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?



How does the grading on the cohomology of a flag variety break up the regular representation of W?
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 $c(T)=n(n-1)/2 -i$.


