3
$\begingroup$

I'm working through Fulton and Harris's Representation Theory, and I'm stuck on Exercise 8.27.

I'm trying to show that if $A$ is an algebra and $G$ is the Lie group of algebra automorphisms (interpreted as a subgroup of $GL(A)$) then the Lie algebra associated to $G$ is the algebra of derivations on $A$ ($Der(A)$).

I've shown that if $\gamma : [0,1] \to G$ is some path (writing $\gamma(t)$ as $\gamma_t$) with $\gamma_0 = id$ and $\gamma'_0 = X$ then just by differentiating with respect to $t$ $$ \gamma_t(ab) = \gamma_t(a)\gamma_t(b) $$ we get $$ \gamma'_t(ab) = \gamma'_t(a)\gamma_t(b) + \gamma_t(a)\gamma'_t(b), $$ which at $t=0$ gives $$ X(ab) = X(a)b + aX(b), $$ so $X$ is a derivation. But I can't show that any derivation must be a tangent to the identity, so that the associated Lie algebra to $G$ is actually $Der(A)$.

Any help with this would be great, thanks!

$\endgroup$
1
$\begingroup$

I guess this should work. Take $X \in \text{Der}(A)$ and consider $\gamma_t = \exp(tX)$. $\gamma$ is a curve on $GL(A)$ such that $\gamma_0=\text{Id}$ and $\gamma'_0=X$. We want to show that $\gamma$ is actually $G$-valued. Remark that $\gamma'_t=X \circ \gamma_t=\gamma_t \circ X$.

Let's fix $a,b \in A$ and consider $\phi : t \mapsto \gamma_t(ab)$ and $\psi : t \mapsto \gamma_t(a)\gamma_t(b)$.

Deriving $\phi$, we get : $$\begin{array}{rcl} \phi'(t) & = & \gamma'_t(ab)\\ & = & X\circ\gamma_t(ab) \\ & = & X \left[ \phi(t) \right] \end{array}$$

In the same way : $$\begin{array}{rcl} \psi'(t) & = & \gamma'_t(a)\gamma_t(b)+\gamma_t(a)\gamma'_t(b)\\ & = & X\left[\gamma_t(a) \right] \gamma_t(b) + \gamma_t(a) X \left[ \gamma_t(b) \right] \\ & = & X \left[ \gamma_t(a)\gamma_t(b) \right] \; \; \; \; \; \text{as} \; X \; \text{is a derivation}\\ & = & X \left[ \psi(t) \right] \end{array}$$

Then, $\phi$ and $\psi$ satisfy the same ODE. Yet, $\phi(0)=\psi(0)=ab$, hence $\phi=\psi$.

Thus, for all $a,b \in A$, $\gamma_t(ab)=\gamma_t(a)\gamma_t(b)$, i.e. $\gamma_t$ is in $G$.

$\endgroup$
3
  • $\begingroup$ Great, thanks very much! $\endgroup$ – CameronJWhitehead Jun 22 '15 at 21:10
  • 1
    $\begingroup$ Quite often for Lie groups, when you know the tangent vector at the identity and you're looking for the curve in the group, the exponential is often a good candidate. $\endgroup$ – Sylvain L. Jun 22 '15 at 21:14
  • $\begingroup$ Thanks, this is the sort of thing I'm struggling with! $\endgroup$ – CameronJWhitehead Jun 22 '15 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.