# Contraction map on tensor product of symmetric powers is surjective.

The context is the representation of $\mathfrak{s}\mathfrak{l}_3$ as per Fulton and Harris: The contraction map $i_{a,b}:\mathrm{Sym}^aV\otimes \mathrm{Sym}^bV^*\rightarrow \mathrm{Sym}^{a-1}V\otimes \mathrm{Sym}^{b-1}V^*$ given by

$$(v_1\dots v_a)\otimes(v_1^*\dots v_b^*)\mapsto\sum \langle v_i,v_j^*\rangle (v_1\dots \hat{v_i}\dots v_a)\otimes (v_1^*\dots \hat{v^*_j}\dots v^*_b)$$

is said by Fulton to be clearly surjective.

Given his use of "clearly" I am obviously missing something trivial as it is not "clear" to me. e.g. what would be in the inverse image in $\mathrm{Sym}^4 V \otimes \mathrm{Sym}^3 V^*$ of say

$$v_1^2v_2\otimes v_1^*v_2^*$$

if $\dim V=2$?

I need to believe this to obtain the decomposition $$\mathrm{Sym}^aV\otimes \mathrm{Sym}^bV^*=\bigoplus_{i=0}^b \Gamma_{a-i,b-i}$$

• If I abbreviate $(v_1^a v_2^b \otimes (v_1^*)^c (v_2^*)^d)$ by $(ab,cd)$ then $i(31,21)=6(21,11)+(30,20)$ and $i(40,20)=8(30,20)$ so $6(21,11) = i(31,21) - 1/8\,i(40,20)$. I guess the general case can be proven by induction but I guess Fulton and Harris must have a nicer reason. Mar 14, 2016 at 17:03
• er.. $i(40,20)=8(30,10)$ perhaps? I guess that there is some combinatorial way of demonstrating surjectivity but I was hoping to unpick "clearly". Mar 14, 2016 at 17:25
• Note that $\langle v_i,v_j*\rangle$ is standard notation and $<v_i,v_j*>$ is not. I also changed $Sym$ to $\mathrm{Sym}$ and did some other more minor copy-editing. $\qquad$ Mar 14, 2016 at 17:32

The comment first provided is essentially correct (but with a small typo which threw me off). For completeness The solution to the particular question I asked is $$6(21,11)=i(31,21)-\frac{1}{12}i(40,30).$$ The general demonstration could be established by induction, but it is not too much of a stretch to see that carefully adding terms will produce any required image.