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I asked this question on Math Stack Exchange https://math.stackexchange.com/questions/4789924/schur-functors-weyl-functors-in-characteristic-zero, but I got no answers, so I ask the same question here.

In the paper `Schur functors and Schur complexes' by Akin et al., the notion of a Schur functor had been defined for the first time over an arbitrary commutative ring $R$.

To recall the definition (I use Efimov's `Derived categories of Grassmannians over integers and modular representation theory' as a reference), for a finitely generated projective module $V$ over a commutative ring $R$ and a partition $\lambda = (\lambda_{1},\dots,\lambda_{n})$ of degree $d$, the Schur functor $S_{\lambda}$ is defined as

$S_{\lambda}(V) = \text{Im}(\Lambda^{\lambda'}(V) \rightarrow V^{\otimes d} \xrightarrow{s_{\lambda}} V^{\otimes d} \rightarrow Sym^{\lambda}(V))$.

Here, $\lambda'$ is the conjugate partition of $\lambda$; $\Lambda^{\lambda}(V) := \Lambda^{\lambda_{1}}(V)\otimes\cdots\otimes \Lambda^{\lambda_{n}}(V)$; $Sym^{\lambda}(V) :=Sym^{\lambda_{1}}(V)\otimes\cdots\otimes Sym^{\lambda_{n}}(V)$; $\Lambda^{\lambda'}(V) \rightarrow V^{\otimes d}$ and $V^{\otimes d} \rightarrow Sym^{\lambda}(V)$ are the canonical inclusions and projections respectively; and $V^{\otimes d} \xrightarrow{s_{\lambda}} V^{\otimes d}$ is defined via $s_{\lambda}(v_{1}\otimes\cdots\otimes v_{d}):= v_{\sigma_{\lambda}(1)}\otimes\cdots\otimes v_{\sigma_{\lambda}(d)}$, where $\sigma_{\lambda} \in S_{d}$ is the permutation associated to $\lambda$. (For more details, Efimov has a readable description in his paper).

Now, one can take the `dual' of the above construction, which gives you the notion of a Weyl functor (this is what Akin et al. call coSchur functors).

Formally, we have the Weyl functor $W_{\lambda}$ defined as

$W_{\lambda}(V) = \text{Im}(\Gamma^{\lambda}(V) \rightarrow V^{\otimes d} \xrightarrow{s_{\lambda'}} V^{\otimes d} \rightarrow \Lambda^{\lambda'}(V))$,

where $\Gamma^{\lambda}(V) := \Gamma^{\lambda_{1}}(V) \otimes \cdots \otimes \Gamma^{\lambda_{n}}(V)$, the tensor product of the divided powers of $V$.

Note that $S_{\lambda}(V)^{*} \cong W_{\lambda}(V^{*})$.

My question is: if $R$ is assumed to be a field of characteristic zero, then $S_{\lambda} = W_{\lambda}$ (up to natural isomorphism). Why is this true?

It is not at all obvious to me (at least from the definitions) why this should be the case. Unfortunately, my knowledge of Representation theory is not very advanced, and my feeling is that this follows trivially from a well-known Representation theory fact, which is perhaps why others don't bother clarifying this.

I became interested in Schur functors as I want to understand the semi-orthogonal decomposition on the bounded derived category of Grassmannians, as given in Efimov's paper.

Any help would be much appreciated!

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1 Answer 1

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Assume that $R$ is a field of characteristic $0$.

Recall the standard embeddings \begin{align*} \operatorname{Sym}^{m}V & \rightarrow V^{\otimes m},\\ v_{1}v_{2}\cdots v_{m} & \mapsto\dfrac{1}{m!}\sum_{\sigma\in S_{m}} v_{\sigma\left( 1\right) }\otimes v_{\sigma\left( 2\right) }\otimes \cdots\otimes v_{\sigma\left( m\right) } \end{align*} and \begin{align*} \Lambda^{m}V & \rightarrow V^{\otimes m},\\ v_{1}\wedge v_{2}\wedge\cdots\wedge v_{m} & \mapsto\dfrac{1}{m!}\sum _{\sigma\in S_{m}}\left( -1\right) ^{\sigma}v_{\sigma\left( 1\right) }\otimes v_{\sigma\left( 2\right) }\otimes\cdots\otimes v_{\sigma\left( m\right) } \end{align*} (for each $m \in \mathbb{N}$), which are injective $\operatorname*{GL}\left( V\right) $-module morphisms whenever $R$ is a field of characteristic $0$. Using these embeddings, we can view both $\operatorname*{Sym}\nolimits^{\lambda}\left( V\right) $ and $\Lambda^{\lambda^{\prime}}\left( V\right) $ as $\operatorname*{GL}\left( V\right) $-submodules of $V^{\otimes d}$ (for example, $\operatorname*{Sym} \nolimits^{\lambda}\left( V\right) =\operatorname*{Sym}\nolimits^{\lambda _{1}}V\otimes\cdots\otimes\operatorname*{Sym}\nolimits^{\lambda_{n}}V\subseteq V^{\otimes\lambda_{1}}\otimes\cdots\otimes V^{\otimes\lambda_{n}}\rightarrow V^{\otimes d}$). Thus, $S^{\lambda}\left( V\right) $ and $W^{\lambda}\left( V\right) $ are $\operatorname*{GL}\left( V\right) $-submodules of $V^{\otimes d}$ as well.

But even better, they can be written out very explicitly in terms of the right $S_{d}$-action on $V^{\otimes d}$. Namely, let $S_{0}$ be the standard Young tableau of shape $\lambda$ that has the entries $1,2,\ldots,\lambda_{1}$ in its first row, the entries $\lambda_{1}+1,\lambda_{1}+2,\ldots,\lambda _{1}+\lambda_{2}$ in its second row, and so on (filling the shape row by row with the entries $1,2,\ldots,d$). Let $a_{\lambda}$ be the row-symmetrizer of $S_{0}$ (that is, the sum of all permutations that fix each row of $S_{0}$), and let $b_{\lambda}$ be the column-antisymmetrizer of $S_{0}$ (that is, the sign-alternating sum of all permutations that fix each column of $S_{0}$). Now, $S_{\lambda}\left( V\right) $ (when regarded as a $\operatorname*{GL} \left( V\right) $-submodule of $V^{\otimes d}$ as explained above) is $V^{\otimes d}b_{\lambda}a_{\lambda}$ (because the image of $\Lambda ^{\lambda^{\prime}}V\rightarrow V^{\otimes d}\overset{s_{\lambda} }{\rightarrow}V^{\otimes d}$ is $V^{\otimes d}b_{\lambda}$, and applying $V^{\otimes d}\rightarrow\operatorname*{Sym}\nolimits^{\lambda}V\rightarrow V^{\otimes d}$ to it multiplies it further by $a_{\lambda}$). Likewise, $W_{\lambda}\left( V\right) $ is $V^{\otimes d}a_{\lambda}b_{\lambda }s_{\lambda^{\prime}}$ up to a permutation of tensorands (since the image of $\Gamma^{\lambda}\left( V\right) \rightarrow V^{\otimes d}$ in characteristic $0$ is $V^{\otimes d}a_{\lambda}$, and then the map $V^{\otimes d}\overset{s_{\lambda^{\prime}}}{\rightarrow}V^{\otimes d}\rightarrow \Lambda^{\lambda^{\prime}}\left( V\right) \rightarrow V^{\otimes d}$ is just multiplying it by $b_{\lambda}s_{\lambda^{\prime}}$). Hence, we must prove that \begin{align*} V^{\otimes d}b_{\lambda}a_{\lambda}\cong V^{\otimes d}a_{\lambda}b_{\lambda }s_{\lambda^{\prime}}\ \ \ \ \ \ \ \ \ \ \text{as }\operatorname*{GL}\left( V\right) \text{-modules.} \end{align*} Since $s_{\lambda^{\prime}}$ is invertible (being just a permutation), we can remove the $s_{\lambda^{\prime}}$ factor, so it remains to show that \begin{align*} V^{\otimes d}b_{\lambda}a_{\lambda}\cong V^{\otimes d}a_{\lambda}b_{\lambda }\ \ \ \ \ \ \ \ \ \ \text{as }\operatorname*{GL}\left( V\right) \text{-modules.} \end{align*}

But this is easy, if you recall the classical fact (e.g., Lemma 5.13.3 in Pavel Etingof et al., Introduction to representation theory) that the product $a_{\lambda}b_{\lambda}$ is a nonzero scalar multiple of an idempotent in the group algebra of $S_{n}$ (once again: in characteristic $0$; otherwise, things are more complicated). Once you know this, you immediately realize that the maps \begin{align*} V^{\otimes d}b_{\lambda}a_{\lambda} & \rightarrow V^{\otimes d}a_{\lambda }b_{\lambda},\\ x & \mapsto xb_{\lambda} \end{align*} and \begin{align*} V^{\otimes d}a_{\lambda}b_{\lambda} & \rightarrow V^{\otimes d}b_{\lambda }a_{\lambda},\\ x & \mapsto xa_{\lambda} \end{align*} are mutually inverse up to a nonzero scalar factor. (More precisely, you need the facts that both $a_{\lambda}b_{\lambda}$ and $b_{\lambda}a_{\lambda}$ are scalar multiples of idempotents; but the two products are rather analogous.) Thus, these two maps are invertible, and hence are $\operatorname*{GL}\left( V\right) $-module isomorphisms (since they are clearly $\operatorname*{GL} \left( V\right) $-linear). Hence, \begin{align*} V^{\otimes d}b_{\lambda}a_{\lambda}\cong V^{\otimes d}a_{\lambda}b_{\lambda }\ \ \ \ \ \ \ \ \ \ \text{as }\operatorname*{GL}\left( V\right) \text{-modules,} \end{align*} qed.

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  • $\begingroup$ Thank you! It will take me some time to go through this, but I appreciate you taking the time to write this answer. $\endgroup$
    – Sunny Sood
    Nov 13, 2023 at 12:49

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