3

Let $V,W$ be $\mathfrak g$-modules with representations $\rho_V: \mathfrak g \rightarrow End(V), \rho_W: \mathfrak g \rightarrow End(W).$ I know that $V\otimes W$ is also a $\mathfrak g$-module, with representation $\rho: \mathfrak g \rightarrow End(V\otimes W)$ mapping $ x \mapsto \rho_V(x)\otimes 1 + 1 \otimes \rho_W(x), \,x\in \mathfrak g.$

Since $End(V\otimes W)$ is an associative unital algebra, from the universal property of $U(\mathfrak g)$, there exists an homomorphism of unital algebras $\tilde \rho : U(\mathfrak g) \rightarrow End(V\otimes W)$ such that $\rho = \tilde\rho \circ i$, where $i: \mathfrak g \rightarrow U(\mathfrak g)$ is the canonical inclusion.

This way, given $xy\in U(\mathfrak g)$, we have $$\tilde \rho(xy) = \tilde \rho(x) \tilde \rho(y) = (\rho_V(x)\otimes 1 + 1 \otimes \rho_W(x)) (\rho_V(y)\otimes 1 + 1 \otimes \rho_W(y)) = \rho_V(x) \rho_V(y)\otimes 1+ \rho_V(x) \otimes \rho_W(y) + \rho_V(y)\otimes \rho_W(x)+ 1\otimes \rho_V(x)\rho_W(y).$$

In language of modules, this is the same thing as saying that $\tilde \rho(xy)(v\otimes w) = x\cdot (y\cdot (v\otimes w))$ for all $v\in V,w\in W$.

From the answer in this post here: tensor product of modules of Lie algebras, it is said that $V\otimes W$ is a $U(\mathfrak g)$-module via the comultiplication $\Delta: U(\mathfrak g) \rightarrow U(\mathfrak g)\otimes U(\mathfrak g)$ which maps $g \mapsto g\otimes 1 + 1 \otimes g,$ whence the composition $$\phi: U(\mathfrak g) \xrightarrow{ \Delta} U(\mathfrak g)\otimes U(\mathfrak g) \xrightarrow{ \tilde \rho_V \otimes \tilde \rho_W} End(V)\otimes End(W) \rightarrow End(V\otimes W)$$ gives the representation of $U(\mathfrak g)$ on $V\otimes W$. However, when we evaluate this representation on $xy\in U(\mathfrak g)$, it seems that I am not obtaining the same result above. In fact, we have $$xy \mapsto xy\otimes 1 + 1 \otimes xy \mapsto \tilde \rho_V(x)\tilde \rho_V(y) \otimes 1 + 1 \otimes \tilde \rho_V(x)\tilde\rho(y),$$ so $\phi(xy)(v\otimes w) = (x\cdot (y \cdot v) ) \otimes w + v \otimes (x\cdot (y\cdot w))$ for all $v\in V,w \in W$.

What is wrong here? Which one of the above is the correct interpretation for the action of $xy\in U(\mathfrak g)$ on $V\otimes W$?

Eran
  • 2,641
user2345678
  • 2,885

2 Answers2

7

The comultiplication $\Delta:U(\mathfrak{g})\to U(\mathfrak{g})\otimes U(\mathfrak{g})$ is given by $x\mapsto x\otimes 1+1\otimes x$ only for $x\in\mathfrak{g}$, not for arbitrary $x\in U(\mathfrak{g})$. This map is then extended to all of $U(\mathfrak{g})$ to be an algebra homomorphism (using the universal property of $U(\mathfrak{g})$). So this means that for $x,y\in\mathfrak{g}$, $\Delta(xy)$ is actually defined as $\Delta(xy)=\Delta(x)\Delta(y)=(x\otimes 1+1\otimes x)(y\otimes 1+1\otimes y)$, not as $xy\otimes 1+1\otimes xy$.

Eric Wofsey
  • 330,363
  • 1
    Great. So indeed $xy\in U(g)$ acts on $ v\otimes w \in V\otimes W$ as first applying $y$, then $x$, ie $(x⊗1+1⊗x)(y⊗1+1⊗y)(v\otimes w)$ – user2345678 Aug 10 '21 at 23:25
  • 2
    I'm glad for this question and answer because I've had this exact confusion before lol. – pancini Aug 10 '21 at 23:59
3

Just to add to Eric Wofsey's answer, it's worth seeing the general formula for arbitrary products. Letting $[n]=\{1,2,\dots,n\}$ and $x_1,\dots,x_n\in\mathfrak{g}$, then $$\Delta(x_1x_2\cdots x_n) = \sum_{S\subseteq[n]} \left(\prod_{i \in S} x_i\right)\otimes \left(\prod_{j\not\in S} x_j\right).$$ where the products are done in ascending order. For example, \begin{align*} \Delta(x_1x_2x_3) &= x_1x_2x_3\otimes 1 + x_1x_2\otimes x_3 \\ &\phantom{=}+ x_1x_3\otimes x_2 + x_1\otimes x_2x_3 \\ &\phantom{=}+ x_2x_3\otimes x_1 + x_2\otimes x_1x_3 \\ &\phantom{=}+ x_3\otimes x_1x_2 + 1\otimes x_1x_2x_3. \end{align*} Something I found easy to miss at first is that $$\Delta(1) = 1\otimes 1.$$

May as well also mention what the action is on tensor products of more than two representations. Since $\Delta$ is coassociative, let $\Delta^k:U(\mathfrak{g}) \to U(\mathfrak{g})^{\otimes k}$ be the $k$-fold comultiplication (so $\Delta^2=\Delta$ and $\Delta^{k} = (\operatorname{id}^{\otimes (k-2)}\otimes \Delta)\circ\Delta$ for $k>2$). For example, when $x\in\mathfrak{g}$, $$\Delta^3(x) = x\otimes 1\otimes 1 + 1 \otimes x \otimes 1 + 1 \otimes 1\otimes x.$$ The general formula can be written as $$\Delta^k(x_1x_2\cdots x_n) = \sum_{f:[n]\to [k]} \left(\prod_{\substack{i\in[n]\\f(i)=1}} x_i\right)\otimes \left(\prod_{\substack{i\in[n]\\f(i)=2}} x_i\right)\otimes \cdots\otimes \left(\prod_{\substack{i\in[n]\\f(i)=k}} x_i\right).$$ This is meant to indicate a summation over all ordered partitions of the set of indices $\{1,\dots,n\}$ into $k$ subsets. This quickly becomes unworkable since it has $k^n$ terms.

Kyle Miller
  • 19,353