Determinant of exponential and exponent of adjoint
27 Dec 2016
Lie theory is beautiful. In this post,
we’ll take a brief look at two marvelous equalities arising from the theory.
Determinant of exponential
Let $X$ be an $n \times n$ real or complex matrix. Then
$$
\newcommand{\me}{\mathrm{e}}
\newcommand{\md}{\mathrm{d}}
\newcommand{\at}[2][]{#1|_{#2}}
\DeclareMathOperator{\tr}{tr}
\newcommand{\GL}{\mathrm{GL}}
\newcommand{\gl}{\mathrm{gl}}
\newcommand{\bC}{\mathbb{C}}
\newcommand{\kg}{\mathfrak{g}}
\DeclareMathOperator{\Ad}{Ad}
\DeclareMathOperator{\ad}{ad}
\begin{equation}
\label{det}
\det \left( \me^X \right) = \me^{\tr \left( X \right)}.
\end{equation}
$$
Exponent of adjoint
We need a couple of definitions to appreciate the main result.
Representation
Adjoint representation
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 $A \in G$.
Connection between adjoint representations
Here is the remarkable result announced in the beginning.
For all $X \in \kg$,
$$
\begin{equation}
\label{adj}
\Ad \left( \me^X \right) = \me^{\ad X}.
\end{equation}
$$
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.