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!


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 $G$.

  • $\begingroup$ Great, thanks very much! $\endgroup$ – CameronJWhitehead Jun 22 '15 at 21:10
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    $\begingroup$ 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

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