20

Your question translates into the language of orbifolds as saying: what is known about spherical $n$-orbifolds with underlying space homeomorphic to $S^n$?
In $S^2$, the examples you give are all there are.
Orbifolds with the geometry of $S^3$ were enumerated by William Dunbar in his thesis. His published paper MR1118824 contains the enumeration of the ...

19

One can use Whitney's theorem to show that the characteristic polynomial is
$$ \sum_{i=0}^k{n\choose i}q^{k-i}(q-2)^{n-i} +
\sum_{i=k+1}^n{n\choose i}(q-2)^{n-i}. $$
I doubt that this can be simplified.

answered Oct 31 '17 at 1:11

Richard Stanley

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17

Suppose your polygon has angles $\alpha_i=\frac{\pi}{k_i}$ where $k_i\geq 2$ are integers. Then they must satisfy
$$\frac{1}{k_1}+\frac{1}{k_2}+\cdots+\frac{1}{k_n}=n-2,$$
which means $n-2\leq \frac{n}{2}$, so $n\leq 4$.
When $n=3$, one finds the solutions $(k_1,k_2,k_3)=(2,4,4),(3,3,3),(2,3,6)$ which correspond to the 45-45-90, equilateral, and 30-60-90 ...

answered Nov 14 '13 at 7:39

Gjergji Zaimi

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Actually, every Coxeter system of rank four is either euclidean or hyperbolic. That is, the canonical bilinear form has at most one negative eigenvalue.
It follows from this paper of Maxwell. He categorize the Coxeter systems by "level": finite or affine Coxeter systems are of level 0 (some may be more comfortable with terms like "spherical/euclidean" etc....

12

There is an intrinsic characterisation which is probably more complicated than what you are looking for. As Ben says, Soergel bimodules are pretty subtle things ...
Because Soergel bimodules are (finitely generated) $R$-bimodules one can think about them as coherent sheaves on $V \times V$ (where $V = Spec R$). Inside $V \times V$ one has for any $w \in S_n$...

12

I think I have completely answered the question in the following form:
Theorem. For a finite subgroup $\Gamma < O(n)$ the quotient space $S^{n-1}/\Gamma$ is homeomorphic to $S^{n-1}$ if and only if $\Gamma$ has the form
\begin{eqnarray*}
\Gamma = \Gamma_{ps} \times P_1 \times \ldots \times P_k
\end{eqnarray*}
for a pseudoreflection group $\Gamma_{ps}$ ...

12

Let the root system be $v_1$, …, $v_n$ with all elements normalized to be length $1$. So $\langle v_i, v_i \rangle =1$, we have $\langle v_i, v_j \rangle \leq - \cos (\pi/3) = -1/2$ for at least $n$ pairs $(i,j)$, and we have $\langle v_i, v_j \rangle \leq 0$ for all $i \neq j$.
But then
$$\left\langle \sum_{i=1}^n v_i, \sum_{i=1}^n v_i \right\rangle \leq ...

answered Aug 26 '14 at 19:33

David E Speyer

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12

The history is definitely somewhat convoluted. Note first that the term "Coxeter group" itself was introduced by Bourbaki in their 1968 volume containing chapters 4-6 of Groupes et algebres de Lie. The first section of Chapter 4 studies Coxeter systems $(W,S)$ (with $S$ typically finite) in great generality, inspired in part by Coxeter's work on ...

answered May 24 '14 at 11:44

Jim Humphreys

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12

Perhaps this, for now, is more an issue of perspective. Yes, for matroids, spheres and Coxeter groups the realizable cases were known before using results in algebraic geometry, but this is natural as our understanding of the cohomology of algebraic varietes was much better, historically. And so we think of this as strange because we are used to think of ...

12

Let $U$ be the uniform matroid of rank $k$ on $n$. Since $U$ is orientable one can consider the Lawrence oriented matroid $\Lambda(U)$ associated with any orientation of $U$ (the Lawrence construction doesn't care about which orientation you take). Then $M(n,k)$ is precisely the underlying unoriented matroid $\underline{\Lambda(U)}$ of $\Lambda(U)$.
Also, ...

11

An algorithm for reducing arbitrary words in any Coxeter group to their lexicographically least reduced representatives is described in the paper:
B. Brink and R.B. Howlett,
``A finiteness property and an automatic structure for Coxeter groups'',
Math. Ann. 296, (1993), 179-190.
It doesn't do exactly what you asked for, in that it doesn't do the reductions ...

10

I have placed data files containing the polynomials for $S_n$ with $n\in \{4,5,6,7,8,9\}$ here. The corresponding file for $S_{10}$ is a few gigabytes as a plain text file; I could certainly send that if you're interested.
This belongs as a response to Jim Humphrey's post, but I don't think I have the reputation for that, so: Patrick Polo's result (for ...

10

As a quasi-minuscule representation has a zero weight, its weights must be in the root lattice. Since every non-zero dominant weight is positive on at least one simple root, the quasi-minuscule condition implies that non-zero weights are in fact roots. There is only one orbit of them, so in case there are different root lengths, the long roots are excluded. ...

10

As you may know, in the family of groups
$H^{m,n,p} := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^m, (bc)^n, (abc)^p \rangle$
considered by Coxeter, there is only one remaining case for which finiteness has not been decided, $H^{3,7,19}$, which is your group $H$. It is possible that $H \cong J_1 \times {\rm PSL}(2,113)$, since only two finite simple ...

10

Yes. By Chevalley-Shepard-Todd, $S(V)^G$ and $S(V)^H$ are polynomial rings. Let $S(V)^G=\mathbb{R}[g_1, \ldots, g_n]$ and $S(V)^H = \mathbb{R}[h_1,\ldots, h_n]$ where the $g_i$ and $h_i$ are homogenous. Then
$$S(V)_G^H = \mathbb{R}[h_1,\ldots,h_n]/\langle g_1,\ldots, g_n \rangle.$$
Here the denominator is the ideal of $\mathbb{R}[h_1,\ldots,h_n]$ generated ...

answered Nov 22 '17 at 21:19

David E Speyer

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10

If you type "help" immediately on entering the program, you'll get a fairly long and useful introductory message. At whatever level you are, you are supposed to be able to type "help," and then the name of any command accessible at that level, to get a message about it. (Well, not all these help files exist.) The help files exist as text files in the ...

9

I don't know about Hamilton cycles in this Cayley graph (although someone surely does, and I have a sneaking suspicion that I have heard about them and forgotten). So I'm not answering the question really, but I think this is the answer you want:
To efficiently "traverse" a finite Coxeter group (i.e. visit every element with low memory overhead), then you ...

9

In fact, for any tree of transpositions in $S_n$ the corresponding Cayley graph is Hamiltonian. Start with my mini-survey with Radoicic which is relatively recent. The type of Hamiltonian cycles you are interested in are best explained in Don Knuth's "Art of Computer Programming", Vol. 4, Fascicle 2b ("Generating all permutations") (preliminary version can ...

9

The conjugacy class of a reflection in an infinite irreducible Coxeter group is always infinite. This follows from a result of mine, which was earlier proved by Kleiner and Pelley in the case of a symmetrizable integer Cartan matrix:
Let $W$ be an infinite irreducible Coxeter group, with generating set $S = \{s_1, s_2, \cdots, s_n\}$. Then the word $s_1 ...

answered Dec 8 '11 at 0:02

David E Speyer

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9

The following theorem was proved independently by Deodhar and by Dyer:
Let $(W, S)$ be a Coxeter group. Let $V$ be a subgroup of $W$ generated by reflections.
Then there is a set $R$ of reflections in $V$ so that $(V, R)$ is a Coxeter group.
$R$ can be characterized uniquely as follows: Let $T$ be the set of reflections in $W$
and, for $w \in W$, ...

answered Aug 20 '13 at 20:17

David E Speyer

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9

Yes, at least if your chosen element is the lexicographically first reduced expression for your permutation. This is how people usually prove that any two reduced expressions for a given permutation are connected by a series of long and short braid moves -- by reducing both to the lexicographically first reduced expression for that permutation.
You can ...

9

For the case $N=3$, this is the random walk on the honeycomb lattice:
(source)
Lemma 2.1 in this paper computes the number of walks of length $2n$ on
the honeycomb lattice
which return to the origin,
so $$g_3(t)=\sum_{n=0}^{\infty}\sum_{k=0}^n \binom{2k}{k}\binom{n}{k}^2 t^{2n}.$$

9

Schur-Weyl duality tells you that under the commuting actions of $GL(V)$ and $S_n$ on $V^{\otimes n}$, it is a direct sum $\bigoplus_{\lambda} S^{\lambda} \otimes V_\lambda$ where $S^\lambda$ is some irreducible representation of $S_n$ and $V_\lambda$ is some irreducible representation of $GL(V)$ and where the $\lambda$ range over some indexing set. ...

8

Not an answer, but too long for a comment. There is actually a general recipe for computing the growth function of any Coxeter group (which implies, in particular, that it is always a rational function). I haven't done the calculation, but your purported calculation of the growth function for the Coxeter group you wrote down can be derived from this (...

8

Just a partial answer, for dimension 2 and 3, the answer is yes, there are infinitely many cocompact hyperbolic Coxeter groups. Furthermore, in these dimensions the commensurability classes of the groups can be distinguished by their respective invariant trace fields.
For $n=3$, it follows from $\S 4.7.3$ of Maclachlan and Reid's "The Arithmetic of ...

8

Suppose we have an action of the 6,3,7 reflection group on some nondegenerate real quadratic space by reflections in positive norm vectors. Then there is a 4-tuple of vectors for which the reflections in the perpendicular hyperplanes generate the group according to the Coxeter relations. We may assume these vectors have length 1, so the inner products are ...

8

In http://www.sciencedirect.com/science/article/pii/S0167278998900057 (Remarks on quasicrystallic symmetries) Arnold so describes the idea of Shcherbak's proof:
Now the proof of this theorem depends on the relation of the icosahedron symmetry group $H_3$ to the so-called crystallographic (or Weyl) group $D_6$
(associated with the simple Lie algebra $...

8

As Uri Bader remarked one has to be careful with the term "combinatorial abstraction". In the cases mentioned by Sam and in other cases the geometric objects are certain algebraic varieties but the abstractions often refers to other geometric or topological objects. Let me give three examples. (I will add links later.)
A) The $g$-theorem and KL-polynomials ...

8

The answer is yes. I claim there are at most $2^n$ elements of $W$ for which $\lambda(w) \neq 0$. Specifically, let $\vee$ be the join in weak order, which is a semi-lattice. If $\{ s_1, s_2, \ldots, s_j \} \subseteq S$ and $s_1 \vee s_2 \vee \cdots \vee s_j$ is defined, then I claim that $\lambda(s_1 \vee s_2 \vee \cdots \vee s_j)=(-1)^j$, and otherwise I ...

answered Jun 20 '19 at 19:32

David E Speyer

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8

Depending on what you mean by a "good way", maybe there is and maybe there isn't. If you want to do this all in terms of the combinatorics of reduced words, probably the following is the best you can do:
Since $\beta$ is a root, there is an element $w\in W$ and a simple root $\alpha$ such that $\beta=w\alpha$. If $s_\alpha$ is the simple reflection for $\...

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