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Let us assume we have a definition of the tangent space (e.g. as in Proof: Tangent space of the general linear group is the set of all squared matrices). Furthermore, we already verified that the tangent space $\mathfrak{gl}(n)$ of the general linear group $GL(n)$ -- the group of invertible matrices -- is the set of all square matrices $\mathbb{R}^{n\times n}$.

Afterwards, we introduce the matrix exponential: \begin{equation} \exp(\mathtt{X}):\mathbb{R}^{n\times n}\rightarrow \mathbb{R}^{n\times n},\quad \exp(\mathtt{X})= \sum_{k=0}^\infty\frac{\mathtt{X}^n}{n!}~. \end{equation} We can prove that $\exp(\mathtt{X})$ is invertible [indeed $\exp(\mathtt{X})^{-1}=\exp(\mathtt{-X})$], thus $\exp(\cdot)$ maps element from the tangent space $\mathfrak{gl}(n)$ to the general linear group $GL(n)$.

Now, we are interested in all subgroups of $GL(n)$, and we will simply call them matrix Lie groups $G\subset GL(n)$:

How can we show in general that $\exp(\cdot)$ maps elements from the tangent space $\mathfrak{g}$ to the corresponding matrix Lie group $G$?


Formal:

Let G be a subgroup of $GL(n)$ and $\mathfrak{g}$ be the tangent space of $G$.

Show: $x\in \mathfrak{g} \Rightarrow \exp(x)\in G.$

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You should correct the textual version of your question. In general the exponential will not be onto (it's image is connected). Depending on your source the question you are asking 'formally' is the definition of the Lie algebra (e.g. in Hall's book). The proof you are looking for is in chapter 3 of Warner's book on differentiable manifolds and Lie groups. –  user20266 Jan 16 '12 at 13:21
    
Sure, I was not intending to imply surjectivity. I'll correct/improve the question. I am looking for a formal, but naive proofs which does not use advanced concepts such as manifold and diffeomorphism, but restrict itself to Matrix Lie groups. –  Hauke Strasdat Jan 16 '12 at 13:45

1 Answer 1

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Let $\gamma$ be the integral curve in $G$ of the left invariant vector field associated to $X$, i.e. $V(Y) = YX$. This means $\gamma(0) = e$ and $\gamma'(t) = \gamma(t) X$ for $Y \in G$. Then $\gamma$ is unique (even as considered a curve in $GL(n)$) by theorems of ODE but it is easy to see that the curve $\exp(tX)$ satisfies the defining properties of $\gamma$ so must be $\gamma$. In particular, $\exp X \in G$.

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So, we can say that tangent spaces of matrix groups are in general defined by the curves $P(t)=\exp(t\mathtt{X})$, right? –  Hauke Strasdat Jan 16 '12 at 14:21
    
That's correct. –  Eric O. Korman Jan 16 '12 at 14:26

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