# Primitive elements in the universal enveloping algebra of Lie superalgebra

Let $$\mathfrak{g}$$ be a Lie superalgebra over $$\mathbb{C}$$. Denote by $$U(\mathfrak{g})$$ the universal enveloping algebra of $$\mathfrak{g}$$. We know that there is a natural super Hopf algebra structure on $$U(\mathfrak{g})$$. Is $$1$$ the only group-like element of $$U(\mathfrak{g})$$? Is the set of primitive elements of $$U(\mathfrak{g})$$ equal to $$\mathfrak{g}$$?

We know that the above propositions are true in the Lie algebra case.

$$\DeclareMathOperator\chr{char}$$Yes this is true: Under your assumptions $$\mathcal{P}(U(g))=g$$.

Also, since for any primitive element $$x$$ we have $$\epsilon(x)=0$$, for any grouplike element $$y$$ we have $$\epsilon(y)=1$$ and given that $$g$$ generates $$U(g)$$ —as a superalgebra— there can be no grouplike element of $$U(g)$$ other than $$1$$.

Actually, the above is a consequence of a much more general result, for colour Lie (super)algebras over any field of $$\chr\neq 2, 3$$:

Let $$k$$ a field, $$g$$ a colour Lie superalgebra over $$k$$ and $$U(g)$$ its universal enveloping algebra.

• If $$\chr k=0$$, then $$\mathcal{P}(U(g))=g$$.
• If $$\chr k=p>3$$, then $$\mathcal{P}(U(g))$$ coincides with the colour Lie $$p$$-superalgebra generated by $$g$$ in $$U(g)$$.

For a detailed proof of the above see theorems 2.10, 2.11, Ch. 3 of: Infinite dimensional Lie superalgebras, by Y.A. Bahturin, A.A. Mikhalev, V.M. Petrogradksy, M.V. Zaicev.

P.S.: If you are interested in a more in-depth study of the role and the properties of (skew-)primitive elements you can see the article: "Skew-primitive elements of Quantum groups and Braided Lie algebras", by B. Pareigis in: Rings, Hopf Algebras, and Brauer Groups (Lect. Notes in Pure and Appl. Math., v. 197)