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It is well known that for $n=2$, this holds. The polar decomposition provides the topology of $\operatorname{SL}(n,\mathbb R)$ as the product of symmetric matrices and orthogonal matrices, which can be written as the product of exponentials of skew symmetric and symmetric traceless matrices. However I could not find out the proof that $\exp: \mathfrak{sl}(n,\mathbb R)\to\operatorname{SL}(n,\mathbb R)$ is not surjective for $n\geq 3$.

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Lie algebras are commonly written with lower case fraktur letters. Hence it should be \mathfrak{sl} instead of \text{sl}. –  kahen May 13 '13 at 20:21
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3 Answers 3

Over $\mathbb{R}$, for a general $n$, a real matrix has a real logarithm if and only if it is nonsingular and in its (complex) Jordan normal form, every Jordan block corresponding to a negative eigenvalue occurs an even number of times. So, you may verify that $\pmatrix{-1&1\\ 0&-1}$ (as given by rschwieb's answer) and $\operatorname{diag}(-2,-\frac12,1,\ldots,1)$ (as given by Gokler's answer) are not matrix exponentials of a real traceless matrix.

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I came up with an example. Consider diag(-2,-1/2, 1, ..., 1) in SL(n,R). Clearly the lattice of logarithm of this does not intersect sl(n,R).

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I remember learning somewhere that every matrix in the image of the exponential from $\mathfrak{sl}(n,\Bbb R)$ is either diagonalizable or is unipotent. So, any defective element which is not unipotent would be missed.

Defective matrices with determinant 1 can be built by installing a defective block. For example:

$$ \begin{bmatrix}-1&1\\0&-1\end{bmatrix},\begin{bmatrix}-1&1&0\\0&-1&0\\0&0&1\end{bmatrix},\begin{bmatrix}-1&1&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}\text,{etc.} $$

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Your second example has determinant $-1$. Also, what you describe in the first paragraph is true for $n=2$, but are you sure that it is true for a general $n$? –  user1551 May 14 '13 at 4:05
    
@user1551 The comment: Doh good point... I managed to forget about the determinant. I originally had in mind just to give the matrix a defective block, but yeah, what I wrote did not do that. The question: I don't remember it depending on $n$, I thought it depended upon nondiagonalizable elements of $\mathfrak{sl}(n,\Bbb R)$ being nilpotent. Do you know if that conclusion is confined to $n=2$? –  rschwieb May 14 '13 at 11:13
    
It doesn't seem to be true for a general $n$. Consider $R=\pmatrix{0&-1\\ 1&0}$, $e^R=\pmatrix{\cos1&-\sin1\\ \sin1&\cos1}$ and $X=\pmatrix{R&I\\ 0&R}$. We have $e^X=\pmatrix{e^R&e^R\\ 0&e^R}$, which is neither diagonalisable nor unipotent. –  user1551 May 14 '13 at 14:22
    
@user1551 OK, thanks for letting me know. I found the original place I learned this, and it was ambiguous about whether it was true in general. It looks like you still believe the example I gave, however, so I'll have to leave this solution in place for your other solution to reference it properly –  rschwieb May 14 '13 at 15:10
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