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I am having a difficult time showing that the exponential map $\mathrm{exp}: \mathfrak{sl}(2, \mathbb{R}) \rightarrow \mathrm{SL}(2, \mathbb{R})$ is not surjective. I have, however, worked out that $\mathfrak{sl}(2, \mathbb{R})$ is given by $\{A \in M(n, \mathbb{R}) \mid \mathrm{trace}(A) = 0\}$.

I will be appreciative of any help. Thank you

Mariano Suárez-Álvarez
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Susan
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2 Answers2

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Can you find a matrix $X\in\mathfrak{sl}(2,\mathbb R)$ such that $\exp X=\left(\begin{smallmatrix}-1&1\\0&-1\end{smallmatrix}\right)$?

Suppose we have such a matrix $X$. Since $\exp X$ is not diagonalizable, $X$ is also not diagonalizable, and it must have a double eigenvalue. Since it has trace zero, that eigenvalue is necessarily zero. Now, if $\lambda$ is an eigenvalue of $X$, $\exp(\lambda)$ is then an eigenvalue of $\left(\begin{smallmatrix}-1&1\\0&-1\end{smallmatrix}\right)$, so that $\lambda=\pi i+2\pi i k$ for some $k\in\mathbb Z$. Yet this cannot be zero!

Mariano Suárez-Álvarez
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  • Why should X have a double eigenvalue? If the eigenvalues are $\lambda_1$ and $\lambda_2$, then exp(X) has eigenvalues $e^{\lambda_1}$ and $e^{\lambda_2}$ they have to be both = -1 which is the double eigenvalue of exp(X), but two different complex numbers $\pi i +2\pi i k_1$ and $\pi i + 2\pi i k_2 $may accomplish this. What am I doing wrong? why should $k_1 =k_2$ ? – some_math_guy Sep 26 '23 at 10:28
  • I am missing what implies that there is a double eigenvalue? By double eigenvalue I take that you mean that it's an eigenvalue with algebraic multiplicity = 2. It seems like it's "X cannot be diagonalized" $\implies$ X has a double eigenvalue ", but why? Is it valid only for 2x2 matrices? – some_math_guy Sep 26 '23 at 16:30
  • I am asking if a 2x2 matrix with two distinct eigenvalues is always diagonalizable, in other words – some_math_guy Sep 26 '23 at 16:36
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The same answer as Mariano's, but with a different explanation:

Let $J=\pmatrix{-1&1\\ 0&-1}$. We claim that there does not exist a real matrix $A$ (traceless or not) such that $\exp(A)=J$. Suppose the contrary. Then $JA=(A+A^2+\frac{A^3}{2!}+\ldots)=AJ$. In general, every matrix that commutes with an upper triangular Jordan block must be an upper triangular Toeplitz matrix Therefore $A=\pmatrix{a&b\\ 0&a}$ for some $a,b\in\mathbb R$. In turn, $J$ has a repeated eigenvalue $e^a>0$, which is a contradiction.

user1551
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