# 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 $\mu=(3,1)$.

For any two partitions $\lambda$ and $\mu$ contained in the Important Box,

$$\sigma_\lambda\cdot\sigma_\mu =\sum c^\nu_{\lambda\mu} \sigma_\nu,$$ where the sum ranges over all $\nu$ in the Important Box, and $c^\nu_{\lambda\mu}$ is the number of semistandard Young tableaux of shape $\nu/\lambda$ having content $\mu$ and whose reading word is lattice.

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.

Hey Maria, thanks for posting this! I’ve been looking to learn a little bit about Schubert calculus, and this looks like a fantastic introduction – I’ll try to get around to reading it in detail this weekend.

Anyway – do you suggest any references for a systematic treatment of the subject? Preferably something that isn’t too technical (everything you use in your post would be fine, I think; I think I can fake an understanding of algebraic geometry up to a point).

Hey Carl, glad you found this useful! I have a few references scattered throughout the post, but I’d say the main one I learned from is Fulton’s book, “Young tableaux”. I’d say that one is pretty accessible for those who don’t have a strong background in algebraic geometry/topology, because he neatly pushes all the technical details about cohomology into the appendix. But then, the appendix is also detailed enough if you do want to understand those details.

I’ve also heard good things about the new book by Eisenbud and Harris, “3264 and all that.” It can be found online here: http://isites.harvard.edu/fs/docs/icb.topic720403.files/book.pdf

and it certainly goes into much more detail on the geometric side of things, because it’s building up a more general theory than just intersections of Schubert varieties. I plan to read (parts of) it at some point soon.

One combinatorial way to look into the Schubert decomposition is:

1.We associate a matrix to each k-subset as you said.

2.Then in each matrix consider the set of subsets of columns (numbered from left to right) which are bases.

3.Put a total order on such sets by considering the induce lexicographic order.

Each schubert cell then correspond to all subspaces with an specified subset of the columns being the maximal basis. So in the example of page 2 we want all matrices with the subset (2,4,7) being the biggest basis lexicographically. This is a complicated way of saying the same but now:

– If we consider ALL possible orders of the columns (schubert is very dependent on ordering from left to right) and intersect them all, in the common refinement we would be grouping subspaces by specifying the complete set of bases. This is the Matroid Strata decomposition. Bad thing is, Mnevs Universality theorem says that such cells can be “as bad as possible”, and while Im never sure of the precise meaning, it certainly means that working with them is much worse than working with the schubert cells which are just affine spaces.

-If we consider all cyclic orders and intersect them, the cells in the refinement are called the Positroid decomposition, and there are lots of combinatorics associated to them. In particular they have something called a Le diagram associated

-Lauren Williams and Kelli Talaska consider something similar, the Go diagrams, to give a refinement of the positroid decomposition. The open cells are not bad, but their closures and how the intersect was (or maybe still is) a bit mysterious.

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Hi Maria, thanks for the post and the link of the video lecture about Schubert calculus in your other post, they have been so helpful!

I have one question which might be stupid. When we want to find which Schubert cell a point of the Grassmannian belongs to, we cut off the upside-down staircase from the matrix, all entries in the staircase should be 0, right? Why must the first column of the matrix be a zero column?