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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.

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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.

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More explicitly, you are multiplying the class of the $k_1 \times (n_2-k_2)$ rectangle and the class of the $k_2 \times (n_1-k_1)$ rectangle, and discarding all terms that don't fit inside $k \times (n-k)$. I don't know that this Littlewood-Richardson product can be described in any way simpler than a general LR product. –  David Speyer Aug 28 '12 at 17:26
    
It was my intuition that it couldn't be made simpler. –  Will Sawin Aug 28 '12 at 17:44
    
Actually, the answer is much simpler than a general LR product. I'll write up the explanation below. –  David Speyer Aug 28 '12 at 17:44
    
Running low on time, and confused by some details, so not today. But I claim that every coefficient in the Schubert basis is either 0 or 1, and there is a simple rule for which is which. –  David Speyer Aug 28 '12 at 17:46
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In fact this simplification is described on wikipedia: en.wikipedia.org/wiki/… –  Will Sawin Aug 28 '12 at 18:02
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Here is a slightly different way to do this: the cohomology of $G(k,n)$ is generated by the Chern classes $c_1,\ldots,c_k$ of the universal bundle $\gamma (k,n)$ and the Chern classes $c'_1,\ldots,c'_{n-k}$ of the universal quotient bundle, subject to one relation,

$$(1+c_1+\cdots +c_k)(1+c_1'+\cdots +c'_{n-k})=1.$$

Using this we can express $c'_i$'s in terrms of $c_i$'s or vice versa.

Now if we embed $G(k_1,n_1)\times G(k_2,n_2)$ in $G(k,n)$ as above $\gamma(k,n)$ restricts to $p^*_1(\gamma(k_1,n_1))\oplus p^*_2(\gamma(k_2,n_2))$ where $p_1$ and $p_2$ are the projections to the first, respectively, second factor. This allows one to compute the cohomology map $H^*(G(k,n))\to H^*(G(k_1,n_1)\times G(k_2,n_2))$ induced by the embedding, and hence also the homology map in the opposite direction. This gives an explicit way of calculating the image of the fundamental class modulo the values of the Chern classes on the Schubert cells.

(If necessary I can add more details.)

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It's embarrassing to say this, but I wasn't be able to be convinced that that's the only relation. We have relations up to degree $2n$, but the dimension of Grassmannian is $2k(n−k)$. Where do relations of higher degrees come from? For instance, for $G(2,5)$, I couldn't get the obvious relation $c_1^7=0$ from $(1+c_1 +c_2)(1+c_1 ' +c_2 ' +c_3 ')=1$. What am I missing? –  Hwang Aug 29 '12 at 0:04
    
Hwang -- I don't know how to get specifically $c_1^7=0$ but you are quite right: it is not completely obvious that there are no other relations. This is explained e.g. in Bott-Tu, Differential forms in algebraic topology, Proposition 23.2. (Bott and Tu consider the real cohomology but the proof works for any coefficients.) See also my answer to this question: mathoverflow.net/questions/11614/grassmannian-bundle-theorem/… –  algori Aug 29 '12 at 1:42
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