how to compute the Euler characters of a Grassmannian?

Question

Let $G(n,m)$ be the Grassmannian of all n-dim subspaces of an m-dim vector space over $\mathbb{C}$. How to compute the Euler characters of $G(n,m)$? For example, $G(1, 2)$ is $\mathbb{C}P^1$ which is $S^2$. So the Euler characters of $G(1,2)$ is $2$. But how to compute $G(n,m)$ in general? Thank you.

Answer 1

The Euler characteristic for real Grassmannians is discussed at the wikipedia page here.

The complex case can be handled similarly.

EDIT (To include details in the complex case):

Let $\chi_{n,m}=\chi(G(n,m))$. Then the recursion relation is $\chi_{n,m}=\chi_{n-1,m-1}+\chi_{n,m-1}$ and we know that $\chi_{0,m}=1$ for all $m$.

Claim: $\chi_{n,m}=\binom{m}{n}$.

This claim can be verified by (double) induction. The inductive step boils down to verifying that $\binom{m-1}{n-1}+\binom{m-1}{n}=\binom{m}{n}$.

Note: This gives the same result as considering row-echelon forms of matrices corresponding to Schubert cells, as Jyrki Lahtonen points out.

Answer 2

I may be wrong, but isn't this much like the calculation of the homology of the complex projective space? The cells will occur only in even dimensions, so the boundary maps are all trivial. This time a useful coordinate system is obtained by representing a subspace with an $n\times m$ complex matrix in the reduced row echelon form (=leading entry equal to $1$, only zeros on top of a leading one, et cetera). For example in the case of $G(2,4)$ the cells consist of matrices of the form $$ \left(\begin{array}{cccc} 1&0&*&*\\ 0&1&*&* \end{array}\right), $$ $$ \left(\begin{array}{cccc} 1&*&0&*\\ 0&0&1&* \end{array}\right), $$ $$ \left(\begin{array}{cccc} 1&*&*&0\\ 0&0&0&1 \end{array}\right), $$ $$ \left(\begin{array}{cccc} 0&1&0&*\\ 0&0&1&* \end{array}\right), $$ $$ \left(\begin{array}{cccc} 0&1&*&0\\ 0&0&0&1 \end{array}\right), $$ and $$ \left(\begin{array}{cccc} 0&0&1&0\\ 0&0&0&1 \end{array}\right). $$ Each asterisk (*) stands for an unkown complex number, so these are $r$-cells, with $r=$ 8,6,4,4,2,0 respectively. Thus the Euler characteristic of $G(2,4)$ would be $\chi=6$.

As a reality check we note that the complex projective space $P^m$ has cells consisting of vectors of length $m+1$ with $k$ ( $0\le k\le m$) leading zeros, followed by a single leading $1$, followed by $m-k$ asterisks. We get one cell in all the even dimensions in the range from $0$ to $2m$ as we should.

It should be easy to generalize this. The cells in $G(n,m)$ are fully determined by the increasing sequence of positions of the $n$ leading $1$s. I take it you know in how many ways we can choose these.

Answer 3

This is a computation based on a straightforward application of the the Atiyah-Bott localization theorem.

Using the version given for example in the paper by Aleksey Zinger:

Given a torus acting smoothly on a on a manifold with isolated fixed points, then the Euler characteristic is equal to thethe number of the fixed points of its action.

The Grassmannian $Gr(n,m)$ can be identified with the space of n-dimensional orthogonal projectors in $\mathbb{R}^m$. The projectors can be represented by Hermitian $m \times m $ projection matrices of rank $n$. The unitary roup $U(m)$ acts by conjugation on the set of projectors. The fixed points of the subgroup $\mathbb{T}^m$ of diagonal unitary matrices are the diagonal $m \times m $ projection matrices of rank $n$. Since these matrices have $n$ units and $m-n$ zeros along the diagonal. Their number is equal to the number of distinct ways we can arrange $n$ units and $m-n$ zeros along the diagonal which is equal to: $\binom{m}{n}$.

Answer 4

Along the lines of David Bar-Moshe's answer above: Let $Gr_\mathbb{C}(k,n)$ denote the Grassmanian of k-dimensional subspaces of $\mathbb{C}^n$. Then $Gr(k,n)$ admits a natural action of $GL(n,\mathbb{C})$, along with an action of $\mathbb{T}$, diagonal $n\times n$ matrices over $\mathbb{C}$ which are non-singular. Now the only $k$-dimensional subspaces of $\mathbb{C}^n$ (i.e., points of the Grassmanian) fixed by $\mathbb{T}$ are those which admit a basis which I'll write as $\{e_I\}$ for $I \subset \{1, 2, \dots, n\}$ with $|I|=k$. That is, $\chi(Gr_\mathbb{C}(k,n)) = \binom{n}{k}$.