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. :/
 A: 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. 
A: 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.
