Let $G$ be a simply connected algebraic group over $C$. We know that a representation of an algebraic group $$\phi : G \to GL(V)$$

induces a representation of its lie algebra (taking the differential of this map $\phi$) .

Now, if $G$ is simply connected and we are given a representation $$\psi : Lie(G) \to gl(W)$$

Is that true that this induces a representation of the group $G$ ? This fact comes up in a paper I am reading and I am not able to realise it.

Please help me in realising how the simple connectedness is used here to get the representation of the group itsef.


  • $\begingroup$ Are you assuming that the representation of the Lie algebra is finite dimensional? Are you only assuming $G$ to be simply connected, or also semisimple? $\endgroup$ – Tobias Kildetoft Feb 18 '16 at 7:07
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    $\begingroup$ I don't know much about algebraic groups, but this is a general fact about Lie groups: a lie algebra homomorphism can be lifted to the groups if the domain is simply connected. You can find a proof in Lee's smooth manifolds book or Warner's book. If you don't have access to those, I can write a proof, but again I'm not sure if this is exactly what you're looking for. $\endgroup$ – Tim kinsella Feb 18 '16 at 7:26
  • $\begingroup$ @Timkinsella I think the proof you are referring to will be really helpful to me. I would appreciate if you could give the proof here or atleast explicitly define the representation of the group induced by the representation of its lie algebra. Thank you ! $\endgroup$ – Jagdeep Singh Feb 18 '16 at 8:43
  • $\begingroup$ So, looking inJantzen, this certainly holds if $G$ is semisimple and the Lie algebra representation is locally finite (as a module for the enveloping algebra). But as far as I can tell, it should be more straightforward in characteristic $0$ (based on some remarks in the book). $\endgroup$ – Tobias Kildetoft Feb 18 '16 at 9:18
  • $\begingroup$ @TobiasKildetoft Lol, so evidently my suggestion was pretty naive! I will steer clear of the algebraic-groups tag :) $\endgroup$ – Tim kinsella Feb 18 '16 at 9:33

Let $G$ and $H$ be Lie groups, $G$ simply connected, and suppose a lie algebra homomorphism $\phi : \frak{g}\rightarrow \frak{h}$. The graph of $\phi$ is a lie sub algebra of $\frak{g}\oplus \frak{h}$. Let $L$ be the corresponding Lie subgroup in $G\times H$. The composition $L\subset G\times H \rightarrow G$ (where the second map is projection) is a Lie group homomorphism and a local diffeomorphism at $e_L$. It is therefore a covering map. Since $G$ is simply connected, the map is invertible: $G\rightarrow L$. Post-composing with the projection to $H$, we have lifted $\phi$.

  • $\begingroup$ I would appreciate if you could more clearly define the graph of the map $\phi$. $\endgroup$ – Jagdeep Singh Feb 18 '16 at 9:09
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    $\begingroup$ Oh sure. Its just $\{(g, \phi(g)):g\in \frak{g}\}$. It's a subalgebra because $\phi $ is an algebra homomorphism $\endgroup$ – Tim kinsella Feb 18 '16 at 9:10
  • $\begingroup$ @JagdeepSingh Note that while you requested this answer in a comment, it does not actually answer the stated question, as this is about Lie groups, rather than about algebraic groups (and there are some subtle differences). $\endgroup$ – Tobias Kildetoft Feb 18 '16 at 9:13
  • $\begingroup$ @TobiasKildetoft Okay. I'll try to move this to a comment, or what? $\endgroup$ – Tim kinsella Feb 18 '16 at 9:14
  • $\begingroup$ I think it looks fine as an answer, if accompanied by a note that this is for the analoguous situation for Lie groups. $\endgroup$ – Tobias Kildetoft Feb 18 '16 at 9:16

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