MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $V$ be a finite-dimensional $K$-vector space. Then, the symmetric power $\mathrm{Sym}\left(V\oplus \wedge^2 V\right)$ is isomorphic to the direct sum of all Schur functors applied to $V$ (each one time only) as a Schur functor.

This follows by comparison of characters using the so-called Schur identity

$\sum\limits_{\lambda\text{ is a partition}} s_{\lambda} = \prod\limits_i \left(1-\xi_i\right)^{-1} \cdot \prod\limits_{i < j} \left(1-\xi_i\xi_j\right)^{-1}$,

where $s_{\lambda}$ denote the Schur "polynomials" and $\xi_i$ are countably many indeterminates (see, for instance, chapter 5.4 in "M. Lothaire", Algebraic Combinatorics on Words). While it is easy to like the Schur identity, it is hard not to dislike the proof of the "strong" isomorphism $\mathrm{Sym}\left(V\oplus \wedge^2 V\right)\cong \bigoplus\limits_{\lambda\text{ is a partition}} \mathrm{Schur}_{\lambda}\left(V\right)$ using the "weak" identity for Schur polynomials. Is there any better argument known? Maybe even one yielding a canonical isomorphism?

share|cite|improve this question

Here is how I think it should go. First

$\mathrm{Sym}\left(V\oplus \wedge^2 V\right)\cong \mathrm{Sym}\left(V\right)\otimes \mathrm{Sym}\left(\wedge^2 V\right)$

Then we have

$\mathrm{Sym}\left(\wedge^2 V\right)\cong \sum\limits_{\lambda} \mathrm{Schur}_{\lambda}\left(V\right)$

where the sum is over partitions such that all parts of the conjugate partition are even. This came up in Symmetric tensor products of irreducible representations

The tensor product $\mathrm{Sym}\left(V\right)\otimes \mathrm{Schur}_{\lambda}\left(V\right)$ is known by Pieri's rule.

Now given a partition we take a maximal subdiagram such that every column has an even number of boxes. The complement is skew shape with at most one box in each column.

Further comment In response to the request for representation theoretic proofs of the results used see

MR1606831 (99b:20073) Goodman, Roe ; Wallach, Nolan R. Representations and invariants of the classical groups. Encyclopedia of Mathematics and its Applications, 68. Cambridge University Press, Cambridge, 1998. xvi+685 pp. ISBN: 0-521-58273-3; 0-521-66348-2

In particular see 9.2.2 Reciprocity rules for Pieri's rule and see 5.2.6 to see the decomposition of $\mathrm{Sym}\left(\wedge^2 V\right)$. The highest weight vectors are constructed using Pfaffians.

share|cite|improve this answer
Cool! Remains to get a character-free grip on the $\mathrm{Sym}^2\left(V\right)$ isomorphism and on Pieri's rule (again, both I can only prove using characters...). – darij grinberg Nov 6 '10 at 15:18
With $\mathrm{Sym}^2\left(V\right)$ I meant $\mathrm{Sym}\left(\wedge^2 V\right)$... – darij grinberg Nov 6 '10 at 23:06
Proving Pieri without characters is more or less equivalent to proving the reduction rule for representations of $S_n$ without characters. This is in some literature, I think, although I have never read such a proof. – darij grinberg Nov 6 '10 at 23:07
You can find a character-free proof of the branching rule for $S_n$, also known as the reduction rule, in Theorem 2.8.3 (p. 77) of Sagan, "The Symmetric Group", GTM 203. – Tom Church Nov 12 '10 at 21:26

Another way to think about it. It's equivalent to show that each dominant character of the Borel occurs with multiplicity one in the polynomial algebra on wedge-2(V) + V. Now multiplicity <= 1 because there is a Zariski dense orbit. To see every one occurs: write down explicit polynomials that correspond to fundamental weights (fun).

share|cite|improve this answer
As for "explicit polynomials", have you done this or are you just assuming that it will be fun? – darij grinberg Nov 8 '10 at 13:06
Time passes, anon still doesn't deliver? – darij grinberg Dec 10 '10 at 11:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.