increpare games

Thinking in 4D is hard? Yes, but so is thinking in 3D! Even picturing things in 1D invariantly (that is to say, without thinking of things in terms of an absolute scale), I posit, is a challenge. Or to think in three dimensions without some sort of “gravitational field” or other absolute notion of up-down. Or to think in three dimensions without having one’s self as an “observer”, a fixed point/basis in the space. Also, all the possible ways of thinking of things in terms of fibering, or “foliating” 4-dimensional objects into “sequences” of three dimensional ones are intrinsically (though not necessarily irreparably) flawed*, as they destroy a lot of the invariance of the space (for instance, a line in one direction may end up appearing as a sequence of points <- this is a very ungeometrical distinction) which is really what you're trying to capture anyway, or? *to go in the reverse way, to construct a space-time from space, requires similar considerations - thankfully, Lorentz transformations do the job fabulously and give us a notion of invariance on the full space. *sigh*, invariant thoughts are so hard to come by ("but", says Cartan, "maybe if you can learn to think of the associated G-bundle of your choiceful thoughts, then you can render your thoughts then invariant". I replied, "Oh, what mystical words you speak Cartan!". He had gone, by then, though). I have gotten a lot more comfortable in thinking about riemannian manifolds (spatially) though, recently.

I’m giving a talk with the mathsoc this Wednesday, in the seminar room in the maths department at noon, bearing the title

“Cauchy Riemann Manifolds.“

Oh yes, this will be some fantastic class of lecture, no pots about that. It probably won’t be about CR manifolds though. I mean, it will be, but not really; it’ll actually probably be only a little bit about them. And a little bit about Cartan’s Supremely Cool method of moving frames.

As I made them warn in the official blurb, there is the possibility that there
probably,somewhat necessarilly, will be some technicalities involved. But, unlike in previous years, there will be a superabundance of evocative/risqué sketches and plenty of suggestive gestures, so you may be able to convince yourself that you understand something that’s going on even if you don’t. Depending on the (absence of) departmento-heirarchical status of the audience, I might well insert some toilet into the proceedings.

There will be tea and biscuits at it as well, and hopefully everything will be pretty informal.

Here’s the official notice (the Cartan-attributed quote is, naturally, mine, of course).

Here we go:
This week the mathsoc proudly welcomes back Mr. Stephen Lavelle who will
be giving a talk entitiled “Cauchy Riemann Manifolds”, which might include
a little bit of something to do with Cauchy Riemann Manifolds. Now you may
be asking yourself or possibly someone else, other than an utterance when
some one recieves a bad hand in poker, what indeed is a manifold, for that
matter who is this Cauchy (pronounced cow-key) Riemann bloke and what does
this “Mr. Lavelle” character know about him that I don’t? Well its not my
place to answer those questions, however I will give you a hint, Stephen
knows. This is a very special once in a lifetime presentation and is not
to be missed.
SO at 12 noon in the smeinar room of the mathematics dept. on Wednesday
5th of April bring your own soups, sandwiches, crisps and cappucini along
and enjoy one of the most unique events we’ll have this year.
This talk may also include a bit of stuff about Cartan’s “utterly
fablicious” (his words, not mine (only he said them in french)) methode du
repere mobile.
Warning: this may get very technical

Crammer Time: Our librarian, has been working very hard over the
midterm, so if you need a book, just ask and he’ll do his best to get it
for you (even if its not in our library)

Next Week Bert Toturo of Cambridge will be toturing along to give us a
talk. This will be feature lecture of the year so we intend to make it
brilliant.

Hope to see you all there on Wednesday,
Col

In mild preparation for my project write up, here’s an attempt at an explanation of Cartan’s (rather cool) method of equivalence.

Firstly, a vector space problem. Say you have two bases E={e_i} and E’={e’_i} in an inner product space. You say that they are equivalent (with respect to the inner product) if there is an orthogonal (i.e. angle preserving) transformation A such that E’=AE

In general, you have equivalence problems with vector spaces, where you have a space V, and a group G of linear transformations of the space that preserve the structure you care about (in the last case G was the orthogonal group O, and it preserves in inner product). And you say that two bases E={e_i} and E’={e’_i} are equivalnet if there is a g ∈ G such that

E’=gE

Now, take a linear transformation of vector spaces L:V→V’, where V,V’ have bases E, E’ respectively. We say that L preserves whatever structure it is that G preserves if LE=gE’.

So, here’s the initial question: We say that V and V’ (with bases E, E’) are G-equivalent iff there is a linear bijection L:V→V’ such that LE=gE’ for some g∈G.

Notice how it involves two different sorts of steps, you have to find both a linear bijection L and a transformation g.

Cartan figured out a way to construct a space so that the question because just about linear bijections instead, reducing the problem to a 1-step thing.

So, the idea (in the degenerate case of vector spaces) involves considering the family of vector spaces VxG, which is a family of vector spaces, one for each g ∈ G.

We will extend a basis E in V to a basis F in the vector space corresponding to the element g ∈ G as gE (I will denote that vector space (V,g)). We have F(g)=gE. Fine..

Now, the cool thing is that a transformation L₁:VxG→V’xG (that maps (V,g) into (V,g) ) such that L₁F’=F is the same as a mapping L:V→V’ with LE’=gE, for some g∈G.

To show it one way, take L₁(v,g)=(L(v),T(g)), with

L₁(F(g),g)=(F'(g),g)

As F(g)=gE, L₁(F(g),g)=L₁(gE,g)=(T(g)L(E),g). This gives

(T(g)L(E),g)=(gE’,g)

i.e. T(g)L(E)=gE’, or

L(E)=T(g)^-1gE’

So this gives L(E)=hE’, where h=T(g)^-1g (which is independent of g).

Conversely, given an equivalence LE=hE’, you can construct a mapping

L₁:VxG→V’xG
L₁(v,g)=(L(v),gh^-1)

which does the trick.

Okay. Now, the general case is *exactly* analagous to the one above, and I will outline it presently (if you don’t know about differential forms, you’ll have to be satisfied with the above).

So, instead of a vector space, you have a manifold M, and instead of a basis E={e_i} you have a coframe ω={ω_i}. Now you still have your group G that acts on tangent spaces, but this time the equivalence is a tangentspace-wise thing; that is g can be different at each point.

So, we say two manifolds M and M’ with coframes ω and ω’ are equivlant if there is a morphism Ψ:M→M’ with Ψ^*ω’ (p,v) = g(p) ω for some g:M→G.

The solution is exactly along the same lines. You look at a principal G-bundle (with projection π(g,p)=p ), and extend the coframe ω to the bundle (though it’s not a full coframe in the bundle- you generally have to add more 1-forms to it by differentiation and stuff to get things to work out) as

Ω(v,g)=gπ^*ω

Now, in exactly the same way as with the vector spaces, you can show that two manifolds M and M’ with given coframes are equivalent if and only if there is a map between their respective G-bundles that pull back the induced 1-forms correctly.

tada!

Well, that probably wasn’t worth of a “tada” given that I skipped the last bit. But still, I think it’s a really nice construction, and it’s the starting point of a more general method/algorithm for determining invariants of manifolds with various structures.

It hapens to me not infrequently, but it happened today that I saw the following image

and read it as “Yoneda Lemma”.

Which, class, leads me smoothly onto my next topic: classifying spaces. [warning: requires topology, and, maybe, some category theory]

A topological space has open sets. Great. But say we are limiting ourselves to looking at continuous maps of topological spaces and not explicity of the open sets that form the topology. How do we talk about open sets then?

Well, there’s a really easy way: You can find it by looking for a classfying space for the open sets.

Basically, a space which classifies open sets in terms of continuous maps is a topological space S such that each open subset of any other topology T is given by a continuous map from T to S.

That is, instead of looking at the open sets of a space, you look at continuous maps from that space to S.

[Why *to* S instead of? Well, because the “open sets” functor is contravariant… (see later) ].

And what is the classifying space for open sets? Why it’s the glorious Sierpinski space, of course! This is a topological space with two points, one open, and one closed. You can verify easily yourself that each continuous map to S picks out one special open set.

So, S is a really handy tool!

Many many other things can be classified; for instance, the set of all vector bundles over a manifold can be given in terms of spaces formed from Stieltjes varieties and Grassmanians; for a covariant example, the set of points of a topological space can be classified in terms of mappings from the one point topological space, and, well, loads of things! Seriously, just try it!

To go through the transitional phase:

The assignment to each space it’s set of open subsets actually constitutes a contravariant functor F from Top to Set.

The classifying space for this functor is the sierpinski space S.

To say that the set of all open sets of a space is the same as the set of continuous maps to S is the same as saying that F(T) is isomorphic to hom(T,S) for all topologies T. (hom(T,S) is the set of all continuous maps from T to S, or, in general, the set of all morphisms). In proper category theory, we require this to be something like a bijective natural transformation between the functor F and the functor hom(-,S).

And for the final phase:
A classifying space for a contravariant set-valued functor F:C->Set, is an object S \in C such that there is an isomorphism from F to hom(-,S).

Even more nicely: A set-valued contravariant functor F is called representable if it is isomorphic to hom(-,S) for some S an object of it’s domain. And S is called the classifying space or representing object of F :)

(if F is covariant, you take hom(S,-) instead).

Now, isn’t *that* nice?

Now I’m in a place to talk about the yoneda embedding. A natural transformation is like a mapping between functors (I was talking about the implicitly when I mentioned isomorphisms in the above definition).

Even when you can’t find a classifying space, you can still go up another level and use the Yoneda Lemma, which says that each open set of a space T, say, is the same as a natural transformation from hom(-,T) to F(T). Or rather more nicely Nat( hom(-,T) , F) is isomorphic to F(T) for all spaces T. Of course, you can go up another level and get rid of a T, talking about an isomorphism between some functor that maps T onto the set Nat( hom(-,T) , F) and the functor F.

If you’re not troubled by vertigo, you can go up another level and get rid of the F (because it holds for all set-valued functors) and talk about a…wait…I’m not going to say it all, you should be able to get the idea, and if you can’t, it probably won’t do you much good for me to say it because I’d be leaving out to many details by that point.

But I will point out a special case of the yoneda lemma, namely the case where F is itself a functor like hom(-,A). This then says that

hom(A,B) is isomorphic to Nat( hom(-,B), hom(-,A) )

This is an even more extreme example of what I was talking about above, wanting to move up a level somehow – in the first case talking about morphisms (continuous maps in my example) instead of elements (open subsets) , and in this case you can stop talking about morphisms and start talking about natural transformations.

So where you can talk about open subsets as morphisms to the classifying space, you can now use Yoneda to talk about them as natural transformations as well :)

[Why does the functor have to be Set-valued? Well, because hom(T,S) is usually a set – except in toposes. But this post is already too long for talking about them. And you can even try to construct classifying spaces for functors from the category of categories, and do stuff like that, I’ve been told. Actually, I’m going to have to take advantage of this parenthetical situation to say that, in a similar vein to the above (but a different direction) you can also view continuous maps between topological spaces as adjoint functors between the categories of sheaves on those spaces. Okay. There. I’ve said it. I actually have another post that I’ve never put on the site written about these that]

Also note that I should have put in a bunch of “naturality conditions” above, but I didn’t, because I’m sloppy that way.

Wow. I don’t know/care about you, but that was certainly good for me to get it off my chest. Hmmm… .

Addendum: I know that Hom is trivially representable in each variable separately, but are there any categories where it is representable as a functor from C*xC->Set ? (this corresponds to there being two objects X and Y such that any morphism f:A->B is (naturally) the same as a pair of morphisms <g :A->X,H:Y->B>, right? ) Doesn’t work for most categories that I’ve tried it for – nor when I use the diagonalized version of it [hom(X,X)].

Okay, if Stephen Walsh can put notes on his website, so can I damnit!

I’m still not even half-finished them, but I thought it no harm to stick them up in their still unfinished state (as of the start of the summer) – nobody ever reads the second half of anything anymore anyway, right? right.

So, yep, here they are:

Constructions in Category Theory Postscript | Acrobat

If you spot any typos, feel free to lynch me or mail me or stuff.

Whatever.

Yeah; I’ve updated way too many times today.