# equivalence of Lie group and Lie algebra intertwiner

I encountered this problem while working on my research. Let $G$ be a Lie group, and consider an intertwiner of the complex representations (possibly infinite-dimensional) $$\pi:G\rightarrow \mathrm{GL}(V),\quad \pi':G\rightarrow \mathrm{GL}(V'),$$ that is, a linear map $T:V\rightarrow V'$ such that $$T\circ\pi(g)=\pi'(g)\circ T,\quad \forall g \in G.$$ By differentiating on both sides, we see that $T$ is an intertwiner for the associated Lie algebra representations as well, since $$T\circ\pi_*(X)=\pi_*'(X)\circ T,\quad \forall X \in \mathfrak{g}.$$

Suppose now that, given $\pi$ and $\pi'$, only the Lie algebra condition is given, i.e., $T$ is an intertwiner for $\pi_*$ and $\pi_*'$. Will it be an intertwiner for the group representations?

The only reference I could find on this says (without proof) that an intertwiner of Lie algebra representations that can be integrated to representations of the groups is also an intertwiner for the group representations. If this is is indeed true, how would one prove it?

## migrated from mathoverflow.netJun 30 '15 at 22:11

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## 1 Answer

It's enough if the group is connected; a finite group is a Lie group, and this definitely won't work in that case. The Lie algebra condition implies that the map commutes with the exponential of any Lie algebra element, and thus with the subgroup generated by these. Since the group is connected, that's the whole group.

EDIT: Even if you are working in a situation where there's no guarantee the exponential map exists, you can still make this work. The other way to think about this is that for any $v\in V,u\in (V')^*$, then you need that $\langle u,(\pi'(g)\circ T-T\circ \pi(g))v\rangle=0$ (I'm assuming you're working on, say, a complex Hilbert space). Now, take the differential of this function on g, and identify $TG\cong G\times \mathfrak{g}$ using right invariant vector fields. In this case, the differential is constant, with value given by $\langle u,(\pi'_*(X)\circ T-T\circ \pi_*(X))v\rangle\colon T_gG\cong \mathfrak{g}\to \mathbb{C}$. Being a Lie algebra representation is the fact that the above map is 0, so this function is constant. It's obviously 0 at the identity, so it's 0 everywhere.

• Isn't more than just being connected needed? $\mathrm{SL}(2,\mathbb{R})$ is connected, but $\exp:\mathfrak{sl}(2,\mathbb{R})\rightarrow \mathrm{SL}(2,\mathbb{R})$ is not surjective. Moreover, is it so obvious that the map commutes with the intertwiner in the case $V$ is infinite-dimensional? what if $\pi_*$ is unbounded? – Giuseppe Sellaroli Jul 1 '15 at 0:26
• @GiuseppeSellaroli I didn't say the exponential map was surjective, I said it generated the group (as I explained here: math.stackexchange.com/questions/1335619/…). But if you commute with a set of generators of the group, you commute with the whole group. – Ben Webster Jul 1 '15 at 1:22