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• CommentRowNumber1.
• CommentAuthorJohn Baez
• CommentTimeJun 24th 2010

I’m struggling to further develop the page on Schur functors, which Todd and I were building. But so far I’ve only done a tiny bit of polishing. I deleted the discussion Todd and I were having near the top of the page, replacing it by a short warning that the definition of Schur functors given here needs to be checked to see if it matches the standard one. I created a page on linear functor and a page on tensor power, so people could learn what those are. And, I wound up spending a lot of time polishing the page on exterior algebra. I would like to do the same thing for tensor algebra and symmetric algebra, but I got worn out.

In that page, I switched Alt to $\Lambda$ as the default notation for exterior algebra. I hope that’s okay. I think it would be nice to be consistent, and I think $\Lambda$ is most widely used. Some people prefer $\bigwedge$.

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeJun 24th 2010
• (edited Jun 24th 2010)

I like ‘$Alt$’ because it matches ‘$Sym$’ for the symmetric algebra, and there doesn’t seem to be any other good symbol for that. I think that ‘$\bigwedge$’ is all right, since it reminds us of the wedge product. It seems to me that ‘Λ’ (which iTeX refuses to print, so I have put it in text mode) is justified only because a Lambda looks like a wedge, so why not just use a wedge? And ‘$\mathit{\Lambda}$’ is even worse. Note that TeX intereprets \Lambda as \mathrm{\Lambda} (that is, more or less, ‘Λ’) by default, but iTeX (via MathML) interprets it as \mathit{\Lambda} (that is ‘$\mathit{\Lambda}$’).

• CommentRowNumber3.
• CommentAuthorJohn Baez
• CommentTimeJun 24th 2010

I’ve always thought the standard symbol for the symmetric algebra was S, with Sym being for people who are a bit slow on the uptake.

I don’t like slightly italicized Greek letters either, but I just lean slightly to the right and they’re okay.

• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeJun 24th 2010
• (edited Jun 24th 2010)

By the way, something is mixed up about characteristic $2$. Based on the relations (1) and (2), these facts are certain:

• (1) implies (2),
• (2) implies (1) over a field whose characteristic is not $2$ (or over any any commutative ring in which $2$ is cancellable).

So you don’t need two definitions, one for characteristic $2$ and one for everything else. You just need one definition, depending on which relation you want to impose in characteristic $2$. And I’m pretty sure that you want to impose the stronger condition, (1).

Ultimately, this comes down to what you think ‘graded commutative’ means; restricting to odd grades, it seems that you think that it means (2), while I think that it means (1). (We agree that it means commutativity in even grades and between odd and even grades.)

Edit: Incidentally, the name for (1) is ‘alternating’, while the name for (2) is ‘antisymmetric’.

FWIW, Wikipedia agrees with me, although it does not cite a reference. I plan to check Fulton & Harris, which I believe agrees with me.

• CommentRowNumber5.
• CommentAuthorTobyBartels
• CommentTimeJun 24th 2010

I’ve always thought the standard symbol for the symmetric algebra was S

I guess that you can do that. But ‘S’ is a pretty overloaded letter. And it doesn’t match ‘Λ’; you’d want to use ‘A’ (or ‘E’) instead, which (I hope!) nobody does. So I’m still convinced that ‘Sym’ and ‘Alt’ (not ‘Ext’, of course) are the best pair.

If you don’t like ‘$\Alt$’ and I don’t like ‘$\Lambda$’, can we compromise on ‘$\bigwedge$’?

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeJun 24th 2010

Here’s another problem with ‘$\Lambda$’, which applies also to ‘$\bigwedge$’. The motivation for this notation is, again, that the symbol is (or looks like) a large version of the symbol for the wedge product, so we think of $\bigwedge^n V$ as a kind of power of $V$ (the alternating power, or exterior power, of course). But I think that it is better to denote the exterior power as either $V^{\wedge n}$ (an explicit power, matching $V^{\otimes n}$ for the tensor power) or as $Alt_n V$, with a subscript.

The reason is that I came into exterior algebra from differential topology, and there superscripts mean contravariant functors, while the exterior power is a covariant functor. More specifically, if $M$ is a manifold and $x$ is a point in $M$, then we have the tangent vector space $T_x M$ whose $n$th alternating power is (using my preferred notation) $Alt_n T_x M$. But the space of degree-$n$ differential forms on $M$ near $x$ is the $n$th alternating power of the dual of $T_x M$, which is (using obvious notation) $Alt^n T_x M$. (We can then abbreviate this as ‘$\Omega^n_x M$’; change ‘$n$’ to ‘$*$’ for the direct sum over all degrees.) Writing ‘$\Lambda^n T_x M$’ for $Alt_n T_x M$ could be confusing. (Well, it confused me once, if I remember correctly, although that was years ago.)

OK, that’s kind of obscure, but it’s one more reason that I feel uncomfortable with ‘$\Lambda$’.

• CommentRowNumber7.
• CommentAuthorHarry Gindi
• CommentTimeJun 24th 2010
• (edited Jun 24th 2010)

@Toby: $V\wedge V=0$ is the correct definition for all characteristics. The second definition reduces to the symmetric case in characteristic 2. Alternating <=> Skew-symmetric for characteristic not 2, but symmetric <=> skew symmetric in characteristic 2. Since one never (intentionally) means to have skew symmetry be equivalent to symmetry, it’s clear that alternation is the correct notion in general.

• CommentRowNumber8.
• CommentAuthorTobyBartels
• CommentTimeJun 24th 2010
• (edited Jun 24th 2010)

Since one never (intentionally) means to have skew symmetry be equivalent to symmetry, it’s clear that alternation is the correct notion in general.

Thanks, Harry, good point! That is my position as well.

(Of course, if you’re working over a commutative ring in which $2$ is neither zero nor cancellable, then the antisymmetric power is neither the alternating power nor the symmetric power.)

• CommentRowNumber9.
• CommentAuthorHarry Gindi
• CommentTimeJun 24th 2010
• (edited Jun 24th 2010)

Meanwhile, Lang has a very nice categorical treatment of Schur functors descending to bundles on Banach manifolds in his book “Fundamentals of Differential Geometry”.

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeJun 24th 2010
• (edited Jun 24th 2010)

My memory of the basic overall conjecture is that you proposed that the high conceptual way to view (or define) ’Schur functor’ is as a pseudonatural transformation from $U$ to itself, where

$U: SymMonLinCat \to Cat$

is the evident forgetful 2-functor. (On our page we were being a little bit vague what $SymMonLinCat$ is, but one test case considered was Cauchy-complete symmetric monoidal categories enriched in finite-dimensional vector spaces over $\mathbb{Q}$.) Anyway, to continue, the conjecture would be that every such pseudonatural is of the form $S_\lambda$ for some Young diagram $\lambda$, thus justifying the correctness of the high-level description. (Was that the conjecture?)

If so, it’s quite an extraordinary and bold conjecture. These $S_\lambda$ are very restricted and “bounded” somehow, and last time I thought about this, I was wondering how pseudonaturality enforced that boundedness. For example, the full exterior algebra construction is defined on (let’s say) the category of finitely generated modules $Mod_R$ over any finite-dimensional $\mathbb{Q}$-algebra $R$, but is not “bounded”: is not given by a Young diagram $\lambda$. However, you saved the day by observing that it isn’t definable on finite-dimensional super vector spaces.

Any plan of attack on this conjecture?

Edit: Silly me. The conjecture is rather, I guess, that the pseudonaturals form a semisimple category for which the classical Schur functors $S_\lambda$ are the simple objects?

• CommentRowNumber11.
• CommentAuthorJohn Baez
• CommentTimeJun 24th 2010
• (edited Jun 25th 2010)

Hi, Todd!

Yes, I’ve again been wanting to tackle the big project of ’doing linear algebra right’, mainly because I’ve been talking with Jim about his concept of ’algebraic stacks’, also called ’algebraic geometric theories’. Maybe you know it already? They’re simply symmetric monoidal $k$-linear categories with finite colimits. So, Schur functors act on them. So, I’d like to clean up the theory of Schur functors a little.

But right now, instead of a grandiose frontal assault on my (probably overly) bold conjecture, I feel like

1) polishing up the nice description of Schur functors that you were giving in terms of representations of the symmetric group,

2) checking that this is equivalent to the definition of Schur functors given at the beginning of the article, and

3) showing that the monoidal category of Schur functors does indeed embed in the monoidal category of endomorphisms of

$U:SymMonLinCat \to Cat$

This mild-mannered project would let us think about a few delicate issues, e.g.:

1) What’s the right definition of ’symmetric monoidal linear category’ for the purposes of this theorem? Yesterday I realized that your switch from ’having finite colimits’ to ’Cauchy complete’ (shouldn’t that be Cauchy cocomplete??) was indeed well-motivated, not just centipedal. Why? Because we definitely want Schur functors to act on categories of vector bundles!

2) What are some more slick, but still not grandiose, characterizations of the category of Schur functors?

3) What are the most important nice relations between the category of Schur functors, the ring of symmetric functions (often called $\Lambda$) , and $\lambda$-rings in general?

As for 1), I really want to switch to the minimal assumptions you proposed. But right now in the nLab article on Schur functors you say something “all the above, except perhaps the Kan extension, can be done using just Cauchy completeness…” But surely we can get $FinVect$ embedded in $C$ even without that assumption, no?

Maybe that’s a small concrete step forwards…

• CommentRowNumber12.
• CommentAuthorJohn Baez
• CommentTimeJun 24th 2010

I assume someone will fix my slip about characteristic 2. I think it’s good to include a (correct!) explanation of the implications relating

$v \wedge v = 0$

and

$v \wedge w = -w \wedge v$

and how they’re equivalent away from characteristic 2.

• CommentRowNumber13.
• CommentAuthorTobyBartels
• CommentTimeJun 24th 2010
• (edited Jun 24th 2010)

I assume someone will fix my slip about characteristic 2.

Assuming that you agree with me and Harry that the the first equation (1) is what we want in general, then I have already fixed it.

• CommentRowNumber14.
• CommentAuthorHarry Gindi
• CommentTimeJun 24th 2010
• (edited Jun 24th 2010)

I actually wrote up the section on alternating forms on wikipedia a year ago and nobody has changed it.

@John: I’m also the one responsible for moving the article “associative algebra” to its current spot, replacing the page “algebra over a field” with the page “algebra over a commutative ring”. I only mention this to you because I saw that you edited that page (associative algebra) as well =D!

• CommentRowNumber15.
• CommentAuthorTodd_Trimble
• CommentTimeJun 24th 2010
• (edited Jun 24th 2010)

(1) It is a marvelous fact that Cauchy-complete = Cauchy-cocomplete! I think that’s correct, and someone (me, perhaps) should prove it at the Lab.

(2) Yes, that business about left Kan needing more than Cauchy completeness was probably quite silly. I can tell you what I had in mind.

I meant that whenever $i: S \to T$ is the inclusion of a skeletal subcategory [here $i: Sk \to FinVect$] and $C$ is “sufficiently cocomplete”, we can define the left Kan extension $Lan_i: C^S \to C^T$ canonically, without making any choices so to speak, by the coend formula

$Lan_i(F)(t) = \int^{s: S} F(s) \otimes hom(s, t)$

(Here “choosing” universal constructions such as coends doesn’t really count as a choice, because any choice is unique up to unique isomorphism.) What I was trying to avoid here is defining an extension from $C^S$ to $C^T$ in some non-canonical way, by choosing e.g. a retraction $r: T \to S$ and defining the extension by pulling back along $r$. But it wasn’t clear at the time of writing how to give an honest sense to that coend without extra cocompleteness assumptions on $C$. So I was just hedging my bets.

I’m pretty sure this is a complete non-issue though, and I think I know how to satisfy myself on this point. Long story short: I think there is nothing at all to worry about, and we can proceed full steam ahead under the minimal Cauchy-complete hypothesis.

• CommentRowNumber16.
• CommentAuthorTodd_Trimble
• CommentTimeJun 24th 2010
• (edited Jun 24th 2010)

@John #11: back to step 2 of the milder-mannered project – what definition of Schur functor are we using at the beginning? There’s a condition that the structure maps of a Schur functor $F$, in other words the maps $\hom(V, W) \to \hom(F V , F W)$, must be polynomial functions (originally $V$, $W$ are complex vector spaces, but then one can make a polymorphic move to module categories over other rings…). But I’m not sure I quite got a crisp definition from this (and if we restrict to module categories, I still worry about my exterior algebra example).

So could we go over this again?

• CommentRowNumber17.
• CommentAuthorEric
• CommentTimeJun 25th 2010
• (edited Jun 25th 2010)

@John #1:

I deleted the discussion Todd and I were having near the top of the page

Ouch! I always think it is a shame to see a discussion disappear. The discussion is often more valuable than the result because we can see the development.

I haven’t gone back to look at the actual discussion, but if it was meaningful, I would prefer to move it to the bottom of the page under the heading “Discussion” than delete material.

PS: Alternatively, we could move old discussions on the nLab to a new thread here and drop a link at the nLab.

• CommentRowNumber18.
• CommentAuthorJohn Baez
• CommentTimeJun 25th 2010
• (edited Jun 25th 2010)

Hi, Todd.

1) So Cauchy complete = Cauchy cocomplete! I guessed my true that after posting my silly comment: “biproducts” and “splitting idempotents” are just as much cocompleteness conditions as completeness conditions. Cool! I don’t feel I understand this stuff very well, but it reminds me of Jim’s remark that “completeness” and “cocompleteness” are really almost the same except for “set-theoretic fine print”.

2) There’s an essentially unique symmetric monoidal linear functor from $FinVect_k$ to $C$ whenever $C$ is a symmetric monoidal Cauchy-complete $k$-linear category, right? Where “essentially unique” means unique up to symmetric monoidal natural isomorphism. This might be nice to mention (if it’s true).

The idea is that any such functor must send the tensor unit to the tensor unit, and preserve biproducts, etc.

3) By “the definition of Schur functors given at the beginning of the article”, I meant “an endofunctor of $FinVect_{\mathbb{C}}$ that’s polynomial on homsets.” This seems to strike a nice compromise between elegance and comprehensibility, but it would be nice to prove that it’s equivalent to the “standard” definition. That’s what step 2 of the mild-mannered project was supposed to be.

4) Another aspect of the mild-mannered project: I also think that the “standard” definition of Schur functors should be freed from the clutches of Young diagrams. At this stage Young diagrams are just a trick for constructing representations of $S_n$. So, I’m thinking we can skip them and say that given our symmetric monoidal functor $FinVect_k \to C$ and any object $x \in C$, any irrep $R$ of $S_n$ gives an idempotent on $x^{\otimes n}$, which we can use to define a subobject $R(x)$ of $x$ - and thus irreps of $S_n$ acts as endofunctors on $C$. I guess we’re almost doing this now, but somehow it would be nice to say all this without bringing in Young diagrams, which are really irrelevant: all you need to know is that irreps of a finite group $G$ give idempotents in its group algebra, right?

This is a stylistic thing: Young diagrams are wonderful things, and I love them, but they give the usual approach to Schur functors a nitty-gritty combinatorial feel which sort of hides their conceptual meaning. So I think it’s best that they appear later, in a theorem about Schur functors, rather than the definition. Theorem: the category of Schur functors is a semisimple abelian $k$-linear category whose simple objects are Young diagrams!

You’ll note James Borger’s comment where he said he liked our explanation more than he expected to. Presumably he suffered from a traumatic experience with a heavily combinatorial approach to Schur functors in early childhood.

In fact I’d like to go a bit further and do something similar for reps that aren’t irreps. I guess these are easy to handle by taking direct sums of irreps, but again this seems like putting the cart before the horse: we usually define group reps first and then later decompose them, and I’d like to do this with Schur functors too.

But I don’t right now see how to do this nicely by “splitting idempotents”. Maybe it really does use both splitting idempotents and biproducts - another nice advertisement for Cauchy completeness!

5) On a pathetically minor note, I replaced your mention of $\mathbb{Q}$ by that famous ’arbitrary field of characteristic zero’ that you always hear about. I did this mainly to soften the sudden switch from $\mathbb{C}$ to $\mathbb{Q}$ that we had in an earlier version of the article.

• CommentRowNumber19.
• CommentAuthorJohn Baez
• CommentTimeJun 25th 2010

I have deleted this query box on the Schur functor page, since I believe these comments are obsolete… but I’ll put these comments here lest anyone think I’m a nasty guy who sneaks around deleting queries:

Todd Trimble: I wimped out and chose rational vector spaces as the base of enrichment. For one thing, the blog commentary seems to suggest that there are delicate issues in nonzero characteristic. Even in cases where there is no integer torsion, it seems to me that integer divisibility makes certain things come out a lot more cleanly, and (if I am not mistaken) means that certain finite cocompleteness conditions can be relaxed in favor of Cauchy completeness (in the enriched category sense of Lawvere). More on this later.

John Baez: You’re right that there are special tricky endofunctors

$F: FinVect_{k} \to FinVect_{k}$

that can be defined only when $k$ has characteristic $p$. So, using categories enriched over $Vect_k$ with $k$ having characteristic zero is probably a wise idea, at least for starters. But I’m really hoping that we can drop that in the more sophisticated approach where we work with all symmetric monoidal abelian categories simultaneously and demand pseudonaturality. I’m hoping this will ’wash out’ the tricky functors that only work in characteristic $p$, leaving us with just the Schur functors we know and love.

Todd Trimble: Could be!

John Baez: By the way, I believe one upshot from the blog discussion so far is that the Schur functors we know and love do work in characteristic $p$… but only if we define them using coinvariants, not invariants!

Ben Webster: I don’t understand pseudo-naturality all that well, but it seems like working with all symmetric monoidal categories simultaneously should just mean taking $\mathbb{Z}$-representations of $S_n$, which certainly will not wash away all the characteristic p problems, but will just hand them all to you simultaneously. For example, over $\mathbb{Z}$, the functor $X\mapsto X^{\otimes 2}$ doesn’t decompose into symmetric and anti-symmetric parts. How does this fit with your picture?

There is a version of the original Schur functors over $\mathbb{Z}$, given by tensor product with Specht modules, but this isn’t very canonical. You could just as easily take dual Specht modules.

• CommentRowNumber20.
• CommentAuthorJohn Baez
• CommentTimeJun 25th 2010

More queries I’m removing and immortalizing here:

Todd Trimble: With the possible exception of the left Kan extension alluded to above, all these constructions may be carried out if we weaken the abelian assumption on $C$ to mere Cauchy completeness (relative to enrichment in rational vector spaces), which more concretely means that $C$ admits finite coproducts and splittings of idempotent projections. This observation was made by Noah Snyder on the blog.

(Jamie: Doesn’t Cauchy completeness for abelian categories in fact correspond to biproducts and split idempotents?)

(Todd: Coproducts are biproducts in the $Ab$-enriched case! That is, I was assuming as given that $C$ was enriched in something like $Ab$ or $Vect$.)

It would actually feel more natural to me to speak of the object of coinvariants rather than the object of invariants, but in the present context it should come to the same thing as either is the splitting of the idempotent operator $\frac1{n!}\sum g$. That is to say: there is a natural splitting of the idempotent natural transformation

$e_X = \frac1{n!} \sum_{g \in S_n} g: V_\lambda \otimes X^{\otimes n} \to V_\lambda \otimes X^{\otimes n}$

which can be viewed either as invariants (equalizer of $e$ and the identity) or coinvariants (coequalizer of $e$ and the identity).

John Baez: we need to use coinvariants when we get to the more sophisticated approach where our constructions are supposed to preserved by right exact functors. Also, coinvariants work even in characteristic $p$, while invariants involve dividing by $n!$. So, coinvariants rule, and I’ve switched to working with them above. Also, following Jamie Vicary’s correction, I’ve switched to using Young tableaux instead of Young diagrams: for each Young diagram there are many isomorphic Schur functors, one for each Young tableau of that shape.

Todd Trimble: That’s fine to switch to coinvariants; I completely agree that’s the correct conceptual way to go in general. As far as Young diagrams vs. Young tableaux, point taken, but see also my response to Jamie (which I’ve incorporated above).

Of course, I’ve already noted that “right exact” could be overkill in certain contexts. When the object of coinvariants arises by splitting an idempotent (as in the case of enrichment over rational vector spaces), we don’t need full right exactness to preserve the construction, although we might want to retain additivity: preservation of direct sums. I hope we can be a bit flexible about hypotheses until we’re further along in this.

Rod McGuire Isn’t a switch from Young diagrams to Young tableaux essentially the same as changing from groups to groupoids?

• CommentRowNumber21.
• CommentAuthorMike Shulman
• CommentTimeJun 25th 2010

The fact that Cauchy-complete = Cauchy-cocomplete is indeed marvelous! And it’s really not hard to see when you write things in terms of profunctors: a Cauchy colimit (resp. limit) is a colimit (resp. limit) whose weight is a profunctor with a right (resp. left) adjoint, and from the definition of weighted limits in terms of profunctors you can check that if $J\dashv K$ are adjoint profunctors, then J-weighted colimits are the same as K-weighted limits (this is probably more or less about mates). In fact, I believe this limit-colimit coincidence characterizes Cauchy (co)limits: if you have two weights J and K such that J-weighted limits and K-weighted colimits coincide, then J and K are necessarily adjoint and therefore both are Cauchy.

• CommentRowNumber22.
• CommentAuthorTodd_Trimble
• CommentTimeJun 25th 2010

By “the definition of Schur functors given at the beginning of the article”, I meant “an endofunctor of $FinVect_{\mathbb{C}}$ that’s polynomial on homsets.”

Sorry I keep harping on this, but it seems this would imply that the exterior algebra, as a functor on $FinVect$, is a Schur functor. But the exterior algebra functor does not decompose as a finite sum of irreducible Schur functors $S_\lambda$. Am I misunderstanding something?

• CommentRowNumber23.
• CommentAuthorHarry Gindi
• CommentTimeJun 25th 2010
The nth exterior power is a schur functor, though, right?

By the way, Todd, I learned in a commutative algebra class (a topics course) that doing linear algebra over a Noetherian local ring is very similar to doing linear algebra over its residue field. If you're trying to give a definition of a schur functor in some generality, maybe that would be a good place to start?

Also of note, Schur functors have a natural generalization to topological vector spaces, so you may want to factor that in as well.
• CommentRowNumber24.
• CommentAuthorTodd_Trimble
• CommentTimeJun 25th 2010

The nth exterior power is a schur functor, though, right?

Yes. And?

As for the rest of the comment: have you taken a look at Schur functor? You can get an idea of what sort of generality we’re contemplating there.

• CommentRowNumber25.
• CommentAuthorHarry Gindi
• CommentTimeJun 25th 2010
• (edited Jun 25th 2010)
Well, for your definition of "linear", if you mean Q-Vect-enriched, then it seems like you lose out on a lot of noetherian local rings as well as finite fields. Correct me if I'm wrong.
• CommentRowNumber26.
• CommentAuthorTodd_Trimble
• CommentTimeJun 25th 2010

Yes, this is a first pass at a general story. Part of the story seems to be that there are numerous subtleties for example in nonzero characteristic. So enrichment in rational vector spaces is (I hope) a simplifying assumption which will hopefully allow us to focus on rather more core conceptual concerns at this initial stage of the game.

• CommentRowNumber27.
• CommentAuthorTodd_Trimble
• CommentTimeJun 25th 2010

I’ve now added a fair amount of detail to Schur functor (detailed description in context of symmetric monoidal Cauchy complete $k$-linear category). For the moment we’re sticking to the characteristic zero game plan.

• CommentRowNumber28.
• CommentAuthorJohn Baez
• CommentTimeJun 25th 2010

John wrote:

By “the definition of Schur functors given at the beginning of the article”, I meant “an endofunctor of $FinVect_{\mathbb{C}}$ that’s polynomial on homsets.”

Todd wrote:

Sorry I keep harping on this, but it seems this would imply that the exterior algebra, as a functor on FinVect, is a Schur functor.

Oh, duh. Somehow you hadn’t gotten me to realize that this particular definition was invalidated by that example.

Okay, so let’s toss this definition into the waste bin! We should start out talking about Schur functors using some definition that’s equivalent to the usual one.

Maybe later we can study infinite direct sums of Schur functors like

$\Lambda = \bigoplus_{n \ge 0} \Lambda^n$

Maybe such functors, with the special property that they converge to a finite-dimensional answer on any specific vector space, are precisely the same as endofunctors on $FinVect_{\mathbb{C}}$ that are polynomial on homsets. But maybe not… there are lots of interesting conjectures, but having been beaten down repeatedly, I’m now in the mood to start on safe ground and slowly march forwards.

• CommentRowNumber29.
• CommentAuthorTodd_Trimble
• CommentTimeJun 25th 2010

I have to say though that I’m very attracted by your original bold-mannered proposal, since it seems to evade this problem.

To plant a little seed that might be worth nurturing, we might consider your 2-functor

$U: SymMonLinCat_k \to Cat$

from the standpoint of representability. I haven’t checked this carefully, but at the moment it looks as though this is represented by the Cauchy completion of the free $k$-algebroid on the groupoid given by the permutation category, so that (calling this $\widebar{k \mathbf{P}}$) we have

$U \cong SymMonLinCat_k(\widebar{k \mathbf{P}}, -)$

This observation, if true, would appear to simplify certain certain calculations immensely.

• CommentRowNumber30.
• CommentAuthorJohn Baez
• CommentTimeJun 25th 2010

Wow! This “little seed” could grow into a huge tree, if true!

I don’t really understand why you made your guess, but isn’t this thing you’re calling $\widebar{k \mathbf{P}}$ just a very plausible definition of the ’category of Schur functors over $k$’? I mean, when $k = \mathbb{Q}$ it looks like the usual category of Schur functors, right?

So is your idea somehow that this is a good way of proving my big bold claim? Maybe not quite as big: just ’over $k$’. But still…

Do you really need to do that Cauchy completion? Isn’t every representation of a permutation group completely reducible?

By the way, isn’t it true that one special thing about the groups $S_n$, unlike other finite groups, is that when we take an irreducible representation over $\mathbb{Q}$, and then extend our scalars to some larger field of characteristic zero, it doesn’t split into smaller pieces? This would be false for $\mathbb{Z}/3$, for example. So when we talk about the representation theory of $S_n$ over a field $k$ of characteristic zero, it doesn’t really matter much which field we’re using. Somehow that’s part of what’s going on.

• CommentRowNumber31.
• CommentAuthorJohn Baez
• CommentTimeJun 25th 2010

Btw, I certainly do want to pursue my original bold-mannered proposal! I just feel that this article might do better starting out with a familiar definition of Schur functors and some true facts about them, and gradually working towards something new and interesting, rather than starting out with a false definition and some bold claims.

So, just now I have:

1) corrected and polished the introduction to Schur functor, sketching out a lot of ideas that would be fun to develop…

2) added a new page called rig category

3) slightly expanded your remarks about the bimonoid $k[S_n]$. I feel sure it’ll become important that this guy is cocommutative.

By the way: congratulations on polishing up that result about how $FinVect_k$ sits canonically inside $C$. Somehow this is the key to how the group algebra of $S_n$ acts on any object $x^{\otimes n}$. But it should also imply that that $GL(n)$ acts on any object $x^{\oplus n}$, right? So I think there’s gonna be some very nice Schur-Weyl duality going on, quite generally…

• CommentRowNumber32.
• CommentAuthorUrs
• CommentTimeJun 25th 2010

Glad to see the nForum get to a point where after being away for half a day, I can no longer follow all the discussions easily… :-)

• CommentRowNumber33.
• CommentAuthorTobyBartels
• CommentTimeJun 25th 2010

How do rig categories fit into the ideas at 2-rig?

• CommentRowNumber34.
• CommentAuthorTodd_Trimble
• CommentTimeJun 25th 2010
• (edited Jun 25th 2010)

@John #30: it just smelled like the right thing to think about. Not long ago there was a discussion on the nForum about abstract Tannaka-Krein duality, and the idea is very much in that vein. (The claim turns out to be easy to prove, and I’ll write up the proof later.)

The proof will make taking the Cauchy completion seem like exactly the right thing to do (because if we want the representing object in $SymMonLinCat_k$, we have to Cauchy-complete). The linearization $k\mathbb{P}$ of the permutation category isn’t itself Cauchy complete because the idempotent splitting will add in all the $V_\lambda$ which are direct summands of the group algebras, and close under direct sums. When we do this in characteristic zero, we get all finite-dimensional $S_n$-modules this way, so the Cauchy completion is none other than the category of finite-dimensional species

$\mathbb{P} \to FinVect_k$

Let’s call this $Spec$. Then the pseudonatural transformations $U \to U$ should, by Yoneda, be given by maps $Spec \to Spec$ in $SymMonLinCat$, or to symmetric monoidal linear endofunctors on Spec. The monoidal category of such endofunctors is (via the universal property of $Spec$) just $Spec$ again, but with the plethystic monoidal product (in other words, the substitution product on Joyal species).

So you see, more or less the whole story does seem to be contained in this remark. :-)

• CommentRowNumber35.
• CommentAuthorJohn Baez
• CommentTimeJun 25th 2010

By the way, some of what I was just saying can be said a trifle more elegantly: $C$ gets tensored over $FinVect_k$ in a canonical way! So, we can talk about an algebra in $FinVect_k$ acting on an object in $C$. For example, $k[S_n]$ acting on $k^{\otimes n}$. And we can talk about an object like $k^n \otimes x \cong x \oplus \cdots \oplus x$ for any object $x \in C$. And $GL(n,k)$ acts on $k^n \otimes x$ in an obvious way.

Nothing really new here, but this way of talking might make certain things seem more like ’ordinary linear algebra’. Which is good.

How did you decide that Cauchy completeness was enough to get everything to work?

• CommentRowNumber36.
• CommentAuthorJohn Baez
• CommentTimeJun 25th 2010

By the way, we’ll need some name for the groupoid of finite sets. I see you favor $\mathbf{P}$, for ’permutation’. I’ve seen $\mathbf{B}$ for ’bijection’, and I kind of like $S$, since then

$S \simeq \bigsqcup_{n \ge 0} S_n$

I’ve also used $E$, since the cardinality of this groupoid is $e$, and $E$ should make you think ’ensemble’.

But I also kind of like Marcelo Aguiar’s notation $C^\times$ for the underlying groupoid of a category $C$ - a nice generalization of the multiplicative group of a field, or ring. So, in the introduction, I decided to call this groupoid $FinSet^{\times}$. One advantage is that anyone with a modicum of knowledge can guess what this might mean.

I don’t really care, though… just need something. Is there a $n$Lab standard symbol for this groupoid?

• CommentRowNumber37.
• CommentAuthorHarry Gindi
• CommentTimeJun 25th 2010
• (edited Jun 25th 2010)

What are you using the finiteness hypotheses for?

• CommentRowNumber38.
• CommentAuthorTobyBartels
• CommentTimeJun 26th 2010
• (edited Jun 26th 2010)

I often use $C_{\sim}$ for the underlying groupoid of $C$; that’s standard in the sense that I’ve already used it on several pages. But if you’re going to use this often on your page, then a short name like $\mathbb{P}$ or $S$ would be handy.

• CommentRowNumber39.
• CommentAuthorzskoda
• CommentTimeJun 26th 2010

11 etc. remind me a bit of Nikolai Durov’s video lecture on vectoids and algebrads which can be found online. The link is somewhere on the algebrad page.

• CommentRowNumber40.
• CommentAuthorTodd_Trimble
• CommentTimeJun 26th 2010

My own custom is to reserve $\mathbf{B}$ or $\mathbb{B}$ for ’braid group’, but I’m not really fussed about this.

Forestalling a possible complaint about notation in my latest revision of Schur functor, where I sketch a proof of representability of $U$: I already know it’s god-awful. I just wanted to get the proof down. The notation can be prettied up.

Yes, I am completely satisfied that in the characteristic zero case, Cauchy completeness is a sufficient cocompleteness condition.

Harry: good question. The initial set-up with enrichment in $FinVect_k$ can certainly be relaxed to enrichment in $Vect_k$, and in fact some things may go down smoother that way. The finite-dimensionality really pops out the other end with the representability result, where the representing object is none other than species valued in finite-dimensional vector spaces. (Cf. the result that the Cauchy completion of a ring $R$ is the category of finitely generated projective right $R$-modules). (If this comment seems a little telegraphic, it should become clearer later as results get written up.)

• CommentRowNumber41.
• CommentAuthorJohn Baez
• CommentTimeJun 26th 2010
• (edited Jun 26th 2010)

Todd wrote:

The linearization $k \mathbb{P}$ of the permutation category isn’t itself Cauchy complete…

Oh, sure. For some bizarro reason I read $k \mathbb{P}$ as “the category of representations of $\mathbb{P}$ in $FinVect_k$“… which is precisely the Cauchy completion of what you meant! So yeah, Cauchy completion is just right.

It looks like you’re really on a roll here… I’m staring at the new stuff you’re writing, and it’s great!

• CommentRowNumber42.
• CommentAuthorJohn Baez
• CommentTimeJun 26th 2010

Eric wrote:

I haven’t gone back to look at the actual discussion, but if it was meaningful, I would prefer to move it to the bottom of the page under the heading “Discussion” than delete material.

Anyone who wants can restore the discussions that I deleted (and brought over here). Please put them at the end. I don’t personally have the desire to do this, so I’ll let you. The way I see it, Todd and I are writing a kind of paper, and it gets a bit tiring to look at old layers of confusion after we’ve worked our way through those problems. But if they’re at the bottom somewhere, it’s fine.

• CommentRowNumber43.
• CommentAuthorEric
• CommentTimeJun 26th 2010

I’m just a groupie around here anyway, but I wish every nLab page had a link pointing to a discussion here at the nForum. Maybe even have a special “Discussion” tab that directs you here like at Wikipedia.

Now that we have the nForum, the need to have discussions on the nLab itself has diminished and possibly disappeared.

I’d like to see nLab and nForum more integrated, but that is just me.

• CommentRowNumber44.
• CommentAuthorEric
• CommentTimeJun 26th 2010

@John: For what it is worth, I am super excited to see you participating more actively around here these days. It is awesome to watch from the sidelines :)

• CommentRowNumber45.
• CommentAuthorMike Shulman
• CommentTimeJun 26th 2010

I kind of like the idea of $C^\times$ for the core of a category C; it’s something that could use a good general notation, and doesn’t suggest anything particularly to me. One disadvantage might be that $\times$ makes me think of something to do with the cartesian product of C.

Along the lines of “you can tell something is important because it has too many names,” I think the core of FinSet has also been called $\Sigma$, on the theory that it is the disjoint union of the symmetric groups $\Sigma_n$.

• CommentRowNumber46.
• CommentAuthorUrs
• CommentTimeJun 26th 2010

Is there a nLab standard symbol for this groupoid?

At some entries I write $core(C)$ for the core of a category $C$.

• CommentRowNumber47.
• CommentAuthorHarry Gindi
• CommentTimeJun 26th 2010
• (edited Jun 26th 2010)

The symbol $\mathbf{C}^\times$ will be extremely confusing, as this is the multiplicative group of the complex numbers. In particular, this is Bourbaki’s notation specifically for that (he did not use blackboard bold).

On that note, I agree with Urs.

• CommentRowNumber48.
• CommentAuthorJohn Baez
• CommentTimeJun 26th 2010

Harry writes:

The symbol $\mathbf{C}^{\times}$ will be extremely confusing, as this is the multiplicative group of the complex numbers. In particular, this is Bourbaki’s notation specifically for that (he did not use blackboard bold).

But the complex numbers are a 1-object category with multiplication as composition of morphisms, and the underlying groupoid of this category consists of the invertible complex numbers. So, this use of notation is a special case of using $C^{\times}$ for the underlying groupoid of a category $C$. This is the whole point of the $C^{\times}$ notation. It’s supposed to be a cute generalization of a familiar usage.

The confusion here, if any, comes from the fact that some people use $\mathbf{C}$ to mean the complex numbers while others use it to mean an arbitrary category. One widely adopted solution is to use $\mathbb{C}$ for the complex numbers and never for anything else.

• CommentRowNumber49.
• CommentAuthorUrs
• CommentTimeJun 26th 2010

But the complex numbers are a 1-object category

In fact a monoidal 0-category. Which is almost, but not quite the same.

• CommentRowNumber50.
• CommentAuthorJohn Baez
• CommentTimeJun 26th 2010
• (edited Jun 26th 2010)

Okay, monoidal 0-category. I’m never going to change my ways, but I don’t feel any desire to claim my way of talking is “right”. It’s just shorter and easier for layfolk to roughly understand, and that’s what matters most to me.

• CommentRowNumber51.
• CommentAuthorJohn Baez
• CommentTimeJun 26th 2010
• (edited Jun 26th 2010)

I put in a query here for Todd. Briefly: I think I can give a shorter definition of Schur functor that doesn’t require people to know about Young diagrams (or tableaux) to understand what a Schur functor is. I think this would be a very good thing: less grunge, more concept. Should I try it?

• CommentRowNumber52.
• CommentAuthorTodd_Trimble
• CommentTimeJun 26th 2010

John, I think that sounds good. I think it would be okay to mention in passing that each primitive central idempotent in the group algebra corresponds to a so-called Young diagram or partition, but certainly we don’t need to get into the grunge at this point. :-)

• CommentRowNumber53.
• CommentAuthorJohn Baez
• CommentTimeJun 27th 2010

Great. I think it’s very important to mention Young diagrams in the introduction, so people will see the standard stuff (and realize we aren’t insane). But in the formal development we can put them off until they’re needed.

I’ll give it a try now…

• CommentRowNumber54.
• CommentAuthorJohn Baez
• CommentTimeJun 27th 2010

Okay, I eliminated the Young diagrams, except for the introduction, where I actually show one.

I would be very happy if some reader would contribute a picture of a Young diagram that doesn’t have 17 boxes. I stole the 17-box example from Young diagram. it would be nice to have a few different, less grandiose examples.

The construction of Schur functors still doesn’t sound as simple as it actually is, because one really needs to get used to the fact that an ’element’ of a monoid in $C$ can act on anything on which $C$ acts, and it acts as an idempotent when the element is idempotent. This stuff should be pathetically obvious, but developing it en passant makes it sound a bit complicated…

• CommentRowNumber55.
• CommentAuthorTodd_Trimble
• CommentTimeJun 27th 2010

John, I tried answering to your list of queries at Schur functor.

• CommentRowNumber56.
• CommentAuthorTodd_Trimble
• CommentTimeJun 27th 2010
• (edited Jun 27th 2010)

Okay, I should admit to a silly mistake I made in Schur functor, on the identification of $\overline{k\mathbb{P}}$. It is certainly not $[\mathbb{P}^{op}, FinVect_k]$. Rather, it is the linear category of “polynomial species” $F: \mathbb{P}^{op} \to FinVect_k$, where $F(n) = 0$ for all sufficiently large $n$.

This is easily seen because the linear Cauchy completion of $k\mathbb{P}$ consists of retracts of finite biproducts of representables $k hom(n, -): \mathbb{P}^{op} \to Vect_k$.

I’m going to fix the mistake now.

• CommentRowNumber57.
• CommentAuthorJohn Baez
• CommentTimeJun 27th 2010

Todd wrote:

Okay, I should admit to a silly mistake I made in Schur functor, on the identification of $\overline{k\mathbb{P}}$. It is certainly not $[\mathbb{P}^{op}, FinVect_k]$.

Whoops — that mistake runs through the paper, starting from near the beginning. It’s a bit annoying that there’s no super-slick notation for the functors satisfying the finiteness condition $F(n) = 0$ for all sufficiently large $n$. Well, I guess $\overline{k\mathbb{P}}$ counts as super-slick.

If we did a version of your theorem with symmetric monoidal cocomplete $k$-linear categories (with tensor product distributing over colimits) replacing $SymMonLinCauch$, would the endomorphisms of the forgetful 2-functor $U$ form the category $[\mathbb{P}^{op}, Vect_k]$? This category is a nice version of linear species, without any finiteness condition.

But I think using Cauchy complete categories has a certain stripped-down elegance…

Btw, on a lesser note, we start out talking about functors $F: \mathrm{P} \to \FinVect_k$ and then switch to $F: \mathrm{P}^{op} \to \FinVect_k$. It’s no big deal, but maybe we should either be consistent or point out that it doesn’t matter.

Your comments about my queries were very cool but I’ll need to think about them; they go in a different direction than I was imagining.

• CommentRowNumber58.
• CommentAuthorTodd_Trimble
• CommentTimeJun 27th 2010

If we did a version of your theorem with symmetric monoidal cocomplete $k$-linear categories (with tensor product distributing over colimits) replacing $SymMonLinCauch$, would the endomorphisms of the forgetful 2-functor $U$ form the category $[\mathbb{P}^{op}, Vect_k]$?

Yes, indeed. I’m much more used to contemplating that, actually. But I think it’s polynomial Schur functors that we actually want from the get-go, inasmuch as those are what people know and love as classical Schur functors.

I’ll look into your lesser note at some point.

• CommentRowNumber59.
• CommentAuthorTodd_Trimble
• CommentTimeJun 27th 2010

I’ve added more content to Schur functor, both correcting the mistake and writing a section on the plethystic monoidal product. I didn’t try to polish it much.

• CommentRowNumber60.
• CommentAuthorUrs
• CommentTimeJun 27th 2010

It’s just shorter and easier for layfolk to roughly understand, and that’s what matters most to me.

The reason I keep pestering you about this is that my feeling is that instead of making things easier to understand it hides to an important distinction and makes it harder to see what’s going on. Delooping is not something that one disregards without getting into trouble later on.

I would think good notation would be for $C$ a monoidal $n$-categeory to write $C^\times$ for the full subcategory on which the product is invertible. (That also restricts to invertible morphisms, but is not generally the same as restricting to all invertible morphisms).

The other operation, that of passing to maximal $\infty$-groupoids / cores should go by some other symbol, I’d say. If we represent out $n$-categories as $n$-fold complete Segal spaces $C$, then it is simply passing to $C_0$.

• CommentRowNumber61.
• CommentAuthorJohn Baez
• CommentTimeJun 27th 2010
• (edited Jun 27th 2010)

I don’t really “disregard” delooping; in fact my paper “Categorification” may have been the first place where delooping was described as part of a collection of systematic processes that hop between different points in the periodic table. I think there are times when it’s good to relax and times when it’s good to be careful, and I think I can tell the difference. It’s just like how I know when it’s okay to identify a groupoid with its opposite and when to be careful, or when it’s okay to pretend all functions are differentiable and when that’s a bad mistake. And I think everyone can learn to make these judgement calls. But it’s probably good if you keep on “correcting” me, so that more people learn more math.

• CommentRowNumber62.
• CommentAuthorJohn Baez
• CommentTimeJun 27th 2010

Todd: you’re working on Schur functor now so I can’t, so here are a couple reminders to myself about small mistakes to fix.

1) Right now there are still some incorrect appearances of $[\mathbb{P}, FinVect_k]$ lurking around near the beginning of the paper.

2) Thanks for drawing those diagrams for the definition of pseudonatural transformation! Maybe I’ll copy them to pseudonatural transformation.

3) “these are the polynomial functors $F: \mathbb{P}^{op} \to FinVect_k$ where $F(n)=0$ for large enough $n$.” should be changed to “these are the functors $F: \mathbb{P}^{op} \to FinVect_k$ that are polynomial, meaning $F(n)=0$ for large enough $n$.” But in fact I should introduce this concept right near the start.

Where do we want to go next? Here are some things on my mind:

1) It would be nice to eliminate the dependence of $Schur_k$ on $k$ and get our hands on the ’integral’ version $Schur_{\mathbb{Z}}$. In other words, the original bold project.

2) The Grothendieck group of $\Schur_k$ (or I guess even $Schur_{\mathbb{Z}}$) is the free $\lambda$-ring on one generator. This gadget has a lot of interesting structure (see David’s remark here), and most of it should just be a decategorified version of the structure on $\Schur_k$. We could think about this, and think about whether we have anything new to say. E.g., how much of this structure is easy to see using the idea of Schur functors as endomorphisms of the forgetful functor $U$?

3) We could even try to find a better explanation for why Young diagrams form a basis of simple objects for $Schur_k$. Here is where the representation theory of the symmetric group really matters. But the usual study of Young symmetrizers seems a bit grungy to me. It’s not very bad… but for such a fundamental result, I’d hope for a proof that was downright mouthwatering.

• CommentRowNumber63.
• CommentAuthorTodd_Trimble
• CommentTimeJun 27th 2010

I added some more to Schur functor. One thing I did is fill in the coherence conditions in the definition of pseudonatural transformation. However, various squares in that section need to be prettied up; there’s some trick for aligning labeled arrows which I don’t know how to do, but which would definitely improve the appearance. I would be much obliged if someone would step in and do that.

John, check out the proof that plethysm is a monoidal category structure. I think you might find it sort of cool!

At this point I may step back a little and let John work on this and revise and so forth. One mathematical thing I’d like to understand better is this fact mentioned by John, that it makes no difference what field of characteristic zero you use, you get basically the same irreducible representations of the symmetric group. Does a similar statement hold in other characteristics?

• CommentRowNumber64.
• CommentAuthorJohn Baez
• CommentTimeJun 27th 2010

Okay, I’ll work on it a bit next. Contrary to my big plans here, I also have an urge to polish this into a small manageable paper and publish it. Not every paper needs to be a woolly mammoth! I keep trying to remind myself of this, lately: short papers are fun to read.

You get different representations of the symmetric group when you work over $\mathbb{F}_p$. I don’t understand this stuff! It could be fun, but perhaps it’s another paper. Here is an easy introduction. The interesting stuff starts on page 69.

• CommentRowNumber65.
• CommentAuthorJohn Baez
• CommentTimeJun 27th 2010

From Wikipedia:

the irreducible representations of the symmetric group are not known in arbitrary characteristic.

and

The determination of the irreducible modules for the symmetric group over an arbitrary field is widely regarded as one of the most important open problems in representation theory.

I know I still haven’t answered your question ’does it make no difference which field of characteristic $p$ you use?’ But these quotes are sufficiently scary that I feel my only chance is to hope that the requirement of pseudonaturality ’washes out’ the peculiarities of different fields and only leaves the functors you can systematically define for all symmetric monoidal abelian categories_ with the help of the tensor product, symmetry, and finite colimits. And I hope are the usual Schur functors.

(Some jargon: Schur functors come from Specht modules, which are certain representations of $S_n$ defined using colimits. If we let ourselves use limits instead of colimits, we’d get “dual Specht modules”. Or maybe it’s the other way round. In characteristic zero there’s no difference between Specht modules and dual Specht modules: they’re both just the usual irreps of $S_n$.)

• CommentRowNumber66.
• CommentAuthorTodd_Trimble
• CommentTimeJun 27th 2010
• (edited Jun 27th 2010)

I’m down with a small manageable paper.

Okay, here’s one thing that would definitely be pleasant to work into the picture: that permutation representations attached to transitive actions of the symmetric group on finite sets form a natural basis of the representation ring. I first learned this trick from Jim, but I personally don’t have a proof for it. Do you?

This might help address some of the “grunge factor”.

• CommentRowNumber67.
• CommentAuthorJohn Baez
• CommentTimeJun 27th 2010

Todd wrote:

I’m down with a small manageable paper.

Cool! Okay, I’ll go in that direction. It should have different introductory stuff than this nLab entry. But for now I’ll just keep working on the nLab entry.

I first learned this trick from Jim, but I personally don’t have a proof for it. Do you?

No, but he’s told me about it, and I have the feeling I saw it stated as a theorem somewhere. Now I can’t find it. But there’s a lot of tantalizing stuff out there about Burnside rings of symmetric groups.

Indeed, I was thinking this might be a nicer way to connect Young diagrams to what we’re doing so far… not sure how it would work, exactly.

• CommentRowNumber68.
• CommentAuthorTodd_Trimble
• CommentTimeJun 27th 2010

Indeed, I was thinking this might be a nicer way to connect Young diagrams to what we’re doing so far…

That was my thought as well. Incidentally, some time ago I put an illustrative example of part of what I learned from Jim at Gram-Schmidt process (thinking of $S_n$-reps as forming a 2-Hilbert space).

• CommentRowNumber69.
• CommentAuthorJohn Baez
• CommentTimeJun 28th 2010
• (edited Jun 28th 2010)

Todd wrote:

Incidentally, some time ago I put an illustrative example of part of what I learned from Jim at Gram-Schmidt process (thinking of $S_n$-reps as forming a 2-Hilbert space).

Cool - I’ll check it out.

I’ve put another bunch of work into Schur functor - check it out, starting from the beginning!

I broke down and admitted what Young symmetrizers are!

I tried to make the process whereby the article morphs from an elementary exposition into a research paper much more continuous.

I removed lots of subscripts ${}_k$, which don’t seem all that necessary.

I got rid of the stuff on bimonoids, since we don’t seem to use it. (I can put it back in, though, if I’m wrong.)

I did lots of polishing of all sorts.

And I think the description of how Young symmetrizers act as endofunctors of $C$ is getting nicer and nicer…

• CommentRowNumber70.
• CommentAuthorTodd_Trimble
• CommentTimeJun 28th 2010

I responded to your query about cokernel. I’m going to erase one of my counter-queries in a minute because it was based on a momentary confusion.

And I think the description of how Young symmetrizers act as endofunctors of $C$ is getting nicer and nicer…

Yes, I agree it looks reasonably nice and simple.

As far as getting rid of the bimonoid stuff, here’s a reason I might want to see it rolled back. A Schur functor is a polynomial functor, which should be thought of as a categorification of an ordinary polynomial, so that a finite direct sum

$\sum_n R(n) \otimes_{S_n} X^{\otimes n}$

is analogous to a polynomial $\sum_n r_n x^n / n!$. Here $R$ is a species $\mathbb{P} \to FinVect_k$ for which all but finitely many coefficient objects $R(n)$ are zero. Now your approach to defining Schur functors is fine, but as I see it, it masks that we are constructing the monomial $R(n) \otimes_{S_n} X^{\otimes n}$. In other words: logically, your approach gives the same thing, but the way I had it before makes it manifestly obvious that we are indeed constructing the object of $S_n$-coinvariants of the representation $R(n) \otimes X^{\otimes n}$. (And we use the bimonoid to construct that representation.)

Perhaps it would make sense to move the bimonoid stuff elsewhere, where we actually bring in polynomial functors per se. I see you are trying to make a gradual ascent from simple to sophisticated, so it would be fine to keep the Schur functor construction as you now have it, but make a strong connection to polynomials later when we get into the $\overline{k\mathbb{P}}$ stuff.

• CommentRowNumber71.
• CommentAuthorTodd_Trimble
• CommentTimeJun 28th 2010
• (edited Jun 28th 2010)

Okay, here’s another thing that crept into a revision which I think we should discuss: it’s unclear what exactly is meant by a Schur functor. We have two distinct (but closely related) notions.

(1) A Schur functor is a certain type of functor $C \to C$ ($C$ symmetric monoidal linearly Cauchy complete).

(2) A Schur functor is a functor $F: \mathbb{P}^{op} \to FinVect_k$ such that almost all $F(n)$ are zero.

(I called (2) a “Schur object”, but then I think you got rid of this and replaced it with “Schur functor”.) The distinction is analogous to, and in fact a categorification of, the familiar distinction between a formal polynomial [a finite list of coefficients] and a polynomial function $p(x) = \sum_k a_k x^k$.

Personally, I think we ought to be calling (1) a Schur functor or polynomial functor, and (2) something else like “Schur object” or “polynomial species” or something. You probably know the old Zen parable:

Before I began Zen study, rivers were rivers and mountains were mountains. When I began to study Zen, rivers were no longer rivers and mountains were no longer mountains. Now, many years later, I understand that rivers are rivers, and mountains are mountains.

So when one first encounters polynomials as a schoolboy, polynomials are just one thing (polynomial functions). Then, when one begins study of abstract algebra, polynomials are no longer polynomial functions; there’s a distinction to be made. Years later, after a number of mathematical satori, one can relax again and call them all just polynomials, because one knows what’s what (cf. your exchange above with Urs about delooping). But I think it would be presuming a bit much of the reader to assume the “enlightened” point of view and call them all just Schur functors or whatever.

• CommentRowNumber72.
• CommentAuthorJohn Baez
• CommentTimeJun 28th 2010

Todd wrote:

(I called (2) a “Schur object”, but then I think you got rid of this and replaced it with “Schur functor”.)

I agree that it’s good to distinguish 1) and 2). Actually I tried to consistently call 2) a “polynomial functor”, but now what you mention it, I think I screwed up somewhere - so feel free to fix it! Somehow I feel “polynomial functor” is more evocative than “Schur object”.

(Sometimes you had written merely “polynomial” instead of ’polynomial functor’, but I’d like to religiously maintain this distinction - especially since we may talk a bit about decategorifying polynomial functors to get polynomials.)

I’m not sure how much we should worry about the fact that in his magnum opus, Joachim Kock is using “polynomial functor” in a closely related but disturbingly different sense, namely something like an analytic functor that doesn’t use the symmetry structure of a symmetric monoidal category. His work looks very interesting, but I think that we, not he, are following the right analogy in our use of “polynomial’ versus ’analytic” functors. But we may need to mention this…

• CommentRowNumber73.
• CommentAuthorJohn Baez
• CommentTimeJun 28th 2010
• (edited Jun 28th 2010)

If you prefer callling 2) a “polynomial species” or “polynomial linear species” or something, that’s fine. Maybe “polynomial functor” is the right name for 1)! - but let’s stick with “Schur functor” for that.

• CommentRowNumber74.
• CommentAuthorTodd_Trimble
• CommentTimeJun 28th 2010

I was thinking that “polynomial functor” ought to categorify “polynomial function”, so that’s why I wanted to attach it to (1), not (2)!

Anyway, “Schur functor” for (1) and “polynomial species” for (2) would be okay with me. The disagreement with Kock is interesting, and it’s hard to argue 100% convincingly because categorification isn’t exactly a well-defined process! But “polynomial species” makes it quite clear which sense we have in mind.

Any thoughts about what I said in #70 (which is closely tied to this discussion)?

• CommentRowNumber75.
• CommentAuthorJohn Baez
• CommentTimeJun 29th 2010

Todd wrote:

Anyway, “Schur functor” for (1) and “polynomial species” for (2) would be okay with me.

Okay, let’s try that.

I don’t know which of the millions of things Jim Dolan has told me about over the years he’s also told you… but he’s really big on “coefficients versus values” when it comes to thinking about polynomials. The “value-centered view’ of a polynomial is to think of it as a list of values

$x \mapsto \sum_n a_n x^n / n!$

and this categorifies to your 1). The “coefficient-centered view” of a polynomial is to think of it as a list of coefficients

$n \mapsto a_n$

and this is your 2). He calls the map from 2) to 1) the “Taylor transform”, which is really a prototype of the more glamorous “Fourier series” (where $x$ is renamed $e^{i x}$) or still more glamorous “Fourier transform”.

Anyway, I know you know all this… I just like Jim’s way of talking about it, and wish I’d learned it early in college. It would have helped demystify Fourier series.

Any thoughts about what I said in #70 (which is closely tied to this discussion)?

Yup! Sorry, I’m alternating between doing math and packing, so I’m being a bit disorganized. (At least that’s my excuse this week.)

Now your approach to defining Schur functors is fine, but as I see it, it masks that we are constructing the monomial $R(n) \otimes_{S_n} X^{\otimes n}$. In other words: logically, your approach gives the same thing, but the way I had it before makes it manifestly obvious that we are indeed constructing the object of $S_n$-coinvariants of the representation $R(n) \otimes X^{\otimes n}$. (And we use the bimonoid to construct that representation.)

Hmm, good point. I was focused on trying to implement as efficiently as possible the idea that Young symmetrizers are idempotents and we can split idempotents in $C$… because I wanted to get those Schur functors as quick as possible! So I was thinking “why worry about a bimonoid when a monoid will do?”, etc.

But you’re completely right that your approach nicely categorifies the expression $\sum_n a_n x^n / n!$

It also should be important at some later stage to recognize that $k[S_n]$ is a bimonoid (even a Hopf monoid), to see that its representations form a symmetric monoidal category (with duals for objects). Where does this come into the Schur functor story???

Perhaps it would make sense to move the bimonoid stuff elsewhere, where we actually bring in polynomial functors per se. I see you are trying to make a gradual ascent from simple to sophisticated, so it would be fine to keep the Schur functor construction as you now have it, but make a strong connection to polynomials later when we get into the $\overline{k\mathbb{P}}$ stuff.

Yeah, maybe that’s the best way to have our cake and eat it too! That’s a great idea. We can prove the two approaches are equivalent.

Btw, speaking of a “gradual ascent from simple to sophisticated”, I want to insert some stuff that demystifies your big theorem a bit by mentioning a decategorified version - a little baby version. Namely: the adjunction between $Cat$ and $SymmMonLinCauch$ is like an overgrown version of, say, the adjunction between $Set$ and $CommRing$. In both cases, the forgetful functor $U$ is representable by $F(1)$ where $F$ is the left adjoint of $U$. And I think that in the baby version, just as in the big version, the endomorphism thingie of $U$ should be $U(F(1))$. Etcetera. In the baby case $F(1)$ is the ring of polynomials in one variable; in the grownup case it’s the symmetric monoidal linear Cauchy-complete category of polynomial species!

It’s also true that in the baby version $U$ factors as

$CommRing \to CommMon \to Set ,$

much as in the grown-up version it factors as

$SymMonLinCauch \to SymMonLin \to SymMonCat \to Cat.$

Not a perfect analogy, and the analogy might be a wee bit better if instead of commutative rings we used commutative $k$-algebras, to get a field $k$ into the picture. But either way, I think it’s illuminating.

By the way, not necessarily for this paper: do we dare start thinking about the example

$U: SymMonAbCat \to Set,$

where $SymMonAbCat$ has symmetric monoidal abelian categories (with tensoring by any object right exact) as objects, symmetric monoidal right exact functors as morphisms, and symmetric monoidal natural transformations as 2-morphisms? Do we have any ideas about a left adjoint $F$ to this $U$? The stuff about “right exact” makes things a bit slippery compared to your setup, which is nicely self-dual. But I guess if my original wild guess was right, we should want $F(1)$ to be a version of $Schur$ defined over the integers instead of $k$. Same objects, fewer morphisms.

• CommentRowNumber76.
• CommentAuthorJohn Baez
• CommentTimeJun 29th 2010

John wrote:

By the way, not necessarily for this paper: do we dare start thinking about the example

$U: SymMonAbCat \to Set,$

where $SymMonAbCat$ has symmetric monoidal abelian categories (with tensoring by any object right exact) as objects, symmetric monoidal right exact functors as morphisms, and symmetric monoidal natural transformations as 2-morphisms?

Perhaps ’abelian’ is not what we want. Maybe we want symmetric monoidal finitely cocomplete $Ab$-enriched categories (with tensoring by any object right exact). Schur functors (and Specht modules) can be thought of as being built just from colimits.

• CommentRowNumber77.
• CommentAuthorTobyBartels
• CommentTimeJun 29th 2010

Just so’s y’all know: ‘polynomial functor’ is also used in the theory of recursion (and corecursion); see W-type.

• CommentRowNumber78.
• CommentAuthorJohn Baez
• CommentTimeJun 29th 2010

We’re going with ’polynomial species’, which should lessen the conflict.

Hey! We need a ’tin can diagram’ in the definition of modification in our Schur functor page. This is a cylinder-shaped diagram that’s the basic equation in the definition of a modification. What’s the easiest way to get a ’tin can diagram’ on the $n$Lab? I have a bunch in papers of mine, and Urs has a bunch in papers of his, but I don’t see any on the $n$Lab or $n$Cafe.

• CommentRowNumber79.
• CommentAuthorJohn Baez
• CommentTimeJun 30th 2010

I did a bit more work on the paper, Todd. There are a few queries where I could either use some help and/or just want you to see if you’re happy about something. Then you can delete ’em.

I still haven’t restored the stuff about bimonoids, which would go nicely in the section ’Structure of the representing object’. But I made it available in a temporary final section called ’Old stuff’.

I think it’s close to the point where further work will be best done in a separate paper, not this $n$Lab entry, because the introduction to the $n$Lab entry is supposed to be a broad overview of Schur functors, while our paper will be quite different, so the introduction and overall flow of the paper will need to be a bit different.

• CommentRowNumber80.
• CommentAuthorTobyBartels
• CommentTimeJun 30th 2010

What’s the easiest way to get a ’tin can diagram’ on the $n$Lab? I have a bunch in papers of mine, and Urs has a bunch in papers of his, but I don’t see any on the $n$Lab or $n$Cafe.

Point me to the source of some of these and I’ll see what I can do. If they’re drawn in standard Xy, then we can get it in the article using codecogs. If they’re EPS, then I can probably manage to convert it something useful.

• CommentRowNumber81.
• CommentAuthorJohn Baez
• CommentTimeJun 30th 2010

Toby wrote:

If they’re drawn in standard Xy, then we can get it in the article using codecogs.

Will xymatrix do? Does that count as part of ’standard Xy’? I think it’d be good if I draw a picture from scratch using xymatrix.

• CommentRowNumber82.
• CommentAuthorTobyBartels
• CommentTimeJun 30th 2010

Yes, it counts. Send the xymatrix code to me, or follow the instructions at #4 (or any of them, really) here.

• CommentRowNumber83.
• CommentAuthorJohn Baez
• CommentTimeJun 30th 2010

I just sent you the xymatrix code, Toby.

The labels are optimized for the nLab page Schur functor, but it would be great to have a different version for the page on modification.

For anyone else who likes to tinker with diagrams, the xymatrix code is here, surrounded by a little TeX. It produces this output.

• CommentRowNumber84.
• CommentAuthorJohn Baez
• CommentTimeJul 9th 2010

I spent a couple hours today polishing the first two sections of the Schur functor page - 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! If they don’t, please either make changes or ask questions here.

The later sections still need a lot of work.

• CommentRowNumber85.
• CommentAuthorDavid_Corfield
• CommentTimeJul 9th 2010
• (edited Jul 9th 2010)

That page is looking very nice now!

You run the analogy:

$Schur$ composed of Schur functors which are pseudonatural transformations $S: U \to U$, for forgetful $U: SymMonLinCauch \to Cat$.

$\mathbb{Z}[X]$ composed of polynomials which are natural transformations $S: U \to U$, for forgetful $U: CommRing \to Set$.

So the symmetric functions can be extracted as natural transformations of the underlying functor $\Lambda-Ring \to Set$.

Is it a rule of thumb that natural transformations of a representable right adjoint are interesting (if the representing object is interesting)? Let me dip into the table of free Xs on one generator on p. 25 of categorification. Natural pseudotransformations of the underlying functor from $k$-tuply monoidal $n$-categories to $n$-categories form $n Braid_k$? Would this mean that these braids ’act on’ those $k$-tuply monoidal $n$-categories?

And this isn’t that close to extracting the fundamental group of a space from the (reversible) transformations of a fibre functor: Locally constant sheaves over $X \to Set$, because this fibre functor isn’t a representable right adjoint?

• CommentRowNumber86.
• CommentAuthorUrs
• CommentTimeJul 9th 2010

A hint for your square diagrams further down the entry: if you include the labels of your vertical arrows in

  \mathllap{...}


and

 \mathrlap{....}


respectively, then they will “stick out of their invisible position box” to the left and right and make the diagrams align as intended.

• CommentRowNumber87.
• CommentAuthorTodd_Trimble
• CommentTimeJul 9th 2010
• (edited Jul 9th 2010)

@David #85:

Yes! But in addition to studying the actions of the monoid of unary operations $U \to U$ via the representing object (as the free thing on one generator), it’s also very interesting to study (actions of) the category of general operations $U^m \to U^n$ via the opposite of the category of free objects, as emphasized by Lawvere in his approach to universal algebra. This will be important to us as well, as we venture into categorified birings.

When I get a little more time today or this weekend, I think I’ll address the query boxes John left me within Schur functor, and drop some of my own based on some suggestions/”complaints” I left in the plethysm thread. I guess my answers to queries can be copied and pasted into one or the other drafts that John is writing up (Schur Functors I and II), although that’s a little hard to predict until I take a closer look at what needs to be done.

@Urs #86: thank you very much!