User simon willerton - MathOverflow

How do you see that higher genus surfaces are not homogeneous?

2012-08-06T12:04:36Z

I am trying to get some intuition about why the torus and the sphere are the only surfaces which can be realised as homogeneous spaces. On the one hand, I know this is true because there is the result that homogeneous spaces must have non-negative Euler characteristic:

A Structure Theorem for Homogeneous Spaces, Mostow, G, Geometriae Dedicata, (114) 2005, 87-102

However, on the other hand, a higher genus surface can be realised as a quotient of hyperbolic space by a group of isometries. The latter would seem (in my head) to give rise to a hyperbolic surface where the points look the same; all of the points have the same curvature, for instance.

My question is then: How do you distinguish the points of such a hyperbolic surface?

Answer by Simon Willerton for Using TikZ in papers

2010-04-15T17:06:58Z

It is not exactly answering either of your two questions, but here is another work-around. I had problems putting papers on the arxiv which used pgf/tikz because the version of pgf/tikz they used at the arxiv was not as up to date as my version. The admin at the arxiv told me to do the following. LaTeX your file with the option -recorder. This will create a .fls file containing a list of all of the FiLeS used by LaTeX when typesetting your document. Choose all of the files in the list containing "pgf" or "tikz" and move them into the directory containing your document. You can then send that directory to your collaborator/the arxiv/the journal without worrying about how up to date their set up is.

[Unfortunately, I then had the problem that the tikz graphics I produced required an enormous amount of working memory which was greater than the allocation on the arxiv server, so I resorted to using 'grab' on my mac to take a high resolution snap-shot of the graphic, which I then incorporated into the LaTeX file :-(. However, I have used this including-all-the-files technique for subsequent uploads to the arxiv.]

Tropical mathematics and enriched category theory

2009-12-01T11:31:02Z

Is there a connection between tropical mathematics and the Lawvere enriched category theory approach to metric spaces? I guess I will give a partial answer to this below, but I mean can they be formally be put on the same level in some sense?

In the Lawverian point of view one does category theory with the extended non-negative real numbers, [0,∞] or R_≥0∪∞, equipped with + as the 'tensor' product and max as the 'categorical' product or sum. In tropical mathematics you work (it seems) with the the extended reals R∪∞ equipped with the 'product' + and the 'sum' max (or min depending on your point of view I think).

In the enriched category theory approach to metric spaces, one has the notion of a kernel (or bimodule or profunctor depending on your point of view) between two metric spaces X and Y which is just a distance non-increasing function K:X×Y->[0,∞]. The correct notion of function on a metric space here is a distance non-increasing function φ:X->[0,∞]. Then the transform of a function φ by a kernel K is a function on Y defined by

K(φ)(y):= inf_xεX ( φ(x) + K(x,y) ).

There is similarly a dual notion which takes functions on Y to functions on X.

K^{^}(ψ)(x):= sup_yεY ( ψ(y) - K(x,y) ).

This is explained in a bit more detail in a post in at the n-Category Café.

It was pointed out to me that these look similar to the Legendre transform. And looking on the internet I found that tropical mathematics is one way to interpret the Legendre transform as an 'integral transform'.

So has anyone ever considered any formal connections between these two points of view?

Answer by Simon Willerton for What is a monoidal metric space?

2009-12-01T19:08:55Z

Here are a couple of answers to your wider question.

1) Tom Leinster defined the notion of Euler characteristic for a finite categories, generalizing things like cardinality of sets, Euler characteristic of posets and Euler characteristic of finite groups. This can be generalized to enriched categories and specialized to metric spaces, giving rise to an (occasionally undefined) invariant of metric spaces called the magnitude. (The names cardinality and Euler characteristic were deemed to be too confusing.) Interestingly, this was discovered in the nineties by some ecologists interested in measuring biodiversity. See our paper http://arxiv.org/abs/0908.1582 for more details.

2) Given an endofunctor F:C->C of a category enriched over V there are two ways of taking the 'trace' of F that I know of, both leading to an object of V. One is the end $\int_c C(c,F(c))$ and the other is the coend $\int^c C(c,F(c))$. In the context of metric spaces this means that for a distance non-increasing function f:X->X the two traces are sup_xd(x,f(x)) and inf_xd(x,f(x)) - which can be thought of as the furthest distance that f moves points and the least distance that f moves points.

Answer by Simon Willerton for Learning LaTeX properly

2009-10-24T08:49:09Z

Of course it depends what those bad habits are. Sometimes people's bad habits involve trying to do things traditional typesetters or designers would never do - for example, setting very narrow margins. I would recommend reading a bit about the craft of typesetting in the traditional sense and reading about the ideas of traditional style design. The TeXBook is a good source for references.

Sometimes bad habits in LaTeX involve hard-coding things like theorems instead of using environments, for example, \textbf{Theorem 1} {\it This is my theorem.}

However, it doesn't sound like your bad habits are of this nature. Maybe you should give us some examples of the kind of bad habits you have in mind.

Answer by Simon Willerton for Induction and Coinduction of Representations

2009-10-21T10:49:23Z

For 1, you don't need H to be a subgroup of G. If you have a morphism of finite groups, f:H->G, then this gives rise to induction, coinduction and restriction functors between the categories of finite dimensional representations. You can try to write down a natural isomorphism between the induction and coinduction functors and you will find that it involves inverting |ker(f)|, the order of the kernel of f. So provided this order is invertible in your ring then induction and coinduction are isomorphic functors.

In particular if the order of the kernel is 1, meaning that f is an inclusion, then you get isomorphic functors regardless of the ring. Also, if the characteristic of the ring is zero then the order of the kernel is automatically invertible and you get isomorphic induction and coinduction functors.

Comment by Simon Willerton

2013-04-03T09:41:34Z

This is just the Kuratowski embedding or $L_\infty$ embedding alluded to in Tom Leinster's and David Eppstein's comments. (In enriched category theory terms this is the Yoneda embedding.) I would say that there is nothing specifically tropical being used here.

Comment by Simon Willerton

2012-08-09T08:47:26Z

Thanks for the useful replies everyone. The key pointers for me were the terms "geodesics" and "Weierstrass points". I've now stitched together much better intuition from what people have said.

Comment by Simon Willerton

2012-08-09T08:25:23Z

+1 for mentioning the Volume Conjecture

Comment by Simon Willerton

2010-04-22T01:28:51Z

You think that's disastrous? I had a referee reject a completely different paper of mine! I assume they must have downloaded a random paper of mine off the arxiv. They wrote quite a scathing rejection of the paper without being very specific - fortunately I noticed that they alluded to certain things that were not mentioned in the submitted paper so I was able to point this out to the editor. (And it all ended happily for both papers.)

Comment by Simon Willerton

2010-04-20T16:01:02Z

I had problems with the arxiv not having an up-to-date version of pgf-plots. The arxiv people said "pgf/tikz simply develops at too fast of a pace for us to keep up." It was suggested that I use the work-around I mention further down this page.

Comment by Simon Willerton

2010-04-20T15:54:38Z

I don't understand your question. Where might I be assuming that?

Comment by Simon Willerton

2009-12-02T08:44:08Z

Not that I'm aware of.

Comment by Simon Willerton

2009-12-01T18:32:09Z

Don't know what the etiquette is here. Should I just leave David's link or should I copy some of the stuff over? I'm inclined towards the former.