I am currently reading a text about the construction of the standard cyclic $L$-module of highest weight $\lambda$, where $L$ is a semisimple Lie algebra, and I am having trouble understanding the principle of the “induced module construction”. I'd be very glad if someone could help me.
The construction goes as follows:
Let $L$ be a semisimple Lie algebra with Borel subalgebra $B = B(Δ) = H + \coprod_{\alpha \succ 0} L_\alpha$.
A standard cyclic module, viewed as a $B$-module, contains a one-dimensional submodule spanned by the given maximal vector $v^+$.
Let $D_\lambda$ be a one-dimensional vector space with $v^+$ as basis.
Define an action of $B$ on $D_\lambda$ by the rule $(h + \sum_{\alpha \succ 0} x_\alpha) \mathbin{.} v^+ = h \mathbin{.} v^+ = \lambda(h) v^+$, for fixed $\lambda \in H^*$.
This makes $D_\lambda$ a $B$-module. Of course, $D_\lambda$ is equally well a $U(B)$-module, so it makes sense to form the tensor product $Z(\lambda) = U(L) \otimes_{U(B)} D_\lambda$, which becomes a $U(L)$-module under the natural (left) action of $U(L)$.
We claim that $Z(\lambda)$ is standard cyclic for weight $\lambda$. The vector $1 \otimes v^+$ evidently generates $Z(\lambda)$. On the other hand, $1 \otimes v^+$ is nonzero, because $U(L)$ is a free $U(B)$-module with a basis consisting of $1$ along with the various monomials $y_{\beta_1}^{i_1} \dotsm y_{\beta_m}^{i_m}$.
Therefore $1 \otimes v^+$ is a maximal vector of weight $\lambda$.
This construction also makes it clear that, if $N^- = \coprod_{\alpha \succ 0} L_\alpha$, then $Z(\lambda)$ viewed as $U(N^-)$-module is isomorphic to $U(N^-)$ itself.
To be precise, $U(L)$ is isomorphic to $U(N^-) \otimes U(B)$, so that $Z(\lambda)$ is isomorphic to $U(N^-) \otimes F$ (as left $U(N)$-modules).
Paul Garrett gave the key universal property of induction in his comment, so I will only address the question in Jim's last comment: How do we see that $D_\lambda$ is a $B$-module?
We can assign any monomial $h_1^{a_1}h_2^{a_2}\cdots h_m^{a_m}x_{\alpha_1}^{b_1}x_{\alpha_2}^{b_2}\cdots x_{\alpha_n}^{b_n}$ in $U(B)$ a weight $\sum_i b_i\alpha_i$. The product of monomials will only have terms of weight that is the sum of the weights of the factors. In the standard ordering of the root lattice this weight is always $\succ0$ and it is equal to $0$ only if $b_1=b_2=\cdots=b_n=0$. In other words, the monomials with the property: $b_i>0$ for at least one $i$, span an ideal $I\subset U(B)$. Furthermore $U(B)/I\cong U(H)$. Therefore we can turn any $H$-module $M$ into a $B$-module in the standard way. If $m\in M$ and $b\in U(B)$, then we define $bm=\phi(b)m$, where $\phi:U(B)\rightarrow U(H)$ is the projection map.