Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The problem: Compute $H_\ast (G(n,k); \mathbb{Z}/m\mathbb{Z})$.

Define $G(n,k)$ to be the space of $k$-dimensional vector subspaces of $\mathbb{R}^n$. Define the Schubert cell $e(m_1,...,m_s)$, so $m_j=m_j(\pi)=\inf\{m\mid \dim(\pi\cap \mathbb{R}^m)\geq j\}$ for $1\leq j\leq k$, where $\pi$ is a $k$-plane in $e(m_1,...,m_s)$. Then the dimension of $e(m_1,...,m_s)$ is $\sum_i m_i-i$.

The boundary of a Schubert cell has the form $\partial e(m_1,...,m_s)=\sum_i n_i e(m_1,...,m_{i-1},m_i-1,...,m_s)$ where $m_{i-1}<m_i-1$, and $n_i$ is the incidence number.

Let $D_i$ be the unit $m_i-1$-dimensional disk. Let $d$ be a Schubert cell belonging to the boundary, and since $\dim(\pi\cap \mathbb{R}^{m_i})=i$ we have that $d$ is spanned by $s$ unit vectors, $(u_1,...,u_s)$. Only one of the $u_i$ may belong to the boundary of the corresponding $D_i$ -- otherwise, since $\dim(\pi\cap \mathbb{R}^{m_i-1})\geq i$ and $\dim(\pi\cap\mathbb{R}^{m_j-1})\geq j$, we would have that $m_i(\pi)\leq m_i-i$ and $m_j(\pi)\leq m_j-j$, which gives a too-low dimension for this boundary component. Therefore every incidence number is either 2 or 0 (or -2, if you're mindful of orientations).

This means that if $m$ is odd, $H_i(G(n,k);\mathbb{Z}_m)=0$ for $k(n-k)>i>0$, $\mathbb{Z}_m$ for $i=0$, and either $\mathbb{Z}_m$ or $0$ at $i=k(n-k)$ depending on the parity of $k(n-k)$. If $m=2$, $H_i(G(n,k);\mathbb{Z}_2)=\mathbb{Z}_2^{N_i(n,k)}$, where $N_i(n,k)$ is the number of Schubert cells of dimension $i$ in $G(n,k)$.

The general case of $m$ even eludes me, however. This problem was assigned before we covered the universal coefficient theorem, so I would like to solve it without using the theorem. It seems as though this would require a way to count the number of Schubert cells in each boundary with coefficient $\pm2$, call it $\mu$, and then the homology would be $\mathbb{Z}_2^\mu\oplus\mathbb{Z}_m^{N_i-\mu}$. If I am handed a concrete Schubert cell, I can of course write down its boundary, but have not been able to write down a more precise/general solution, and would appreciate a hint in this direction.

share|cite|improve this question

I haven't explicitly worked out the answer yet (and will not do so soon), but a likely route seems related to Littlewood-Richardson numbers. It is known that the intersection of two cells $e(\lambda)e(\mu)=\sum c^\nu_{\lambda\mu}e(\nu)$.

Moreover, my professor gave me the following hint (well, it's more of an answer than a hint): if one considers the Young tableaux corresponding to a Schubert cell, an element of the boundary of this cell is a Young tableaux with one box removed (one which can be removed). From consideration of the explicit inclusion map of $D^m=e(...)\rightarrow G(n,k)$ (I do not want to write it out, but it can be found in J. T Schwartz's book Differential Geometry and Topology), it becomes evident that removing a box on an even-numbered diagonal of the Young tableaux will result in an incidence number of 2, and removing a box on an odd-numbered diagonal will result in an incidence number of 0. Then one can compute the (co)homology with coefficients in whatever $\mathbb{Z}_m$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.