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Is spin cobordism an invariant for surgery of codimension $n$?

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 3 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 3? Two directions:

  1. Both $M$ and $N$ are spin. If $N$ can be obtained from $M$ by performing surgery of codimension 3, 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 3?

wonderich
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