Complex Grassmannian as orthogonal projections ($P=P^{\ast }=P^2,\mathrm{tr}P=k$)

The complex Grassmannian of $k$-planes in ${\mathbb{C}}^n$ identifies with the displayed Hermitian orthogonal projection matrices. A plane is recovered as the image of its projection. Near a plane $E$, a neighboring plane is the graph of a complex linear map $Z:E\to E^{\perp }$. Its projection is $W(W^{\ast }W{)}^{-1}{W}^{\ast }$ with $W=(I,Z{)}^T$. The unitary group acts transitively by conjugation, and its tangent directions are $[A,P]$ with $A^{\ast }=-A$.

For real $c_1<\cdots <c_n$ and $C=\mathrm{diag}(c_1,\dots ,c_n)$, the critical points are the coordinate projections $P_I$, indexed by $k$-subsets $I=\{i_1<\cdots <i_k\}$. Indeed $df([A,P])=\mathrm{tr}([P,C]A)$, which vanishes for every skew-Hermitian $A$ exactly when $[P,C]=0$. In the graph chart at $P_I$,Every coefficient is nonzero, so the Hessian matrix is nondegenerate. Each pair $j<i$ supplies two negative real directions, giving the displayed Morse index. The function is a Morse function even if distinct critical points happen to have equal critical values.

Here $N_r$ counts $k$-subsets $i_1<\cdots <i_k$ with ${\sum }_a(i_a-a)=r$, equivalently partitions of an integer $r$ contained in a $k$ by $(n-k)$ rectangle. The diagonal trace Morse function on a complex Grassmannian has only even Morse indices. Its Morse chain complex therefore has zero differentials, proving the displayed torsion-free integral homology. For $\mathrm{Gr}(2,4)$ the even-degree ranks are $(1,1,2,1,1)$ in degrees $0,2,4,6,8$.