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I am just learning about universal enveloping algebras, and I am wondering about the following.

Question: Is the universal enveloping algebras of $\mathfrak{gl}(n,\Bbb R)$ just $\mathfrak{gl}(n,\Bbb R)$ itself?

It is an associative algebra with matrix multiplication. But I am not sure that it satisfies the universal property.

Given an $\Bbb R$-algebra $A$ and a Lie algebra homomorphism $$\varphi:\mathfrak{gl}(n,\Bbb R)\to A$$ we must show that it is also an $\Bbb R$-algebra homomorphism. That is $$\varphi([X,Y])=\varphi(X)\varphi(Y)-\varphi(Y)\varphi(X),\quad\forall X,Y\quad\implies\quad \varphi(XY)=\varphi(X)\varphi(Y),\quad\forall X,Y.$$ I am not sure this is true.

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    $\begingroup$ note that the universal enveloping algebra of a lie algebra will typically be infinite dimensional as an $\mathbb{R}$-vector space $\endgroup$
    – hunter
    Dec 14, 2015 at 10:42

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It is not true. For example, consider the trace operator $$\mathrm{tr}:\mathfrak{gl}(n,\Bbb R)\to \Bbb R.$$ Then, $$\mathrm{tr}([X,Y])=\mathrm{tr}(XY)-\mathrm{tr}(YX)=0=\mathrm{tr}(X)\mathrm{tr}(Y)-\mathrm{tr}(Y)\mathrm{tr}(X).$$ However, it is not true in general that $$\mathrm{tr}(XY)=\mathrm{tr}(X)\mathrm{tr}(Y).$$

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