Let $\rho$ be the half sum of the positive roots (also the sum of fundmental weights) for $SL_n(\mathbb C)$. Then $2ρ$ is in the root lattice. Then what is the dimension of the zero weight space for the irreducible representation $V_{2\rho}$? What do the weights look like in this case ?
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1$\begingroup$ It seems like this should be possible to calculate using Freudenthal's formula (see for example Theorem 22.3 in Humphreys' introductory book, which also has this as an exercise for the $n=3$ case). Not sure if there is some more direct way for this special case. $\endgroup$– Tobias KildetoftCommented Oct 14, 2016 at 7:17
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$\begingroup$ As far as I know it's difficult to say anything general about this. A couple of years ago I posted on my webpage some notes on the zero weight spaces, which include what I could find then in the literature: people.math.umass.edu/~jeh/pub/zero.pdf (But I haven't attempted any further computations of my own.) It's a reasonable-looking question for all Lie types, but little seems to be known in general about either the dimension or the full set of weights when the highest weight is $2\rho$ (though the zero weight space is definitely nonzero). $\endgroup$– Jim HumphreysCommented Oct 14, 2016 at 12:55
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$\begingroup$ @Tobias: Apparently Freudenthal's method is still the basis for most computer calculations of weights, but it has the drawback of being highly recursive. In large ranks, the zero weight space is hard to reach that way for a regular highest weight such as $2\rho$.. $\endgroup$– Jim HumphreysCommented Oct 14, 2016 at 13:00
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1$\begingroup$ If I understand the second question correctly, you are asking about the multiplicities of all weights. They can be described either in terms of the semistandard Young tableaux of shape $(2n-2,2n-4,\ldots,2) $ or the corresponding Gelfand-Tsetlin patterns. The weights themselves are considerably easier to describe: they are just all elements of the root lattice in the convex hull of $W (2\rho) $, where $W=S_n $ is the symmetric group. $\endgroup$– Victor ProtsakCommented Oct 14, 2016 at 13:00
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$\begingroup$ As Victor points out, it's fairly easy to give at least a qualitative description of the weights occurring: given any dominant highest weight $\lambda$, you get all weights in the convex hull of $W\lambda$ which lie in the same coset modulo the root lattice. In case $\lambda = 2\rho$ in high rank, the zero weight occurs at the center of the picture with a large multiplicity which is hard to calculate in isolation. In general, all weight multiplicities can be hard to determine here, except recursively. $\endgroup$– Jim HumphreysCommented Oct 17, 2016 at 14:36
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In general, $V_{k\rho} \cong \bigotimes\limits_{\beta\in \Delta_+} (\mathbb C_{k\beta/2} \oplus \mathbb C_{(k-2)\beta/2} \oplus \ldots \oplus \mathbb C_{-k\beta/2})$ as $T$-representations, provable via WCF. Hence the dimension of the zero weight space, for $k=2$, is the number of labelings of $\Delta_+$ with coefficients $\{\pm 1,0\}$ and total vector sum zero, if you consider that an answer.
answered Oct 15, 2016 at 23:35
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$\begingroup$ There is a natural action of the Weyl group (In this case its $S_n$) on the zero weight space that is the $T$ invariants of the Global sections of of the line bundle $L_{2\rho}$ on $G/B$ which has a basis of standard tableaux with some restriction on entries. What is this action ? $\endgroup$– MathewCommented Nov 14, 2016 at 16:46