Let $\mathfrak g$ be a simple complex Lie algebra and let $V(\lambda)$ be a finite-dimensional representation with highest weight $\lambda$. Let $v$ be the highest weight vector. Then the action of $\mathfrak n_-$ on $v$ defines an isomorphism $V(\lambda)=U(\mathfrak n_-)/I_{\lambda}$ where $I_{\lambda}$ is a left ideal of $U(\mathfrak n_-)$. It is known that this ideal is generated by elements $f_i^{\langle \lambda,\alpha_i^{\vee}\rangle+1}$ (I am using the standard notation). I wonder what other (non-obvious) explicit elements in the ideal $I_{\lambda}$ are known (of course, Lusztig's canonical basis gives a basis in $I_{\lambda}$ but I would like to have something more explicit).
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asked Oct 23, 2014 at 19:19
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$\begingroup$ Probably it's difficult to say much more in general, since the interaction of negative root vectors in this annihilator reflects the complexity of the weight multiplicities in $V(\lambda)$. What you start with is essentially the first step of the BGG resolution, which is itself explicit but conceals the intricacy of later cancellations as you go along. The formulas for weight multiplicities are really complicated for arbitrary dominant highest weights, and even more so for non-dominant (infinite-dimensional) irreducibles. $\endgroup$– Jim HumphreysOct 23, 2014 at 20:44
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$\begingroup$ I am not looking for a general answer, I need just some non-trivial elements. For example, I expect that it should be possible to write one canonical element for every degree. $\endgroup$– Alexander BravermanOct 23, 2014 at 20:56
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$\begingroup$ To clarify, "general" refers in my comment to arbitrary dominant highest weights; sometimes one can say more about special types of weights. By the way, can you say what you mean here by "degree"? $\endgroup$– Jim HumphreysOct 23, 2014 at 21:31
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$\begingroup$ P.S. Going back to Verma's 1966 thesis work, one can try to replace the $f_i$ here with elements $f_\alpha$ corresponding to arbitrary negative roots to get further elements in the annihilator (see for instance 1.4 and 4.3 in my 2008 book). It's not clear how much information this adds to the description, but it's explicit (though complicated) for dominant weights. $\endgroup$– Jim HumphreysOct 24, 2014 at 12:08
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$\begingroup$ "Degree" actually means weight. In fact, I want something more general. Let $\lambda$ be any weight. Then we can consider the Verma module $M(\lambda)$ and the simple module $L(\lambda)$. We know exactly for which $\lambda$ these are not isomorphic and in what weights they are different for every $\lambda$ (this is given by the Shapovalov determinant formula). I'd like to see this explicitly - namely, for every such weight I'd like to have an explicit (one) element of $M(\lambda)$ which vanishes in $L(\lambda)$. $\endgroup$– Alexander BravermanOct 24, 2014 at 12:48
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