Given a $\mathbb{S}$-module, we can construct a Schur functor, which is an endo functor in the category of vectorspaces. But given a Schur functor, how can we get back the $\mathbb{S}$-module? In Loday/Vallette's book on Operads, they say that if $P$ is the Schur functor, look at $P(V)$, where $V$ is an $n$-dimensional vectorspace. The $n$-multilinear part of $P(V)$ is isomorphic to $P(n)$, as an $S_n$-module. Can someone explain what is meant by the 'multilinear part'?
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Let $e_1,\dots,e_n$ be a basis of $V$. There is a torus $T=(\mathbf{C}^*)^n$ acting on $V$ as follows: it multiplies $e_i$ by $\lambda_i$ (where $\lambda_1, \dots,\lambda_n$ are coordinates on the torus $T$).
Then given a Schur functor $F$ and its value $F(V)$ on $V$, the multi-linear part is the subspace of $F(V)$ spanned by simultaneous eigen-vectors of the torus $T$, which is acting on $F(V)$ in a natural way.
To understand why the torus $T$ acts on $F(V)$, you can think of elements of $F(V)$ as linear combinations of expressions in $e_1,\dots,e_n$ with some prescribed symmetries.
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4$\begingroup$ By "simultaneous eigen-vectors" you probably mean "simultaneous eigenvectors for eigenvalue $\lambda_1 \lambda_2 \cdots \lambda_n$"? And the action of the torus on $F\left(V\right)$ should just follows from functoriality, via the embedding of the torus $T$ into $\operatorname{GL}\left(V\right)$ as diagonal matrices. $\endgroup$ Commented Aug 13, 2015 at 14:34