MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The complex cohomology H^* of the manifold of flags in C^n is a quotient of C[x1,...,xn] by the ideal generated by symmetric polynomials with no constant term. In particular it has an action of the symmetric group, by permuting the variables x. If you ignore the grading on H^*, this is the regular representation of S_n.

For a fixed i < n(n-1)/2, is there a simple rule that says how the character of S_n on H^2i decomposes into irreducible characters?

share|cite|improve this question
It would help to add more tags, though I'm not sure what works best here. Maybe ag.algebraic-geometry, co.combinatorics, – Jim Humphreys Aug 23 '13 at 22:46
up vote 5 down vote accepted

I'm unsure if this would be termed a 'simple rule' but perhaps this will be of some use: the Kostka-Foulkes polynomials $K_{\lambda\mu}(q)$ ($\lambda,\mu$ partitions of $n$) determine the change of basis matrix in $\Lambda(q)$, the one variable ring of symmetric functions, between the Hall-Littlewood polynomials and the Schur polynomials (described in Macdonald's 'Symmetric Functions and Hall Polynomials', for example). Garsia-Procesi proved (Adv. in Math, 94, p.82-138) that the Kostka-Foulkes polynomials determine the graded decomposition of the cohomology of of Springer fibres

$$X_{\lambda} = \{ (F_{\cdot})\in Fl(n)\;|\; U_{\lambda}(F_{i})\subset F_{i-1}\}$$

where $U_{\lambda}$ is a nilpotent matrix of Jordan type $\lambda$: they show that

$$K_{\mu\lambda}(q) = q^{n(\lambda)}\sum_{i} \langle H^{2i}(X_{\lambda},\mathbb{Q}), \chi^{\mu}\rangle q^{-i}.$$

Here $\chi^{\mu}$ is the character of the Specht module associated to the partition $\mu$, $\langle P, Q\rangle$ is the multiplicity of an irreducible $S_{n}$-module $Q$ appearing in an $S_{n}$-module $P$ and $n(\lambda)$ is the (complex) dimension of $X_{\lambda}$ (see Garsia-Procesi, formula (I.9)).

In earlier work, Lascoux and Schutzenberger proved that the Kostka-Foulkes polynomials $K_{\mu\lambda}(q)$ can be determined using the charge statistic $c$: this associates to a tableaux of shape $\mu$ and weight $\lambda$ some non-negative integer - this is described in Macdonald's book, p.242. They showed that

$$K_{\mu\lambda}(q) = \sum_{T\in SSYT(\lambda;\mu)} q^{c(T)}$$

where $SSYT(\lambda;\mu)$ is the set of semistandard Young tableau of shape $\mu$ and weight $\lambda$

In your case, we have $\lambda=(1^{n})$, so that the charge statistic associates to a standard tableau $T$ some integer $c(T)$. Thus, to determine the multiplicity of $\chi^{\mu}$ in $H^{2i}(Fl(n))$ you can determine the coefficient of $q^{n(n-1)/2-i}$ in $K_{\mu(1^{n})}(q)$ using the charge statistic; in short, the multiplicity of $\chi^{\mu}$ in $H^{2i}(Fl(n))$ equals the number of standard tableau $T$ of shape $\mu$ with charge $c(T)=n(n-1)/2 -i$.

share|cite|improve this answer
It's may be worth mentioning that Springer's ideas apply to all Weyl groups and flag varieties of simple algebraic groups over algebraically closed fields (of good characteristic, automatic in type A). However, the combinatorics for type A is better behaved and has a more classical flavor as you point out. – Jim Humphreys Aug 23 '13 at 20:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.