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Let $\Sigma=G/P$ be a flag variety of type $A$,i.e. $G=SL_{n+1}$ and is a stabilizer of the partial flag $0 \subset V_{d_1}\subset \cdots\subset V_{d_r}\subset V_{n+1}$ of length $r$. Let $W$ be its Weyl group which is the symmetric group $S_n$. Then $W_P=<\sigma_i: i \notin \{d_1,\cdots,d_r\}> $ the subgroup of $W$ corresponding to $P$. Let $W^P=W/W_P$ be the set of minimal coset representatives. It is know that the set $W^P$ parametrizes the Schubert varieties in $\Sigma$.

Also recall, that if $\omega_k=\sum_{i=1}^k e_i$ denote the dominant fundamental weights of the weight lattice of $G$ then $\Sigma$ is a projective subvariety of $\mathbb P V_\lambda$. Here $V_\lambda$ is the highest weight representation with of $G$ with highest weight $$\lambda=\sum_{i=j}^r\omega_{d_j}$$.

$\textbf{Question}$: Does there exist a formula to calculate the number of elements in $W^P$ in terms of the above data or in terms of some kind of tableau associated to the highest weight representation $V_\lambda$.

To make the question more clear, we know that for the Grassmannian $Gr(k,n)$ the number of cosets $W^P$ are $n\choose{k}$. Moreover they are the number of partitions $\mu$ in $k \times (n-k)$ boxes of type $$n-k\ge \mu_1\ge \cdots \ge \mu_k\ge 0. $$ Does there exist similar descriptions in the case of general $G/P$? If yes, can we calculate the number of coset representatives of various lengths in $W^P$ combinatorially.

The answer to same question in types $B, C, D$ will indeed be a bonus.

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    $\begingroup$ The number of elements in $W^P$ is the number of elements in $W$ divided by the number of elements in $W_P$. $\endgroup$ Commented Jul 18, 2017 at 20:02
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    $\begingroup$ As Hugh points out, you just need to know the order of $W$ and the orders of its various "parabolic" subgroups $W_P$, which are all known rather explicitly (cf. Bourbaki etc.). The answer is similar for any root system and its Weyl group, but of course it's more combinatorial-looking for type $A$: here a parabolic subgroup is a Young subgroup of the symmetric group, i.e., a product of smaller symmetric groups. $\endgroup$ Commented Jul 18, 2017 at 20:27
  • $\begingroup$ @JimHumphreys I couldn't find that in the Bourbaki of Lie Algebras CH4-6 (English version, don't understand French). Could you please provide me with a more precise reference? Moreover, I have further made my question clearer by adding more details. Thanks $\endgroup$
    – MIQ
    Commented Jul 19, 2017 at 9:15
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    $\begingroup$ I don't know why you think tableaux should be involved, since they don't appear in the Grassmannian case that you seem to already understand. If you want an analogue of "partitions in a $k$ by $(n-k)$ rectangle" for general G/P, you can replace each element of $W^P$ by its set of inversions in the positive root poset. In the case of Grassmannian permutations, this is exactly giving you the partitions you expect. $\endgroup$
    – Oliver
    Commented Jul 19, 2017 at 14:00
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    $\begingroup$ @MIQ: The orders of irreducible Weyl groups are well-known and given case-by-case near the end of the Bourbaki volume, as well as in the Wikipedia article linked, and in my books, etc. In turn, a proper parabolic subgroup is just a direct product of such Weyl groups, so the quotient giving the number of cosets is easy to calculate. (This can be done for any finite reflection group and is unrelated to representation theory.) $\endgroup$ Commented Jul 19, 2017 at 17:21

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Every standard parabolic subgroup can be encoded by crossing out some nodes in the Dynkin diagram of $\mathfrak{g}$. The subgroup $W_P$ is then the Weyl group of a semisimple Lie algebra whose Dynkin diagram is that of $\mathfrak{g}$ with the corssed nodes deleted. By the Lagrange theorem, the number of cosets (and hence the number of coset representative) is then just $|W|/|W_P|$ and this can be computed directly using formulas for number of elements of $W$ for any simple Lie algebra $\mathfrak{g}$. See e.g. Wikipedia. So for example for Grassmannian you get that $W_P$ is the Weyl group of a Lie algebra isomorphic to $\mathfrak{sl}(k)\oplus\mathfrak{sl}(n-k)$ and thus it has $k! (n-k)!$ elements.

Description in terms of tableaus or some such is a trickier business because if done properly, it should actually count not only $|W^P|$ but elements of $W^P$ of certain length, i.e. the Betti numbers of $\Sigma$. I think that's a difficult problem in general.

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