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

2010-10-25T00:37:04Z

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,

Answer by Theo Johnson-Freyd for How to define the action of $U(G)$ in this situation?

2010-10-25T05:46:02Z

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!

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

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

Answer by Theo Johnson-Freyd for How to define the action of $U(G)$ in this situation?