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I'm having trouble understanding part of a proof of the following statement

Let $L,L'$ be Lie algebras and $L \oplus L'$ their direct sum. Then $$ U(L \oplus L') \cong U(L) \otimes U(L')$$

Let $i_L : L \to U(L)$ denote the natural inclusion into the enveloping algebra, and similarly for $i_{L'}$ and $i_{L \oplus L'}$. The proof begins by defining a morphism $\varphi:U(L \oplus L') \to U(L) \otimes U(L')$ by first defining a Lie algebra morphism $f:L \oplus L' \to U(L) \otimes U(L')$ by $$f(x,x') = i_L(x) \otimes 1 + 1 \otimes i_{L'}(x')$$ Then, $\varphi$ is defined so that $\varphi \circ i_{L \oplus L'} = f$ using the universality of the enveloping algebra.

The inverse map is defined $\psi:U(L) \otimes U(L') \to U(L \oplus L')$ by $$\psi(a \otimes a') = \psi_1(a)\psi_2(a')$$ where $\psi_1(x) = i_{L \oplus L'}(x,0)$ and $\psi_2(x) = i_{L \oplus L'}(0,x)$

The part I have the problem with is when they show that $\psi \circ \varphi$ is the identity on the image of $L \oplus L'$ in $U(L \oplus L')$. The calculation given is, for $x \in L, x' \in L'$,

$$\psi(\varphi(x,x')) = \psi(x \otimes 1) + \psi(1 \otimes x') = i_{L \oplus L'}((x,0) + (0,x')) = i_{L \oplus L'}(x,x')$$

I don't really get how the calculation of $\psi(x \otimes 1)$ is done. If I attempt myself, it comes out $$\psi(x \otimes 1) = \psi_1(x)\psi_2(1) = i_{L \oplus L'}((x,0) \otimes (0,1)) = ~?$$ so I guess I don't understand how the tensor product $(x,0) \otimes (0,1)$ is simplified in $U(L \oplus L')$. I'm sure it's something completely basic, but I'd appreciate if someone could explain it.

Carl
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  • @QiaochuYuan, notice, though, that the tensor product is not the coproduct of the codomain of the universal enveloping algebra functor, because it takes values in non-comm. algebras. Nor is the direct product of Lie algebras the coproduct in their category, of course :-) – Mariano Suárez-Álvarez Jun 17 '13 at 05:21
  • @Mariano: whoops, my apologies. But both concepts are a kind of "commutative coproduct," namely the direct sum is the universal Lie algebra admitting a map from two Lie algebras whose images commute, and similarly for the tensor product, so one can show that the LHS and RHS naturally represent the same functor on algebras (this avoids PBW). – Qiaochu Yuan Jun 17 '13 at 05:23
  • No apologies necessary! I take the tensor product for the coproduct of algebras twice every week, at least :-) – Mariano Suárez-Álvarez Jun 17 '13 at 05:29

2 Answers2

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While general nonsense, as indicated by Qiaochu, does prove the statement, one can do something meaningful instead (as usual :-) )

For example, suppose that $B=\{v_i\}_{i\in I}$ and $B'=\{w_j\}_{j\in J}$ are vector-space bases of $L$ and $L'$, respectively, and that we pick arbitrary total orderings on the index sets $I$ and $J$. Then the Poincaré-Birkhoff-Witt theorem tells us that the set $\tilde B$ of elements of $U(L)$ of the form $$v_{i_1}v_{i_2}\cdots v_{i_p}$$ with $p\geq0$, all the $i_k$ in $I$ and $i_1\leq i_2\leq\cdots\leq i_p$, is a basis of $U(L)$. Of course, the same theorem gives us a similar basis $\tilde B'$ for $U(L)$, constructed from $L'$.

Now obviously $B''=B\cup B'$ is a basis for $L\oplus L'$, which we can totally order extending the orders of $B$ and $B'$ with $B'$ larger than $B$ (in other words, doing the «ordinal sum») and we can do the same construction to construct a basis $\tilde B''$ for $U(L\otimes L')$.

This makes it clear how to define a map $U(L\oplus L')\to U(L)\otimes U(L')$ which is an isomorphism of vector spaces, and it is easy to see that it is in fact a map of algebras.

A working general non-sense argument is the following: $\def\Lie{\operatorname{Lie}}$

  • an algebra map $U(L)\otimes U(L')\to A$ to some algebra $A$ is the same (in view of universal properties of the tensor product of algebras0 as a pair of algebra maps $U(L)\to A$ and $U(L')\to A$ whose images commute,
  • which is the same (in view of the universal property of the enveloping algebra) as a pair of Lie algebra maps $L\to \Lie(A)$ and $L'\to \Lie(A)$ (where $\Lie(A)$ is $A$ viewed as a Lie algebra with its commutator bracket) whose images commute
  • which is the same as a Lie algebra map $L\oplus L'\to \Lie(A)$, in view of the universal product of the direct product of Lie algebras,
  • which is the same as an associative algebra map $U(L\oplus L')\to A$.

From this one gets an isomorphism of functors from $\hom_{\mathrm{Alg}}(U(L)\otimes U(L'),\mathord-)$ to $\hom_{\mathrm{Alg}}(U(L\oplus L'),\mathord-)$. Yoneda's lemma then gives us an isomorphism like you want.

Mariano Suárez-Álvarez
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5

The maps $\psi_1$ and $\psi_2$ are supposed to be thought of as the obvious inclusions

$$\begin{cases}\psi_1:U(L)\,\hookrightarrow U(L\oplus L') \\ \psi_2: U(L')\hookrightarrow U(L\oplus L')\end{cases}$$

Thus e.g. $\psi(x\otimes 1_{U(L')})=\psi_1(x)\psi_2(1_{U(L')})=i_{L\oplus L'}(x,0)1_{U(L\oplus L')}=i_{L\oplus L'}(x,0)$.

Of course, the isomorphism $U(L\oplus L')\cong U(L)\otimes U(L')$ should be "obvious": an element in the first of the form $a_1\cdots a_n$ can be decomposed as $(u_1+v_1)\cdots(u_n+v_n)$ with $u_i\in L$, $v_i\in L'$, which can be expanded and rearranged into monomials of the form $u_{i_1}\cdots u_{i_l} v_{j_1}\cdots v_{j_r}$ (because we have the relation $[u,v]=0$ for all $u\in L,v\in L'$, allowing $u$s and $v$s to be slid across each other without issue), which corresponds to $u_{i_1}\cdots u_{i_l}\otimes v_{j_1}\cdots v_{j_r}$ in $U(L)\otimes U(L')$. How to go back from $U(L)\otimes U(L')$ to $U(L\oplus L')$ should be even more clear: just erase the $\otimes$ symbol everywhere. (We have been slightly abusive, viewing $L$ and $L'$ as subsets of $L\oplus L'$.)

Note this is very near to the fact that direct sums distribute through tensor products (why?).

anon
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