Schubert classes and the cohomology ring of the Grassmannian

Now that we have defined our Schubert varieties, we wish to compute their intersection. The handy fact here is that their intersection corresponds to the cup product of certain classes in the cohomology ring of the Grassmannian.

We first take a look at the homology $\newcommand{\PP}{\mathbb{P}}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\ZZ}{\mathbb{Z}}
\DeclareMathOperator{\Gr}{Gr}
\DeclareMathOperator{\Fl}{Fl}
\DeclareMathOperator{\GL}{GL}H_\ast(\Gr^n(\CC^m))$. Fix a flag and consider the resulting CW complex structure as above. Since we are working over $\CC$, we only have cellular chains in even degrees, and so the homology is equal to the chain groups in even degrees and is $0$ in odd degrees. In particular, the Schubert varieties $\Omega_\lambda$ determine a unique homology class $[\Omega_\lambda]$, as they are elements of some chain group in cellular homology.

Since $\GL_n$ acts transitively on complete flags and sends $\Omega_\lambda$ for one flag to $\Omega_\lambda$ for another, it is not hard to see that each $\Omega_\lambda$ will determine the same homology class independent of the flag.

Now, by Poincaré duality, the homology class $[\Omega_\lambda]\in H_{2k}(\Gr^n(\CC^m))$ corresponds to a unique cohomology class in $H^{2nr-2k}(\Gr^n(\CC^m))$. This too is independent of the choice of flag, so we simply write $\sigma_\lambda=[\Omega_\lambda]$. We call $\sigma_\lambda$ a Schubert class.

It is known (see Fulton) that in a CW complex structure in which $X_{2k}\setminus X_{2(k-1)}$ is a disjoint union of open cells, as it is in this case, the cohomology classes of the closures of these open cells form a basis for the cohomology. Thus the $\sigma_\lambda$ generate $H^\ast(\Gr^n(\CC^m))$.

Finally, in nice cases the cup product in cohomology corresponds to intersection of the closed subvarieties defining them. This is true of the Schubert varieties for generic flags, i.e. for most choices of flags $F_\bullet$ and $E_\bullet$, $$\sigma_\lambda\cdot\sigma_\mu=\sum [Y_i]$$ where $Y_i$ are the irreducible components of $\Omega_\lambda(E_\bullet)\cap \Omega_\mu(F_\bullet)$.

“It holds generically”

To make the notion of genericity more precise, we define the complete flag variety $\Fl(\CC^m)$ to be the collection of all complete flags in $\CC^m$. We can view its elements as $m\times m$ full-rank matrices by thinking of the first $i$ vectors as spanning the $i$th flag. Then using similar reasoning to the row equivalence in the Grassmannian case, we find that the matrices defining a complete flag are equivalent up to the action of $B$, the group of upper triangular matrices in $\GL_n$.

Therefore, $\Fl(\CC^m)\cong \GL_n(\CC)/B$, which naturally has the structure of an algebraic variety.

Finally, we say that a property holds for a “generic” collection of flags if it holds for all tuples of flags in some (nonempty, dense) Zariski open subset of the product variety $$\Fl(\CC^m)\times \Fl(\CC^m)\times \cdots\times \Fl(\CC^m).$$

The Littlewood-Richardson rule

Since the $\sigma_\lambda$’s generate $H^\ast(\Gr^n(\CC^m))$, we can express the product of two Schubert classes as a sum of Schubert classes. The LR rule gives a formula for their coefficients.

We first introduce some notation and terminology. Given two partitions $\lambda$ and $\nu$ with the Young diagram of $\lambda$ contained in that of $\nu$, we define $\nu/\lambda$ to be the skew shape formed by removing $\lambda$’s boxes from $\nu$. A semistandard Young tableau (SSYT) of shape $\nu/\lambda$ is a way of filling these boxes with positive integers so that the numbers are weakly increasing across rows and strictly increasing down columns. We say the SSYT has content $\mu$ if there are $\mu_i$ boxes labeled $i$ in the tableau for each $i$. The reading word of the tableau is the word formed by reading the entries in each row from left to right, starting with the bottom row and working towards the top. A word is lattice if every suffix has at least as many $i$’s as $i+1$’s for all $i$.

The following example shows a semistandard Young tableau of shape $\nu/\lambda$ where $\lambda=(2,2)$ and $\nu=(4,3,1)$. Its reading word is $1211$, which is lattice. Its content is $\mu=(3,1)$.

In his book, Fulton gives a full proof of this rule, by first proving the following special case.

The Pieri rule is not hard to prove using some basic linear algebra (see Fulton, section 9.4), but the Littlewood-Richardson rule is much harder. For this, we turn to the hammer of symmetric function theory.