Schur Functor

Preliminaries

We begin with the definition of the \(n\)-th tensor product.

\(n\)-th tensor product

Definition. For a finite dimensional complex vector space \(V\), and an natural number \(n\), we define the complex \(n\)-th tensor product as follows

\[ V^{\otimes n}:=F\left(V^{n}\right) / \sim \]
where \(F\left(V^{n}\right)\) is the free abelian group on \(V^{n}\), considered as a set, and \(\sim\) is the equivalence relation defined for \(v_{1}, \ldots, v_{n}, \hat{v}_{1}, \ldots, \hat{v}_{n}, w_{1}, \ldots, w_{n} \in V\), as
\[ \left(v_{1}, \ldots, v_{n}\right) \sim\left(w_{1}, \ldots, w_{n}\right) \]
whenever there exists a \(k \in \mathbb{C}\) such that \(k\left(v_{1}, . ., v_{n}\right)=\left(w_{1}, \ldots, w_{n}\right)\), and
\[ \left(v_{1}, . ., v_{i}, \ldots v_{n}\right)+\left(v_{1}, \ldots, \hat{v}_{i}, \ldots, v_{n}\right) \sim\left(v_{1}, . ., v_{i}+\hat{v}_{i}, . ., v_{n}\right) \]

Remark. One can easily verify that this is indeed an equivalence relation.

Next we will use the following definitions of some interesting subvector spaces of the \(n\)-th tensor product, which will be of the most interest in this paper.

\(n\)-th symmetric and alternating spaces (powers)

Definition. For a complex finite dimensional vector space \(V\), and a natural number \(n\), we define the complex \(n\)-th symmetric space denoted \(\operatorname{Sym}^{n}(V)\), as the subspace of \(V^{\otimes n}\), spanned by the following set,

\[ \left\{\sum_{\sigma \in \mathfrak{S}_{n}} v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)} \mid v_{i} \in V\right\} \]
and denote a generator \(\sum\limits_{\sigma \in \mathfrak{S}_{n}} v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)}=v_{1} \cdots v_{n}\).

Definition. For a finite dimensional complex vector space \(V\), and an natural number \(n\), we define the complex \(n\)-th alternating space denoted \(\bigwedge^{n} V\), as the subspace of \(V^{\otimes n}\), spanned by the following set

\[ \left\{\sum_{\sigma \in \mathfrak{S}_{n}} \operatorname{sgn}(\sigma) v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)} \mid v_{i} \in V\right\} \]
and denote a generator \(\sum_{\sigma \in \mathfrak{S}_{n}} \operatorname{sgn}(\sigma) v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)}=v_{1} \wedge \cdots \wedge v_{n}\).

Remark. The experienced and curious reader may be asking themselves why we chose these definitions, rather than more accepted ones, while these definitions, over \(\mathbb{C}\), are equivalent it takes some work to show this non-trivial fact, so to avoid the discussion we chose these definitions.

Complex Group Algebra

Definition. For a group \(G\), we define the complex group algebra of \(G\) denoted as \(\mathbb{C} G\) as, the \(|G|\)-dimensional complex vector space with the canonical basis indexed by elements of \(G\), that is for \(g \in G\) we have a basis element \(e_{g} \in \mathbb{C} G\), where we define the multiplication in \(\mathbb{C} G\) on basis elments, for \(g_{1}, g_{2} \in G\) as \[ e_{g_{1}} \cdot e_{g_{2}}=e_{g_{1} \cdot g_{2}} \] and expand the definiton linearly for the remaining elements.

Remark. We can see that if \(G\) is non-abelian then so is \(\mathbb{C} G\) and if \(|G|=\infty\) then \(\mathbb{C} G\) has inifinte dimension.

Young Symmetrizer

A Young symmetrizer is defined using a Young tableau associated with a partition of \( n \). It involves two subgroups of the symmetric group \( \mathfrak{S}_{n} \):

Two elements in the group algebra \( \mathbb{C}\mathfrak{S}_{n} \) are constructed:

\[ a_{\lambda} = \sum_{g \in P_{\lambda}} e_g \] \[ b_{\lambda} = \sum_{g \in Q_{\lambda}} \text{sgn}(g) e_g \]

The Young symmetrizer is the product \( c_{\lambda} = a_{\lambda} b_{\lambda} \), which corresponds to an irreducible representation of \( \mathfrak{S}_{n} \).

For instance, if \( n = 4 \) and \( \lambda = (2, 2) \), with the canonical Young tableau \( \{ \{1, 2\}, \{3, 4\} \} \). Then the corresponding \( a_{\lambda} \) is given by \[ a_{\lambda} = e_{\text{id}} + e_{(1,2)} + e_{(3,4)} + e_{(1,2)(3,4)}. \] For any product vector \( v_{1,2,3,4} := v_1 \otimes v_2 \otimes v_3 \otimes v_4 \) of \( V^{\otimes 4} \) we then have

\[ v_{1,2,3,4} a_{\lambda} = v_{1,2,3,4} + v_{2,1,3,4} + v_{1,2,4,3} + v_{2,1,4,3} = (v_1 \otimes v_2 + v_2 \otimes v_1) \otimes (v_3 \otimes v_4 + v_4 \otimes v_3). \]
Thus the set of all \( a_{\lambda} v_{1,2,3,4} \) clearly spans \( \operatorname{Sym}^2 V \otimes \operatorname{Sym}^2 V \) and since the \( v_{1,2,3,4} \) span \( V^{\otimes 4} \) we obtain \[ V^{\otimes 4} a_{\lambda} \equiv \operatorname{Im}(a_{\lambda}). \] Notice also how this construction can be reduced to the construction for \( n = 2 \). Let \( 1 \in \operatorname{End}(V^{\otimes 2}) \) be the identity operator and \( S \in \operatorname{End}(V^{\otimes 2}) \) the swap operator defined by \( S(v \otimes w) = w \otimes v \), thus \( 1 = e_{\text{id}} \) and \( S = e_{(1,2)} \). We have that \[ e_{\text{id}} + e_{(1,2)} = 1 + S \] maps into \( \operatorname{Sym}^2 V \), more precisely \[ \frac{1}{2} (1 + S) \] is the projector onto \( \operatorname{Sym}^2 V \). Then
\[ \frac{1}{4} a_{\lambda} = \frac{1}{4} (e_{\text{id}} + e_{(1,2)} + e_{(3,4)} + e_{(1,2)(3,4)}) = \frac{1}{4} (1 \otimes 1 + S \otimes 1 + 1 \otimes S + S \otimes S) = \frac{1}{2} (1 + S) \otimes \frac{1}{2} (1 + S) \]
which is the projector onto \(\operatorname{Sym}^2 V \otimes \operatorname{Sym}^2 V \).

Lemma 3.1. With the definition above \(P\) and \(Q\) are both subgroups of \(\mathfrak{S}_n\), in particular, \(P \cong \bigoplus_{i=1}^k \mathfrak{S}_{\lambda_i}\), and \(Q \cong \bigoplus_{i=1}^{k^{\prime}} \mathfrak{S}_{\lambda_i^{\prime}}\), where \(k^{\prime}\) denotes the number of columns in the Young's diagram.

Lemma 3.2. Let \(V\) be a finite dimensional complex vector space, \(\lambda\) be a partition of \(n \in \mathbb{N}\), and consider the canonical Young tableau. Then

\[ a_{\lambda}\left(V^{\otimes n}\right) \cong \operatorname{Sym}^{\lambda_{1}} V \otimes \operatorname{Sym}^{\lambda_{2}}(V) \otimes \ldots \otimes \operatorname{Sym}^{\lambda_{d}} V \]
and
\[ b_{\lambda}\left(V^{\otimes n}\right) \cong \bigwedge^{\lambda_{1}^{\prime}} V \otimes \bigwedge^{\lambda_{2}^{\prime}} V \otimes \ldots \otimes \bigwedge^{\lambda_{d^{\prime}}^{\prime}} V \]
where \(\lambda^{\prime}\) is the conjugate partition to \(\lambda\).

Schur's Functor

We will denote the image of the Young symmetrizer on \(V^{\otimes n}\) as \(\mathbb{S}_{\lambda} V\), that is \[ \mathbb{S}_{\lambda} V=c_{\lambda}\left(V^{\otimes n}\right) \]

Definition. We consider \(\mathbb{S}_{\lambda}:\) Vect \(_{\mathbb{C}} \rightarrow\) Vect \(_{\mathbb{C}}\), as a functor from the category of complex vector spaces to itself. We call this functor the Schur functor corresponding to \(\lambda\).

Remark. To see that this is a well defined functor recall that the functors \(\bigwedge^{n}:\) Vect \(_{\mathbb{C}} \rightarrow\) Vect \(_{\mathbb{C}}\), Sym \(^{n}:\) Vect \(_{\mathbb{C}} \rightarrow\) Vect \(_{\mathbb{C}}\), and \(\bigotimes^{n}:\) Vect \(_{\mathbb{C}} \rightarrow\) Vect \(_{\mathbb{C}}\), are all well defined functors. Thus for any vector spaces \(U, V, W\) and \(\mathbb{C}\)-linear maps, \(\varphi: V \rightarrow W, \psi: W \rightarrow U\) we have, that the following diagram commutes.

\[ \begin{array}{cccccc} & V^{\otimes n} & \xrightarrow{\quad\varphi^{\otimes n}\quad} & W^{\otimes n} & \xrightarrow{\;\psi^{\otimes n}\;} & U^{\otimes n} \\ & \downarrow{b_\lambda} & & \downarrow{b_\lambda} & & \downarrow{b_\lambda} \\ & b_\lambda(V^{\otimes n}) & \to & b_\lambda(W^{\otimes n}) & \to & b_\lambda(U^{\otimes n}) \\ & \downarrow{a_\lambda} & & \downarrow{a_\lambda} & & \downarrow{a_\lambda} \\ & a_\lambda\!\bigl(b_\lambda(V^{\otimes n})\bigr) & \to & a_\lambda\!\bigl(b_\lambda(W^{\otimes n})\bigr) & \to & a_\lambda\!\bigl(b_\lambda(U^{\otimes n})\bigr) \\ & \downarrow{=} & & \downarrow{=} & & \downarrow{=} \\ & S_\lambda(V) & \xrightarrow{S_\lambda(\varphi)} & S_\lambda(W) & \xrightarrow{S_\lambda(\psi)} & S_\lambda(U) \\ \end{array} \]
Thus \(\mathbb{S}_{\lambda}:\) Vect \(_{\mathbb{C}} \rightarrow\) Vect \(_{\mathbb{C}}\) is well defined.

To get a better grip of these we will first look at the two easiest examples and then return to example 3.3 from the last section to see a more involved example of \(\mathbb{S}_{\lambda}\).

Example. Let \(n=4\) and consider the two partitions let \(\lambda_{1}=(1,1,1,1)\) and \(\lambda_{2}=(4)\), that is when we consider the canonical Young tableau of \(\lambda_{1}\), and \(\lambda_{2}\) we have

\[ \lambda_1 \;=\; \begin{array}{|c|} \hline 1 \\ \hline 2 \\ \hline 3 \\ \hline 4 \\ \hline \end{array}\,, \quad \lambda_{2}=\begin{array}{|l|l|l|l|} \hline 1 & 2 & 3 & 4 \\ \hline \end{array} \]
By definitions, we have that \[ P_{\lambda_{1}}=\{()\}, \quad Q_{\lambda_{1}}=\mathfrak{S}_{4} \] and \[ P_{\lambda_{2}}=\mathfrak{S}_{4}, \quad Q_{\lambda_{2}}=\{()\}. \] Then, we can quickly calculate that
\[ c_{\lambda_{1}}=a_{\lambda_{1}}b_{\lambda_{1}}=b_{\lambda_{1}}=\sum_{\sigma \in \mathfrak{S}_{n}} \operatorname{sgn}(\sigma) e_{\sigma}, \quad c_{\lambda_{2}}=a_{\lambda_{2}}b_{\lambda_{2}}=a_{\lambda_{2}}=\sum_{\sigma \in \mathfrak{S}_{n}} e_{\sigma}. \]
Hence, if we consider any vector space \(V\) over \(\mathbb{C}\), by Lemma 3.2 we have that \[ \mathbb{S}_{\lambda_{1}}(V)=\bigwedge^{4}(V), \quad \mathbb{S}_{\lambda_{2}}(V)=\operatorname{Sym}^{4}(V) \]

References