Primitive elements in the universal enveloping algebra are Lie elements

Question

Given a Lie algebra $\mathfrak{g}$, we can define a bialgebra structure on its universal enveloping algebra $U(\mathfrak{g})$ by $$\Delta(x)=x\otimes 1 + 1\otimes x$$ for all $x\in\mathfrak{g}$. I know it is true that the the space of primitive elements of $U(\mathfrak{g})$, those satisfying the above equation, is equal to $\mathfrak{g}$, at least for $\mathfrak{g}$ a free Lie algebra. However, I cannot see how to prove it. Could anyone provide a reference/sketch a proof? I assume it has to do with showing primitive elements lie in the kernel of the abelianisation map, but I cannot see how to proceed from there

Answer 1

Denote $\mu: \mathcal{U}(\mathfrak{g})\otimes \mathcal{U}(\mathfrak{g})\rightarrow \mathcal{U}(\mathfrak{g})$ is the multiplication map $a\otimes b\mapsto ab$. For each $x\in \mathfrak{g}$ one has $\mu\Delta(x)=2x$, and more generally $\Delta$ acts as multiplication by $2^n$ on $\mathcal{U}(\mathfrak{g})^{(n)}$ by the PBW theorem. Since for a primitive element $x\in \mathcal{U}(\mathfrak{g})$ we have $\mu \Delta(x)=2x$, it follows that $x$ must be homogeneous of degree $1$.

However, if the characteristic is $p\neq 0$, then $x^p$ is primitive for every $x \in \mathfrak{g}$.

Answer 2

$ \def\gr{\operatorname{gr}} \def\g{\mathfrak{g}} $ I've come up with the following solution myself. I'm not completely sure if it's what Alvaro Martinez already meant on his answer. In any case, I'll give more details than his.

Unless otherwise stated, $k$ denotes an arbitrary field. All filtrations and graduations are increasing over $\mathbb{N}$ ($\ni 0$) (so for me an $\mathbb{N}$-filtration is of the form $X_0\subset X_1\subset\cdots$).

There is a functor $\gr:\mathsf{FiltAlg}_k\to\mathsf{GrAlg}_k$ from the category of filtered $k$-algebras (with filtered $k$-algebra homomorphisms) to the category of graded $k$-algebras (with graded $k$-algebra homomorphisms). This functor is called the associated graded algebra to a filtered algebra. Note that if $F:\mathsf{GrAlg}_k\to\mathsf{FiltAlg}_k$ is the forgetful functor (if $A\in\mathsf{GrAlg}_k$, then $F(A)_{\leq n}=\bigoplus_{m=0}^nA_n$), then $\gr\circ F\cong 1_{\mathsf{GrAlg}_k}$.

Lemma 1.

(a) If $A,B\in\mathsf{FiltAlg}_k$, then $A\otimes B$ can be endowed with a filtered algebra structure $(A\otimes B)_{\leq n}=\sum_{m+\ell=n}A_{m}\otimes B_\ell$.

(b) If $A,B\in\mathsf{GrAlg}_k$, then $A\otimes B$ can be endowed with a graded algebra structure $(A\otimes B)_{n}=\bigoplus_{m+\ell=n}A_{m}\otimes B_\ell$.

(c) The tensor product commutes with the associated graded functor in the sense that there is a natural isomorphism of graded algebras $\gr(A\otimes B)\cong\gr(A)\otimes\gr(B)$.

Proof. The first two points are routine verifications. The third point can be done following the proof done here. Note that in the linked post the filtration index goes in the opposite direction as the subspace contaiments, whereas we are assuming the filtration index to grow in the same direction as the containments, $A_{\leq 0}\subset A_{\leq 1}\subset A_{\leq 2}\subset\cdots$. That proof is only done for filtered vector spaces, so one should also verify that the canonical map becomes an algebra homomorphism (preserves products) in our case. $\square$

Given a $k$-vector space $V$, we denote $T(V)$ and $S(V)$ to the tensor and symmetric algebras over $V$, respectively.

Lemma 2. Let $\g$ be a $k$-Lie algebra and consider the morphism $T(\g)\cong\gr(F(T(\g)))\to\gr(U(\g))$ that comes from the morphism $T(\g)\to U(\g)$. Then there is a unique graded algebra isomorphism $S(\g)\to \gr(U(\g))$ that makes the diagram $$ \begin{array}{ccc} T(\g)&\to& \gr(U(\g))\\ \downarrow&\nearrow&\\ S(\g) \end{array} $$ commutative.

Proof. Although not completely immediate and some work is necessary, the proof follows from PBW theorem. $\square$

We can explain now better what I think it is Alvaro's trick.

The comultiplication $$ \Delta:U(\g)\to U(\g)\otimes U(\g) $$ is an algebra homomorphism ($U(\g)$ is a bialgebra), and actually a filtered algebra homomorphism. Thus, we get a graded $k$-algebra morphism $$ \tag{1}\label{eq} \gr(\Delta):\gr(U(\g))\to \gr(U(\g)\otimes U(\g))\cong\gr(U(\g))\otimes \gr(U(\g)), $$ where the isomorphism is from Lemma 1.c.

Crucial claim. The map \eqref{eq} identifies with the comultiplication in $S(\g)$.

I won't give details, but to show this claim, one should understand well (i) the isomorphism from Lemma 1.c, (ii) the isomorphism from Lemma 2, and (iii) how the comultiplication is defined on $U(\g)$ and on $S(\g)$.

As I think Alvaro Martinez said, one should verify that in the bialgebra $S(\g)$ the identity $\mu(\Delta(x))=2^nx$ holds for $x\in S(\g)_n$ homogeneous of degree $n$.

Finally, we are ready to classify the primitive elements of $U(\g)$.

Suppose $x\in U(\g)$ is primitive and nonzero. There is $n\geq 0$ such that $x\in U(\g)_{\leq n}\setminus U(\g)_{\leq n-1}$. Since $\Delta(x)=x\otimes 1+1\otimes x$ in $U(\g)$, we have $\mu(\Delta(x))=2x$ in $U(\g)$. On the other hand, we also have that $x$ is a nonzero element of $\gr(U(\g))_n$. Now, the identity $2x=\mu(\Delta(x))=2^nx$ holds on $\gr(U(\g))_n$. Since $x$ is nonzero here, then if $\operatorname{char} k=0$ (and only if $\operatorname{char} k=0$), we conclude $n=1$.

This means that $x=\lambda+y$ inside $U(\g)$, where $\lambda\in k$ and $y\in\g$. But \begin{align*} 2\lambda (1\otimes 1) + 1\otimes y + y\otimes 1 &=1\otimes x+x\otimes 1\\ &=\Delta(x)\\ &=\Delta(\lambda+y)\\ &=\lambda (1\otimes 1)+1\otimes y + y\otimes 1. \end{align*} So we deduce $\lambda(1\otimes 1)=0$, and therefore $\lambda=0$ and $x=y\in\g$.