# Injectivity of Lie group exponential function

If $G$ is a (finite-dimensional) Lie group, then the exponential function $\exp\colon\mathfrak{g}\to G$ is injective on some identity neighbourhood. If, moreover, $\mathfrak{g}$ is semi-simple and $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}$ is a Cartan-decomposition, then the restriction of $\exp$ to $\mathfrak{p}$ is a diffeomorphism onto its closed image. An example thereof is $G=SL_n(\mathbb{R})$, where $\mathfrak{k}$ are the skew-symmetric matrices $\mathfrak{p}$ are the symmetric traceless matrices. Then $\exp$ is certainly injective on a neighbourhood of the form $U\times V$ with $U\subseteq\mathfrak{k}$ and $V\subseteq\mathfrak{p}$ open.

My Question is now: does there exist some open $W\subseteq\mathfrak{k}$ such that $\exp$ is injective on $W\times \mathfrak{p}$? From the way the Cartan decomposition is built I would expect that one can achieve injectivity on an open neighbourhood of $0\times\mathfrak{p}$ (although I cannot write down a precise argument for this at the moment), so the question is whether this open neighbourhood is globally of product type.

• In your question, what is $G$? any finite-dimensional Lie group? Any semisimple? $SL_n(\mathbb{R})$? Also I don't understand what you mean by "the restriction to $\mathfrak{p}$ is a diffeomorphism", since it is certainly not surjective. Do you mean injective? immersion? closed immersion?... – YCor Jan 24 '14 at 22:25
• $G$ is any semi-simple finite-dimensional Lie group. $\exp$ is a diffeomorphism onto a closed submanifold of $G$ (in the case of $SL_n(\mathbb{R})$ we have that $\exp$ maps symmetric matrices to positive definite symmetric matices). – Christoph Wockel Jan 25 '14 at 8:00
• OK. Have you tried $SL_2(\mathbf{R})$? – YCor Jan 26 '14 at 21:55
• I tried $SL_n(\mathbb{R})$ in general, but didn't come to any conclusion. Do you think there is something special about $SL_2(\mathbb{R})$? – Christoph Wockel Jan 28 '14 at 7:25
• I don't expect anything special, it's just a test case: most likely $SL_2$ is easier, and if the result fails for $SL_2$ then it will certainly fail for $SL_n$. – YCor Jan 28 '14 at 7:54