It is clear that if $K_1$ and $K_2$ are two concordant knots by a concordance that only present ambient isotopic phenomena (no saddles, maxima, or minima) they are invertible concordant from both ends. I conjecture that this could also be a necessary condition, that is, if two knots $K_1$ and $K_2$ are invertible concordant from both ends it should exist a concordance $C$ between $K_1$ and $K_2$ that does not present saddles, maxima, or minima. Is this true? Thanks in advanced!
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$\begingroup$ What does your "invertible concordant from both ends" terminology refer to? $\endgroup$– Ryan BudneyCommented Oct 16 at 15:01
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$\begingroup$ My guess as to what your question is asking is that if you have concordences C_1 and C_2 from K_1 to K_2 and K_2 to K_1 respectively such that if you stack them $C_1 * C_2$ it is isotopic to an isotopy, similarly $C_2 * C_1$ is isotopic to an isotopy, then you are trying to conclude that $K_1$ and $K_2$ are isotopic. $\endgroup$– Ryan BudneyCommented Oct 16 at 17:11
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$\begingroup$ By invertible concordant from both ends I mean that there exists concordances C_1 from K_1 to K_2 and C_2 and C_3 from K_2 to K_1 such that C_1 U C_2 is homeomorphic to K_1 x [0,1] and C_3 U C_1 is homeomorphic to K_2 x [0,1]. $\endgroup$– jampCommented Oct 16 at 18:01
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1$\begingroup$ Yes, this is true. Invertibility of a concordance from $K_1$ to $K_2$ at $K_2$ implies that $K_2$ dominates $K_1$ in a sense that there is a one-degree map $\mathbb S^3\setminus K_2 \to \mathbb S^3\setminus K_1$ which maps boundary to boundary (see Agol's answer mathoverflow.net/questions/462336/inverse-of-a-smooth-concordance-of-smooth-knots for details). But domination is a partial order, see arxiv.org/abs/1511.07073 for example. So invertibility at both ends implies that the knots are the same. $\endgroup$– Maxim PrasolovCommented Oct 16 at 22:42
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$\begingroup$ I suppose I don't understand your definition. Invertible concordant would appear to be a tautology, i.e. any two concordent knots are invertible concordent with that definition. $\endgroup$– Ryan BudneyCommented Oct 16 at 22:52
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2 Answers
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No, this is not true: any doubly-slice knots gives a counterexample to your statement.
A knot $K$ is doubly-slice if there is an unknotted 2-sphere $F$ in $S^4$ such that $(S^3,K)$ is diffeomorphic to $(S^3_{\rm eq}, F\cap S^3_{\rm eq})$, where $S^3_{\rm eq}$ is the equatorial 3-sphere (and we assume that $F$ and $S^3_{\rm eq}$ intersect transversely).
By removing two small balls centered at points $F$ in the two hemispheres in which $S^3_{\rm eq}$ divides $S^4$, by definition we have the trivial cobordism from the unknot to itself, which decomposes into two cobordisms, one from the unknot to $K$, and the other from $K$ to the unknot.
To give a negative answer to your question, it only remains to show that there exist non-trivial doubly-slice knots: $9_{46}$ (in Rolfsen's knot table---see knotinfo) is the smallest one, but more generally $J \# {-J}$ is doubly-slice for any knot $J$ (I know of a proof by Zeeman, but maybe this was known earlier). (Here $-J$ denotes the mirror of $J$ with its orientation reversed.)
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$\begingroup$ Thanks a lot for your answer. But, If I understood well your example, you have only an invertible concordance, not from both ends. What makes me believe this statement is true is the proved fact that if two knots are invertible concordant from both ends, they are equivalent. $\endgroup$– jampCommented Oct 16 at 12:54
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$\begingroup$ Oh, sorry—I misunderstood your question. $\endgroup$ Commented Oct 16 at 13:13
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Let $C$ denote a concordance from $K_1$ to $K_2$ in $S^3\times [0,1]$, and let $C_1, C_2$ be concordances from $K_2$ to $K_1$. Let $C_1 \cdot C \sim K_2 \times [0,1]$, $C\cdot C_2\sim K_1\times [0,1]$ (so $C_1$ is a left inverse and $C_2$ a right inverse of $C$). Then we see that $$C_1\cdot C \cdot C_2 \sim C_1\cdot (C \cdot C_2)\sim C_1 \sim (C_1\cdot C ) \cdot C_2 \sim C_2,$$ so $C_1 \sim C_2$. Let $C’$ denote the open concordance $C\cap S^3 \times [0,1)$.
Now one may perform the Mazur Swindle to show that $C’$ is a product:
$$C\cdot C_1\cdot C \cdot C_1 \cdot C \cdots =C\cdot (C_1\cdot C) \cdot (C_1 \cdot C) \cdots \sim C’ \sim (C\cdot C_1) \cdot (C \cdot C_1) \cdot (C\cdot C_1) \cdots \sim K_1\times [0,1). $$
I think it should be true as well that $C \sim K_1\times [0,1]$ in the topological category, but I haven’t through it through carefully.
