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.

Adjoint

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].

The rest of the lecture we covered thm 2.27 through cor. 2.33 ( which says matrix lie groups are lie groups).

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