Why $GL(n+1,\mathbb{C})$ is compact?

Question

I'm trying to prove that:

The set of all lines in $\mathbb{C}^{n+1}$ ($\mathbb{C}\mathbb{P}(n)$) is a complex manifold.

I'm knowing that:

If a compact group $G$ acts on $X$ transitively and for one $e\in X$ we have $H=\{g\in G\mid g.e=e\}$ then $X$ is a complex manifold.

So I just need compactness of $GL(n+1,\mathbb{C})$.

Answer 1

There is a way to salvage your proof. The unitary group $$U(n+1) = \{ A \in GL(n+1,\mathbb{C}): A^* A = A A^* = I \}$$ is a compact group and acts transitively on $\mathbb{CP}^n$. (Here, $A^*$ denotes the conjugate transpose of $A$.) Recall that an $n \times n$ complex matrix is unitary if and only if its columns give an orthonormal basis of $\mathbb{C}^n$.

To show $U(n+1)$ is compact, show that is a closed subset of $\mathbb{C}^{(n+1)^2}$ and contained in the compact set $$\{A = (a_{ij}) \in \mathbb{C}^{(n+1)^2} : \sum_{i,j} a_{ij} a_{ij}^* \leq n + 1\}.$$

To see that it acts transitively on $\mathbb{CP}^n$, let $\ell_1$ and $\ell_2$ be two lines in $\mathbb{CP}^n$ spanned by the unit vectors $v_1$ and $v_2$, respectively. Complete each vector to an orthonormal basis of $\mathbb{C}^n$ using the Gram-Schmidt algorithm to produce unitary matrices $U_1$ and $U_2$ such that $v_i$ is the first column of $U_i$ for $i = 1,2$. Then the unitary matrix $U = U_2 U_1^{-1}$ sends $v_1$ to $v_2$ and hence $\ell_1$ to $\ell_2$.

Answer 2

Hint: For the first question, use $n+1$ charts $U_i:=\{x_i \neq 0\}$ of $\mathbb{P}^n$ where $[x_0: \cdots: x_n]$ is the homogeneous coordinates on $\mathbb{P}^n$ and $U_i \cong \mathbb{C}^n$ given by $[x_0: \cdots: x_n] \mapsto (x_0/x_i,\cdots,x_{i-1}/x_i, x_{i+1}/x_i, \cdots , x_n/x_i).$

Certainly, $GL_n(\mathbb{C})$ is not compact, as you can find a sequence of elements without any converging subsequences, or show that it is an unbounded (open) subset of $\mathbb{C}^{n^2}$ and the result follows from Heine-Borel theorem.

Answer 3

Hint for what you need:

The determinant map $\;\det\;$ is a continuous one and $\,\;GL(n+1,\Bbb C)=\det^{-1}(\Bbb C-\{0\})\,\;$ , so indeed as you were already said: the group of all invertible maps of order $\,n+1\,$ is an open set.

Answer 4

Hint take the invertible matrix $A_k=\mathrm{diag}(k,1,\ldots,1)$ and choose any norm $||.||$ for $\mathcal{M}_{n+1}(\mathbb{C})$, let $k\to\infty$, what about $||A_k||$?