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November 6

Cyclic actions on manifolds

Does anyone know if the following conjecture is true?

If a connected smooth manifold (without boundary) X has Euler characteristic divisible by n, then there exists a smooth map X → X where every point has period exactly n.

I'm sure there are lots of more elegant ways to phrase that. I tend to think it's false, but it unfortunately holds for:

Circle bundles
Spheres
Spaces with Euler characteristic +/-1
Simple products of the above spaces, e.g. (S x S) / 4
Simple punctured spaces, e.g. (S - 5) / 3

I can't think of a period-4 map on T#T#T or a period-8 map on the 2-sphere cubed, but that's hardly proof that they don't exist. And now I'm going to try to get back to sleep. Melchoir 15:26, 6 November 2007 (UTC)[reply]

...Maybe the Mobius strip? Melchoir 15:45, 6 November 2007 (UTC)[reply]

The moebius strip without boundary? – b_jonas 16:27, 6 November 2007 (UTC)[reply]

Right. And I'm not so sure about (S x S) / 4 anymore, either; I thought I had a map, but I didn't. Melchoir 19:09, 6 November 2007 (UTC)[reply]

This fails for e.g. the open n–disk, or in general any finite-dimensional contractible space X. Since every point has period n, the action is free; since our action is by the finite group G=Z/nZ, it is properly discontinuous. Together these imply that the quotient map X → G\X is a covering map, and so G\X is a finite dimensional K(G,1). But no such thing exists (since groups with torsion have nontrivial group cohomology in infinitely many dimensions). [See Algebraic Topology by Hatcher.] Tesseran 01:56, 7 November 2007 (UTC)[reply]

I think a contractible space has Euler characteristic 1 (reduced is 0), so the original poster's n has to divide 1. Then we are looking for a K(G,1) for G = Z/Z = 1. I think any open ball is one, though a single point is probably an easier example. I'm guessing you were using reduced Euler characteristic, since you seem to assume n is arbitrary, and the only integer divisible by arbitrary n is 0. JackSchmidt 02:07, 7 November 2007 (UTC)[reply]

It's true for all (closed, at least) surfaces. It's easy for the non-orientable surfaces. To show it for an orientable surface with genus g > 1 (hence euler characteristic 2-2g), it suffices to show there's an orientable normal covering space of T#P (i.e. the non-orientable surface with euler characteristic -1) with deck group

\mathbb {Z} /(2g-2)\mathbb {Z}

. For this we need a (surjective) homomorphism

\pi _{1}(T\#P)=<a,b,z|aba^{-1}b^{-1}=z^{2}>\rightarrow \mathbb {Z} /(2g-2)\mathbb {Z}

(and then we need to check that the associated cover is orientable). Something like a -> 1, b -> 0, z -> g-1 will do the trick (and the associated cover is orientable essentially because z is sent to something nonzero). It's a pleasant exercise to visualize the associated map on the surface of genus g.

I'm not really sure about dimension 3 and up, but this doesn't seem like the sort of statement that would be true in general.

Actually, it appears that there are 3-manifolds with no symmetries at all. This is mentioned in "Do manifolds have little symmetry?" by V. Puppe (it's online and can be found by searching for it, but is fairly technical). All 3-manifolds have euler characteristic zero, so this contradicts the conjecture. I'm not sure about examples, but I'd guess one could get a [hyperbolic] 3-manifold to do it. kfgauss 08:49, 7 November 2007 (UTC)[reply]

Very interesting, thanks! Although I recognize pieces of your argument for surfaces, I must say that I don't understand the whole thing. I'll have to spend some time thinking about this.

Yeah, when I made the original post I had forgotten about closed, oriented 3-manifolds having zero Euler characteristic. It certainly seems like the chances are slim. I still wonder if there's a single simple, compelling counterexample... Melchoir 17:10, 7 November 2007 (UTC)[reply]

Quiz prize

Say you have a quiz that asks four questions on different subjects (Say History, Geography, Science, French literature). You get, sequentially, £1, £2, £4, £8 for answering a question correctly. You lose everything on a wrong answer. You can opt out at any point keeping what you've won so far. You can choose the order of the subjects. If you have a 20% likelihood of answering History correctly, 40% for Geography, 60% for Science and 80% for French lit, what is the best order to request the subjects in to maximise your winnings? -- SGBailey 23:07, 6 November 2007 (UTC)[reply]

Take the Science and French lit questions (in either order), then (if you're not already out) opt out. --Lambiam 23:57, 6 November 2007 (UTC)[reply]

Strictly speaking, the condition 'maximizing your winnings' is not quite well defined, since your winnings are not certain, but rather a random variable depending on your strategy. Lambiam's answer is the strategy that maximises your expected winnings, but this is not the only possible approach to take. For example, you could adopt a minimax approach, which would lead to taking the French lit question and nothing else. This would be appropriate if you need some money but are fairly indifferent whether it's £1 or £2. In this case, maximizing expected gain is reasonable, but not always: see St. Petersburg paradox for a case when it can lead to extremely stupid decisions. Algebraist 20:36, 7 November 2007 (UTC)[reply]

However, for this particular game, it turns out that asking for the questions to be given in the order from easiest to hardest is always optimal, regardless of the player's utility function — only the choice of when to stop varies. There can, and usually will, be other orders that are equally good, but none that are better. This conclusion, however, does not necessarily generalize to more complex games where, for example, the player might be allowed to continue after losing. —Ilmari Karonen (talk) 21:57, 7 November 2007 (UTC)[reply]