ac.commutative algebra - Is there a non-Gorenstein ring but locally Gorenstein?

In all of the standard references I know, Gorenstein is either only defined for local rings or a not necessarily local ring is defined to be Gorenstein if all of its localizations at maximal ideals are Gorenstein (which implies that its localizations at all prime ideals are Gorenstein). So I am guessing you really want to ask: is there a commutative Noetherian ring $R$ all of whose localizations have finite injective dimension but such that $R$ itself does not?

I believe that the answer is "yes" and that a counterexample is given by Nagata's (in)famous example of a Noetherian ring of infinite Krull dimension. See (5.96) in Lam's Lectures on modules and rings for an explanation why Nagata's example is regular, hence locally Gorenstein, i.e., locally of finite injective dimension.

Furthermore, the proof of the Theorem at the bottom of p. 7 of

http://www.math.hawaii.edu/~lee/homolog/Goren.pdf

states that the injective dimension of a ring is the supremum of the injective dimensions of its local rings $R_{mathfrak{m}}$. In this case each $R_{mathfrak{m}}$ is regular, hence Gorenstein, hence its injective dimension is simply equal to its Krull dimension, i.e., to the height of $mathfrak{m}$. It follows that the injective dimension of $R$ itself is infinite.

Conversely, Lee's handout contains a proof that the answer is "no" if $R$ is Noetherian of finite Krull dimension.

soft question - Learning Roadmap for Software Engineer

Hi,

I did some browsing around here on MathOverflow, but I feel that my question is different enough from others that it warrants its own question.

Background:

I'm a computer science major, with a math minor, and have been working professionally for a couple years now. When I went through my undergraduate degree, I was basically trying to get out as fast as possible, and suffice it to say that I didn't take the time to fully digest or appreciate what I was learning, especially on the math front. I got the minor mostly because I came in with a lot of math credits, and because it would set me apart from those who just had a CS degree, not because of any particular passion for math.

While in school, I took courses through calc III, diff eq, lin alg, and then moved through analysis, topology, and algebra (one semester each) in a blur. I also got additional exposure to number theory through cryptology-focused courses, as well as some probability and statistics from Natural Language Processing and AI courses.

The Problem:

After working for a couple years, I decided to pursue a part-time master's in computer science, because this time around I genuinely was interested in pursuing material at a deeper level. Additionally, I've come to appreciate the value of math (both pure and applied), since much of computer science is based around mathematical concepts and underlying structures, and much of the "real world" is based on applied math. I know I wasted some of my time as an undergrad (I like to think of it as I wasn't mature enough to appreciate the math I was learning), but now I'd really like to learn more about math in all its abstraction, beauty, frustration, and je ne sais quoi.

The Question:

Can you recommend a roadmap for re-engaging someone with a passion for software engineering, a desire to learn about the hidden beauties in the mathematical world, and the ability to draw connections between the two? I know I can't dedicate myself full-time to this venture, but I want to learn (or re-learn) things that will make me a better mathematician, and engineer.

Thanks!

singularity theory - What strict resolutions of singularities are needed?

For S = {rational singularities} (assuming I've understood the statement right) this is an open and presumably hard question. It's something that Koll'ar wanted (see Chapter 12, I think, of Kollar's Shaferavich Maps and Automorphic forms). Essentially this would let you know that if $X$ has rational singularities, then it also has a compactification $bar{X}$ with rational singularities (which is what Kollar wanted it for I think).

On the other hand, if you are doing more MMP type singularities (for example, S = log terminal singularities), then this kind of thing is often possible due to the MMP. See for example BCHM, Corollary 1.4.3 and also KK, Theorem 3.1. (I've been told more general statements exist also, but I don't know a reference).

EDIT1: For S = Cohen-Macaulay, this is also an open question (Macaulayfication is not strict).

EDIT2: Also see the literature on semi-resolutions. While this isn't quite the same thing, it's very closely related (the idea there is to leave the codim 1 singularities alone if they are nice enough, ie SNC or pinch points). This is a fundamental construction in the study of moduli spaces of higher dimensional varities.

at.algebraic topology - Spectra and localizations of the category of topological spaces

[Removed a paragraph relating to an earlier version of the question]

You can construct Spectra categorically by adjoining an inverse to the endofunctor Σ of Top as a presentable (∞,1)-category. Inverting an endofunctor is a very different operation than inverting maps! It's like the difference between forming ℤ[1/p] and ℤ/(p).

Here is one way to verify the claim. To invert the endomorphism Σ of Top we should form the colimit, in the (∞,1)-category Pres of presentable categories and colimit-preserving functors, of the sequence Top → Top → ... where all the functors in the diagram are Σ. A basic fact about Pres is that we can compute such a colimit by forming the diagram (on the opposite index category) formed by the right adjoints of these functors, and taking its limit as a diagram of underlying (∞,1)-categories [HTT 5.5.3.18]. The functors in the limit cone will have left adjoints which are the functors to the colimit in Pres. In our case we obtain the sequence Top ← Top ← ... where the functors are Ω, and the limit of this sequence is precisely the classical definition of (Ω-)spectrum: a sequence of spaces X_n with equivalences X_n → ΩX_n+1.

computational complexity - Is there a promise version of 3-coloring equivalent to Graph Isomorphism?

Hi Steven,

(1) To start with the "duh" observation, you could define an artificial class, namely "those 3-coloring instances that are obtained by starting from Graph Isomorphism and then applying a standard NP-completeness reduction." That would indeed give you a subclass of 3-coloring instances that are provably polynomially equivalent to graph isomorphism, but I'm guessing it's not what you had in mind. :-)

(2) I can't think of a "natural" subclass of 3-coloring instances that are reducible to graph isomorphism and not as a consequence of being in P. I'd be interested and surprised if there was one -- 3-coloring and GI are both about graphs, but in complexity terms they're extremely different! GI has lots of special group-theoretic structure, which in some sense can't be shared by 3-coloring (or any other NP-complete problem) unless the polynomial hierarchy collapses.

(3) On the other hand, if you're willing to talk about nonuniform algorithms (i.e., algorithms that can take "advice" depending on the input length n), then here's another "duh" observation. Suppose you have a fixed graph G on n vertices, together with a 3-coloring of G. Then you can 3-color any graph G' that happens to be isomorphic to G, provided you can compute an isomorphism between G and G'. Or if you prefer decision problems: given as advice a graph G that's 3-colorable and another graph H that isn't, and also given a decision oracle for GI, you can decide 3-colorability for those graphs that happen to be isomorphic to either G or H.

Sorry the above wasn't more helpful... :-)

ag.algebraic geometry - Localisation of sites and associated topoi (some questions)

In SGAIV.1, Exp. III, sec. 5, sites $(C,J)$ are localised with respect to a presheaf $X$ on $C$ (not nec. just a representable one).

Question1: There it is asserted that if $F to X$ is a morphism of presheaves on C, i.e., an object of $PreSh(C)/X$, then its image under the equivalence $PreSh(C)/X to PreSh(hC/X)$ is a sheaf iff F is isomorphic to the cartesian product ${X}x_{ia(X)}ia(F)$ considered over $X$ in $PreSh(C)/X$. Has this been worked out somewhere? While working out the details one encounters some points more difficult than the material in Exp.'s II and III preceding it.

Question2: If G is a sheaf on $(hC/X)$ and $i_X$ denotes the inclusion of $Sh(hC/X)$ into $PreSh(hC/X)$, is ${j_X}_!(i_X(G))$ always a sheaf on $C$? (The topology $J$ is not necessarily subcanonical.)

Some of the details of this section can be extended to a general pair of adjoint functors and their categorical localisation to slice categories, which leads one to pose the following:

Question3: Has the discussion there been extended to arbitrary pairs of adjoint functors (L,R), $L:C to C'$, $R:C' to C$, (L',R'), $L':D to D'$, $R':D' to D$ and an equivalence of categories (A,B), $A:C to D, B:D to C$ (perhaps with R, R' fully faithful)?

fa.functional analysis - For which Lie groups is the convolution of any two nonzero integrable compactly supported functions nonzero?

The Titchmarsh convolution theorem implies that the convolution of two nonzero functions $f,gin L^1(mathbb R)$ with compact support is nonzero. There is a generalization of this theorem to the case of $f,gin L^1(mathbb R^n)$ with compact support.

Now, consider a Lie group $G$ endowed with its Haar measure, and $pgeq1$. We may formulate a property

(T1) If $f,gin L^p(G)$ are nonzero and compactly supported, the convolution $f*g$ is nonzero.

This is true for $G=mathbb R^n$, but fails for abelian groups of the form $G=mathbb R^n times mathbb T^m$, where $mathbb T^m$ is the $m$-dimensional torus, and $m>0$. [Counterexample: let $f$ be the characteristic function of $Ktimes mathbb T^m$, where $Ksubseteq mathbb R^n$ is compact, and $g(x,y)=f(x,y)h(y)$, where $hin L^p(mathbb T^m)$ and $int_{mathbb T^m}h =0$].

However, we may strengthen the assumptions in (T1), obtaining

(T2) There exists an open set $Usubseteq G$ with compact closure, such that if $f,gin L^p(G)$ are nonzero and supported in left translates of $U$, the convolution $f*g$ is nonzero.

This is actually true for any abelian Lie group $G$, since in this case $G$ is locally isomorphic to $mathbb R^k$.

Of course, if (T1) or (T2) hold for some $p$, then they also hold for all $q>p$. The most interesting cases are $p=1$ and $p=2$.

So, my question is:

Which Lie groups are known to satisfy property (T1) or (T2) with $p=1$ or $p=2$?

ac.commutative algebra - Are submodules of free modules free?

To answer ding8191's question: let $A$ be any valuation ring of height $1$ which is not a discrete valuation ring. That is to say, let $F$ be a field and suppose that there is a function $v : F to mathbb{R} cup { infty }$ such that $v(xy) = v(x) + v(y), v(1) = 0, v(0) = infty$ and $v(x + y) geq min $ ${ v(x), v(y) }$ (thus $v$ is a valuation), such that $v(F)$ is not a discrete subgroup of $mathbb{R}$, and let $A = { x in F :v(x)geq 0}$ (the valuation ring of $F$).

Then the maximal ideal $mathfrak{m}$ in this ring is a direct limit of principal ideals and is therefore flat. If $mathfrak{m}$ was projective, then it would have to be free by Kaplansky's Theorem (all projectives over any local ring are free), but since $mathfrak{m}$ is contained in $F$ it has to have rank at most $1$, hence $mathfrak{m}$ is principal. But this would force the valuation on $A$ to be discrete.

You can find this material in Chapter VI of Bourbaki's "Commutative Algebra" (Hermann, 1972). In particular, Lemma $1$ of $S 3.6$ says that if $A$ is a valuation ring (in a more general sense than I defined above), then every torsion-free $A$-module is flat, and Proposition $9$ of the same subsection implies that $A$ is a discrete valuation ring whenever $A$ is a valuation ring of height $1$ with principal maximal ideal.

To give a concrete example, consider the field $F$ of Puiseux series and let $A$ be its valuation ring.

riemann surfaces - Dolbeault cohomology

Chern's book "Complex manifolds without potential theory" is a good book, and it does explain Dolbeault Cohomology. But it's a short book, and it explains it concisely. If you need more details, you could also try Griffiths-Harris (but I greatly prefer Chern's book).

Kodaira's book "Complex manifolds and deformations of complex structures" is much more leisurely, and with great attention paid to exposition and detail (it doesn't appeal to some people, but I enjoyed it a lot).

co.combinatorics - Solving a Diophantine equation related to Algebraic Geometry, Steiner systems and $q$-binomials?

Let $t = 1+q+q^2+dots+q^n $ then each of the equations (1) and (2) implies that $24t+1$ is a square (namely, $24t+1=(12k+1)^2$ and $24t+1=(12k+5)^2$, respectively). For $n=2$ that leads to a Pellian equation (with possibly infinitely many solutions), for $n=3,4$ to an elliptic curve (with finitely many solutions, if any), and for $n>4$ to a hyper-elliptic curve (with no solutions for most $n$).

Cases $n=3,4$ are easy to solve.

For $n=3$, integer solutions are $q=-1, 0, 2, 3, 13, 25, 32, 104, 177$ out of which only $2,3,13,25,32$ are powers of primes.
For $n=4$, integral solutions are $q=-1,0,1,25,132$ out of which only $25$ is a power of prime.
These numerical values are computed in SAGE and MAGMA.

Also, for a fixed value of $k$, it is possible to verify solubility of the given equations by iterating all possible $q$ dividing the l.h.s. minus 1. In particular, equation (1) has solutions only for the following $k$ below $10^6$:
1, 2, 3, 15, 52, 75, 1302, 32552, 813802.
Similarly, equation (2) has solutions only for the following $k$ below $10^6$:
1, 10, 260, 6510, 162760.

gn.general topology - How thinly connected can a closed subset of Hilbert space be?

Some remarks, not an answer.

1) As pointed out by BS, the problem is: does there exist a connected Polish space containing no nondegenerate proper closed connected subspace? In the language of continuum theory this sounds suspiciously simple: does there exist a connected Polish space whose all composants are singletons? Or (still equivalently): does there exist a connected Polish space that is irreducible between every pair of distinct points? Surely continuum theorists have thought about this question, haven't they?

2) As hinted by Garabed, such a space $X$ cannot be locally compact. Indeed, in locally compact spaces, components coincide with quasi-components. Let $F$ be a closed ball of some radius about some $xin X$, so that $Fne X$, and let $C$ be the component of $x$ in $F$.
If $Cne{x}$, then $C$ is a closed connected nondegenerate subset of $X$. If $C={x}$, then ${x}$ is also a quasi-component, so $x$ is contained in arbitrarily small clopen sets in $F$. Then they are also clopen in $X$, so $X$ cannot be connected.

3) There exists a connected Polish space $X$ containing no nondegenerate compact connected subspace. For instance, the graph of $f(x)=sum_{n=1}^infty 2^{-n}sin(frac1x-r_n)$, where $r_n$ is the $n$th rational number (in some order) and $sin(infty)=0$. See Kuratowski's "Topology" (volume II, section 47.IX in the 1968 edition).
In fact, such an $X$ can even be locally connected (of course, it cannot be locally path-connected at any point).

4) The graph of a discontinuous function $f:Bbb RtoBbb R$ satisfying $f(x+y)=f(x)+f(y)$
can be connected, and in that case it contains no nondegenerate bounded connected subset.

mg.metric geometry - How to find the Fermat Point using the construction of the tangent to ellipse?

I have the vague idea that Hadamard is referring to the construction where you erect equilateral triangles BCA', CAB' and ABC' on the sides of the triangle, as described here. The Fermat point is the intersection of the cevians AA', BB' and CC'. It can also be constructed using the various angles of 60 resp. 120 degrees.

In the construction of the tangent from a point P to an ellipse with foci F and F' (in the book you cite), they consider an additional point f. The correspondence should be

F ↔ C,

F' ↔ A,

P ↔ B,

f ↔ P'.

The general philosophy behind both, I think, is to convert a sum of segments into a single segment.

equivariant - G-Modules on X=G/H modules on X/H ?

I think it is true that $G$-equivariant sheaves on $X$ are equal to $G/H$ equivariant sheaves on $X/H$. More precisely I'm interested in the following statement:

Given an algebraic group $G$ with normal subgroup $H$ and an action of $G$ on $X$, such that the quotient $X/H$ exists as a variety and $Xrightarrow X/H$ is an $H$ principal bundle.

Then the category of $G$ equivariant
$mathcal{O}_X$ Modules is equivalent to the category of $G/H$ equivariant modules on the quotient.

The equivalence should be given by pullback along $Xrightarrow X/G$ in one direction and taking invariant sections in the other.

Is there a reference for this?

ag.algebraic geometry - Negative Gromov-Witten invariants

Gromov--Witten invariants are designed to count the "number" of curves in a space in a deformation invariant way. Since the number of curves can change under deformations, the Gromov--Witten invariants won't have a direct interpretation in terms of actual numbers of curves, even taking automorphisms into account.

Here is an example of how a negative number might come up, though strictly speaking it isn't a Gromov--Witten invariant. Let M be the moduli space of maps from P^1 to a the total space of O(-4) on P^1. Call this space X. Note that I said maps from P^1, not a genus zero curve, so the source curve is rigid. That's why this isn't Gromov--Witten theory. Any such map factors through the zero section (since O(-4) has no nonzero sections), so this space is the same as the space of maps from P^1 to itself. I just want to look at degree one maps, so the moduli space is 3 dimensional.

We could also compute the dimension using deformation theory: the deformations of a map f are classified by $H^0(f^ast T)$ where T is the tangent bundle of the target. The target in this case is O(-4), not just P^1, and the tangent bundle restricts to O(2) + O(-4) on the zero section. Thus $H^0(f^ast T)$ is indeed 3-dimensional, as we expected. However, the Euler characteristic of $f^ast T$ is not 3 but 0, which means that the "expected dimension" is zero.

The meaning of expected dimension is rather vague. Roughly speaking, it is the dimension of the moduli space for a "generic" choice of deformation. The trouble is that such a deformation might not actually exist. Nevertheless, we can still pretend that a generic deformation does exist and, if the expected dimension is zero, compute the number of curves that it "should" have.

What makes this possible is the obstruction bundle E on M. Any deformation of X gives rise to a section of E and the vanishing locus of this section is the collection of curves that can be deformed to first order along with X. Even though a generic deformation might not exist, the obstruction bundle does still exist, and we can make sense of the vanishing locus of a generic section by taking the top Chern class.

In our situation, the (fiber of the) obstruction bundle is $H^1(f^ast T)$. Since O(2) does not contribute to H^1, the obstruction bundle is $R^1 p_ast f^ast O_{P^1}(-4) = R^1 p_ast O_{P^3 times P^1}(-4, -4)$ where $p : P^3 times P^1 rightarrow P^3$ is the projection. By the projection formula, this is $O(-4)^{oplus 3}$ and the top Chern class is -64. This is the "Gromov--Witten invariant" of maps from P^1 to $O_{P^1}(-4)$.

Unfortunately, I don't have anything to say about what this -64 means...

differential topology - Morse Theory on Non-smooth Manifolds

Let $X$ be a circle that with one corner (i.e. think of a triangle where we smooth out two of the vertices). Now let us consider the topological torus $M cong mathbb{T}^n$ which is the product of $n$ copies of $X$. Note that $M$ contains $n$ distinct circles along which it is not smooth.

Finally, suppose we are given a function $f:M rightarrow mathbb{R}$ which is continuous on all of $M$ and is smooth wherever $M$ is smooth. Is there any way to conclude that there exists a critical point on the smooth part of $M$? What if we replace some of the non-smooth circles with smooth ones? That is take $M$ to be the product of $k$ smooth circles, and $n -k$ non-smooth ones?

I'm not concerned whether the critical point is non-degenerate.

An easy example is the case $n = 1$: The $min$ or $max$ of $f$ will correspond to a critical point on the smooth part of $X$. This example tells us that generically we should expect $f$ to have a critical point on the smooth part of $M$ since, generically, the $min$ or $max$ should not lie on the measure zero subset of $M$ that is not smooth.

I heard of something called Stratified Morse Theory, but I'm not sure if this applies, or whether there is a more elementary way to think about the problem not using Stratified Morse Theory.

ct.category theory - What are the advantages of phrasing results in terms of exact sequences and commutative diagrams?

If you are asking why very elementary results like the first isomorphism theorem are phrased
in the language of exact sequences/commutative diagrams (rather than why this language is used at all), then there are (at least) two answers: (1) for those who are used to using this language, they frequently think about even those elementary results in terms of it, and so it is natural to write them in that language; (2) we want to train students to learn this language, and have to start somewhere, so we begin by taking elementary results that can be understood in another way, such as the first isomorphism theorem, and then rewrite them in this language for pedagogical purposes.

If you are asking why people use this language at all (which is to say, why are there many people to whom (1) above applies, and why do we want to engage in the educational practice labelled (2) above), then Keith Conrad gives a pretty good answer.

At a slightly broader level of generality, one might cite the old saying "a picture is worth a thousand words", and note that a well-chosen diagram or exact sequence can convey a lot of mathematical information in a succinct and intuitive way (the intuition coming once you have some familiarity with this way of thinking). We have a lot of mathematics to remember, and are always looking for ways to compress our descriptions of things without losing information or becoming unclear. Well-chosen definitions and terminology are one way this is achieved; well-drawn diagrams and exact sequences are another.

Finally, one could note that contemplating diagrams appeals (however slightly) to geometric modes of reasoning. Typically, any method which allows one to import some kind of geometric reasoning into algebra is welcome, since it brings less typically algebraic ways of thinking to bear on algebraic problems.

ca.analysis and odes - Splines, harmonic analysis, singular integrals.

If you want to extend differentiation to all continuous functions, then (provided you have some convenient mathematical properties of the extension) you are FORCED to use distributions or roughly equivalent things; you have no choice! Similarly, to extend the Fourier transform you are forced to consider tempered distributions.

Speaking as a pure mathematician: the main purpose of general distributions is to extend differentiation, not integration (since integration makes things nicer; it is differentiation which is the nastier operation). They are fine as long as you aren't using the Fourier transform.

Thus, every locally integrable function can be regarded as a distribution, and therefore differentiated; so, when you're considering differential equations, this might be all you need (you don't have to worry whether the functions are differentiable or not, because distributions always are). You find distribution solutions, then try to prove that they're actually functions.

It's similar to solving polynomial equations by using complex numbers; even if all the roots are real, it's still sometimes easier to solve them with complex numbers, then try to prove they're real (e.g. by showing they're self-conjugate).

However, if you want to do Fourier Transforms then you have to consider tempered distributions (or Schwartz distributions), since general distributions are sometimes too nasty to have Fourier transforms.

Note that even genuine locally integrable functions need not represent tempered distributions, so general distributions are not appropriate for Fourier transforms even when you only want to consider functions.

But Fourier inversion works perfectly for tempered distributions, no further restrictions are needed, unlike, say, $L^1$. If $f in L^1$ then $widehat{f}$ is usually not in $L^1$, so you can't do Fourier inversion theory nicely on $L^1$ (you would have to assume that also $widehat{f} in L^1$, which is often not true!)

Extension in mathematics is very powerful; when you don't have to worry about restrictions and annoying details, it is easier! For example, complex numbers are easier than real numbers, complex analysis is easier than real analysis, and Lebesgue integration is easier than Riemann integration!! Students never believe this, but it's true if you actually want to use it (rather than do toy problems in books)...

Examples of algebraic stacks without coarse moduli space?

[A^1/G_m] is one example. You can check that any G_m invariant map from A^1 to a scheme is constant. Thus the map from [A^1/G_m] to the point is universal for maps to schemes, but is not a bijection on geometric points (since [A^1/G_m] has two geometric points).

Check out Jarod Alper's thesis to learn more.

set theory - Club sets and substructures

The idea in my previous answer can, I think, be upgraded to solve the whole problem, as follows. Again, fix Skolem functions for $mathcal A$ as given by the well-ordering $<$, and again let $D$ be a set of fewer than $kappa$ ordinals $delta$, each of which is $sup(kappacap M_delta)$ for some $M_deltaprecmathcal A$ with $|M_delta|<kappa$. I need to show that $sup(D)$ is also in $C_{mathcal A}$. For each $deltain D$, define $N_delta$ to be the Skolem hull of $kappacapbigcup_{xiin D, xileqdelta}M_xi$. With appropriate gratitude for the hypothesis that $kappa$ is regular, note that $N_delta$ is an elementary submodel of $mathcal A$ of size $<kappa$ and that the sequence $langle N_deltarangle$ is an elementary chain. Let $N$ be its union, and note that it, too, is an elementary submodel of $mathcal A$ of size $<kappa$. So it suffices to show that $sup(D)=sup(kappacap N)$. The $leq$ direction here is obvious, as $N$ includes $kappacap M_delta$ for each $deltain D$. To complete the proof, suppose the $geq$ direction failed. Then we would have $sup(D)<sup(kappacap N)$, so there would be an ordinal $alphainkappacap N$ with $sup(D)leqalpha$. By construction, we would have some $deltain D$ with $alphain N_delta$, and so $alpha$ would be of the form $f(vecbeta)$ for some Skolem function $f$ and some ordinals $beta_i$ in $kappacap M_{xi_i}$ for certain $xi_ileqdelta$. For each $i$, we have $beta_i<xi_ileqdelta$, and, since there are only finitely many $i$ (as Skolem functions are finitary), we can find $gamma<delta$ with all $beta_i<gamma$. Increasing $gamma$ if necessary, we can arrange that $gammain M_delta$. In $mathcal A$, we can define the function $g$ sending each ordinal $nu<kappa$ to the supremum of all $f(veceta)<kappa$ for $eta$ bounded by $nu$; the values of this function are $<kappa$ by regularity. As an elementary submodel of $mathcal A$, $M_delta$ is closed under $g$ and, in particular, contains $g(gamma)$. But (again by elementarity) $g(gamma)$ majorizes $f(vecbeta)=alpha>sup(D)geqdelta$. That contradicts the fact that $delta$ is the supremum of $kappacap M_delta$, and this contradiction completes the proof.

matrices - Maximize the multiplicity of an eigenvalue

Let $W = M + D$, where $M$ is the original $n times n$ matrix and $D$ is the added diagonal matrix that we want to determine.

$W$ is symmetric, thus diagonalizable by an adjoint action of the orthogonal group.
Larger multiplicities in the eigenvalues of $W$ imply smaller dimensions of the adjoint orbits.

For example, if we have an eigenvalue of multiplicity $n-1$, then the adjoint orbit will be
$O(n)/(O(n-1) times O(1)) = S^{n-1}/Z_2 = RP^{n-1}$ of dimension $n-1$;
while, if all the eigenvalues are distinct, the adjoint orbit is the real flag manifold $Fl_{mathbb{R}}^n = O(n)/(Z_2)^n$ of dimension $ frac{n(n-1)}{2}$. (Of course, in the case of scalar multiple of the unit matrix, the adjoint orbit is just a single point).

Thus in order to achieve large multiplicities, we need to minimize the dimension of the adjoint orbit.
This problem can be reduced to a problem of matrix rank minimization as follows:

Let ${l_i }_{i=1,...,n(n-1)/2}$, be a set of generators of the Lie algebra of $O(n)$ normalized according to:
$textrm{tr}(l_i l_j) = delta_{ij}$. The dimension of the adjoint orbit equals the rank of the Gram matrix $C$ whose elements are given by, $C_{ij} = textrm{tr}([l_i, W][l_j, W])$. The problem is thus reduced to the minimization of the rank of the Gram matrix whose elements depend linearly on the added diagonal matrix elements.

One of the possible methods to solve this problem is through a convex programming
heuristic approach for the solution of matrix rank minimization,
based on replacing the rank by the nuclear norm (the sum of the singular values), as explained in the following lecture notes by: P.A. Parillo. The nuclear norm is a convex envelope of the rank which may explain why this method works well in practice in general.

ag.algebraic geometry - Finding components of a preimage

Let $f:Xto Y$ be a degree $d$ morphism of complex projective varieties, and let $Vsubset Y$ an irreducible subvariety, $W$ its preimage under $f$. I want to find all of the components of $W$.

Suppose that I've already found several components, $W_1,ldots,W_k$, and the components that are known are such that the sum of the degrees of the maps $f|_{W_i}$ adds up to $d$, and I know that there exists at least one component where the map restricts to one that isn't finite.

How can I determine if there are any other components of this nature, that don't contribute to the degree of $f$? And if they exist, is there a good way to identify what they are?

(Note: this is an attempt to redo a question I asked a few days ago and deleted, hopefully, this is better phrased. Roughly, I'm looking for ways to find all the components of the preimage of a variety through a morphism as described above)

ag.algebraic geometry - Stacks determined by their coarse moduli spaces

Yes, the class of all smooth, separated DM stacks over a field of characteristic $0$, with trivial inertia in codimension at most $1$, over a field of characteristic $0$, has the propery you want. The point is that every moduli space of such a stack has quotient singularities; and every variety with quotient singularities is the moduli space of a unique such stack. I believe that this was first proved in Angelo Vistoli: Intersection theory on algebraic stacks and on their moduli spaces. Invent. Math. 97 (1989), no. 3, 613-670, Proposition 2.8 (uniqueness is not stated there, but it follows from the proof).

multi-index Dirichlet series

See P. Deligne, Multizeta values, Notes d'exposes, IAS Princeton, for the deep mathematical aspects of this.

Also for a general relevance philosophy, see Kontsevich and Zagier, Periods, Mathematics Unlimited(2001). An electronic version is available here.

There are various references, including those of Zudilin, Cartier, Zagier, Terasoma, Oesterle(On polylogarithms), Manin(iterated integrals and ....). Please look into mathscinet.

There seem to be many papers by Dorian Goldfeld and collaborators, too.

What is Borel-de Siebenthal theory?

I'm not sure the term "theory" is appropriate here, but the joint paper by Borel and de Siebenthal has had considerable influence in Lie theory over the years: MR0032659 (11,326d)
Borel, A.; De Siebenthal, J.,
Les sous-groupes fermés de rang maximum des groupes de Lie clos.
Comment. Math. Helv. 23, (1949). 200--221. (There was a short Comptes Rendus announcement in 1948.) This is found near the start of the Springer four-volume collected papers of Borel, with a couple of minor corrections appended.

To quote from the review by P.A. Smith: "Let $G$ be a compact Lie group, $G'$ a closed connected subgroup having the same rank as $G$. Let $Z'$ be the center of $G'$. The main object of this paper is to show that $G'$ is a connected component of the normalizator of $Z'$ in $G$." The proof involves
"a necessary and sufficient condition that a subsystem of root vectors of $G$ be the root vectors of a closed subgroup of $G$" and the subgroups of this type are found explicitly for all simple $G$. [Here $G$ is always assumed to be connected.]

The result on subsystems of root systems carries over in a natural way to the study of semisimple complex Lie (or algebraic) groups and their Lie algebras,
for example the determination of subalgebras of maximal rank in the latter.

computational complexity - Does the independence of P = NP imply existence of arbitrarily good super-polynomial upper bound for SAT?

Let me first define what "super-polynomial" means.

Definition. We call a function f super-polynomial if for all k, there exists a constant n such that for all x ≥ n, f(x) > x^k.

Now please judge whether the following claim is true.

Claim. Suppose P = NP is independent of ZFC. Then for any super-polynomial function f, there exists an algorithm for SAT whose worst case running time is bounded by f.

OK. To help your judgement, I will give you my proof of the claim. Of course it may be wrong or missing something.

Proof. Proof by contradiction. Suppose that there is no such SAT algorithm. Then all algorithms for SAT are not bounded by f. Note that f is not bounded by any polynomial function of the form g(x) = x^k for some k. Therefore no SAT algorithm is bounded by a polynomial. This means SAT ∉ P and hence P ≠ NP. We have just shown that P ≠ NP is provable, contradicting with our assumption that P = NP is independent. QED

If my proof is correct, isn't this claim too obvious? OK, but why there are still people publishing weaker claims, for example the main result of this paper? Did I misinterpret their result? Is it actually stronger than my claim?

Also, this claim is an example for the strength of the independence of P = NP. If it's indeed independent, then all statements that imply P = NP are false and all statements that imply P ≠ NP are also false. An example of the former is "There exists a polynomial time SAT solver." Since this is false, then there does not exist any polynomial time SAT solver. And the latter implies whatever super-polynomial time bound you give, there is a SAT solver with this bound. So if independence is true, then the picture is that there is an infinite sequence of (SAT solver, time bound) pairs, with each bound faster than the previous, approaching the limit of P. And yet, it never crosses the boundary. So if independence is true, we can expect to improve the time bound for SAT endlessly in the region of super-polynomial and yet never reach P. Now I hope you have a clearer picture of the possibility that P = NP is independent.

Sorry for the digression, but first of all, is the claim true?

co.combinatorics - Limit Distribution of Topological Form of Polyhedra with Large Number of Edges

Consider the set of all topologically inequivalent polyhedral graphs with $k$ edges, the number of which is given by Sloan sequence A002840 (1,0,1,2,2,4,12,22,58,158,448)).

Now define a topological form parameter as $beta:= (text{number of vertices}=v)/(k+2)$ and consider the distribution of the polyhedral graphs with $k$ edges as a function of $beta$. Due to duality the distribution is symmetric about $beta=1/2$. Due to the fact that for a planar graph $k le 3v-6$, the distribution vanishes outside the interval $beta in [1/3, 2/3]$.

Now a natural question is whether this distribution tends to a limiting distribution when the number of edges tends to infinity. Is it known whether such a limiting distribution exists - or will it be singular, i.e. concentrated with smaller and smaller variance around $beta=1/2$, as numerical data seem to suggest? Is there any nontrivial limit distribution theorem by means of rescaling?

EDIT: Some numerical data can be found under http://www.numericana.com/data/polycount.htm. Using these data gives the following values for the standard deviation of the distribution $p(beta)$ as a function of the number of edges: $sigma(k=21) = 0.029895922, sigma(k=27) = 0,027943943, sigma(k=33) = 0,02625827$.

gn.general topology - Is the pure braid group on three strands generated as a normal subgroup of the braid group by the six-crossing braid?

Artin's presentation of braid group on three strands is:
$$ B_3 = langle l,r : lrl = rlr rangle $$
where you should think of "$l$" as the positive crossing between the left and middle strands and "$r$" as the positive crossing of the right and middle strands:

     |   |   |         |   |   |
       /    |         |     /
 l =        |     r = |     
      /     |         |    / 
     |   |   |         |   |   |

Then there is a surjection $B_3 to S_3$ given by $l mapsto (12)$ and $r mapsto (23)$. ($S_3$ is the symmetric group on three letters: it is generated by $(12)^2 = 1 = (23)^2$ and the braid relation above.) The pure braid group $PB_3$ is the kernel of this surjection.

The six-crossing braid $b = lr^{-1}lr^{-1}lr^{-1}$ is an element of the pure braid group. Let $N$ be the minimal normal subgroup of $B_3$ that contains $b$. Certainly $N subseteq PB_3$.

Question: Do we have $N = PB_3$?

Motivation

The motivation for my question comes from a neat trick that Conway showed us years ago. It leads to a more nuanced question than what I asked that I will pose as its own question if the answer above is "no". My memory is that at the time Conway did not know the answer to the more nuanced question, which suggests that the answer above cannot be "yes".

Take a long and reasonably thin rectangle of paper, and score it with two cuts, so that you have three strips of paper that are connected at both ends, so that you end up with a (framed, oriented, ...) "theta graph":

 |--------|
 |        |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |        |
 |--------|

Then with some finagling, you can in fact "tie" the six-crossing braid in those three strands, without ripping the paper further, by passing the bottom through itself a few times. (The trick is that it's easier to unbraid than to braid, so make your paper long enough that you can put $bb^{-1}$ into it, and then unbraid the $b^{-1}$.) Put another way: you can first put a hair-tie at the end of your ponytail, and then braid your hair.

The harder question is to characterize all braids that you can put on the above "theta graph". The following facts are essentially obvious:

• Any "braiding" of the theta graph is pure.


• The set of braidings of the theta graph is a subgroup of $B_3$.


• The set of braidings of the theta graph is closed under conjugating by arbitrary braids.

Therefore, the set $T$ of braidings of the theta graph is a normal subgroup of $B_3$, with $N subseteq T subseteq PB_3$. In particular, a positive answer to the question above characterizes $T$.

dg.differential geometry - Basic question about differential forms and physics

This question is long dead, but I think there's one aspect of it that wasn't really addressed, which is how there can be such a direct relationship between the antisymmetric object $mathrm{d}x wedge mathrm{d}y$ and the symmetric object $dxdy$.

The reason is that ever-important proviso when integrating a differential form: the result depends on an arbitrary choice of orientation. So for a manifold $M$ admitting an orientation $o$, the relationship is actually $int_{(M,o)} dx wedge dy = begin{cases} int_M dxdy = int_M dydx & text{if } [mathrm{d}x wedge mathrm{d}y] = o\
-int_M dxdy = -int_M dydx & text{if } [mathrm{d}x wedge mathrm{d}y] neq oend{cases}$

Here the integral on the left is integration of forms and the integral on the right is a measure-theoretic integral. Also, the brackets denote taking the orientation-equivalence classes, where $omega sim omega'$ iff $omega = fomega'$ for some everywhere-positive function $f$. Actually, this only makes sense if $omega$ is nonvanishing; for a general 2-form $omega = f dxwedge dy$ we extend by locality of the integral / linearity of the measure and have $int_{(M,o)} f dxwedge dy = int_{M_+} f dxdy - int_{M_-}fdxdy$ where $M_{pm} = {p in M mid [f_p mathrm{d}x wedge mathrm{d}y] = pm o_p}$ (and $M_0$ gets thrown out).

Now, what is going on when we swap the order of symbols under an integral sign? Well, from the equation above, we see that when we swap $mathrm{d}x wedge mathrm{d}y$ with $mathrm{d}y wedge mathrm{d}x$, we're multiplying our integrals by $-1$ (assuming we intend on keeping the same orientation). On the other hand, when we swap $dxdy$ for $dydx$, we're just using a notational variant as far as the product measure is concerned. But there's an ambiguity in standard integral notation between integration over a product measure and "Fubini'ed" integration, and usually if we swap $dxdy$ for $dydx$, what we mean is that we're changing the order in which we want to Fubini! What is the counterpart in terms of integrals of differential forms? Well, the equations, with $omega = f mathrm{d}xwedge mathrm{d}y$, will be

$int(int fdx)dy = int fdxdy = int (int fdy)dx$

$int(int i_{partial_{y|x}} omega) mathrm{d}y = int omega = int(int i_{partial_{x|_y}} omega) mathrm{d}x$

Here $i_X$ is the insertion operator for the vector field $X$, which feeds $X$ into one of $omega$'s input slots (first or last, by convention, I'm not sure which), lowering the degree by 1. Also, $partial_{x|y}$ is short for $frac{partial}{partial x}|_y$. Orientations: suppose that $int omega$ is oriented like $mathrm{d}x wedge mathrm{d}y$. Then we can take the other integrals to be oriented like $mathrm{d}x$ or $mathrm{d}y$ as appropriate, as long as we take the insertion operator to be inserting in the last slot on the LHS, and the first slot on the RHS...

Independently of what happens when you swap the order of symbols, we can talk about the weird role of orientation in integration of forms. A form like $mathrm{d}x wedge mathrm{d}y$ doesn't integrate to volume exactly, because its integral changes sign with orientation whereas volume doesn't. A quantity like this is usually called (somewhat pejoritively) a pseudoform. Whereas a "true form" requires a choice of orientation on its domain of integration, a pseudoform requires a choice of orientation on the normal bundle of its domain of integration (despite what it sounds like, this does NOT require a metric). One reference for this material is in Theodore Frankel's Geometry of Physics, which Steve Huntsman (cryptically) linked to in his comment above.

soft question - What are some examples of colorful language in serious mathematics papers?

André Weil uses some very colourful language in the introduction of his 1946 book Foundations of Algebraic Geometry. I recommend any mathematician to read it. Here are some excerpts:

"As in other kinds of war, so in this bloodless battle with an ever retreating foe which it is our good luck to be waging, it is possible for the advancing army to outrun its services of supply and incur disaster unless it waits for the quartermaster to perform his inglorious but indispensable task."

"Of course every mathematician has a right to his own language---at the risk of not being understood; and the use sometimes made of this right by our contemporaries almost suggests that the same fate is being prepared for mathematics as once befell, at Babel, another of man's great achievements."

"... however grateful we algebraic geometers should be to the modern algebraic school for lending us temporary accommodation, makeshift constructions full of rings, ideals and valuations, in which some of us feel in constant danger of getting lost, our wish and aim must be to return at the earliest possible moment to the palaces which are ours by birthright, to consolidate shaky foundations, to provide roofs where they are missing, to finish, in harmony with the portions already existing, what has been left undone."

"...it is hoped that these may be helpful to the reader, to whom the author, having acted as his pilot until this point, heartily wishes Godspeed on his sailing away from the axiomatic shore, further and further into open sea."

geometry - Finding the union of N random circles arbitrarily (or conspiratorially) placed on a two-dimensional surface

Please consider a two-dimensional surface populated with a set of Cartesian coordinates $(x_i, y_i)$ for $N$ circles with individual radii $r_i$, where $r_{min} < r_i < r_{max}$.

Here, the number of circles, $N$, may be large - ranging from hundreds to tens of thousands. The circles may sparsely populate the plane in some places, and in others, be 'conspiratorially' packed together. Furthermore, $r_{min}$ / $r_{max}$ are not necessary defined in such a way to allow for accurate convex hull, spline interpolation, etc.

While we can always perform a monte carlo sampling of coordinates on the plane (or over some defined lattice), is there an efficient deterministic method of calculating the exact area given by the union of all $N$ circles?

---

Update - After a more extensive literature search (and thanks to "jc" for mentioning Edelsbrunner!), I was able to find a few relevant papers in the literature. First, the problem was of finding the union of 'N' discs was first proposed by M. I. Shamos in his 1978 thesis:

Shamos, M. I. “Computational Geometry” Ph.D. thesis, Yale Univ., New Haven, CT 1978.

In 1985 Micha Sharir presented an O(n $log^2n$) time and O(n) space deterministic algorithm for the disc intersection/union problem (based on modified Voronoi diagrams):
Sharir, M. Intersection and closest-pair problems for a set of planar discs. SIAM .J Comput. 14 (1985), pp. 448-468.

In 1988, Franz Aurenhammer presented another, more efficient O(n log n) time and O(n) space algorithm for circle (disc) intersection/union using power diagrams (generalizations of Voronoi diagrams):
Aurenhammer, F. Improved algorithms for discs and balls using power diagrams. Journal of Algorithms 9 (1985), pp. 151-161.

It would be really neat if anyone could be point me to an implementation of one of the two determistic algorithms above, perhaps in a computational geometry package (neither appear trivial to put into practice)...

computational complexity - Is every input gate of a Boolean Circuit (to decide a language) on a path to the output gate?

It's not necessarily the case that each input gate is on a path to the output gate. Determining if there is a path in a directed (acyclic) graph from one node s to another node t is NLOGSPACE-complete, so it is not a condition that you can arbitrarily enforce on (say) LOGSPACE-uniform circuits. It is easy to enforce this condition without loss of generality on P-uniform circuits: if you only require that the circuit be constructed in polynomial time, then after building the circuit it is easy to compute the strongly connected components of the graph representing the circuit, and throw out all gates that do not have a path to the output gate. Hence the inputs that do not eventually connect to the output would simply not be present. (Note I am not sure if this is precisely what you want.)

Still, I can't think of any typical usages of circuits where some input gates may not have a path to the output. Most reductions involving circuits are from machines to circuits, and we typically assume that a machine reads all its input bits. And when one is trying to prove a circuit lower bound, one always considers functions that depend on every bit.

UPDATE: Here's a silly way to enforce the condition you want, but I don't know if it will help you in your particular case. Suppose for simplicity that the number of inputs $n$ is a power of $2$. Make a complete binary tree of $2n-1$ new gates (where all edges lead towards the root of the tree), so we have $n$ leaves. Label all nodes in the tree as $AND$ gates. For the $i$th leaf in the tree, lead in the inputs $x_i$ and $neg x_i$. Now $OR$ the root of this binary tree with the output gate of your original circuit, and make this $OR$ your new output gate. Clearly every input now has a path to an output gate and the functionality of the circuit has not changed. Moreover this transformation is extremely uniform; I am pretty sure it can be made DLOGTIME-uniform. But is this really what you need for your problem?

oa.operator algebras - Ideals in Factors

One can easily prove that factors have no nontrivial ultraweakly closed 2-sided ideals as these are equivalent to nontrivial central projections. One can also show type $I_n$, type $II_1$, and type $III$ factors are algebraically simple (any 2-sided ideal must contain a projection. All projections are comparable in a factor, so you can show 1 is in the ideal). Ideals in $B(H)$ ($dim(H)=infty$, $H$ separable) have been studied extensively. What about ideals in $II_infty$ factors?

One might think, since every $II_infty$ factor $M$ can be written as $Noverline{otimes} B(H)$ for $N$ a $II_1$ factor, if $Isubset B(H)$ is an ideal, then $Notimes I$ is a 2-sided ideal. This is false. One needs to take the ideal generated by $Notimes I$. What does that mean from a von Neumann algebra viewpoint? Is it the same as taking the norm closure?

We can also describe some ideals in terms of the trace. One has the equivalent of the Hilbert-Schmidt operators: $$I_2={xin M | tr(x^ast x)<infty}$$ and the trace class operators:
$$I_1={xin M | tr(|x|)<infty}=I_2^ast I_2 =left{sum^n_{i=1} x_i^ast y_i | x_i, y_iin I_2right}.$$
What is the relation of $I_j$ to $Notimes L^j(H)$ for $j=1,2$ (where $L^2(H)$ is the Hilbert-Schmidt operators and $L^1(H)$ is the trace class operators in $B(H)$)? Is $I_j$ the norm closure of $Notimes L^j(H)$?

algebraic groups - The normalizer of a reductive subgroup

While the comments by Angelo and Wilberd go a long way toward an answer, I'd prefer to start with a more precise formulation of the question. The group $G$ and its (closed!) subgroup $H$ may be assumed to be connected with $H$ moreover reductive. The structure theory involved here is essentially independent of the characteristic of the ground field, but on the other hand a field of definition apparently plays no role and could be assumed to be algebraically closed.

Up to isogeny, a connected reductive group is a direct product of closed connected subgroups which can be tori or products of (quasi-)simple groups of a fixed type. Due to the standard theorem on "rigidity" of tori, the
centralizer of a torus in $G$ has finite index in the normalizer. On the other hand, the outer automorphism group of a simple group is finite (coming from the automorphisms of the Dynkin diagram) while other outer automorphisms of a direct product of groups of the same type come from just finitely many permutations of isomorphic simple factors.

Since an automorphism of $H$ preserves the center and the various products of simple factors of the same type (whose overall product is the derived group of $H$), these pieces combine to show that $N_G(H)/H , Z_G(H)$ is finite. (Though I haven't written it all down myself.)
I don't recall a specific reference for such a general result in the literature; it may only come up in special situations.

ca.analysis and odes - what is summation in the sense of a principal value?

In one paper I saw this equality:

$$sum_{eta=-infty}^{infty}frac{z}{(z+eta)}=pi zcot(pi z)$$
which is the same as
$$sum_{eta=-infty}^{infty}frac{1}{(z+eta)}=pi cot(pi z)$$
where summation is understood in the sense of a principal value. What does it mean?

In another paper I found the next expression:

$$frac{exp(2pi iaz)}{exp(2pi iz)-1}=frac{1}{2pi i}sum_{n=-infty}^{infty}frac{exp(2pi ina)}{z-n}$$
for $a=0$ it is equivalent to
$$frac{1}{exp(2pi iz)-1}=frac{1}{2pi i}sum_{n=-infty}^{infty}frac{1}{z+n}$$
which is not exactly the same expression like in the first case.
$$sum_{n=-infty}^{infty}frac{1}{z+n}=pi Cot[pi z]-ipi$$

Where is my mistake?

If the second formula is wrong, what is the correct formula for the second case?
$$sum_{n=-infty}^{infty}frac{exp(2pi ina)}{z+n}=?$$

von neumann algebras - When does a conditional expectation preserve some trace?

In developing a theory of index for inclusions of finite von Neumann algebras, several authors ([Kosaki, 1986], [Fidaleo & Isola,1996], etc.) define the index of a conditional expectation of a von Neumann algebra M onto a vN-subalgebra N (here, a conditional expectation is a normal, faithful N-N bimodule map fixing the subalgebra pointwise). An inclusion is said to have finite index if there exists a conditional expectation that has finite index. However, in the case where M is finite we might be interested in restricting ourselves to the conditional expectations that preserve some trace on M.

This leads us to the question: For a given (normal, faithful, finite) trace on M, Umegaki gives us a unique trace preserving conditional expectation E:M->N. Are there any nice necessary and sufficient conditions for a conditional expectation to arise in this manner? What if we allow the trace to be semifinite?

Since subfactors give rise to more than one conditional expectation, it is certainly not the case that all conditional expectations come from traces. A necessary condition is that E(xy)=E(yx) whenever x or y is an element of the relative commutant $N^prime cap M$. This is not sufficient, however.

chern classes - Can one bound the todd class of a 3-dimensional variety polynomially in c_3

This question is on bounding the degree of the Todd class. Let me explain where this comes from.

Suppose that $X$ is a smooth compact connected complex surface. Let $c_i=c_i(TX)$ be its $i$-th Chern class.

1. If $X$ is not of general type, we have that $c_1^2$ is bounded absolutely from above by 9. See Table 10 of Chapter VI of Compact complex surfaces by Barth, Hulek, Peters and van de Ven.

2. If $X$ is of general type, the Bogomolov-Miyaoka-Yau inequality states that $$c_1^2 leq 3c_2.$$

Conclusion. Since the degree of the Todd class of $X$ is the degree of $$frac{c_1^2+c_2}{12},$$ we basically get an upper bound for this if we show that the degree of $c_2$ is bounded from above. Now, the degree of $c_2$ is the topological Euler characteristic. Thus, since my goal was to bound the Todd class from above (in the set-up of my problem), I reduce to bounding the topological Euler characteristic. (That's good!)

Now, I am really interested in a similar result for 3-dimensional smooth compact connected complex varieties.

In this case, the degree of the Todd class of $X$ is the degree of $$frac{c_1 c_2}{24}.$$ Since the topological Euler characteristic of $X$ is (the degree of) $c_3$, I would like to reduce the problem of bounding this quantity from above to bounding $c_3$. So now for my question.

Question. Do there exist any absolute upper bounds on $c_1c_2$ or any bounds for that matter which are polynomial in $c_3$?

ergodic theory - a.e. convergence of the powers of an operator built from rotations

Ok, I rethought my old comment. I believe it is better with $Af (x) = f(x+a)$ and $B f(x) = f(x+b)$ to think about
$$
T^n = frac{1}{2^n} (A + B)^n = frac{1}{2^{n}} sum_{k=0}^{n} binom{n}{k} A^{k} B^{n-k}
= frac{1}{2^{n}} B^{n} sum_{k=0}^{n} binom{n}{k} C^{k},
$$
where $C = AB^{-1}$ so that $Cf(x) = f(x + a - b)$. It think that one should be able to show that this converges relatively easily ... (one somehow needs to deal with the weights).

---

Old Post

Let me rephrase the answer of Fabrizio Polo first:

Consider all words $w$ in A, B of length $n$. Call this set $mathcal{W}_n$. Now define $Af (x) = f(x+a)$ and $B f(x) = f(x+b)$. Then $T^n$ from the original post is equal to
$$
frac{1}{|mathcal{W}_n|}
sum w,
$$
where the sum is taken over all elements of $mathcal{W}_n$. I am somehow unable to make that display properly.
Here $w$ stands for the appropriate product of operators. E.g. for $n - 2$, we have
$mathcal{W}_n = {AA, AB, BA, BB}$ so that the expression above becomes
$$
frac{1}{4} (AA + AB + BA + BB),
$$
which is the $T^2$ from the original post.

Now if $a - b$ is irrational, I believe that $(mathbb Z_+) ast (mathbb Z_+)$ action defined above is ergodic, so one should have almost sure convergence. However, I am not sure if this holds, since the group $(mathbb Z_+)ast(mathbb Z_+)$ is not ameanable. So you will probably have to look into ergodic theorems for non ameanable actions to answer this question.

Another hope could be to somehow resum the expression for $T^n$ and be able to use more classical ergodic theorems ...