# Exponential of a derivation on a Lie algebra

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.

• 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 $e^\delta$. – Berci Mar 8 '20 at 23:47

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$$.