What kind of geometric operations "scale up" cohomology? - MathOverflow

What kind of geometric operations "scale up" cohomology?

2009-10-29T03:51:33Z

There's an obvious operation on the category of graded rings, given by "scaling up," multiplying the grading of every element by some fixed constant.

Does anyone know of an operation on the level of spaces that will do this to the cohomology rings?

Answer by Reid Barton for What kind of geometric operations "scale up" cohomology?

2009-10-29T05:22:26Z

Even if you don't ask for the ring structure to be preserved, this seems quite unlikely, at least if you require the functor to do the same "scaling up" operation to maps on cohomology induced by maps of spaces. I believe all the Goodwillie derivatives of such a functor would be zero, whereas "geometric" operations typically have some nontrivial radius of analyticity.

Edit: For the argument I was thinking of, I need to assume that π₁(F(any sufficiently highly connected space)) = 0, so that for instance F(•) = •.

Answer by Urs Schreiber for What kind of geometric operations "scale up" cohomology?

2009-10-29T07:12:27Z

The scaling up that this is about is the operation of the shift functor in a triangulated category I suppose?

The archetypical geometric example of this would be the shift functor acting on spectra.

(But possibly I am misunderstanding the answer?)

Answer by Andrew Stacey for What kind of geometric operations "scale up" cohomology?

2009-10-29T08:07:55Z

There isn't one. The scaling operation that you describe (which would have to be an odd scale factor, by the way) doesn't commute with suspension. Cohomology is a "suspension-preserving" functor (there's probably some fancy way of saying this using triangulated categories) so when you consider a cohomology theory you should really think of the target as "graded algebras over the coefficient ring with a suspension operation" only that's not quite as snappy as "graded rings" so we tend to be a bit sloppy about saying it.

Edit: This answer was a bit of a "throwaway" answer since the question felt like idle speculation (I'm a bit embarrassed that it got so many votes!). With a little more thought I'd've concentrated on the fact that the target category isn't graded rings, or even graded algebras, or even graded algebras with suspension, but there's the action of the cohomology operations to take into account. Simply regrading everything will completely mess up those actions. Suspensions will come into play, though, when one considers maps between spaces and I suspect that one could show that all such maps were null-homologous.

I like Tyler's answer best. I'm voting for that one, and I recommend it be accepted as the answer.

Answer by James Griffin for What kind of geometric operations "scale up" cohomology?

2009-10-29T12:32:49Z

To get from spaces to graded rings you go via dg-rings. That is,

Spaces → dg-Rings → graded Rings.

where the first arrow is the functor taking simplicial cochains, and the second is the functor taking cohomology. So any "scaling up" of spaces should exist in dg-Rings. But how do you scale up a cochain complex? You can't because because that would break the degree of the codifferential.

So I don't think that there is a sensible "scaling up" for spaces.

Nice idea though, I don't know how you could classify functors that act on graded rings that can be lifted to spaces. An obvious example might be that of setting all groups of degree greater than i say to 0. That would correspond to filling in things in the space to kill the cohomology groups (I don't know the correct terminology off the top of my head).

Answer by Tyler Lawson for What kind of geometric operations "scale up" cohomology?

2009-10-29T17:16:25Z

There is no such operation. Here is one reason why.

Take your space to be CP². Its cohomology ring is a truncated polynomial algebra ℤ[x]/(x³), where x is an element in degree 2.

Suppose we had such a shift functor S on spaces that multiplies cohomology degrees out by n. Then S(CP²) would have mod-2 cohomology ring a truncated polynomial algebra ℤ/2[y]/(y^3) on a generator y in degree 2n.

This violates the Hopf invariant 1 conjecture, proved by Adams. There are no spaces whose cohomology is such a truncated polynomial ring ℤ/2[y]/(y^3) on a generator in degree k unless k is 1 (real projective 2-space), 2 (complex projective 2-space), 4 (quaternionic projective 2-space), or 8 (something associated to the octonions that I feel uncomfortable describing as projective space).

If n is not 2 or 4, you immediately get a contradiction here. If n is 2 or 4, you could apply your shift operator multiple times and get a contradiction.

Answer by Charles Rezk for What kind of geometric operations "scale up" cohomology?

2009-10-29T18:37:54Z

How about we change the question to this: for what kind of spaces X are there a family of continuous maps ψ^k: X->X which act on cohomology by "scaling up"? Here is a famous and bewildering example of something sort of like this:

As you can read in the notes of Sullivan's 1970 MIT course (see esp. chapter 5), if you have an algebraic variety X defined over Q(the rationals), then you get an action of G(=absolute Galois group of Q) on a space X_et=the etale homotopy type of X.

If X is a Grassmanian variety, then

* X_et has the homotopy type of the usual complex Grassmannian (up to a "profinite completion"),

* G acts on the profinite cohomology of X_et through its abelianization G_ab=Z^*(=the units of the profinite integers),

* this action of G_ab on cohomology is by scaling.

I don't know what the motivation of the original question is, so I don't know if this kind of thing has any relation for what you want. (I really posted this to advertise the Sullivan notes, which everybody should read.)