I'm reading Green's book ''Polynomial Representations of GL_n. with an Appendix on Schensted Correspondence and Littelmann Paths''.

Consider theorem 2.4b) part (i) on page 14:

Consider the map $e : K\Gamma \rightarrow S_K(n,r)$, defines as follows: For each $g\in \Gamma$ we define the element $e_g\in S_K(n,r)$ by $e_g(c):=c(g)$ for all $c\in A_K(n,r)$. We extend the map $g \mapsto e_g$ linearly and get a map $e : K\Gamma \rightarrow S_K(n,r)$ which is a morphism of $K$-algebras. Any function $f\in K^{\Gamma}$ has a unique extension to a linear map $f : K\Gamma → K$. With this convention, the image under $e$ of an element $\kappa=\sum \kappa_gg\in K\Gamma$ is ''evaluation at $\kappa$''; i.e. $e(\kappa) : c \mapsto c(\kappa)$, for all $c\in A_K(n,r)$.

Question: Why is $e$ surjective? Assume the contrary. Then I don't know, why there would exist some $0\neq c\in A_K(n,r)$, such that $e_g(c)=c(g)=0\ \forall\ g\in \Gamma$, if $\text{Im}(e)$ were a proper subspace of $S_K(n,r)={A_K(n,r)}^{*}$.

Any hints would be appreciated.

Thanks for the help!

  • 1
    $\begingroup$ What are $\Gamma$ and $S_K(n, r)$? Is $K$ an arbitrary field? $\endgroup$
    – anomaly
    Commented Jul 15, 2015 at 19:23
  • $\begingroup$ $\Gamma:=\text{GL}_n(k)$, $k=K$ is an infinite field of arbitrary characteristic and $S_K(n,r)$ is the Schur algebra, i.e.: $S_k(n,r):={A_k(n,r)}^{∗}=\text{Hom}_k(A_k(n,r),k)$ and $A:=A_k(n):=\text{polynomial functions on}\ \Gamma$ and $A_k(n,r):=$ the subspace of $A$ consisting of the elements expressible as polynomials homogeneous of degree $r$. $\endgroup$
    – Stein Chen
    Commented Jul 15, 2015 at 20:02

1 Answer 1


The proposition is mostly a result about finite-dimensional vector spaces, and has very little to do with the Schur algebra itself. Let $V$ be a finite-dimensional vector space over some fixed field $K$, and let $A \subset V^*$ be a proper subspace. Then we can choose a basis $\{v_1, \dots, v_n\}$ of $V$ such that $v_1^*, \dots, v_m^*$ are a basis of $A$ for $m = \dim A$, where the $v_i^*$ are the dual elements $v_i^*(v_j) = \delta_{ij}$. In particular, there exists some nonzero $v\in V$ such that $f(v) = 0$ for all $f\in A$.

Thus if $e(K\Gamma)$ is a proper subspace of $S_K(n, r)$, then we can find some nonzero $f\in A_K(n, r)$ such that $f(g) = 0$ for all $g\in \Gamma$. But $K$ is infinite, so any such $f$ must vanish.

  • $\begingroup$ Thank you very much for your answer! $\endgroup$
    – Stein Chen
    Commented Aug 6, 2015 at 18:12

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