How is Schur-Weyl duality (specifically, the fact that the actions of the group ring $\mathbb{K}\left[ S_{n}\right] $ and the monoid ring $\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] $ on the tensor power $V^{\otimes n}$ are each other's centralizers) for a field $\mathbb{K}$ of characteristic $0$ proven constructively?

Let me now define the notions and explain what I mean by "constructively" and what I want to avoid.

## Notations

Let $\mathbb{K}$ be a field of characteristic $0$. Fix $n\in\mathbb{N}$, and let $S_{n}$ be the symmetric group of the set $\left\{ 1,2,\ldots,n\right\} $.

Let $V$ be a finite-dimensional $\mathbb{K}$-vector space. The symmetric group $S_{n}$ acts on the $n$-th tensor power $V^{\otimes n}$ by permuting the tensorands:

$\sigma\left( v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) =v_{\sigma^{-1}\left( 1\right) }\otimes v_{\sigma^{-1}\left( 2\right) }\otimes\cdots\otimes v_{\sigma^{-1}\left( n\right) }$ for all $\sigma\in S_{n}$ and $v_{1},v_{2},\ldots,v_{n}\in V$.

Thus, the group ring $\mathbb{K}\left[ S_{n}\right] $ acts on $V^{\otimes n}$ as well (by linearity). This makes $V^{\otimes n}$ into a $\mathbb{K} \left[ S_{n}\right] $-module.

On the other hand, the monoid $\left( \operatorname*{End}V,\cdot\right) $ acts on $V^{\otimes n}$ as follows:

$M\left( v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) =Mv_{1}\otimes Mv_{2}\otimes\cdots\otimes Mv_{n}$ for all $M\in\operatorname*{End}V$ and $v_{1},v_{2},\ldots,v_{n}\in V$.

Thus, the monoid ring $\mathbb{K}\left[ \left( \operatorname*{End} V,\cdot\right) \right] $ acts on $V^{\otimes n}$ as well. This makes $V^{\otimes n}$ into a $\mathbb{K}\left[ \left( \operatorname*{End} V,\cdot\right) \right] $-module.

(Many authors tend to restrict this module to a $\mathbb{K}\left[ \operatorname*{GL}V\right] $-module, but this doesn't feel particularly natural to me. Either way, these things behave pretty much interchangeably.)

**Schur-Weyl duality** makes the following two claims:

(a)Each endomorphism of the $\mathbb{K}\left[ S_{n}\right] $-module $V^{\otimes n}$ is the action of some element of $\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] $.

(b)Each endomorphism of the $\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] $-module $V^{\otimes n}$ is the action of some element of $\mathbb{K}\left[ S_{n}\right] $.

In general, "some element" is not uniquely determined, as none of the two module structures is faithful. The $\mathbb{K}\left[ S_{n}\right] $-module structure is faithful when $n\leq\dim V$; the $\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] $-module structure is probably never faithful. The quotients that do act faithfully can be described, but this is a different story.

## How is this usually proven?

For a theorem that appears in every other book on representation theory,
Schur-Weyl duality seems to have a shortage of actually distinct proofs. The
argument (as given, e.g., in §4.18 and §4.19 of Pavel Etingof et al,
*Introduction to representation theory*, arXiv:0901.0827v5) proceeds,
roughly, as follows: [**EDIT:** The proof outlined in the following is neither the simplest nor the slickest version of the standard argument. The Etingof-et-al text does it in a much clearer way, by factoring out some of the semisimple-modules arguments into a general lemma. As pointed out by commenters, David Speyer and Mark Wildon (in MO question #90094) have further elementarized the argument, but their versions are still not as lightweight as I'd like them to be (e.g., they still use Schur's lemma, requiring proof of absolute irreducibility).]

First prove part

**(a)**using fairly elementary methods. (Outline: Let $f$ be an endomorphism of the $\mathbb{K}\left[ S_{n}\right] $-module $V^{\otimes n}$. Write $f$ as a $\mathbb{K}$-linear combination of endomorphisms of the form $f_{1}\otimes f_{2}\otimes\cdots\otimes f_{n}$, where each $f_{i}$ is in $\operatorname*{End}V$. Since $f$ is $\mathbb{K} \left[ S_{n}\right] $-equivariant, we can symmetrize it, so that $f$ also becomes a $\mathbb{K}$-linear combination of endomorphisms of the form $\dfrac{1}{n!}\sum_{\sigma\in S_{n}}f_{\sigma\left( 1\right) }\otimes f_{\sigma\left( 2\right) }\otimes\cdots\otimes f_{\sigma\left( n\right) } $, where each $f_{i}$ is in $\operatorname*{End}V$. It remains to show that each endomorphism of the latter form is the action of some element of $\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] $. This is done using some polarization identity, e.g., $\sum_{\sigma\in S_{n} }f_{\sigma\left( 1\right) }\otimes f_{\sigma\left( 2\right) }\otimes \cdots\otimes f_{\sigma\left( n\right) }=\sum_{I\subseteq\left\{ 1,2,\ldots,n\right\} }\left( -1\right) ^{n-\left\vert I\right\vert }\left( \sum_{i\in I}f_{i}\right) ^{\otimes n}$.)Let $B$ be the $\mathbb{K}$-algebra $\operatorname*{End} \nolimits_{\mathbb{K}\left[ S_{n}\right] }\left( V^{\otimes n}\right) $. Then, $B$ is a quotient of $\mathbb{K}\left[ \left( \operatorname*{End} V,\cdot\right) \right] $ because of part

**(a)**. Thus, $\operatorname*{End} \nolimits_{\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] }\left( V^{\otimes n}\right) =\operatorname*{End}\nolimits_{B} \left( V^{\otimes n}\right) $.Recall that $\mathbb{K}\left[ S_{n}\right] $ is a semisimple algebra (by Maschke's theorem), and thus $V^{\otimes n}$ decomposes as $V^{\otimes n}=\bigoplus_{\lambda\in\Lambda}V_{\lambda}\otimes L_{\lambda}$ for some finite set $\Lambda$, some nonzero $\mathbb{K}$-vector spaces $V_{\lambda}$ and some pairwise non-isomorphic simple $\mathbb{K}\left[ S_{n}\right] $-modules $L_{\lambda}$. Conclude that $\operatorname*{End} \nolimits_{\mathbb{K}\left[ S_{n}\right] }\left( V^{\otimes n}\right) \cong\prod_{\lambda\in\Lambda}\operatorname*{End}\left( V_{\lambda}\right) $. This is a particularly tricky step, since several things are happening at once here: First of all, we need to know that $\operatorname*{End} \nolimits_{\mathbb{K}\left[ S_{n}\right] }\left( L_{\lambda}\right) \cong\mathbb{K}$, which would be a consequence of Schur's lemma if we assumed that $\mathbb{K}$ is algebraically closed, but as we don't, requires some knowledge of the representation theory of $S_{n}$ (namely, of the fact that the simple $\mathbb{K}\left[ S_{n}\right] $-modules are the Specht modules, and are absolutely simple). But even knowing that, we need to know that the endomorphism ring of a direct sum of irreducible $\mathbb{K}\left[ S_{n}\right] $-modules decomposes as a direct product according to the isotypic components, and on each component is a matrix ring. This is standard theory of semisimple algebras, but also requires a nontrivial amount of work.

Now, $B=\operatorname*{End}\nolimits_{\mathbb{K}\left[ S_{n}\right] }\left( V^{\otimes n}\right) \cong\prod_{\lambda\in\Lambda} \operatorname*{End}\left( V_{\lambda}\right) $, so the $V_{\lambda}$ for $\lambda\in\Lambda$ are the simple $B$-modules. Hence, the decomposition $V^{\otimes n}=\bigoplus_{\lambda\in\Lambda}V_{\lambda}\otimes L_{\lambda}$ can be viewed as a decomposition of the $B$-module $V^{\otimes n}$ into simples. Hence, the endomorphisms of the $B$-module $V^{\otimes n}$ are direct sums of the form $\bigoplus_{\lambda\in\Lambda}\operatorname*{id} \nolimits_{V_{\lambda}}\otimes f_{\lambda}$, where each $f_{\lambda}$ lies in $\operatorname*{End}\left( L_{\lambda}\right) $. (This, again, requires some basic semisimple module theory.) It is now straightforward to show that all such isomorphisms are actions of elements of $\mathbb{K}\left[ S_{n}\right] $ (indeed, the $L_{\lambda}$ are pairwise non-isomorphic simple $\mathbb{K} \left[ S_{n}\right] $-modules, and thus $\prod_{\lambda\in\Lambda }\operatorname*{End}\left( L_{\lambda}\right) $ is a quotient of $\mathbb{K}\left[ S_{n}\right] $). Due to $\operatorname*{End} \nolimits_{\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] }\left( V^{\otimes n}\right) =\operatorname*{End}\nolimits_{B} \left( V^{\otimes n}\right) $, this yields part

**(b)**.

## What do I want?

I am fairly happy with the proof of part **(a)** given above, but the proof of
part **(b)** is exactly the kind of argument I shun. It is implicit,
non-constructive and relies on half a semester's worth of representation
theory. Probably my biggest problem with it is aesthetical -- I see **(b)** as
a combinatorial problem (at least a lot of its invariant-theoretical
applications are combinatorial in nature), but the proof is combing this cat
completely against the grain (if not pulling it along its tail). But asking
for a combinatorial or explicit proof is not a well-defined problem, whereas
asking for a constructive one, at least, is clear-cut.

That said, I suspect that a constructive proof can be obtained by some straightforward manipulations of the above argument. The representation theory of $S_{n}$ can be done constructively (see, e.g., Adriano Garsia's notes on Young's seminormal form), and most semisimple-algebra arguments can probably be emulated by plain linear algebra (albeit losing what little intuitive meaning they carry). I would much prefer something that avoids this and either significantly simplifies the representation theory or replaces it by something completely different.

## What has been done?

My hopes for a better proof have a reason: Schur-Weyl duality actually works
in far greater generality than the above proof. Theorem 1 in Steven Doty's
*Schur-Weyl duality in positive characteristic* (arXiv:math/0610591v3)
claims that both **(a)** and **(b)** hold for any infinite field $\mathbb{K}$,
no matter what the characteristic is! The proof in that paper, however, goes
way over my head (it isn't self-contained either, so the 17 pages are not an
upper bound). Another paper that might contain answers is Roger W. Carter and
George Lusztig, *On the Modular Representations of the General Linear and
Symmetric Groups*, but that one looks even less approachable.

Of course, I would love to see a proof that works for any infinite field $\mathbb{K}$, or maybe even more generally for any commutative ring $\mathbb{K}$, assuming that we replace the endomorphisms of the $\mathbb{K} \left[ \left( \operatorname*{End}V,\cdot\right) \right] $-module $V^{\otimes n}$ by a more reasonable notion of $\operatorname*{GL} $-equivariant maps (namely, endomorphisms of $V^{\otimes n}$ that commute with the action of a "generic $n\times n$-matrix" adjoined freely to the base ring). But I would be happy enough to see just vanilla Schur-Weyl duality proven in a neat way.

One step that can be done easily is a proof of part **(b)** in the case when
$\dim V\geq n$. Namely, in this case, we can argue as follows: Let $\left(
e_{1},e_{2},\ldots,e_{d}\right) $ be the standard basis of $V$; thus, $d=\dim
V\geq n$. Let $F$ be an endomorphism of the $\mathbb{K}\left[ \left(
\operatorname*{End}V,\cdot\right) \right] $-module $V^{\otimes n}$. Let
$\eta=F\left( e_{1}\otimes e_{2}\otimes\cdots\otimes e_{n}\right) $. For
every $n$ vectors $v_{1},v_{2},\ldots,v_{n}\in V$, we can find a linear map
$M\in\operatorname*{End}V$ satisfying $v_{i}=Me_{i}$ for all $i\in\left\{
1,2,\ldots,n\right\} $, and thus we have

$F\left( v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) =F\left( Me_{1}\otimes Me_{2}\otimes\cdots\otimes Me_{n}\right) $

$=\left( M\otimes M\otimes\cdots\otimes M\right) \underbrace{F\left( e_{1}\otimes e_{2}\otimes\cdots\otimes e_{n}\right) }_{=\eta}$ (since $F$ is $\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] $-equivariant)

$=\left( M\otimes M\otimes\cdots\otimes M\right) \eta$.

Thus, the value of $\eta$ uniquely determines the endomorphism $F$. Furthermore, we can write $\eta$ as a $\mathbb{K}$-linear combination of pure tensors of the form $e_{i_{1}}\otimes e_{i_{2}}\otimes\cdots\otimes e_{i_{n}}$ and show that, for each such pure tensor that actually occurs in this linear combination (with nonzero coefficient), the $n$-tuple $\left( i_{1} ,i_{2},\ldots,i_{n}\right) $ must be a permutation of $\left( 1,2,\ldots ,n\right) $. (To prove this, we assume the contrary; i.e., assume that the $n$-tuple $\left( i_{1},i_{2},\ldots,i_{n}\right) $ is not a permutation of $\left( 1,2,\ldots,n\right) $, but the tensor $e_{i_{1}}\otimes e_{i_{2} }\otimes\cdots\otimes e_{i_{n}}$ does occur in $\eta$. Thus, either one of the numbers $i_{1},i_{2},\ldots,i_{n}$ is $>n$, or two of these numbers are equal. In the first case, pick an $M\in\operatorname*{End}V$ that sends the corresponding $e_{i_{k}}$ to $0$; in the second, pick an $M\in \operatorname*{End}V$ that multiplies the corresponding $e_{i_{k}}$ by a generic $\lambda$. Either way, again use the $\mathbb{K}\left[ \left( \operatorname*{End}V,\cdot\right) \right] $-equivariance of $F$ to obtain something absurd. Sorry for the lack of details.) The result is that $\eta$ is a $\mathbb{K}$-linear combination of various permutations of $e_1 \otimes e_2 \otimes \cdots \otimes e_n$; and therefore, $F$ (being determined by this $\eta$) is the action of some element of $\mathbb{K}\left[S_n\right]$.

However, this argument completely breaks down when $\dim V<n$, since $e_{1}\otimes e_{2}\otimes\cdots\otimes e_{n}$ does not exist any more. Is there any way to fix it, or is it a dead end?

(*Remark:* This argument for part **(b)** in the case $\dim V \geq n$ is quite similar to the proof of Theorem 3.6 in Tom Halverson, Arun Ram, *Partition Algebras*, arXiv:math/0401314v2, which in itself is a kind of Schur-Weyl duality (but where the symmetric group acts on each tensorand instead of permuting the tensorands!).)

A few more pointers:

Maybe Weyl's unitary trick gives another proof of Schur-Weyl duality, but it would probably be neither constructive nor combinatorical in my book (Haar measure!).

There are various papers on a combinatorial approach to invariant theory (e.g., D. Eisenbud, D. De Concini, C. Procesi,

*Young diagrams and determinantal varieties*, which appears to be one of the most readable). However, it is not clear whether they give an answer. The invariant theory of the $\operatorname{GL}$-action on tuples of vectors and covectors (by left/right multiplication) is usually derived (in characteristic $0$, in the non-combinatorial approach) from Schur-Weyl duality; however, I don't know how one would go in the reverse direction (after all, the projection from the tensor algebra to the symmetric algebra is not injective). The FFT for the invariant theory of the $\operatorname{GL}$-action on tuples of matrices by (simultaneous) conjugation is actually equivalent to Schur-Weyl duality, but I haven't seen anyone claim a combinatorial approach to that one.Schur's thesis

*Ueber eine Klasse von Matrizen, die sich einer gegebenen Matrix zuordnen lassen*is available online from two places (EUDML/GDZ and archive.org/Harvard), but I am not sure if Schur-Weyl duality is actually in there. (Its notations are sufficiently dated that even searching for the statement is a nontrivial task.)The notion of "Schur-Weyl duality" is not standardized across literature; some authors use this name for different assertions. For example, Daniel Bump, in Chapter 34 of his

*Lie Groups*(2nd edition), proves something he calls "Frobenius-Schur duality", and claims that this is exactly Schur-Weyl duality. But it is not what I call Schur-Weyl duality above; it is just the one-to-one correspondence between representations of symmetric groups and Schur functors.

**UPDATE:** In comments to this post, Frieder Ladisch has alerted me to the fact that the double centralizer theorem (or, rather, the part of the double centralizer theorem that is relevant to the proof of part **b)**) can be proven constructively (provided that the input is sufficiently explicit). And now I am seeing that essentially his proof appears in Section 11.1 of Jan Draisma and Dion Gijswijt, *Invariant Theory with Applications*. (Jan: I took the freedom to guess the URL of the PDF file, seeing that the hyperlink was broken due to an incorrect relative path. If you actually don't want these notes to be linked, please let me know!) Some parts of their argument need to be slightly modified to ensure constructivity: The use of continuity in the proof of Theorem 11.1.1 should be replaced by a straightforward argument using Zariski density. The group $H$ in Theorem 11.1.2 should be required to be finite. The vector space $W$ in Theorem 11.1.2 should be required to be finite-dimensional. The requirement in Theorem 11.1.2 that the representation $\lambda$ be completely reducible should be replaced by a requirement that $\left|H\right|$ is invertible in the ground field. The direct complement $U$ of $M$ in the proof of Theorem 11.1.2 should be constructed using Maschke's theorem, which has a well-known proof relying merely on linear algebra (viz., the existence of a complement of an explicitly-defined subspace of a finite-dimensional vector space).

Of course, this beautiful argument still "feels inexplicit" in the sense that it uses some representation-theoretical ideas. But the worst offenders (Artin-Wedderburn theory, passing to algebraic closure, analysis/geometry etc.) are gone. Had I known this argument in advance, I wouldn't have asked this question. Nevertheless, I am leaving this question open, since I have yet to digest various other answers, some of which appear to lead to more general proofs, maybe even in positive characteristic (for whatever parts of Schur-Weyl duality hold there).

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