# Help please from non-Baez people on twoHilberts!

1. Jun 27, 2006

### marcus

It's in everybody's interest here not to use too much of JB's time, so I am anxious to get help from someone who is not JB-----selfAdjoint, Hurkyl, arivero, Kea and probably several others, anyone who could help me understand twoHilberts.

I am stuck with a detail of the categorified version of vector addition.

Has to do with not understanding the idea of "canonical isomorphism".

I understand the idea of an object being "universal" for a certain property.

We are defining COPRODUCTS. (direct sum of vector spaces AND union of sets are BOTH examples of coproducts----coproducts is a good idea)

in a category with coproducts, you have objects A and B and their coproduct which I will lazily write (A+B, i, j) is a triple consisting of this other object A+B and morphisms A--i-->A+B and B--j-->A+B
which is UNIVERSAL for that double-stoker situation in the sense that for any other such triple (Z, i', j') there exists a unique morphism f that factors i' and j' so that A+B can come first.

A+B can always act as a way-station in any other double-stoker.

So there is a unique f with i' = if and j' = jf

A --i--> A+B --f--> Z
B --j--> A+B --f--> Z

the thing I sense is true but dont see how to show is the UNIQUENESS of the coproduct A+B "up to canonical isomorphism"

AAAHHH! maybe this is how you get the uniqueness. You suppose there is another coproduct (A&B, i', j')

then because each is universal exist unique f and g with, so THESE morphisms constitute the "canonical isomorphism" they were talking about

A --i--> A+B --f--> A&B
B --j--> A+B --f--> A&B

A --i'--> A&B --g--> A+B
B --j'--> A&B --g--> A+B

and we need to check fg = 1A+B but that is because we can factor the A+B double stoke THRU ITSELF and the definition says there must be a UNIQUE factor morphism that works and both fg works and also 1A+B works, so they have to be equal.

I dont really need to write it out but copy/paste makes it easy so why not

A --i--> A+B --f--> A&B --g--> A+B
B --j--> A+B --f--> A&B --g--> A+B

that shows that fg works and also 1A+B works (is that obvious?) so they are the same. Likewise for gf = 1A&B

So up to "canonical isomorphism" which mean unavoidable preordained that you bark your shins on isomorphism there ISN'T ANY A&B. There is only the one coproduct A+B.

OK. That went all right. Let us see what else there is

what we have to do is to see how to REALIZE EVERY CONDITION IN THE DEFINITION OF A HILBERT SPACE in terms of category protocols

Like a usual Hilbert has a ZERO, so now we are going to have a category where the objects are Hilberts and there is actually a ZERO HILBERT

Fantastic, there is actually a Hilbert space that ACTS LIKE A ZERO VECTOR amongst other Hilbert spaces.

Since on a usual Hilbert space we have the "inner product" of two vectors, we have to figure out what is an "inner product" of two Hilberts.
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this is apt to be amusing. Please check in to see how I am doing, in case I get stuck or overlook something.

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the motivation is something JB said about spin foams

Just as Feynman diagrams and spin networks are diagrammatic ways of thinking about hookups between ordinary Hilberts,
JB said that spinfoams are diagrammatic ways of thinking about hookups between TWOHILBERTS. what I am trying to understand is the definition of a twoHilbert.

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being a highly skeptical sober and restrained person, i naturally doubt that such an abstract gizmo could be useful and have steadfastly resisted up til now understanding anything about them. But ORDINARY Hilberts have proven very useful----I picture an ordinary Hilbertspace as a handy device like a PalmPilot, or a ProgrammableRemote, or a PersonalDigitalAssistant that one constantly uses to keep track of the information one has about something. you can punch in stuff, or you can just let it run on autopilot (it has a clock in it and can unitarily evolve stuff when you arent watching). So it is one of these things about the size of a candybar which they make in Malaysia, with buttons, and a nice liquidcrystal display. That is what a Hilbert is.
And these things are everywhere and indispensible.

Hard to resist, right?

Mozart C-minor is playing in the kitchen. Almost time for lunch.

Last edited: Jun 27, 2006
2. Jun 27, 2006

### marcus

zero hilbert

the ZERO HILBERT SPACE has got to be an inherently laughable idea.

but we are implenting the IDEA of a "higher" hilbert space which is a CATEGORY whose objects are themselves hilbertspaces. (!) (part of me can't believe this, there must be some mistake!)

well a hilbertspace is a vectorspace and a vectorspace has a zero. so this Category whose objects are hilberts must include a special object
("unique up to canonical isomorphism" ) which acts like a zero.

Does anyone want to take over and explain? JB did this in the other thread, but there is no harm in trying to recite the lesson just for review.

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and then, because in the sensible workaday world we know how to SUBTRACT VECTORS we have to understand how to SUBTRACT HILBERT SPACES. neat that should actually be trickier than adding them by the universal double-stoker "coproduct" method. subtracting might involve using the inner product, at least as a shortcut.

anyway, anyone who likes this idea is invited to take over any part of it.

it is like revisiting the day you sat in class and Teacher defined a vector space, and an innerproduct, and a hilbertspace except now it looks just slightly larger than life

I will take a break in case someone wants to do zero hilbert, or subtraction A-B, or any of that other.

Last edited: Jun 27, 2006
3. Jun 27, 2006

### Hurkyl

Staff Emeritus
I think this is one of those things at just looks dense and unpenetrable until you realize you already know what they are. (Disclaimer: I'm not too comfortable with these things yet)

This is convenient -- I get to bring up my favorite example of a category: the category of matrices over the complex numbers!

(The matrices are the morphisms -- I think there's value in all but ignoring the objects... you can simply take them as nonnegative integers, or you can call them complex vector spaces if you really want to)

There are, I think, only two tricky parts to seeing that this is a 2-Hilbert space:
(1) Don't forget that there are 0xn and mx0 (and 0x0) matrices!
(2) The inner product of two mxn matrices is <A, B> = tr(A^* B)

(And this is, of course, an equivalent category to that of finite dimensional Hilbert spaces and linear transformations)

I won't work out the details -- I think it's a good (and fun!) exercise for the reader! But you'll discover that you've already been doing higher algebra, you were just never told that you were.

The zero object in Hilb is an easy one; you're just overthinking it. It's nothing more than the zero vector space... a.k.a. the zero-dimensional vector space... a.k.a. the vector space that contains the zero vector and nothing else.

Subtraction in Hilb isn't that bad either. You have to subtract along some morphism:

T : V ---> W

And there are at least three ways to think of it:
(1) The difference is the quotient space (W / im T)
(2) The difference is the orthogonal complement of (im T)
(3) The difference is the image of some map S:W ---> X where the nullspace of S is exactly the image of T.

If T is injective and you look at the dimensions of everything involved, it really does look like subtraction. Of course, you could also see that by noting that if X is the orthogonal complement of (im T), then:

X + (im T) = W

One way I've seen to make more interesting 2-Hilbert spaces is by "painting" things. So, you would have a category that might have "red" and "green" spaces -- any object would be the sum of a "red" part and a "green" part... and your morphisms must be color-preserving. I think with n-colors, you get the (!) n-dimensional 2-Hilbert space

Last edited: Jun 27, 2006