# Is spin cobordism an invariant for surgery of codimension $q\ge3$?

Recall that a surgery of codimension $$q$$ on an $$n$$-manifold $$X$$ is a modification of $$X$$ of the following type. Let $$\Sigma^{n-q}\subset X$$ be a smoothly embedded $$(n-q)$$-sphere with a trivialized tubular neighborhood. By this we mean a neighborhood $$U$$ of $$\Sigma^{n-q}$$ together with a diffeomorphism $$f:U\to S^{n-q}\times D^q$$ such that $$f(\Sigma^{n-q})=S^{n-q}\times\{0\}$$. The surgery operation now consists of removing the neighborhood $$U\cong S^{n-q}\times D^q$$ and replacing it with the product $$D^{n-q+1}\times S^{q-1}$$ by gluing in the obvious, canonical way along the boundary $$S^{n-q}\times S^{q-1}$$.

Gromov-Lawson and Schoen-Yau proved that if $$N$$ can be obtained from $$M$$ by performing surgery of codimension $$q\ge3$$ and $$M$$ carries a metric of positive scalar curvature, then $$N$$ also carries a metric of positive scalar curvature.

Question: Is spin cobordism an invariant for surgery of codimension $$q\ge3$$? Two directions:

1. Both $$M$$ and $$N$$ are spin. If $$N$$ can be obtained from $$M$$ by performing surgery of codimension $$q\ge3$$, are $$M$$ and $$N$$ spin cobordant?

2. Both $$M$$ and $$N$$ are spin. If $$M$$ and $$N$$ are spin cobordant, can $$N$$ be obtained from $$M$$ by performing surgery of codimension $$q\ge3$$?

## 1 Answer

1. Yes. Consider the trace $$tr$$ of the surgery: Take $$D^{n-q+1}\times D^q$$ with boundary $$\partial(D^{n-q+1}\times D^q) = (S^{n-q}\times D^q)\cup (D^{n-q+1}\times S^{q-1})$$ and glue the $$(S^{n-q}\times D^q)$$-part of the boundary to the upper boundary of $$M\times [0,1]$$ along the identification $$f$$. This is a cobordism from $$M$$ to $$N$$. Since the codimension of the surgery is bigger than $$3$$, the inclusion of $$N$$ into the trace is $$2$$-connected (by general position), so the induced map $$H_2(N;\mathbb Z/2)\to H_2(tr;\mathbb Z/2)$$ is surjective. By the universal coefficient theorem for cohomology, the map $$H^2(tr;\mathbb Z/2)\to H^2(N;\mathbb Z/2)$$ is injective. The second Stiefel--Whitney of $$tr$$ is in the kernel of this map since $$N$$ is Spin and hence $$tr$$ is spin.

2. No. A counterexample is the following: the sphere $$S^n$$ and the torus $$T^n$$ are both spin null-cobordant, in particular they are spin-cobordant. $$S^n$$ admits positive scalar curvature, but $$T^n$$ does not, so it cannot be obtained from $$S^n$$ by surgeries of codimension at least $$3$$. (Which is of course an overkill proof.)

If you additionally assume that $$N$$ is simply connected and the dimension is large enough ($$\ge5$$), then the answer to question $$2$$ is yes. For arbitrary spin manifolds (of dimension at least 5), the answer is yes if you look at $$spin\times B\pi_1(N)$$-cobordism. This is a standard argument in the study of positive scalar curvature metrics and follows from the handle cancellation lemma from the $$h$$- or $$s$$-cobordism theorem (cf. Appendix of https://arxiv.org/pdf/1311.3164.pdf or Proposition 6.3 of https://arxiv.org/pdf/1807.06311.pdf)