Weyl's construction for symplectic groups--an exercise in Fulton and Harris's book

This is an exercise in section 17.3 in Fulton and Harris's book:Representation theory-a first course.

Let $V=\mathbb{C}^{2n}$ and $Sp(2n)$ be the symplectic group w.r.t the nondegenerate bilinear form $$\left( \begin{array}{ccc} 0 & I_n\\ -I_n & 0\\ \end{array} \right).$$ For each pair $I=\{p<q\}$ of integers between $1$ and $d$, the symplectic form determines a contraction $$\Phi_I: V^{\otimes d}\rightarrow V^{\otimes d-2}$$ which sends $v_1\otimes ... \otimes v_d$ to $(v_p,v_q)v_1\otimes ...\otimes \hat{v_p}\otimes...\otimes \hat{v_q}\otimes... \otimes v_d$. Conversely, let $\psi=\sum_{i=1}^{n}(e_i\otimes e_{n+i}-e_{n+i}\otimes e_i)$, where ${e_i}$ is the basis for $V=\mathbb{C}^{2n}$, we can define the expansion operator $$\Psi_I: V^{\otimes d-2}\rightarrow V^{\otimes d}$$ by inserting $\psi$ in the $p,q$ factors. We have the following observation: $$V^{\otimes d}=V^{<d>}\oplus \sum_{I}\Psi_{I}(V^{\otimes(d-2)}),$$ where $V^{<d>}$ is the intersection of the kernels of all contractions.\

Now, in the exercise 17.18, the authors claim that if $s>n$, then $\wedge^{s}V\otimes V^{\otimes(d-s)}$ is contained in $\sum_{I}\Psi_{I}(V^{\otimes(d-2)})$. In the hint given by the authors, they say that it suffices to show that $$\Psi_{1,2}: \wedge^{s-2}V \rightarrow \wedge^{s}V$$ is surjective if $s>n$. However, I don't know how to understand $\Psi_{1,2}$ on $\wedge^{s-2}V$. They have NOT given the definition of expansion operator on exterior power of $V$, but only on tensor products. In fact, $\wedge^{s-2}V$ can be imbedded into $V^{\otimes(s-2)}$ by sending $v_1\wedge v_2 \wedge...\wedge v_{s-2}$ to $\sum_{\sigma \in \mathfrak{S}_{s-2}}sgn(\sigma)v_{\sigma(1)}\otimes...\otimes v_{\sigma(s-2)}$ and in this sense, the image of $\Psi_{1,2}$ will lies in $\wedge^{2}V\otimes \wedge^{s-2}V$, NOT in $\wedge^{s}V$.\

In my opinion, if we can show that $\wedge^{s}V \subseteqq \sum_{I}\Psi_{I}(\wedge^{s-2}V )$, then we are done. This is true when $dim V=4$ and $s=3$ or $4$. In fact, $$e_1\wedge e_2\wedge e_3=\Psi_{1,3}(e_2)+\Psi_{2,3}(-e_2)+\Psi_{1,2}(-e_2)$$ and \begin{aligned}e_1\wedge e_2\wedge e_3\wedge e_4=&\Psi_{1,3}(e_2\wedge e_4)+\Psi_{2,3}(-e_2\wedge e_4)+\Psi_{1,2}(-e_2\wedge e_4)+\\ &\Psi_{3,4}(-e_2\wedge e_4)+\Psi_{1,4}(-e_2\wedge e_4)+\Psi_{2,4}(e_2\wedge e_4). \end{aligned} But when $dim V=6$ and $s=4$, there are some problems with $e_1\wedge e_2\wedge e_4\wedge e_5$. In fact, $$\Sigma_{I}\Psi_{I}(\pm e_2\wedge e_5)=e_1\wedge e_2\wedge e_4\wedge e_5+e_3\wedge e_2\wedge e_6\wedge e_5,$$ $$\Sigma_{I}\Psi_{I}(\pm e_1\wedge e_4)=e_1\wedge e_2\wedge e_4\wedge e_5+e_1\wedge e_3\wedge e_4\wedge e_6,$$ $$\Sigma_{I}\Psi_{I}(\pm e_3\wedge e_6)=e_3\wedge e_2\wedge e_6\wedge e_5+e_3\wedge e_1\wedge e_6\wedge e_4,$$ hence we have $$e_1\wedge e_2\wedge e_4\wedge e_5=\frac{1}{2}(\Sigma_{I}\Psi_{I}(\pm e_2\wedge e_5)+\Sigma_{I}\Psi_{I}(\pm e_1\wedge e_4))-\Sigma_{I}\Psi_{I}(\pm e_3\wedge e_6)).$$\

Now, my question is

1) How to understand $\Psi_{1,2}: \wedge^{s-2}V \rightarrow \wedge^{s}V$? \ 2) Is $\wedge^{s}V$ contained in $\sum_{I}\Psi_{I}(\wedge^{s-2}V)$ in general?

• 1) I share your doubts about the "solution" in Fulton-Harris. In particular, I agree that $\Psi_{\left\{ 1,2 \right\}}$ does not map $\wedge^{s-2} V$ into $\wedge^s V$ for $s > n$, as witnessed by the example $n = 2$ (so $\dim V = 4$) and $s = 1$. Maybe you are supposed to compose this map with the projection $V^{\otimes s} \to \wedge^s V$ ? Either way, I wouldn't be surprised if Fulton-Harris plainly is mistaken here; that wouldn't be the first time. – darij grinberg Sep 14 '14 at 22:09
• I'd add to what darij says the suggestion to try contacting one of the authors directly. Keep in mind that the book is really a set of written-up lectures, somewhat informal at times, and exercises (as I know well from experience) sometimes tempt the authors to take shortcuts. So it's a good idea not to expect every detail of an exercise to be precise, even if the general idea makes sense. – Jim Humphreys Sep 14 '14 at 23:18
• Clearly there is a map from $\wedge ^{s-2}V \otimes \wedge ^2 V$ into $\wedge ^s V$. Since $\wedge ^2 V$ contains a vector invariant under the symplectic group (namely the invariant symplectic form in the dual $V^*$, we have a map from $\wedge ^{s-2}V$ into $\wedge ^s V$. – Venkataramana Sep 18 '14 at 1:18
• Moreover, the quotient $\wedge ^s V /\Omega ^* \wedge ^{s-2}V$ is irreducible, with highest weight $x_1+\cdots+x_s$ in the obvious notation – Venkataramana Sep 18 '14 at 3:58
• I know this post is old, but yes, I agree with darij grinberg in that the authors probably assumed that they were following the map $\Psi_{1,2}$ with complete skew-symmetrization. They probably expected the reader to start with an element of $\Lambda^sV$, then apply the contraction map, to get an element in $\Lambda^{s-2}V$ which maps to the original element by the wedging map $\Psi_{1,2}$ (and you probably need the condition $s>n$ for this to work). – Malkoun Dec 11 '16 at 12:37