Relating smooth concordance and homology cobordism via integral surgeries

Question

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?

Answer 1

I will call $X_n(K)$ the trace of $n$-surgery along $K$, that is a 4-manifold diffeomorphic to the union of $B^4$ and an $n$-framed 2-handle attached along $K \subset S^3 = \partial B^4$.

Call $A \subset S^3 \times I$ the concordance from $K_0$ to $K_1$. Consider $X_1 := X_n(K_1)$, viewed as $B^4 \cup S^3\times I \cup H$, where $H$ is the 2-handle. For convenience, I will call $C$ the core of $H$. I claim that $X_n(K_0)$ embeds in $X_n(K_1)$ as a regular neighbourhood, that I'll call $X_0$, of $B^4 \cup A \cup C$. This is because a regular neighbourhood of $A \cup C$ (which is a disc) is just a 2-handle $H'$; the framing along which $H'$ is attached is determined by the intersection form, and is bound to be $n$.

Now the second claim is that $W := X_1 \setminus {\rm Int\,} X_0$ is an integral homology cobordism from $Y_0 := S^3_n(K_0)$ to $Y_1 := S^3_n(K_1)$. I will use excision, which tells us that $H_i(W, Y) = H_i(X_1, X_0)$ for each $i$. Since $H_i(X_0) = H_i(X_1)$ is trivial when $i \neq 0,2$, and since at the level of $H_0$ nothing really happens, we only need to look at $H_2$.

Now, $H_2(X_0)$ is generated by a class represented by a Seifert surface for $K_0$ capped with the core of the 2-handle, that is $A \cup C$. This surface intersects geometrically the co-core $D$ of the 2-handle $H$ of $X_1$ once (since this intersection takes place in $H$, it's exactly $D\cap C$, which is one point), so the generator of $H_2(X_0)\simeq \mathbb Z$ is sent to a generator of $H_2(X_1) \simeq \mathbb Z$. It follows that the relative homology is trivial, as we wanted to show.

As for the addition: any integral homology cobordism invariant now gives a wealth of knot invariants. The Rokhlin invariant, for instance, gives you the concordance invariance of the Arf invariant. I am very partial to Heegaard Floer homology, so correction terms there give you a wealth of concordance invariants. (It should be pointed out that correction terms in Heegaard Floer homology were inspired by work of Frøyshov in Seiberg–Witten theory.)

Answer 2

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: $$d: \Theta^3_\mathbb Z \to 2 \mathbb Z,\ \ \ \ \ \ \ \ h: \Theta^3_\mathbb Z \to \mathbb Z.$$