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

\begin{matrix} (1) & det (e^{X}) = e^{tr (X)} . \end{matrix}

Exponent of adjoint¶

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

Π : G \to GL (V)

where

V

is a finite-dimensional complex vector space.

Let

g

be a Lie algebra . Then a finite-dimensional complex representation of

g

is a Lie algebra homomorphism

π : g \to gl (V)

where

V

is a finite-dimensional complex vector space.

Adjoint representation¶

Let

G

be a matrix Lie group with Lie algebra

g

. For each

A \in G

, define a linear map

Ad A : g \to g

by the formula

Ad A (X) = A X A^{- 1} .

The map

Ad : G \to GL (g)

is called the adjoint representation of

G

Let

g

be a Lie algebra. For each

X \in g

, define a linear map

ad X : g \to g

by the formula

ad X (Y) = [X, Y],

where

[X, Y]

is the Lie bracket . The map

ad : g \to gl (g)

is called the adjoint representation of

g

Connection between representations¶

Let

G

be a matrix Lie group with Lie algebra

g

, and let

Π

be a (finite-dimensional real or complex) representation of

G

acting on the space

V

. Then there exists a unique representation

π

g

acting on the same space such that

Π (e^{X}) = e^{π (X)}

for all

X \in g

. The representation

π

can be computed as

π (X) = \frac{d}{d t} |_{t = 0} Π (e^{t X})

and satisfies

π (A X A^{- 1}) = Π (A) π (X) Π (A)^{- 1}

for all

X \in g

and all

A \in G

Connection between adjoint representations¶

Here is the remarkable result announced in the beginning.

For all

X \in g

\begin{matrix} (2) & Ad (e^{X}) = e^{ad X} . \end{matrix}

Conclusion¶

It would be interesting to know if there is a connection between $(1)$ and $(2)$ . 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.

Boris Belousov

Determinant of exponential and exponent of adjoint