User david sprehn - MathOverflow

Answer by David Sprehn for On the cohomology of a finite covering map

2011-03-02T04:35:27Z

The niceness condition you want is on the action, not on the space $X$. Specifically, you want to have that $X\to X/G$ is a principle $G$-bundle, so that we have a Serre spectral sequence for $G\to X\to X/G$. Of course, since you're assuming that $G$ is a finite discrete group, the singular cohomology of $G$ is free, and only in degree 0. In fact, the requirement of being a "principle $G$-bundle" is the same as $X\to X/G$ being a covering space.

The problem: though the spectral sequence looks simple, and collapses immediately at the $E_2$ page, it's not really very useful, since all the interesting data is hidden in the local coefficient system (which is absolutely not trivial unless $G=0$.)

However, we can perhaps get the relationship you want in a much easier way if you're willing to modify the coefficient ring a bit. In particular, the answer is much simpler if you use a ring in which the order of $G$ is a unit. In that case, it's not hard to show directly (using covering space theory) that $H^*(X/G)\to H^*(X)$ is an isomorphism onto the invariants of $G$, i.e. the subring $H^*(X)^G$ of classes which are invariant under the action of $G$. This is an exercise in Milnor's Characteristic Classes, and I believe some form of it appears in Hatcher as well.

Answer by David Sprehn for Theorems that are 'obvious' but hard to prove

2011-01-10T07:39:40Z

All isometries of the plane are affine linear.

Answer by David Sprehn for Equality vs. isomorphism vs. specific isomorphism

2010-05-12T19:08:33Z

Whitehead's Theorem states that a map between two connected CW complexes which induces isomorphisms on all homotopy groups is a homotopy equivalence. The naive reformulation which forgets to keep track of the specific isomorphisms (i.e. "Two connected CW complexes with all homotopy groups isomorphic are homotopy equivalent") is definitely false.

Answer by David Sprehn for Is there any geometry where the triangle inquality fails?

2010-05-01T20:30:36Z

There are a number of good answers from differential geometry, analysis, etc. But I suspect the questioner had axiomatic geometries in mind. So here's another take from that perspective.

There are two statements of the triangle inequality in plane geometry. (1) If A,B,C are noncollinear points, then $AC\lt AB+BC$; (2) If A,B,C are any three points, then $AC\leq AB+BC$.

In any system which includes a Ruler Postulate, (1) is stronger than (2). In neutral geometry (which includes all of the axioms of Euclidean geometry except the parallel postulate), statement (1) can be proven pretty directly from the SAS congruence postulate. (Does anyone know if they're equivalent in the presence of the other neutral geometry axioms?) Statement (2), on the other hand, is the one needed to show that the plane is a metric space, and is strictly weaker, as shown by the "taxicab geometry" mentioned above, which satisfies all of the postulates for neutral geometry except for SAS, and has property (2) but not property (1).

So, to summarize, the triangle inequality is true in neutral geometry, so any model of it (including the Euclidean and hyperbolic planes, etc.) will satisfy the triangle inequality. But of course we can consider weaker axiom systems in which models do not satisfy it (like taxicab).

Comment by David Sprehn

2011-03-03T18:38:31Z

Yes, that's a much quicker way to see it than the chain-level argument I had in mind. Of course spectral sequence for $X\to X/G\to BG$ will not collapse without some assumptions... if you want a totally general answer (i.e. with integer coefficients) there will be no way to avoid doing a spectral sequence. I pointed out in my answer (because I thought the questioner was thinking of it) that the spectral sequence for $G\to X\to X/G$ does collapse, but without giving any useful information -- sorry that wasn't clear.

Comment by David Sprehn

2011-03-02T05:14:00Z

Chris: the confusion here is that some people are interpreting $H^*(G)$ to be the singular cohomology of $G$, and others the group cohomology. I don't know which one the questioner intended (but group cohomology would make more sense).

Comment by David Sprehn

2011-03-02T05:11:26Z

Right, perfect! Of course, in this case (free action) the Borel construction is homotopy-equivalent to the orbit space $X/G$. A small quibble: it's not a principle bundle (the fiber is $X$, not $G$) but merely a fiber bundle with structure group $G$. Of course, it's still a fibration so we get a Serre spectral sequence as desired.

Comment by David Sprehn

2011-03-02T04:46:03Z

(added) The thing that goes wrong in that example is that 2, the order of $G$, is not a unit in the coefficient ring.

Comment by David Sprehn

2011-03-02T04:44:22Z

Careful, hypotheses are needed here (to show that `$H^*(X/G)$ is the $G$ invariants of $H^*(X)$ `). Example: take the covering space `$S^2\to\mathbb{R}P^2$` with integer coefficients. Then $H^2(X)=0$ but $H^2(X/G)\neq0$.

Comment by David Sprehn

2011-03-02T04:39:44Z

Also: I have a vague memory that there's another sequence you can use, in case you want the more general case (I haven't checked this!): If memory serves, there is a different fibration, $X\to X/G\to BG$. The latter map is the classifying map of the principle bundle $X\to X/G$. But studying this would give a relationship involving the group cohomology of $G$ (maybe that's what you meant in the question though....)

Comment by David Sprehn

2010-05-18T22:25:41Z

Also: absolutely anything involving vector bundles. :)