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Let $L$ be a Lie algebra over a field $k$ and $U(L)$ its universal enveloping algebra. I want to show that there is a $1$-$1$-correspondence of representations of $L$ and $U(L)$.

A representation of $U(L)$ is a $k$-linear map $f:U(L)\rightarrow\text{gl}(V)$ such that $f(a\otimes b)=f(a)f(b)$.

Now I want to get an associated Lie algebrahomomorphism $g_f:L\rightarrow\text{gl}(V)$. Since $U(L)$ comes with a Lie algebrahomomorphism $u:L\rightarrow U(L)$, $g_f=f\circ u$ should work.

This is a Lie algebrahomomorphism since $u$ is so and $f$ also, because of $f([a,b])=f(a\otimes b-b\otimes a)=f(a)f(b)-f(b)f(a)=[f(a),f(b)]$.

But what is the inverse map of $f\mapsto g_f$?

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I think the question is how does one derive a Lie algebra representation of $L$ from an associative algebra representation of $U(L)$.

There's a canonical map $i:L\to U(L)$ (injective by Poincare-Birkhoff-Witt). If $h:U(L)\to \text{End}(V)$ is a representation (an algebra homomorphism) then $h\circ i$ is a Lie algebra representation of $L$. (We can identify $\text{End}(V)$ with $\text{gl}(V)$ by giving it the commutator as the Lie algebra. This map $h\mapsto h\circ i$ is the inverse to your map.

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If you have a Lie algebra representation $V$, you can regard it as a map from $L$ to $\text{End}(V)$. This extends to a algebra homomorphism from the tensor algebra $T(L)$ to $\text{End}(V)$ (universal property of tensor algebra). By Lie-ness this vanishes on all $x\otimes y-y\otimes x-[x\, y]$, and so on the ideal $I$ these generate, so it factors through $T(L)/I$. But that is $U(L)$.

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  • $\begingroup$ But this is what I've done so far, isn't it? I want to derive a algebra repres. of a Lie algebra representation. $\endgroup$ – user369147 May 4 '17 at 17:28

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