User hugh thomas - MathOverflow

What math institutes offer research in pairs/research in teams?

2011-04-23T06:57:19Z

Some math institutes offer programs in which a small number of researchers are enabled to meet at the institute for a week or more. A list seemed as if it could be useful.

Answer by Hugh Thomas for Expected distance of a random point to the convex hull of N other points

2013-04-02T21:27:59Z

This answers the question about comparing the expected value of the distance from $X_0$ to the convex hull of the $X_i$ (for $i>1$) and the expected value of the distance from $Y$ to the convex hull.

Suppose the distribution of $X_1,\dots,X_n$ is rotationally symmetric about some origin $O$, and so is that of $X_0$. (I don't need $X_0$ to be distributed according to the same distribution as the $X_i$ with $i\geq 1$.) If we consider the function $F(X_1,\dots,X_n;X_0)$ which is the shortest vector from $X_0$ into the convex hull of the $X_i$'s, then $F$ will be rotationally symmetric too.

Pick $X_1,\dots,X_n$ according to their distribution. Let $P$ be their convex hull.

Now condition the choice of $X_0$ on the length of $F$ equalling $r$; so $F(X_1,\dots,X_n;X_0)$ is uniformly distributed on a sphere of radius $r$. Suppose first that $X_0$ is not in $P$, so $r$ is strictly positive. By symmetry, and after changing co-ordinates, we can put $X_0$ at $(0,\dots,0)$ and the closest point in $P$ at $re_1=(r,0,\dots,0)$. $P$ lies outside the sphere of radius $r$, so it lies in the half-plane $x_1\geq r$.

Let $m$ be the difference between the mean of $Y$ and $O$, the mean of $X_0$. Define $Y=X_0+m$.

We now want to consider what happens when $X_0$ is replaced by $Y$. In the co-ordinate system we are using, $Y$ is moved a uniformly distributed random direction from $X_0$, and the distance is the length of the vector $m$.

Let $v$ be such a randomly chosen vector. Suppose that the first co-ordinate of $v$ is non-negative. We note that the distance from $X_0+v$ to $P$ is decreased by at most the first coordinate of $v$, while the distance from $X_0-v$ to $P$ is increased by at least the first coordinate. Thus, the average over all possible $v$ will not decrease the distance (since $v$ and $-v$ were equally likely amounts by which to perturb $X_0$ to obtain $Y$).

Now consider the case that $X_0$ is in $P$ (i.e. r=0). In this case, it is obviously impossible for the average distance from $Y$ to $P$ to be smaller than that of $X_0$ to $P$, since the latter is 0. So we are done in this case also.

Answer by Hugh Thomas for Number of Permutations with k-inversions and with a single clamped value

2013-03-12T03:45:02Z

It follows from the Knuth-Netto formula that the asymptotics of $I_n(k)$, for $k$ fixed, is $n^k/k!$ to first order.

I claim the asymptotic behaviour of $I_n^{\sigma(y)=x}(k)$ is as $n^{k-|x-y|}/(k-|x-y|)!$.

For convenience I am going to assume that $x\geq y$. The other case can be treated the same way.

First, we construct a set of permutations of the desired order inside $I_n^{\sigma(y)=x}(k)$. Consider permutations such that

$\sigma(j)=j$ for $1\leq j\lt y$,
$\sigma(y)=x$,
$\sigma(j)=j-1$ for $y+1\leq j \leq x$.

I.e., I have fixed the behaviour of the permutation on $1,\dots,x$, and this has introduced $x-y$ inversions. Now determine the rest of $\sigma$ by choosing an arbitrary permutation on $x+1,\dots,n$ with $k-(x-y)$ inversions. This gives us enough permutations in $I_n^{\sigma(y)=x}(k)$.

Now we show that there are no more than this many (to first order).

It's helpful to think of permutations of $n$ as planar matchings between a top row of $n$ dots and a bottom row of $n$ dots. Then inversions are just points where two arcs cross. (We assume the diagram is drawn in a sensible way so three arcs don't cross at a point, and arcs which don't need to cross, don't cross.)

When looking at permutations which satisfy that $\sigma(y)=x$, we have fixed one arc. If we erase that arc, the resulting diagram has $n-1$ dots left on top and bottom, so it can be viewed as a permutation in $S_{n-1}$.

Suppose we take a permutation in $I_n^{\sigma(y)=x}(k)$ and remove the arc from $y$ to $x$. This arc must be crossed by at least $x-y$ arcs (because it has that many more starting points than ending points to its right). Therefore, erasing this arc removes at least $x-y$ inversions. Thus, we get an injective map $$I_n^{\sigma(y)=x}(k) \rightarrow I_{n-1}(k-(x-y)) \cup I_{n-1}(k-(x-y)-1) \dots .$$

By the asymptotics already mentioned, the second and later terms on the RHS are lower order, and the first term has the desired asymptotics. This shows that the claimed formula is an upper bound to first order as well.

Answer by Hugh Thomas for Why is there a unique increasing maximal path in any Bruhat interval under any reflection order?

2013-01-02T21:24:33Z

Clearly there is at most one increasing path, so the only problem is to find it.

Take some path, and suppose it is not increasing. So there is some length 2 subpath which is not increasing. Replace it by an increasing path. The fact that this is possible only requires that you work inside a rank 2 root system.

Repeat. We need to check that we can't get caught in an infinite loop. To do this, we need to make sure that each replacement step can be done in such a way as to reduce some statistic on chains. For example, we could use lexicographic order on the labels of the chains (using the given reflection order).

Having written this down, I now realize that this and more is contained in the classic paper by Björner and Wachs introducing CL-shellability: Bruhat order of Coxeter groups and shellability. Adv. in Math. 43 (1982), no. 1, 87–100.

EDIT: Oops. It's not obvious that there is only one increasing path. A priori, it could be possible that you could take your fixed word for $v$ and, by removing two different increasing subsequences of reflections, obtain a word for $u$.

The argument that this doesn't happen, as given in Björner-Wachs, is as follows. Suppose the result is already established for chains of length less than $\ell(v)-\ell(u)$. Let $v=s_{1}\dots s_{q}$ be your fixed word for $v$, and suppose that you can remove reflections $s_{i_1}<\dots<s_{i_r}$ or $s_{j_1}<\dots<s_{j_r}$ to get a word for $u$. Suppose further that $i_r<j_r$ . Let $t$ be the reflection $s_qs_{q-1}\dots s_{j_r}\dots s_{q-1}s_q$ . Then $\ell(ut)<\ell(u)$ . But $ut$ must be the first step up the chain corresponding to the reflections labelled by $j$'s. Contradiction. So we must not have $i_r<j_r$ . By symmetry, we can't have $j_r<i_r$ . So $i_r=j_r$ , and we have reduced the problem to a shorter interval.

Answer by Hugh Thomas for Why is the representation dimension of an Artin algebra never equal to 1?

2012-12-17T02:30:07Z

First of all, you have to assume that $A$ is non-semi-simple. For a semi-simple Artin algebra, the representation dimension is defined to be 1.

For a non-semi-simple algebra, the representation dimension is, by definition, the smallest $d$ such that there exists $M$ an $A$-module which is both a generator and a co-generator, and such that the global dimension of the endomorphism ring of $M$ is $d$.

To show the representation dimension of $M$ is not 1, we need to show that the endomorphism ring of $M$ is not hereditary.

Since $M$ is a generator and a co-generator, it contains all the projective indecomposables and all the injective decomposables as direct summands.

If $A$ is non-semi-simple, then it has a projective indecomposable module $P$ which is not simple. Let $Q$ be another projective which has a non-zero map to $P$. Suppose $P$ is the projective cover of the simple $S$, and let $I$ be its injective hull. Then the composition of the maps from $Q$ to $P$ and from $P$ to $I$ is zero. This shows that there are relations among the elements the endomorphism ring of $M$. It follows that the endomorphism ring is not hereditary.

Answer by Hugh Thomas for Representation dimension of a special algebra

2012-12-15T20:38:08Z

I don't think (*) is correctly copied from the paper. The corresponding claim in the paper is that every morphism from an indecomposable summand of $M$ except for the identity morphism from $T$ to $T$ factors through $N_1$.

I think a precisely correct statement is that, for any direct summand $T$ of $M$, Hom($T$,$T$) can be split as a vector space into the multiples of the identity map and a complementary subspace consisting of morphisms which factor through $N_1$. This seems to be sufficient for the argument; I expect it is what the authors meant by what they wrote.

Answer by Hugh Thomas for The highest root of an ADE quiver

2012-10-13T23:09:32Z

Part of what makes this question more interesting outside type $A$ is that the highest root can't be projective or injective. A test case to consider is $D_4$, say with all three arrows pointing to the central vertex. A representation of $D_4$ with dimension vector (1,1,1,2) will be indecomposable provided you don't choose any zero maps or have any two of the maps be multiples.

In your setting, this translates into saying that you shouldn't choose the zero vector at any of the three vertices, and they shouldn't map to the same line in the 2-dimensional vector space over the central vertex either. It's not quite clear to me how to express this as a condition on the choice of vectors at the three vertices, but independent generic choices would work.

It's a general fact about quiver representations that if the dimensions come from a real root, then if you choose the maps generically, you will get the indecomposable representation. So one way to view the question is whether choosing a vector $v$ for each vertex is sufficiently generic.

For $D_n$ with $n>4$, I find it plausible that someone a bit more comfortable with representation theory of finite groups than I am, could convince themselves that this works fine as well. The situation for $E$ seems more complicated.

The following is not an answer to the question, but if it's unfamiliar to you, it might be of interest. This setup is called the "algebraic McKay correspondence", and is due to Auslander.

Let $S=\mathbb C[x,y]$. Let $R=S^G$. As an $R$-module, $S$ is a direct sum of the maximal Cohen-Macaulay $R$-modules. The maximal CM $R$-modules can also be constructed from the irreps of $\Gamma$ as follows: for $V$ an irrep, take $(V\otimes_R S)^G$.

The endomorphism ring of $S$ as an $R$-module is the preprojective algebra of the corresponding affine type. (If we throw away the node for $R$, we get the finite type.)

From this point of view, the arrows of the McKay quiver come with maps for free. In the $A_n$ case, they are just multiplication by $x$ and $y$, so the preprojective relation is just the fact that $xy=yx$.

However, I don't see how to get back to finite-dimensional $\mathbb C$-vector spaces in a natural way now.

Answer by Hugh Thomas for What do cluster algebras tell us about Grassmannians?

2012-10-13T18:49:34Z

One simple answer is to talk about the totally positive part of $(G_{k,n})_{> 0}$, the part of the Grassmannian where all the maximal minors (=Plücker coordinates) are real and positive. Naively, if you want to test whether a point is in the totally positive part, you would check all the Plücker coordinates. Since there is a cluster structure which includes all the Plücker coordinates as cluster variables, though, it suffices to check positivity with respect to any single cluster of them.

This might be a bit more geometrical than you were looking for, but I think it's a nice, and easy-to-explain, idea, which is also consistent with the roots of the development of cluster algebras.

Answer by Hugh Thomas for The Fukaya category of a simple singularity (reference request)

2012-09-15T19:15:48Z

This sounds wrong to me. I think $D^b(Q)$ should be replaced by the derived category of finite length modules over the corresponding preprojective algebra of affine type.

Homological mirror symmetry would say the derived Fukaya category should be equivalent to the bounded derived category of quasi-coherent sheaves over the mirror manifold (in this case, over the exceptional fibre of the resolution of the singularity). I don't understand this in any detail, and I don't know a reference for it, so I may not be getting this quite right.

The quasi-coherent sheaves supported on the exceptional fibre form an abelian category equivalent to the finite length modules of the preprojective algebra of the corresponding affine type. This is discussed in the introduction Bridgeland's Stability conditions and Kleinian singularities. The reference he gives for the link between the two is Crawley-Boevey and Holland, Noncommutative deformations of Kleinian singularities, Duke Math. J. 92 (1998), no. 3, 605–635.

Answer by Hugh Thomas for College (Euclidean) geometry textbook recommendations

2012-07-11T16:28:42Z

I really like Isaacs' "Geometry for College Students", which I have taught from several times. It is proof-focussed but not pedantic. It's also pretty cheap ($62).

http://www.ams.org/bookstore-getitem/item=amstext-8

Answer by Hugh Thomas for How can pushing a vertex in a polytope lead to merging facets?

2012-06-07T22:13:38Z

What happens as you push $v$, is that you can break a face into two. Start with a cube in $\mathbb R^3$, choose a vertex, and start to push it into the interior of the cube. Each of the three facets of the cube containing $v$ will be broken into two triangles.

The "merging" happens if you think about reversing this procedure (moving $v'$ back to $v$). Santos says that he is thinking about it this way at the beginning of the proof.

Answer by Hugh Thomas for Two curious asymptotic results for dimensions of type A objects

2012-04-03T14:42:48Z

This is an answer to Alexander's combinatorial reformulation of the question in comments to Bruce's answer.

dim $V_\lambda$/$n$! is the chance that you will get a standard Young tableau if you assign the values 1 to $n$ to the boxes of a tableau of shape $\lambda$ according to a random permutation.

dim $W_\lambda/N^n$ is the chance that you will get a semi-standard Young tableau if you assign a value in $[1,N]$ to each box in a tableau of shape $\lambda$.

Think of the second procedure in the following way: first choose a set of $n$ numbers from 1 to $N$ to serve as entries, and then assign them to boxes. If the entries are all different, then the chances that what you get is a semistandard tableau is the same as the chance that you get a standard tableau starting with 1..$n$.

As $N$ tends to infiniity, the chance that you will choose two entries the same becomes vanishingly small, so in the limit, dim $W_\lambda/N^n$ tends to dim $V_\lambda$/$n$!.

Catalan objects associated to a univariate polynomial

2012-02-04T21:19:31Z

Given a monic degree $n$ polynomial $f(z)$ with no double roots, and a phase $0\leq \theta < \pi$, there are natural constructions which associate to this data:

a noncrossing matching on $2n$ vertices, and

a triangulation of an $n+2$-gon.

These objects are both counted by Catalan numbers, leading me to feel that there should be some connection. It is not the case that the fibres of the two maps are the same, and I will say more below about how they are clearly different. My question is: what is the connection?

Let me now explain the constructions. Consider the points $z$ in $\mathbb C$ such that arg($\pm f(z)$)$=\theta$. Generically, this consists of $n$ unbounded curves, which, for large $|z|$, are evenly-spaces spokes of a wheel (with $2n$ spokes), and therefore can be interpreted as giving rise to a noncrossing matching. See Martin, Savitt, Singer and Savitt. As you tune $\theta$, the matchings change in a nicely controlled way, as pairs of components meet and reconnect. Tuning from $\theta=0$ to $\theta=\pi$ results in $n-1$ of these reconnections, and the total effect is to rotate the diagram by one step.

Now we consider how to produce the triangulation. I found out about this from a sequence of papers by Cecotti, Vafa, and a varying list of others, studying a connection between BPS states of 4d supersymmetric quantum field theories and cluster algebras. For the present topic see, in particular, Sections 4 and 5 of Cecotti, Córdova, Vafa. I found background provided in Section 4 of Mulase and Penkava helpful. The idea is to define a foliation of $\mathbb C$, where a curve $p(t)$ lies on a leaf of the foliation if $(p'(t))^2f(p(t))e^{2\theta i}$ is always real and positive. From a zero of $f$, there will be three equally-spaced curves coming out, and generically they will connect to points at infinity. For $|z|$ very large, the trajectories approach spokes of a wheel, with $n+2$ spokes. By connecting into a triangle the endpoints (on the circle at infinity) of the three curves coming out from each zero of $f$, you get a triangulation of an $n+2$-gon. As you tune $\theta$, the triangulation changes by diagonal flips. Again, the total effect of tuning from 0 to $\pi$ is to rotate the diagram by one step (though note that the "step" is a different fraction of $2\pi$ from the step that appears in the noncrossing matching).

(Edited to add: I should mention that, in the terminology of the people who study these flows, $f(z)dz$ is called a "quadratic differential", though these are often considered in more complicated situations, where there might be poles, for instance.)

If you prefer to complicate matters rather than simplifying them, you could also consider the question of whether there are any other Catalan objects associated to a monic polynomial and a phase (with different fibres than the above maps).

Answer by Hugh Thomas for What can be said about number-theoretic properties of the solid angle measures of polytopal cones in the weight lattice of sl(n)?

2012-01-03T14:58:47Z

$S_n$ acts on the set of cones, and preserves the solid angle measure. If you consider the $n!$ rotations of the cone by elements of $S_n$, and except on the union of some hyperplanes, the number of such cones covering an arbitrary point $x$ in $\mathfrak h$ is constant, say $N$, then the solid angle measure of the cone is $N/n!$.

This argument applies to the cones of the Coxeter arrangement (i.e. the cone generated by the fundamental weights $(1,0,\dots,0), (1,1,0,\dots,0),\dots,(1,\dots,1,0)$ and its $S_n$ translates); the $n!$ rotations are disjoint except on their boundaries and cover $\mathfrak h$, so their volumes are $1/n!$.

Less obviously, this argument also applies to the cone generated by a set of simple roots. This is worked out in Graham Denham's short paper "A note on De Concini and Procesi's curious identity".

Answer by Hugh Thomas for Does the following condition imply the homotopy type of a wedge of spheres?

2011-12-16T02:39:27Z

You can take $S^1\vee S^2\vee S^3$ and then use the Hopf fibration from $S^3$ to $S^2$ as the attaching map for a 4-cell onto the 2-cell. This has to have the cohomology (and homotopy) as you described.

(Having now seen Vitali's answer, I guess this is the same as his, but I thought I might as well give this answer anyway.)

Answer by Hugh Thomas for Breaking down an impartial game into Nim equivalent

2011-12-12T17:43:46Z

The fact that an impartial game is equivalent to a nim-heap doesn't necessarily mean that it's easy to find the heap in the combinatorics of the game. If all you want to do is determine whether a position is a first-player win or a second-player win, though, you don't need the nim-heap size anyway: a position is a second-player win if there are no moves from it to a second-player win position; otherwise, it's a first-player win.

I'd suggest looking at a bunch of permutations, determining which is a first-player win and which is a second-player win, and then trying to guess and prove a pattern.

Answer by Hugh Thomas for Permutations of Grid Colorings

2011-07-01T23:26:18Z

O(1) is impossible even if you drop the condition of no monochromatic rectangles, and even if you know that the two cells are always chosen within a given row.

Suppose the length of the rows, $m$, is very large, say $c^k$. Fix a row. You need at least k colourings to ensure that any two cells in the row have some colouring with respect to which they differ.

Answer by Hugh Thomas for Which cluster algebras have been categorified?

2011-05-25T22:24:50Z

Jan's answer includes many excellent references. I will try to give a few quick comments.

First of all, although the original Buan-Marsh-Reineke-Reiten-Todorov paper contained some results which were restricted to finite type cluster algebras, in subsequent work of them and others, notably Caldero-Keller, it has been shown that the BMRRT cluster category does indeed categorify any acyclic cluster algebra without coefficients (in particular, including those of tame type).

Although categorifications are often done in coefficient-free settings, one can also "cheat" by categorifying the cluster algebra in which the coefficients are treated as variables. For example, this allows one to treat the case of principal coefficients (in some sense, the most important case) for an acyclic cluster algebra using the BMRRT technology.

I should also point out that the categorifications in question here are what are now sometimes called additive categorifications; Hernandez-Leclerc and Nakajima have a multiplicative categorification which is more in the spirit of what people mean by categorification in other settings: they construct a category with direct sum and tensor product, and show the ring structure induced on the Grothendieck group of their categories is naturally a cluster algebra.

Answer by Hugh Thomas for What math institutes offer research in pairs/research in teams?

2011-04-23T07:07:10Z

BIRS in Banff, Alberta, Canada, offers "research in teams" (2-4 people, 1-2 weeks) and "focussed research groups" (up to eight researchers, 1-2 weeks).

Answer by Hugh Thomas for Nonfree projective module over a regular UFD?

2011-02-04T20:38:47Z

What about $R = \mathbb R[x,y,z]/\langle x^2+y^2+z^2-1\rangle$ and $M$ the projective module corresponding to the tangent bundle? This seems to satisfy your criteria except probably for factorialness (for which I have no intuition).

Answer by Hugh Thomas for orbit of a Dynkin diagram automorphism

2011-01-10T22:03:17Z

In a bunch of important cases, the orbits (of the group $\langle f \rangle$) acting on $\Delta$, correspond to roots in another root system. This is the procedure called "folding", which gives a way to reduce non simply-laced root systems to simply-laced ones. For example, $A_{2n-1}$ folds to $C_n$, $D_{n+1}$ folds to $B_n$, $E_6$ folds to $F_4$. $D_4$ folds to $G_2$ (if you take an automorphim of order 3).

If you start with $A_{2n}$, I guess the result is a non-reduced root system (i.e., one in which $\alpha$ and $2\alpha$ can both be roots for some $\alpha$), but I don't have a reference that includes non-reduced root systems handy, so I haven't confirmed that.

Answer by Hugh Thomas for How do I stop worrying about root systems and decompostion theorems (for reductive groups)?

2010-10-21T04:27:22Z

I had a problem which may be similar when I encountered root systems for the first time. In retrospect, I think that the problem was that I was reading about specific realizations of root systems (eg, the $A_n$ roots embedded into $\mathbb R^{n+1}$ as $e_i-e_j$ , etc.). Those specific realizations looked very ad hoc to me. The point, I would say, is that, while it's good to spend some time with some of those specific realizations, you should get used to thinking about root systems in a uniform way, at which point their power becomes pretty evident. (But maybe this is really a different problem from yours.)

Answer by Hugh Thomas for Finitely many arithmetic progressions

2010-05-20T04:13:41Z

This is "the same" as the generating function proof, but it doesn't use generating functions explicitly. Take the largest common difference in any of the sequences, say n, and pick $\zeta$ a primitive n-th root of unity. To each positive integer $m$, associate the complex number $\zeta^m$. Note that in any arithmetic sequence with common difference less than n, the sum over its entries of $\zeta^m$ stays bounded, while for an arithmetic sequence of common difference n, it grows unboundedly. Since the sum over all integers stays bounded, there has to be a second sequence of common difference n to balance out the first one.

Answer by Hugh Thomas for Resources for graphical languages / Penrose notation / Feynman diagrams / birdtracks?

2010-05-19T00:39:41Z

Part III of Kassel's "Quantum Groups" is an almost completely self-contained introduction to monoidal categories and knot theory. Diagrams of the sort you are describing play an increasing role throughout. I taught a course based of Part III this past term, and really enjoyed it.

Part III assumes some sophistication on the part of the reader, but the specific requirements are pretty slight. In particular, you definitely don't need to have read the first two parts of the book, and you don't need to know what a quantum group is. (On the other hand, if you want to know what a quantum group is, maybe you would want to read the whole book. I've put much less effort into understanding parts I and II, and haven't looked at part IV, so I'm not going to comment on how successful those parts are.)

Answer by Hugh Thomas for How to prove that a set of facets are all the facets of a convex polytope.

2010-05-15T03:48:42Z

Let V be the set of vertices, P their convex hull, and Q the polytope defined by the hyperplanes. If P is assumed to be simple, it suffices to check that each x in V is also a vertex of Q.

By your assumptions, P is contained in Q. Let x be a vertex of P. Assume that P is n-dimensional. By my assumption of simpleness, x has n facet-supporting hyperplanes of P through it. These hyperplanes must all be hyperplanes of Q; otherwise, x would not be a vertex of Q. Every facet of P includes some vertex of P (rather trivially) so this shows that all the facet-supporting hyperplanes of P are among the defining hyperplanes of Q. Thus Q is contained in P, and so Q=P.

Answer by Hugh Thomas for Statistics of irreps of S_n that can be read off the Young diagram, and consequences of Kerov-Vershik

2009-12-14T01:56:03Z

There is a beautiful interpretation of f(chi) (that is to say, of the length of the first column of the partition), though it isn't very representation-theoretic.

One way to generate Plancherel measure on partitions is to take uniformly at random a permutation of ${1,\dots,n}$ and apply Robinson-Schensted-Knuth to get a pair of Young tableaux of the same shape, and then take the partition encoded by that shape.

The length of the first column in the shape corresponding to a permutation $\pi$ is the length of the longest decreasing sequence in $\pi$, while the length of the first row is the length of the longest increasing sequence in $\pi$.

So Kerov-Vershik says (among other things) that the length of the longest decreasing sequence in a random permutation of ${1,\dots,n}$ is $2\sqrt{n}$. For more along these lines, see Richard Stanley's 2006 ICM talk.

Answer by Hugh Thomas for Do there exist non-PIDs in which every countably generated ideal is principal?

2009-12-07T02:48:22Z

No such ring exists.

Suppose otherwise. Let $I$ be a non-principal ideal, generated by a collection of elements $f_\alpha$ indexed by the set of ordinals $\alpha<\gamma$ for some $\gamma$. Consider the set $S$ of ordinals $\beta$ with the property that the ideal generated by $f_\alpha$ with $\alpha<\beta$ is not equal to the ideal generated by $f_\alpha$ with $\alpha\leq \beta$.

$I$ is generated by the $f_\beta$ with $\beta \in S$, so if $S$ is finite, then $I$ is finitely generated and thus is principal.

On the other hand, if $S$ is infinite, then take a countable subset $T= \{\beta_1<\beta_2<\dots\}$ of $S$. If the ideal generated by the corresponding set of $f_\beta$'s were principal, its generator would have to be in some $\langle f_{\beta_k} \mid k\leq i \rangle$ for some $i$ (since any element of $\langle f_{\beta}\mid \beta \in T\rangle$ is a finite combination of $f_\beta$'s and therefore lies in some such ideal). Now no $\beta_j$ with $j>i$ could be in $T$.

The same argument shows that all rings for which any countably generated ideal is finitely generated, have all their ideals finitely generated.

Corrected thanks to David's questions.

Answer by Hugh Thomas for Is there a poset with 0 with countable automorphism group?

2009-12-06T03:50:51Z

It seems unlikely (once you assume d.c.c.). Define the height of an element $x$ in $P$ to be the length of the shortest unrefinable chain from $x$ to $0$.

Let $P_n$ denote the elements of $P$ whose height is at most $n$. Since each element has a finite number of covers, the number of elements in $P_n$ is finite.

By d.c.c., every element of $P$ is in some $P_n$.

Let $G$ denote the automorphisms of $P$ and let $G_n$ denote the automorphisms of $P_n$. $G$ is the inverse limit of the system $G_n$. Let $H_n$ denote the image of $G$ inside $G_n$. (Note that this might not be all of $G_n$, since there could be automorphisms of $P_n$ that don't extend to $P$.) $G$ is also the inverse limit of the system $H_n$.

If the system $H_n$ stabilizes, then $G$ is finite. On the other hand, if $H_n$ doesn't stabilize, then the cardinality of $G$ is an infinite product, i.e. uncountable.

Answer by Hugh Thomas for Break Polyhedron into Tetrahedron

2009-12-03T14:36:57Z

If I understand your question correctly, you're saying that the given information is the face structure of a 3-dimensional convex polytope, and you would like a subdivision of the polytope into tetrahedra.

Here is one way to proceed. First, subdivide all the faces into triangles. Then pick your favourite vertex $v_0$. Connect $v_0$ to each triangle belonging to a face not containing $v_0$. This subdivides your polytope into tetrahedra.

Answer by Hugh Thomas for Math journal for high school students?

2009-12-01T02:00:58Z

Mathematical Mayhem was a math journal intended for and run by high school and university students. It now runs as a section within the journal Crux Mathematicorum, published by the Canadian Math Society. It looks like the Math Mayhem section is accessible without a subscription. However, the focus seems to be on problem-solving, so maybe it isn't what you're looking for, but I thought I'd mention it just in case.

Comment by Hugh Thomas

2013-04-03T19:41:54Z

This might be more of a question if you say more about what you need for your proof assistant. The fact that the suspension of an (n-1)-sphere is an n-sphere is clear if you think of a sphere as a cube with its boundary identified to a single point.

Comment by Hugh Thomas

2013-03-16T00:42:49Z

"Go on like that" is not entirely clear to me. Do we pick a new random $k$ at each step? Or do we proceed $k$, $k+1$, ..., which is consistent with the example you gave? Also, I suggest that it will probably help to visualize the problem to think of these 0,1 strings as lattice paths from $(0,0)$ to $(t,n-t)$, where we read 1's as horizontal steps and 0's as vertical steps. The basic swap move which you describe looks at two adjacent steps. If they are both horizontal or both vertical, nothing happens; otherwise, a single box is added or subtracted from the region under the path.

Comment by Hugh Thomas

2013-03-14T18:56:27Z

I don't see why you're removing a point. As stated, the Lemma applies even before you remove the point.

Comment by Hugh Thomas

2013-03-14T15:24:38Z

I don't think you have stated the problem in a sufficiently precise form in order for it to have a solution. Also, I still cannot understand the sentence "Now for each sequence $S_i$, there is an element that is used to replace only one element in $S_i$."

Comment by Hugh Thomas

2013-03-14T15:15:34Z

From what I can see it's a counter-example. Though I would go back and check the statement of the conjecture as given by Nash-Williams (the source of which doesn't seem to be online). The check of non-Hamiltonicity is pretty easy to do by hand: a Hamiltonian cycle has to alternate between vertices in {0,1,2} and {3,4,5} since there are no edges among {0,1,2}. Now it's clear that 1 has to fall between 4 and 5, so the two possibilities are 3,$x$,5,1,4,$y$ and 3,$x$,4,1,5,$y$ and they can both be ruled out.

Comment by Hugh Thomas

2013-03-14T13:27:03Z

I cannot understand your question. In particular, I don't know what the sentence that starts "Now for each sequence..." means. If we know nothing about $f$, how can we minimize it except by checking all the possible inputs?

Comment by Hugh Thomas

2013-01-14T01:58:52Z

Thanks for the additional explicit description. I think I now understand your definition (on a superficial level). Am I right that the tensor product of an oriented path with m arrows and one with n arrows gives you an oriented path with n+m-1 arrows (where this works properly if one of m,n is zero, but not if both are)?

Comment by Hugh Thomas

2013-01-10T23:48:43Z

Related to your last paragraph: there are examples of non-isomorphic varieties which are derived equivalent. The derived equivalences are then called Fourier-Mukai transforms. For example, in dimension 3, all crepant resolutions of a variety with terminal singularities are derived equivalent. There is lots of information in the chapter "Derived categories of coherent sheaves on algebraic varieties" by Yukinobu Toda in Triangulated Categories, edited by Holm, Joergensen, and Rouquier, including some discussion of tilting objects, but an answer to your question didn't obviously follow for me.

Comment by Hugh Thomas

2013-01-10T22:04:02Z

Another tensor product that has been somewhat studied is the pointwise tensor product, see Ryan Kinser (arXiv:0711.1135) and Martin Herschend ("Tensor products on quiver representations", Journal of Pure and Applied Algebra 212 (2008) 452-469. This is different from yours, though. In Martin's paper he also discusses some other notions of tensor product (but I didn't read carefully). Also, I don't understand exactly what you've written: for e in E and e' in E', what is s(e\otimes e')?

Comment by Hugh Thomas

2013-01-09T01:15:28Z

I agree that if W is not finite, the argument I gave potentially runs into trouble, since nothing guarantees that the process I described will terminate (and thus you might never obtain an increasing path). However, I think an argument like the following should work: among all reflections t such that u < vt <. v (where <. means "is covered by") take the first possible t (wrt the reflection order). Keep doing this. I bet if this wasn't increasing, the existence of an earlier rank 2 replacement would give a contradiction.

Comment by Hugh Thomas

2012-12-19T05:57:57Z

Thanks. So the equivalent statement is that a kG-module M admits a resolution of that form iff M is projective as a kG-module. Is that right? Further up in the question, you say that O(M) will be 1 or infinity. I'm confused. I would expect the projective dimension of a projective to be zero, so I would expect the projective dimension of O(M) to be zero or one. Can you clarify this? I should apologize in advance that, even once I understand the question, I may well not be able to help a lot in solving it.

Comment by Hugh Thomas

2012-12-16T20:59:58Z

I think the best approach is staring at the module and convincing yourself it can't be written non-trivially as a direct sum.

Comment by Hugh Thomas

2012-12-16T20:51:07Z

I think mainly what's going on is that a vector space of morphisms is implicitly being replaced with a suitable basis for the vector space. One basis element is the identity map, and the other basis elements (i.e., all the basis elements "except for" the first one) are chosen so that they do factor as desired.

Comment by Hugh Thomas

2012-12-15T21:36:18Z

I do not understand your equivalent reformulation. What are the morphisms in the sequence P -> P -> M? Or what do we know about this sequence? We must know something (which is not clear to me from what you've written) if there is to be any hope that a property exclusively of M will tell us whether or not the sequence is exact. Also, are all the P's in that paragraph isomorphic?

Comment by Hugh Thomas

2012-12-14T19:39:34Z

Let me mention that the Lyndon words of length n obtained by prepending an x, form a basis of the subspace of FL_n of the form [x,FL_{n-1}]. This seems as if it could be helpful for showing that only terms of this form appear in the n-even case.