Universal enveloping algebra

Remarks

The map $\iota$ is linear and satisfies $\iota([x, y]) = \iota(x)\iota(y) - \iota(y)\iota(x)$. Universal property: for every associative algebra $A$ over $k$ with identity and every linear map $\varphi \colon \mathfrak g \to A$ with $\varphi([x, y]) = \varphi(x)\varphi(y) - \varphi(y)\varphi(x)$, there is a unique algebra homomorphism $\tilde\varphi \colon U(\mathfrak g) \to A$ with $\tilde\varphi \circ \iota = \varphi$. Hence representations of $\mathfrak g$ on a vector space $V$ are the same as left $U(\mathfrak g)$-module structures on $V$. Example: if $\mathfrak g$ is abelian, all brackets are $0$, so $I$ is generated by the elements $x \otimes y - y \otimes x$. Then $U(\mathfrak g)$ is the symmetric algebra $S(\mathfrak g)$, a polynomial ring in a basis of $\mathfrak g$. Non-example: for the Lie algebra $\mathfrak g$ with basis $x, y$ and $[x, y] = y$, the tensor algebra $T(\mathfrak g)$ itself does not satisfy $xy - yx = [x, y]$, since $x \otimes y - y \otimes x \ne y$ there; one must pass to the quotient.