# Determinant of exponential and exponent of adjoint

Lie theory is beautiful. In this post, we’ll take a brief look at two marvelous equalities arising from the theory.

## Determinant of exponential


We need a couple of definitions to appreciate the main result.

### Representation

Let $G$ be a Lie group . Then a finite-dimensional complex representation of $G$ is a Lie group homomorphism $$\begin{equation*} \Pi : G \rightarrow \GL(V) \end{equation*}$$ where $V$ is a finite-dimensional complex vector space.
Let $\kg$ be a Lie algebra . Then a finite-dimensional complex representation of $\kg$ is a Lie algebra homomorphism $$\begin{equation*} \pi : \kg \rightarrow \gl(V) \end{equation*}$$ where $V$ is a finite-dimensional complex vector space.

Let $G$ be a matrix Lie group with Lie algebra $\kg$. For each $A \in G$, define a linear map $\Ad A : \kg \rightarrow \kg$ by the formula $$\begin{equation*} \Ad A(X) = AXA^{-1}. \end{equation*}$$ The map $$\begin{equation*} \Ad : G \rightarrow \GL(\kg) \end{equation*}$$ is called the adjoint representation of $G$.
Let $\kg$ be a Lie algebra. For each $X \in \kg$, define a linear map $\ad X : \kg \rightarrow \kg$ by the formula $$\begin{equation*} \ad X(Y) = \left[ X, Y \right], \end{equation*}$$ where $\left[ X, Y \right]$ is the Lie bracket . The map $$\begin{equation*} \ad : \kg \rightarrow \gl(\kg) \end{equation*}$$ is called the adjoint representation of $\kg$.

### Connection between representations

Let $G$ be a matrix Lie group with Lie algebra $\kg$, and let $\Pi$ be a (finite-dimensional real or complex) representation of $G$ acting on the space $V$. Then there exists a unique representation $\pi$ of $\kg$ acting on the same space such that $$\begin{equation*} \Pi \left( \me^X \right) = \me^{\pi \left( X \right)} \end{equation*}$$ for all $X \in \kg$. The representation $\pi$ can be computed as $$\begin{equation*} \pi(X) = \frac{\md}{\md t}\at[\bigg]{t=0} \Pi \left( \me^{tX} \right) \end{equation*}$$ and satisfies $$\begin{equation*} \pi\left( AXA^{-1} \right) = \Pi(A)\pi(X)\Pi(A)^{-1} \end{equation*}$$ for all $X \in \kg$ and all $A \in G$.

Here is the remarkable result announced in the beginning.

For all $X \in \kg$, $$$$\label{adj} \Ad \left( \me^X \right) = \me^{\ad X}.$$$$

## Conclusion

It would be interesting to know if there is a connection between \eqref{det} and \eqref{adj}. They look similar, but note that $\det$ is not a Lie group homomorphism because it is not invertible. Definitions and propositions presented here can be found in An Elementary Introduction to Groups and Representations.