I'm trying to understand the proof of Proposition 7 part (iii) from this paper of Dadok https://www.ams.org/journals/tran/1985-288-01/S0002-9947-1985-0773051-1/S0002-9947-1985-0773051-1.pdf (Statement of the proposition on page 128, proof on page 131), but getting stuck on one line in the the middle.

The set-up is as follows: We have an irreducible self-dual (complex) representation $\pi_{\lambda}:\mathfrak{g}\to \mathfrak{gl}(V)$ of highest weight $\lambda$ (say $v_{\lambda}\in V$ is a heighest weight vector). There is a distinguished set $\{\beta_1,\dots, \beta_l\}$ of strongly orthogonal positive roots of $\mathfrak{g}$ (their specific definition isn't important for this part), and we are assuming that $\lambda$ is in the (real) span of the $\beta_i$. We are considering the subalgebra $\mathfrak{u}\subseteq \mathfrak{g}$ generated by the rootspaces of the $\pm \beta_i$ (note $\mathfrak{u}\cong \mathfrak{sl}_2(\mathbb{C})^{\oplus l}$).

Self-duality of $\pi_{\lambda}$ implies that there is a non-degenerate invariant bilinear form $B:V\times V\to \mathbb{C}$, and to determine whether $\pi_{\lambda}$ is orthogonal or symplectic we need to figure out whether $B$ is symmetric or skew-symmetric. Now he defines $U_{\lambda}\subseteq V_{\lambda}$ as the $U(\mathfrak{u})$-submodule generated by the highest weight vector $v_{\lambda}$ (the module $U_{\lambda}$ will be simple by the theorem of highest weight, using the fact that $\lambda$ is in the span of the $\beta_i$). The idea of the argument is that he wants to show that one decide whether $B$ is symmetric or skew-symmetric by looking at the restriction $B:U_{\lambda}\times U_{\lambda}\to \mathbb{C}$. To that end, he writes:

"If $B:V\to V\to \mathbb{C}$ is the nondegenerate bilinear form invariant under $\pi_{\lambda}$, we see that $B$ must remain non-degenerate on $U_{\lambda}\times U_{\lambda}\to\mathbb{C}$ since the $\mathfrak{u}$-module $U_{\lambda}$ appears in $V_{\lambda}$ with multiplicity one."

Can anyone help me make sense of this line? I can't even figure out what he means by multiplicity in this context.


  • $\begingroup$ Why is $\mathfrak{u}\cong\mathfrak{sl}_2$? Isn't it generated by multiple root spaces, or just $\pm\beta_i$ for some $i$? $\endgroup$ – David Hill Mar 22 at 15:50
  • 1
    $\begingroup$ Ah, yes thank you, that was a typo! It should have said $u\cong \mathfrak{sl}_2(\mathbb{C})^{\oplus l}$. $\endgroup$ – itinerantleopard Mar 22 at 19:41

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