# Splitting the injection that you get from the Poincaré-Birkhoff-Witt theorem

Let $$\mathfrak g$$ be a Lie algebra over a field of characteristic zero, with universal enveloping algebra $$U\mathfrak g$$. By the Poincaré-Birkhoff-Witt theorem one knows that $$i:\mathfrak g \to U\mathfrak g$$ is injective. In fact there is $$s :U\mathfrak g \to \mathfrak g$$ such that $$s \circ i = \mathrm{id}$$ and $$s$$ is a morphism of $$\mathfrak g$$-modules.

Can one choose such a splitting $$s$$ which is a morphism of Lie algebras?

My guess would be "no" (since if this were possible I should find something about it by googling). But I would appreciate a counterexample.

A remark is that there is a retraction of Lie algebras $$\operatorname{Gr} U\mathfrak g \to \mathfrak g$$. Indeed, the PBW isomorphism $$\operatorname{Gr} U\mathfrak g \cong S\mathfrak g$$ of algebras is compatible with the Poisson structure on both sides, where the Poisson bracket on $$S\mathfrak g$$ is $$\{x_1\ldots x_n,y_1\ldots y_m\} = \sum_{i,j} [x_i,y_j] x_1 \ldots \widehat x_i \ldots x_n y_1 \ldots \widehat y_j \ldots y_m$$ and the bracket $$\operatorname{Gr}^n U\mathfrak g \otimes \operatorname{Gr}^m U\mathfrak g \to \operatorname{Gr}^{n+m-1} U\mathfrak g$$ is given by $$x \otimes y \mapsto [x,y]$$, and $$S\mathfrak g$$ retracts onto $$\mathfrak g$$.

• Are there any instances in the literature where one has looked at universal enveloping algebras as Lie algebras themselves? – YCor Nov 11 '18 at 22:00
• @YCor Not that I know of. The question came up naturally in something I was thinking about - you could prove a quite nice result if you could prove the answer is yes. – Dan Petersen Nov 12 '18 at 6:45

Edit: here's now a computation-free way to prove the existence for $$\mathfrak{sl}_n$$ and a few more, and discussion.

First, let me start with a simple observation, for an arbitrary Lie algebra $$\mathfrak{g}$$ over an arbitrary (associative unital) commutative ring $$R$$, and $$\mathfrak{g}\to U(\mathfrak{g})$$ the universal enveloping algebra: we have the equivalence of

(i) There is a Lie $$R$$-algebra retraction $$U(\mathfrak{g})\to\mathfrak{g}$$;

(ii) there is a unital associative $$R$$-algebra $$A$$ and Lie $$R$$-algebra homomorphisms $$\mathfrak{g}\to (A,[\cdot,\cdot])\to\mathfrak{g}$$ composing to $$\mathrm{Id}_{\mathfrak{g}}$$.

Indeed, (i)$$\Rightarrow$$(ii): just take $$A=U(\mathfrak{g})$$. For the converse, use the universal property to get a $$R$$-algebra homomorphism $$U(\mathfrak{g})\to A$$, and composing with $$A\to\mathfrak{g}$$ yields the desired retraction.

So all we need is to find $$A$$. Fix a field $$K$$ and denote by $$p\ge 0$$ its characteristic. Namely for $$\mathfrak{g}=\mathfrak{sl}_n(K)$$ when $$p$$ does not divide $$n$$, we take $$A=M_n(K)$$ and the retraction there is given by $$\nu(v)=v-\frac1{n}\mathrm{Tr}(v)$$. (The lengthy computation of my initial post is actually what this retraction $$U(\mathfrak{sl}_2)\to M_2\to \mathfrak{sl}_2$$ looks like!)

To obtain more examples, one observes that if $$\mathfrak{g}$$ is endowed with a Cartan grading then any graded Lie subalgebra also inherits a retraction. Let me explain in the case of a diagonalizable Cartan grading. The assumption is that we have a certain subspace $$\mathfrak{g}_0$$ (with linear dual denoted $$\mathfrak{g}_0^*$$) Lie algebra grading $$\mathfrak{g}=\bigoplus_{\alpha\in\mathfrak{g}_0^*}\mathfrak{g}_\alpha$$, such that $$\mathfrak{g}_\alpha=\{x\in\mathfrak{g}:[h,x]=\alpha(h)x,\forall h\in\mathfrak{g}_0\}$$ for each $$\alpha\in\mathfrak{g}_0^*$$. If so, for any (unital associative) algebra $$A$$ endowed with a Lie algebra homomorphism $$i:\mathfrak{g}\to (A,[\cdot,\cdot])$$ one can define: $$A_\alpha=\{x\in A:i(h)x-xi(h)=\alpha(h)x,\forall h\in\mathfrak{g}_0\}$$ for each $$\alpha\in\mathfrak{g}_0^*$$; this is multiplicative in the sense that $$A_{\alpha}A_\beta\subset A_{\alpha+\beta}$$ for all $$\alpha,\beta$$, and hence $$\bigoplus_{\alpha\in\mathfrak{g}_0^*}A_\alpha$$ is a unital subalgebra: in particular, if $$i(\mathfrak{g})$$ generates $$A$$ as a subalgebra (which is the only case of interest here since we consider quotients of the universal enveloping algebra), this is an algebra grading of $$A$$. The retraction has to be grading-preserving. Hence it maps every graded subalgebra to itself.

This applies to graded subalgebras of $$\mathfrak{sl}_n$$ (with $$n$$ not divisible by the characteristic $$p$$): for instance, the 2-dimensional non-abelian Lie algebra for $$p\neq 2$$, the 3-dimensional Heisenberg Lie algebra (viewed inside $$\mathfrak{sl}_3$$, or inside $$\mathfrak{sl}_4$$ to remove the restriction $$p\neq 3$$), etc, and all products $$\prod_i\mathfrak{sl}_{n_i}(K)$$.

This also applies under disguised occurrences of $$\mathfrak{sl}_n$$: for instance, over $$\mathbf{R}$$, $$\mathfrak{so}_3\simeq \mathfrak{sl}_1(\mathbf{H})$$ and we have a similar retraction.

This does not apply to other semisimple Lie algebras (i.e., not of type $$A_n$$ or products thereof). Indeed, assume for simplicity that $$K$$ is algebraically closed of characteristic zero and that $$\mathfrak{g}$$ is simple. Then if $$\mathfrak{g}$$ has the property that some $$A$$ as in (ii) exists with $$A$$ finite-dimensional, then $$\mathfrak{g}$$ is isomorphic to $$\prod_i\mathfrak{sl}_{n_i}(K)$$ for some family $$(n_i)$$. Indeed, since $$\mathfrak{g}$$ is semisimple, one easily deduces that $$A$$ is semisimple, so the underlying Lie algebra is isomorphic to a direct product $$K^\ell\times\prod\mathfrak{sl}_{m_j}(K)$$, and all its semisimple quotients have the required form.

So a test-case would be the 10-dimensional Lie algebra $$\mathfrak{sp}_4\simeq\mathfrak{so}_5$$. As I just said, $$A$$ in (ii) should be infinite-dimensional. But possibly just a few computations are enough to show that there is no retraction at all.

The question is also reasonable for $$\mathfrak{g}$$ nilpotent (say, over an algebraically closed field of characteristic zero. Here (ii'), defined as (ii) but without "unital" is a convenient criterion. ((ii) trivially implies (ii') and (ii') implies (ii) by adding a unit: $$A'=A\oplus R$$, observing that the projection onto $$A$$ is a Lie algebra homomorphism.) The question is then rephrased as: when is a Lie algebra retract of a Lie algebra whose law is the commutator bracket of some associative law?

Initial post:

Yes, there's such a retraction when $$\mathfrak{g}=\mathfrak{sl}_2$$.

Write the basis $$(h,x,y)$$, $$[h,x]=2x$$, $$[h,y]=-2y$$, $$[x,y]=h$$. Denote the Casimir element $$c=(h+1)^2+4yx$$. I only assume that the ground field $$K$$ has characteristic $$\neq 2$$.

The enveloping algebra $$U$$ has its usual grading $$U=\bigoplus_{n\in 2\mathbf{Z}}$$. Here $$U_0$$ is the unital subalgebra generated by $$h$$ and $$c$$: it is commutative, and actually a polynomial algebra $$K[h,c]$$, freely generated by $$h$$ and $$c$$, $$U_{2n}=x^nU_0$$ and $$U_{-2n}=y^nU_0$$ for $$n\ge 0$$. In characteristic zero, one can characterize $$U_{2n}$$ as the $$2n$$-eigenspace for the derivation $$v\mapsto hv-vh$$ of $$U$$. (In arbitrary characteristic, one has a similar description using a 1-dimensional torus of automorphisms instead, but this does not matter.) It is known that there is no zero divisor in $$U$$, and in particular the multiplication by $$x^n$$ or $$y^n$$ from $$U_0$$ to $$U_{\pm 2n}$$ is a linear isomorphism.

Define a linear map $$r:U\to\mathfrak{sl}_2(K)$$ by

• on $$U_0$$ by $$r(P(c,h))=\frac{P(4,1)-P(4,-1)}2h$$;

• on $$U_2$$ by $$r(xP(c,h))=P(4,-1)x$$;

• on $$U_{-2}$$ by $$r(yP(c,h))=P(4,1)y$$;

• on $$U_{2n}$$, $$|2n|\ge 4$$, as zero.

It maps each of $$x,y,h$$ to itself, so it is a linear retraction.

Theorem. This is a Lie algebra homomorphism.

Before checking it, let me insist that the value 4 in the evaluation at the $$c$$ variable is absolutely not random. The above retraction actually factors through the quotient $$U/(c-4)U$$, which therefore retracts onto $$\mathfrak{sl}_2$$, but this is not the case of other quotients $$U/(c-t)U$$ for $$t\neq 4$$. Also notice that $$c=4$$ corresponds to the 2-dimensional representation...

Proof of the theorem. I'll use the formula $$[A,BC]=[A,B]C+B[A,C]$$, and its consequence $$=[A,C]BD+A[B,C]D-C[D,A]B+CA[B,D].$$ Observe that $$r$$ vanishes on the 2-sided ideal $$(c-4)U$$, so that $$r$$ factors through $$V=U/(c-4)U$$.

Since $$r$$ preserves the grading, by linearity, we have to show that it preserves the bracket at every given degree. This is trivial in degree $$\notin\{-2,0,2\}$$.

In degree zero, by linearity (and using the vanishing on $$(c-4)U$$) we have to check that $$[r(x^\ell h^n),r(y^\ell h^m)]=r([x^\ell h^n,y^\ell h^m])$$ for all $$\ell,n,m\ge 0$$.

This is clear if $$\ell=0$$ since both brackets lie in degree 0 where everything commutes. If $$\ell\ge 2$$, the left-hand term is clearly $$0$$. If $$\ell=1$$, the left-hand term is $$[r(xh^n),r(yh^m)]=[(-1)^nx,y]=(-1)^nh.$$ The right-hand term is the evaluation of $$r$$ at
$$[x^\ell h^n,y^\ell h^m]=[x^\ell,y^\ell]h^{n+m}+x^\ell[h^n,y^\ell]h^m-y^\ell[h^m,x^\ell]h^n+0.$$ We have $$hx^\ell=x^\ell(h+2\ell)$$, and hence $$h^mx^\ell=x^\ell(h+2\ell)^m$$, and thus $$[h^m,x^\ell]=x^\ell((h+2\ell)^m-h^m)$$. Similarly, $$[h^n,y^\ell]=y^\ell((h-2\ell)^n-h^n)$$. Hence $$[x^\ell h^n,y^\ell h^m]=[x^\ell,y^\ell]h^{n+m}+x^\ell y^\ell((h-2\ell)^n-h^n)h^m-y^\ell x^\ell((h+2\ell)^m-h^m)h^n$$ $$=x^\ell y^\ell (h-2\ell)^nh^m-y^\ell x^\ell (h+2\ell)^mh^n.$$

A computation yields $$4^\ell x^\ell y^\ell=(c-(h-1)^2)(c-(h-3)^2)\dots (c-(h-2\ell+1)^2),$$ and similarly $$4^\ell y^\ell x^\ell=(c-(h+1)^2)(c-(h+3)^2)\dots (c-(h+2\ell-1)^2).$$ At $$(c,h)=(4,1)$$, evaluation of the polynomial $$4^\ell y^\ell x^\ell$$ vanishes (because of the term $$(c-(h+1)^2)$$, because $$\ell\ge 1$$, while evaluation of the polynomial $$4^\ell x^\ell y^\ell$$ vanishes, because of the term $$(c-(h-3)^2)$$... as soon as $$\ell\ge 2$$. If $$\ell\ge 2$$, similarly both vanish at $$(4,-1)$$, and we have $$r([x^\ell h^n,y^\ell h^m])=0$$ as required.

Now concentrate on $$\ell=1$$, and write $$4xy=c-(h-1)^2$$, $$4yx=c-(h+1)^2$$. We have $$4[x h^n,y h^m]=(c-(h-1)^2)(h-2)^nh^m-(c-(h+1)^2) (h+2)^mh^n.$$ Evaluation at both $$(c,h)=(4,1)$$ yields $$4(-1)^n$$, and at $$(4,-1)$$ yields $$-4(-1)^n$$. Thus $$r([xh^n,y h^m])=(-1)^nh=[r(xh^n),r(y h^m)]$$.

Now let us turn to degree $$-2$$ (degree $$2$$ is similar). We have to show that $$[r(x^{\ell} h^n),r(y^{\ell+1} h^m)]=r([x^{\ell} h^n,y^{\ell+1} h^m])$$ for all $$\ell,n,m\ge 0$$. The left-hand term is $$0$$ for $$\ell\ge 1$$, and for $$\ell=0$$ equals $$-(1-(-1)^n)y$$.

Let us pass to the right-hand term. We have $$[x^\ell h^n,y^{\ell+1}h^m]=[x^\ell,y^{\ell+1}]h^{n+m}+x^\ell[h^n,y^{\ell+1}]h^m-y^{\ell+1}[h^m,x^\ell]h^n+0$$ $$=[x^\ell,y^{\ell+1}]h^{n+m}+x^\ell y^{\ell+1}((h-2\ell-2)^n-h^n)h^m-y^{\ell+1}x^\ell((h+2\ell)^m-h^m)h^n$$ $$=x^\ell y^{\ell+1}(h-2\ell-2)^nh^m-y^{\ell+1}x^\ell(h+2\ell)^mh^n.$$

We need to write everything as $$yP(c,h)$$. Anticipating on the result, we define $$w_\ell=(c-(h+3)^2)(c-(h+5)^2)\dots (c-(h-2\ell+1)^2)$$ and $$w'_\ell=(c-(h-5)^2)(c-(h-7)^2)\dots (c-(h-2\ell-1)^2)$$, when $$\ell\ge 1$$. Both are products of $$\ell-1$$ terms. Using that $$P(c,h)y=yP(c,h-2)$$ for every polynomial $$P$$, one has $$(x^\ell y^\ell)y=(c-(h-1)^2)(c-(h-3)^2)\dots (c-(h-2\ell+1)^2)y$$ $$=y(c-(h-3)^2)(c-(h-5)^2)\dots (c-(h-2\ell-1)^2),$$ which for $$\ell\ge 1$$ equals $$yw'_\ell (c-(h-3)^2);$$ for $$\ell=0$$ this is $$y$$. Also, one writes $$y^{\ell+1}x^\ell=yw_\ell(c-(h+1)^2).$$ Then, for $$\ell\ge 1$$, one gets $$[x^\ell h^n,y^{\ell+1}h^m]=y\Big(w'_\ell (c-(h-3)^2)(h-2\ell-2)^nh^m+w_\ell(c-(h+1)^2)(h+2\ell)^mh^n\Big).$$ Then, because of the factors $$(c-(h-3)^2)$$ and $$(c-(h+1)^2)$$, evaluation at $$(c,h)=(4,1)$$ yields zero. Hence $$r([x^\ell h^n,y^{\ell+1}h^m])=0$$ for all $$\ell\ge 1$$. It remains $$\ell=0$$.

$$[h^n,yh^m] =y\big((h-2)^n-h^n\big)h^m.$$ Then $$r([h^n,yh^m])=((-1)^n-1)y$$. This is the desired value.

(Note that by restriction, we also obtain a retraction for the 2-dimensional Lie algebra.)

• Is this covered in any sources you know, or is this something you just figured out yourself / learned as folklore? Like Dan, I'm surprised I haven't encountered this before. – Kevin Casto Nov 11 '18 at 8:27
• @KevinCasto no, I actually tried to prove that there's no splitting (as was initially claimed in a now deleted post) and gradually converged to being convinced that there's one (possibly unique) and eventually to this proof. – YCor Nov 11 '18 at 9:12
• The basic facts (first few lines), namely that $U(\mathfrak{sl}_2)$ has no zero divisor, and that its 0-component is a polynomial algebra $K[c,h]$, is classical and can be found in Mazorchuk's book "lectures on $\mathfrak{sl}_2(\mathbb{C})$-modules". The computation of $x^ky^k$ as a product $(c-(h-1)^2)(c-(h-3)^2)\dots (c-(h-2k+1)^2)$ is probably classical but I rediscovered it (initially, for this MathSE post math.stackexchange.com/questions/2966276). – YCor Nov 11 '18 at 9:12
• Wow! What a tour de force. Thanks, that's very interesting. – Dan Petersen Nov 11 '18 at 16:10
• Is there an obvious basis-independent definition of the kernel of your map?I'd found a description for the 2nd filtered subspace (the kernel there is the subspace generated by the Casimir and the squares), but it becomes more complex for higher filtered subspaces. – user44191 Nov 12 '18 at 3:09