on the abstract structure, not on graph representations such as particular labellings or drawings of the graph" (from Wikipedia), I have the feeling that Harrsion's question for complete graph invariants remained basically unanswered, since Greg's answer is mainly about (canonical) labellings.

Thinking - as Harrison did - of "the usual ones (the Tutte polynomial, the spectrum, whatever)", I'd like to repeat Harrison's question, but restrict it to trees:

Are there any known complete tree invariants?

ac.commutative algebra - Which rings are subrings of matrix rings?

Kevin Buzzard informs me, by e-mail and in the above comment, that what he cares about is the case where $A$ is regular. In particular, it would be good to know whether every $T$ is a matrix ring in this case. I thought about it but didn't solve this, so here is a record of my ideas. Let $n$ be the dimension of $A$.

(1) A subring of a matrix is obviously a matrix ring. So, if we knew that $T$ was a matrix ring whenever $T$ was normal, this would establish that $T$ was always a matrix ring.

(2) As explained in my previous answer, if $T$ is free as an $A$-module, then $T$ is a matrix ring. We have the following implications: if $T$ is Cohen-Macaulay then $T$ torsion-free and finite over $A$ implies $T$ flat over $A$; if $A$ is a polynomial ring then $T$ flat over $A$ implies $T$ free over $A$. So, if $A$ is a polynomial ring and $T$ is Cohen-Macaulay, then $T$ is a matrix ring.

In particular, if $n=1$ or $2$, then $T$ normal implies $T$ Cohen-Macaulay. So, in these cases, and with $A$ a polynomial ring, $T$ is a matrix ring.

(3) As explained above, if $T^*$ is free over $A$, we also get to conclude that $T$ is a matrix ring. Unfortunately, this can fail when $n geq 3$.

(4) If there is some $T$-module $M$ on which $T$ acts without kernel, and $T$ is free as an $A$-module, then $T$ is a matrix algebra. Restated geometrically, if there is any coherent sheaf on $mathrm{Spec}(T)$, with support on the whole of $mathrm{Spec}(T)$, whose pushforward to $mathrm{Spec}(A)$ is a trivial vector bundle, then $T$ is a matrix algebra. If we restrict our attention to $A$ a local ring, or a polynomial ring, then the adjective "trivial" comes for free.

So, if there were to be a counter-example, we would want $n geq 3$ and we would want $T$ to be normal but not Cohen-Macaulay. Moreover, we would want that basically any $T$-module is not free as an $A$-module.

Any ideas?

mathematics education - Reference letters for graduate school after a couple years in the industry

How does one return to graduate school after spending a couple years in the industry? In particular, what are ways of getting good recommendations? I'm not concerned about the "adjustment" to the grad student lifestyle, but rather about the application process if the goal is a top school.

I was a CS/math major at MIT for undergrad, but wasn't really sure at the time if I wanted to go into academia, so I ended up doing more software and machine learning. For a while now, though, I've been realizing that I miss the academic life, so I've been thinking about my original goals of going to grad school in pure math or computer science theory.

So how should I go about getting letters of recommendation? It's been a while since undergrad, so letters from professors and research advisors aren't really feasible (I didn't have much interaction with them in any case), and while my work now is pretty quantitative (in machine learning), my supervisors aren't really qualified to write for me.

Does taking classes help much? I live in the Bay Area, and I know Berkeley -- and possibly Stanford? -- allows people to enroll in courses. Or should I be trying somehow to do research? I'm guessing there aren't REU-type things available for me and professors willing to take the time to talk with non-affiliated students are probably pretty hard to find, so I'm not sure how to go about this.

I've seen somewhat similar questions on MO, so hopefully this isn't too soft a question!

Are these notions of strongly equivariant D-modules equivalent?

It seems that there are two notions of strongly equivaraint $D_X$- Modules and I would like to know if they are equivalent, or at least how they are related.
Let $rho: Gtimes X rightarrow X$ be an action of an algebraic group on a smooth variety over the complex numbers.
The first definition goes like this:

An equivariant $D_X$ Module is just a $D_X$ module $M$ together with an isomorphism
$$rho^* Mrightarrow pi^* M$$ of $D_{Gtimes X}$ -modules. That isomorphism has to satisfy some cocycle condition.

The other definition is a bit more cumbersome to write down. First it requires just an isomorphism of $O_{Gtimes X}$ modules, not necessarily of $D_{Gtimes X}$-modules $$rho^* Mrightarrow pi^* M$$ modules, which again satisfies the cocycle conditon.
In addition it requires the action map $$D_Xotimes M rightarrow M$$ to be equivariant.
Finally there is another condition to be satisfied:
Observe that we get two operations of the liealgera on $M$:

One operation, by directly differentiating the action of $G$ on $M$.

Another operation in the following way: First we differentiate the action of $G$ on $X$, and get a map
$$Lie(G)rightarrow Der_X$$
from the liealgebra into vectorfields on $X$. Because $M$ is a $D_X$ module we can compose this map with the action of vectorfields on $M$ and get our second operation.

We require these operations to coincide.

A more precise definition of the second kind is given here on pages 48-49:
http://www.math.harvard.edu/~gaitsgde/267y/catO.pdf

So the question is, are these two notions equivalent?

Edit: If anybody else needs these facts, I found a reference which gives a proof:
http://alpha.uhasselt.be/Research/Algebra/Publications/Geq.ps

soft question - japanese/chinese for mathematicians?

This may not be the answer you're looking for, but I thought I'd share my experience as someone who was born in Japan but was transplanted quickly into the United States. My Japanese is not nearly as good as it should be, but is certainly good enough to read math.

A beautiful part of reading Japanese/Chinese math is that you can grasp the meaning without knowing how to pronounce anything. I don't know any technical Chinese, but in Japanese,

写像

is the word for "mapping" or "function", and the literal meaning of its characters hints at this. Let me explain.

The first character means to transcribe, to picture, or to give a visual form--poetically, it can mean to simply give an abstract form to something, rather than a visual one. (For instance, the word 写真 means photograph, where the second character in this particular word means "truth". It might be silly to think the word for photograph is to "picture something truly/in its true form", but that's a beautiful translation to ponder on another occasion.)

The final character in 写像 means figure, or image, or an embodiment. For instance, the word 画像 means "image" in the computer sense of file type. In fact the character 像 alone can mean "image" in the sense of mathematics, as in the image of something under a map.

In short, the word for "function" or "map" can be literally and clumsily translated back into English as "forming an image" or "creating a figure" or "realizing a form", most abstractly. I doubt any Japanese person ever thinks in these terms, no more than we think of the word "projection" deeply in terms of its Latin roots. But to harzard a guess at the meanings of these words can be a beautiful experience, and one unique to those weirdos who know the meanings of things without knowing how to say them.

So it may be a really interesting experience to simply learn the meaning of each commonly occurring (math) character---I'll list a few below---and to get a feel for the mathematical meanings of their combinations via intuition. When I've read Japanese math books, the feeling of knowing the meaning on a page without knowing how to pronounce a word has been the most rewarding and beautiful part. If you choose to do this, the best tip I have is to simply write: Make sure you copy and write the characters over and over again, so you begin to distinguish subtle differences between them.

For the enjoyment of some, here are examples of Japanese math words and the meanings of their constituent characters. I'll list some irrelevant meanings of some characters--though characters often only take on one of many meanings based on context, I still think it's fun to know their other possible meanings.

空間 (space)

空 = sky, emptiness, space, air

間 = between, the space between, an interval of time

位相 (topology)

位 = rank (as in seniority or importance in an organization), a word for counting dead souls, decimal place, position. As a verb, it can mean to locate--i.e., to determine the position of.

相 = form, shape, appearance, the relationship of one thing to another.

Strangely enough, 位相　can also mean the phase of something, as in the angle or phase of a complex number or a wave. It also mean the phase of something as in "solid/liquid/gaseous". I would assume that the term first came to use to describe the states of matter, was tangentially used to describe the phase of wave-like phenomena since the English term "phase" was used in both instances.

微分 (derivative, to take the derivative of)

微 = infinitesimal, tiny, slight

分 = to divide, an amount of something.

In learning language so much emphasis is placed on the sounds of things, rather than on the abstract units of meaning. I suppose Chinese characters were developed exactly to avoid this aural emphasis, but it is always a joy to have zero verbal understanding with a Chinese or Korean person, but to be able to communicate by writing characters in the air.

Well, perhaps this was not helpful in the least, but maybe it will at least entertain some non-Japanese-speakers. (By the way, I'd be very curious to hear if the Chinese technical terms are the same, as almost all technical terms in Japanese utilize kanji, or Chinese characters.)

ramsey theory - Bound on cardinality of a union

A better solution than my previous one is

max_{1leq i leq n} iN - {i choose 2}N_2

(That is to say, we can simply consider only i of the sets instead of all n of them, and then apply my previous argument to obtain a lower bound on the size of the union of those i sets, which is also a lower bound on the size of the total union.)

In fact, one can figure out the value of i which maximizes this bound: it will be the largest i for which N - (i-1)N_2 is positive (since this is the difference between the bound obtained using i and that obtained using i-1). This is i = lfloor N/N_2 rfloor +1.

So, taking that value for i, the best solution I can see is

nN - {nchoose 2} N_2 if n< i and otherwise iN - {i choose 2} N_2.

This is, at least, non-negative. In the latter case, it's on the order of N^2/(2N_2).

**

Of course, I'm still not using all the information. I don't see how to get anything further out of inclusion-exclusion when all we have are upper bounds on the sizes of the triple and higher intersections. I encourage anyone who thinks there is an argument in there to post it.

(Edited to correct my arithmetic for the order of the bound in the i <= n case.)

ag.algebraic geometry - Is there triangulated category version of Barr-Beck's theorem?

There isn't a descent theory for derived categories per se - one can't glue objects in the derived category of a cover together to define an object in the base. (Trying to apply the usual Barr-Beck to the underlying plain category doesn't help.)

But I think the right answer to your question is to use an enriched version of triangulated categories (differential graded or $A_infty$ or stable $infty$-categories), for which there is a beautiful Barr-Beck and descent theory, due to Jacob Lurie. (This is discussed at length in the n-lab I believe, and came up recently on the n-category cafe (where I wrote basically the same comment here..)
This is proved in DAG II: Noncommutative algebra. In the comonadic form it goes like this. Given an adjunction between $infty$-categories (let's call the functors pullback and pushforward, to mimic descent), if we have

1. pullback is conservative (it respects isomorphisms), and


2. pullback respects certain limits (namely totalizations of cosimplicial objects,
   which are split after pullback)

then the $infty$-category downstairs is equivalent to comodules over the comonad
(pullback of pushforward). (There's an opposite monadic form as well)
This can be verified in the usual settings where you expect descent to hold.
In other words if you think of derived categories as being refined to $infty$-categories (which have the derived category as their homotopy category), then everything you might want to hold does.

So while derived categories don't form a sheaf (stack), their refinements do:
you can recover a complex (up to quasiisomorphism) from a collection of complexes on a cover, identification on overlaps, coherences on double overlaps, coherences of coherences on triple overlaps etc.
More formally: define a sheaf as a presheaf $F$ which has the property that
for an open cover $Uto X$, defining a Cech simplicial object $U_bullet={Utimes_X Utimes_X Ucdotstimes_X U}$, then $F(X)$ is the totalization of the cosimplicial object $F(U_bullet)$. Then enhanced derived categories form sheaves (in appropriate topologies) as you would expect. This is of course essential to having a good theory of noncommutative algebraic geometry!

lie algebras - Constructing Affine Kac-Moody Groups

As you said, the main thing is to construct the central extension.
The story is relatively straightforward for groups of type $A$ and gets more complicated in the general case.
First, let $mathcal K=k((t)), mathcal O= k[[t]]$.
As usual, let the affine Grassmannian $Gr_G$ be the quotient $G(mathcal K)/G(mathcal O)$. Then in order to construct a central extension of $G(mathcal K)$ it is enough to construct a line bundle $mathcal L$ on $Gr_Gtimes Gr_G$ which is

a) $G(mathcal K)$-equivariant (note that this is the same as to specify a $G(mathcal O)$-equivariant line bundle on $Gr_G$)

b) Has the property that for any $x_1,x_2,x_3in Gr_G$ we have an isomorphism $mathcal L(x_1,x_3)=mathcal L(x_1,x_2)otimes mathcal L(x_2,x_3)$ satisfying obvious associativity relation (of course, all this is actually an additional structure - a careful way to say it is to identify two lifts of $mathcal L$ to
$Gr_Gtimes Gr_Gtimes Gr_G$ such that two identifications on $Gr_G^4$ coincide, but this is standard).

Indeed given such $mathcal L$ we can define the central extension $hat{G}$ as the set of pairs $(g,alphain mathcal L(1,g)), alphaneq 0$ (here I identify elements of $G$ with their image in $Gr_G$) and the multiplication is easy from b) above.

Now we need to construct $mathcal L$ as above. This is easy if $G=GL(n)$ (or $G=SL(n)$). Namely, in this case $Gr_G$ is the same as the space of lattices $Lambda subset mathcal K^n$ and for any two such lattices $Lambda_1,Lambda_2$ we can set $mathcal L(Lambda_1,Lambda_2)=det(Lambda_1/Lambda_1cap Lambda_2)otimes det(Lambda_2/Lambda_1capLambda_2)^{-1}$. Properties a), b) are obvious.

For general $G$ the story is more complicated. First, canonically central extensions are in one-two-one correspondences with even invariant bilinear forms
on the Cartan of $Lie(G)$. For simple $G$ there is a minimal such form and it is enough to construct the extension corresponding to this minimal form (all others are "powers" of it in the appropriate sense). For $G=SL(n)$ the above construction gives exactly the minimal extension. For general $G$ you can easily adapt the above construction once you choose a representation $V$ of $G$. The problem is that when $G$ is not of type $A$, there is no $V$ that gives the minimal form (any $V$ defines an even invariant form on the Lie algebra of $G$ but usually it is not minimal). So, in this way you are going to get powers of the correct line bundle (and thus powers of the correct central extension). For instance, if you take $V$ to the be the adjoint representation, you get the $2h^{vee}$-power of the minimal bundle, where $h^{vee}$ is the dual Coxeter number. Constructing minimal $mathcal L$ for general $G$ is a relatively tricky business - the best treatment of this that I know was given by Faltings in

Gerd Faltings, Algebraic loop groups and moduli spaces of bundles. J. Eur. Math. Soc. 5 (2003), 41-68. doi: 10.1007/s10097-002-0045-x.

ag.algebraic geometry - Triviality of the Hodge bundle for a special family of semistable curves

Let g,h be positive integers. Let E be an elliptic curve, C be a genus h curve, and D be a genus g-h-1 curve. Let c,d,e be points on (resp.) C,D, and E.

Let f:CC --> E-e be the family whose fiber over a point e' is the curve obtained by glueing C to E together at the points c and e and D to E at the points d and e'.

Question: what is the pushforward f_* ω_f
?

It should be trivial, and David Speyer's answer to my question here should answer this question, but I (and a few others I asked earlier) couldn't get it to work.

If a tensor product of modules is semi-simple, are the tensor factors semi-simple ?

I can't resist providing the following comment and references, even though it is likely not really relevant to the (already old by now) original post.

Quick background: Let $k$ be a field of char. $p>0$ and let $G$ be a(ny) group.
Serre proved [Invent. Math. 116 (1994), no. 1-3, 513--530] that if $V$ and $W$ are finite dimensional, semisimple $kG$-modules and $$(dim V - 1) + (dim W -1) < p$$ then $V otimes_k W$ is a semisimple $kG$-module. (The argument is quite nice - one reduces to alg. closed $k$, and replaces $G$ by a certain algebraic group
over $k$ whose identity component is reductive. One then has to argue that $[G:G^0]$ has order prime to $p$, so one is reduced to consideration of connected reductive $G$. And that case is handled by some "weight combinatorics" and the linkage principle.)

In a subsequent paper [Semisimplicity and tensor products of group representations: converse theorems. With an appendix by Walter Feit. J. Algebra 194 (1997), no. 2, 496--520] Serre proved some "converse theorems". For example, he shows that
$$V otimes_k W quad text{semisimple} implies V text{semisimple if $dim W not equiv 0 pmod{p}$}$$
Examples (due to Feit, and included in the paper) show that one can't get rid of the assumption on $dim W$.

ca.analysis and odes - Transformation from domains to half-spaces

The half spaces R^n are the inward-pointing halves of the tangent spaces to the points on the boundary.
Starting with a local boundary value problem you end up with a model equation on the half space associated to each point on the boundary by "localizing" at the point.
A solution to the elliptic boundary value problem would give you also a solution to each of these model problems.
Conversely, you can construct a formal solution to the elliptic boundary value problem at the boundary by using these model problems.
Then you can improve this formal solution to an `honest' solution by solving away the error; this is easier because you no longer have to worry about the boundary.

This system of ODEs is used to state the Lopatinski-Shapiro condition, and searching for this will yield many references. My favorite is Hormander's analysis of partial differential operators -- this topic in particular is best explained in the first edition.

ac.commutative algebra - An "Elementary" Math Question Generalized (Ring Theory Perhaps)

The following question is posed in the book "The USSR Olympiad Problem Book: Selected Problems and Theorems of Elementary Mathematics"

"Prove that if integers a_1, ..., a_n are all distinct, then the polynomial

(x-a_1)^2(x-a_2)^2...(x-a_n)^2 + 1

cannot be written as the product of two other polynomials with integral coefficients"

I still haven't solved this in the elementary case, but I want to pose it in a more general setting. (I'll post the elementary answer in a bit, I want to try a bit more to figure it out)

EDIT: Now I have solved it, it's MUCH more trivial than I thought, major brain fart on my part

Suppose we have a ring R[x], and the polynomial written above is factorable in this ring. Suppose further that the coefficients are members of a subset of the field that the ring sits in, and that none of the a_i are equal (which means that some finite fields are of course out). Under what conditions IS the polynomial factorable into a product of two polynomials such that their coefficients sit inside the subset? Does the subset have to be algebraically closed?

(Is factorability even remotely easy in a noncommutative ring? I don't see a priori why a factorization would be unique either in a noncommutative ring).

Motivation: I'm doing research in mathematics education, and am interested in the metacognitive faculties of early college and high school pupils, which, for those not versed in metacognition, is the ability to separate oneself from "nitty-gritty" of the problem and think in more general terms. Alan Schoenfeld is the standard reference on this. I'm looking for problems that are good to pose to students to try and understand their thinking skills, and so am looking through particularly hard problems that do not require a strong background in mathematics. In this particular case I'd like to understand the problem in greater depth myself, and hopefully use my more general knowledge of the situation to aid in my study of how students think about such problems.

Hope this is interesting to someone, and that it isn't too specific.

linear algebra - (Path) connected set of matrices?

Edit: This is just half an answer: I can only show that the sets matrices with $X^2=N$ and fixed Jordan type are path connected.

Every nilpotent matrix is conjugate to a nilpotent matrix in Jordan form, which is unique up to permutation of Jordam blocks. So we have a bijection

$$mathrm{Nilp}_n(mathbb C)/mathrm{conjugation} quad cong quad mbox{integer partitions of }n$$

associating with a conjugacy class of a nilpotent matrix $X$ the sizes of its Jordan blocks $(a_1, ldots, a_r)$ which sum up to $n$. The max of the $a_i$'s is the nilpotency-degree of $X$. To the class of $X^2$ is associated the partition
$$(lfloor (a_1 +1)/2 rfloor, lfloor a_1/2rfloor ,lfloor (a_2 +1)/2 rfloor ,lfloor a_2/2rfloor, ldots, lfloor a_r/2 rfloor)$$

From here we can derive a necessary and sufficient condition on a nilpotent matrix to be a square.

Now fix your preferred nilpotent matrix $N$. Let $X$ be a matrix with $X^2=N$ and Jordan type $(a_1, ldots, a_r)$. Conjugating the whole setup, we may assume $X$ is in Jordan form.

Let $Y$ be a matrix with $Y^2=X^2 = N$ having the same Jordan type as $X$, and let us construct a path from $X$ to $Y$. Since $X$ and $Y$ have the same Jordan type, there exists an invertible matrix $S$ with $Y=SXS^{-1}$. Because $Y^2=X^2=N$ the matrix $S$ commutes with $N$. It is enough to construct a path from the identity matrix to $S$ in the set $mathcal C_N$ of invertible matrices that commute with $N$.

I claim $mathcal C_N$ is path connected (for just any $N$). Indeed, the set $[N]$ of all commutators of $N$ is linear subspace of the vector space $mathrm M_n(mathbb C)$. The determinant, as a function on $[N]$ is a polynomial function which is not identically zero since the identity matrix belongs to $[N]$. Thus, the zero set of the determinant is Zariski closed, so $mathcal C_N$ is Zariski open in $[N]$. Any Zariski-open in a complex vector space is path connected.

What remains is to connect different Jordan types. We certainly can connect $(5,2)$ with $(5,1,1)$ by the 1 in the $2times 2$--block with $t$ and vary $t$ from $1$ to zero. The problem that remains is to connect, say, type $(4,2)$ with type $(3,3)$ as pointed out in the comments.

set theory - Connection between the axiom of universes and Tarski's axiom

Over ZFC, the theories are equivalent. Both Grothendieck's Universe Axiom and Tarski's Axiom are equivalent to the assertion that there are unboundedly many inaccessible cardinals.

Grothendieck universes are exactly the sets $H_kappa$ for an inaccessible cardinal $kappa$, consisting of all sets hereditarily of size less than $kappa$. They can also be described as $V_kappa$ for inaccessible $kappa$, using the cumulative Levy hierarchy. The Tarski universes, in contrast, needn't be transitive, and so the notions are not equivalent.

The issue is that over ZF, they lose their equivalence. Solovay explained on the FOM list that this is due to the way that cardinalities are treated in TG, making ZF+TG imply AC, whereas ZF+GU does not imply AC. You can read Solovay's interesting post here.

lo.logic - Does the exact pair phenomenon for partial orders occur in your area of mathematics?

In trying to understand better a certain weak choice axiom, I once concocted an example in sheaf theory which involves an exact pair in the sense of this question, without knowing it was a "thing" elsewhere.

In constructive mathematics, one choice principle which has been found useful is called the Presentation Axiom, Pax for short, which says roughly that any object $A$ (of one's given category) can be covered by an object $P$ which is projective: any further covering $Q twoheadrightarrow P$ must split. You can imagine this applying for example to certain categories of sheaves of modules where one might want to consider projective resolutions. Or you can imagine it applying to certain categories of Heyting-valued sets or other toposes where the full axiom of choice is out of the question.

In constructive mathematics, it is also known as "CoSHEP" (Category of Sets Has Enough Projectives); a more philosophical justification (quoting the nLab) is

every set $A$ should have a ‘completely presented’ set of ‘canonical’ elements, that is elements given directly as they are constructed without regard for the equality relation imposed upon them. For canonical elements, equality is identity, so the BHK interpretation of logic justifies the axiom of choice for a completely presented set. This set is $P$, and $A$ is obtained from it as a quotient by the relation of equality on $A$. This argument can be made precise in many forms of type theory (including those of Martin-Löf and Thierry Coquand), which thus justify CoSHEP, much as they are widely known to justify dependent choice.

Now, Pax as stated above (for let's say a topos $E$) was actually the external version: an object $P$ is externally projective if the external hom-functor $hom(P, -): E to Set$ takes epimorphisms to epimorphisms. There's also an internal version of Pax which is often more appropriate: if $(-)^P: E to E$ denotes the internal hom-functor, then we say $P$ is internally projective if $(-)^P$ preserves epimorphisms, and internal Pax would say that every object $A$ is covered by an internally projective $P$. What I had wanted was an example of a topos which validates external Pax but not internal Pax.

Presheaf toposes $E = Set^{C^{op}}$ (the category of functors $C^{op} to Set$, where $C$ is a small category and $Set$ is just a vanilla category of sets in a ZFC background) always validate external Pax. It's easy to show that representable functors $hom_C(-, a)$ are externally projective, as are coproducts of representable functors, and any object $F: C^{op} to Set$ is covered by such a coproduct via a canonical map $sum_{A in Ob(C)} sum_{x in F A} hom_C(-, A) twoheadrightarrow F$. So I wanted a $C$ which would force internal Pax to fail for $Set^{C^{op}}$.

The example I came up with is where $C$ is a generic exact pair, i.e., the poset $mathbb{N} cup {a, b}$ where the natural numbers $mathbb{N}$ is ordered as usual and $a, b$ are incomparable elements that dominate every element of $mathbb{N}$. We consider $C$ as a 'thin' category as is customary for posets. I claim that $A = hom_C(-, a)$, or indeed any object $P$ covering $A$, cannot be internally projective. Specifically, that such $(-)^P: E to E$ cannot preserve an evident epimorphism $F twoheadrightarrow G$ where $F(n) = {m in mathbb{N}: m geq n}$ and $F(a), F(b)$ are empty, where the restriction maps are given by the inclusions $F(n+1) hookrightarrow F(n)$, and where $G$ is the presheaf with $G(n) = ast$ (a one-point set) and $G(a), G(b)$ are empty.

To show the induced natural transformation $F^P to G^P$ isn't an epimorphism, it suffices to show that the component $F^P(b) to G^P(b)$ isn't a surjection between sets; in fact one may calculate that $F^P(b)$ is empty and $G^P(b)$ has exactly one element. Such calculations aren't exactly a spectator sport, so I'll wrap it up by making a few observations:

• For general $F$ one may identify $F^P(b)$ with $Nat(hom_C(-, b) times P, F)$, where the right side indicates a set of natural transformations.




• In the case where $P = hom_C(-, a)$, we have in fact $hom_C(-, b) times hom_C(-, a) cong G$, and a natural transformation $G to F$ boils down to having an element that belongs to the intersection of all the $F(n)$. Of course no such element exists since the intersection is empty.




• So $F^{hom_C(-, a)}(b)$ is empty, and $G^{hom_C(-, a)}(b)$ has an element (in fact just one).




• For general $P$ covering $hom_C(-, a)$, there is a section $s: hom_C(-, a) to P$ (Yoneda lemma), and this section induces a map $F^P to F^{hom_C(-, a)}$, and thence a function $F^P(b) to F^{hom_C(-, a)}(b)$. The domain $F^P(b)$ has to be empty since the codomain is. On the other hand, $G^P(b)$ is nonempty because there is a map $hom_C(-, b) times P to hom_C(-, b) times hom_C(-, a) cong G$.

tensor products - Special subalgebras of central simple algebras

In this question F is a field and all algebras are finite dimensional F algebras.

Let X be the set of all F algebras A for which there exist an F algebra B and an F division algebra D such that F is the center of D and the tensor product of A and B over F is isomorphic to M_n(D) for some n. Can we find all the elements of X?

(M_n(D) is the algebra of all n-by-n matrices with entries from D.)

It is Obvious that every central simple F algebra is in X. Are there some interesting elements of X?

cheers

sociology of math - How connected are you?

I apologize if this question seems frivolous, but the motivation for it is quite serious.
When I encounter the endless topic of the 'relevance' of mathematics, I am rather
fond of referring to a network of knowledge. Before explaining this term, I should say
that I'm not one who considers it very difficult to make a plausible direct case for the importance of mathematics to the general public, provided the context and tone of such a discussion is chosen with some care. Nevertheless, a little bit of thought makes it clear that the realistic value of any research assessed over time depends very heavily on the integrity of a network of related activity within which it sits. That is to say, a pure mathematician might occasionally speak to an applied mathematician who will speak to a physicist who will speak to an engineer who will speak to a biologist who will speak to a computer scientist who will speak to a mathematician, and so on. If we find sufficient coherence in the overall network of interaction, the individual components will frequently take on deeper significance than may have been noticed in isolated observation. Clearly, this isn't to imply that any obscure activity is as important as any other, but it seems to me an awareness of the network is critical in any discussion of relevance.

After making this point recently to some graduate students, it occurred to me to investigate more
than casually a rather amusing measure of connectedness, the collaboration distance calculator. So here is a small but somewhat surprising list of (finite!) distances from myself I've found using mathscinet:

Albert Einstein 5

James Clerk Maxwell 6

Richard Feynman 5

Stephen Hawking 4

J.B.S. Haldane 7

Noam Chomsky 4

Willard Van Orman Quine 6

Nelson Goodman 5

Amartya Sen 7

Ilya Prigogine 4

Jean Piaget 6

Needless to say, this array of celebrity intellectuals says nothing about me in particular. Among mathematicians of
comparable seniority, I have relatively few collaborators, and examining the specific
bridges that make up the paths will quickly reveal that they have nothing much to do with
the signficance of my own research. It's obvious then that the diversity of this list reflects nothing less than the centrality of mathematics, and perhaps the coherence of human scholarly endeavour as a whole.

Now to the question: Could you investigate a bit yourself now, and let me know of interesting research connections you find to people working outside mathematics, using mathscinet or otherwise? For the purposes of this question, what I would like to know about are concrete sequences of research links as might appear in the collaboration distance calculator, rather than an anecdotal discussion of some application of mathematics. Perhaps you could also add a word about your own area of research, so I can get some sense of surprise (or lack of it). By the way, I work on arithmetical algebraic geometry.

As mentioned at the beginning, I do think my motivation is serious: In grandiose terms, a
general awareness of the connections is pretty important not just for reassuring students, but for cultivating ourselves a reasonably sophisticated sense of where we stand in the scheme of things.

---

Added:

As mentioned by Sonia Balagopalan in the comments below, it would be also interesting to hear of unlikely connections Mathematician A--Non-mathematician B. However, I did think people would be pleasantly surprised to experiment a bit and find out specifically about their personal connections. That is, I thought it would be fun to concretely illustrate my own generic answer to

'How connected are you?':

More than you expect!

I agree with most of the other comments, especially what Professor Milne writes about inaccuracy. But even with the few weak links that show up, the paths still seem to be interesting, and mostly illustrate my point. (Of course I can think of examples where they would be nearly uninteresting, such as a link created by two articles appearing in an encyclopedia.)

Felipe: Actually, I have the impression that my more unusual connections go through you!