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Let $G$ be a lie group and $\mathfrak{g}$ its lie algebra. Consider the category $Rep(G)$ of finite dimensional representations of $G$ and the category $Rep(\mathfrak{g})$ of finite dimensional representations of $\mathfrak{g}$. Morphisms in those categories are just $G$- resp. $\mathfrak{g}$ equivariant linear maps. There is an obvious functor $$d\colon Rep(G)\rightarrow Rep(\mathfrak{g}),\pi\mapsto d_e\pi$$ which maps a representation of $G$ to a representation of $\mathfrak{g}$ by taking the derivative at the neutral element. On morphisms, $d$ is just the identity because $G$ equivariant maps are also $\mathfrak{g}$ equivariant, what can be seen by putting 1-parameter subgroups of $G$ in the defining definition of beeing $G$ equivariant and taking the derivative.

If $G$ is simply connected, $d$ is bijective on objects, because Lie group homomoprhism are in bijection to the morphisms of their lie algebras, if the domain is simply connected.

Is $d$ bijective on morphisms too? So $d$ is an isomorphism of categories?

The question reduces to the question, whether every $\mathfrak{g}$ equivariant linear map between vector spaces is $G$ equivariant.

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    $\begingroup$ Welcome to Math.SE Tina $\endgroup$
    – magma
    Jan 17, 2014 at 10:00
  • $\begingroup$ If you consider real Lie groups, $Rep(G)$ are unitary representations, then not every representation of $\mathfrak g$ is equal to $d\pi,\ \pi\in Rep(G).$ See e.g. the book by Barut,Raczka. $\endgroup$
    – yurius
    Jan 17, 2014 at 11:05
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    $\begingroup$ If $G$ is a simply connected lie group and $H$ and arbitrary Lie group, then for each lie algebra homomoprhism $\alpha\colon\mathfrak{g}\rightarrow\mathfrak{h}$ there is a unique Lie homomorphism $f\colon G\rightarrow H$, such that $\alpha=d_ef$. See the introduction of tinyurl.com/m6qhl6a for example. Representations of $G$ are just Lie-homomorphisms $G\rightarrow GL(V)$, where $V$ is a real or complex vector space, so it is indeed correct, that representations of $G$ are in 1:1 correspondance to the one of $\mathfrak{g}$ in the simply connected case. $\endgroup$
    – Tina
    Jan 17, 2014 at 11:18
  • $\begingroup$ Okay maybe I should add that I restrict me to finite dimensional representations, because I want GL(V) to be a (finite dimensional) Lie group. $\endgroup$
    – Tina
    Jan 17, 2014 at 11:42

1 Answer 1

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Let $V$ and $W$ be two representations of $G$ and $f : V \rightarrow W$ be invariant by $\frak{g}$, i.e. for all $X \in \frak{g}$ and $v \in V$: $$f(X.u) = X.f(u).$$ Let $X \in \frak{g}$ and $g=\exp(X)$. Define $$ \phi : \mathbb{R} \longrightarrow \mathrm{End}(V,W), \quad t \mapsto f \circ \exp(tX).$$ and $$ \psi : \mathbb{R} \longrightarrow \mathrm{End}(V,W), \quad t \mapsto \exp(tX) \circ f.$$ Then $$\frac{d}{dt} \phi(t) = f \circ X \circ \exp(tX) = X \circ f \circ \exp(tX) = X.\phi(t).$$ and $$\frac{d}{dt} \psi(t) = X \circ \exp(tX) \circ f = X.\psi(t).$$ Hence $\phi$ and $\psi$ both satisfy the differential equation. $$\begin{cases} \frac{d}{dt} y(t) &= X.y(t) \\ y(0) & = f \end{cases} $$ So $\phi=\psi$ by Cauchy-Lipschitz, and $f.g = g.f$.

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    $\begingroup$ Note that this argument makes no use of the hypothesis that $G$ is simply connected; it suffices that $G$ be connected. $\endgroup$ Aug 25, 2015 at 9:11
  • $\begingroup$ What is $t \mapsto f \circ \exp(tX)$ a one-parameter subgroup of? I thought one-parameter subgroups were Lie group homomorphisms from $\mathbb{R}$ to another Lie group $G$, but obviously this cannot be a homomorphism $\mathbb{R} \to GL(W)$. $\endgroup$
    – ಠ_ಠ
    Nov 9, 2016 at 0:53
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    $\begingroup$ @ಠ_ಠ You are perfectly right. I have fixed the proof. $\endgroup$
    – user10676
    Nov 30, 2016 at 17:00

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