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I am currently reading a text about the construction of a standard cyclic L-module of highest weight $ \lambda$ (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:

A standard cyclic module, viewed as a B-module ($\ B=B(\Delta)$, contains a one dimensional submodule spanned by the given maximal vector $\ v^+$. ( $ B(\Delta)=H+ \coprod_{\alpha \succ 0}L_\alpha $ ) is the Borel subalgebra.

Let $\ D_\lambda$ be a one dimensional vector space with $\ v^+$ as basis.

Define an action of B on $\ D_\lambda$ by the rule $\ (h+ \sum_{\alpha \succ 0}x\alpha).v^+=h.v^+=\lambda(h)v^+$, 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 $\ Z(\lambda)=U(L) \bigotimes_{U(B)}D_\lambda$ which becomes a U(L)-module under the natural (left) action of U(L).

We claim that $\ Z(\lambda)$ is standard cyclic for weight $\ \lambda$. $\ 1 \bigotimes v^+$ evidently generates $\ Z(\lambda)$. On the other hand, $\ 1 \bigotimes v^+$ 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}...y_{\beta_m}^{i_m} $.

Therefore $\ 1 \bigotimes v^+$ 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 $U(N^-) \bigotimes U(B)$, so that $Z(\lambda)$ is isomorphic to $U(N^-) \bigotimes F$ (as left U(N)-modules).

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A standard cyclic module of what? –  Qiaochu Yuan Jul 12 '11 at 19:59
    
Hello, I meant a standard cyclic L-module of highest weight λ (L being a semisimple Lie algebra). –  Jim Helbert Jul 12 '11 at 20:06
    
Have you studied representation theory of finite groups? The recipe of induction is very similar there, and I assume that this has motivated the author of your source to include this bit also. The basic recipe is: $R$ a ring, $S$ a subring, $M$ a (left) $S$-module. Then $R$ is an (R,S)-bimodule (R acting from the left, S from the right), and we can form the tensor product $R\otimes_S M$. Tensoring 'consumes' the $S$-action, but we still have $R$ acting from the left, so we end up with an $R$-module. –  Jyrki Lahtonen Jul 12 '11 at 20:18
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It might be useful to note that induced modules/repns can be characterized (without necessarily exhibiting a construction) as fitting into an "adjunction relation" (Frobenius Reciprocity or similar name): $Hom_R(R\otimes_S M,N)\approx Hom_S(M,Res^R_S N)$ for R-module N and S-module M. And with left-right reversed for the "other" adjoint/induction, namely $M\rightarrow Hom_S(R,M)$. One reason I like this is that it conveys the intent/property of the thing, and in the usual categorical way proves its uniqueness, so that quirks of any particular (successful) construction are not primary... –  paul garrett Jul 12 '11 at 20:26
    
@Jyrki Lahtonen Hi, no unfortunately I'm not familiar with representation theory of groups. The above construction was given as one of two possible ways to construct a standard cyclic module of highest weight. The first one is the induced construction shown in this passage and the other one is by "generators and relations". Your recipe has helped me a lot in understanding the general principle, thanks! –  Jim Helbert Jul 12 '11 at 21:14

1 Answer 1

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.

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