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Is the matrix exponential $exp:M_n(\mathbb C) \to GL_n(\mathbb C)$ injective? Can it be that $e^A=I$ where $A$ is not the zero matrix?

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Consider $n=1$. Is $e^z$ injective on $\mathbb C$? –  Jonas Meyer Jan 9 '12 at 7:50
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Slightly less trivial is that it is not injective on $M_n({\mathbb R})$ for $n \ge 2$: consider $\exp\left( \pmatrix{0 & t\cr -t & 0\cr}\right)$ –  Robert Israel Jan 9 '12 at 7:57
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up vote 8 down vote accepted

Well, let me answer this in the obvious way. If $n=1$ then this is asking whether $\exp:\mathbb{C}\to\mathbb{C}^\times$ is injective. Once you see how to prove this isn't the case, you can generalized by using the fact that

$$\exp(\text{diag}(\lambda_1,\cdots,\lambda_n))=\text{diag}(\exp(\lambda_1),\cdots,\exp(\lambda_n))$$

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All the nice Lie Group elements look like $g=\exp(iA)$, for some "angle" $A$. Lets here think of finite dimensional and compact ones, which are connected to the unit. The $i$ is somewhat redundant, this is physicists notation to make $A$ have real eigenvalues if $g$ is unitary.

A good example would be the rotation group $SO(\mathbb{R},n)$ in any dimension. In three dimensions it's obvious that any rotation around $360°$ is represented by the unit element $I$.

In one dimension, you have $SO(2)$ or equivalently $U(1)$ with elements $e^{i\varphi}$ acting on the circle $S$ and for $\varphi=2\pi$ you get $e^{i\varphi}=1$. Alex Youcis's example acts on the cross product of circles $S^n$, with for example $S^2$ topologically being a torus.

If the group is compact as a manifold, then I think it's pretty reasonable that if you follow specific trajectories, you might come back to the unit.

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