If we have two Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ over a field $k$, and if we have a Lie algebra homomorphism $\mathfrak{g}\rightarrow \text{Der}_k(\mathfrak{h})$, then we can define the semi-direct product $\mathfrak{g}\ltimes \mathfrak{h}$: as a $k$-linear space it is just $\mathfrak{g}\oplus\mathfrak{h}$ and the Lie bracket is given by $$ [(g_1,h_1),(g_2,h_2)]=([g_1,g_2],[h_1,h_2]+g_1\cdot h_2-g_2\cdot h_1). $$

Now we have the universal enveloping algebras $U(\mathfrak{g})$, $U(\mathfrak{h})$ and $U(\mathfrak{g}\ltimes \mathfrak{h})$. $\textbf{My question}$ is: could we form a semi-direct product $U(\mathfrak{g})\ltimes U(\mathfrak{h})$ such that $$ U(\mathfrak{g})\ltimes U(\mathfrak{h})\cong U(\mathfrak{g}\ltimes \mathfrak{h})? $$

  • $\begingroup$ In your last line you might just define the left side to equal the right side. Obviously this isn't helpful, but what kind of intrinsic properties would you expect a "semi-direct product" of such Hopf algebras to have? $\endgroup$ Sep 19, 2013 at 19:53
  • $\begingroup$ The definition should extend the special case of the direct product, i.e., $U(\mathfrak{g}\times \mathfrak{h})\simeq U(\mathfrak{h})\otimes U(\mathfrak{g})$. $\endgroup$ Sep 19, 2013 at 20:49
  • $\begingroup$ @JimHumphreys Yes, you're right. However, I'm still not clear how to define semi-direct product through universal properties, even the semi-direct product of two groups. The only way I know is to construct it by hand. Maybe that's the reason that why I cannot define the semi-direct product of two universal enveloping algebras. $\endgroup$ Sep 19, 2013 at 21:31
  • $\begingroup$ @DietrichBurde Sure! We should expect that the semi-direct product extends the tensor product. But this also leads to a problem: in the category of $k$-algebras, the tensor product does not have good universal properties, as far as I know. For example, $A\otimes B$ is not the coproduct of $A$ and $B$ if we do not restrict to commutative algebras. $\endgroup$ Sep 19, 2013 at 21:36
  • $\begingroup$ $A \otimes B$ is the "commutative coproduct": it's the universal thing with a map from $A$ and $B$ whose images commute. $\endgroup$ Sep 19, 2013 at 22:34

2 Answers 2


I guess this is what is usually called the cross-product of Hopf algebras, restricted to the case of cocommutative Hopf algebras. The starting relevant paper should be this one, by Susan Montgomery:


and of course also her book "Hopf algebras and their actions on rings", for the chapter in which she deals with crossed products (the statement you need should be the one at page 110).

Limiting oneself to the associative algebra part of the story is described in McConnell-Robson's book on Noncommutative Noetherian Rings at page 34: (crossed product of an associative algebra to a universal enveloping algebra, together with the case of the crossed product of two universal enveloping algebras).

(The definition of crossed product admits an incredible amount of generalzations, not always very easy to deal with, also because of overlapping terminolgies: smash product, bismash product etc...)

  • $\begingroup$ Thank you very much! I think that's what I'm looking for. By the way, is there any difference between the cross product as Hopf algebras (and then forget the comultiplication) and the cross product as associative algebras? $\endgroup$ Sep 21, 2013 at 5:50
  • $\begingroup$ If you want that the resulting structure is a Hopf algebra you should require more on the action. I guess you want your algebra acted upon by a Hopf algebra to be a comodule algebra or something of the kind... $\endgroup$ Sep 21, 2013 at 14:44

You could have a look at this MO question, where the explicit formula is given (within the question, at the end).


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