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Stream: deprecated: mathematics

Topic: R[Sn]-R[Sp]-bimodules

Jean-Baptiste Vienney (Jun 02 2023 at 03:12):

In $Mod_{R}$, every module of the form $(A_{1} \oplus ... \oplus A_{n})^{\otimes p}$ is a $R[\mathfrak{S}_{n}]$-$R[\mathfrak{S}_{p}]$-bimodule by making:

• $\mathfrak{S}_{n}$ acting by permutation of the summands $A_{1},...,A_{n}$ and
• $\mathfrak{S}_{p}$ by permutation of the $1^{st}$, ..., $p^{th}$ factors $A_{1} \otimes ... \otimes A_{n}$.

You can imagine that in terms of noncommutative homogeneous polynomials. If you have a non commutative homogeneous polynomial of degree $p$, $P \in R_{p} \langle X_{1},...,X_{n}\rangle$, then $\mathfrak{S}_{n}$ permutes $X_{1},...,X_{n}$ whereas $\mathfrak{S}_{p}$ permutes the factors in every summand of the polynomial which are all of the form $X_{i_{1}}...X_{i_{p}}$ for $1 \le i_{1},...,i_{p} \le n$.

More generally, I guess that $\mathcal{C}[(A_{1} \oplus ... \oplus A_{n})^{\otimes p},(A_{1} \oplus ... \oplus A_{n})^{\otimes p}]$ is a $R[\mathfrak{S}_{n}]$-$R[\mathfrak{S}_{p}]$-bimodule in the same way in a symmetric bimonoidal $R$-linear category $(\mathcal{C},\otimes,\oplus)$.

A $R[\mathfrak{S}_{n}]$-module is the same thing that a $R$-linear representation of $\mathfrak{S}_{n}$, in the same way a $R[\mathfrak{S}_{n}]$-$R[\mathfrak{S}_{p}]$-bimodule is the same thing that a $R$-module which is at the same time a representation of $\mathfrak{S}_{n}$ and a representation of $\mathfrak{S}_{p}$, in a compatible way. Is there a name for such a "module which is a linear representation of two symmetric groups in a compatible way", or more generally for a module which is a linear representation of two groups in a compatible way? If $R$ is a field of characteristic $0$, then Schur functors or Young diagrams classify irreducible representations of a symmetric group. What about the same facts for such double representations?

Jean-Baptiste Vienney (Jun 02 2023 at 04:15):

Well, I've thought a bit. If instead of a bimodule, you look at left-left-bimodule ie. with the two actions on the left which are compatible, then a $R[\mathfrak{S}_{n}],R[\mathfrak{S}_{p}]$-left-left-bimodule is the same thing than a $R[\mathfrak{S}_{n}] \otimes R[\mathfrak{S}_{p}]$-module which is the same that a $R[\mathfrak{S}_{n} \times \mathfrak{S}_{p}]$-module. So what I'm thinking to are the linear representations of the groups of the form $\mathfrak{S}_{n} \times \mathfrak{S}_{p}$ I guess.

Jean-Baptiste Vienney (Jun 02 2023 at 04:16):

The compatibility of the two actions is reflected in the fact that $(\sigma,id).(id,\tau) = (id,\tau).(\sigma,id)$.

Jean-Baptiste Vienney (Jun 02 2023 at 04:19):

And there is a question about these representations on MathOverflow

Jean-Baptiste Vienney (Jun 02 2023 at 04:20):

And the answer is the irreducible representations (on a field of characteristic 0) of $\mathfrak{S}_{n} \times \mathfrak{S}_{p}$ are the tensor product of an irreducible representation of $\mathfrak{S}_{n}$ and an irreducible representation of $\mathfrak{S}_{p}$

Jean-Baptiste Vienney (Jun 02 2023 at 04:20):

So my problem is solved :sweat_smile:

Notification Bot (Jun 02 2023 at 04:20):

Jean-Baptiste Vienney has marked this topic as resolved.

John Baez (Jun 02 2023 at 05:03):

By the way:

And the answer is the irreducible representations (on a field of characteristic 0) of $\mathfrak{S}_{n} \times \mathfrak{S}_{p}$ are the tensor product of an irreducible representation of $\mathfrak{S}_{n}$ and an irreducible representation of $\mathfrak{S}_{p}$

this last (well-known) result is a thing @Joe Moeller, @Todd Trimble and I needed in our paper on Schur functors. It follows from the fact that the rationals are a splitting field for all the symmetric groups.

John Baez (Jun 02 2023 at 05:05):

By contrast, unless I'm getting mixed up, not every tensor product of two irreducible representations of $\mathbb{Z}/3$ over the rationals gives an irreducible representation of $\mathbb{Z}/3 \times \mathbb{Z}/3$.

John Baez (Jun 02 2023 at 05:08):

To see this, note that even though $\mathbb{Z}/3$ is abelian, over the rationals its regular representation splits into just two irreducible representations, one 2-dimensonal and one 1-dimensional (the trivial representation).

John Baez (Jun 02 2023 at 05:10):

Then tensor that 2d rep with itself; the resulting 4d rep of $\mathbb{Z}/3 \times \mathbb{Z}/3$ is not irreducible!

John Baez (Jun 02 2023 at 05:11):

None of this sort of nonsense happens with the symmetric groups.

Jean-Baptiste Vienney (Jun 02 2023 at 14:16):

Thanks, it's interesting. But I don't know what means "the rationals are a splitting field for all the symmetric group". I only know what is a splitting field of a polynomial.

John Baez (Jun 02 2023 at 14:22):

It's explained in our paper Schur functors in the proof of Lemma 6.5 on page 41:

Lemma 6.5. Let k be a field of characteristic zero and let $n_1, . . . , n_p$ be a collection
of natural numbers. Then every irreducible representation of $k[S_{n_1} \times \cdots \times S_{n_p}]$ is a tensor product $\rho_1 \otimes \cdots \otimes \rho_p$ of irreducible representations $\rho_i$ of $k[S_{n_i} ]$ that are determined uniquely up to isomorphism.

John Baez (Jun 02 2023 at 14:23):

Briefly, $k$ is a splitting field for a finite group $G$ if each irreducible representation of $G$ has $k$ as its endomorphism algebra.

Jean-Baptiste Vienney (Jun 02 2023 at 14:24):

Ok, I'm going to read that.

John Baez (Jun 02 2023 at 14:30):

With a polynomial over a field $k$, when you extend the field the polynomial can factor into more factors. But when this doesn't happen - when the polynomial over $k$ doesn't get more factors as you extend the field - we say $k$ is a splitting field for the polynomial.

With a representation of a group over a field $k$, when you extend the field the representation can break up into a sum of more irreducible representations. But when this never happens - when any representation over $k$ doesn't break into more irreducible representations as you extend the field - we say $k$ is a splitting field for the group.

John Baez (Jun 02 2023 at 14:30):

We could also talk about the splitting field of a particular representation but I haven't seen people do that.

John Baez (Jun 02 2023 at 14:31):

The concept of splitting field is very general, and it always means "the field is big enough that extending it further doesn't let you do more."

John Baez (Jun 02 2023 at 14:37):

So, the algebraic closure of a field is always a splitting field for everything, but you usually don't need to extend that far to get a splitting field.

Notification Bot (Jun 02 2023 at 16:08):

Jean-Baptiste Vienney has marked this topic as unresolved.

Jean-Baptiste Vienney (Jun 02 2023 at 16:11):

That's clear to me that if $P \in k[X]$, if $l$ is an extension of $k$, then $P \in l[X]$ and so you can look at new roots in $l$. But that's not so clear to me how if you have a $k$-representation of $G$, it is going to become a $l$-representation of $G$. If $(\rho,V)$ is a $k$-representation of $G$, how do you make it a $l$-representation of $G$?

Jean-Baptiste Vienney (Jun 02 2023 at 16:13):

Is it related to extension of scalars?

Jean-Baptiste Vienney (Jun 02 2023 at 17:26):

Not extension, I guess it uses coextension of scalars.

John Baez (Jun 02 2023 at 18:00):

It's extension of scalars. Form $V' = l \otimes_k V$ and you get a representation $(V', \rho')$ of $G$ via $\rho'(g) = 1_l \otimes \rho(g)$.

John Baez (Jun 02 2023 at 18:03):

In general almost anything you have defined over a field of $k$, like an associative algebra or a Lie algebra or an algebraic variety or a scheme, becomes the same sort of thing over any field extension of $k$, via extension of scalars.

Jean-Baptiste Vienney (Jun 02 2023 at 18:04):

Ok, thanks

Jorge Soto-Andrade (Jun 03 2023 at 13:58):

Jean-Baptiste Vienney said:

In $Mod_{R}$, every module of the form $(A_{1} \oplus ... \oplus A_{n})^{\otimes p}$ is a $R[\mathfrak{S}_{n}]$-$R[\mathfrak{S}_{p}]$-bimodule by making:

• $\mathfrak{S}_{n}$ acting by permutation of the summands $A_{1},...,A_{n}$ and
• $\mathfrak{S}_{p}$ by permutation of the $1^{st}$, ..., $p^{th}$ factors $A_{1} \otimes ... \otimes A_{n}$.

You can imagine that in terms of noncommutative homogeneous polynomials. If you have a non commutative homogeneous polynomial of degree $p$, $P \in R_{p} \langle X_{1},...,X_{n}\rangle$, then $\mathfrak{S}_{n}$ permutes $X_{1},...,X_{n}$ whereas $\mathfrak{S}_{p}$ permutes the factors in every summand of the polynomial which are all of the form $X_{i_{1}}...X_{i_{p}}$ for $1 \le i_{1},...,i_{p} \le n$.

More generally, I guess that $\mathcal{C}[(A_{1} \oplus ... \oplus A_{n})^{\otimes p},(A_{1} \oplus ... \oplus A_{n})^{\otimes p}]$ is a $R[\mathfrak{S}_{n}]$-$R[\mathfrak{S}_{p}]$-bimodule in the same way in a symmetric bimonoidal $R$-linear category $(\mathcal{C},\otimes,\oplus)$.

A $R[\mathfrak{S}_{n}]$-module is the same thing that a $R$-linear representation of $\mathfrak{S}_{n}$, in the same way a $R[\mathfrak{S}_{n}]$-$R[\mathfrak{S}_{p}]$-bimodule is the same thing that a $R$-module which is at the same time a representation of $\mathfrak{S}_{n}$ and a representation of $\mathfrak{S}_{p}$, in a compatible way. Is there a name for such a "module which is a linear representation of two symmetric groups in a compatible way", or more generally for a module which is a linear representation of two groups in a compatible way? If $R$ is a field of characteristic $0$, then Schur functors or Young diagrams classify irreducible representations of a symmetric group. What about the same facts for such double representations?

Bonjour! A late comment on your general question above about two groups acting in a compatible way in the same vector space: you may have a look at Howe duality (easy to google, also as "dual reductive pairs"), named after my colleague Roger Howe at Yale, who was inspired by the case of a symplectic group, and an orthogonal group $O(Q)$ associated to an even dimensional non-degenerate quadratic space $(E, Q)$, acting in $L^2(E)$. The orthogonal groups acts here in a natural way (Anglosaxons say "by a permutation representation") and the symplectic group acts via the so called Weil representation, which is not so "natural" (it comes actually from the oscillator representation in Quantum Mechanics, via the Stone-von Neumann theorem for the Heisenberg group). The two actions commute (this is what you mean, I guess, when you say that they are compatible). A baby example of this would be the case of the natural representation of $O(Q)$ in $L^2(E)$ for a rank 2 quadratic form $Q$. If you have a keen eye you can see a group of intertwining operators of this representation which behaves exactly like $SL(2,K)$ ($K$ denotes the base field, which could be a local or finite field!). The original Howe duality has unfolded in many ways and the question arises whether this is a case of a more general phenomenon, including the classical Schur-Weyl duality, quantum analogues, etc, which could be nicely expressed in categorical terms. I guess John may have some categorical insights on this...
PS For an (old) elementary account of this (no quantum mechanics in sight) you may have a look at http://www.numdam.org/item/10.24033/msmf.240.pdf (chapitre 1) or also JSA_PSPUM_Gelfand-Weil

Jorge Soto-Andrade (Jun 03 2023 at 14:51):

http://JSA_PSPUM_Gelfand-Weil

Jean-Baptiste Vienney (Jun 03 2023 at 15:35):

Thanks, I didn't know about Howe duality. I'm interested by Schur-Weyl duality, in the original version you look at two actions which commute on $(\mathbb{C}^{\oplus n})^{\otimes p}$. First $\mathfrak{S}_p$ acts by permutation of the factors $\mathbb{C}^{\oplus n}$, secondly $GL(n,\mathbb{C})$ acts simultaneously in the same way on the $p$ factors, by $g.(x_{1}\otimes...\otimes x_{p}) = (g.x_{1})\otimes ... \otimes (g.x_{p})$. Now you can look only at the action of $\mathfrak{S}_{n} \subseteq GL(n,\mathbb{C})$ and then you obtain exactly the two compatible actions of the symmetric group I was talking about. I don't know about Howe duality, but there is maybe a way to state different versions of Schur-Weyl duality in a same categorical way. Maybe could we start with something like: in a symmetric monoidal (to get $\otimes$) $\mathbb{C}$-linear (to get $\mathbb{C}$) category with biproducts (to get $\oplus$), Schur-Weyl duality states that ... .

Jean-Baptiste Vienney (Jun 03 2023 at 15:39):

So a question I'm interested about now is: can we generalize Schur-Weyl duality to symmetric monoidal $\mathbb{C}$-linear categories with biproducts such that the monoidal unit $I$ generates the objects ie. every object is isomorphic to $I^{\oplus n}$ for some $n \ge 1$ ?

Jean-Baptiste Vienney (Jun 03 2023 at 15:40):

Jorge Soto-Andrade said:

http://JSA_PSPUM_Gelfand-Weil

There is a problem with this link, doesn't work

Jean-Baptiste Vienney (Jun 03 2023 at 15:42):

Jean-Baptiste Vienney said:

So a question I'm interested about now is: can we generalize Schur-Weyl duality to symmetric monoidal $\mathbb{C}$-linear categories with biproducts such that the monoidal unit $I$ generates the objects ie. every object is isomorphic to $I^{\oplus n}$ for some $n \ge 1$ ?

It is not quite correct, because here you could have $I=\mathbb{H}$ for instance.

John Baez (Jun 03 2023 at 15:46):

How would you make the quaternions into the monoidal unit of C-linear symmetric monoidal category with biproducts?

Jean-Baptiste Vienney (Jun 03 2023 at 15:47):

$Vec_{\mathbb{H}}$?

Jean-Baptiste Vienney (Jun 03 2023 at 15:48):

And using $\mathbb{C} \rightarrow \mathbb{H}$ for the $\mathbb{C}$-linear part

John Baez (Jun 03 2023 at 15:49):

Since the quaternions aren't a field you'll need to say what you mean by Vec_H.

Jean-Baptiste Vienney (Jun 03 2023 at 15:49):

Ok, $Mod_{\mathbb{H}}$ so.

John Baez (Jun 03 2023 at 15:50):

How are you tensoring modules of this noncommutative ring?

Jean-Baptiste Vienney (Jun 03 2023 at 15:50):

Ok, you can't, I didn't think to that.

Jean-Baptiste Vienney (Jun 03 2023 at 15:50):

In fact, that's nice, because I don't want $I \cong \mathbb{H}$.

Jean-Baptiste Vienney (Jun 03 2023 at 15:53):

Now, I think such a category which is not $FVec_{\mathbb{C}}$ would be the finite-dimensional $\mathbb{C}$-vector bundles over a fixed smooth manifold maybe?

Jorge Soto-Andrade (Jun 03 2023 at 16:00):

Jean-Baptiste Vienney said:

Jorge Soto-Andrade said:

http://JSA_PSPUM_Gelfand-Weil

There is a problem with this link, doesn't work

hope it works now ...
JSA_PSPUM_Arcata_Gelfand-Weil.pdf

John Baez (Jun 03 2023 at 16:08):

Jean-Baptiste Vienney said:

Now, I think such a category which is not $FVec_{\mathbb{C}}$ would be the finite-dimensional $\mathbb{C}$-vector bundles over a fixed smooth manifold maybe?

Not every vector bundle is a biproduct of copies of the unit object (the trivial line bundle), but maybe the category of trivial finite-dimensional complex vector bundles will have all the properties you seek.

Jean-Baptiste Vienney (Jun 03 2023 at 16:12):

Ok, nice. That's good for me if such categories are barely more than $FVec_{\mathbb{C}}$.

Jean-Baptiste Vienney (Jun 03 2023 at 16:27):

In such a category, for every objects $A,B$, if $n \ge 1$ is the integer such that $A \cong I^{\oplus n}$, you can look at $\mathcal{C}[A^{\otimes p},B] \cong \mathcal{C}[(I^{\oplus n})^{\otimes p},B]$

Jean-Baptiste Vienney (Jun 03 2023 at 16:29):

You have an action of $\mathfrak{S}_{p}$ on this finite-dimensional $\mathbb{C}$-vector space by composing on the left by the symmetry $\sigma_{I^{\oplus n}}:(I^{\oplus n})^{\otimes p} \rightarrow (I^{\oplus n})^{\otimes p}$ associated to $\sigma \in \mathfrak{S}_{p}$

Jean-Baptiste Vienney (Jun 03 2023 at 16:30):

ie. $\sigma.f := \sigma_{I^{\oplus n}};f$

Jean-Baptiste Vienney (Jun 03 2023 at 16:32):

You also have an action of the group of isomorphisms $I^{\oplus n} \rightarrow I^{\oplus n}$ on this hom-set I think

Jean-Baptiste Vienney (Jun 03 2023 at 16:32):

If $\rho:I^{\oplus n} \rightarrow I^{\oplus n}$ is an isomorphism

Jean-Baptiste Vienney (Jun 03 2023 at 16:33):

You can put $\rho.f := \rho^{\otimes p};f$

John Baez (Jun 03 2023 at 17:39):

Jean-Baptiste Vienney said:

... symmetric monoidal $\mathbb{C}$-linear categories with biproducts such that the monoidal unit $I$ generates the objects ie. every object is isomorphic to $I^{\oplus n}$ for some $n \ge 1$ ?

By the way, the category of finitely generated free modules over any commutative algebra A over $\mathbb{C}$ is a symmetric monoidal category of this sort. And the category of trivial complex vector bundles over a space is an example of this, where we take A to be the algebra of functions on this space.

John Baez (Jun 03 2023 at 17:44):

You may even be able to classify categories of this sort.

Jean-Baptiste Vienney (Jun 03 2023 at 17:47):

John Baez said:

You may even be able to classify categories of this sort.

How would you want to do that?

Jean-Baptiste Vienney (Jun 03 2023 at 17:48):

I mean, do you have some idea?

John Baez (Jun 03 2023 at 17:52):

Well, I think I almost listed them all: remember that End(I) is a commutative algebra over $\mathbb{C}$, and that algebra is A when our category was the category if f.g. free A-modules.

So, it seems likely that the algebra End(I) determines the category up to equivalence, for the sort of category we're discussing.

John Baez (Jun 03 2023 at 17:55):

The only flexibility I see is that there can be various choices of symmetry (and associator, maybe).

Jean-Baptiste Vienney (Jun 03 2023 at 18:02):

So maybe every such category is equivalent (maybe with a symmetric monoidal equivalence that preserves biproducts and also linear combination of morphisms) to the category of finitely generated free modules over $End(I)$

John Baez (Jun 03 2023 at 18:06):

Yes, I now think that's true - except you mean finitely genererated free modules. There are usually lots of other f.g. modules

Jorge Soto-Andrade (Jun 03 2023 at 22:51):

Jean-Baptiste Vienney said:

Jean-Baptiste Vienney said:

So a question I'm interested about now is: can we generalize Schur-Weyl duality to symmetric monoidal $\mathbb{C}$-linear categories with biproducts such that the monoidal unit $I$ generates the objects ie. every object is isomorphic to $I^{\oplus n}$ for some $n \ge 1$ ?

It is not quite correct, because here you could have $I=\mathbb{H}$ for instance.

Interesting, but I am afraid that your intended setting does not include the known cases of Howe duality. Notice also that we have also an analogue of Schur-Weyl where you replace complex $GL_n$ by finite $GL_n$ and $\mathbb C^n$ by $L^2(\mathbb F_q^n)$ ($n=p$, in your notations)This is sometimes called Juyumaya duality, after Jesus Juyumaya (which is not a cousin of Azumaya, but a former Ph D student of mine, of Aymara ascent). Check https://arxiv.org/pdf/0801.3633.pdf Again, a baby example is finite $GL_2$ and $\mathfrak S_2$ where $GL_2$ acts via the Weil representation in complex function space on the finite plane and the flip in $\mathfrak S_2$ acts as the Fourier - Grassmann transform J, associated to the unique alternating non-degenerate bilinear form on the finite plane (notice that we have $J^2 = Id$). The concrete question is, however, whether you can find other examples of "dual groups" as spin-offs of your categorical approach.