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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 $$\mathrm{Sym}^aV\otimes \mathrm{Sym}^bV^*=\bigoplus_{i=0}^b \Gamma_{a-i,b-i}$$

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    $\begingroup$ 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. $\endgroup$
    – Myself
    Mar 14, 2016 at 17:03
  • $\begingroup$ 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". $\endgroup$ Mar 14, 2016 at 17:25
  • $\begingroup$ 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. $\qquad$ $\endgroup$ Mar 14, 2016 at 17:32

1 Answer 1

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After 5 months with no further response, I decided to contact Professor Fulton directly. He very kindly (and promptly) responded that "clearly" was possibly an overstatement (although they might have had some clever idea at the time which he has since forgotten!). So I have possibly been chasing a chimera.

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.

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