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This question concerns lower bounds of certain weight multiplicities in finite dimensional representations of algebraic groups (or Lie groups, Lie algebras).

Let's say $G$ is a simple algebraic group of rank $r$ over $\mathbb{C}$ and $\lambda$ a dominant integral weight. Consider the irreducible finite dimensional representation $V_\lambda$ with highest weight $\lambda$ and let $\mu$ be a weight occuring in $V_\lambda$. Let $W$ be the Wely group of $G$.

Suppose that $\lambda$ is regular (lies in interior of Weyl chamber), $\mu$ is dominant, and $\lambda-\mu$ lies in the interior of the positive root cone (the "wide cone"), then it seems to be true that

$m_{\lambda\mu}\ge 2^{r-1}$

Here $m_{\lambda\mu}$ is the dimension of $\mu$ weight space in $V_\lambda$.

Question: Does anyone know an elementary or straightforward proof (or maybe counterexample) of this lower bound? For example using some multiplicity formula or combinatorial models.

I came to this possible lower bound in a very indirect way when studying geometry of certain affine Springer fibers. But I guess there should be a straightforward way to see this, for example by some weight multiplicity formula, which I'm not quite familiar with. It would also be great if this could be seen by using some combinatorial models of algebraic representations (crystals, path models, MV polytopes etc). But since I'm not familiar with all these, any comments or suggestions are welcome.

Remarks on the number $2^{r-1}$:

Recall $G$ is simple with rank $r$. Fix a set of simple reflections $s_1,...,s_r$ in the Weyl group $W$, then $2^{r-1}$ is the number of elements in $W$ that can be written as products of these $r$ simple reflections, each occurring precisely once. Previously I have abused terminology and called this the "number of Coxeter elements", which lead to some confusion. As far as I know, this abuse of terminology is also present in literature, so it's better to pay attention to the context when seeing these words. Thanks to Jim Humphreys for clarifying this in his answer below. Simply speaking, $2^{r-1}$ only counts those Coxeter elements that can be expressed as products of a given set of simple reflections, which is in general smaller than the number of all Coxeter elements. For $W=S_n$, we have $2^{r-1}=2^{n-2}$ while the Coxeter elements in $S_n$ are the $n$-cycles, so there are $(n-1)!$ of them in total.

Update (Nov 1st, 2017): The very indirect way I found this inequality is in here. See Corollary 4.5.2.

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  • $\begingroup$ Usually a "Coxeter element" is any conjugate of the ones you specify, without keeping a fixed set of simple reflections. An example such as $W=S_n$ would clarify what you mean here by "number of Coxeter elements". Aside from that, I'm unclear about your estimate when $\lambda = \mu$ if this is permitted. (Your remarks at the end suggest you intend the characteristic to be 0, but the question may arise in any characteristic.) $\endgroup$ – Jim Humphreys Jul 18 '17 at 18:03
  • $\begingroup$ @Jim Humphreys: I have edited according to your suggestions. I'm mainly interested in char 0. Here my definition of "Coxeter element" depends on choice of simple roots, because otherwise I cannot get the correct number. Also the $\lambda=\mu$ case is not permitted as I said $\lambda-\mu$ is in the interior of positive root cone. $\endgroup$ – Jingren Chi Jul 18 '17 at 18:21
  • $\begingroup$ The number of Coxeter elements is bounded above by $2^{r-1}$. Could this also be a lower bound for the multiplicities? It seems like $2^{r-1}$ is a lower bound for the Kostant partition function of $\lambda-\mu$ once we know that all simple roots appear in its simple root expansion, but on the face of it that isn't quite enough. $\endgroup$ – Hugh Thomas Jul 18 '17 at 19:57
  • $\begingroup$ @HughThomas Could you briefly indicate why number of Coxeter elements is bounded above by $2^{r-1}$? In type A, $2^{r-1}$ seems correct lower bound, not sure for general case. I feel that the Kostant multiplicity formula is inconvenient for such bounds since it's an alternating sum. $\endgroup$ – Jingren Chi Jul 19 '17 at 2:37
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    $\begingroup$ Oops, actually $2^{r-1}$ simply is the number of Coxeter elements. A choice of Coxeter element requires precisely that you specify, for each pair of non-commuting simple reflections, which one comes first. (The order of commuting reflections doesn't matter since they commute.) There are $r-1$ such pairs, and since the edges of the Dynkin diagram form a tree, any choice of orders is obviously realizable. Eg for A_3, the four choices are $s_1s_2s_3$, $s_3s_2s_1$, $s_2s_1s_3=s_2s_3s_1$, $s_1s_3s_2=s_3s_1s_2$. In the last of these, for example, we chose $s_2$ to come after $s_1$ and $s_3$. $\endgroup$ – Hugh Thomas Jul 19 '17 at 2:54
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I have a full proof in type $A$, and most of a proof in the other types. Notation: $\alpha_1$, $\alpha_2$, ..., $\alpha_n$ are the (positive) simple roots, $\Phi^{+}$ is the set of all positive roots, $\rho$ is determined by the condition $\langle \alpha^{\vee}_i, \rho \rangle =1$ for all $i$. The element $\sigma$ is defined as $\sum_i \alpha_i$

I'll write $\alpha \geq \beta$ to indicate that $\alpha -\beta = \sum c_i \alpha_i$ for $c_i \in \mathbb{Z}_{\geq 0}$. So the condition that $\lambda-\mu$ is in the interior of the positive cone and occurs in $V_{\lambda}$ states that $\mu \leq \lambda - \sigma$. Our strategy is to show that $$K_{\lambda \mu} \geq K_{\lambda (\lambda - \sigma)} = \# (\mbox{number Coxeter elements}).$$

The gap in this argument is to give a proof in all types of

The Annoying Lemma: Suppose that $\beta \geq \delta$ and both $\beta$ and $\delta$ are dominant. Then there is a sequence of dominant weights $\gamma_0$, $\gamma_1$, ..., $\gamma_N$ such that $\beta = \gamma_0$, $\gamma_N=\delta$ and $\gamma_i - \gamma_{i+1} \in \Phi^{+}$.

Note that the Annoying Lemma is false if we ask for $\gamma_i - \gamma_{i+1}$ to be a simple root. For example, write dominant $GL_3$ weights as partitions in the usual way and take $\beta = (3,2,1)$ and $\delta = (2,2,2)$. Then the Lemma is true, because $\beta - \delta = (1,0,-1)= \alpha_1 + \alpha_2$ is a positive root. However, either of the two sequences $(\beta, \beta-\alpha_1, \delta)$ or $(\beta, \beta-\alpha_2, \delta)$ has a non-dominant element in the middle (namely, $(2,3,1)$ and $(3,1,2)$ respectively.)

Proof of $K_{\lambda \mu} \geq K_{\lambda (\lambda - \sigma)}$ assuming the Annoying Lemma: We immediately reduce to the case that the Dynkin diagram is connected. Also, the $SL_2$ case is immediate, so we assume we are not in it. With those reductions made, $\rho-\sigma$ is dominant. Since $\lambda$ is regular dominant, we have that $\lambda - \rho$ is domininant, so $\lambda - \sigma$ is dominant. We may therefore apply the Annoying Lemma to obtain a chain $\gamma_0 = \lambda - \sigma$, $\gamma_1$, $\gamma_2$, ..., $\gamma_N = \mu$ where $\gamma_i - \gamma_{i+1} \in \Phi^+$.

It is enough to show that $K_{\lambda \gamma_{i+1}} \geq K_{\lambda \gamma_{i}}$. Restrict to the $SL_2$ corresponding to $\pm (\gamma_i - \gamma_{i+1})$. Then the weights $\gamma_i$ and $\gamma_{i+1}$ lie on the same $SL_2$ string, and the condition that they are both dominant says that $\gamma_{i+1}$ lies nearer the center than $\gamma_i$, so $K_{\lambda \gamma_{i+1}} \geq K_{\lambda \gamma_{i}}$ as desired. $\square$

Proof that $K_{\lambda (\lambda - \sigma)} = \#(\mbox{number of Coxeter elements})$.

We recall the BGG resolution $$0 \leftarrow V_{\lambda} \leftarrow M_{\lambda} \leftarrow \bigoplus_{i} M_{\lambda - (\langle \alpha_i^{\vee}, \lambda \rangle +1) \alpha_i} \leftarrow \cdots$$ where $M_{\kappa}$ is the Verma module with highest weight $\kappa$. Since $\lambda$ is regular dominant, $\langle \alpha_i^{\vee}, \lambda \rangle \geq 1$ and thus $\lambda - \sigma = \lambda - \sum_j \alpha_j$ is not a weight of $M_{\lambda - (\langle \alpha_i^{\vee}, \lambda \rangle +1) \alpha_i}$. So the multiplicity of $\lambda - \sigma$ in $V_{\lambda}$ is the same as in $M_{\lambda}$; that is to say, it is the multiplicity of $-\sigma$ as a weight of the universal enveloping algebra of $\mathfrak{n}_-$.

Let $e_1$, $e_2$, ..., $e_n$ be the Chevalley generators of $\mathfrak{n}_-$. Then $U(\mathfrak{n}_-)$ is generated by the $e_i$ modulo the Chevalley-Serre relations. We see that a monomial in the $e_i$ is of degree $\sigma$ if and only if it uses each $e_i$ once, and the only Serre relations in such low degree are those of the form $e_i e_j = e_j e_i$ when $A_{ij} =0$. So the multiplicity of $-\sigma$ as a weight of $U(\mathfrak{n}_+)$ is the number of permutations of $e_1$, ..., $e_n$, up to interchanging commuting elements; this is precisely the description of the Coxeter elements. (It is also not hard to directly prove, using for example PBW bases, that the number is $2^{n-1}$, but I thought this was more fun.)

The annoying Lemma in Type A This paper by Matthew Fayers asserts the Annoying Lemma in type A as Proposition 2.3 but leaves the proof to the reader. Prop 1.2 of this paper is the same application that I intended -- showing that if $\mu \geq \nu$ and $\mu$ and $\nu$ are both dominant then $K_{\lambda \mu} \leq K_{\lambda \nu}$. I have completed Fayer's exercise but am delaying posting the solution in hopes that I'll find a less messy one that works for all types.

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  • $\begingroup$ It would be easy to edit this argument to be much more combinatorial, at least in type A, but my impression is that you like representation theory better than combinatorics. $\endgroup$ – David E Speyer Jul 20 '17 at 1:10
  • $\begingroup$ One combinatorial way to get a handle on the annoying lemma- to find from a given dominant weight $\sum c_i \omega_i$ all the $\alpha \in \Phi^{+}$ I can add and stay dominant: write the $c_i$ on the Dynkin diagram; look at all connected subdiagrams $I$; if each vertex not in $I$ has a label at least as big as the number of arrows coming into it, then I can add the highest root or highest short root corresponding to the parabolic $I$, which has the effect of subtracting one from the end of each arrow leaving $I$. $\endgroup$ – Sam Hopkins Jul 20 '17 at 1:20
  • $\begingroup$ Ok, sorry, of course that cannot give you all the possible moves (there are e.g. too many positive roots in $G_2$) but at least gives you a lot of them? $\endgroup$ – Sam Hopkins Jul 20 '17 at 1:32
  • $\begingroup$ Thanks for this enlightening answer. I just realized that $2^{n-1}$ is the number of ways of writing $\sigma$ as sum of positive roots, but your argument is also interesting. I would also enjoy down-to-earth combinatorial arguments. $\endgroup$ – Jingren Chi Jul 20 '17 at 5:06
  • $\begingroup$ @HCCH Right. In terms of PBW bases, each way of writing $\sigma = \beta_1 + \beta_2 + \cdots + \beta_r$ gives a PBW basis element $e_{\beta_1} e_{\beta_2} \cdots e_{\beta_r}$ in $U(\mathfrak{n}_+)$ of weight $\sigma$. $\endgroup$ – David E Speyer Jul 20 '17 at 9:56
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Though I can't answer the question directly, it may be helpful to clarify some of the issues here. First, it's usually best to focus on the case when $W$ (or the root system of the Lie algebra) is irreducible. For any irreducible finite reflection group (such as $W$), Coxeter found a remarkable conjugacy class consisting of elements now called Coxeter elements (of order $h$ equal to the unique largest degree of the $W$-invariant polynomials in the reflection representation of $W$. These are just products in any order of the simple reflections in $W$ (for any simple system). Recall that the order of $W$ is just the product of the degrees, which in the example $W=S_n$ are $2, \dots, n$, the Lie rank being $n-1$.

What is the number of Coxeter elements in $W$? By basic finite group theory, this number is the index in $W$ of the centralizer of any fixed Coxeter element $c$ (the size of the conjugacy class of $c$). In turn, the centralizer is just the cyclic subgroup of $W$ generated by $c$, whose order is $h$. This assertion is worked out more generally for all "regular" elements in a finite real or complex reflection group, in a 1974 (Invent. Math. 25) paper by T.A. Springer here: see especially 4.2 and 4.4.

For example, when $W = S_n$ there are $(n-1)!$ Coxeter elements.

Concerning finite dimensional irreducible representations of a simple Lie algebra (say in characteristic 0), these are nicely parametrized by the dominant weights. But it's nontrivial as a rule to work out the explicit weight multiplicities, though it's enough to do this for dominant weights (and then apply $W$-conjugacy). Typically computer computations rely on some version of Freudenthal's recursive formula. While Kostant's formula is more explicit (and equivalent to Weyl's character formula), it isn't usually useful for computations.

[ADDED] To make the details and references more explicit, I've written up some notes here (comments welcome).

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  • $\begingroup$ Are you sure that all conjugates of a given Coxeter element are Coxeter elements themselves? This seems false in type $A_3$. Maybe I misunderstand your definition. $\endgroup$ – John Shareshian Jul 19 '17 at 15:52
  • $\begingroup$ Note that the OP is using a different definition of Coxeter element from the one in this answer. $\endgroup$ – Hugh Thomas Jul 19 '17 at 17:15
  • $\begingroup$ Sorry, I didn't read the comments above. Everything is clear after reading them. $\endgroup$ – John Shareshian Jul 19 '17 at 18:07
  • $\begingroup$ Thanks for your clarification and references. I apologize for the confusion of terminology. The number in my lower bound only counts those Coxeter elements that can be written as products of a fixed set of simple reflections, so it's much smaller than the number of all Coxeter elements, for $S_n$, it's $2^{n-2}$ vs $(n-1)!$. $\endgroup$ – Jingren Chi Jul 20 '17 at 5:13
  • $\begingroup$ @HCCH: To clarify further, there is just one definition of "Coxeter element". Usually a Coxeter system includes specification of the set of simple reflections, but in fact it doesn't matter whether you use another such generating set since all of these are conjugate. (This isn't emphasized in the literature, but I think an elementary argument takes care of it.) The crucial fact is that all Coxeter elements are conjugate, and thus the number of them is just the order of $W$ divided by the order $h$ of the centralizer of one such element. $\endgroup$ – Jim Humphreys Jul 20 '17 at 15:21

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