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I am attending a course on elementary Lie algebra theory and I have trouble understanding something that we mentioned in class: If $\mathfrak{g}$ is a Lie-algebra and $\delta\in\text{Der}(\mathfrak{g})$ is a derivation, then $e^\delta$ is an automorphism of $\mathfrak{g}$ and this was left as an easy exercise.

There is something that I don't get here. How is $e^\delta$ defined? I guessed through $$\sum_{k=0}^\infty\frac{\delta^k}{k!}$$ but since we have not talked about any topology, I figured that this is not the case. What baffles me is that I cannot find any reference of this on any text I've been through. The most relevant thing I came across was exponential map of a Lie group, but we have not even talked about the notion of a Lie group.

Any help and any reference is greatly appreciated.

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  • $\begingroup$ A finite dimensional vector space induces a natural topology by (any) norm, so the infinite series of linear transformations does make sense, and it is actually the definition of $e^\delta$. $\endgroup$ – Berci Mar 8 '20 at 23:47
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I assume, $\def\g{\mathfrak g} \g$ is finite dimensional as a vector space. Then we can fix a norm on it, and consider the operator norm on ${\rm der}\,\g\subseteq{\rm end}_{\Bbb R}\, \g$, which allows to speak about convergence of sequences. Also, every linear or bilinear map must be continuous.
In particular the exponential series $e^\varphi:=\sum_{n=0}^\infty\frac{\varphi^n}{n!}\ $ is convergent for all linear transformations $\,\varphi\,$ on a finite dimensional vector space.

Now, the proof that $e^\delta$ preserves the Lie bracket, is analogous to the proof of $e^{a+b}=e^ae^b$ for numbers (or commuting linear transformations):

Hint: Observe that $\delta^n([a,b])=\displaystyle\sum_{i=0}^n{n \choose i}[\delta^ia,\,\delta^{n-i}b]$.

Thus we have $$ \sum_{n=0}^\infty\frac{\delta^n([a,b])}{n!}\ =\ \sum_{n=0}^\infty\sum_{i=0}^n\frac1{n!}{n\choose i}\, [\delta^ia,\,\delta^{n-i}b]\ =\\ =\ \sum_{i,j=0}^\infty \frac1{i!\cdot j!}\, [\delta^ia,\,\delta^jb]\ =\ \left[\sum_{i=0}^\infty \frac1{i!}\,\delta^ia,\sum_{j=0}^\infty \frac1{j!}\,\delta^jb \right]$$ because of bilinearity and continuity of the Lie bracket.

Finally, for invertibility of $e^\delta$, note that its eigenvalues are just $e^{\lambda_i}\ \ne 0$, where $\lambda_i$ are the eigenvalues of $\delta$.

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  • $\begingroup$ Thank you for a detailed, thorough and organized answer. +1 $\endgroup$ – JustDroppedIn Mar 9 '20 at 8:23

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