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Recall that the nilCoxeter algebra $\mathcal{N}_W$ for a Coxeter group $W$ is given by the $\mathbf{k}$-basis $x_w$ for each $w\in W$ and multiplication $x_ux_v=x_{uv}$ if $\ell(uv)=\ell(u)+\ell(v)$ and $=0$ otherwise.

Q: What is known about finite dimensional modules over $\mathcal{N}_W$?

When I google this question, I find specifically this paper, but I don't know which explicit information it gives about the structure of such modules - and if there's more about it.

Context: For the special situation my research is about, I am only considering the case $W=\mathbb{S}_6$ - you can well assume this in your answer - and I am trying to understand the structure of a specific constructed finite dimensional $\mathcal{N}_{\mathbb{S}_6}$-module.

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This algebra has just one isomorphism class of simple module - let's call it $S$. Its projective cover is the regular representation, and is also the injective hull. The socle of the regular representation is the longest word in $W$. The Loewy length of the algebra is one more than the length of the longest word, and the Loewy and socle series coincide. In particular, an element $x_u$ is in the $j$th power of the radical if and only if $\ell(u)=j$. The algebra is Frobenius, but usually not symmetric.

The space $\operatorname{\rm Ext}^1(S,S)$ has dimension equal to the rank of $W$. So this $n-1$ in the case of $S_n$. Also in this case, the representation type is finite for $n=2$, tame for $n=3$, and wild for $n\geqslant 4$. If you want to imagine what the regular representation "looks like" in this case, take a permutohedron and dangle it from a vertex. So for $n=3$ a hexagon, and for $n=4$ a truncated octahedron.

What more would you like to know? What is the module you would like to describe?

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  • $\begingroup$ If you need a description of the Ext algebra as an $A_\infty$ algebra, I can probably supply that. $\endgroup$ Apr 15 at 22:03
  • $\begingroup$ Thank you! I will add a description of the module in my question with more context when I have time. It's just a vague (and probably fruitless) idea: But in order to show that the Fomin-Kirillov algebra $\mathcal{E}_6$ is infinite dimensional, I constructed a module over $\mathcal{N}_{\mathbb{S}_6}$ which has to be shown to be zero. Anyways, there's more known than I expected. Can you give a reference for the (probably nontrivial) properties in your answer? $\endgroup$ Apr 16 at 7:35
  • $\begingroup$ Especially that there is only one simple module seems not obvious to me... $\endgroup$ Apr 16 at 7:41
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    $\begingroup$ The honest truth is that I knew nothing about this algebra before reading your question, and spent an hour investigating it. I know of no literature. The reason why there's only one simple module is that any long enough product of the generators is zero, so they're all in the nil radical. Long enough, here, means longer than the length of the longest word in $W$. $\endgroup$ Apr 16 at 8:21
  • $\begingroup$ For the Ext algebra, we have $\sum_{i=0}^\infty t^i\dim\operatorname{\rm Ext}^i(S,S) = 1/(1-t)^{n-1}$. $\endgroup$ Apr 16 at 9:18
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@DaveBenson, has already given a beautiful answer to this question. I just wanted to point out that a number of the things he says (although not his full computation of the dimension of the Ext-algebra in the comments) can be deduced from the general theory of semigroup algebras.

A finite semigroup $S$ is nilpotent if it has an absorbing element $0$ such that $S^k=0$ for some $k\geq 1$. The nonidentity elements of your nilCoxeter monoid form a nilpotent semigroup and the smallest $k$, as pointed out by @DaveBenson, is the length of the longest word plus 1. If $M=S\cup \{1\}$, then the contracted monoid algebra $K_0M$ is the $K$-algebra with basis $M\setminus \{0\}$ and product defined by extending the product of $M$, but where the zeroes of $M$ and $K$ are identified.

In this set up, there is always one simple module $L$, the one-dimensional representation sending $1$ to $1$ and $S$ to $0$. The radical of $K_0M$ is the $K$-span of $S$ and therefore the Loewy length is the smallest power $k$ with $S^k=0$. The quiver of $K_0M$ has a single vertex and the number of loops is the number of elements of $S\setminus S^2$, which is also the dimension of $\mathrm{Ext}^1(L,L)$. For the nilxCoxeter monoid, this is the set of $x_s$ with $s$ a Coxeter generator, and so has size the rank of $S$ as pointed out by @DaveBenson. All this is a elementary but is a special case of results than can be found in my book on the representation theory of finite monoids (for instance the quiver computation follows because this is a special case of a $\mathcal J$-trivial monoid and the quiver is computed for such in Chapter 17, but the nilpotent semigroup case is much easier). Note that quiver presentation for $K_0M$ is obtained by taking a semigroup presentation of $S$ with respect to the minimal generating set $S\setminus S^2$. For the nilCoxeter monoid this is well known to be obtained from the Coxeter presentation of $W$ by replacing each relation $s^2=1$ by $x_s^2=0$. So for $S_3$, this is generators $a,b$ and relations $a^2=b^2=aba-bab=0$.

The fact that the algebras is Frobenius can be seen also from the general theory of such semigroups. There is a general criterion due to Wenger that can be found in Section 5 of my paper Factoring the Dedekind-Frobenius determinant of a semigroup. It is easy to see that if $S^k$ is the smallest power which is $0$, then each element of $S^{k-1}$ is in the socle. So to have a chance to be Frobenius (as the algebra has a single one-dimensional simple), one needs $S^{k-1}$ to be a singleton $\{z_0\}$. In your case of the nilCoxeter monoid, $k-1$ is the length of the longest element $w_0$ and $S^{k-1}=\{x_{w_0}\}$. But this is not enough to be Frobenius. Wenger says take the $M\setminus \{0\}\times M\setminus \{0\}$ matrix $A$ where $A_{s,t}$ is $1$ if $st=z_0$ and $0$ else, and the algebra $K_0M$ is Frobenius with socle $Kz_0$ iff the determinant of $A$ is nonzero. This matrix gives the isomorphism of the left regular representation with the dual of the right regular representation (if there is such an isomorphism).

In your case, this matrix is $W\times W$ and the entry $A_{x_u,x_v}$ is $1$ is $uv=w_0$ and $0$ else. This is a permutation matrix since for each $u$ there is a unique $v$ with $uv=w_0$ and $\ell(u)+\ell(v)=\ell(w_0)$ by Coxeter group theory. Therefore, the algebra is Frobenius with socle $Kw_0$.

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  • $\begingroup$ That's a nice interpretation! Thanks for adding it. $\endgroup$ Apr 16 at 10:16
  • $\begingroup$ @DaveBenson, sadly it took us semigroup people more than one hour to figure this out :) I'd be curious to see how your higher Ext computations look. In general, we often struggle to get past $Ext^1$ except in very nice instances but maybe for nilpotent semigroups with Frobenius monoid algebras things are easier. I'm guessing from your nice formula that some Coxeter stuff comes into play. People also have studied the 0-Hecke monoid where instead of $s^2=0$ you put $s^2=s$. So instead of falling off the permutahedron you stay put when length doesn't go up. $\endgroup$ Apr 16 at 10:21
  • $\begingroup$ (ctd.) This again has a basic algebra and the quiver is known (maybe the relations I am not sure). $\endgroup$ Apr 16 at 10:21
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    $\begingroup$ @DaveBenson, that's good to know. I always feel deficient that I can't get a good feel for computing higher Ext for most monoids. $\endgroup$ Apr 16 at 10:37
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    $\begingroup$ I feel that I should mention the paper of David Anick, "On the homology of associative algebras" and a subsequent paper by Anick and Green. That's where I really learned how to compute Ext over algebras given by quiver and relations. The method quickly gives you a headache, but it is practical for doing computations by hand, even in quite complicated examples. $\endgroup$ Apr 21 at 20:46
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Motivated by the answers I tried to check when it is symmetric and I got the result that it symmetric is if and only if conjugation by the longest element acts as the identity in the corresponding Coxeter group. The order of the Nakayama automorphism should be equal to the order of the automorphism induced by conjugation by the longest element in the Coxeter group. So it should be non-symmetric in type $A_n$, $D_n$ for $n$ odd and $E_6$ and $I_2(p)$ for $p$ odd.

Sadly local Frobenius algebras rarely have interesting homological algebra.

One interesting open problem is whether we always have $Ext^1(M,M) \neq 0$ for any non-projective $M$ for local Frobenius algebras. I wonder whether one can say anything about this problem for this class of algebras.

(To be sure, we talk about the same algebras: I think this is just the non-commutative polynomial ring in variables $x_i$ modulo braid relations and relations $x_i^2$ (see https://math.mit.edu/~rstan/pubs/pubfiles/94.pdf ). So it seems obvious that a $K$-basis corresponds to the group elements of the corresponding Coxeter group and there is a unique longest element up to scalars, which means the algebra is Frobenius. So being Frobenius is more or less equivalent to the fact that finite Coxeter groups have a unique longest element.)

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  • $\begingroup$ Yes, we talk about the same algebras. $\endgroup$ Apr 16 at 13:47
  • $\begingroup$ Another way to say this is that $w_o$ lies in the center or acts as $-1$ on the roots $\endgroup$ Apr 16 at 14:02
  • $\begingroup$ This can also be seen from my answer. The matrix I described is a Frobenius form and it is symmetric iff conjugating by the longest element is trivial. $\endgroup$ Apr 22 at 11:32

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