# Relating smooth concordance and homology cobordism via integral surgeries

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?

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.)

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