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• CommentRowNumber1.
• CommentAuthorEric
• CommentTimeJul 4th 2010
• (edited Jul 4th 2010)

Wikipedia has a nice article on quantum operations.

The nLab also had a page quantum operations and channels (cache bug?), but I’ve renamed this to simply quantum operation since a quantum channel seems to be nothing but a quantum operation when viewed from the perspective of quantum information theory. Eventually, this page might need some disambiguation since there may be several uses of the term, but for now I think it is “ok”.

I think this page can be cleaned up. I started, but don’t think I will be able to finish.

In particular, there is some background material that might be better on separate pages. I’ll continue trying to clean things up, but family might be calling soon and I’ll need to run quickly whatever state it is in.

I also made the simple statement

In quantum mechanics, a quantum operation is a morphism in the category of density matrices

at the beginning of the Idea section motivated by O’Loan’s comment

A quantum channel is a mapping which sends density matrices to density matrices.

This seems innocent enough, but someone might check the statement. For one, I’ve never seen a category of density matrices, but the idea seems obvious enough. Maybe a word on density matrix would be good.

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeJul 4th 2010
• (edited Jul 4th 2010)

I fixed the cache bug.

As for the category of density matrices, the main question is to ask, given two density matrices $D$ and $E$ (presumably on the same Hilbert space), what is a morphism from $D$ to $E$. I cannot guess that.

Furthermore, I do not think that this is what O’Lean’s comment suggests at all.

It is a capital mistake to think that a morphism from $X$ to $Y$ takes $X$ to $Y$. It goes from $X$ to $Y$, but it does not take $X$ anywhere. Instead, a morphism from $X$ to $Y$ will (sometimes) take a structure on $X$ to a structure on $Y$. (For example, it will take elements of $X$ to elements of $Y$, if that is a concept that makes sense in the relevant category.)

I interpret O’Lean’s comment as: a quantum channel is a mapping between Hilbert spaces (that is, a morphism in some category of Hilbert spaces) that has a particular property: namely, that it takes density matrices to density matrices. Note that a density matrix is a structure on an object in the relevant category (since it is a structure on a Hilbert space), not an object itself.

What do you think about the reference to Ian’s work? You removed the commentary about that.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeJul 4th 2010

I once did spend half an hour to type out the correct statements of what a quantum channel is at the entry. Please read that. It’s a linear map between vector spaces of matrices that preseves positive matrices and traces.

• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeJul 4th 2010

I’ve rewritten the Idea section and restored the discussion at the bottom; I haven’t touched anything in between.

• CommentRowNumber5.
• CommentAuthorEric
• CommentTimeJul 5th 2010
• (edited Jul 5th 2010)

Thanks Toby. I think it looks much better now. I generally don’t like to “remove” stuff from pages, but that one seemed like it could use quite a bit of trimming. That stuff from the Idea section (that apparently Urs wrote) was good, but I thought it might be better on a different page, so I moved it to the bottom in a Discussion section. At the end, I removed it at the last minute.

I still think that bit can be moved to another page and replaced with a link, but it seems fine for now. For example, on the train this morning, I was skimming Peter Selinger’s

• Dagger compact closed categories and completely positive maps (extended abstract)

where it says

We show that the category of completely positive maps considered in [9] is actually a biproduct dagger compact closed category. We further introduce a general construction, called the CPM construction, that associates to any dagger compact closed category $C$ its “category of completely positive maps” $CPM(C)$. In this way, we obtain a canonical and very general way of passing from “Hilbert-style” to “von Neumann-style” setting.

That one little bit in the Idea section might eventually be moved to a page on category of completely positive maps and replaced with a link there.

PS: It sounds like this “CPM construction” is a functor $CPM: \dagger CC\to CPM$.

• CommentRowNumber6.
• CommentAuthorJohn Baez
• CommentTimeJul 5th 2010

I think it’s a functor

$CP: \dagger CC \to \dagger CC$

Maybe that’s what you meant to write?

• CommentRowNumber7.
• CommentAuthorEric
• CommentTimeJul 5th 2010
• (edited Jul 5th 2010)

Maybe that’s what you meant to write?

I’m not 100% sure since I’m still just gathering references, but $\dagger CC$ is supposed to be the category of dagger closed categories. My notation didn’t compile when I wrote $CPM: \dagger CC\to CPM$ due to the variable names. The second instance of $CPM$ is supposed to be the category of completely positive maps. The first instance was inspired by the wording in Selinger’s paragraph where he used $CPM(C)$ to denote the category of completely positive maps corresponding to the dagger closed category $C$. It probably would have been better to just write the functor (if it is one) as

$F:\dagger CC\to CPM.$

I’m not sure if any of what I’m saying makes sense yet, so don’t take it too seriously.

Edit: On second thought, it could be that $CPM$ is the category of categories of completely positive maps :)

• CommentRowNumber8.
• CommentAuthorIan_Durham
• CommentTimeJul 5th 2010

What do you think about the reference to Ian's work? You removed the commentary about that.

That was just an APS presentation which basically contained things I had stripped straight from nLab. I had put it there since I thought it was a good starting point for a discussion of merging the ideas of categories with those of quantum channels, but, otherwise, it was probably crap (APS talks are usually limited to 10 minutes with 2 minutes for questions and so it's often hard to squeeze in everything).
• CommentRowNumber9.
• CommentAuthorJohn Baez
• CommentTimeJul 6th 2010
• (edited Jul 6th 2010)

Anyway, I’m sure that Selinger’s “completely positive map” construction takes a compact dagger-category and builds a new one, giving a functor

$CPM : \dagger CC \to \dagger CC$

• CommentRowNumber10.
• CommentAuthorEric
• CommentTimeJul 6th 2010

Ok. Yes. I was able to read a bit more on the train home last night and learned that a “category of completely positive maps” is also a “dagger compact closed category” so your notation is best (surprise! :))

The idea is that given a dagger compact closed category, the CPM construction gives a category of completely positive maps (but not vice versa as far as I know) and a category of completely positive maps is also a dagger compact closed category.

I wonder what happens if you iterate the procedure? The diagrams become pretty gnarly fairly quickly so it is a little difficult to see what is going on.

I like Selinger’s paper a lot. I started reading Coecke et al “Classical and Quantum Structures” last night as well. This all seems like a lot of fun. I wish I could make a career doing fun stuff like this :)

In Urs and my paper, we worked with directed graphs with “adjoint edges”, so I suspect we were somehow dealing with dagger categories somehow. Maybe even dagger compact closed categories, so all this is close to my heart.

I liked the definition of a positive map:

Definition 4.1 (Positive map) A morphism $f:A\to A$ in a dagger category is positive if there exists an object $B$ and a morphism $g:A\to B$ such that $f = g^\dagger\circ g$.

In our paper, we would think of $g^\dagger \circ g$ as the norm of the edge $g$, so a positive map is the norm of some edge. That is an interesting way to define it that avoids introducing coefficients in some field, i.e. things like “forming the free algebra on nodes and edges”.

• CommentRowNumber11.
• CommentAuthorIan_Durham
• CommentTimeJul 6th 2010

I wonder what happens if you iterate the procedure?

That is an interesting question. My immediate reaction is to wonder if that would have any applicability to the Birkhoff-von Neumann theorem, though I suspect your original suggestion concerning functor categories might be the way to go there.

I wish I could make a career doing fun stuff like this :)

Well, I may be hiring in a year or so...
• CommentRowNumber12.
• CommentAuthorEric
• CommentTimeJul 6th 2010

Speaking of fun stuff, on the train ride home I read a good chunk of Jamie Vicary’s:

Completeness and the complex numbers

Ian, I think you will love this paper. I loved it!

All this stuff is so close to what we were doing.

One question I had though, when Jamie defined a $\dagger$-limit, the diagrams seem to be missing some morphisms. He only draws morphisms to some of the objects, but should there be morphisms to all objects? For example, when he draws the $\dagger$-limit for an equalizer, there appears to be only one morphism from the limit object. But there are two objects. Shouldn’t there be two morphisms from the limit?

For example, in Figure (2), he only draws morphisms from the limit to the “leafs”, but there should be morphisms to all objects shouldn’t there? I guess those can be implied due to composition? This makes a difference for $\dagger$-limits I think. For example, when you have two objects $A$ and $B$ and two parallel morphisms $f,g:A\to B$, then there are really two morphisms from the limit object $l_A$ and $l_B$, but Jamie seems to only count one of them, i.e. $l_A$ in the definition of $\dagger$-limit. See Figure (6).

I don’t doubt that what he did is fine, it is just a little different than any other limit diagram I’ve ever seen. Could someone help reassure me that everything is fine by explaining why he does not need to include all morphisms in the definition of $\dagger$-limit?

It almost seems like when you form a cone over a diagram, some of the morphisms are superfluous as far as $\dagger$-limits are concerned.

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeJul 6th 2010

It almost seems like when you form a cone over a diagram, some of the morphisms are superfluous

Yes, because the cone is suppose to commute, so

$\array{ && t \\ & \swarrow && \searrow \\ a &&\to&& b }$

$\array{ && t \\ & \swarrow && \\ a &&\to&& b }$

The morphism on the right has to be the composite of the other two.

In particular an equalizer cone is often just written

$t \to a \stackrel{\to}{\to} b$

because that already fixes the morphism $t \to b$ to be the composite $t \to a \to b$ (either of the two, since they are required to be equal).

• CommentRowNumber14.
• CommentAuthorIan_Durham
• CommentTimeJul 6th 2010
Interesting. OK, I've downloaded it and am about to start reading it. From what you said above, it does look very interesting. I have read several other papers concerning the necessity of complex numbers in quantum physics. As an aside, it's fun to blow undergraduate minds by showing them the necessity of complex numbers.
• CommentRowNumber15.
• CommentAuthorMike Shulman
• CommentTimeJul 6th 2010

It does, however, I believe, make a difference which objects you choose to be the “leaves” in the definition of a $\dagger$-limit. Jamie’s paper is perhaps not maximally clear about this, but his equation (5) defining a $\dagger$-limit:

$\sum_S l_S; l_S^\dagger = id_L$

explicitly sums only over the leaves $S$. At the beginning of that section, he says something about considering only diagrams over “forest-shaped multigraphs,” which I’m not exactly sure of the definition of, but it seems to be some shape of diagram which canonically determines a set of objects called the “leaves.” However, you could write down equation (5) with the sum over any subset of the objects $S$ in the diagram, and it seems to me you’d get a different notion of “$\dagger$-limit.”

• CommentRowNumber16.
• CommentAuthorEric
• CommentTimeJul 7th 2010

Thanks guys.

Mike, yeah, that is what was on my mind. The sum is only over a subset of objects in the diagram so we have to be careful about which objects are included.

I guess he is only talking about finite diagrams so in any finite diagram, there should be objects that are not the target of any morphism. I suppose these are the “leaves”.

Combining this with what Urs says, for any leaf there should be a path to a root (an object that is not the source of some morphism) and since all morphisms from the limit form commuting triangles, all but the morphisms to the leaves are superfluous.

• CommentRowNumber17.
• CommentAuthorEric
• CommentTimeJul 7th 2010
• (edited Jul 7th 2010)

I said:

Combining this with what Urs says, for any leaf there should be a path to a root

WHoaOaOa, wait a second.

This is not true if your diagram contains loops. What is a leaf of a 3 object loop diagram? What is its $\dagger$-limit?

• CommentRowNumber18.
• CommentAuthorIan_Durham
• CommentTimeJul 7th 2010

OK, so, first of all, you are absolutely right - I LOVE this paper. This is so simple and intuitive it’s not even funny. Second of all, regarding your question, couldn’t a loop diagram serve as its own $\dagger$-limit?

• CommentRowNumber19.
• CommentAuthorEric
• CommentTimeJul 7th 2010

I knew you would love it Ian :)

A limit of a diagram is a special (universal) cone. A cone is an object together with morphisms to all objects in the original diagram such that all newly formed diagrams commute.

When I was trying to learn about limits, I wrote some stuff you might find helpful at

Understanding Limits in Set

So, a limit of a diagram is a special “object” together with some special morphisms to the diagram. A $\dagger$-limit is a “normalized” limit where the $\dagger$ is used to normalize the limit.

I like some other comments Jamie made about limits in general, but it will have to wait. I am being sucked into the vortex of my other life.

• CommentRowNumber20.
• CommentAuthorMike Shulman
• CommentTimeJul 7th 2010

My impression is that the “shape” of a $\dagger$-limit is not just “a finite diagram” but “a finite diagram equipped with some chosen subset of the objects called ’leaves’.” Jamie considers not arbitrary finite diagrams but only diagrams whose shape is a “forest-shaped multigraph,” in which case there is a canonical (but not unique) definition of the “leaves” purely in terms of the diagram structure. I think the definition makes sense if you take an arbitrary finite diagram and pick some subset of its objects and declare that you want to call them “leaves,” but probably in a lot of cases that won’t give you a useful notion. For instance, you could declare that there are no leaves at all, in which case equation (5) says that the limit is a null object – not very useful! But it seems that for the particular shapes of diagrams Jamie is interested in, the particular choice of the “leaves” that he makes for those diagrams results in a useful notion.

• CommentRowNumber21.
• CommentAuthorEric
• CommentTimeJul 7th 2010
• (edited Jul 7th 2010)

Interesting.

I suspect that there are some (causal) categories for which all finite diagrams have $\dagger$-limits. If you consider a process as requiring a finite time interval, then the composition of two processes should require the sum of their time intervals. Loops would be ruled out for causal reasons, i.e. a process cannot have a negative time interval.

There seems to be some causality built into the definition of $\dagger$-limit.

This reminds me of the (vague) idea I’ve talked about regarding “extruding” groupoids. I’ll need to do a search to find that conversation…

• CommentRowNumber22.
• CommentAuthorIan_Durham
• CommentTimeJul 7th 2010
• (edited Jul 7th 2010)

A limit of a diagram is a special (universal) cone. A cone is an object together with morphisms to all objects in the original diagram such that all newly formed diagrams commute.

Right, but I still don’t know why a loop diagram whose loops all begin and end at a single object couldn’t serve as its own $\dagger$-limit (and there is a method to my “madness” concerning this :-)).

Edit: Oh, unless I’m mis-interpreting “loop” diagram. I bet I am. Doh! I was thinking the morphisms are all from an object to itself (so they look like “loops”), but I’m guessing that what you’re talking about it some kind of cyclic thing involving multiple objects and morphisms.

• CommentRowNumber23.
• CommentAuthorEric
• CommentTimeJul 7th 2010
• (edited Jul 7th 2010)

Yep yep. By “3 object loop diagram”, I meant a diagram with 3 objects and 3 morphisms tracing around them in a loop (which implies the existence of a loop in the opposite direction… do you know why? :))

Edit: Argh. I keep falling into the trap of thinking a diagram needs to be a category, but that is not the case. If you have a category with 3 objects and 3 morphisms tracing around them in a loop, you get “for free” 3 more morphisms tracing around the loop in the opposite direction. Do you know why? It is a very simple thing, but I thought it was kind of cute when I first stumbled on it.

• CommentRowNumber24.
• CommentAuthorIan_Durham
• CommentTimeJul 7th 2010

Do you know why? It is a very simple thing, but I thought it was kind of cute when I first stumbled on it.

Hmmm. Oh, does it have to do with the composition of the morphisms, e.g. going from A to B to C is the same as going from A to C to B?

• CommentRowNumber25.
• CommentAuthorEric
• CommentTimeJul 7th 2010

Something like that, yeah.

If

• $f:A\to B$,
• $g:B\to C$, and
• $h:C\to A$

are morphisms in a category, the composites should also be in the category, but the composites go around the other way, i.e.

• $f\circ h: C\to B$
• $h\circ g: B\to A$
• $g\circ f: A\to C$

So in a category, if morphisms traverse a triangle, then the composites traverse the same triangle in the opposite direction. Nothing earth shattering :)

It is tempting to speculate about geometrical implications…

• CommentRowNumber26.
• CommentAuthorIan_Durham
• CommentTimeJul 7th 2010
Yeah, that's what I was essentially trying to say. It's amazingly simple (but there's a part of me that just doesn't want to believe it! :)).
• CommentRowNumber27.
• CommentAuthorUrs
• CommentTimeJul 7th 2010

So in a category, if morphisms traverse a triangle, then the composites traverse the same triangle in the opposite direction.

i don’t see what you might mean by that.

• CommentRowNumber28.
• CommentAuthorEric
• CommentTimeJul 7th 2010

i don’t see what you might mean by that.

Nothing deep at all. I just mean if you have a category with three objects $A$, $B$, and $C$ with three morphisms $f:A\to B$, $g:B\to C$, and $h:C\to A$, these morphisms can be thought of as representing a path around a loop $A\to B\to C\to A$. The composites $g\circ f$, $f\circ h$, and $h\circ g$ would then represent a path around the same loop in the opposite direction $A\to C\to B\to A$.

This means (I think) that triangles in any category come in two flavors:

• Directed
• Undirected

Composition produces a simple directed triangle, but if there is more than one composition for any given three objects, this implies there are actually 3 compositions and the triangle is undirected.

This is similar to the way that people often think of an undirected edge in a graph as a pair of oppositely directed edges.

In a category, an undirected triangle is a triple of distinct directed triangles sharing the same three objects.

• CommentRowNumber29.
• CommentAuthorEric
• CommentTimeJul 7th 2010

By the way, I was reading Vicary’s paper again on the train this evening and was thinking about these $\dagger$-limits and how they seem to imply causality.

$\dagger$-limits are only defined for finite diagrams for which there is a causal flow of time. This keeps reminding me of extruding categories. I even forgot I created a page to try to further develop the idea:

Putting these ideas together made me wonder if a $\dagger$-category admits all finite limits, then would the “extruded” version of this category have all $\dagger$-limits?

Extruding a category introduces a “time” on it so that all morphisms require some finite, maybe even unit, time interval. The extrusion of an adjoint morphism would have a negative time interval. For example, given

$f:A\to B$

and

$f^\dagger:B\to A$,

then when we extrude them we get

$f(i,i+1):A(i)\to B(i+1)$

and

$f^\dagger(i+1,i):B(i+1)\to A(i).$

Only adjoint morphisms are allowed to travel back in time. Maybe we should call them anti-morphisms that describe anti-processes of anti-systems :)

• CommentRowNumber30.
• CommentAuthorUrs
• CommentTimeJul 7th 2010

The composites g∘f, f∘h, and h∘g would then represent a path around the same loop in the opposite direction

Ah, you are thinking of the order in which the source and target objects are run through, ignoring the morphism itself.

That’s the same trick by which wheels in a movie sometimes seem to rotate backwards.

• CommentRowNumber31.
• CommentAuthorIan_Durham
• CommentTimeJul 7th 2010

OK, so I’m working through the Vicary paper and have gotten stuck on something. I’m having trouble understanding the following (found on the top of p.7):

Let $F:\mathbf{J} \to \mathbf{C}$ define such a diagram in the category $\mathbf{C}$. A cone for this diagram is an object, $X$, in $\mathbf{C}$, equipped with cone maps $x_{S}:X \to F(S)$ for all objects $S$ of $\mathbf{J}$, such that for any map $f:A \to B$ in $\mathbb{J}$ the equation $x_{A};F(f)=x_{B}$ holds.

So let’s take this one step at a time. I get that the first sentence says (I think) that a diagram like Figure 1 can be represented as some kind of functor between categories but I don’t get what it means when he says “in the category $\mathbf{C}$.” Does that mean that the diagram is itself an object in $\mathbf{C}$? If so, does the next sentence then define the cone maps as being morphisms from other objects in $\mathbf{C}$ to this diagram (given the stipulations defined by the end of that second sentence)? If that is true (i.e. I’m interpreting things correctly so far), what is a cone map in the “physical” terms of systems and processes, i.e. can you give me an example?

• CommentRowNumber32.
• CommentAuthorKevin Lin
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)
Ian:

1. Let's take this one step at a time: I suggest that you go back and review the relevant definitions (e.g. "category", "diagram", "functor") before proceeding. The passage you quote is completely straightforward and patently clear. Your questions indicate that you do not understand and are not yet familiar with the basic definitions.

2. You should first understand the words you're saying before you try to apply them to anything.

3. If I invented a new field of mathematics and did not intend for it to be applied to "systems and processes", would you expect everything in my field to be applicable to "systems and processes"?
• CommentRowNumber33.
• CommentAuthorEric
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

Hey Ian,

I hear you. I’ve been through this and it takes some work to get things straight. Like Kevin said, it is all just about the basic definitions.

First, like MacLane describes, a functor $F:C\to D$ tells you how $C$ sets “in” $D$. If you look at diagram, you see that a diagram is a functor $F:C\to D$. Now, think of $J$ as just some abstract collection of objects and morphisms satisfying the definition of a category with no further meaning assigned to them. Then $F:J\to C$ tells you how that abstract shape sits inside $C$. Usually, $J$ is finite and relatively small because we like to be able to draw it, but it doesn’t have to be.

So…

Does that mean that the diagram is itself an object in C?

No, it means a diagram is a bunch of objects and morphisms in $C$. Just as you would draw it on paper. $C$ is more than just a set of objects (as you know).

If so, does the next sentence then define the cone maps

No, since the “if” condition was not met :)

The second part of the statement says that the new triangles introduced by the cone maps commute. When you form a cone, each morphism in the original diagram becomes a leg of a triangle and all such triangles formed this way commute.

Just to be clear, if $f:A\to B$ and $g:B\to C$ and $h:A\to C$, we can think of these morphisms as forming a triangle. Saying this trinagle commutes is to say that $h = g\circ f$. Now compare this to the quoted paragraph.

Gotta run!

• CommentRowNumber34.
• CommentAuthorIan_Durham
• CommentTimeJul 8th 2010

Let’s take this one step at a time: I suggest that you go back and review the relevant definitions (e.g. “category”, “diagram”, “functor”) before proceeding. The passage you quote is completely straightforward and patently clear. Your questions indicate that you do not understand and are not yet familiar with the basic definitions.

Well, I’ve read through Mac Lane and Awodey several times over the past year-and-a-half and I beg to differ with you regarding that particular passage in Vicary’s paper. While it may look straightforward to someone who has worked with categories for a long time, it is not clear to someone whose knowledge only includes the basic definitions. I don’t want to get into an argument about this, but I have strong feelings about this method of “teaching” (i.e. the “go-back-and-reread-the-book” method - if you’d like to debate its merits or drawbacks, e-mail me or post to my blog).

That said, I think we’re talking past each other here. Here’s what I’m reading:

From the idea of diagrams, “Informally, a diagram in a category C consists of some objects of C connected by some morphisms of C.”

In Vicary’s paper, his notation seems to indicate (and perhaps this is because I am used to a certain notation for categories) that both J and C are categories. So does that make J a sub-category of C? What the heck is J?

• CommentRowNumber35.
• CommentAuthorEric
• CommentTimeJul 8th 2010

Despite the timing of the post, I am pretty sure Ian wrote his comment prior to seeing mine (an example of how it would be nice if the page refreshed itself while previewing so you can see any subsequent comments).

Ian, I hope my comments help. Vicary’s statement is really just a statement of the some standard definitions. For comparison, have a look at the corresponding nLab pages, e.g. diagram, cone, functor.

Reading MacLane is a lot different than understanding MacLane. I’ve read MacLane many times myself, but still only understand a very small portion of it. I assure you however, that Vicary’s definition of cone is the same as the nLab’s.

I really really liked Vicary’s definition of limit and hope to incorporate it into the nLab someday.

Kevin and Ian, I think papers like this present a very good opportunity for physicists to “learn” some basic category theory by relating it to things familiar to them. I don’t think the standard definitions should be a prerequisite for reading this paper. Rather, this paper (and papers like it) can be an introduction to the standard definitions.

• CommentRowNumber36.
• CommentAuthorIan_Durham
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

Now, think of $J$ as just some abstract collection of objects and morphisms satisfying the definition of a category with no further meaning assigned to them. Then $F: J \to C$ tells you how that abstract shape sits inside $C$. Usually, $J$ is finite and relatively small because we like to be able to draw it, but it doesn’t have to be.

See, now that makes complete sense and now I think I understand Vicary’s sentence. Many thanks.

And I’m never doubted Vicary’s definitions of anything. I just have a brain that processes things in very strange ways and had trouble parsing that particular sentence (despite the fact that I am familiar with what the basic definitions are supposed to be - again, sometimes I think what looks basic to an expert is a lot different than what truly is basic and people have a hard time recognizing that). And you’re absolutely right about understanding Mac Lane. I don’t claim to understand much of it (which is why I really like Awodey, but there are gaps between the level Awodey is at and the level that Mac Lane is at).

• CommentRowNumber37.
• CommentAuthorEric
• CommentTimeJul 8th 2010

Right right. And by “basic”, Kevin did not necessarily mean “easy” or “obvious”, he meant it more like “standard”. That is why I used “standard” instead of “basic”, but said basically the same thing.

Beware of any math texts with the words “Basic” in the title. You can pretty much be assured the text is NOT “Easy” :)

• CommentRowNumber38.
• CommentAuthorzskoda
• CommentTimeJul 8th 2010

When speaking of algebraic geometry, Grothendieck sometimes used the term “elementary methods” meaning any proof not using cohomology theory. This usage is now standard in his school.

• CommentRowNumber39.
• CommentAuthorUrs
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

Concerning the exchange between Kevin and Ian and Eric:

I have said the following before, but I’ll say it again and again, as long as the issue keeps coming up

Being ignorant of something, having to learn something and especially learning something is no shame. If discussion of even simple or “elementary” facts here leads eventually to an improvement of the expositional and pedagogical parts of the $n$Lab, it is more than welcome and in fact invited and encouraged.

While it surprises me to see just how much time one can spend on digesting the concept of “functor” and “diagram”, if that’s useful to more than one lay reader then it’s nice if the nLab eventually provides these explanations. I am thinking of the nLab as being eventually a useful resource for readers at all possible levels. There is plenty of room, namely unlimited room, so we can accomodate everybody’s needs.

Having said all this, here comes the but : but what I find problematic, and what maybe Kevin finds problematic, is to see ignorance of ignorance. I think it is fine to struggle with the concepts of functor and diagram. But I think it is a problem if somebody who struggles with the concepts of functors and diagrams insists on putting his ideas about category theory applied to physics into nLab entries.

• CommentRowNumber40.
• CommentAuthorjamievicary
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

Hi! Let me make a few comments on things that have been said in this thread.

Completely positive maps are really important. It’s fun to play around with the category of Hilbert spaces as if that’s all that quantum mechanics has to offer, but you could argue that this category isn’t particularly physical. First, if you have two unitary maps describing the evolution of an entire quantum system, and they only differ by a complex phase, they’re physically indistinguishable – so shouldn’t they really be the same morphism? Also, we know that pure states can sometimes evolve into mixed states – when we perform a measurement, or when we have certain sorts of event horizons in our spacetime – so shouldn’t our category include morphisms that turn pure states into mixed states?

The category of completely positive maps has these properties. It’s objects are Hilbert spaces $A$, $B$, …, and a morphism from $A$ to $B$ is a completely positive map from the space of bounded linear maps on $A$ to the space of bounded linear maps on $B$. You can build this category from an abstract compact dagger-category by looking at Peter’s construction in the paper cited by John. You can even build this category in the absence of compact structure, using a procedure described by Bob Coecke.

If you want to axiomatize quantum mechanics category-theoretically, you have to decide pretty quickly whether you want to embrace ’CPM-style’ categories or ’Hilb-style’ categories. The former is perhaps more ’physical’, while the latter has nicer properties category-theoretically. Also, you can build CPM(Hilb) from Hilb, but not the other way around.

This leads us nicely into $\dagger$-limits – Hilb has them, but CPM(Hilb) doesn’t! I came up with $\dagger$-limits by thinking about the completeness properties of Hilb. The category CPM(Hilb) has bad completeness properties; in particular, it doesn’t have biproducts.

Mike: As usual, you’re right on the money. It’s crucial that, for a standard $\dagger$-limit, the sum is taken over the set of leaves. Sorry that the definition of a “forest-shaped multigraph” wasn’t clear – I’ve just uploaded a new version of the paper to the arXiv (available here as well) which makes that, and a few other things, a bit clearer.

It’s fun to consider other subsets of objects for normalizing a $\dagger$-limit, and if you take this approach, it’s not so crucial any more that your diagram has a particular forest-like shape. I investigate this a bit in the paper, and describe how you can define generalized $\dagger$-limits along these lines. For these, any finite diagram is allowed (and some infinite ones would work, too.) The great thing is that having all forest-shaped $\dagger$-limits, for which the sum is just performed over the leaves, is actually equivalent to having all limits of the more general sort! (Actually, I need one extra property for this equivalence to be true: for every pair of isomorphic objects, there has to be a ’unitary’ isomorphism going between them, which is a morphism $f$ satisfying $f \circ f ^\dagger = \mathrm{id}$ and $f ^\dagger \circ f = \mathrm{id}$.) So, I concentrate on the forest-shaped $\dagger$-limits, as they’re a simpler class.

Eric: You’re exactly right, the leaves are the subset of objects which are not the target of any morphism (apart from their identity morphism, of course.)

Ian: I agree, cones and limits are a bit confusing when you first meet them! But persevere, they are everywhere :).

• CommentRowNumber41.
• CommentAuthorEric
• CommentTimeJul 8th 2010

Hi Jamie! :)

I was hoping you would find your way here. I am really thoroughly enjoying reading your paper on my train rides home. On the train this evening I got to the section where you discussed generalized $\dagger$-limits. It seems like each question I have turns out to be the next section of your paper :)

I’m curious about one thing. When you described limit, is the insight/description you gave your own? Where did you learn that? After reading your description a huge light bulb lit up. In other words, if the morphisms in the diagram are like constraints, the limit is the initial condition that satisfies those constraints. That makes a lot of sense to me. Nice :) Or… the limit is the object from which each path to each root results in the same final state at that root. Roughly. That is a very nice way to think about it and I’d love to know how you came to that understanding (although once spoken, it seems obvious and I can’t now see how anyone would think of a limit in any other way :))

This also reminds me of mathematical finance. Being very vague and speculative here, if you think of each object as a financial instrument having a value and morphisms represent different economic evolutions leaving the object with different values, then the limit would be the “fair price” (or more precisely, the “no arbitrage” price) of an instrument somehow. The limit is the initial price such that no matter what economic outcome occurs, there is no “free lunch” and the value always ends up the same at the same root node regardless of the path taken. I know that is vague and probably doesn’t make much sense on first glance, but my gut (which often serves me well) tells me there is something true about it.

An interesting playground for this would the free category on a directed binary tree graph.

Why is my brain suddenly thinking about path integrals?

Oh oh… I wish I was smarter…

• CommentRowNumber42.
• CommentAuthorJohn Baez
• CommentTimeJul 8th 2010

Eric wrote:

After reading your description a huge light bulb lit up. In other words, if the morphisms in the diagram are like constraints, the limit is the initial condition that satisfies those constraints. That makes a lot of sense to me.

Category theorists say this all the time. Unfortunately they say it in a way that only category theorists enjoy. They say: finite limits can all be built from products and equalizers. (When you hear “equalizer”, you should think about constraints.)

• CommentRowNumber43.
• CommentAuthorKevin Lin
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

Ian:

I think all of this should be straightforward to anyone who is familiar with the basic definitions. It’s not a matter of being an expert (I am certainly not an expert), nor a matter of exposition or pedagogy. I don’t intend to be an arrogant jerk. I’m just being honest — I don’t think you’re familiar with the basic definitions. I think you should just sit down and think these things over carefully. I think you’re asking for an unreasonable amount of hand-holding.

When I say “basic”, I really mean “basic” — I mean page 1 of any category theory book. Literally page 1. If that’s not basic, then what’s basic?

And yes, I agree with what Urs says. I find offensive the notion that one might have “ideas” about applying category theory to physics (or to anything else) before one even understands page 1 of any category theory book.

• CommentRowNumber44.
• CommentAuthorjamievicary
• CommentTimeJul 8th 2010

I’m glad to enliven your train journey, Eric! You ask about where I encountered the idea of a limit as the solution to a constraint. As with all my work, it was intoned to me directly from Baal. However, I think that the moment this idea really clicked into place for me was when I read Joseph Goguen’s “A Categorical Manifesto”, where he describes limits in very similar terms. That should be cited in the $\dagger$-limits paper, actually.

Your idea about relating this to finance sounds excellent. You need to get to the bottom of that! Can you make a more specific guess about how you would define your objects and morphisms?

• CommentRowNumber45.
• CommentAuthorjamievicary
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

Hi Ian. I think your questions are great – keep them coming! When I was learning category theory I remember being a bit confused about the definition of diagram as a functor, especially in the case that the same object in $C$ is appearing in different places in the diagram. But once you get it, it’s so elegant, it sticks.

• CommentRowNumber46.
• CommentAuthorIan_Durham
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

Kevin: I don’t really know what your problem is. I asked a question. Eric answered it. End of story. If you thought it was too simple a question, then ignore it.

While it surprises me to see just how much time one can spend on digesting the concept of “functor” and “diagram”, if that’s useful to more than one lay reader then it’s nice if the nLab eventually provides these explanations.

Oliver Wendall Holmes once said that “nothing is as rewarding as a stubborn examination of the obvious.”

I find offensive the notion that one might have “ideas” about applying category theory to physics (or to anything else) before one even understands page 1 of any category theory book.

And I find it offensive that you can’t seem to accept that there are levels of understanding. The world isn’t black and white.

But I think it is a problem if somebody who struggles with the concepts of functors and diagrams insists on putting his ideas about category theory applied to physics into nLab entries.

I haven’t put up anything on the nLab that directly deals with categories in months. I asked a question. This is a forum. You are free to ignore my postings.

• CommentRowNumber47.
• CommentAuthorIan_Durham
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

Hi Ian. I think your questions are great – keep them coming! When I was learning category theory I remember being a bit confused about the definition of diagram as a functor, especially in the case that the same object in is appearing in different places in the diagram. But once you get it, it’s so elegant, it sticks.

Thanks.

• CommentRowNumber48.
• CommentAuthorIan_Durham
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

OK, so here’s a question, still on cone maps. Could a cone map be described somewhat like the following?

You’ve got a bunch of objects $S$ in $\mathbf{J}$. This diagram (which is a functor, right?) takes them into $\mathbf{C}$, i.e. its action is to take $S$ to $F(S)$ (right?). Then there are these objects $X$ in $\mathbf{C}$ that are the cones. A cone map assigns a cone to each $F(S)$.

Is that a reasonable description of a cone map?

• CommentRowNumber49.
• CommentAuthorTobyBartels
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

• CommentRowNumber50.
• CommentAuthorjamievicary
• CommentTimeJul 8th 2010
• (edited Jul 8th 2010)

Ian, that’s sort of right, but I think you’re missing some detail. As you say, the action of the functor $F:J \to C$ is to take an object $S \in J$ to an object $F(S)$ in $C$. A cone for the diagram $F$ is an object $X$ of $C$, equipped with a map $x_S:X \to F(S)$ for each object $S$ in $J$, such that for each morphism $f:U \to V$ in $J$, the equation $F(f) \circ x_U = x_V$ holds. You should draw yourself some pictures of this, it will make it clearer!

• CommentRowNumber51.
• CommentAuthorUrs
• CommentTimeJul 8th 2010

What do you think about the reference to Ian’s work? You removed the commentary about that.

That was just an APS presentation which basically contained things I had stripped straight from nLab. I had put it there since I thought it was a good starting point for a discussion of merging the ideas of categories with those of quantum channels, but, otherwise, it was probably crap

Okay, so I removed it.

• CommentRowNumber52.
• CommentAuthorIan_Durham
• CommentTimeJul 8th 2010

Jamie,

Yeah, thanks, I got that there is the added requirement that $x_{A};F(f)=x_{B}$ which (I think) basically says that when you put $x_{A}: X \to F(A)$ together with $F(f): F(A) \to F(B)$ you must get $x_{B}: X \to F(B)$.

What would a cone map be in terms of systems and processes? Is there some kind of physical example?

• CommentRowNumber53.
• CommentAuthorMike Shulman
• CommentTimeJul 9th 2010

Thanks Jamie! I find it mind-blowing that for a given shape J, there are multiple possible choices of “supporting subset” $\Omega$ (replacing the leaves) such that $\dagger$-limits relative to all of them can exist simultaneously. My initial reaction was “no, that can’t possibly be right” – it looks at first as though you’re asking the identity $id_L$ to be equal to two different sums, one of which is strictly larger than the other. But the example with complex vector spaces helped. It sure takes some practice to get used to the fact that non-unitary isomorphisms aren’t really a “sameness” notion in a $\dagger$-category.

I’ve heard people say that $\dagger$-categories are something of a “decategorification” of 2-categories in which all 1-morphisms have adjoints. In particular, I seem to recall from one of Street’s papers that limits in bicategories with local colimits have some similar properties to $\dagger$-limits. Is the notion of a “unitary” $\dagger$-category (in which any pair of isomorphic objects are unitarily isomorphic) at all related to the fact that in a 2-category, any equivalence can be improved to an adjoint equivalence?

• CommentRowNumber54.
• CommentAuthorEric
• CommentTimeJul 9th 2010
• (edited Jul 9th 2010)

@Mike #53:

Regarding “sameness” in a $\dagger$-category…

The way I think of it, which could be completely wrong, is that a $\dagger$-category is a “category with an inner product”. So morphisms have a “size”. Dagger categories seem to represent a shift from topology to geometry.

In topology, things are the same if they are the same shape.

In geometry, things are the same if they are the same shape AND the same size.

So if the meaning of “same” in a dagger category is somehow different than what you are used to, it doesn’t surprise me because, for things to be the same in a dagger category, they have to be the same size as well as the same shape.

If that makes any sense…

It sure takes some practice to get used to the fact that non-unitary isomorphisms aren’t really a “sameness” notion in a $\dagger$-category.

My reading on this is that a non-unitary isomorphism changes the “size”.

• CommentRowNumber55.
• CommentAuthorMike Shulman
• CommentTimeJul 9th 2010

Topology/geometry doesn’t capture it for me, because in the ordinary category of (say) Riemannian manifolds, an ordinary isomorphism preserves both shape and size. Adding a $\dagger$-structure and talking about unitary things isn’t necessary in order for isomorphisms to say something about size.

• CommentRowNumber56.
• CommentAuthorEric
• CommentTimeJul 9th 2010
• (edited Jul 9th 2010)

Like I said, I could be and probably am completely wrong, but let me try to explain myself better in the hopes something good might come out of it.

When you have a morphism $f:A\to B$, then the morphism $f^\dagger\circ f:A\to A$ is (or so I think) the “norm of $f$”, i.e. its size. This is something slightly different than categories of Riemannian manifold and inner product preserving maps where you’re talking about sizes of stuff inside the objects or maybe the size of the objects. Here we’re talking about the size of the category itself and its pieces.

Note: Speaking of which, it would be fun to relate this to cardinality of a category!

Using this, we can start talking about “angles between morphisms” since

$(f+g)^\dagger\circ (f+g) = f^\dagger\circ f + g^\dagger\circ g + (f^\dagger\circ g + g^\dagger\circ f)$

and we can define

$cos\theta = \frac{1}{2} (f^\dagger\circ g + g^\dagger\circ f): A\to A$.

So geometry is being built into the category rather than the category being used to study geometry.

I like this idea a lot. Add some graded algebras and we can relate this to Urs and my stuff and my dream will be fulfilled :)

• CommentRowNumber57.
• CommentAuthorMike Shulman
• CommentTimeJul 9th 2010

Ah, I misunderstood. Yes, that seems good – talking about the “size” of a morphism may be a good way to think about it. Thanks!

Once again I am feeling like $\dagger$-categories are very similar to categories enriched over something. For instance, you get similar phenomena to this if you think about categories enriched over graded sets – you can have two objects that are “isomorphic” via an “isomorphism” of degree $\neq 0$, but such isomorphisms aren’t “true” isomorphisms; in particular they don’t live in the underlying ordinary category (as standardly defined for enriched categories) which consists only of degree-0 morphisms. Here again the morphisms have some sort of “size” and it is only the “unit-size” isomorphisms that really exhibit two objects as “the same.”

• CommentRowNumber58.
• CommentAuthorjamievicary
• CommentTimeJul 9th 2010
• (edited Jul 9th 2010)

Ian: Let $H$ be the Hilbert space for a single bosonic particle (a photon, for example), and $\gamma: H \otimes H \to H \otimes H$ be the standard symmetry isomorphism. Then we can consider the parallel pair consisting of $\gamma$ and $\id_{H \otimes H}$. The $\dagger$-equalizer for this pair will be the subspace of $H \otimes H$ consisting of those states for which exchanging the photons does the same thing as leaving them alone! This is physically important, as it gives us the physical Hilbert space for a pair of bosons. But in fact, our pair of bosons could certainly have a state space which is strictly smaller than this – in which case their state space wouldn’t be a $\dagger$-equalizer of the diagram, but merely a cone. So, the idea is that the cones are all the different ways to satisfy the constraints, whereas the $\dagger$-equalizer gives the maximal way to satisfy the constraints.

Another case, which might seem a bit degenerate, is a finite set of Hilbert spaces, with no processes going between them. The $\dagger$-limit for this diagram is just the direct sum of all the Hilbert spaces. A cone, on the other hand, is formed by choosing a particular subspace of each Hilbert space, and taking the direct sum of these. So again, the $\dagger$-limit gives the the maximal way to join the systems together (in general modulo the action of processes, but we don’t have any here), whereas a cone just gives some particular way to view them all as a single object.

Now, these might both seem like slightly boring examples – but in the paper, it’s shown that if you have these two types of $\dagger$-limit and a zero object, then as long as your $\dagger$-category is unitary, you actually have all finite $\dagger$-limits! So between them, these cases really are universal.

Eric and Mike: I think it’s right to consider morphisms as having ’sizes’. In the category of finite-dimensional Hilbert spaces, a good measure of the ’size’ of a morphism $f:A \to B$ is Tr$(f;f^\dagger)=$Tr$(f^\dagger;f)$. Mike, I know what you mean when you say it initially seems a bit weird to have different $\dagger$-limits of the same diagram for different choices of the supporting subset! As you’ve noticed, the reason this is possible is that, in general, the more objects you choose in the supporting subset, the smaller the ’sizes’ of each of the individual $\dagger$-limit maps to these objects, so that the ’total size’ of these maps stays the same. The surprising thing is that, for each particular choice of supporting subset, there’s a unique way to divide up the ’total size’ of the limit object among the different supporting objects! I still find that a bit spooky.

Eric: Yes, I like the idea of ’geometry’ somehow being introduced here. I show in a theorem towards the end of the paper that putting a continuous $\dagger$-functor on FinVect, such that with this $\dagger$-functor FinVect has all finite $\dagger$-limits, is exactly the same as putting an inner product on each object in FinVect, which just turns it into FinHilb. (Here, ’continuous’ means that the $\dagger$-functor acts continuously on the hom-sets of FinVect, with their standard topology.) So the slogan is “VECTOR SPACES + $\dagger$-LIMITS = HILBERT SPACES”, and Hilbert spaces are precisely vector spaces with added ’geometry’.

Mike: I would love to make a good connection between this stuff and higher categorical limits. These $\dagger$-limits are based on axiomatizing the properties of Hilb, so a good 2-category to start with should be 2Hilb. However, I have tried and failed to get a good understanding of what the limits look like here, let alone what $\dagger$-limits should be.

By the polar decomposition theorem, in Hilb, any isomorphism can be ’rescaled’ to a unitary isomorphism by precomposing with a positive operator. I think that this should be seen as related to the fact you mention, that any equivalence can be ’rescaled’ to an adjoint equivalence. However, there are differences – given an equivalence, the adjoint equivalence can be directly constructed algebraically, whereas given an isomorphism in Hilb, it’s not quite so easy.

I would hope that the $\dagger$-limits in Hilb could be seen as a decategorification of some sort of higher $\dagger$-limit structure in 2Hilb. But I don’t think it would be the type of decategorification that takes 1-morphisms of a 2-category into 1-morphisms of a 1-category, which is the sort of thing you’re suggesting. Rather, the most immediate way to see Hilb as a decategorification of 2Hilb is that Hilb form the scalars of 2Hilb. I know you like traces of categories – from this perspective, I’m saying we should decategorify by taking the end of the identity on 2Hilb, rather than the coend!

• CommentRowNumber59.
• CommentAuthorMike Shulman
• CommentTimeJul 9th 2010

Yes, and to make an equivalence into an adjoint one, you only need to change either the unit or the counit, but you can leave both 1-morphisms unchanged. That also seems like a notable difference.

• CommentRowNumber60.
• CommentAuthorIan_Durham
• CommentTimeJul 9th 2010
• (edited Jul 9th 2010)

@Jamie: Wow! Now that’s cool! It also helps me see a little bit better what purpose a cone and cone map has (sometimes I just learn better seeing something applied - maybe it’s the engineer in me).

whereas a cone just gives some particular way to view them all as a single object.

So like an entangled pair for instance?

• CommentRowNumber61.
• CommentAuthorEric
• CommentTimeJul 10th 2010
• (edited Jul 10th 2010)

Hi Jamie!

As far as taking this to higher dimensions, I have some speculative thoughts that are no where near solid form, but I’ll say somethings and hope something makes sense, i.e. I’m thinking out loud here.

A little background…

I spent every moment (waking and asleep) for 6 years in grad school thinking about Maxwell’s equations and how one might formulate them in a finitary/discrete universe. I disguised this desire under the phrase “designing numerical algorithms” :) But all along, I was trying to think about what nature could really be like if not based on smooth continuum manifolds. You can probably imagine that it didn’t take long for me to become drawn to category theory. My dissertation did not contain any breakthroughs, but a little later Urs and I wrote a paper together that I’m pretty happy with.

Anyway, our paper deals with dagger graphs instead of dagger categories, but not just $\dagger$-graphs, they were $n$-$\dagger$-graphs.

So I’m thinking that maybe some of the insights we obtained there might be relevant here as well since the methodology was not completely dissimilar.

Maybe, given a $\dagger$-category, you can look for a special class of morphisms:

1. Creation morphisms $a_i$
2. Annihilation morphisms $a_i^\dagger$

Note: I reversed the usual convention out of laziness below.

The way I might think of these (and remember, I’m speculating, thinking out loud) would be as morphisms “creating pieces of spacetime” beginning with spacetime points or nodes, e.g.

$a_i|0\rangle$

creates a node we can label “$i$”.

Similarly

$a_{i j}|0\rangle$

creates a directed edge from node “$i$” to node “$j$”.

These satisfy

$a_i a_j = \delta_{i j} a_i$

and

$a_i a_{j k} = \delta_{i j} a_{j k}$

and

$a_{j k} a_i = \delta_{i k} a_{j k}$.

Note that

$a_i a_{j k} \ne a_{j k} a_i,$

i.e. nodes and directed edges do not commute.

To get higher degree cells we simply repeatedly apply the creation morphisms

$a_{i j} a_{k l} = \delta_{j k} a_{i j l}$.

This is a 2-cell and 2-cells are subject to some rules of differential graded algebras. To see this, we introduced a “graph operator”

$G = \sum_{i j} a_{i j}$

which sums over all directed edges and define (on a certain class of graphs) a differential

$d(-) = [G,-]$

where $[,]$ is a graded commutator.

I could blabber about this to infinity, but I think you get the general direction I’m trying to go with this.

All along, I’ve always assumed that the morphisms of my category should correspond to directed edges of my spacetime, but after reading your paper and writing the above, now I’m thinking that the spacetime should emerge from the category but may not “be” the category in the sense I’ve always assumed. For example, if you could find a special class of creation morphisms in a $\dagger$-category as outlined above, these creation morphisms may represent a superposition of many morphisms in the $\dagger$-category, but would give rise to a single directed edge in spacetime.

• CommentRowNumber62.
• CommentAuthorIan_Durham
• CommentTimeJul 10th 2010
• (edited Jul 10th 2010)

@Eric: Since I’m still in the newbie stage, I stopped understanding when you said “2-cell.” But, I do think that the notion of an operator (at least as I am familiar with it) as a morphism seems completely natural to me. One of the images I use when I teach - and I’ve used this in pure math classes as well as physics classes, though it is admittedly not a perfect analogy - is of one of those old-fashioned meat grinders (I actually think I got the idea from a Monty Python animation, but can’t remember): something goes in and (usually) something else comes out. I’ve applied this to maps, transformations, operators, functions, and have imagined the morphisms of category theory as fitting this as well, though in some cases I understand that it’s more like a picture of something rather than something actually changing.

Anyway, in short, the creation and annihilation operators from QM seem to naturally fit the category theoretic notion of a morphism (“arrows”).

• CommentRowNumber63.
• CommentAuthorEric
• CommentTimeJul 10th 2010

Sorry for that last distraction. I should try harder to control myself :)

But on the train to the office this morning (yeah, it’s Saturday) I was reading Jamie’s paper again and what do you know? The next section was talking about “sizes” of morphisms.

I promise I didn’t read ahead :) But yeah, that was very intuitive to me. I think it is worth stressing again that I think it is interesting that geometry is being brought into the category itself. The example with $FinVect$ shows how the $\dagger$-functor implies a $\dagger$ on objects, hence you get $FinHilb$. Having inner products on objects is interesting, but we have something slightly different and potentially interesting.

And yeah, sometimes I forget $Tr$. We should throw a trace in there so $Tr(f^\dagger\circ f)$ is the norm of the morphism $f$. Then like Jamie says, $Tr(Id_L)$ really does have an interpretation as the size of $L$.

Also I didn’t read ahead to the part where you mention a “weight” or “measure”. Are you familiar with Leinster measure a.k.a. cardinality of a category? It also involves weights in an interesting way. It would be cool to relate these two concepts of “weight”.

• CommentRowNumber64.
• CommentAuthorEric
• CommentTimeJul 13th 2010

John wrote:

Eric wrote:

After reading your description a huge light bulb lit up. In other words, if the morphisms in the diagram are like constraints, the limit is the initial condition that satisfies those constraints. That makes a lot of sense to me.

Category theorists say this all the time. Unfortunately they say it in a way that only category theorists enjoy. They say: finite limits can all be built from products and equalizers. (When you hear “equalizer”, you should think about constraints.)

Oh I wish I understood this tantalizing statement. It seems like it would help me understand some finer points of Jamie’s paper. To try to make sense of it, I’ve now started reading

Categorical Manifesto

• CommentRowNumber65.
• CommentAuthorIan_Durham
• CommentTimeJul 13th 2010

I think John’s point about constraints makes it a lot clearer (but I could just be deluding myself). My understanding is that an equalizer is just a type of limit. Is that right?

• CommentRowNumber66.
• CommentAuthorDavidRoberts
• CommentTimeJul 13th 2010

My understanding is that an equalizer is just a type of limit. Is that right?

That is correct. A simple example is if you have two real-valued functions $f,g$ on $\mathbb{R}^n$. Then their equaliser in $Set$ is (isomorphic to) the subset $E$ of $\mathbb{R}^n$ on which $f\big|_E = g\big|_E$. This works for continuous functions and topological spaces, but if the functions are smooth, then the equaliser may not exist (i.e. it exists as a space but it may not be a submanifold). Replacing the reals by something else, and then generalising to a general category and you have an intuitive description of what an equaliser is (the formal definition is not like this, but for concrete categories, where the objects have ’underlying sets’, it boils down to this)

• CommentRowNumber67.
• CommentAuthorTodd_Trimble
• CommentTimeJul 13th 2010
• (edited Jul 13th 2010)

Just to put a further down-to-earth gloss on this: products and equalizers have implicitly been around since Descartes. His GREAT innovation of analytic geometry, which united classical geometry and algebra, is largely about limits in terms of products and equalizers, and the fact that you can do analytic geometry whenever you have products and equalizers.

To thinkers like Galileo, we owe the observation that complicated motions, say for instance the flight of a bird as measured by a function

$f: Time T \to Space$

can be analyzed in terms of ordinary real-valued functions $f_x, f_y, f_z: T \to \mathbb{R}$. (Hence Galilean physics can be analyzed in terms of real-valued functions alone.) In more modern-day parlance, the object $Space$ comes equipped with three coordinate functions, traditionally named

$x: Space \to \mathbb{R}, \qquad y: Space \to \mathbb{R}, \qquad z: Space \to \mathbb{R}$

and any map into $f: T \to Space$ is uniquely characterized by the maps obtained by composing with the coordinate functions: $f_x \coloneqq x \circ f$, $f_y \coloneqq y \circ f$, $f_z \coloneqq z \circ f$. This is exactly what we mean when we say $Space$ is a cartesian product $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$.

Next, let us take a simple equation like $y = x^2$. Its locus as a subset of the plane $\mathbb{R} \times \mathbb{R}$ is the equalizer of two maps

$\mathbb{R} \times \mathbb{R} \stackrel{\overset{y}{\to}}{\underset{x^2}{\to}} \mathbb{R}$

where again ’$x$’ and ’$y$’ denote coordinate projection functions. The equalizer just means we are considering the constraint on $x$, $y$ imposed by the equation.

What is $x^2$ here? Well, we know what it is ($x x$), but it is instructive to work it out in terms of product data. It is the composition of the projection $x: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ with the following composition:

$\mathbb{R} \stackrel{\Delta}{\to} \mathbb{R} \times \mathbb{R} \stackrel{mult}{\to} \mathbb{R}$

Here $\Delta$ is the unique map such that $\Delta_x = x \circ \Delta = id_{\mathbb{R}}$ and $\Delta_y = y \circ \Delta = id_{\mathbb{R}}$; in effect it creates two copies of an element $r$, i.e., $\Delta r = (r, r)$. The map $mult$ is the multiplication map sending $(x, y)$ to the product $x y$. The composition takes $r$ to $r r = r^2$.

So in summary, provided we have basic functions $+: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$, $mult: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ to get us off the ground, together with an arrow $r: 1 \to \mathbb{R}$ for each constant, we can carry out all constructions of classical analytic geometry by using basic product and equalizer constructions (glossing over the need to divide sometimes).

It is a great insight of category theory to see the generality of such constructions. For example, the $5$-adic integers are constructed by means of a limit whose elements, such as say

$\ldots, 4312, 312, 12, 2$

are infinite ’tuples’ subject to the constraint that the truncation of one leads to the next.

Edit: Hey, why isn’t the $\backslash cdot$ command recognized? I was using the Markdown+Itex filter.

• CommentRowNumber68.
• CommentAuthorTobyBartels
• CommentTimeJul 13th 2010

Edit: Hey, why isn’t the $\backslash cdot$ command recognized? I was using the Markdown+Itex filter.

Actually, it is recognised! If it weren’t recognised, then it would come out as ‘$cdot$’, but instead it comes out as ‘$\cdot$’. In other words, it is recognised but broken somewhere further along.

• CommentRowNumber69.
• CommentAuthorIan_Durham
• CommentTimeJul 13th 2010
• (edited Jul 13th 2010)

Todd,

Thanks for that great response! That was really enlightening. You should write a book (seriously - it was very clear). I’m familiar with the notion of space as a Cartesian product but, for whatever reason, had never heard the term equalizer applied to two maps before (note: I’m largely self-taught in most things so there are strange gaps here and there).

Everyone,

OK, so now on the the next couple of questions/observations. If Jamie’s lurking around out there, perhaps he can shed light on one or two of these:

1) There’s a line on p.7 in the paper that says

Since there are only a finite number of leaves, this is a finite sum, and will always be well-defined.

This made me realize that I’m still a little foggy on how you’re relating Hilbert spaces to categories here. Is each Hilbert space a category? And, if so, what are their objects (leaves, etc.)? And are we limiting ourselves to finite-dimensional Hilbert spaces here?

2) The description of finite $\dagger$-equalizers made me think of Feynman’s path integral formulation of QM. Is there a relation?

3) While reading the bottom of p.9 I had the thought (and maybe Eric had the same thought, which is why this discussion began) that maybe $\dagger$-equalizers could also be used to formulate a quantum version of Birkhoff’s theorem (which is what got me interested in category theory to begin with). Eric: is this indeed what you were thinking?

• CommentRowNumber70.
• CommentAuthorTodd_Trimble
• CommentTimeJul 13th 2010

Picky, picky, picky!

Okayyyy… I meant “why isn’t it being rendered correctly?” But, you knew that! :-)

• CommentRowNumber71.
• CommentAuthorTobyBartels
• CommentTimeJul 13th 2010

@ Ian

You should write a book (seriously - it was very clear).

Have you read Conceptual Mathematics by Lawvere and Schanuel? It is full of stuff like Todd’s comment #67. Now that I think about it, it would probably be a very good introduction to category theory for many physicists who already know how to think about physics conceptually. (And when reading this book, it would be good to set aside any preconceptions about what mathematics, as opposed to physics, is like.)

• CommentRowNumber72.
• CommentAuthorTodd_Trimble
• CommentTimeJul 13th 2010

Admittedly, I did crib Galileo’s flying bird example from my memory of Conceptual Mathematics. For all I know, the equalizer example I gave might be in there too.

But I think there I was more influenced by a memory of something in Categories, Allegories by Freyd and Scedrov, where they defend their use of ’cartesian category’ to mean what we call a finitely complete category, by emphasizing that Descartes wasn’t interested in just products, but in equalizers too.

• CommentRowNumber73.
• CommentAuthorIan_Durham
• CommentTimeJul 13th 2010

Thanks for the reference! I actually did my undergraduate work at Buffalo (where Lawvere is) but, despite minoring in Mathematics, I never knew him. If his book is as good as Todd’s comment I will probably wish I had known him and taken a course from him.

• CommentRowNumber74.
• CommentAuthorEric
• CommentTimeJul 14th 2010

You should write a book (seriously - it was very clear).

I second that! Todd, you do have an amazing ability to explain this stuff in such a way that even I can understand :)

Yep yep. The way that equalizers are constraints is starting to sink in. Progress.

Since there are only a finite number of leaves, this is a finite sum, and will always be well-defined.

This made me realize that I’m still a little foggy on how you’re relating Hilbert spaces to categories here. Is each Hilbert space a category? And, if so, what are their objects (leaves, etc.)? And are we limiting ourselves to finite-dimensional Hilbert spaces here?

Well, first, Jamie’s comment is basically a simple statement that since the sums are finite, then we do not need to worry about certain issues that plague infinite sums, e.g. we can freely interchange summation symbols where it makes sense without worry. Nothing deep, just a reassuring comment.

Regarding your question, it is probably good trying NOT to think about Hilbert spaces yet. Of course everything here is motivated by thinking about Hilbert spaces, but it is more general. $FinHilb$ does provide an example of a $\dagger$-category. There, objects are finite-dimensional Hilbert spaces.

And are we limiting ourselves to finite-dimensional Hilbert spaces here?

Yes.

My understanding is that an equalizer is just a type of limit. Is that right?

If you’ve ever worked with system administrators, you’ll appreciate this acronym. RTFM :) equalizer

By the way Ian, if you haven’t seen the Categorical Manifesto, you should check it out. Very very nice. I would like to follow up on a reference that looks very good.

[27] Joseph Goguen and Susanna Ginali, A categorical approach to general systems theory. In George Klir, editor, Applied General Systems Research, pages 257-270. Plenum, 1978.

I’d LOVE to get a hold of this. When I asked Jamie where he learned that intuition about limits as solutions to constraints, he pointed to the Manifesto. When you read the Manifesto, he points to this paper.

Goguen seems very cool. Too bad I didn’t know about him earlier, I would have pestered him during grad school. Sad to learn of his recent passing.

• CommentRowNumber75.
• CommentAuthorJohn Baez
• CommentTimeJul 16th 2010
• (edited Jul 16th 2010)

Eric wrote:

Oh I wish I understood this tantalizing statement.

To understand the concept of equalizer just realize that it’s the category theorist’s way of generalizing “the subset of $X$ on which two functions $f, g : X \to Y$ are equal”. We use equalizers all the time in math: that’s what we’re doing whenever we set two functions equal and look for the set of solutions!

Equalizers are one of simplest kinds of limits after binary products. You should take the definition of limit, apply it to the case of a diagram like

$X \stackrel{\to}{\to} Y$

and see that in the category of sets you get what I said:

$\{x \in X : f(x) = g(x) \}$

There’s no possible way to understand limits without doing this exercise!

All the basics like products, coproducts, equalizers and coequalizers are nicely explained in Goldblatt’s intro to category theory, which you can get for free online or for cheap from Dover. Don’t be scared by the title: the main complaint most category theorists have about this book is that it’s too easy.

• CommentRowNumber76.
• CommentAuthorEric
• CommentTimeJul 16th 2010

Thanks! :)

There’s no possible way to understand limits without doing this exercise!

And having done that exercise doesn’t ensure the understanding of limits, but it hopefully helps :) I went through several of the limits in $Set$ here

Understanding Limits in Set

but nothing ever clicked until the words “constraints” and “solution” came into the picture.

The tantalizing thing I still hope to understand (and limited time is the biggest impediment at the moment) is how

finite limits can all be built from products and equalizers

Also

All the basics like products, coproducts, equalizers and coequalizers are nicely explained in Goldblatt’s intro to category theory

I’ll have another look at Goldblatt with this new understanding and hope I get further than last time :)

Thanks!

PS: As someone who spent all his grad student days thinking about numerical solutions to PDEs with constraints, this brings up all kinds of speculative thoughts.

• CommentRowNumber77.
• CommentAuthorDavidRoberts
• CommentTimeJul 16th 2010

One way to think about pullbacks $A \times_B C$ in categories with product and equalisers is to take the product $A \times C$ and then look at the equaliser – in this case it is the biggest subobject (subset if you are in $Set$) on which the obvious square commutes. If one looks at wide pullbacks (look this up on the nlab - I’m being lazy and in a rush) then one takes a bigger product, like $A_1 \times A_2 \times \ldots \times A_n$ (perhaps $n$ is infinite - it can be done if enough products exist) and then looks at the iterated equalisers which makes all the squares commute. You get a cone by using the projections on the various factors. Forming a general limit from products and equalisers is like this: form a big product, then use the projections to get a cone, then form an iterated equaliser so as to make all the bits of the diagram commute. Think of it as multiple interacting systems, which form a big system, on which there are constraints so as to make everything play along nicely. I’m not sure if there is a rigorous way to implement this analogy, but it may help.

Coproducts + coequalisers = colimits works the same way: take a big coproduct (=disjoint union) then form a big equivalence relation using the cocone, and then form the quotient of this equivalence relation. If you know the formula for geometric realisation then this is an example: it is a coend, which can be calculated as some colimit, which is then written as a codproduct quotiented by an equivalence relation. Easy ;-) (sarcasm notwithstanding, it is a good example to think through, ignoring the bit about coends for now)

• CommentRowNumber78.
• CommentAuthorIan_Durham
• CommentTimeJul 17th 2010

To understand the concept of equalizer just realize that it’s the category theorist’s way of generalizing “the subset of on which two functions are equal”. We use equalizers all the time in math: that’s what we’re doing whenever we set two functions equal and look for the set of solutions!

Yet another crystal-clear explanation! Thanks John!

And having done that exercise doesn’t ensure the understanding of limits, but it hopefully helps :)

I’m probably deluding myself, but Jamie’s explanation of cones seemed to help me understand limits somehow. His description on p.7 makes me think of limits as kind of like a special “pathway” for getting $X$ to $F(S)$ (I can visualize it in my head better than I seem to be able to put it into English, but I really do think I get it the more I look at it).

• CommentRowNumber79.
• CommentAuthorTobyBartels
• CommentTimeJul 20th 2010

If one looks at wide pullbacks (look this up on the nlab - I’m being lazy and in a rush) then […]

That’s wide pullbacks. But that page doesn’t really explain things in the detail that Eric may need.

• CommentRowNumber80.
• CommentAuthorTodd_Trimble
• CommentTimeOct 2nd 2011
• (edited Oct 2nd 2011)

I’m looking at the page quantum operation, at a sentence written by Urs (I think):

“Note that Coecke has shown that FdHilb, the category of finite dimensional Hilbert spaces and linear maps, has the completely positive maps as morphisms.”

This sentence is confusing, since frequently the words ’maps’ and ’morphisms’ are used synonymously. I didn’t see a hint in the preceding text that gives warning about this. I am guessing, after looking briefly at Coecke’s paper that was linked to, that the ’morphisms’ here are arrows of a monoidal category $Mix(C)$ that Coecke constructs out of a $\dagger$-symmetric monoidal category $C$. For the present purposes, it looks like the $\dagger$-symmetric monoidal category is FDHilb.

I could try to insert clarifications myself, but this is not really my area. I was only led to that page while doing some preliminary looking around before I attempt a discussion with Ben over here.

• CommentRowNumber81.
• CommentAuthorTodd_Trimble
• CommentTimeOct 2nd 2011

I hope my previous message is readable. I haven’t the foggiest idea what I did wrong.

• CommentRowNumber82.
• CommentAuthorUrs
• CommentTimeOct 2nd 2011

Todd,

I didn’t write this. Instead, Ian Durham wrote this, in revision 48.

But I can fix it. I have fixed already many things in this entry, but seem to have missed this one.

• CommentRowNumber83.
• CommentAuthorUrs
• CommentTimeOct 2nd 2011

Okay, I have streamlined the whole paragraph after the statement of Choi’s theorem by making all pointers to the literature be actual pointers to the References-section and then adding the lines

This is originally due to (Choi, theorem 1). A proof in terms of †-categories is given in (Selinger). A characterization of completely positive maps entirely in terms of †-categories is given in (Coecke).

• CommentRowNumber84.
• CommentAuthorTobyBartels
• CommentTimeOct 2nd 2011

Todd, what you did wrong was to put HTML in a Markdown comment. (I don’t know why Markdown doesn’t fail gracefully, munging the link but rendering the rest.) Change

to

to fix it.

• CommentRowNumber85.
• CommentAuthorTodd_Trimble
• CommentTimeOct 2nd 2011

Thanks, Toby! I’ll go back and fix it.

• CommentRowNumber86.
• CommentAuthorTodd_Trimble
• CommentTimeOct 2nd 2011

Thanks, Urs – I should have figured it wasn’t you. I remembered you did a lot of work fixing things like quantum channel and quantum operation, and since the whole entry reads much more like something you would have written, I made an incorrect assumption.

• CommentRowNumber87.
• CommentAuthorUrs
• CommentTimeJan 23rd 2014
• (edited Jan 23rd 2014)

I have added to quantum operation pointers to the historical articles by Stinespring and Kraus, predating Choi’s. See the new list of citations right below the theorem.

The entry as a whole still needs more improvement. Maybe later.

• CommentRowNumber88.
• CommentAuthorSam Staton
• CommentTimeJun 25th 2018

I wrote about the universal property of the category of quantum channels. Here I am writing about my own recent work, because I think it is a good way to see this category from the nlab perspective. Of course, feel free to demote it if you like.

• CommentRowNumber89.
• CommentAuthorMike Shulman
• CommentTimeJun 26th 2018

I think that is a nice contribution.

Speaking as an outsider, though, it is somewhat confusing that the “Universal Property” section starts out by referring to “the category of natural numbers and quantum channels” when the page up until that point has not actually defined “a quantum channel”. The “Definition” section only defines “positive” and “completely positive” linear maps. The “Idea” section suggests that “quantum operations” should be, I guess, the trace-preserving and completely positive maps (although the connection is nowhere made to the Definition section). Is a “quantum operation” the same as a “quantum channel”? This is suggested by the fact that the “Properties” section defines a category $QChan$ (though it still never uses the word “quantum channel”) whose morphisms are the completely positive trace-preserving maps.

1. I spelled out the definition of a quantum operation, and made it clear that “quantum channel” is a synonym.