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

Question

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 $q\ge3$?

Answer 1

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)