User victor - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T09:01:49Z http://mathoverflow.net/feeds/user/12964 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55368/norm-in-the-fundamental-representations-of-lie-algebras Norm in the fundamental representations of Lie algebras Victor 2011-02-14T00:24:20Z 2011-02-16T01:30:15Z <p>Let $\mathfrak{g}$ be a complex simple Lie algebra, $\omega_0$ a dominant weight, $\rho$ a fundamental representation with highest weight $\Lambda$.</p> <p>Fix some weight $w$ in this representation. Let $\sigma$ enumerate ways in which weight space $w$ can be reached from the highest weight, i.e. each $\sigma$ corresponds to the sequence of the form $$E_{i_n}^-\dots E_{i_1}^-|\Lambda\rangle$$ where $\alpha_{i_n}+\dots+\alpha_{i_1}=\Lambda-w$. On the way from $\Lambda$ to $w$ we encounter weights $$(\Lambda=w_1, w_2, \dots, w_n, w_{n+1}=w)$$ Form the following vector in the weight space of $w$: $$|v(w)\rangle=\sum_\sigma \prod_{a=1}^n \frac{1}{\langle w-w_a,\omega_0\rangle} E_{i_n}^-\dots E_{i_1}^-|\Lambda\rangle$$ <strong>Question one</strong> is: does there exist any simple formula for the norm $\langle v(w)|v(w)\rangle$?</p> <p><strong>Question two</strong> is: prove the identity $$\sum_w (-1)^n \langle v(w)|v(w)\rangle \prod_{\beta>0} \langle \omega_0,\beta\rangle^{-2\langle w,\beta>/&lt;\beta,\beta>}=0$$ Some experimentation shows that the identity is correct, and when the weight space is one-dimensional the formula is $$\langle v(w)|v(w)\rangle=\left(\frac{constant}{\prod_{\beta_a>0} \langle \beta_a,\omega_0\rangle^{n_a}}\right)^2$$ where $\beta_a$ are positive roots and for each of them $n_a$ is the length of the sequence $$w-n_a\beta_a,\dots,w-\beta_a,w$$ How can something like that be proved? There should be some nice formula for weight space with dimension >1 as well, but the brute force method didnt help... Any ideas would be appreciated. Do there exist any similar expressions in the literature? Thank you!</p> http://mathoverflow.net/questions/55368/norm-in-the-fundamental-representations-of-lie-algebras Comment by Victor Victor 2011-02-14T04:25:49Z 2011-02-14T04:25:49Z 'which appear in $\langle v|v\rangle$' http://mathoverflow.net/questions/55368/norm-in-the-fundamental-representations-of-lie-algebras Comment by Victor Victor 2011-02-14T04:25:14Z 2011-02-14T04:25:14Z Sorry, I cant write a full example here in comments, it is too lengthy, but you can check how it works for example for the lowest root $\Lambda-\alpha_1-2\alpha_2-\alpha_3$ of (010) of $A_3$. All the scalar products $\langle E^+\dots E^+E^-\dots E^-\rangle$ which appear in $\langle v|\rangle$ here are equal to 1. http://mathoverflow.net/questions/55368/norm-in-the-fundamental-representations-of-lie-algebras Comment by Victor Victor 2011-02-14T04:12:22Z 2011-02-14T04:12:22Z But for the 5th and 6th weights the sum over $\sigma$ has 2 terms, and $\langle v|v\rangle$ has 4 terms. The denominators here include factors which are not roots, because for example (1st weight - 6th weight) is not a root. But after we summ the result again agrees with the formula. For $G_2$ the sums for some weighst include 16 terms, the scalar products $\langle E^+\dots E^+E^-\dots E^-\rangle are up to like 5184, but it all organizes again into a neat formula from question1. http://mathoverflow.net/questions/55368/norm-in-the-fundamental-representations-of-lie-algebras Comment by Victor Victor 2011-02-14T04:09:07Z 2011-02-14T04:09:07Z I've checked it for all fundamental irreps of$A_{1,2,3}$,$B_2$and$G_2$. For example, for$A_3$(0,1,0) representation the weights are$(\Lambda, \Lambda-\alpha_2, \Lambda-\alpha_2-\alpha_1, \Lambda-\alpha_2-\alpha_3, \Lambda-\alpha_1-\alpha_2-\alpha_3, \Lambda-\alpha_1-2\alpha_2-\alpha_3$. All the scalar products$\langle \Lambda| E^+\dots E^+E^-\dots E^-|\Lambda\rangle$are equal to 1 for this irrep. For the first 4 weights the equation is trivial, because for$|v\rangle$has there is only 1 term in the sum, and it gives exactly the conjectured formula, because$w-w_i$is equal to a root http://mathoverflow.net/questions/55368/norm-in-the-fundamental-representations-of-lie-algebras Comment by Victor Victor 2011-02-14T03:29:54Z 2011-02-14T03:29:54Z Oh, sorry, if you were asking about numeration, I should clarify:$|\Lambda\rangle$corresponds to$w_1$,$E_{i_1}^-|\Lambda\rangle$to$w_2$and so on, and$E^-_{i_n}\dots E^-_{i_1}|\Lambda\rangle$to$w_{i_{n+1}}\equiv w$http://mathoverflow.net/questions/55368/norm-in-the-fundamental-representations-of-lie-algebras Comment by Victor Victor 2011-02-14T02:38:38Z 2011-02-14T02:38:38Z No,$\omega_0$is just some fixed weight which appears in scalar products, it doesn't even belong to the considered representation. And$\Lambda\$ is the heighest weight of this representation.