# Tensor algebra and universal enveloping algebra

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. $$U(\mathfrak g)$$).

## 2 Answers

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.

• Why is PBW a morphism of $\mathfrak g$-modules? Commented Aug 12 at 6:46
• One reference for this is Loday, Cyclic homology, §3.3.4. Commented Aug 12 at 7:10
• Are you assuming characteristic zero? Commented Aug 12 at 7:16
• Loday's §3.3.4 assumes char. zero.
– YCor
Commented Aug 12 at 7:30
• Yes, I'm assuming characteristic zero. Commented Aug 12 at 7:36

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 $$f\in T(L)^L$$.