John Baez Schur notes

This material from Schur functors I may still be useful.

The first thing that should be understood from the beginning is that a general Schur functor $F$ is nonlinear: the action on hom-sets

$$hom(V, W) \to hom(F(V), F(W))$$

is not assumed to respect the linear structure. In fact, linear Schur functors are rather uninteresting: because every finite-dimensional space is a finite direct sum of copies of the $1$-dimensional space $\mathbb{C}$, and because linear functors preserve finite direct sums (that is, biproducts), it turns out that every linear Schur functor $F$ is representable as $hom(X, -)$ where $X = F(\mathbb{C})$. So, the category of linear Schur functors is equivalent to $\Fin\Vect$.

=–

John Baez: This stuff should get worked into the discussion near the end of how Schur functors are like polynomials…

Modules over a bimonoid

Next we exploit the fact that, just like any group algebra, $k[S_n]$ is a bialgebra — or in fancier language, a bimonoid in the symmetric monoidal category $\Fin\Vect_k$. Since $i : \Fin\Vect_k \to C$ is a symmetric monoidal functor, this means that $i$ carries $k[S_n]$ to a bimonoid in $C$. As noted above, we call this bimonoid by the same name, $k[S_n]$.

The category of modules over a bimonoid is a monoidal category. More explicitly, in the case of the bimonoid $k[S_n]$ in $C$ with comultiplication

$$\delta: k[S_n] \to k[S_n] \otimes k[S_n] \,,$$

the tensor product $V \otimes W$ of two $k[S_n]$-modules in $C$ carries a module structure where the action is defined by

$$k[S_n] \otimes V \otimes W \stackrel{\delta \otimes 1_{V \otimes W}}{\to} k[S_n] \otimes k[S_n] \otimes V \otimes W \stackrel{1 \otimes \sigma \otimes 1}{\to} k[S_n] \otimes V \otimes k[S_n] \otimes W \stackrel{\alpha_V \otimes \alpha_W}{\to} V \otimes W$$

where $\sigma$ is a symmetry isomorphism and $\alpha_V$, $\alpha_W$ are the actions on $V$ and $W$.

Now we consider a particular case of tensor product representations. If $X$ is an object of $C$, the symmetric group $S_n$ has a representation on $X^{\otimes n}$. (Indeed, for each $\sigma \in S_n$, there is a corresponding symmetry isomorphism $X^{\otimes n} \to X^{\otimes n}$. From this one may construct an action

$$k[S_n] \otimes X^{\otimes n} \cong \bigoplus_{\sigma \in S_n} I \otimes X^{\otimes n} \to X^{\otimes n}$$

which is the required representation.) So, if $V_\nu$ is a Young tableau representation of $k[S_n]$ in $C$, we obtain a tensor product representation

$$V_\nu \otimes X^{\otimes n}$$

of $k[S_n]$ in $C$.

Consider next the averaging operator $e = \frac1{n!} \sum_{\sigma \in S_n} \sigma$:

$$e: V_\nu \otimes X^{\otimes n} \to V_\nu \otimes X^{\otimes n}$$

This operator makes sense since $k$ has characteristic zero, and crucially, this operator is idempotent (because $e = \sigma e$ for all $\sigma \in S_n$). Because we assume idempotents split in $C$, we have a (split) coequalizer

$$V_\nu \otimes X^{\otimes n} \stackrel{\overset{e}{\to}}{\underset{1}{\to}} V_\nu \otimes X^{\otimes n} \to V_\nu \otimes_{S_n} X^{\otimes n}$$

This coequalizer is indeed the object of $S_n$-coinvariants of $V_\nu \otimes X^{\otimes n}$, i.e., the joint coequalizer of the diagram consisting of all arrows

$$V_\nu \otimes X^{\otimes n} \stackrel{\sigma \cdot-}{\to} V_\nu \otimes X^{\otimes n}$$

ranging over $\sigma \in S_n$ (it is the joint coequalizer because $e = \sigma e$ for all $\sigma$).

We may now define the Schur functor $S_\nu$ on $C$ attached to a Young tableau $\nu$.