4 Linear Lie groups and their Lie algebras 4.4 Lie algebras 4.6 Topological properties

4.5 Lie group and Lie algebra homomorphisms

Definition 4.32.

A Lie group homomorphism $\phi:G\to G^{\prime}$ between two linear Lie groups $G$ and $G^{\prime}$ is a continuous group homomorphism.

An isomorphism is a bijective Lie group homomorphism whose inverse is also continuous.

Remark 4.33.

In fact, a continuous homorphism between linear Lie groups is automatically a smooth map of smooth manifolds, and if it is bijective then the inverse is automatically continuous.

Definition 4.34.

A homomorphism $\varphi:\mathfrak{g}\to\mathfrak{h}$ of Lie algebras is an $\mathbb{R}$ -linear map such that

\varphi([X,Y])=[\varphi(X),\varphi(Y)]

for all $X,Y\in\mathfrak{g}$ .

An isomorphism is an invertible homomorphism.

Definition 4.35.

Let $\phi:G\to G^{\prime}$ be a Lie group homomorphism. Define the derivative (or differential or derived homomorphism)

D\phi:\mathfrak{g}\to\mathfrak{g}^{\prime}

D\phi(X)=\left.\frac{d}{dt}\phi(\exp(tX))\right|_{t=0}

for $X\in\mathfrak{g}$ . ³³ 3 We can justify taking the derivative by appealing to Remark 4.17.

Remark 4.36.

In fact, $D\phi$ is the derivative of $\phi$ at the identity in the sense of smooth manifolds; recall that $\mathfrak{g}$ and $gf^{\prime}$ are the tangent spaces to $G$ and $G^{\prime}$ at the identity.

Theorem 4.37.

Let $\phi:G\rightarrow G^{\prime}$ be a Lie group homomorphism with derivative $D\phi$ . Then

1.

The following diagram commutes:

$\begin{CD}\mathfrak{g}@>{D\phi}>{}>\mathfrak{g}^{\prime}\\ @V{\exp}V{}V@V{}V{\exp}V\\ G@>{}>{\phi}>G^{\prime}.\end{CD}$

That is, for $X\in\mathfrak{g}$ we have

$\phi(\exp(X))=\exp(D\phi(X)).$
2.

For all $g\in G,X\in\mathfrak{g}$ ,

$D\phi(gXg^{-1})=\phi(g)D\phi(X)\phi(g)^{-1}.$
3.

The map $D\phi:\mathfrak{g}\to\mathfrak{g}^{\prime}$ is a Lie algebra homomorphism.

Proof.

1.

Consider the one parameter subgroup $f:\mathbb{R}\to G^{\prime}$ defined by $f(t)=\phi(\exp(tX))$ . By construction, $f^{\prime}(0)=D\phi(X)$ . By Proposition 4.18, one parameter subgroups are determined by their derivative at 0, so that we must have

$\phi(\exp(tX))=f(t)=\exp(tD\phi(X)).$

We have

	$\displaystyle D\phi(gXg^{-1})$	$\displaystyle=\left.\frac{d}{dt}\phi(\exp(tgXg^{-1}))\right\|_{t=0}$
		$\displaystyle=\left.\frac{d}{dt}\phi(g\exp(tX)g^{-1})\right\|_{t=0}$
		$\displaystyle=\left.\frac{d}{dt}\phi(g)\phi(\exp(tX))\phi(g^{-1})\right\|_{t=0}$
		$\displaystyle=\phi(g)\left.\frac{d}{dt}\phi(\exp(tX))\right\|_{t=0}\phi(g^{-1})$
		$\displaystyle=\phi(g)D\phi(X)\phi(g)^{-1},$

as claimed.

To show that $D\phi$ is a Lie algebra homomorphism, we need to show that

•

$D\phi$ is $\mathbb{R}$ -linear; and
•

$D\phi([X,Y])=[D\phi(X),D\phi(Y)]$ .

So let $X,Y\in\mathfrak{g}$ and $s\in\mathbb{R}$ . By definition,

	$\displaystyle D\phi(sX)$	$\displaystyle=\left.\frac{d}{dt}\phi(\exp(tsX))\right\|_{t=0}.$
If we now set $\mu=st$ , we can rewrite this as:
	$\displaystyle\left.\frac{d}{dt}\phi(\exp(tsX))\right\|_{t=0}$	$\displaystyle=s\left.\frac{d}{d\mu}\phi(\exp(\mu X))\right\|_{\mu=0}$
		$\displaystyle=sD\phi(X).$

So $D\phi$ commutes with scalar multiplication. For additivity, we have

D\phi(X+Y)=\left.\frac{d}{dt}\phi(\exp(t(X+Y)))\right|_{t=0}.

On the other hand, by Corollary 4.14 and using part (1)

	$\displaystyle\phi(\exp(t(X+Y)))$	$\displaystyle=\lim\limits_{k\to\infty}\phi\left(\left(\exp\left(\frac{t}{k}X% \right)\exp\left(\frac{t}{k}Y\right)\right)^{k}\right)$
		$\displaystyle=\lim\limits_{k\to\infty}\phi\left(\exp\left(\frac{t}{k}X\right)% \right)\phi\left(\exp\left(\frac{t}{k}Y\right)\right)^{k}$
		$\displaystyle=\lim\limits_{k\to\infty}\left(\exp\left(\frac{t}{k}D\phi(X)% \right)\exp\left(\frac{t}{k}D\phi(Y)\right)\right)^{k}$
		$\displaystyle=\exp(t(D\phi(X)+D\phi(Y))).$

Taking the derivative at $t=0$ , we conclude that

D\phi(X+Y)=D\phi(X)+D\phi(Y),

showing additivity.

Finally we show that $D\phi$ respects the Lie bracket. Let $X,Y\in G$ . By parts (1) and (2) we have

D\phi\left(\exp(-tY)X\exp(tY)\right)=\exp(-tD\phi(Y))D\phi(X)\exp(tD\phi(Y)).

Take the derivative for both sides at $t=0$ : the derivative of the RHS is $[D\phi(X),D\phi(Y)]$ , and the derivative of the LHS is $D\phi([X,Y])$ (as $D\phi$ is linear).∎

Definition 4.38.

Suppose that $G$ and $G^{\prime}$ are complex Lie groups and $\phi:G\rightarrow G^{\prime}$ is a homomorphism. Then $\phi$ is holomorphic if $D\phi:\mathfrak{g}\to\mathfrak{g}^{\prime}$ is $\mathbb{C}$ -linear.

(This implies that $\phi$ is a holomorphic map of complex manifolds.)

Example 4.39.

The map $\det:\operatorname{GL}_{2,\mathbb{C}}\rightarrow\operatorname{GL}_{1,\mathbb{C}}$ is holomorphic. The map $\operatorname{GL}_{2,\mathbb{C}}\rightarrow\operatorname{GL}_{2,\mathbb{C}}$ sending $g$ to $\overline{g}$ is not holomorphic.

Back to the real case. We have shown that the differential $\phi\mapsto D\phi$ gives a map

D:\operatorname{Hom}(G,G^{\prime})\longrightarrow\operatorname{Hom}(\mathfrak{% g},\mathfrak{g}^{\prime}).

This raises two natural questions:

1.

Is the map injective? Does the derivative $D\phi$ uniquely determine the Lie group homomorphism $\phi$ ?
2.

Is the map surjective? Or in other words, does every Lie algebra homomorphism $\varphi$ ’exponentiate’ (or ’lift’) to a Lie group homomorphism $\phi$ such that $D\phi=\varphi$ ? We say ’exponentiate’ since if yes, then $\phi$ would need to satisfy $\phi(\exp(X))=\exp(\varphi(X))$ . So this gives a formula for $\phi$ , at least on the image of $\exp$ . The question is whether this is well-defined (the exponential map is neither injective nor surjective in general) and whether this defines a homomorphism.

The answer to these questions is actually of topological nature, which we discuss in the next subsection.

	$\displaystyle D\phi(gXg^{-1})$	$\displaystyle=\left.\frac{d}{dt}\phi(\exp(tgXg^{-1}))\right\|_{t=0}$
		$\displaystyle=\left.\frac{d}{dt}\phi(g\exp(tX)g^{-1})\right\|_{t=0}$
		$\displaystyle=\left.\frac{d}{dt}\phi(g)\phi(\exp(tX))\phi(g^{-1})\right\|_{t=0}$
		$\displaystyle=\phi(g)\left.\frac{d}{dt}\phi(\exp(tX))\right\|_{t=0}\phi(g^{-1})$
		$\displaystyle=\phi(g)D\phi(X)\phi(g)^{-1},$