Hi. I'm Gareth McCaughan. I've been a consistent reader and occasional commenter since the Overcoming Bias days. My LW username is "gjm" (not "Gjm" despite the wiki software's preference for that capitalization). Elsewehere I generally go by one of "g", "gjm", or "gjm11". The URL listed here is for my website and blog, neither of which has been substantially updated in about the last four years. I live near Cambridge (UK) and work for a small technology company in Cambridge. My business cards say "mathematician" but in practice my work is a mixture of simulation, data analysis, algorithm design, software development, problem-solving, and whatever random engineering no one else is doing. I am married and have a daughter born in mid-2006. The best way to contact me is by email: firstname dot lastname at pobox dot com. I am happy to be emailed out of the blue by interesting people. If you are an LW regular you are probably an interesting person in the relevant sense even if you think you aren't.

If you're wondering why some of my old posts and comments are at surprisingly negative scores, it's because for some time I was the favourite target of old-LW's resident neoreactionary troll, sockpuppeteer and mass-downvoter.


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Topological Fixed Point Exercises

Inappropriately highbrow proof of #4 (2d Sperner's lemma):

This proves a generalization: any number of dimensions, and any triangulation of the simplex in question. So, the setup is as follows. We have an n-dimensional simplex, defined by n+1 points in n-dimensional space. We colour the vertices with n+1 different colours. Then we triangulate it -- chop it up into smaller simplexes -- and we extend our colouring somehow in such a way that the vertices on any face (note: a face is the thing spanned by any subset of the vertices) of the big simplex are coloured using only the colours from the vertices that span that face. And the task is to prove that there are an odd number of little simplexes whose vertices have all n+1 colours.

This colouring defines a map from the vertices of the triangulation to the vertices of the big simplex: map each triangulation-vertex to the simplex-vertex that's the same colour. We can extend this map to the rest of each little simplex by linear interpolation. The resulting thing is continuous on the whole of the big simplex, so we have a continuous map (call it f) from the big simplex to itself. And we want to prove that we have an odd number of little simplices whose image under f spans the whole thing. (Call these "good" simplices.)

We'll do it with two ingredients. The easy one is induction: when proving this in n dimensions we shall assume we already proved it for smaller numbers of dimensions. The harder one is homology, a standard tool in algebraic topology. More precisely we'll do homology mod 2. It associates with each topological space X and each dimension d an abelian group Hd(X), and the key things you need to know are (1) that if you have f : X -> Y then you get an associated group homomorphism f* : Hd(X) -> Hd(Y), (2) that Hd(a simplex) is the cyclic group of order 2 if d=0, and the trivial group otherwise, and (3) that Hd(the boundary of a simplex) is the cyclic group of order 2 if d=0 or d = (dimension of simplex - 1) and the trivial group otherwise. Oh, and one other crucial thing: if you have f : X -> Y and g : Y -> Z then (gf)* = g*f*: composition of maps between topological space corresponds to composition of homomorphisms between their homology groups.

(You can do homology "over" any commutative ring. The groups you get are actually modules over that ring. It happens that the ring of integers mod 2 is what we want to use. A simplex is, topologically, the same thing as a ball, and its boundary the same thing as a sphere.)

OK. So, first of all suppose not only that the number of good simplices isn't odd, but that it's actually zero. Then f maps the whole of our simplex to its boundary. Let's also consider the rather boring map g from the boundary to the whole simplex that just leaves every point where it is. Now, if the thing we're trying to prove is true in lower dimensions then in particular the map gf -- start on the boundary of the simplex, stay where you are using g, and then map to the boundary of the simplex again using f -- has an image that, so to speak, covers each boundary face of the simplex an odd number of times. This guarantees -- sorry, I'm eliding some details here -- that (gf)* (from the cyclic group of order 2 to the cyclic group of order 2) doesn't map everything to the identity. But that's impossible, because (gf)*=g*f* and the map f* maps to Hn(whole simplex) which is the trivial group.

Unfortunately, what we actually need to assume in order to prove this thing by contradiction is something weaker: merely that the number of good simplices is even. We can basically do the same thing, because homology mod 2 "can't see" things that happen an even number of times, but to see that we need to look a bit further into how homology works. I'm not going to lay it all out here, but the idea is that to build the Hd(X) we begin with a space of things called "chains" which are like linear combinations (in this case over the field with two elements) of bits of X, we define a "boundary" operator which takes combinations of d-dimensional bits of X and turns them into combinations of (d-1)-dimensional bits in such a way that the boundary of the boundary of anything is always zero, and then we define Hd(x) as a quotient object: (d-dimensional things with zero boundary) / (boundaries of d+1-dimensional things). Then the way we go from f (a map of topological spaces) to f* (a homomorphism of homology groups) is that f extends in a natural way to a map between chains, and then it turns out that this map interacts with the boundary operator in the "right" way for this to yield a map between homology groups. And (getting, finally, to the point) if in our situation the number of good simplices is even, then this means that the map of chains corresponding to f sends anything in n dimensions to zero (essentially because it means that the interior of the simplex gets covered an even number of times and when working mod 2, even numbers are zero), which means that we can think of f* as mapping not to the homology groups of the whole simplex but to those of its boundary -- and then the argument above goes through the same as before.

I apologize for the handwaving above. (Specifically, the sentence beginning "This guarantees".) If you're familiar with this stuff, it will be apparent how to fill in the details. If not, trying to fill them in will only add to the pain of what's already too long a comment :-).

This is clearly much too much machinery to use here. I suspect that if we took the argument above, figured out exactly what bits of machinery it uses, and then optimized ruthlessly we might end up with a neat purely-combinatorial proof, but I regret that I am too lazy to try right now.

Buridan's ass in coordination games

This doesn't (I think) really have much to do with randomness as such. The relevant thing about R is that it's shared information that a hypothetical adversary doesn't get to see.

If isn't chosen adversarially, then our players don't care about pessimizing over but about something like an average over , and then R isn't needed. Or, if they are ultra-cautious people who universally care about worst cases, then they don't care about expectation w.r.t. R but about the worst case as R varies, and then R doesn't help. R only helps them when is chosen adversarially and R isn't; or when they care about pessimizing w.r.t but not w.r.t. R.

So the real conclusion here is not "sometimes shared randomness helps", it's "if some people are trying to coordinate in the presence of an adversary, anything they share and the adversary has no access to helps".