TITLE: Relating smooth concordance and homology cobordism via integral surgeries
QUESTION [12 upvotes]: Let $K_0$ and $ K_1$ be knots in $S^3$. They are called smoothly concordant if there is a smoothly properly embedded cylinder $S^1 \times [0,1]$ in $S^3 \times [0,1]$ such that $\partial (S^1 \times [0,1]) = -(K_0) \cup K_1$.
Let $Y_0$ and $ Y_1$ be integral homology spheres, i.e., $H_*(Y_i; \mathbb Z) = H_*(S^3; \mathbb Z)$. They are called homology cobordant if there exists a smooth compact oriented $4$-manifold such that $\partial X = -(Y_0) \cup Y_1$ and $H_*(X,Y_i; \mathbb Z)=0$ for $i=0,1$.
I cannot explicitly figure out but I made some progress. How can we concretely prove that the following well-known theorem: Let $S_n^3(K)$ denotes $3$-manifold obtained by the $n$-surgery on the knot $K$ in $S^3$.
Theorem: If $K_0$ is smoothly concordant to $K_1$ in $S^3$, then for all $n$, $S_n^3(K_0)$ is homology cobordant to $S_n^3(K_1)$.
Addition: Can we use this theorem to obtain "strong" obstructions for knots being smoothly concordant?

REPLY [3 votes]: I am not quite sure it is a “strong” obstruction but it is “nice” at least to me:
Observation: The left-handed trefoil and the right-handed trefoil are not smoothly concordant in $S^3$.
Let $K_0$ and $K_1$ respectively denote the left-handed trefoil and right-handed trefoil. Assume that $K_0$ and $K_1$ are smoothly concordant in $S^3$. Then by theorem, we know that $S^3_{-1}(K_0)$ and $S^3_{-1}(K_1)$ are homology cobordant.
Observe that $S^3_{-1}(K_0)$ is the Brieskorn sphere $\Sigma(2,3,5)$ while $S^3_{-1}(K_1)$ is the Brieskorn sphere $\Sigma(2,3,7)$. This can be done by Kirby calculus. For example, see Chapter 3 in Saveliev's book.
But Fintushel-Stern $R$-invariants of $\Sigma(2,3,5)$ and $\Sigma(2,3,7)$ are not same and Fintushel-Stern $R$-invariant provides a homology cobordism invariant. Hence we have reached a contradiction. It is worthy to note that this invariant can be easily computed due to Neumann-Zagier’s shortcut.
This conclusion also can be derived Ozsváth-Szabó $d$-invariant because $d(\Sigma(2,3,5))=-2$ and $d(\Sigma(2,3,7))=0$, see the example section in their paper. As Golla emphasized, this obstruction also comes from Frøyshov's $h$-invariant.
Further note: Let $\Theta^3_\mathbb Z$ denote integral homology cobordism group. It is the set of integral homology spheres modulo smooth homology cobordism. Then $d$- and $h$-invariants provide the following surjective group homomorphisms: $$d: \Theta^3_\mathbb Z \to 2 \mathbb Z,\ \ \ \ \ \ \ \ h: \Theta^3_\mathbb Z \to \mathbb Z.$$