The set of $k$-dimensional complex-linear subspaces of $\mathbb{C^n}$ is denoted by
$G_k(\mathbb{C}^n)$. Show that $G_k(\mathbb{C}^n)$ has a unique smooth manifold structure making
it into a homogeneous $GL(n,\mathbb{C})$-space, with the action of $GL(n,\mathbb{C})$
induced from its usual action on $\mathbb{C}^n$. Show that $G_k(\mathbb{C}^n)$ is compact, and
determine its dimension.
I already solved this for real numbers could anybody help me solve this for complex numbers?
Best Answer
Define the homogeneous coordinate of $x\in G_k(\mathbb{C}^n)$ as
$$x=\left[\begin{array}{} x_{11} & x_{12} & \cdots & x_{1n}\\ x_{21} & x_{22} & \cdots & x_{2n}\\ \cdots & & \cdots & \\ x_{k1} & x_{k2} & \cdots & x_{kn} \end{array}\right] $$ where the matrix is of full rank $k$.
Next, we define an equivalence relation. Two element $x,y\in G_k(\mathbb{C}^n)$ are said to be equivalent if there exists $M\in GL(k, \mathbb{C})$, such that $Mx=y$. The quotient space under this equivalence is clearly $G_k(\mathbb{C}^n)$, since each element consists of $k$ row vectors that together span a $k$-dimensional subspace, and two set of vectors represent the same subspace if and only if they linearly represent each other.
Clearly $G_k(\mathbb{C}^n$), given in this manner, is a $GL(n, \mathbb{C})$-space. Let $x, \in G_k(\mathbb{C}^n), N\in GL(n, \mathbb{C})$, define $N(x)=xN$, where the RHS is matrix multiplication. Since $N$ is nonsingular, $xN$ is of full rank, hence in $G_k(\mathbb{C}^n)$.
To show that $G_k(\mathbb{C}^n)$ is compact, let $S=\{m\in M(k\times n, \mathbb{C}): \text{the rows are orthogonal to each other and of length 1}\}$. $S$ is closed and bounded, hence compact. Observe that every matrix $m\in S$ is of full rank. Hence the quotient space of $S$ is $G_k(\mathbb{C}^n)$, which is compact.
Finally, the dimension of $G_k(\mathbb{C}^n)$ is $k(n-k)$, because the system of equations $Mx=y$ has $k^2$ independent equations. There is a trick to observe this:
Suppose $\Sigma_1,\Sigma_2$ are two $k$-dimensional differentiable manifolds, $p\in \Sigma_1,q\in \Sigma_2$, the rows of $x$ is a basis of $T_p\Sigma_1$ and the rows of $y$ is a basis of $T_q\Sigma_2$. Then $v\mapsto Mv$ is a map from $T_p\Sigma_1$ to $T_q\Sigma_2$, which has $k^2$ degrees of freedom.
From what I have achieved so far, it seems there's no big change if we subsitute $\mathbb{C}$ with $\mathbb{R}$.