The Two Cultures of Mathematics Revisited

May 22, 2007

The Two Cultures of Mathematics Revisited

Posted by David Corfield

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Where did we get to in our discussion of the two cultures of mathematics? To explore the possibility that interaction may be possible between what Gowers called ‘combinatorics’ and our Café subculture we were set the challenge of categorifying instances of the Cauchy–Schwarz inequality, which, unless I missed something, didn’t result in any noticeable success.

Now, an extreme wing of our subculture would take Urs’ remark

One knows one is getting to the heart of the matter when the definitions in terms of which one conceives the objects under consideration categorify effortlessly.

and replace the when by when and only when.

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I was probing in this direction in my paper, Categorification as a Heuristic Device, when I put:

A happenstantial equation is one which cannot be categorified productively.

The term ‘happenstantial’ is supposed to contrast with ‘law-like’. I had written something on the idea of mathematical laws inspired by reading Poincaré say

Les faits mathématiques dignes d’être étudiés, ce sont ceux qui, par analogie avec d’autres faits, sont susceptibles de nous conduire à la connaissance d’une loi mathématique de la même façon que les faits expérimentaux nous conduisent à la connaissance d’une loi physique. Ce sont ceux qui nous révèlent des parentés insoupçonnées entre d’autres faits, connus depuis longtemps, mais qu’on croyait à tord étrangers les uns aux autres.

Now, if we can’t categorify the Cauchy–Schwarz inequality, this would seriously weaken the extreme view. But perhaps we can think of more subtle ways of bridging the gap between the cultures. Terence Tao reports on Shing-Tung Yau’s lectures in UCLA’s Distinguished Lecture Series in a field “adjacent” to his own areas of expertise. Plenty of Yau’s themes over the three lectures are adjacent to Café interests too.

Posted at May 22, 2007 9:49 AM UTC

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Re: The Two Cultures of Mathematics Revisited

Yau’s extraordinarily verbal lecture notes are here: 1, 2, 3.

Posted by: David Corfield on May 22, 2007 11:56 AM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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David wrote:

To explore the possibility that interaction may be possible between what Gowers called “combinatorics” and our Café subculture we were set the challenge of categorifying instances of the Cauchy–Schwar inequality, which, unless I missed something, didn’t result in any noticeable success.

Hmm.

Here’s a preliminary attempt to categorify the Cauchy–Schwarz inequality.

Let $Rep (G)$ be the category of finite-dimensional representations of a compact topological group $G$ — for example, a finite group.

For any representations $A, B \in Rep (G)$ , there’s an obvious inclusion of vector spaces:

$hom (A, B) ↪ A^{*} \otimes B$

Here:

$hom (A, B)$ is the space of intertwining operators from the representation $A$ to the representation $B$ . This is the external hom in $Rep (G)$ .
$A^{*} \otimes B$ is the space of all operators from the vector space $A$ to the vector space $B$ . This is the internal hom in $Rep (G)$ .

The inclusion says, ironically: the external hom is smaller than the internal hom. Some things look bigger from the inside!

If we take dimensions, we get

$\dim (hom (A, B)) \leq \dim (A^{*} \otimes B)$

This looks vaguely Cauchy–Schwarzian. Let’s see what it says!

Every representation is of the form

$A = \oplus_{i} ℂ^{a_{i}} \otimes e_{i}$

where $e_{i}$ ranges over a basis of irreducible representations of $G$ .

By Schur’s lemma we have this isomorphism of vector spaces:

$hom (A, B) ≅ \oplus_{i} (ℂ^{a_{i}} \otimes ℂ^{b_{i}})$

but we also clearly have this isomorphism of vector spaces:

$A^{*} \otimes B ≅ (\oplus_{i} ℂ^{a_{i}}) \otimes (\oplus_{j} ℂ^{b_{j}})$

The inclusion

$hom (A, B) ↪ A^{*} \otimes B$

is just the obvious diagonal inclusion

$\oplus_{i} (ℂ^{a_{i}} \otimes ℂ^{b_{i}}) ↪ (\oplus_{i} ℂ^{a_{i}}) \otimes (\oplus_{j} ℂ^{b_{j}})$

If we decategorify by taking dimensions, we thus get

$\sum_{i} a_{i} b_{i} \leq (\sum_{i} a_{i}) (\sum_{j} b_{j})$

Alas, this ain’t the Cauchy–Schwarz inequality! Cauchy–Schwarz says:

$(\sum_{i} a_{i} b_{i})^{2} \leq (\sum_{i} a_{i}^{2}) (\sum_{j} b_{j}^{2})$

Unlike Cauchy–Schwarz, the inequality I got doesn’t hold for all real numbers $a_{i}$ and $b_{j}$ . My proof only shows it holds for all natural numbers — which is also obvious directly.

But, all hope is not lost! Maybe we can get the Cauchy–Schwarz inequality using similar tricks. For example, by finding an inclusion

$hom (A, B) \otimes hom (B, A) ↪ hom (A, A) \otimes hom (B, B)$

Anyone want to take a stab at it?

I think, though, that we should call this:

$\sum_{i} a_{i} b_{i} \leq (\sum_{i} a_{i}) (\sum_{j} b_{j})$

the TARDIS inequality.

Posted by: John Baez on May 23, 2007 1:57 AM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

Looks similar to quantum probabilities: there are many evidences why quantum amplitude (with quadratic law) is most natural generalization of classic probability, but no distinct categorified evidence, as I can see. The problem of quadratic laws categorification, ah?

Posted by: osman on May 23, 2007 7:49 AM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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A dumb question from someone who hasn’t taken the time to understand very much of this category stuff: all the “basic” instances of Cauchy-Schwarz start by looking at $(f + g)^{2}$ (or $〈 f + g, f + g 〉$ ) and then splitting it into bits like $f^{2}$ , $fg$ , etc. So why isn’t the natural place to start to be looking for some way (lets call it $\lor$ just because this operator hasn’t been used elsewhere) of combining $A$ and $B$ such that $hom (A \lor B, A \lor B)$ can be broken into pieces? Ie, I don’t see why one would expect the space corresponding to the “squared cross terms” to be a included in the “product of square terms” rather than “complementary to it with respect to some combined structure”?

Posted by: dave tweed on May 23, 2007 9:47 AM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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Minor correction: Cauchy-Schwarz actually begins considering $(f + λ g)^{2}$ for all $λ$ . So the analogous combined space will be even more complicated than just $A \lor B$ (for some $\lor$ ), but I think the basic question why one wouldn’t expect to start with some superstructure still arises?

Posted by: dave tweed on May 23, 2007 10:06 AM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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In case it’s helpful to explicitly state for other neophytes here (gleaned from other comments), the “implicit” block to coming up with an categorical analogy to the $∣ f + λ g ∣^{2}$ proof of Cauchy-Schwarz is that you can’t use the notion of subtraction anywhere in the proof (because you want to work in contexts where subtraction may not be applicable); I’d only thought about not using it in the first step. This seems unlikely to be acheivable in an analogy to the quadratic form proof of C-S.

Posted by: dave tweed on May 23, 2007 7:33 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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Why do I get the feeling that there won’t be a natural way of doing this? Back here when we were talking about Cauchy-Schwarz for sets, Terry Tao pointed out a reliance on Choice.

I’m hearing a faint tinkling of that phenomenon encountered when categorifying series to species. We saw that

$\cosh (2 x) \equiv \sinh^{2} (x) + \cosh^{2} (x)$

categorifies to

The species ‘2-coloured even numbered set’ is isomorphic to the species ‘either an ordered pair of odd numbered sets or an ordered pair of even numbered sets’.

However, trying to categorify

$\cosh^{2} (x) \equiv \sinh^{2} (x) + 1$ ,

one can’t arrive at an isomorphism. (Ah, found it in a trice (p. 65), using Mike Stay’s excellent index.)

I seem to remember an e-mail chat we had about this that to have an isomorphism we would need extra structure on the sets. Placing the structure ‘pair of even numbered sets’ on a pointed set, one could flick the pointed element into the other set to make ‘pair of odd numbered sets’.

I also remember a piece of wanton speculation to the effect that this difference reflected why the second equivalence is more interesting than the first, with its link to the hyperbola, etc.

Posted by: David Corfield on May 23, 2007 12:24 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

Analysis as a field is almost completely “nonsense”-free, and I think that this is another reason you get the sense that there’s not a natural way of doing this. Maybe, if one could articulate what the obstacle is a little more clearly, this is why there’s so little algebra in analysis.

Or maybe analysts are doing it wrong.

Posted by: Changbao on May 23, 2007 4:58 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

How should we view Sato’s work in relation to this question of algebra in analysis? Pierre Schapira writes:

Looking back, forty years later, we realize that Sato’s approach to mathematics is not so different from that of Grothendieck, that Sato did have the incredible temerity to treat analysis as algebraic geometry, and that he was also able to build the algebraic and geometric tools adapted to his problems.

Would some analysts see the reach of Sato’s concepts as limited?

Posted by: David Corfield on May 24, 2007 8:58 AM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

This identity is not true for species, but it is for K-species with K with enough structure (in this case, I seem to remember that you need Q-species where the usual species are N-species [rationals versus the rig of non-negative integers]). This can be found in a paper of Yeh, in LNM 1234.

Posted by: Jacques Carette on May 24, 2007 3:24 AM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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David wrote:

Why do I get the feeling that there won’t be a natural way of doing this? Back here when we were talking about Cauchy-Schwarz for sets, Terry Tao pointed out a reliance on Choice.

Hmm! That’s usually a sign that one is doing something hugely non-canonical, which doesn’t deserve to be categorified.

The first time you guys had this discussion of categorifying Cauchy–Schwarz, I wasn’t paying much attention. Now I think I see what all the fuss is about.

In ordinary uncategorified math, to get an inequality, you usually need an inequality. Cauchy–Schwarz is really just a spinoff of the basic inequality

$A \cdot A \geq 0$

where you see that we can only have

$(A + λ B) \cdot (A + λ B) \geq 0$

for all $λ$ if $A \cdot B$ doesn’t get too big.

All of what I just wrote is perfectly easy to categorify; the problem — I think — is that to derive Cauchy–Schwarz you need to subtract.

While I’ve spent a lot of time trying to categorify subtraction and get a good theory of negative sets, it hasn’t fully gelled, so often I’m willing to follow Nancy Reagan’s advice and just say no when it comes to the use of subtraction in categorified settings: when tempted to use subtraction to make a result look prettier, just don’t do it!

If we followed this policy with regards Cauchy–Schwarz, we’d just try to categorify

$(A + λ B) \cdot (A + λ B) \geq 0$

and this would be perfectly easy to do; for example, in many contexts we have a monomorphism

$0 ↪ (A + λ B) \otimes (A + λ B)$

$0 ↪ hom ((A + λ B), (A + λ B))$

(an inner product being more like an internal hom than a tensor product).

I’m hearing a faint tinkling of that phenomenon encountered when categorifying series to species.

Yes, there are some nice identities that don’t categorify unless you bend the rules, and surely this is trying to tell us something.

I seem to remember an e-mail chat we had about this that to have an isomorphism we would need extra structure on the sets. Placing the structure ‘pair of even numbered sets’ on a pointed set, one could flick the pointed element into the other set to make ‘pair of odd numbered sets’.

I seem to remember that conversation. Right. You seem to have just proved $\cosh^{2} (x) = \sinh^{2} (x)$ , but of course this trick of equipping our set with a distinguished point doesn’t work for the empty set, which contributes the ‘ $+ 1$ ’ to the identity $\cosh^{2} (x) = \sinh^{2} (x) + 1$ .

Equipping sets with extra structure, we increase our power to categorify identities, but diminish the interest of the results.

Posted by: John Baez on May 24, 2007 2:35 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

Are there lessons to be learned about the difficulty of a good theory of negative sets?

By the way, with regard to your page on counting, I found a better translation of the Weyl quotation you have there:

Mathematics is not the rigid and petrifying schema, as the layman so much likes to view it; with it, we rather stand precisely at the point of intersection of restraint and freedom that makes up the essence of man itself.

Posted by: David Corfield on May 24, 2007 3:09 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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Here’s a preliminary attempt to categorify the Cauchy-Schwarz inequality.

It seems to me that the John’s approach above has a somewhat different flavor than Terry Tao’s approach:

in one case the product is sought to be categorified by a categorical product – in the other case one attempts to identify it with a Hom-space!

I like John’s approach for the following reason: we know from examples arising in concrete applications that it is often indeed the case that the Hom-functor plays the role of the inner product on a vector space.

So, sticking with the culture of problem-solving, I now decide to be interested in the formal game of categorification here only in as far as I know it relates to concrete applications.

So: inner product should go to Hom.

Moreover, we know that complex conjugation should go to reversal of arrows, in this context.

More precisely, the existence of a structure that allows complex conjugation should translate to a category with duals, such that we have an operation $(\cdot)^{†} : (f : a \to b) \mapsto (f^{†} : b \to a) .$

This should mean that we read $〈 a, b 〉$ as $Hom (a, b)$ and $〈 a, b 〉 〈 b, a 〉$ as $Hom (a, b) \otimes Hom (b, a) .$

I notice that we have a canonical morphism $c : Hom (a, b) \otimes Hom (b, a) \to End (a) \otimes End (b)$ which acts as $c : (f : a \to b) \otimes (g : b \to a) \mapsto (g \circ f) \otimes (f \circ g) .$

This is in general not epi.

Of course it is also in general not mono, so this might not be quite the right answer yet, but I thought it might be worth mentioning nevertheless.

Posted by: urs on May 23, 2007 7:32 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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I think I can strenghten that statement, exhibiting the entire natural transformation:

Consider a category $C$ with duals such as $Hilb$ , such that $a \overset{f}{\to} b \overset{f^{†}}{\to} a = a \overset{Id}{\to} a$ for all morphisms $f$ .

In this case we have on top of the canonical Hom-functor $Hom : C^{op} \times C \to Set$ (I don’t want to be too specific about where Hom takes values) an endomorphism functor $End : C \to Set$ which acts as $End : (a \overset{f}{\to} b) \mapsto (End (a) \overset{{Ad}_{f}}{\to} End (b)) .$

Then, I claim, we have a canonical natural transformation

$Hom ((\cdot)^{†}, \cdot) \times σ^{*} Hom ((\cdot)^{†}, \cdot) \to End (\cdot) \times End (\cdot)$

between functors $C \times C \to Set$ , where $σ$ is the standard braiding on $Cat$ (i.e. just swaps the arguments).

Its component map is that from my previous comment.

Read

Hom is inner product

End is norm square

to see something like Cauchy-Schwarz.

Posted by: urs on May 23, 2007 8:21 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

This gave me a sense of deja vu all over again :)

Simply reversing the arrows leads to some inner product that may ultimately not be what you want, but you can get what you want by introducing deformations once the mechanics are worked out.

Just a quick (and random) comment from the sidelines…

Posted by: Eric on May 23, 2007 10:08 PM | Permalink | Reply to this

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(1) I must be missing something, it seems to me that the map c isn’t linear. The definition given is $c : (f : a \to b) \otimes (g : b \to a) \mapsto (g \circ f) \otimes (f \circ g)$ So $c (f_{1} \otimes g + f_{2} \otimes g) = c ((f_{1} + f_{2}) \otimes g) = ((f_{1} + f_{2}) \circ g) \otimes (g \circ (f_{1} + f_{2}))$ . Since our category is abelian, this is $(f_{1} \circ g + f_{2} \circ g) \otimes (g \circ f_{1} + g \circ f_{2})$ . This is not $c (f_{1} \otimes g) + c (f_{2} \otimes g) = (f_{1} \circ g) \otimes (g \circ f_{1}) + (f_{2} \circ g) \otimes (g \circ f_{2})$ .

(2) There are very reasonable abelian categories, with finite dimensional Hom spaces, in which dim Hom(A,A) dim Hom(B,B) can be less than dim Hom(A,B) dim Hom(B,A). For example, consider the category of coherent sheaves on a disjoint union of two ${CP}^{2}$ ’s. Take A to be O(1) on the first ${CP}^{2}$ and O(0) on the second, and take B to be the reverse. Then Hom(A,A) consists of functions which are constant on each component, and hence Hom(A,A) is two dimensional. The same holds for Hom(B,B). But Hom(A,B) consists of functions which are zero on the first component but can be any section of O(1) on the second component, and hence Hom(A,B) is three dimensional. The same holds for Hom(B,A). 2*2 is not greater than 3*3.

I suspect that I can build a similar example on a connected variety which is proper but not projective. All I need is a line bundle L such that $H^{0} (L)$ and $H^{0} (L^{- 1})$ both have dimension bigger than 1. (Then I take A to be the structure sheaf and B to be L.) But I can’t think of such an example right now.

Posted by: David Speyer on May 23, 2007 11:46 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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David Speyer commented my suggestion with

I must be missing something,

Well, as I said:

this might not be quite the right answer yet, but I thought it might be worth mentioning nevertheless.

One reason is

There are very reasonable abelian categories, with finite dimensional Hom spaces, in which dim Hom(A,A) dim Hom(B,B) can be less than dim Hom(A,B) dim Hom(B,A).

Yes, as I remarked, $c$ is in general not epi but also not mono.

So if you decategorify $c$ in “the obvious way” by passing to isomorphism classes, it doesn’t really yield Cauchy-Schwarz.

I am not sure, though, that this is the right way to look at it: if we were talking about an ordinary product, then yes. But it might be relevant that we are talking about an inner product (“scalar product”). It’s not obvious to me that this, as a concept, should be thought of, as seems to be the premise of other comments here, as nothing but an abbreviation for the component formula $〈 v, w 〉 = \sum_{i} {\bar{v}}_{i} w_{i},$ and hence just as an abbreviation for a combination of ordinary products.

Posted by: urs on May 24, 2007 9:06 AM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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Here is a variation on Urs’ approach, which has the advantage of being linear (so it is immune to David Speyer’s first objection), since one can write it down with the graphical calculus, which I don’t think you can do for Urs’ original formula (if you could , Urs’ formula would have been linear).

In other words, I’m continuing in the spirit of Urs’ approach - following John’s suggestion - and interpreting “categorifying the Cauchy-Schwarz inequality” as finding an injective morphism

(1)

hom (A, B) \otimes hom (B, A) ↪ hom (A, A) \otimes hom (B, B) .

Here I’m assuming I’m working in a 2-Hilbert space. This means we’re working in a linear category which is enriched in Hilb (i.e. there are inner products on the hom-sets), and which has an involution $* : hom (A, B) \to hom (B, A)$ .

As in John’s post, the best example of a 2-Hilbert space is the category $Rep (G)$ of (possibly projective) unitary representations of a finite group, or groupoid, where the involution is the adjoint of a linear map.

As David essentially showed in his second objection, things like “coherent sheaves” aren’t 2-Hilbert spaces (though they’re closely related), so this interpretation of the Cauchy-Schwarz inequality won’t work there.

On the other hand, if one thinks of the Cauchy-Schwarz inequality as having to do with inner products in a vector space ( is that the right way to think of it?), and then one tries to categorify this concept, it is natural one should work in a “2-inner product space”, i.e. a 2-Hilbert space.

After this long introduction, here’s the graphical definition of the morphism. Remember these ribbon-like diagrams take place in the monoidal category Hilb. To be reminded how they work, look at this post of mine or John’s notes on classical vs quantum computation.

Here is the morphism (top-to-bottom):

Recall that the “half-twist” maps are $*$ -operations, while the upside-down Y’s are the adjoints of the composition maps.

One can show this map is isometric (and hence injective) by graphical manipulations. I’m running out of space, so let’s just use brute force, and see what the map actually does in terms of a chosen basis.

So let $a_{i}^{p}$ be a basis for $A$ , where $i$ runs through the simple objects (irreducible components), and $p$ runs through the $i$ ‘th isotypic part of $A$ . Similarly let $b_{i}^{q}$ be a basis for $B$ .

We then get a basis $(AB)_{i}^{p, q}$ for the hom-set $hom (A, B)$ , where we are writing $(AB)_{i}^{p, q}$ for the map $A \to B$ which sends

(2)

a_{i}^{p} \mapsto b_{i}^{q} .

Similarly we get bases $(AA)_{i}^{p, p'}$ and $(BB)_{i}^{q, q'}$ for $hom (A, A)$ and $hom (B, B)$ .

In terms of these bases, the map

(3)

hom (A, B) \otimes hom (B, A) ↪ hom (A, A) \otimes hom (B, B)

sends

(4)

(AB)_{i}^{p, q} \otimes (BA)_{j}^{q', p'} \mapsto δ_{ij} (AA)_{i}^{p, p'} \otimes (BB)_{i}^{q, q'},

which is clearly injective. Another way to think of it graphically is as a kind of “crossing” duality $pic$ :

Posted by: Bruce Bartlett on May 24, 2007 3:46 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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Thanks Bruce, great! I was hoping you would look into this and see how to do it.

I am glad this blog exists and has contributors like you.

Posted by: urs on May 24, 2007 4:51 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

Wow, that’s really pretty. So the point is, if we want dim Hom to act like a positive definite inner product, we need to already have some sort of positive definite structure (a Hilbert space, in this case) on Hom(V,W).

Thanks!

Posted by: David Speyer on May 24, 2007 5:35 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

Dear Bruce, is the map (4) really injective? When i is not equal to j it looks like it’s mapping to zero, to me.

Posted by: Terry Tao on May 24, 2007 6:03 PM | Permalink | Reply to this

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Ah, you’re right, it isn’t injective! This oversight on my part comes from working things out “one isotypic part at a time”, and then adding together all the contributions at the end, without thinking carefully enough about how the final result pans out.

By the way, if our 2-Hilbert space is $Rep (G)$ for a finite group $G$ , and we have given the hom-sets the “trace inner product”,

(1)

(f, g) = Tr (f^{*} g),

then there should be a factor of $\dim (e_{i})$ appearing (where $e_{i}$ is the i’th irreducible representation) which I missed before,

(2)

(AB)_{i}^{p, q} \otimes (BA)_{j}^{p', q'} \mapsto δ_{ij} \dim (e_{i}) (AA)_{i}^{p, p'} \otimes (BB)_{i}^{q, q'} .

I guess on the diagonal, i.e. where $i = j$ , the map is just a simple bijective rearrangement of the indices (multiplied by a scalar factor), while off the diagonal it’s zero. And that’s not exactly Cauchy-Schwarz $pic$ .

I guess the map I presented in graphical notation can’t be of fundamental importance : it looks a little ugly.

Another thing that’s lacking in the approach I gave is a ‘TARDIS-like’ internal hom vs external hom dichotomy, which I found quite appealing.

While I’ve written this David Speyer has written down some interesting points.

Posted by: Bruce Bartlett on May 24, 2007 10:48 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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Darn, Terry is right.

Moreover, there are representation theoretic reasons that we can’t do this in a natural way. Let C be the category of finite dimensional representations of the group Z/2. If we want to fit this into the 2-Hilbert space formalism, I think we can do this by having the representations come equipped with an invariant positive definite inner product, so that we get a Hilbert space structure on Hom(A,B) by $〈 f, g 〉 = Tr (g f^{*})$ . We’ll write an object A of C as $(A (+) \oplus A (-))$ , where $A (\pm)$ are the isotypic components of A. Then $Aut (A) = GL (A (+)) \times GL (A (-))$ . If representations come equipped with invariant inner products, then the automorphism group of A is $O (A (+)) \times O (A (-))$ (where O(V) is the orthogonal group on an inner product space V).

The proposed goal is to find an injection

${Hom}_{C} (A, B) \otimes {Hom}_{C} (B, A) ↪ {Hom}_{C} (A, A) \otimes {Hom}_{C} (B, B) .$

If this injection were natural, it would in particular be $Aut (A) \times Aut (B)$ equivariant. But the irreducible representation $A (+) \otimes A (-)^{*} \otimes B (+) \otimes B (-)^{*}$ (where $*$ is dualization) appears on the left but not the right.

A similar argument with symmetric groups shows that there is no natural way to make the argument work with the finite set version either.

Posted by: David Speyer on May 24, 2007 9:40 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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There’s a slightly nicer, more invariant way to write down the calculations John is performing above, as John is probably aware. It’s true that every representation is of the form

(1)

A = \oplus_{i} ℂ^{a_{i}} \otimes e_{i}

where $e_{i}$ ranges over a basis of irreducible representations of $G$ , but this is a noncanonical isomorphism… it is only canonical to split $A$ up into its isotypic factors, but further splitting requires choices, as I’m sure we all agree.

A more invariant way to portray the noncanonical isomorphism

(2)

hom (A, B) ≅ \oplus_{i} ℂ^{a_{i}} \otimes ℂ^{b_{i}}

is to write

(3)

hom (A, B) ≅ \oplus_{i} hom (A, e_{i}) \otimes hom (e_{i}, B)

which is canonical, since it is the inverse of the canonical composition map,

(4)

hom (A, e_{i}) \otimes hom (e_{i}, B) \to hom (A, B) .

I mentioned this in a previous post, where I also showed how one can write it in ribbon-like diagrams… which all of us at the n-cafe (hopefully!) learnt from John’s classical versus quantum computation notes :-).

Posted by: Bruce Bartlett on May 24, 2007 1:38 AM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

I’ve just naively checked following code in Python:
ss = lambda x, y: (pow(x,x)*pow(y,y)) - (pow(x,y)*pow(y,x)) for i in xrange(1,1000): ....for j in xrange(i,1000): ........if 0 > ss(i,j): print i, j

In [1,1000] it’s always > 0!

Posted by: osman on May 25, 2007 12:37 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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In other words, I’ve tried to calculate $x^{x} y^{y} - x^{y} y^{x}$ for integer $x$ and $y$ in interval from 1 to 1000. And in this interval always $0 \leq x^{x} y^{y} - x^{y} y^{x}$

Posted by: osman on May 25, 2007 4:12 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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Oh, I’m so stupid! If $y = x + n$ then $x^{x} y^{y} - y^{x} x^{y} = y^{n} - x^{n} \geq 0$ ! :(

Posted by: osman on May 25, 2007 9:33 PM | Permalink | Reply to this

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Nobody answers to me, so I feel I say something irrelevant or something well-known. Anyway I must complete the proof whithout errors. So, excuse me for one more post.

I want to prove:

in category of finite sets $Hom (A, B) \times Hom (B, A)$ is isomorphic to some $C \subseteq Hom (A, A) \times Hom (B, B)$ .

In this case it’s enough to prove that for any natural $x$ and $y$ we have $x^{x} y^{y} - x^{y} y^{x} \geq 0$ . This expression is symmetric for $x$ and $y$ , so we can choose $y = x + n$ for some natural $n$ . So we have:

$x^{x} y^{y} - x^{y} y^{x} = x^{x} y^{x} (y^{n} - x^{n}) \geq (y^{n} - x^{n}) \geq 0$ .

Posted by: osman on May 26, 2007 10:33 AM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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Sorry that no-one replied to your comments; they were difficult to understand at first.

I think it’s clear that there exists an injection

(1)

Hom (A, B) \times Hom (B, A) ↪ Hom (A, A) \times Hom (B, B)

in the category of finite sets (or using $\otimes$ in a linear category), merely by a counting argument, as you’ve shown. I think the point of this discussion was to try and find a natural injection, as originally suggested by John.

Urs tried one way, and I tried another way, but both methods basically failed, for one reason or another.

Then David Speyer essentially proved a “no-go” theorem, which suggested that any attempt to find a natural injection

(2)

Hom (A, B) \times Hom (B, A) ↪ Hom (A, A) \times Hom (B, B)

would fail (either in finite sets or in some category like $Rep (G)$ ), because it would necessarily violate symmetry.

Thus it seems we reached a dead-end in this direction, and I think people have moved on now to try more exotic methods. That’s the way I interpret this discussion anyway.

Posted by: Bruce Bartlett on May 26, 2007 5:26 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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Thank you Bruce! At leats I understood how to use itex here. :) I’m sorry for my first obscure posts.

It’s a little difficult for me to watch you conversations, because I still don’t know enough mathematics. But now, thanks to your explanations, I know the main result of this thread.

Posted by: osman on May 27, 2007 11:58 AM | Permalink | Reply to this

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Hi again! I’m happy to see you take up this puzzle again - welcome to the “problem solving” culture of mathematics :-).

I have a few comments. Firstly, I recommend reading Gowers’ thoughts on the Cauchy-Schwarz inequality. He mostly discusses the $∥ v + λ w ∥^{2}$ approach to proving the Cauchy-Schwarz inequality, and only briefly mentions the Lagrange identity approach at the end. Of the two I feel the Lagrange identity approach has a slightly better chance of categorifying, since it is after all an identity. (A third approach is algorithmic, manipulating v and w by incremental changes to narrow the gap in the inequality until equality is attained, but this probably has the least chance of being categorified - on the sphere of mathematics, I would consider the mathematics of algorithms is being almost the antipode of the mathematics of categories (try defining the category Algorithm and you’ll see what I mean).)

Secondly, one thing about the “problem solving” culture as opposed to the “theory building” is that we tend to make progress by working upward from special cases, rather than downward from more general formulations. In the last post on this topic, Robin was able to categorify the arithmetic mean-geometric inequality

$A \times B + B \times A \to A \times A + B \times B$

for finite sets A, B by introducing reasonable axioms on Set. I think the next step is to obtain a relative version of this injection, namely

$A \times_{C} B + B \times_{C} A \to A \times_{C} A + B \times_{C} B$

where we are given maps $A \to C$ and $B \to C$ , thus A and B are basically bundles over C. If we think non-categorically, the latter injection is a trivial consequence of the former, as we can work on each fibre separately and sum up; but I believe you can’t get away with that categorically, and you probably need to introduce relativised versions of Robin’s axioms (or maybe create an abstract axiom of relativisation).

The key difficulty here seems to me that whereas on the non-relative setting, we either had A larger than B or B larger than A, in the relative setting, the relative size of the A-fibre and the B-fibre depends on the choice of fibre, and so you have to make that part of the argument (i.e. the splitting into two cases) fibre-dependent, and thus embedded inside the category instead of being external to it.

I believe that if the above relative AM-GM inequality (which is basically trying to capture the fact that “an arbitrary sum of squares of integers is positive”) is categorified then it will be straightforward to get Cauchy-Schwarz in the form

$(A \times_{C} B) \times (B \times_{C} A) \to (A \times_{C} A) \times (B \times_{C} B)$

by categorifying Lagrange’s identity appropriately, since the error term in that identity is also a sum of squares. Indeed, from relative AM-GM we have

$(A \times B) \times_{C \times C} (B \times A) + (B \times A) \times_{C \times C} (A \times B) \to (A \times B) \times_{C \times C} (A \times B) + (B \times A) \times_{C \times C} (B \times A)$

which rearranges to form a doubled version of Cauchy-Schwarz. (One then has to “divide by 2”, i.e. deduce $A \to B$ from $A + A \to B + B$ , but that seems to be an easier problem.)

Posted by: Terry Tao on May 23, 2007 5:08 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

“… trying to capture the fact that an arbitrary sum of squares of integers is positive is categorified…”

BUT there can be extra structure. For example, we should not neglect that every nonnegative integer can be represented as the sum of 4 squares of nonnegative integers, when we categorize. Right?

Posted by: Jonathan Vos Post on May 23, 2007 5:17 PM | Permalink | Reply to this

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when we categorize

Once upon a time I used to make that mistake, too. The right word is categorify.

Posted by: urs on May 23, 2007 6:41 PM | Permalink | Reply to this

right word is categorify; Re: The Two Cultures of Mathematics Revisited

I apolify.

Posted by: Jonathan Vos Post on May 24, 2007 12:59 AM | Permalink | Reply to this

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Terry Tao wrote:

I recommend reading Gowers’ thoughts on the Cauchy-Schwarz inequality.

I enjoyed them too. Here is the link.

Posted by: Tom Leinster on May 23, 2007 6:42 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

…try defining the category Algorithm…

Noson Yanofsky gave it a go here.

We define an algorithm to be the set of programs that implement or express that algorithm. The set of all programs is partitioned into equivalence classes. Two programs are equivalent if they are “essentially” the same program. The set of all equivalence classes is the category of all algorithms.

Perhaps John has seen other attempts in his reading for his ‘Cohomology and Computation’ course.

Posted by: David Corfield on May 23, 2007 7:07 PM | Permalink | Reply to this

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Could anything be done in a “K-theoretic” way? So in the unrelativized case, for any pair of sets, $A$ and $B$ , there are $D$ and $E$ such that $A + D$ and $B + E$ are isomorphic.

A relativized version would similarly postulate $D$ and $E$ over $C$ .

If, as noted here,

FINITE SETS ARE LIKE ‘VECTOR SPACES OVER THE FIELD WITHOUT ELEMENTS’,

there is a hint of something literally K-theoretic in the air, sums of vector bundles anyway.

Posted by: David Corfield on May 24, 2007 11:11 AM | Permalink | Reply to this

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Terry writes:

Secondly, one thing about the “problem solving” culture as opposed to the “theory building” is that we tend to make progress by working upward from special cases, rather than downward from more general formulations.

It’s all very complicated.

Theory-builders may act as if general theories come full-blown into their heads, just waiting to be applied to special cases. But often the most exciting part is finding those general theories in the first place. For this, one has to dream up patterns that would make sense of obscure clues. This often progresses from special to general.

I think of my style as ‘finding paths of least resistance’, or ‘following the tao of mathematics’. I’m more of a hiker than a rock-climber. I prefer to know lots of fun trails, and occasionally bump into a new one while I’m strolling about, than spend my day trying to climb up a sheer cliff. Instead of trying to do anything hard, I prefer to keep lots of facts, patterns and questions in mind, and make obvious guesses and deductions when I notice patterns turning up.

Given this rather lazy approach to mathematics, the only way to discover anything new is by knowing a little about lots of things, and knowing some good patterns that haven’t been fully exploited. ‘Categorification’ is my favorite pattern, because it’s extremely broad, and nowhere near being tapped out. But, it’s just one of many.

We can imagine two extremes. At one extreme we have someone who tries very hard to open a specific door, perhaps melting the lock with an acetylene torch if necessary, or even blowing the door open with dynamite. At the other extreme, we have someone who carries a huge ring of keys and tries them all on any door they happen to walk by.

The Wikipedia article on lock picking makes for interesting metaphoric reading:

Lock picking is the ideal way of opening a lock without the correct key, while not damaging the lock, allowing it to be rekeyed for later use, which is especially important with antique locks that would be impossible to replace if destructive entry methods are used.

Usually it is possible to bypass a lock without picking it. Most common locks can be quickly and easily opened using a drill, bolt cutters, or a hydraulic jack. The hasp, door, or fixture they are attached to can be cut or broken.

A lock that offers high resistance to picking does not necessarily make unauthorized access more difficult, but it will make surreptitious unauthorized access more difficult. They are often used in combination with alarms to provide layered security.

Some people enjoy picking locks recreationally, because they find it brings high satisfaction and hack value.

Posted by: John Baez on May 24, 2007 8:32 PM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

“Some people enjoy picking locks recreationally…”

[Insert canonical Feynman Los Alamos safecracking anecdote here]

I had the distinct impression that Feynman followed the Tao of safecracking and lockpicking in his applications of mathematics, where he was wildly creative and sneaky, and willing to let others add rigor later, after he got the right answer elegantly.

In this, he was a consummate member of the ‘problem solving’ culture – but left apparatus of interest to the ‘theory building’ culture.

Like Terrence Tao, he went out of his way to show how he did it, and did not hide behind formal paper-writing “behold, consider this bizarre equation; see, it works; never mind how I found it” variety.

Posted by: Jonathan Vos Post on May 25, 2007 1:19 AM | Permalink | Reply to this

Re: The Two Cultures of Mathematics Revisited

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By, the way, I’ll mention a take on this that I had last night. It might be better to take the K-theoretic approach and consider our inner product to be $E (A, B) : = \sum (- 1)^{i} {Ext}^{i} (A, B)$ , rather than just ${Ext}^{0} (A, B) = {Hom}^{0} (A, B)$ . Now, $E (A, B)$ won’t be symmetric. There are three reasonable ways out: define Cauchy-Schwartz to be the claim $E (A, B) E (B, A) \leq E (A, A) E (B, B)$ whether or not $E$ is symmetric; define $Q (A, B) = E (A, B) + E (B, A)$ and investigate the claim for $Q$ ; use Tor instead of Ext.

No matter what approach we take, our problem breaks into two halves. First, figure out whether the non-categorical, plain numeric Cauchy-Schwartz is true. (Or, more likely, what hypotheses we need to add to make it true.) Second, find the categorical statement explaining the numerical result.

For the second option, $Q$ is not always positive definite on the $K$ -group of our category; it is not even true that $Q (A, A) \geq 0$ for actual objects of our category. In the category of (finite dimensional) representations of (acyclic) quivers, there is a beautiful result that $Q$ is positive definite on the $K$ -group precisely for the Dynkin quivers. So it might be worth seeing whether there are some more general good properties of abelian categories (with various finiteness conditions) where $Q$ is positive definite.

Option $3$ is reminiscent of Serre’s multiplicity conjectures. If I get a chance, I’ll check with some commutative algebraists and see whether anyone has thought about it. A question for the category theorists – why is Hom a better analogue for inner products than $\otimes$ ?

Option 1 I have no thoughts about at all.

Posted by: David Speyer on May 24, 2007 9:59 PM | Permalink | Reply to this

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May 22, 2007