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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.

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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 $\mathbb{Z}_m$.

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