Representation of the lie algebra of a simply connected algebraic group $G$ induces a representation of the group itself

Question

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.

Thanks

Answer 1

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 $\phi$.

Answer 2

The existing answer by Tim kinsella explains why it induces a Lie group representation but does not address algebraicity.

Algebraicity is false without a robustness assumption such as that $G$ is semisimple. Take for example $G = \mathbb C$ and a $1$-dimensional Lie algebra representation $G \to \mathfrak{gl}_1(\mathbb C) \cong \mathbb C$ given by $z \mapsto z$. The induced Lie group homomorphism $\mathbb C \to \mathbb C^\times$ is the exponential map, which is not rational.

Since you ask about simple connectedness, this is needed even for the Lie group statement: there is an isomorphism $\mathfrak{pgl}_n(\mathbb C) \cong \mathfrak{sl}_n(\mathbb C)$ as Lie-algebras (there is a natural map from the latter to the former and it is an isomorphism) but it does not come from a representation $\operatorname{PGL}_n(\mathbb C) \to \operatorname{SL}_n(\mathbb C)$.

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Now for a positive answer: I find the book Lie groups and algebraic groups by Onishchik and Vinberg helpful for Lie group vs algebraic group questions.

Assume $G$ is a complex (linear, connected) algebraic group that is semisimple and simply connected. Connected means Zariski-connected, but it is connected for the Euclidean topology either way: A connected algebraic group over any field is irreducible, and the complex points of an irreducible complex variety are connected in the Euclidean topology (https://mathoverflow.net/a/12896/50929). I identify $G$ with its complex points.

Take a Lie algebra homomorphism $\mathfrak g \to \mathfrak{gl}(V)$. Simply connected refers to the Euclidean topology, so by topological considerations there is a Lie group lift $G \to \operatorname{GL}(V)$. Now we conclude using the following:

Fact. With $G$ connected semisimple and $H$ any complex algebraic group, any Lie group homomorphism $f : G \to H$ is algebraic.

Proof: we have a Lie group homomorphism $G \to G \times H$ given by $(1, f)$. Its image, call it $G'$, is a semisimple Lie group. It is therefore automatically closed; this follows from Proposition 7.9 in Knapp, Lie groups beyond an introduction and also from §1.4.2, Theorem 3 in Onishchik-Vinberg. Moreover, it is automatically a closed algebraic subgroup; this follows from §3.3.3 in Onishchik-Vinberg. Now the projection $G' \to G$ is algebraic, and a bijection, and therefore the inverse is algebraic so that $f$ is algebraic. (This last part of the argument comes from §3.3.4, Theorem 4 in Onishchik-Vinberg, if you want references.) $\square$