Sorry if this one is already asked - couldnt find anything about it.

If I take the irreducible representation of $GL_n$ whose highest weight is the half-sum $\rho$ of positive roots, it has dimension $2^k$ where $k$ is the number of positive roots. Even better the weights are precisely (with multiplicity) $\sum_{\alpha} \pm \alpha$, where the sum is over all positive roots $\alpha$, and each sign can be either $+1$ or $-1$.

Is this true for other groups? Is there more known about these representations, e.g. a nice weight basis indexed by choices of sign for each positive root?

(At least for $GL_n$ there looks to be a similar story if you took the highest weight to be $2\rho, 3\rho$ etc.)

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    $\begingroup$ Is it the observation attributed to Kostant in the first two paragraphs here? mathoverflow.net/questions/14770/… $\endgroup$ Jul 23, 2015 at 22:46
  • $\begingroup$ Oh yes, that's exactly it! Thanks. I found it by accident while checking the computations in a paper I'm refereeing. $\endgroup$ Jul 23, 2015 at 23:09

1 Answer 1


Weyl's dimension formula (for semisimple groups) shows immediately that the irreducible representation with highest weight $\rho$ has the dimension you state: $2^k$ with $k$ the number of positive roots. Of course, you are starting with the reductive group $GL_n$ rather than its semisimple derived group $SL_n$; but these groups have the same irreducible representations up to multiplication by a power of det, which doesn't change the dimension. You are dealing with a particular weight which doesn't involve det.

Similarly, Weyl's formula makes it easy to find the dimensions for multiples of $\rho$. On the other hand, the subweights for the highest weight $\rho$ are fairly easy to work out in a similar way to what you've done in this special case, though I don't have a specific reference to quote.

ADDED: For other irreducible root systems, the subweights are certainly a little more complicated than in your special case: for example, the weight 0 (not a root) occurs for some types, and when there are two root lengths the computations get more complicated. The essential fact is that for a simple root $\alpha$, the reflection $s_\alpha$ sends $\rho$ to $\rho -\alpha$, and these reflections generate the entire Weyl group. But having two root lengths complicates the computations, as one sees already for type $B_2$ (odd orthogonal case) and especially the exceptional type $G_2$. Still, for the "very regular" highest weight $\rho$ there is a clear-cut way to compute everything case-by-case for simple root systems. I'm not sure whether all cases can be handled simultaneously.

UPDATE: As David Treumann points out, there is a specific statement in a long 1997 paper by Kostant here, equation (147) on page 310. See also his previous 1996 paper here. For him it's especially interesting to look at the tensor product of this irreducible (and self-dual) representation with itself, though that can get uncomfortably large to decompose. The significance of the representation itself is complicated to extract from these rather complicated papers.


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