Let $G(k,n)$ be the Grassmannian of complex $k$-planes in $\mathbb{C}^n$. Then for $k\_1+k\_2=k$ and $n\_1+n\_2=n$, $G(k\_1,n\_1)\times G(k\_2,n\_2)$ is a submanifold of $G(k,n)$. So the cohomology class of it should be written as a linear combination of Schubert classes. Is there a method to compute the coefficients?


All I know about it is that each Schubert class corresponds to a partiton(or Young diagram). If this question is trivial or well-known, please let me know what I should read.



product of two sub-Grassmannians
The subvariety of $k$-planes whose intersection with a fixed $n\_1$-plane has dimension at least $k\_1$ is the closure of a specific Schubert cell. This is the intersection of that subavariety with a similar one, the space of $k$-planes whose intersection with a fixed $n\_2$-plane has dimension at least $k\_2$.


If the intersection is transverse, then the class you want is just the cup product of those two classes.


It's clear that it is transverse. The tangent space is $Hom(\mathbb C^k,\mathbb C^n/\mathbb C^k)$ which we can break up into


$Hom(\mathbb C^{k\_1},\mathbb C^{n\_1}/\mathbb C^{k\_1})\oplus Hom(\mathbb C^{k\_1},\mathbb C^{n\_2}/\mathbb C^{k\_2})\oplus Hom(\mathbb C^{k\_2},\mathbb C^{n\_1}/\mathbb C^{k\_1}) \oplus Hom(\mathbb C^{k\_2},\mathbb C^{n\_2}/\mathbb C^{k\_2})$


The tangent space of the first variety is just where the second summand is zero. The tangent space of the second variety is where the third summand is zero. These do indeed intersect transversely.


Thus, you can find the cohomology class with the cup product formula.


