# Lie Groups Lecture 6

This lecture (lie6) was given on 9/15/09

### Excerpts

Given matrix lie groups and a matrix lie group homo., $\Phi \colon G \to H$ there exists a unique map $\mathfrak{g} \to \mathfrak{h}$ that is basically a lie algebra homomorphism compatible with $\Phi$, meaning

1. $\Phi(e^X) = e^\phi(X)$
2. $\phi(AXA^{-1}) = \Phi(A)\phi(X)\Phi(A)^{-1}$
3. $\phi([X,Y]) = [\phi(X),\phi(Y)]$
4. $\phi(X) = \frac{d}{dt} \Phi(e^{tX})|_{t = 0}$

here is how to construct $\phi$: let $K = graph(\Phi) \subset G \times H$, it can be checked that its a lie group whose lie algebra is a subset of $\mathfrak{g} \times \mathfrak{h}$, so any $X \in Lie(K)$ can be written as $X = X_g + X_h$.  Define $\phi$ via $\phi(X_g) = X_h$.  Alternatively, using the result that every one parameter subgroup is $\exp tX$ we have $\Phi(\exp tX)$ is a one parameter subgoup hence $\Phi(tX) = \exp tZ$, so define $\phi(X) = Z$, the desire properties can be checked.

A consequence of the lie product formula is that $Ad \colon G \to GL(\mathfrak{g})$ defined by $A \mapsto A( - )A^{-1} = Ad_A$ is a lie group homomorphism, so by the above result there is $ad \colon \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ such that $Ad_{e^X} = e^{ad_X}$.  It can be checked, using 4 above, that $ad(X).Y = ad_X Y = [X,Y]$.