Tensor algebra and universal enveloping algebra

Question

Let $\mathfrak g$ be a Lie algebra which is not reductive. Let $T(\mathfrak g)$ and $U(\mathfrak g)$ be the tensor algebra and universal enveloping algebra of $\mathfrak g$ respectively. We have a cannonical surjection $p: T(\mathfrak g) \rightarrow U(\mathfrak g)$. Does it give a surjective map from $T(\mathfrak g)^{\mathfrak g}$ to $U(\mathfrak g)^{\mathfrak g}$ ? Here $T(\mathfrak g)^{\mathfrak g}$ (resp. $U(\mathfrak g)^{\mathfrak g}$) are the $\mathfrak g$-invariants of $T(\mathfrak g)$ (resp. $U(\mathfrak g)$).

Answer 1

The projection from the tensor algebra to the symmetric algebra is a split surjection. Therefore so is the map from $T(\mathfrak g)$ to $U(\mathfrak g)$, by the PBW theorem. Now note that PBW is an isomorphism of $\mathfrak g$-modules, and that split surjections are preserved by any functor.

Answer 2

To round things up, let's give a counterexample in positive characteristic.

Let $F=\mathbb{F}_2$ the field of two elements and let $L=F X\oplus FY$ be the Lie algebra with $[X,Y]=X$. We write $\pi$ for the map $T(L)\to U(L)$.

Any $f\in T(L)^L$ can uniquely be written as $f=Xa+Yb+\theta$ with $a,b\in T(L)_k$ and $\theta\in F$. We have $$ 0=\mathrm{ad}_X(f)=X\mathrm{ad}_X(a)+Xb+Y\mathrm{ad}_X(b). $$ Note that, as $F$ has characteristic 2 one has $\mathrm{ad}_X(a)=Xa+aX$ for every $a\in T(L)$. Since $T(L)=XT(L)\oplus YT(L)\oplus F$, it follows $\mathrm{ad}_X(b)=0$, so $0=X\mathrm{ad}_X(a)+Xb$ and therefore $b=\mathrm{ad}_X(a)$, i.e., $$ f=Xa+X\mathrm{ad}_X(a)=Xa+XXa+XaX. $$ By PBW and char($F$)=2 it follows $$ \pi(f)=Xa+\ \mathrm{lower\ order\ terms}. $$ The element $D=Y^2+Y\in U(L)$ satisfies $$ \mathrm{ad}_X(D)=[X,Y]Y+Y[X,Y]+[X,Y]=XY+YX+X=[X,Y]+X=X+X=0. $$ Therefore $D\in U(L)^L$ but it is not of the form $\pi(f)$ for any $f\in T(L)^L$.