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Re: This Week's Finds in Mathematical Physics (Week 209)
Posted:
Feb 1, 2005 2:31 PM
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In article <coisoa$207$1@news2.netvision.net.il>, Squark <fiis5d@yahoo.com> wrote:
>John Baez wrote:
>> The n-categories called "n-groupoids" magically know >> everything there is to know about homotopy theory,
>That part I know...
>> while those called >> "n-categories with duals" know everything there is to know about the >> topology of manifolds.
>But how would this work? What are n-categories with duals anyway?
We don't know a general definition, but we have a good idea in low-dimensional cases. Consider the category Hilb, where the objects are finite-dimensional complex Hilbert spaces and the morphisms are linear operators. This category has a tensor product so we call it a monoidal category. In any monoidal category, we can say what it means for an object to have a "dual". In the case of Hilb, every object H has a dual H* - the dual Hilbert space. We can also say what it means for a morphism to have a "dual". In the case of Hilb, every morphism f: H -> H' has a dual f*: H' -> H - the adjoint operator. So, we say our category has duals for objects and morphisms.
Interestingly, the main source of monoidal categories with duals is topology: categories of cobordisms. This suggests an interesting relation between quantum mechanics (which uses Hilb) and topology. Indeed, this relation is precisely what we exploit in topological quantum field theory! In fact, I claim it will eventually make a lot of mysteries in quantum theory seem less mysterious. The reason we find quantum theory mysterious is that it's based on monoidal categories with duals, and these are fairly unfamiliar - UNTIL we start drawing pictures of cobordisms and see that they formalize simple ideas about spacetime topology!
So, I hope that quantum theory will become a lot clearer when we unify it with a theory of spacetime. I explain this hope in vastly more detail here:
http://math.ucr.edu/home/baez/quantum/
Now, to make a little more progress in explaining n-categories with duals, I should explain why they are a bit like n-groupoids, which are n-categories with inverses.
To start, let me explain why the dual of a finite-dimensional vector space, is similar to the inverse of a number.
I assume you know that multiplying numbers is a lot like tensoring vector spaces. For example, just as multiplication distributes over addition, tensoring distributes over direct sums. Also, just as there is a number called 1 which is the unit for multiplication, there is a 1-dimensional vector space, the ground field itself, which is the unit for tensoring. Let me take the unusual liberty of writing tensor products by juxtaposition, so that xy is the tensor product of the vector space x and the vector space y, and let me call the 1-dimensional vector space that's the unit for tensoring simply "1".
Now, if a number x has an inverse y, we have
yx = 1
and
1 = xy.
Similarly, if a vector space x has a dual y, we have linear maps
e: yx -> 1
and
i: 1 -> xy
What are these linear maps? Well, the whole point of the dual vector space y is that a vector in y is a linear functional from x to 1. This "dual pairing" between vectors in y and those in x defines a linear map e: yx -> 1, often called the "counit". On the other hand, the space xy can be thought of as the space of linear transformations of x. The linear map i: 1 -> xy sends any scalar (i.e., any vector in 1) to the corresponding scalar multiple of the identity transformation of x.
So we see that dual vector spaces are a bit like inverse numbers, except that we don't have yx = 1 and 1 = xy, and we don't even have that yx is *isomorphic* to 1 and 1 is *isomorphic* to xy. We just have some maps going from yx to 1, and from 1 to xy.
These maps satisfy two equations, though. Here's the first. We start with x, use the obvious isomorphism to map to 1x, then use i: 1 -> xy to map this to xyx, then use e: yx -> 1 to map this to x1, and then use the other obvious isomorphism to map back to x. This composite of maps should be the identity on x. What this says is that the identity linear transformation of x really acts as the identity!
We can draw this as follows. Draw the counit e: yx -> 1 as follows:
y x \ / \ / \/
and draw the unit i: 1 -> xy as follows:
/\ / \ / \ x y
Then the above equation says that x x /\ | | / \ | | / \ | | x| y\ x/ = | | \ / | | \/ | x x
Here the left side, which we read from top to bottom, corresponds to the composite x -> 1x -> xyx -> x1 -> x. (The factors of 1 are invisible in the picture, since they don't do much.) The left side corresponds to the identity map x -> x.
The second equation goes like this. We start with y, use the obvious isomorphism to map to y1, then use the unit to map this to yxy, then use the counit to map this to 1y, and then use the other obvious isomorphism to map back to y. This composite should be the identity on y. What this says is that the identity linear transformation of x also acts dually as the identity on y! We can draw this as follows:
y y | /\ | | / \ | | / \ | y\ x/ y| = | \ / | | \/ | | y y
In general, whenever we have a monoidal category, we say an object x has a dual y when it's equipped with morphisms i: 1 -> xy, e: yx -> 1 satisfying these two equations - which are called the "zig-zag identities".
You can see already that curves in the plane give us an example of a monoidal category where all the objects have duals.
So, you can begin to see the relation between monoidal categories with duals and the topology of 1-dimensional manifolds embedded in 2-dimensional space. This is the tip of an enormous iceberg called the "tangle hypothesis", which is explained here:
http://www.arxiv.org/abs/q-alg/9503002
In particular, if this hypothesis could be made precise, we would have a completely algebraic description of smooth manifolds in terms of n-categories with duals. For some progress see this:
http://www.arxiv.org/abs/math.QA/9811139
which gives an algebraic description of 2-dimensional manifolds embedded in 4-dimensional space in terms of braided monoidal 2-categories with duals.
(Maybe you can already guess the pattern!)
There's a lot more to say but maybe this is a start.
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