What is the unit in the universal enveloping algebra?

Question

Let $\mathfrak{g}$ be a Lie algebra over a field $k$ of characteristic zero. Form the tensor algebra $T(\mathfrak{g})$ and let $I(\mathfrak{g})$ be the two-sided ideal generated by elements of the form $X \otimes Y - Y \otimes X -[X,Y]$ for $X,Y \in \mathfrak{g}$. The quotient algebra $T(\mathfrak{g})/I(\mathfrak{g})$ is called the universal enveloping algebra of $\mathfrak{g}$. The book I am reading says that it is an associative algebra with unit. What is the unit element? The multiplication operation is tensor product, so I don't see what the multiplicative unit would be.

Answer 1

Let $V$ be a vector space over a field $k$. Note that $$T(V)=\bigoplus_{n=0}^\infty T^n(V),$$ where the $n$th tensor power $T^n(V)$ of $V$ is given by $T^0(V)=k$ and for $n>0$, $$T^n(V)=\underbrace{V\otimes V\otimes \ldots\otimes V}_{n\ \text{copies of}\ V}.$$ Thus $T(V)$ is a $k$-algebra with $1$ (coming from $k=T^0(V)$ itself), and so any quotient of $T(V)$ is also a $k$-algebra with $1$. Apply this result with $V=\mathfrak{g}$.

As to why $T^0(V)=k$, here is an explanation. We want to honor the relation $$T^{m+n}(V)=T^m(V)\otimes T^n(V).$$ In particular, $$T^n(V)=T^{0+n}(V)=T^0(V)\otimes T^n(V).$$ Therefore the $k$-module $T^0(V)$ should act as the identity of $\otimes$. The only $k$-module that does this job is $k$ itself (up to isomorphism).