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.

  • 3
    $\begingroup$ Have you tried Schubert cells? $\endgroup$
    – Grigory M
    Jun 14, 2011 at 13:27
  • $\begingroup$ I think we can prove that a Grassmannian is a CW-Complex, then by definition.... $\endgroup$
    – user27177
    Mar 20, 2012 at 16:48

3 Answers 3


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.

  • 1
    $\begingroup$ I'd appreciate comments along with the down votes. $\endgroup$
    – wckronholm
    Jun 14, 2011 at 15:23
  • $\begingroup$ @Grigory M Ah! I now see you linked to the same wikipedia article with your comment. Should I delete my answer? $\endgroup$
    – wckronholm
    Jun 14, 2011 at 15:39
  • 2
    $\begingroup$ It would be cool if you generalized the recursion for C and gave the answer. $\endgroup$ Jun 14, 2011 at 15:47
  • $\begingroup$ I've just read new revision of the answer and decided to remove the downvote. $\endgroup$
    – Grigory M
    Jun 14, 2011 at 17:24

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.

  • $\begingroup$ Two comments: 1) Usually the stars come first in the matrix representations of the Schubert cells. This is so that the matrices themselves are in row echelon form. 2) The number of $(2r)$-cells in $G(n,m)$ is equal to the number of partitions of $r$ into at most $n$ integers, each of which is less than or equal to $m-n$. This is not so easy to compute for general $r$, $n$, and $m$. $\endgroup$
    – wckronholm
    Jun 14, 2011 at 15:04
  • $\begingroup$ @wckronholm: Thanks for filling me in about the usual convention. Agree that more work is needed to tally the $(2r)$-cells. $\endgroup$ Jun 14, 2011 at 15:13

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: $\binom{m}{n}$.

  • $\begingroup$ Unless I am mistaken, for $\chi(Gr(1, 2m+1))=\chi(\mathbb{R}P^{2m+1})=0$ despite your claim. I think the problem is that the compact manifolds are not orientable, which is needed to apply the theorem. But the general result is obtained easily from apply the same argument to the orientable double cover (which is just oriented $n$-dimensional subspaces). $\endgroup$
    – Pax
    Feb 9, 2015 at 20:50
  • $\begingroup$ @PaxKivimae The original question is about the complex grassmannian, which is always orientable. So I think where David writes $\mathbb{R}^m$ it should be $\mathbb{C}^m$ insted. Otherwise (I think) $U(m)$ wouldn't act on the real orthogonal projectors. $\endgroup$
    – hjhjhj57
    Aug 2, 2016 at 0:53

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