# Direct sum of an algebra and its opposite

I hate to do this, but I cannot seem to remember/find a particular result that I thought was true. Forgive me if I have some points wrong, since this is the point of my asking.

I thought I remembered that the direct sum of an algebra and its opposite algebra was the universal enveloping algebra.

For a Lie algebra, the opposite Lie algebra is just that with negative bracket. But I don't see why the sum of these should be the universal enveloping algebra of the Lie algebra.

Edit: Qiaochu makes a great point below on the dimension in his comment.

I also don't think this should be true of some associative algebras, i.e. consider an associative algebra and its opposite(assuming the algebra is noncommutative, the opposite algebra is that with reversed multiplication, i.e. $a*b:=ba$), then considering the direct sum of these, it shouldn't be isomorphic to the enveloping algebra of the underlying Lie algebra of $A$, which is isomorphic to $A$.

I thought that I read the result in Dixmier, but I can't seem to find it. :/

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AFAIK this is a different use of the term "universal enveloping algebra," unrelated to its use in Lie algebras, and it's the tensor product, not the direct sum. It appears for example in discussions of Hochschild (co)homology. In general the universal enveloping algebra of a Lie algebra is infinite-dimensional, so it can't be described by such a construction. – Qiaochu Yuan Jul 7 '11 at 17:47
Yes, that is a good point about dimensionality. Can you point me to a statement of this so I can at least remember what I was thinking of? – BBischof Jul 7 '11 at 17:51

## 2 Answers

One can find in Weibel (for example) the following definition: the enveloping algebra of an associative algebra $A$ is the algebra $A \otimes A^{op}$. The significance of this construction is that an $A\text{-}A$ bimodule is the same thing as a left $A \otimes A^{op}$-module, and one can use this to define Hochschild (co)homology in terms of Tor and Ext.

This term is, as far as I know, unrelated to the universal enveloping algebra of a Lie algebra. It is also false that the universal enveloping algebra of the underlying Lie algebra of an associative algebra $A$ is isomorphic to $A$, for the same reason as I pointed out in the comments: for $A$ finite-dimensional you can get infinite-dimensional universal enveloping algebras. In other words, not every representation of $A$ as a Lie algebra extends to a representation of $A$ as an algebra.

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ok great. Thanks. – BBischof Jul 7 '11 at 18:16
I recently finished teaching a summer half course in which I debated whether to mention the enveloping algebra $A^e$ of an algebra $A$ explicitly (I didn't, as it turned out). One of the references I was using -- Pierce's great book on associative algebras -- really likes to phrase things in terms of $A^e$, which was giving me some grief. I mentioned this recently to a colleague, and it came out that I have no idea why $A^e$ is called the enveloping algebra of $A$...other than the possible guess that it "envelops" $A$ by coming at it from both left and right. Does anybody know? – Pete L. Clark Jul 7 '11 at 21:36

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In particular I wonder whether $A^e$ has anything to do with universal enveloping algebras...and presume that it does not. – Pete L. Clark Jul 7 '11 at 21:37
@Pete L. Clark, yes, this is definitely the thrust of my question, why the heck is it the universal enveloping algebra of anything? – BBischof Jul 8 '11 at 13:31
I am pretty sure the two terms are unrelated, as I said. – Qiaochu Yuan Jul 8 '11 at 14:02

Here is an amusing situation where the two are related. This is not an "answer" to my question, just an auxiliary fact that I found interesting(hence the CW).

Consider $\mathfrak{g}$ a Lie algebra over a field $\mathbb{K}$ and $M$ a $\mathfrak{g}$-module. We define Cartan-Eilenberg Cohomology as \begin{equation} H^n_{CE}(\mathfrak{g},M):= Ext^n_{U(\mathfrak{g})}(\mathbb{K},M) \end{equation} for $U(\mathfrak{g})$ the universal enveloping algebra of $\mathfrak{g}$, and $Ext^n_{U(\mathfrak{g})}(\mathbb{K},-)$ the $n$'th derived functor of $Hom_{U(\mathfrak{g})}(\mathbb{K},-)$.

Similarly, as Qiaochu mentions, for $A$ an associative unital algebra with $B$ an $A$-bimodule, we define Hochschild Cohomology as, \begin{equation} HH^n(A,B):= Ext^n_{A^e}(A,B) \end{equation} for $A^e$ the enveloping algebra of $A$.

So the amusing fact is first, that the two "enveloping algebras" appear in the downstairs of the Extension functor for defining respective Cohomology theories.

More amusingly, for $M$ a $U(\mathfrak{g})$-bimodule, we have \begin{equation} HH^n(U(\mathfrak{g}),M)\simeq H^n_{CE}(\mathfrak{g},M_{ad}) \end{equation} i.e. \begin{equation} Ext^n_{U(\mathfrak{g})}(\mathbb{K},M_{ad})\simeq Ext^n_{U(\mathfrak{g})^e}(U(\mathfrak{g}),M). \end{equation}

I claim this is amusing. Tee hee.

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This is true more generally for Hopf algebras, and you can even keep the Mad joke as there is a notion of adjoint action for a Hopf algebra. It's one way to prove that the ordinary cohomology ring $\operatorname{Ext}(k,k)$ is graded commutative for a Hopf algebra: you use the machinery above to inject the ordinary cohomology into the Hochschild cohomology which is always graded commutative. – mt_ Jul 27 '11 at 15:18
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