5
$\begingroup$

I've recently encountered the following cobordism theory modulated by a class $\sigma \in H^{d+1}(B^2\mathbb{Z}/2,U(1))$.

My objects are $d$-dimensional spin manifolds with chosen spin structure. Note that a spin structure, thought of as a null-homotopy of $w_2$, gives a trivialization of $w_2^*\sigma$.

My morphisms are given by cobordisms by manifolds with $w_2^*\sigma = 0$ for which the trivializations of $w_2^*\sigma$ on the boundary components given by the chosen spin structures there all come from restricting a single trivialization of $w_2^*\sigma$ on the cobordism.

Call the resulting group $\Omega^{{\rm spin},\sigma}_d$. There is a natural surjective map $\Omega^{\rm spin}_d \to \Omega^{{\rm spin},\sigma}_d$. I'd like to calculate the kernel of this map, at least for $d=1,2,3,4,5$.

There is another map $\Omega^{{\rm spin},\sigma}_d \to \Omega^{\rm SO}_d$ forgetting all the trivialization business. The composition $\Omega^{\rm spin}_d \to \Omega^{{\rm spin},\sigma}_d \to \Omega^{\rm SO}_d$ is known to have kernel in degrees $8k+1,8k+2$ with rather explicit generators. For $k=0$, these are the circle with antiperiodic spin structure and its Cartesian square, respectively. The kernel we're after is contained in this one, so to do the calculation in the range I'm interested in, it might be possible to do things case by case.

Let's do the circle case. Trivializations of $w_2^*\sigma$ on $S^1$ form a torsor over $H^1(S^1,U(1))$. Trivializations on a surface $\Sigma$ bounding $S^1$ are a torsor over $H^1(\Sigma,U(1))$. Since the class of $\partial \Sigma$ in $\Sigma$ is zero, by universal coefficients the map $H^1(\Sigma,U(1)) \to H^1(\partial\Sigma,U(1))$ is zero. Thus, any two trivializations of $w_2^*\sigma$ on $\Sigma$ restrict to the same trivialization on the boundary. Thus, the two spin structures of the circle are distinct in $\Omega^{{\rm spin},\sigma}_d$ when $\sigma \neq 0$ since in this case the spin structures actually give different trivializations of $w_2^*\sigma$ (as can be checked from the long exact sequence coming from $\sigma:\mathbb{Z}/2 \to U(1)$).

When $\sigma = 0$, then we get the full kernel. This is the case for $d=2$ since $H^3(B^2\mathbb{Z}/2,U(1)) = 0$, but this is the last degree where this cohomology vanishes.

This finishes the calculation in the range I originally considered, so how about the general case?

The next case of interest is $d=9$, for which $H^{10}(B^2\mathbb{Z}/2,U(1))$ is nontrivial but indescribable to me.

A first subproblem for the general attack might be ``when is a cohomology operation injective?" I don't know the answer to this either, except in the case $d=1$ when $\sigma:H^2(-,\mathbb{Z}/2) \to H^2(-,U(1))$ is given by a homomorphism $\mathbb{Z}/2 \to U(1)$ and we have a long exact sequence where the kernel of $\sigma$ is the image of $H^1(-,U(1))$.

Any help or references would be much appreciated.

$\endgroup$
  • $\begingroup$ You have a $B^2$ in the first sentence; is that intended? $\endgroup$ – Mariano Suárez-Álvarez Nov 26 '13 at 5:32
  • $\begingroup$ Yes, since $w_2$ is a degree 2 class it should be a map to the twice delooping of $\mathbb{Z}/2$. $\endgroup$ – Ryan Thorngren Nov 26 '13 at 5:48
  • $\begingroup$ What do you mean by a trivialization of a cohomology class? (Or, how do you view $w_2^\ast\sigma$ as a bundle?) $\endgroup$ – Mark Grant Nov 26 '13 at 7:04
  • 1
    $\begingroup$ A trivialization of $w_2^*\sigma$ is a $U(1)$-cochain whose differential is $w_2^*\sigma$. It's also the same as a null-homotopy of the composite $w_2 \circ \sigma$. $\endgroup$ – Ryan Thorngren Nov 26 '13 at 18:07
  • $\begingroup$ @MarianoSuárez-Alvarez, see math.stackexchange.com/questions/582841/… $\endgroup$ – Will Jagy Nov 27 '13 at 6:28
5
$\begingroup$

To bring this into a classical setting, it seems to me that one can use the bordism groups $\Omega_d^{\sigma}$ where both d-manifolds and (d+1)-bordisms are not necessarily spin, but have only a trivialization of $w_2^*\sigma$. Then there is a natural forgetful map $\Omega_d^{Spin}\to \Omega_d^{\sigma}$ with image $\Omega_d^{spin,\sigma}$. This fits into a long exact sequence $$\dots \to \Omega_{d+1}^{\sigma,spin}\to \Omega_d^{Spin}\to \Omega_d^{\sigma}\to \dots$$ where $\Omega_{d+1}^{\sigma,spin}$ is by definition the bordism group of $(d+1)$-manifolds with boundary equipped with a trivialization of $w_2^*\sigma$, and, on the boundary, a refinement to a spin structure.

For computations, this all translates into homotopy groups of Thom spectra, in particular the Thom spectrum for $\Omega_*^{\sigma}$ is the one for $F\to BO$, where $F$ is the homotopy fiber for $$ \sigma\circ w_2: BSO\to K(\mathbb Z/2,2)\to K(\mathbb Z,d+2).$$

Stong's book is a reference for this. For computations the Adams spectral sequence might be helpful.

$\endgroup$
  • $\begingroup$ Thanks for your answer, nsrt. I am still thinking about it. When you say the bordism group of $(d+1)$-manifolds with boundary, do you mean that a bordism between $(X,A)$ and $(Y,B)$ is a manifold $M$ with boundary $N$ containing $X$ and $Y$ so that $N-(X\cup Y)$ is a bordism from $A$ to $B$? $\endgroup$ – Ryan Thorngren Dec 9 '13 at 23:40
  • $\begingroup$ Yes, this is exactly what a bordism of manifolds with boundary is. $\endgroup$ – nsrt Dec 10 '13 at 12:16
  • $\begingroup$ Okay great. How about the right hand continuation of the sequence above? Is $\Omega^{\sigma,spin}_d$ a cobordism group of $d$-manifolds with boundary? $\endgroup$ – Ryan Thorngren Dec 10 '13 at 21:27
  • $\begingroup$ Yes. Again: you should really get a copy of: Stong, Robert E. Notes on cobordism theory. $\endgroup$ – nsrt Dec 11 '13 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.