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I think I've to show it's path connected, but can't figure out the path functions explicitly. Can anyone give these path maps?

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  • $\begingroup$ the intuition is that almost all matrices are inversibles $\endgroup$
    – Denis
    Oct 15, 2014 at 11:28
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    $\begingroup$ @Denis Well that intuition is also true for $GL(n,\mathbb R)$, which is not connected. $\endgroup$ Oct 15, 2014 at 11:32

1 Answer 1

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I'll give two proofs, a naive one, and a fast one which relies on a special fact about certain Lie groups:

Proof 1 It's not hard to find a path $B(t)$ in $GL(n, \mathbb{C})$ that connects an arbitrary Jordan-form matrix $J$ with nonzero eigenvalues to the identity matrix $I$: For a diagonal matrix $J$ with entries $\lambda_a$, one can simply pick any paths $\gamma_a$ in $\mathbb{C} - \{0\}$ (which in particular is path-connected) such that $$\gamma_a(0) = 1 \qquad\text{and}\qquad \gamma_a(1) = \lambda,$$ so that $$B(t) = \text{diag}(\gamma_1(t), \ldots, \gamma_n(t))$$ satisfies $$B(0) = I \qquad\text{and}\qquad B(1) = J.$$ Note that $\det B(t) = \prod \gamma_a(t) \neq 0$, so the image of $B(t)$ is indeed contained in $GL(n, \mathbb{C})$.

If $J$ is not diagonal, one can form the path $B(t)$ that (a) as above connects $I$ to the matrix $J'$ with the same diagonal entries as $J$ but zeros elsewhere, and then (b) connects $J'$ to $J$ by a line segment in $M(n, \mathbb{C})$. The determinant is constant on the latter segment, and so again $B(t)$ is contained in $GL(n, \mathbb{C})$.

Now given any matrix $A \in GL(n, \mathbb{C})$, form its Jordan decomposition $A = P^{-1} J P$, and take $B(t)$ to be a curve as above. Then, the curve $$\Gamma: t \mapsto P^{-1} B(t) P$$ satisfies $$\Gamma(0) = P^{-1} B(0) P = P^{-1} I P = I\\ \Gamma(1) = P^{-1}B(1) P = P^{-1}JP = A.$$ So, every point in $GL(n, \mathbb{C})$ can be connected to $I$ with a path, and hence the space is path-connected.

Proof 2 One can show that the exponential map $\exp: \mathfrak{gl}(n, \mathbb{C}) \to GL(n, \mathbb{C})$ is surjective. So, for any $A \in GL(n, \mathbb{C})$ there is an element $a \in \mathfrak{gl}(n, \mathbb{C})$ such that $\exp a = A$. Then, the curve $\Gamma: t \mapsto \exp(at)$ satisfies $\Gamma(0) = \exp(a \cdot 0) = \exp 0 = I$ and $\gamma(1) = \exp (a \cdot 1) = \exp a = A$.

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