# How to define the action of $U(G)$ in this situation?

The usual action of $fg$ on $u⊗v$ , where $f,g$ are elements in the Universal Enveloping Algebra $U(G)$ of a Lie algebra $G$ and $u,v$ are elements of a representation $V$ of $G$, is given by $fg(u⊗v)=fgu⊗v+fu⊗gv+gu⊗fv+u⊗fgv$, using the comultiplication, right? How to state this fact for $V^{\otimes n}$, i.e. $fg$ acting on $u⊗v$, where $u=\otimes_{i=1}^{n-k} u_i$ and $v=\otimes_{i=1}^k v_i$, for each $k=1,...,n-1$ ? Thanks,

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The formula you've written is how fg acts on u⊗v. This is just the associativity of tensor product. Honestly, I don't think this is a very appropriate question for MO (you might get a better response on math.stackexchange.com). Actually, I think you would probably be better served by just going and reading a book about Hopf algebras, as all these points will be addressed in any reasonable one. –  Ben Webster Oct 25 '10 at 0:52
@Ben Webster: I have tried many books. But I haven't got any answer about this specific situation. I found in the Jantzen book that it is possible, but not how to do. Anyway, thanks for the link! –  Chris Oct 25 '10 at 1:12
I would also suggest that you ask this on math.stackexchange.com. –  Mariano Suárez-Alvarez Oct 25 '10 at 2:56
This is, I take it, the question you meant to ask at mathoverflow.net/questions/43382 ? –  Theo Johnson-Freyd Oct 25 '10 at 5:35
Chris, your formula only works when $f$ and $g$ are primitive elements. –  S. Carnahan Oct 25 '10 at 5:52

There is nothing deep here. The coproduct $\Delta : U(\mathfrak g) \to U(\mathfrak g)\otimes U(\mathfrak g)$ simply implements Leibniz's product rule: if $v_1\in V_1$ and $v_2\in V_2$, then $x\in \mathfrak g$ acts on $V_1\otimes V_2$ by $x: v_1\otimes v_2 \otimes (xv_1)\otimes v_2 + v_1\otimes (xv_2)$. Extending this, on a large tensor product we have: $$x(v_1 \otimes v_2 \otimes \cdots \otimes v_n) = xv_1 \otimes v_2 \otimes \cdots \otimes v_n + v_1 \otimes xv_2 \otimes \cdots \otimes v_n + \cdots + v_1 \otimes v_2 \otimes \cdots \otimes xv_n$$ If $x,y\in \mathfrak g$, then $xy \in U(\mathfrak g)$ acts via the composition $x\circ y$. For example, $$xy \left( \bigotimes_{k=1}^n v_k\right) = \sum_{k=1}^n (xy\text{ acts on }v_k) + \sum_{j\neq k}(x\text{ acts on }v_j\text{ and }y\text{ acts on }v_k)$$
Note that in general, for $f\in U(\mathfrak g)$, we do not have $f(u\otimes v) = fu \otimes v + u\otimes fv$. For example, the constant $1$ is an element of $U(\mathfrak g)$, and $1(u\otimes v) = u\otimes v \neq 2u\otimes v = 1u\otimes v + u\otimes 1v$. More generally, not all differential operators are derivations: certainly you do not expect $\frac{\partial^2}{\partial x^2}$ to satisfy a Leibniz rule!