When introducing topologically slice knots (i.e. knots $K\subset S^3=\partial D^4$ which bound a locally flat disc in $D^4$) one explains the local flatness condition by noticing that without local flatness, every knot would be topologically slice, since any knot bounds a disc in $D^4$ in the following way:
- Consider the "coned knot" $C(K)=K\times [0,1] /\sim K \times 0$ embedded into $D^4$ via $[(x,t)]\mapsto x\cdot t$. It is obviously homeomorphic to a disc, and the map defines an embedding, but the embedding is not locally flat at its cone point.
I call a topologically embedded surface (possibly with boundary) $\Sigma \subset D^4$ locally flat if:
- for every $x\in \operatorname{int}(\Sigma)$ there is a closed neighborhood $U\subset D^4$ such that $(U,U\cap \Sigma)$ and $(D^4,D^2)$ are homeomorphic as pairs and for every $x\in \partial \Sigma$, there is a neighborhood $U$ such that $(U,U\cap \Sigma)$ is homeomorphic to ($D^4,D^2$).
Somehow I am stuck trying to really finish the proof of why this embedding is not locally flat at its cone point. I am of course aware of the intuition that it doesnt look like the standard embedding there etc. but when trying to show exactly where the definition fails I get stuck. I think I can show that every ball of radius $r$ does not satisfy the condition, since its boundary will be a non-trivial knot. But I need to show this for every neighborhood right? And if a neighborhood $U$ satisfies the locally flat condition, I do not see how every neighborhood contained in it satisfies it too. Because this would allow me to produce a contradiction, since any neighborhood contains a ball of radius $r$ for some small enough $r$. I know this maybe sounds a bit trivial, but I would really appreciate a precise explanation for why this definition fails. I have also read somewhere that there is an argument one can make using the fact that the fundamental group of a knot complement is trivial iff it is the unknot to show that the cone of a knot is locally flat iff it is the unknot.
Thank you for your time.
Progress:
I can show that any ball of radius $r$ around the cone point does not satisfy the local flatness condition I described. One can do this by restricting the homeomorphisms between the pairs to obtain that $S^3\setminus S^1$ is homeomorphic to $\partial U_r(0) \setminus \partial( U_r(0)\cap C(K))$. The first space is the complement of the unknot and the second is the complement of K. These spaces are not homeomorphic since the unknot and K are not isotopic. This follows from the Gordon-Luecke theorem saying that two knots are homeomorphic if and only if their complements are.
I cannot for the moment however, show that any neighborhood of the cone point does not satisfy the local flatness condition. My idea here would to use the fact that any such neighborhood would contain a ball of radius $r$ (which does not satisfy the local flatness condition) and somehow lead this to a contradiction. But I was not successful in doing that.
Best Answer
Let $K\subset S^3$ be a nontrivial tame knot with the knot complement $E_K=S^3-K$. Then the knot group $\pi_1(E_K)$ is noncyclic (this is a consequence of the Dehn Lemma/Loop Theorem). Let $C_K\subset B^4$ denote the cone over $K$ (with the apex equal to the center $0$ of the ball $B^4$). Then $E^4_K:= B^4 -C(K)$ is homeomorphic to $E_K\times [0,1)$, hence, $\pi_1(E^4_K)\cong \pi_1(E_K)$ is noncyclic. The same homeomorphism $E^4_K\to E_K\times [0,1)$ implies that that for arbitrarily neighborhood $U$ of $0$, the induced map $$ \pi_1(U\cap E^4_K)\to \pi_1(E^4_K) $$
is surjective.
Suppose now that the disk $C_K$ is (locally) flat at $0$, i.e. there exists a neighborhood $U$ of $0$ such that the pair $(U, U\cap C_K)$ is homeomorphic to $({\mathbb R}^4, {\mathbb R}^2)$. Then $\pi_1(U\setminus C_K)= \pi_1(U\cap E^4_K) \cong {\mathbb Z}$, is cyclic. Since the image of a cyclic group under a homomorphism is again cyclic, we obtain a contradiction with the observations in the 1st pagraph.
Note that this proof requires tameness of the knot $K$.