Action of enveloping algebra on Verma module How does $U(\mathfrak{g})$ act on a verma module $M(\lambda)$? Say I have $a\otimes b\in U(\mathfrak{g})\otimes \Bbb C_\lambda$ and I want to act some $c\in U(\mathfrak{g})$ on this.
Do I still take $c\cdot (a\otimes b) = (c\cdot a \otimes b) + (a\otimes c\cdot b)$ as in the Lie algebra case?
This confuses me, since $1\otimes 1$ is highest weight in $M(\lambda)$ and say we do act $h\in \mathfrak{h}$ on this.
$$h\cdot(1\otimes 1) =h\cdot1\otimes 1 + 1\otimes h\cdot 1=h\otimes 1 + 1\otimes \lambda(h) = 1\otimes h+ 1\otimes \lambda(h)?$$
I guess $U(\mathfrak{g})\otimes \Bbb{C}_\lambda$ has $\mathfrak{b}$-action by $x\cdot(a\otimes b) = (ax\otimes x^{-1} \cdot b)$, then $M(\lambda)$ is given by the collection of orbits under this action. Still doesn't give what I want though, since then $h\cdot (1\otimes 1)=(h\otimes h^{-1}\cdot 1) = h\otimes \lambda(h^{-1})1=\lambda(h^{-1})^2(1\otimes 1)$?
 A: Given that this is a common source of confusion, let me write a bit more detailed about what is going on here.
If we have two representations $V$ and $W$ of a Lie algebra, we usually define a new representation $V\otimes W$ by setting $g.(v\otimes w) = (g.v)\otimes w + v\otimes (g.w)$. But note that the symbol $\otimes$ there should really be $\otimes_k$ where $k$ is the field we are working over. In general, if we have a $k$-algebra $A$ and two $A$-modules $V$ and $W$, we could not form a new module in this way, but would be restricted to acting only on the left, which would make the entire module structure of the right factor irrelevant. The reason we can do this for Lie algebras is that the enveloping algebra is a Hopf algebra, with comultiplication given by $\Delta(x) = x\otimes 1 + 1\otimes x$ when $x$ is in the Lie algebra.
On the other hand, if we have a $k$-algebra $A$ and a subalgebra $B$ then $A$ is a $B-B$-bimodule, so if we have a $B$-module $M$ we can form the tensor product $A\otimes_B M$ which works because $A$ is a right $B$-module. It also becomes an $A$-module since $A$ is a left $A$-module, and the action really is just applied on the left. However, since we are now tensoring over $B$, the module structure of $M$ is not as irrelevant, since all scalars from $B$ can be moved over to $M$.
This last version is what we do when we define Verma modules, with $A$ being the enveloping algebra of the Lie algebra and $B$ being the enveloping algebra of the Borel subalgebra (and $M$ being any $1$-dimensional representation of the Borel).
A: Let me offer a more concrete description of how the module $M(\lambda)$ is constructed. The $1$-dimensional $\mathfrak{b}$-module $\mathbb{C}_\lambda$ is defined by the formula $hv_\lambda=\lambda(h)v_\lambda$ for $h\in \mathfrak{h}$ and $xv_\lambda=0$ for $x\in \mathfrak{n}^+$. The $U(\mathfrak{g})$-module $M(\lambda)$ is the induced module
$$M(\lambda)=U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\mathbb{C}_\lambda.$$
The action of an element $x\in \mathfrak{g}$ on the vector $1\otimes v_\lambda$ is given by
$$x(1\otimes v_\lambda)=x\otimes v_\lambda=\begin{cases}x\otimes v_\lambda&\mbox{if }x\in \mathfrak{n}^-\\
1\otimes xv_\lambda&\mbox{if }x\in \mathfrak{b}.\end{cases}$$
Using this observation, we note that there is a vector space isomorphism (in fact, a $U(\mathfrak{n}^-)$-isomorphism) $M(\lambda)\cong U(\mathfrak{n}^-)\otimes\mathbb{C}_\lambda$.
Now, to understand the action of $U(\mathfrak{g})$ on an arbitrary element of the form $F\otimes v_\lambda$, where $F\in U(\mathfrak{n}^-)$ is a monomial, we need to recall the PBW theorem. This theorem says (roughly) that every element of $U(\mathfrak{g})$ is a linear combination of elements of the form $FHE$, where $F\in U(\mathfrak{n}^-)$, $H\in U(\mathfrak{h})$, and $E\in U(\mathfrak{n}^+)$ are monomials. In fact, the theorem says more. Recall that $U(\mathfrak{g})$ is filtered by total degree and, for any $x\in \mathfrak{g}$, $$x(FHE)=F'H'E'+(*),$$
where $\deg(F'H'E')=\deg(FHE)+1$, and $(*)$ is a linear combination of monomials in $U(\mathfrak{g})$ of degree $\leq \deg(FHE)$. Therefore, for $x\in \mathfrak{g}$,
\begin{align}
x.(F\otimes v_\lambda)&=(xF)\otimes v_\lambda=(F'H'E')\otimes v_\lambda +(*)\otimes v_\lambda\\
&=F'\otimes(H'E') v_\lambda +(*)\otimes v_\lambda
\end{align}
and we compute $(*)\otimes v_\lambda$ in the same way we do for the leading term.
