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Let G=Gr(k,n) the Grassmannian of $k$-dimensional subspaces of $\mathbb{C}^n$ and denote by $T$ the (rank $k$) tautological bundle over $G$, and by $Sym^p T$ its $p$-th symmetric power. Is there any known formula for computing the exterior powers $$\wedge^q(Sym^pT)$$ in terms of Schur powers $\Sigma^{\lambda}T$?

I am aware of other plethysm formulae (say, for $Sym^k(\wedge^2) $) but I haven't been able to find a reference in the literature for this particular case.

The most relevant case for me is $p=2$, but having a general result would be immensely useful.

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  • $\begingroup$ Just to clarify: you want to apply the Schur functor $\Sigma^\lambda$ to $T$, correct? $\endgroup$ Jun 8 '17 at 13:48
  • $\begingroup$ @JasonStarr, yes exactly. $\endgroup$
    – Enrico
    Jun 8 '17 at 13:49
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For $p=2$, the plethysm is known. A reference is Macdonald's book "Symmetric Functions and Hall Polynomials" (p.138, example 6d in the second edition).

$\wedge^\bullet( Sym^2) = \bigoplus_\lambda \Sigma^\lambda$ where the sum is over all $\lambda$ with the following property: for each box on the main diagonal of its Young diagram, you have exactly one less box in the column directly below it (not counting the diagonal) than there are boxes in the row directly to the right of it (again, not counting the diagonal), and the sum is multiplicity-free. To get the case of $\wedge^q Sym^2$ just take those partitions of size $2q$ in the set above. So for example, the first few which are hooks are:

$0, (2), (3,1), (4,1,1), (5,1,1,1), (6,1,1,1,1), \dots$

but you can also get things which are unions of more hooks. Here are some examples with two hooks:

$(3,3), (4,3,1), (5,3,1,1), (6,3,1,1,1), \dots, (4,4,2)$

Another way to describe it: if you have a partition in the set I described, you are allowed to add a box both to the $i$th row and $i$th column of its Young diagram to get another partition in the set (assuming this creates a valid partition).

For a specific $T$, just take all $\lambda$ with at most $k$ parts (all other Schur functors vanish). The case $p>2$ is essentially a "wild" situation and you probably shouldn't expect any workable formulas except when $q$ or $k$ is small.

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  • $\begingroup$ Steven, one further question. What about $Sym^k(Sym^2)$ instead? $\endgroup$
    – Enrico
    Jun 10 '17 at 0:20
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    $\begingroup$ $Sym^k(Sym^2)$ is easier to describe: it's the sum of all Schur functors indexed by partitions of size $2k$ all of whose parts are even (no multiplicities). This is example 6a from Macdonald (same place). $\endgroup$
    – Steven Sam
    Jun 11 '17 at 3:10

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