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In the book Lie Algebra and Lie Groups by Serre, there is an exercise in Chapter three that reads as follows:

Exercise(Bergman). Prove that $U(\mathfrak{g})=k$ $\iff$ $\mathfrak{g}=0$. (Hint. Use the adjoint representation.)

Here $k$ is a commutative ring and $U(\mathfrak{g})$ is the universal enveloping algebra.

I believe this is wrong since the universal enveloping algebra of say $\mathbb{C}$ is $\mathbb{C}$ which is clearly not zero. Am I right?

Carlos De la Mora
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1 Answers1

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You are incorrect. The enveloping algebra is the quotient of the tensor algebra by the ideal generated by $x\otimes y-y\otimes x-[x,y]$.

Let $x\in\mathbb{C}$ be any nonzero element. Then, a basis for $T(\mathbb{C})$ is $\{1,x,x\otimes x, x\otimes x\otimes x,\ldots\}$. But, as $[x,x]=0$, we immediately have that $U(\mathbb{C})\cong\mathbb{C}[x]$.

David Hill
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