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• CommentRowNumber1.
• CommentAuthorJohn Baez
• CommentTimeJun 28th 2010

I started a stub on plethysm.

Does anyone know how this mathematical term originated? I hear someone suggested it to Littlewood. But who? And why? And what’s the etymology, exactly?

• CommentRowNumber2.
• CommentAuthorDavid_Corfield
• CommentTimeJun 28th 2010

According to Bruce Westbury here:

Plethysm was introduced in MR0010594 (6,41c) Littlewood, D. E. Invariant theory, tensors and group characters. Philos. Trans. Roy. Soc. London. Ser. A. 239, (1944). 305–365

The term “plethysm” was suggested to Littlewood by M. L. Clark after the Greek word plethysmos $\pi\lambda\eta\theta\nu\sigma\mu o\zeta$ for “multiplication”.

(the Greek is an approximation)

• CommentRowNumber3.
• CommentAuthorJohn Baez
• CommentTimeJun 29th 2010

Thanks! Duly noted!

• CommentRowNumber4.
• CommentAuthorJohn Baez
• CommentTimeJun 29th 2010

By the way, David, you asked Jim Borger about categorifying the free lambda-ring on one generator, Hazewinkel’s beloved object $\Lambda$: the lambda-ring of elementary symmetric functions. The obvious categorification of $\Lambda$ is none other than the category $Schur$ which Todd and I are studying here. This has been known for a long time.

I can’t resist saying a few more things, especially for Todd:

While we list 5 monoidal structures on $Schur$, I believe that 2 of these - those involving Day convolution - might be better understood as comonoidal structures. (It’s not obvious what this means, but I think we could figure it out.) Then the other 2 will make $Schur$ into a ring category, while these 2 will make it into a co-ring category… and I think putting these 4 together, we should get a ’biring category’. This should categorify the fact that $\Lambda$ is a biring.

But $\Lambda$ is more than a biring! It’s a monoid object in the monoidal category of birings! in other words, it’s a plethory!

Why this name? Well, the tensor product of birings was developed by Tall and Wraith, precisely for the purpose of understanding $\lambda$-rings - I wrote a quick explanation of it on the $n$Lab. And I think it’s deeply related to plethysm. Indeed, I’m hoping that the reason $\Lambda$ is a monoid object in birings is precisely because $Schur$ has a 5th monoidal structure: the one Todd calls the ’plethystic tensor product’. This is indeed the quintessential feature of $Schur$.

I would really like to work this stuff out! Not in this paper, but soon! It’s so incredibly beautiful, and it all falls out from the simplest features of the interplay between set theory ($FinSet$) and linear algebra ($FinVect$).

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeJun 29th 2010

There is potentially another ingredient in this general story: the fact that cocommutative coalgebras (aha!) form a cartesian closed category with many wonderful properties, and that many categories (e.g. noncommutative rings) are enriched in this closed category. For some time now I’ve had this gut feeling that such facts have been vastly underexploited.

I guess you’ve heard that Andrew Stacey has done research on Tall and Wraith monoids?

• CommentRowNumber6.
• CommentAuthorJohn Baez
• CommentTimeJun 29th 2010
• (edited Jun 29th 2010)

Yes, the $n$Lab article on Tall-Wraith monoids was written by Andrew in response to prodding by some of us. I think it might benefit from a bit of further polishing by someone comfortable with algebraic theories… but perhaps not so comfortable as to make it unreadable to the unwashed masses!

I just started a stubby page called plethory and updated lambda-ring.

• CommentRowNumber7.
• CommentAuthorJohn Baez
• CommentTimeJun 29th 2010
• (edited Jun 29th 2010)

Btw, I think a functor category $[X,Y]$ wants to be a ring category whenever $Y$ is a ring category, a coring category whenever $X$ is a ring category, and a biring category whenever both $X$ and $Y$ are ring categories. Linear species are the case $X = \mathbb{P}$, $Y = Vect$. So it’s only the plethystic tensor product that’s a bit deeper.

All this is a bit of a wild guess, since I’m just in the process of working out what a ’coring category’ should be. But this is my hope.

• CommentRowNumber8.
• CommentAuthorTodd_Trimble
• CommentTimeJun 29th 2010

Okay, this is getting rather interesting! I’m sort of closing my eyes and meditating on what you’ve suggested above, and maybe I’ll try to jot down some thoughts here.

So if I follow your analogy, $Set$ categorifies to $Cat$ and $Ring$, or let’s say $CRing$ (commutative rings) categorifies to something like SymMonLinCauch. I’m not going to worry about precise details, I just want to see where my thoughts take me.

The underlying functor $U: CRing \to Set$ categorifies to our $U: SymMonLinCauch \to Cat$. In both cases we have a representing object, the free thingy on one generator. It’s the polynomial algebra $\mathbb{Z}[x]$ in one case and $\overline{k\mathbb{P}}$ in the categorified case.

The free thingy on zero variables is $\mathbb{Z}$ in one case and $FinVect_k$ in the categorified case. Guided by the analogy, we might as well rename $\overline{k\mathbb{P}}$ as this: $FinVect[x]$: “polynomial species”.

Now the archetypal (perhaps “tautological” is a better adjective) biring has to be $\mathbb{Z}[x]$ itself. Note that the binary coring operations “coadd” and “comultiply” both have the form

$\mathbb{Z}[x] \to \mathbb{Z}[x] \otimes \mathbb{Z}[x] \cong \mathbb{Z}[x_1, x_2]$

where coadd takes $x$ to $x_1 + x_2$ and comultiply takes $x$ to $x_1 x_2$.

Consider categorifying this. The tensor product is a little complicated here I think; obvious guesses don’t seem to work. (The correct tensor product is easier for me to see in the linearly cocomplete case, not the linearly Cauchy complete case. More later.) Safer to work rather with

$FinVect[x] \to FinVect[x_1, x_2]$

where the right side consists of polynomial species $\mathbb{P} \int 2 \to FinVect$. Here $\mathbb{P} \int 2$ is the symmetric monoidal category on two generators, but in fact it is equivalent to $\mathbb{P} \times \mathbb{P}$. So however the tensor product works, we should have

$FinVect[x] \otimes_{FinVect} FinVect[x] \simeq PolySpecies(\mathbb{P} \times \mathbb{P}, FinVect)$

Let me see if I can understand comultiplication. It is the “unique” such symmetric monoidal linear functor of this form induced by the object (element)

$\hom(x_1, -) \otimes \hom(x_2, -) \cong hom(x_1 x_2, -): \mathbb{P} \times \mathbb{P} \to FinVect$

where $\otimes$ is Day convolution. (Throwing all variance issues to the wind here – okay since $\mathbb{P}$ is a groupoid.) Coaddition is induced by

$\hom(x_1, -) \oplus \hom(x_2, -): \mathbb{P} \times \mathbb{P} \to FinVect$

Actually I don’t think the co-operations here are tremendously exciting. But the monoid structure on the categorified biring is! I think it is precisely the plethystic tensor product as you surmise.

I don’t firmly grasp the monoidal product on birings yet (or the generalized version used by Tall-Wraith). But I think if we call it $\odot$, then (let’s say in the commutative ring case) $\mathbb{Z}[x]$ is the unit for $\odot$, and also there might be a canonical function

$U R \times U S \to U(R \odot S)$

(may try to justify this at some point soon) so that in particular, the composite

$\mathbb{Z}[x] \times R \to \mathbb{Z}[x] \odot R \stackrel{unit iso}{\to} R$

sends the pair $(f(x), r)$ to $f(r)$. [Cf. Andrew’s remark that Tall-Wraith monoids are the things that act on algebras of a Lawvere theory.] In the case where $R = \mathbb{Z}[x]$ itself, this gives the monoid structure of polynomial composition. The categorified version of polynomial composition is the plethystic monoidal product, so I feel some confidence in my guess.

So yes, I think the entire story of $\mathbb{Z}[x]$ must categorify in more or less the way you surmise, and the Grothendieck rig gives back $\Lambda$.

This truly is great stuff!

• CommentRowNumber9.
• CommentAuthorJohn Baez
• CommentTimeJun 29th 2010

I’m too sleepy to say anything intelligent about what you just wrote, so I’ll say something unintelligent:

Cool! Let’s work out the details! Probably this is another paper..

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeJun 29th 2010

Yes, let’s!

By the way, I should point out that one of my statements in my long response in a query box at Schur functor (the box comes after the representability result) said that Cauchy completion is strong monoidal wrt the tensor product on $V$-Cat, which is nonsense. It should say it is lax monoidal (and that’s enough for my purposes).

• CommentRowNumber11.
• CommentAuthorDavid_Corfield
• CommentTimeJun 29th 2010

This is great stuff! Did you notice James’ thought that “much of the structure Hazewinkel lists is redundant”?

• CommentRowNumber12.
• CommentAuthorzskoda
• CommentTimeJun 29th 2010

I added a paper by Svrtan and mentioned the book of Fulton and Harris.

• CommentRowNumber13.
• CommentAuthorJohn Baez
• CommentTimeJun 29th 2010

David wrote:

Did you notice James’ thought that “much of the structure Hazewinkel lists is redundant”?

Yes, that’s part of what got me thinking about this business. Though I think “follows from a few key structures” sounds better than “redundant”.

• CommentRowNumber14.
• CommentAuthorJohn Baez
• CommentTimeJun 30th 2010
• (edited Jun 30th 2010)

Actually I don’t think the co-operations here are tremendously exciting.

But isn’t coaddition just another way of describing the usual “product” of species - the product that corresponds to multiplication of generating functions? I don’t know if that counts as tremendously exciting to you, but it is to me. I guess I’m rather excitable.

Just so someone other than Todd might be able to follow what I’m saying: we usually use Day convolution to take disjoint union as an operation on $\mathbb{P} = core(FinSet)$ and turn it into a monoidal structure on $[\mathbb{P},Vect]$. And this is the usual product of $Vect$-valued species. But if we can use it to put a comonoidal structure on $[\mathbb{P},Vect]$, this would be arguably ’more natural’, since operations in the ’source’ $X$ of a mapping space $[X,Y]$ generally give co-operations on $[X,Y]$.

Btw, I’m adding a bit more stuff on the etymology of plethysm.

• CommentRowNumber15.
• CommentAuthorTodd_Trimble
• CommentTimeJul 1st 2010
• (edited Jul 1st 2010)

I’m having some slight difficulty myself following what you’re saying here, since coaddition is a co-operation and the product of species is an operation. I think it might help though if we could just run through this in the simpler decategorified setting first, by focusing on the simplest plethory of all: $\mathbb{Z}[x]$.

Let’s consider coaddition on the biring $\mathbb{Z}[x]$. It is given by a ring map

$\mathbb{Z}[x] \to \mathbb{Z}[x] \otimes \mathbb{Z}[x] \cong \mathbb{Z}[x_1, x_2]$

which takes $x$ to the sum $x_1 + x_2$, or equivalently to $x \otimes 1 + 1 \otimes x$. It takes $x^n$ to

$(x_1 + x_2)^n = \sum_{j + k = n} \binom{n}{j} x_{1}^{j} x_{2}^{k}$

or equivalently to

$x^n \mapsto \sum_{j+k = n} \binom{n}{j} x^j \otimes x^k$

(It is possibly also suggestive to think of this as the map $e_n \mapsto \sum_{j+k = n} e_j \otimes e_k$ where $e_n$ is a symbol for $x^n /n!$.) Now there’s a resemblance here between this formula for coaddition and the formula for convolution product. I’m not sure of the best way of putting this, but perhaps we want to say that under some pairing, we have

$\langle coadd(x^n), f(x) \otimes g(x) \rangle = \langle x^n, f(x)g(x) \rangle \qquad (1)$

As a guess, suppose we define the pairing $\langle x^n, f(x) \rangle$ to be the $n^{th}$ Taylor coefficient $(D^n f)(0)$, so that $\langle x^n, \sum_N \frac{a_N x^N}{N!} \rangle = a_n$, and define

$\langle x^j \otimes x^k, f(x) \otimes g(x) \rangle \coloneqq \langle x^j, f(x) \rangle \langle x^k, g(x) \rangle$

Then we have, for $f(x) = \sum \frac{a_j x^j}{j!}$ and $g(x) = \sum \frac{b_k x^k}{k!}$,

$\array{ \langle coadd(x^n), f(x) \otimes g(x) \rangle & = & \langle \sum_{j+k=n} \frac{n!}{j! k!} x^j \otimes x^k, f(x) \otimes g(x) \rangle \\ & = & \sum_{j+k=n} \frac{n!}{j! k!} \langle x^j, f(x) \rangle \langle x^k, g(x) \rangle \\ & = & \sum_{j+k=n} \frac{n!}{j! k!} a_j b_k }$

and

$\array{ \langle x^n, f(x)g(x) \rangle & = & \langle x^n, \sum_N (\sum_{j+k=N} \frac{N!}{j! k!} a_j b_k) \frac{x^N}{N!} \rangle \\ & = & \sum_{j+k=n} \frac{n!}{j! k!} a_j b_k }$

So under that pairing, $coadd$ would indeed be adjoint to the product of polynomials.

All this looks eminently categorifiable. Is this roughly what you have in mind? Or is it something else?

• CommentRowNumber16.
• CommentAuthorJohn Baez
• CommentTimeJul 1st 2010
• (edited Jul 1st 2010)

Yes, that is what I had in mind.

But also very much in mind was Rota’s famous rant about how combinatorists use operations when they should use co-operations. Eventually people listened to him and Hopf algebras became fundamental to combinatorics…. especially my pal Bill Schmitt, who was a student of Rota and did his thesis on Hopf algebras coming from species. So now that’s old news… but I think it’s still new news in the categorified situation.

In case you haven’t heard it, Rota’s rant goes something like this:

People often use operations on a vector space when they should use co-operations; they usually do this where there’s a blatantly obvious basis for that vector space, which lets them identify that space with its dual without even noticing… at least if it’s finite dimensional. When it’s infinite-dimensional, the difference between the vector space and its dual can make the difference between operations and cooperations painfully significant, even in the presence of a basis. But in the finite-dimensional case it’s almost a matter of taste. And in the categorified situation, where infinite sums (now really colimits) converge more easily, it’s again almost a matter of taste.

For example, if we have a group $G$, we can look at the complex functions on $G$, and we can try to define an operation called ’convolution’ of functions on $G$. If $G$ is finite, it works fine and we get an algebra. But if $G$ is infinite, we have to worry about whether an infinite sum converges to make sure convolution is well-defined. Irksome! But we could avoid this nuisance if we followed the tao of mathematics and realized that functions on $G$ want to form a coalgebra, not an algebra. Its only the presence of the standout basis of ’delta functions’ that lets us unconsciously reinterpret this comultiplication as a multiplication.

If we categorify, and say that $G$ is a monoidal category, the analogue of functions on $G$ is presheaves on $G$. We can multiply these using ’Day convolution’, which is just categorified convolution… and we never need to worry about whether the sums converge, because they always do: the sums are now colimits and a presheaf category has all (small) colimits. But one might still argue that the really moral thing is to use comultiplication. Especially if the comultiplication lets us see that we have a Hopf structure.

The problem is that while an operation like

$multiplication : k[G] \otimes k[G] \to k[G]$

can be reinterpreted as a bilinear function

$multiplication : k[G] \times k[G] \to k[G]$

thus relieving us of the obligation to understand tensor products, for a co-operation like

$comultiplication: k[G] \to k[G] \otimes k[G]$

we really need to understand tensor products. And in the categorified situation not enough people understand tensor products! We either need to know about something like the tensor product of cocomplete categories, or we need to know about something like profunctors. Luckily, you and I do know about this stuff. But most combinatorists don’t, I think.

So, I think that in the study of species people have placed a bit too much emphasis on operations and not enough emphasis on co-operations. They make species into a monoidal category in 5 different ways, but don’t seem to notice that 2 of these operations are defined using Day convolution and could just as well, or even better, be thought of as co-operations.

• CommentRowNumber17.
• CommentAuthorJohn Baez
• CommentTimeJul 1st 2010

Actually I should admit that the above is my imagined version of Rota’s rant - I never actually heard him give his rant, I just picked it up through Bill. I think Rota’s actual rant was a bit different… it could perhaps be found here:

Saj-Nicole Joni and Gian-Carlo Rota, Coalgebras and bialgebras in combinatorics, Studies in Applied Mathematics 61 (1979), 93-139.

Gian-Carlo Rota, Hopf algebras in combinatorics, in Gian-Carlo Rota on Combinatorics: Introductory Papers and Commentaries, ed. J. P. S. Kung, Birkhauser, Boston, 1995.

• CommentRowNumber18.
• CommentAuthorTodd_Trimble
• CommentTimeJul 1st 2010

Just for the record, why don’t you say what those five monoidal category structures are?

• CommentRowNumber19.
• CommentAuthorJohn Baez
• CommentTimeJul 2nd 2010

Todd wrote:

Just for the record, why don’t you say what those five monoidal category structures are?

I did that a while back, over at Schur functor. Two of them arise from $\oplus$ and $\otimes$ in $Vect$: given functors $F,G : \mathbb{P} \to Vect$ we get functors

$x \mapsto F(x) \oplus G(x)$

and

$x \mapsto F(x) \otimes G(x)$

The first is usually called “addition” of linear species while the second is sometimes called the Hadamard product.

Two more arise from $+$ and $\times$ in $\mathbb{P}$ via Day convolution. (Here $+$ means the monoidal structure coming from coproduct of finite sets, while $\times$ means the monoidal structure coming from product; of course these aren’t product and coproduct in the groupoid $\mathbb{P}$.) The first of these is usually called “multiplication” of linear species, or sometimes the Cauchy product, while the second has no famous name that I know. (I plan to say a lot about the second in “week300”.)

The last comes from plethysm, and is often called “substitution” of linear species.

My conjecture is that the first two make linear species into a (suitably weakened) ring object in the bicategory of profunctors, the second two should really be thought of as co-operations and make linear species into a (suitably weakened) coring object, so that the first four together make it into a (suitably weakened) biring object - and then all five taken together make it into a (suitably weakened) plethory object.

By the way, I don’t see anything particularly important about using linear species here; if we used set-valued species the first two monoidal structures applied to $F$ and $G$ would give

$x \mapsto F(x) + G(x)$

and

$x \mapsto F(x) \times G(x) \, .$
• CommentRowNumber20.
• CommentAuthorJohn Baez
• CommentTimeJul 2nd 2010
By the way, Todd all your thoughts at #8 look good to me. I think we're just talking about the same stuff in slightly different ways.
• CommentRowNumber21.
• CommentAuthorJohn Baez
• CommentTimeJul 4th 2010

Todd wrote:

The correct tensor product is easier for me to see in the linearly cocomplete case, not the linearly Cauchy complete case. More later.

Here are two questions. First: is there a known tensor product of cocomplete categories with the property that

$Set^{X \times Y} \simeq Set^X \otimes Set^Y ?$

Second: is there a known tensor product of cocomplete $k$-linear categories with the property that

$Vect^{X \times Y} \simeq Vect^X \otimes Vect^Y ?$

It seems there should be in both cases, but I’m having trouble finding references. I’m writing a bit of stuff under biring under the assumption that these structures exist.

A tensor product for the linearly Cauchy complete case would also be great!

• CommentRowNumber22.
• CommentAuthorTodd_Trimble
• CommentTimeJul 5th 2010

Let me warm up to this by considering first the case of sup-lattices, which are cocomplete categories enriched over the walking arrow $2$ (as a cartesian closed category). The idea is that the story there ought to generalize to more general enriched contexts.

For sup-lattices, the tensor product is constructed by analogy with tensor products of abelian groups: functions $f: X \times Y \to Z$ which are bicocontinuous in each separate argument should give rise to a unique cocontinuous $X \otimes Y \to Z$. It’s fun to work out what this tensor product should be! The bicocontinuous maps of the form $X \otimes Y \to 2$ are in natural bijection with cocontinuous maps of the form $X \to Cocont[Y, 2]$. Now cocontinuous maps $f: Y \to 2$ are, by the adjoint functor theorem, left adjoints whose right adjoints $g: 2 \to Y$ are… well we know what right adjoints do to the terminal element of $2$, so they are uniquely determined by where they take the other element of $2$. Hence the right adjoints $2 \to Y$ are in bijection with elements of $Y$. In addition, inequalities $f \leq f'$ between left adjoints $f, f': Y \to 2$ are mated to reverse inequalities $g' \leq g$ between their right adjoints. Therefore we have

$Cocont[Y, 2] \cong Y^{op}$

And so, $Cocont[X \otimes Y, 2] \cong (X \otimes Y)^{op}$, and also

$Cocont[X \otimes Y, 2] \cong Bicocont[X \times Y, 2] \cong Cocont[X, Cocont[Y, 2]] \cong Cocont[X, Y^{op}]$

And so $X \otimes Y \cong (Cocont[X, Y^{op}])^{op}$. In particular, for posets $A$, $B$, if we let $2^{A^{op}}$ be the 2-enriched presheaf category, then

$2^{A^{op}} \otimes 2^{B^{op}} \cong (Cocont[2^{A^{op}}, (2^{op})^B])^{op} \cong [A, (2^{op})^B] \cong ((2^{op})^{A \times B})^{op} \cong 2^{A^{op} \times B^{op}}$

which is exactly the formula you were hoping for, but in the 2-enriched world.

Now let’s work in the general $V$-enriched case, where $V$ is complete cocomplete symmetric monoidal closed. The correct analogue of “sup-lattice” is called a “(V)-total category”, which is a $V$-category $C$ whose enriched Yoneda embedding has an enriched left adjoint, sort of a best-of-possible-worlds notion of cocompleteness in the enriched world. (I’m going to get tired of saying “enriched” all the time, so it should be inserted where needed. I’m also going to be a bit cavalier about matters of size, because I think it can all be sorted out satisfactorily anyway.)

One of the virtues of total categories is that cocontinuous functors between them are the same as left adjoints between them. I claim that the correct tensor product of total categories $X$, $Y$ is exactly as given in the 2-enriched case, that

$X \otimes_{tot} Y \cong (Cocont[X, Y^{op}])^{op}$

and in particular, for presheaf categories $X = V^{A^{op}}$, $Y = V^{B^{op}}$ where $A$ and $B$ are small, that

$V^{A^{op}} \otimes_{tot} V^{B^{op}} \cong V^{(A \otimes B)^{op}}$

in parallel to the calculation above for the 2-enriched case. Here the tensor product in the exponent on the right is the standard tensor product of small $V$-categories.

If $A$ and $B$ are ordinary categories, then we can “$V$-ify” them in the usual way, so that the $V$-ification $A \mapsto A^\sim$ gives

$(A \times B)^\sim \cong A^\sim \otimes B^\sim$

and also the $V$-category of ordinary functors $A \to V$ is equivalent to the $V$-category of enriched functors $A^\sim \to V$. So for instance, for $V = Vect$ we have

$Vect^A \otimes_{tot} Vect^B \simeq Vect^{A^\sim} \otimes_{tot} Vect^{B^\sim} \simeq Vect^{A^\sim \otimes B^\sim} \simeq Vect^{(A \times B)^\sim} \simeq Vect^{A \times B}$

as you wrote down.

References? Wish I knew. There’s an article total category in the nLab with an online link to an article by Max Kelly, but I don’t think he gets into the tensor product stuff. Someone like Ross Street might be a good source of information on this, or maybe Mike can help us out. I have a good feeling it’s been worked out in the literature. As far as the Cauchy complete stuff goes: let me think about it some more when I get a free moment.

Sorry for not writing more in response to our joint venture recently! I’m playing catch-up on various fronts, and of course life with kids in the summer, and barbeques etc., intervenes in innumerable ways. :-)

• CommentRowNumber23.
• CommentAuthorJohn Baez
• CommentTimeJul 5th 2010

Thanks for all this, Todd! It’s just what I need! I’m glad to hear you’ve been distracted by life instead of, say, annoyed at me for some reason. Since Lisa is in Germany now, and my grad students are all travelling, and I don’t have any barbecues or kids to entertain me, I’ve been spending a lot of time working on these Schur-functor-related papers. But that’ll change tomorrow when Lisa shows up - especially since we’re leaving for Singapore on Friday. We’re close to being packed and having the house ready to rent, but we’ll still be pretty frantic, I bet.

I just created two rough drafts of papers on my personal web, ingeniously entitled Schur Functors I and Schur Functors II. The first is taken from stuff at Schur functor, while the second is taken from my latest stuff at biring. I’m just starting to make them look more like papers and less like nLab entries. Feel free to make changes there. If you make further changes on the nLab entries it may get a bit confusing unless someone remembers to copy them over.

About ’$V$-total categories’: I see from Remark 5.7 in Kelly’s article that not all cocomplete categories are $V$-total categories for $V = Set$. But his Corollary 6.5 seems to imply that $V^{X^{op}}$ is total when $X$ is small… and that should be the most important case for me.

• CommentRowNumber24.
• CommentAuthorTodd_Trimble
• CommentTimeJul 5th 2010
• (edited Jul 5th 2010)

I think the correct tensor product for the linearly Cauchy complete case is nothing more complicated than the Cauchy completion of the ordinary tensor product of linear categories. In fact, I’m going to sketch out a proof.

There’s a general formalism for constructing these various fancy tensor products we’re after (e.g. for sup-lattices, for total categories, and for Cauchy complete categories). The general formalism is just an extrapolation of the construction of the tensor product for abelian groups in abstract categorical language, and it all has to do with monoidal monads. I’ll run through it for abelian groups first, and then generalize.

Let $T: Set \to Set$ be the monad whose algebras are abelian groups, let $U: Ab \to Set$ be the underlying functor, and let $F: Set \to Ab$ be the free functor. Both $U$ and $F$ are monoidal functors (with $F$ strong monoidal), and in fact we have an adjunction $F \dashv U$ in the 2-category of monoidal categories, so $T$ is a monad in the 2-category of monoidal categories, hence a monoidal monad. The object is to reconstruct $\otimes$ on $Ab$ using the monoidal data of $T$, $U$, and $\times$ on $Set$. Then we’ll generalize to other monads $T$, for example the Cauchy completion monad for categories enriched in $Vect$.

The tensor product of abelian groups $A$ and $B$ is a quotient of $T(U A \times U B)$. The canonical basis elements of $T(U A \times U B)$ are the elements in the image of the unit $U A \times U B \to T(U A \times U B)$, taking $(a, b) \in U A \times U B$ to the formal expresssion $a \otimes b$. The monoidal structure on $T$ gives a map

$\array{ \phi_{A B}: T(U A) \times T(U B) & \to & T(U A \times U B) \\ (\sum_i a_i, \sum_j b_j) & \mapsto & \sum_{i, j} a_i \otimes b_j }$

The bilinearity relation $(a + a') \otimes b \sim a \otimes b + a' \otimes b$ requires us to coequalize two functions. The expression on the left corresponds to the function

$T U A \times U B \stackrel{\theta_A \times 1}{\to} U A \times U B \stackrel{unit}{\to} T(U A \times U B)$

where $\theta_A$ is the algebra structure on $A$, and the one on the right to the function

$T U A \times U B \stackrel{1 \times unit}{\to} T U A \times T U B \stackrel{\phi_{A B}}{\to} T(U A \times U B)$

Of course, we really want to coequalize not just two functions, but the two linear functions obtained freely from these functions. For the left expression, the linear function is

$T(T U A \times U B) \stackrel{T(\theta_A \times 1)}{\to} T(U A \times U B)$

and for the one on the right, the linear function is the threefold composite

$T(T U A \times U B) \stackrel{T(1 \times unit)}{\to} T(T U A \times T U B) \stackrel{T(\phi_{A B})}{\to} T T(U A \times U B) \stackrel{m}{\to} T(U A \times U B)$

where $m$ indicates the monad multiplication. So we have two linear maps which we are to coequalize:

$T(T U A \times U B) \stackrel{\to}{\to} T(U A \times U B)$

and the bilinearity relation on the other side (with sums of $b$’s) requires us to coequalize two similarly formed linear maps

$T(U A \times T U B) \stackrel{\to}{\to} T(U A \times U B)$

and in the end, the underlying set of $A \otimes B$ is the colimit of the diagram consisting of these four maps. $U(A \otimes B)$ carries a $T$-algebra structure because we set up the coequalizer in the linear world (in other words, in $T$-$Alg$).

This is perfectly general and gives us the “correct” tensor product on $T$-$Alg$ whenever $T$ is a monad in the 2-category of monoidal categories, (lax) monoidal functors, and monoidal transformations. It also applies to 2-monads $T$ in the 3-category of 2-monoidal categories, and this is where Cauchy completion lives: as a 2-monad of the form $Vect$-$Cat \to Vect$-$Cat$. I think I remarked on the fact that Cauchy completion is a 2-monoidal functor when I was running through the fine print of the proof of representability, within a query box on the Schur functor page.

Thus on general grounds we know how to construct the Cauchy completion tensor product, as a coequalizer of the form

$\overline{A \otimes B} \to A \otimes_{Cauchy} B$

where the bar indicates the Cauchy completion monad. I claim this coequalizer is actually an equivalence.

To show this, I need to mention another general phenomenon in this story that is really quite pleasant. In my previous comment we discussed things like the equivalence

$Vect^{A^{op}} \otimes_{tot} Vect^{B^{op}} \simeq Vect^{(A \otimes B)^{op}}$

Here the 2-functor $A \mapsto Vect^{A^{op}}$ is the free enriched cocompletion (or free totalization, if you will). Similarly we have an isomorphism

$F(A) \otimes F(B) \cong F(A \times B)$

where $F$ is the free abelian group functor $Set \to Ab$. The general phenomenon is that the left adjoint in a monoidal adjunction is always strong monoidal, as indicated in these two examples. This is a very basic lemma which crops up in various guises all over the place. In the present instance, what it means for Cauchy completion is that there is an equivalence

$\overline{A} \otimes_{Cauchy} \overline{B} \simeq \overline{A \otimes B}$

However, remember that Cauchy completion is an idempotent monad. In particular, if $A$, $B$ are already Cauchy complete, then we have $A \simeq \overline{A}$, $B \simeq \overline{B}$, and therefore

$A \otimes_{Cauchy} B \simeq \overline{A \otimes B}$

which brings us full circle to the first sentence of this comment.

So it’s all working out without a hitch (and things are making all kinds of sense at various levels).

• CommentRowNumber25.
• CommentAuthorDavid_Corfield
• CommentTimeJul 5th 2010

If $Schur$ is the categorification of $\Lambda$, how should we categorify that $\Lambda$

turns up as the homology of the classifying space $B U$ and also as the cohomology of that space, illustrating its self-duality. It turns up as the direct sum of the representation spaces of the symmetric group and as the ring of rational representations of the infinite general linear group.

• CommentRowNumber26.
• CommentAuthorTodd_Trimble
• CommentTimeJul 5th 2010

That $\Lambda$ is the cohomology ring of $B U$ is surely related to the fact that $B U$ classifies vector bundles up to isomorphism, and $Schur$ acts naturally on the category of vector bundles (over any space), in fact probably “is” the category of endomorphisms on $Bundle$ in much the same way as it is the category $End(U)$ of endo-modifications on $U$ (as on the page Schur functor). The latter half of the statement I guess comes down to $Schur \simeq \overline{k\mathbb{P}}$, coupled with Schur-Weyl duality.

But this would be a good thing to make precise.

• CommentRowNumber27.
• CommentAuthorDavid_Corfield
• CommentTimeJul 5th 2010

Two more things (maybe that’s enough easy speculation without doing the work myself):

$\Lambda$ : Lambda-ring :: $Schur$ : Schur-???

Is there a Spec(Schur)?

• CommentRowNumber28.
• CommentAuthorJohn Baez
• CommentTimeJul 5th 2010
• (edited Jul 5th 2010)

I keep thinking we should be able to see that the homology of $B U$ is the ring of symmetric functions, $\Lambda$, by combining a few ideas:

1) $B U \simeq B G L$.

2) $B G L$ is obtained functorially from $Fin Vect_{\mathbb{C}}$ using Segal’s construction of spectra from symmetric monoidal categories. (Or is it symmetric monoidal topological categories? I always forget, but I think the topology on the hom-sets is crucial here. Right?)

3) $Schur$ acts on $Fin Vect_{\mathbb{C}}$.

4) $\Lambda$ is the Grothendieck group of $Schur$.

But I’m still missing some puzzle pieces. In particular, how should we describe the homology of a spectrum to make the puzzle pieces fit nicely? And how does the Grothendieck group get into the game? And how does the fact that $Schur$ acts on $Fin Vect_{\mathbb{C}}$ help? It would seem better if $Schur$ sort of were $Fin Vect_{\mathbb{C}}$, since we want a space built from $Fin Vect_{\mathbb{C}}$ to have the Grothendieck group of $Schur$ as its cohomology.

One nice thing is that Segal’s construction of spectra from symmetric monoidal categories is 2-functorial in some sense. So, the fact that $Fin Vect_{\mathbb{C}}$ is a ring category makes $B G L$ into a ring spectrum, and I guess this makes the homology of $B G L$ into a ring. The fact that the tensor product in $Fin Vect_{\mathbb{C}}$ is symmetric makes $B G L$ into an $E_\infty$ ring spectrum, which should make the homology of $B G L$ into a graded-commutative ring. And the fact that $Fin Vect_{\mathbb{C}}$ is a ’$\lambda$-ring category’ should make $B G L$ into a ’$\lambda$-ring spectrum’, which should make its homology into a $\lambda$-ring!

By ’$\lambda$-ring category’ I roughly mean a category on which Schur functors act. Our paper proposes a more precise definition: any symmetric monoidal Cauchy-complete linear category.

I don’t know exactly what a ’$\lambda$-ring spectrum’ is - does anyone ever talk about those? But the idea that $B G L$ is a $\lambda$-ring spectrum is not very deep: it’s simply that any Schur functor gives a map from $FinVect_{\mathbb{C}}$ to itself, and thus a map from $B G L$ to itself. Since $B G L$ classifies (stable) vector bundles, this in turn gives a functorial recipe for turning vector bundles into new vector bundles. But we know what this recipe must be! Simply apply our Schur functor to each fiber!

• CommentRowNumber29.
• CommentAuthorJohn Baez
• CommentTimeJul 5th 2010

Todd wrote:

That $\Lambda$ is the cohomology ring of $B U$ is surely related to the fact that $B U$ classifies vector bundles up to isomorphism, and $Schur$ acts naturally on the category of vector bundles (over any space), in fact probably “is” the category of endomorphisms on $Bundle$ in much the same way as it is the category $End(U)$ of endo-modifications on $U$ (as on the page Schur functor).

Yes! Just a couple of annoying nitpicky comments that don’t detract from your actual point. First, $B U$ classifies vector bundles not up to isomorphism but only ’stably’. The space that classifies vector bundles up to isomorphism is the disjoint union of the $B U(n)$’s, but to form $B U$ we need to squash down that disjoint union a bit. Second, it’s not quite true that all endomorphisms of the category of vector bundles come from Schur functors: as some mean guy once taught me, there are also endomorphisms like the ’exterior algebra of a vector bundle’ functor, which come from infinite direct sums of Schur functors. This can probably be finessed. But it would, in fact, be nice to know someday all the endomorphisms of the category of vector bundles (over arbitrary spaces).

• CommentRowNumber30.
• CommentAuthorJohn Baez
• CommentTimeJul 5th 2010

David wrote:

Two more things (maybe that’s enough easy speculation without doing the work myself):

$\Lambda$ : Lambda-ring :: $Schur$ : Schur-???

Is there a Spec(Schur)?

The first one is not so hard: I’m claiming that a categorified lambda-ring is a category on which Schur functors act. More precisely, it’s a category on which the monoidal category $Schur$ (with its plethystic tensor product) acts.

Or it may be even better to give conditions on a category $C$ that imply that $Schur$ acts on it. That’s what Todd has done: $C$ should be a symmetric monoidal Cauchy-complete $k$-linear category where $k$ is some field of characteristic zero. If we take any category of this sort, $Schur$ acts on it, and its Grothendieck group becomes a $\lambda$-ring. So I’d be happy calling any such category a ’lambda-ring category’.

But there are also some more examples, where we drop the ’characteristic zero’ assumption. So, I don’t think we should limit the term ’lambda-ring category’ to the class we understand so far.

As for

Is there a Spec(Schur)?

do you mean its spectrum as a ring category, or as a lambda-ring category? I could venture a guess on the former. The latter sounds like the sort of thing James Borger might understand. I don’t know what the spectrum of a lambda-ring is.

• CommentRowNumber31.
• CommentAuthorDavid_Corfield
• CommentTimeJul 6th 2010
• (edited Jul 6th 2010)

So, I don’t think we should limit the term ’lambda-ring category’ to the class we understand so far.

Then there’s the more general P-ring to categorify into a P-ring category. Are there standard examples of P other than $\Lambda$? Plethystic Algebra is the place to look I guess:

In section two, we give some examples of plethories. The most basic is the symmetric algebra $S(A)$ of any cocommutative bialgebra $A$; in particular, if $A$ is a group algebra $Z G$, then $S(A)$ is the free polynomial algebra on the set underlying $G$. These plethories are less interesting because their actions on rings can be described entirely in terms of the original bialgebra $A$; for example, an action of the plethory $S(ZG)$ is the same as an action of the group $G$. But even in this case, there can be more maps between two such plethories than there are between the bialgebras, and in some sense, this is ultimately responsible for existence of $\lambda_p$ and hence the $p$-typical Witt ring.

Is $Schur$ then a plethory-category?

Is there a Spec(Schur)?

do you mean its spectrum as a ring category, or as a lambda-ring category?

I meant just as a ring category. There seems to be a whole bunch of extensions of the ring concept to which Spec can be applied: Durov’s generalized rings, Connes and Consani and their hyperrings, and then $E_{\infty}$-rings, and ring categories. No doubt someone will come up with generalized $E_{\infty}$-hyperring categories.

I could venture a guess

• CommentRowNumber32.
• CommentAuthorjdc
• CommentTimeJul 6th 2010

I just looked through the n-lab entry on Schur functors and I’m confused about some of the statements there. Maybe I’m just interpreting things incorrectly.

In the first section, the Young symmetrizer $p_\lambda$ is defined, and then it says that $p_\lambda$ lies in the center of the group algebra. I don’t think this is true. In general, $p_\lambda$ really depends on a tableaux $T$ of shape $\lambda$ (to map the boxes to the factors in the tensor product) and if $\sigma$ is a permutation, $\sigma p_T \sigma^{-1}$ is equal to $p_S$, where $S$ is the tableaux that has the entries of $T$ permutated by $\sigma$. For example, take the Young diagram corresponding to the partition (2,1) labelled with 1 and 2 in the first row and 3 in the second, and take $\sigma$ to be the transposition (2,3). You can work this case out explicitly.

As another example, is $S$ and $T$ are standard tableaux of the same shape but with different entries, then one of $p_S p_T$ and $p_T p_S$ is automatically zero, but the other is sometimes (but rarely) non-zero. (The failure to notice this is a mistake in both the book by James and Kerber and Cvitanovic’s book.)

Also, even if $p_\lambda$ were central, I don’t follow the proof given that claims to show that it is idempotent. (It is, of course, but the proof is a bit harder.)

Later in that section, it says that the image $S_\lambda(V)$ of $p_\lambda : V^{\otimes n} \to V^{\otimes n}$ is invariant under $S_n$, but this is also false. Taking the same partition (2,1) as above, $S_\lambda(V)$ consists of tensors which are symmetric in the first 2 variables and antisymmetric in the first and third. This condition isn’t invariant under exchange of factors.

All of those comments were about the first section. I also have a comment about the section on the action of Young symmetrizers. There it defines $p_\lambda$ as the projection onto a certain matrix algebra inside the group algebra.
I just wanted to point out that this definition doesn’t reduce to the one in the first section. Each $p_\lambda$ from the first section projects onto a single irrep inside the group algebra, while the $p_\lambda$’s here project onto sums of irreps (viewing $C[S_n]$ as an $S_n$-rep under multiplication from one side). You might hope that the $p_\lambda$ from this section is just a sum of the $p_\lambda$’s from the first section indexed by standard tableaux of shape $\lambda$ (and others have thought this too), but the non-orthogonality of the Young symmetrizers makes this false in general. While the images of the $p_\lambda$’s (indexed by all standard tableaux) do give a direct sum decomposition of $C[S_n]$, the projections $p_\lambda$ are not the natural projection maps of that decomposition!

I’ve recently found a way to describe those projections, but I have to rush off now, so I’ll write more later.

• CommentRowNumber33.
• CommentAuthorTodd_Trimble
• CommentTimeJul 7th 2010
• (edited Jul 7th 2010)

Thanks jdc –

John put in the stuff on the explicit description of Young symmetrizers, so I’ll let him bear most of the brunt in addressing these questions (while I run off scot-free (-: ). But in some sense I think it’s no big deal, because the whole approach we’re taking in this paper is based on general abstract theory which doesn’t actually need the explicit combinatorial descriptions. I suspect John put that stuff in just to round out the discussion for those accustomed to the more combinatorial approaches.

One small note about the idempotence of $p_\lambda = p^A p^S$: when I first read that, I puzzled about it too. But the equation

$p^A p^S p^A p^S = p^A p^A p^S p^S$

is clearer if we expand it to say (using centrality of $p_\lambda$):

$p^A p^S p^A p^S = p_\lambda p^A p^S = p^A p_\lambda p^S = p^A p^A p^S p^S$

which reduces to $p^A p^S = p_\lambda$ if $p^A$ and $p^S$ are idempotent. Does this answer to your question there, or was it something else?

As for your penultimate paragraph: what you’re finding fault with again harks back to the descriptions of the first section. But suppose we ignore the first section! Instead, start all over again, by invoking Maschke’s theorem (which implies that the group algebra is a finite product of matrix algebras), and defining elements $p_\lambda$ to be the primitive central idempotent elements of $k[S_n]$ that correspond to identity elements of the matrix algebras. (Don’t worry for now about the Young diagram/tableaux meaning of $\lambda$; just take it to be a symbol that indexes isomorphism classes of irreducible representations, one for each matrix algebra.) Then, in any symmetric monoidal linear category in which idempotents split, you can interpret how $p_\lambda$ acts as an idempotent operator $X^{\otimes n} \to X^{\otimes n}$ for any object $X$. The images of these projections give the $S_\lambda(X)$ that we are defining en route to the Schur functor $S_\lambda$.

This is not to dismiss your criticisms (which are appreciated!), but rather to emphasize that I think they don’t much affect the approach we’re really taking. We could just erase the stuff about Young symmetrizers in the beginning, and I think we’d be okay.

• CommentRowNumber34.
• CommentAuthorjdc
• CommentTimeJul 7th 2010

Thanks for explaining why having $p_\lambda$ central would imply that it is idempotent. Unfortunately (as I said), the usual Young symmetrizers are not idempotent, so this proof doesn’t apply to them. And the later $p_\lambda$’s, being projectors onto isotypic components, are clearly idempotent.

I haven’t read the rest of the Schur functor entry in detail, but it could be that you are right that one could ignore the first section and let the rest stand on its own. However, then you are in the situation that what you write as $S_\lambda$ wouldn’t be the usual Schur functor which (for example) produces irreps of the unitary groups. It would be a much larger gadget which might be isomorphic to a sum of the usual $S_\lambda$’s.

Moreover, I think the true richness of the representation theory of the symmetric group lies precisely in how the isotypic components break up into irreps in a way that has beautiful combinatorics underlying it. If the only ingredient you use is the decomposition into isotypic components (which is canoncial, of course), then I’d be surprised if you can reproduce the theory of Schur functors. But I’m no expert on Schur functors, so maybe I’m wrong.

• CommentRowNumber35.
• CommentAuthorjdc
• CommentTimeJul 7th 2010
• (edited Jul 7th 2010)

By the way, I didn’t mean to post semi-anonymously above. It was my first time posting to the n-forum and I was in a rush. Hopefully this comment shows up correctly.

Also, in my rush, my comments weren’t phrased very carefully. I should have said that I think your approach is very cool, and I’m looking forward to reading it more carefully to understand what’s really going on with Schur functors. Moreover, even if I’m right that your approach produces reducible Schur functors, I suspect that it can be adapted to reproduce the usual story.

… Hmm, it still just shows “jdc” instead of “Dan Christensen”. I ticked the box to let people know my real name, but it still doesn’t show up.

• CommentRowNumber36.
• CommentAuthorTodd_Trimble
• CommentTimeJul 7th 2010
• (edited Jul 7th 2010)

Thanks Dan (glad we got that squared away!). I certainly recognize your name, and it’s nice to hear from you!

Funny – I had had a worry that I suspect corresponds to one of the points that you brought up, and then for some reason thought there was nothing to worry about after all (and retracted my worry), and now I’m worried again! Well, not actually worried – I think the original approach to the Schur functors that I had written down works fine. John had done some rewriting, and the original approach was to be moved to another section, but I didn’t get around to doing that yet.

Here was my original approach. The group algebra $k[S_n]$ is a bimonoid in the symmetric monoidal category of finite-dimensional vector spaces over $k$, $FinVect_k$. Now for any symmetric monoidal Cauchy complete $Vect_k$-enriched category $C$, there is a symmetric monoidal enriched functor $FinVect_k \to C$, uniquely so up to isomorphism. Being a symmetric monoidal functor, it carries bimonoids to bimonoids. Thus, in any such $C$, we have a bimonoid which by abuse of notation we again call $k[S_n]$. Also, if $V_\lambda$ is an irreducible module over $k[S_n]$ in $FinVect_k$, the same symmetric monoidal functor carries $V_\lambda$ to a module over the bimonoid $k[S_n]$ in $C$. Again, by abuse of notation, we call this object $V_\lambda$.

Now, the category of modules $Mod(B)$ over a bimonoid $B$ in a symmetric monoidal category $C$ carries a monoidal category structure. The tensor product on $Mod(B)$ is constructed with help from the comultiplication on $B$, and the monoidal unit with help from the counit, and the underlying functor $U: Mod(B) \to C$ is a strong (symmetric) monoidal functor. Now for any object $X$ of $C$, the $n$-fold tensor $X^{\otimes n}$ is naturally a module over $k[S_n]$. Take the tensor product of $V_\lambda$ and $X^{\otimes n}$ in $Mod(k[S_n])$.

Finally, the average of all the group elements,

$e = \frac1{n!} \sum_{\sigma \in S_n} \sigma,$

defines an idempotent operator on the module $V_\lambda \otimes X^{\otimes n}$. The splitting of this idempotent in $C$ (which we assumed Cauchy complete) gives the object of coinvariants $V_\lambda \otimes_{S_n} X^{\otimes n}$. This gives $S_\lambda(X)$, hence defines the Schur functor $S_\lambda$.

Do you agree?

Okay, now I think the description of $S_\lambda$ that is in the current draft doesn’t work. Here we start with an idempotent $e_\lambda$ corresponding to the identity of the matrix algebra (on the underlying vector space of the isotype $V_\lambda$), and form a corresponding idempotent operator on $X^{\otimes n}$. This operator

$E_\lambda: X^{\otimes n} \to X^{\otimes n}$

could also be written

$X^{\otimes n} \cong k[S_n] \otimes_{S_n} X^{\otimes n} \stackrel{e_\lambda \otimes_{S_n} 1}{\to} k[S_n] \otimes_{S_n} X^{\otimes n} \cong X^{\otimes n}$

and the identity on $X^{\otimes n}$ would be $\sum_\lambda E_\lambda$ where $\lambda$ ranges over Young diagrams. But, left multiplication by the idempotent,

$e_\lambda: k[S_n] \to k[S_n],$

splits through $\hom_{Vect}(V_\lambda, V_\lambda)$ which as a module is isomorphic to $V_{\lambda}^{d(\lambda)}$, where $d(\lambda) \coloneqq \dim(V_\lambda)$. So we’re not getting

$V_\lambda \otimes_{S_n} X^{\otimes n}$

as we want, but rather

$V_{\lambda}^{d(\lambda)} \otimes_{S_n} X^{\otimes n}$

which is too big! So I seem to agree: the current description we have is flawed, and maybe we should revert to the original approach.

John, is this making sense to you?

• CommentRowNumber37.
• CommentAuthorjdc
• CommentTimeJul 7th 2010

While I haven’t checked every detail, I think everything that you wrote is correct.

By the way, instead of $V_\lambda \otimes_{S_n} X^{\otimes n}$, I have seen $Hom_{S_n}(V_\lambda, X^{\otimes n})$ used to describe the Schur functor. This should correspond to dualizing $V_\lambda$, so it would just give a different mapping between irreps and Schur functors.

• CommentRowNumber38.
• CommentAuthorjdc
• CommentTimeJul 7th 2010

Now that I have a bit more time, let me say something about what I’ve been thinking about on this topic.

First, I really like the definition of a Schur functor as $Hom(V, X^{\otimes n})$ (or the dual version), with $V$ an $S_n$-rep, since it avoids using Young tableaux as an intermediate step. (On the other hand, while concise, it doesn’t give the same intuition as the definition using Young symmetrizers.)

But rather than focusing on this as a functor of $X$, I’ve been thinking about it as a functor of $V$. Write $\mathbb{P}$ for the groupoid of finite sets and bijections (or the obvious skeleton, namely the coproduct of the groupoids $S_n$). Then consider at the category of representations of $\mathbb{P}$. As you note, this category has a monoidal structure, which takes representations $V$ and $W$ of $S_p$ and $S_q$ to $V \otimes W$ induced up from $S_p \times S_q$ to $S_{p+q}$, and then extended linearly to general representations of $\mathbb{P}$. (This is called the induction product in the literature, and it puts a ring structure on the direct sum over $n$ of the spaces of class functions on $S_n$.)

If you fix a vector space $X$ (or an object in a category $C$ like you’ve described), you get a functor $Ch$ sending a representation $V$ of $S_n$ to $Hom(V, X^{\otimes n})$, and this extends linearly to the category of representations of $\mathbb{P}$. It’s an interesting and important fact that $Ch$ is monoidal.

In my applications, $X = \mathbb{C}^k$ is the defining representation of the unitary group $U(k)$, so $Ch$ sends representations of $\mathbb{P}$ to representations of $U(k)$. (Decategorified, the character of $Ch(X)$ is called the $U(k)$-characteristic of the character of $V$. That’s where the notation $Ch$ comes from. I’m not trying to name something after myself! :-)

Now, the category of representations of $U(k)$ has duals, so it has a natural trace structure (which just takes the usual trace of a linear map). The category of representations of $\mathbb{P}$ does not have duals, but it nevertheless has an obvious trace structure. (See Selinger’s paper, A survey of graphical languages for monoidal categories, Chapter 5 for what I mean by a trace structure.)

It’s clear that the functor $Ch$ doesn’t preserve traces, since the dimensions of $V$ and $Hom(V, X^{\otimes n})$ usually don’t agree. However, it really makes sense to think of an irrep as being 1-dimensional, and if you adjust the two trace structures to take this into account, you find that $Ch$ does preserve traces. And this leads to an important computational consequence: if you want to evaluate a $U(k)$ spin network which is $Ch$ applied to a $\mathbb{P}$ spin network, then as long as the diagram factors through one irrep (which controls the normalization factors), it suffices to evaluate the $\mathbb{P}$ spin network you started with. The amazing thing is that the value of the $\mathbb{P}$ spin network is independent of $k$! This explains and generalizes some techniques that Cvitanovic describes in his book Group Theory, and turns out to be useful in lattice gauge theory.

One reason I’m writing this to find out if anyone has seen this functor $Ch$ written down before. My searches have turned up nothing.

• CommentRowNumber39.
• CommentAuthorJohn Baez
• CommentTimeJul 7th 2010

I always get confused about central idempotents versus non-central idempotents in group algebras - I worry about this constantly. So it’s quite possible I’m getting mixed up again in the $n$Lab entry. Let me try to explain:

If $G$ is a finite group, the group algebra $\mathbb{C}[G]$ is a direct sum of matrix algebras, one for each irrep $\lambda$ of $G$. The element “1” in each of these matrix algebras gives a central idempotent $p_\lambda$ in $\mathbb{C}(G)$, and these idempotents add up to $1 \in \mathbb{C}(G)$. If $X$ is any rep of $G$ it follows that $p_\lambda X$ is a subrep of $X$, and the direct sum of these subreps is all of $X$.

Now specialize to the case where $G = S_n$ and $X = V^{\otimes n}$. This is what I was trying to talk about. I think this construction gives one Schur functor

$V \mapsto p_\lambda V^{\otimes n}$

for each Young diagram.

(None of this depends on having a specific formula for the central idempotents $p_\lambda$. I actually think my formula for the central idempotents is right, but that’s a separate, detachable question.)

But now I think I see the problem that’s been nagging at me all the time. These Schur functors of mine coming from Young diagrams are direct sums of copies of some other, irreducible Schur functors! And those are the ones that Todd is talking about.

Does that sound right?

• CommentRowNumber40.
• CommentAuthorjdc
• CommentTimeJul 7th 2010
• (edited Jul 7th 2010)

Yes, John, that sounds right. The idempotents coming from the decomposition into matrix algebras are just the projections onto the isotypic components, not the projections onto a decomposition into irreps. (When you view $C[S_n]$ as an $S_n \times S_n$-rep using left and right multiplication, then this is the decomposition into irreps; but not when you view it as an $S_n$-rep with just one of the actions.)

And while the usual Young symmetrizers (indexed by all standard tableaux) do have images that form a direct sum decomposition into irreps, they are not the projections associated to that decomposition because they are not orthogonal.

That’s the other thing I’ve figured out recently: if you order the standard tableaux with $n$ boxes using last-letter order (or maybe the reverse, depending exactly on your conventions), you find that $p_S p_T = 0$ for $S \gt T$. Write the list of all standard tableaux as $T_1, T_2, \ldots, T_r$ in last letter order and write $p_i$ for $p_{T_i}$. For each $i$, define $g_i = p_i (1-p_{i+1}) \cdots (1-p_r)$. Then the $g_i$’s have the same images as the $p_i$’s and are the natural projections onto these images (so, in particular, they are orthogonal).

As I mentioned earlier, it’s a common mistake to assume that the $p_i$’s are orthogonal, but it turns out that some of the work done using the $p_i$’s can be redone with the $g_i$’s (with additional complications due to the extra factors).

So my next question: has anyone seen these $g_i$’s anywhere? I couldn’t find them written down.

• CommentRowNumber41.
• CommentAuthorzskoda
• CommentTimeJul 7th 2010

I wrote a new section Plethysm product and symmetric operads within the entry plethysm. Please check.

• CommentRowNumber42.
• CommentAuthorJohn Baez
• CommentTimeJul 7th 2010

Dan Christensen (aka “jdc”) wrote:

That’s the other thing I’ve figured out recently: if you order the standard tableaux with $n$ boxes using last-letter order (or maybe the reverse, depending exactly on your conventions), you find that $p_S p_T = 0$ for $S \gt T$.

I think the fact you just mentioned is Lemma 4.23 of Fulton and Harris’ Representation Theory. I didn’t know your formula for orthogonal idempotents $g_i$. But clearly I’m not the person who would know this sort of thing.

I’m still a bit confused - so let me ask a question to see if I’m finally catching on. Todd has described a Schur functor for each irrep $V_\lambda$ of $S_n$ as follows:

$X \mapsto V_\lambda \otimes_{k[S_n]} X^{\otimes n}$

Is the following functor naturally isomorphic to his? Take $k[S_n]$ and write it as a sum of matrix algebras, one for each $\lambda$. Take the element “1” in the $\lambda$th matrix algebra and write it as a sum of commuting minimal idempotents - there’s a lot of arbitrariness in how to do this, but the sum will have $dim(V_\lambda)$ terms. Take any one of these minimal idempotents and call it $q_\lambda$. Then consider the functor

$X \mapsto q_\lambda X^{\otimes n}$

Is this naturally isomorphic to Todd’s?

• CommentRowNumber43.
• CommentAuthorJohn Baez
• CommentTimeJul 7th 2010
• (edited Jul 7th 2010)

Dan wrote:

The category of representations of $\mathbb{P}$ does not have duals,

A representation of $\mathbb{P}$ has duals if it’s just a finite direct sum of finite-dimensional reps of $S_n$’s, no?

• CommentRowNumber44.
• CommentAuthorJohn Baez
• CommentTimeJul 8th 2010

Dan wrote:

If you fix a vector space $X$ (or an object in a category $C$ like you’ve described), you get a functor $Ch$ sending a representation $V$ of $S_n$ to $Hom(V, X^{\otimes n})$, and this extends linearly to the category of representations of $\mathbb{P}$. It’s an interesting and important fact that $Ch$ is monoidal.

This is part of the point of our approach to Schur functors. You can think of $Schur$ as the category of representations of $\mathbb{P}$ that are finite direct sums of irreducible representations. In our $n$Lab paper, Todd shows that $Schur$ is the ’free symmetric monoidal $k$-linear Cauchy complete category on one object’. Call this object $x$. So, if you fix any object in any category $C$ of this sort, you get an essentially unique $k$-linear symmetric monoidal functor

$Ch: Schur \to C$

sending $x$ to $X$.

But it probably should be called $Tr$ rather than $Ch$.

I’ve been dreaming of things like this for years, starting in my paper on 2-Hilbert spaces, where I described universal properties for the categories of finite-dimensional unitary representations of $U(k)$, and $SU(k)$, and $O(k)$, and $SO(k)$, and $Sp(k)$. But Todd figured out the right class of symmetric monoidal categories for which $Schur$ is the free guy on one object.

• CommentRowNumber45.
• CommentAuthorjdc
• CommentTimeJul 8th 2010

When I wrote that $p_S p_T = 0$ when $S \gt T$, that was just background that is well known. That was just leading up to what I haven’t seen before, which is the formula for the $g_i$’s.

And when I said that the category of $\mathbb{P}$-representations doesn’t have duals, I meant with respect to the induction product. If $W$ was a dual for a rep $V$ of $S_n$, then it would have to have a non-trivial component of degree $-n$, but these things are graded by the natural numbers. I’m using the induction product throughout, so when I talk about spin networks in the category of $\mathbb{P}$-representations, parallel lines mean induction product! I haven’t seen these sorts of spin networks discussed explicitly before, but they are very natural from a certain point of view. For example, if $V_i$ is the image of an idempotent $e_i$ in $C[S_{n_i}]$, for $i = 1, 2$, then the induction product of $V_1$ with $V_2$ is the image of an idempotent $e_1 . e_2$ which is the element of $C[S_{n_1 + n_2}]$ formed by placing $e_1$ and $e_2$ “side-by-side”.

You asked about choosing a minimal idempotent $q_\lambda$. Yes, if you do, I think you’ll get the usual Schur functor. But I think a lot of the interest in the representation theory comes from the explicit way the various irreps of $S_n$ embed in $C[S_n]$, and how they interact (which is revealed by the non-commuting of the Young symmetrizers).

• CommentRowNumber46.
• CommentAuthorJohn Baez
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

Dan wrote:

And when I said that the category of $\mathbb{P}$-representations doesn’t have duals, I meant with respect to the induction product.

Yeah, sorry, I realized that in the gym while thinking more about your post. My first comment about that was fired off before I had absorbed what you were really saying in that post. I eventually caught on:

It’s with respect to the induction product that $Schur$ is the free symmetric monoidal $k$-linear Cauchy complete category on one object. And that one object can’t have a dual because if it did, there would be an isomorphism

$X^* \cong F(X)$

that held for some fixed Schur functor $F$ and every object $X$ in every symmetric monoidal $k$-linear Cauchy complete category $C$. And this fails already when $C$ is the category of finite-dimensional reps of $U(1)$.

I’m using the induction product throughout, so when I talk about spin networks in the category of $\mathbb{P}$-representations, parallel lines mean induction product! I haven’t seen these sorts of spin networks discussed explicitly before, but they are very natural from a certain point of view.

I guess you’re thinking of them as ’generic’ spin networks that don’t care which representation of which group is the fundamental object $X$ that you’re using to build up fancier objects via Schur functors. And then you’re trying to go as far as you can in this ’generic’ context before specializing to a particular group. Right?

This reminds me of a paper I wrote about 2d Yang-Mills theory with gauge group $SU(N)$ versus 2d Yang-Mills theory with gauge group $SU(\infty)$. This was an attempt to formalize some ideas physicists have about the $N \to \infty$ limit of $SU(N)$ gauge theory. The Hilbert space for the circle in 2d $SU(\infty)$ Yang-Mills theory has a basis given by Young diagrams. To get the Hilbert space for $SU(N)$ Yang-Mills theory you can impose some relations called ’Mandelstam relations’ which cut the basis down to Young diagrams with fewer than $N$ rows. These correspond to the usual basis of class functions on $SU(N)$ - the usual basis of states for $SU(N)$ Yang-Mills on the circle.

At the time I didn’t fully understand how this was related to the functor $Schur \to Rep(SU(N))$, but it was one of the things that got me interested in this…

You asked about choosing a minimal idempotent $q_\lambda$. Yes, if you do, I think you’ll get the usual Schur functor.

Okay, good. I’ve been thinking about that and it’s starting to seem obvious.

But I think a lot of the interest in the representation theory comes from the explicit way the various irreps of $S_n$ embed in $C[S_n]$, and how they interact (which is revealed by the non-commuting of the Young symmetrizers).

I guess so, but this has always struck me as confusing rather than interesting. I wish someone would say some of the key ideas in something resembling plain English, but people always seem to take a highly computational approach. So I’ve been trying and (as you can see) not quite succeeding in doing what I want to do while avoiding a lot of these computations. So, thanks for straighteing me out. I think something like Todd’s original approach, but really downplaying Young tableaux, will make me happiest.

• CommentRowNumber47.
• CommentAuthorjdc
• CommentTimeJul 8th 2010

Dan wrote:

I’m using the induction product throughout, so when I talk about spin networks in the category of $\mathbb{P}$-representations, parallel lines mean induction product!

John wrote:

I guess you’re thinking of them as ’generic’ spin networks that don’t care which representation of which group is the fundamental object $X$ that you’re using to build up fancier objects via Schur functors. And then you’re trying to go as far as you can in this ’generic’ context before specializing to a particular group. Right?

Right! And amazingly, the correction factor in spin network evaluations is just a polynomial in $N$, where $N$ is the $N$ in $SU(N)$!

John wrote:

This reminds me of a paper I wrote about 2d Yang-Mills theory with gauge group $SU(N)$ versus 2d Yang-Mills theory with gauge group $SU(\infty)$.

I’m also keeping in mind that it might be relevant to computations in the $N \to \infty$ limit of lattice gauge theory.

Dan wrote:

But I think a lot of the interest in the representation theory comes from the explicit way the various irreps of $S_n$ embed in $C[S_n]$, and how they interact (which is revealed by the non-commuting of the Young symmetrizers).

John wrote:

I guess so, but this has always struck me as confusing rather than interesting. I wish someone would say some of the key ideas in something resembling plain English, but people always seem to take a highly computational approach.

Well, I don’t think I have enough of an overview to answer this to your satisfaction. All I can say is that many people put large amounts of effort into understanding the properties of the minimal idempotents $p_\lambda$, even though for abstract reasons it’s clear that there exists a set $g_i$ of minimal central idempotents. I’m hoping my work helps bridge the gap here, by revealing that the $g_i$’s can be easily expressed in terms of the $p_\lambda$’s.

Why care about the $p_\lambda$’s at all? Well, they have some very nice properties. For example, as I mentioned earlier, if $\sigma$ is a permutation then $\sigma^{-1} p_S \, \sigma$ is $p_T$, where the tableaux $T$ is obtained from $S$ by permuting the entries using $\sigma$. This doesn’t work for the $g_i$’s. Moreover, by staring at the wiring diagram (maybe for a long time!) you can convince yourself that for any tableaux $S$ and $T$ of the same shape, and any $\sigma$,

$p_S \sigma p_T = m p_S \tau$

where $m$ is in $\{ 0, 1, -1 \}$ and $\tau$ is the permutation sending $T$ to $S$. This is extremely useful for computations involving composites of the Young symmetrizers. So, except for the fact that they aren’t orthogonal, the Young symmetrizers are very nicely behaved.

Another reason why studying $p_S$ for all tableaux $S$ rather than just for one of each shape is that there is no canonical ordering to the boxes, so why should you pick one arbitrarily?

And a third reason why the interrelationships are important is that one is often really working with $V^{\otimes n}$, where $V$ is the defining rep of $SU(N)$, and you have a specific tensor that you want to express as a sum of others with nice symmetry properties. This is a vague way of describing what happens in lattice gauge theory, but I think it’s pretty general.

So I’ve been trying and (as you can see) not quite succeeding in doing what I want to do while avoiding a lot of these computations. So, thanks for straighteing me out. I think something like Todd’s original approach, but really downplaying Young tableaux, will make me happiest.

I’m also all for downplaying the Young tableaux and making the story more functorial. I’m just pointing out that it might not explain the whole story, just a portion of it. (But hopefully I’m wrong!)

• CommentRowNumber48.
• CommentAuthorJohn Baez
• CommentTimeJul 9th 2010

I have tried to fix the mistake Dan noticed in the Schur functor article - at least in the first two sections, namely the idea and definition section, and the section called the category of Schur functors. These two sections are supposed to be nice and readable now - and correct, too.

Please, everyone, see if these two sections look okay!

I have not yet fixed the later sections.

• CommentRowNumber49.
• CommentAuthorjdc
• CommentTimeJul 9th 2010

Those first two sections look good to me. My only quibble is that you don’t explicitly define what you mean by a Schur functor, but instead say things like “may also be called”. This gets a bit confusing near the end where you say that the composite of two Schur functors is a Schur functor, and it’s not completely clear whether this is a theorem or is a further extension of the definition. (I know you mean it to be a theorem. As a hint of the proof, maybe it would be worth saying that a composite of polynomials is a polynomial?)

Incidentally, can Schur functors be characterized as the functors $FinVect \to FinVect$ such that some property holds?

[I’ll be away from Friday afternoon until Sunday night and won’t be able to post during that time. And I think John is leaving tomorrow for Singapore! Have a great time, John!]

• CommentRowNumber50.
• CommentAuthorTodd_Trimble
• CommentTimeJul 9th 2010

As partway toward the answer of characterizing Schur functors, the more general analytic functors (where the $R(n)$ may be nonzero for infinitely many $n$) were characterized by Joyal in an appendix of his paper on the Lie species, Lecture Notes in Mathematics 1234: necessary and sufficient conditions are that they preserve filtered colimits and weak pullbacks, and there may be one other condition that I’m forgetting.

But what is notable about his result is that he shows how to extract the coefficients $R[n]$ of the species from the analytic functor $R(x) = \sum_{n \geq 0} R[n] \otimes_{S_n} X^{\otimes n}$. This is by no means obvious!

This is related to a worry I have: John wrote that there is a category $Schur$ consisting of Schur functors $FinVect \to FinVect$ and natural transformations between them. Then later there is a claimed equivalence between this category and the category of polynomial species. Apparently the claim is that $Schur$ is by definition the essential image of a full embedding

$PolySpecies \to [FinVect, FinVect]$

Now (1) it’s true, but I don’t see it as quite obvious, that this is a full embedding – I suspect one has to grapple with the theory laid out in Joyal’s appendix to prove this result, unless I’m missing something, and (2) as jdc asks, there is still the question of what are the objects of the essential image, intrinsically?

See, this is the beauty of going beyond the conception of Schur functors as certain endofunctors on $FinVect$ to the much wider playing field of endofunctors on general symmetric monoidal Cauchy complete $Vect$-categories (or in fancier but more precise language, pseudonatural transformations $U \to U$ developed later in the paper). There are two important points:

(1) Instead of trying to characterize classical Schur endofunctors just on $FinVect$ (as functors satisfying certain conditions like preserving filtered colimits, weak pullbacks, etc.), we instead concentrate the essence of “Schurness” by saying that they are precisely those gadgets that can be defined on all symmetric monoidal linearly Cauchy complete categories in such a way as to respect “change of base”: the precise way to express it is to say that Schur functors are, seen in the proper light, pseudonatural transformations $U \to U$ (as you, John, had suggested from the get-go). And also,

(2) By expanding to that bigger playing field, we can then invoke the representability result and get the theorem that the category of pseudonatural transformations and modifications really is the category of polynomial species! This is much softer and easier (less technical) than Joyal’s analysis!

I happen to think that’s quite elegant and conceptual, and I think this is what we should be gently hinting at in the introduction. Ultimately we are trying to say that Schur functors are utterly simple and natural and conceptually inevitable, and so are things like plethysm. But somehow I fear that basic message is now getting lost or masked by saying things like

This description of irreducible representations of $S_n$ [in terms of Young symmetrizers] paves the way toward an important generalization of Schur functors…

because it seems to place the Young symmetrizer approach at the forefront of the development, which is not the direction we’re really taking.

• CommentRowNumber51.
• CommentAuthorTodd_Trimble
• CommentTimeJul 9th 2010

One other “complaint” (not sure that’s the right word) is the sentence

Conceptually, the importance of $\mathbb{P}$ is that it is a skeleton of the groupoid of finite sets and bijections.

That’s one importance, but perhaps to me the chief conceptual importance is that $\mathbb{P}$ (or finite sets and bijections) is the free symmetric monoidal category on one generator: that’s the main thing we’re using about $\mathbb{P}$ throughout the analysis.

• CommentRowNumber52.
• CommentAuthorDavid_Corfield
• CommentTimeJul 9th 2010

Does $\Lambda$ have a universal characterisation in terms of the category of plethories? Otherwise, why is it described as the “most famous example”? (Then again, I suppose the most famous finite simple group doesn’t have a universal characterisation.) If it did, that might give a way to characterise $Schur$ in its bicategory.

• CommentRowNumber53.
• CommentAuthorJohn Baez
• CommentTimeJul 10th 2010
• (edited Jul 10th 2010)

It’s funny that you don’t like the emphasis on Young symmetrizers, Todd, because that’s precisely the emphasis that I want to avoid in our actual paper. But the classical treatment of Schur functors relies on Young symmetrizers very heavily. In fact, some people seem to think a Schur functor is a functor coming from a Young symmetrizer! Of course these are just what we’d call the irreducible Schur functors. But I figure the $n$Lab article should explain the whole story. So I wanted to explain Schur functors coming from Young symmetrizers in the introductory section, and then show how direct sums of these ’irreducible’ Schur functors are precisely functors of this form:

$X \mapsto S_R(X) = \sum_{n \ge 0} R_n \otimes_{\mathbb{C}[S_n]} X^{\otimes n}$

where $R = \{R_n\}$ is any list of finite-dimensional reps of $S_n$ with $R_n = 0$ for large enough $n$. Then I wanted to note that such a list $R$ can be thought of as a ’polynomial species’ $R: core(FinSet) \to Vect$. And then we can say “Bye-bye, Young symmetrizers!”

I did all this, more or less, in the first 2 sections of the $n$Lab article - but I guess I didn’t sufficently emphasize the “disposable” nature of the Young symmetrizers.

There may also be a problem - as Dan suggests - with explaining the concept of Schur functor in a way where the definition keeps expanding in scope. I want it to be clear the definitive concept for us was the category $Schur$ (defined over an arbitrary but fixed field of characteristic zero). But it could probably stand to be a lot clearer.

I guess Todd will take a crack at improving everything…

• CommentRowNumber54.
• CommentAuthorJohn Baez
• CommentTimeJul 10th 2010
• (edited Jul 10th 2010)

Todd wrote:

One other “complaint” (not sure that’s the right word) is the sentence

Conceptually, the importance of $\mathbb{P}$ is that it is a skeleton of the groupoid of finite sets and bijections.

That’s one importance, but perhaps to me the chief conceptual importance is that $\mathbb{P}$ (or finite sets and bijections) is the free symmetric monoidal category on one generator: that’s the main thing we’re using about $\mathbb{P}$ throughout the analysis.

I guess I meant to say something like “Hey! Glomming all the symmetric groups to form a groupoid is not just some ad hoc technical trick! This is a groupoid you know and love! Finite sets! And that means we’re talking about linear structure types!”

Whenever I say something like “the importance of…” or “the true meaning of…”, it’s not that I really think there’s just one true significance to anything. I really just mean “hey, this is a really interesting shift of viewpoint, that feels deep!” Indeed I sometimes say the real reason the groupoid of finite sets is important is that it’s the free symmetric monoidal category on one object.

So, feel free to change the wording there. I just wanted to get $core(FinSet)$ into the picture.

• CommentRowNumber55.
• CommentAuthorTodd_Trimble
• CommentTimeJul 10th 2010

@Zoran #41: More or less. You need some cocompleteness assumptions on $C$ to be assured that colimit exists. Or else you need to put some restriction on what sorts of $M$’s are allowed (for example, 0 for all but finitely many $n$, if it deserves to be called polynomial). But basically it’s the right idea, and what is being spelled out over at Schur functor. The plethystic tensor product goes by a number of names, for example “substitution product” of (polynomial) linear species. Cf. also the more general notion of club.