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Let $\mathfrak{g}$ be a complex simple Lie algebra, $\omega_0$ a dominant weight, $\rho$ a fundamental representation with highest weight $\Lambda$.

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 $w$: $$|v(w)\rangle=\sum_\sigma \prod_{a=1}^n \frac{1}{\langle w-w_a,\omega_0\rangle} E_{i_n}^-\dots E_{i_1}^-|\Lambda\rangle$$ Question one is: does there exist any simple formula for the norm $\langle v(w)|v(w)\rangle$?

Question two is: prove the identity $$\sum_w (-1)^n \langle v(w)|v(w)\rangle \prod_{\beta>0} \langle \omega_0,\beta\rangle^{-2\langle w,\beta>/<\beta,\beta>}=0$$ 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!

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  • $\begingroup$ In your list of weights encountered, should $\Lambda$ be $\omega_0$ instead of $\omega_1$? $\endgroup$
    – ARupinski
    Commented Feb 14, 2011 at 1:25
  • $\begingroup$ 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. $\endgroup$
    – Victor
    Commented Feb 14, 2011 at 2:38
  • $\begingroup$ 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 $w_{i_{n+1}}\equiv w$ $\endgroup$
    – Victor
    Commented Feb 14, 2011 at 3:29
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    $\begingroup$ This is an interesting question, to say the least. I suspect that since quite a bit is known about the fundamental representations it shouldn't be hard to prove, but since there are so many variables floating around, would you mind showing a concrete example worked from beginning to your conjectured final form (of say a 1-dimensional weight space)? $\endgroup$
    – ARupinski
    Commented Feb 14, 2011 at 3:46
  • $\begingroup$ 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 $\endgroup$
    – Victor
    Commented Feb 14, 2011 at 4:09

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