Relating smooth concordance and homology cobordism via integral surgeries

MathOverflow Asked by Maxim Ulyanov on August 2, 2020

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 $$B^4 cup S^3times I cup H$$, 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 $$Dcap C$$, 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.)

Correct answer by Marco Golla on August 2, 2020

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

Answered by Oğuz Şavk on August 2, 2020