# Is the connected sum of knots an isometry?

Take $X$ as the set of knots in the 3-sphere (i.e. smooth embeddings of $S^1$ in $S^3$ up to smooth isotopy), endowed with the Gordian distance $d$.

For a fixed knot $K$ we can define the map $\varphi_K : X \rightarrow X$ as $\varphi_K (K^\prime) = K^\prime \sharp K$ .

It is easy to show that $\forall K \in X$ we have $d(K_1, K_2) \ge d(\varphi_K (K_1) , \varphi_K (K_2))$.

My question is: does the equality hold? Or in other words is the map $\varphi_K$ and isometry on $X$ with respect to $d$?

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What is this distance between knots ? (pure curiosity) –  Thomas Richard Apr 21 '13 at 16:45
@Thomas: Distance between knots is the least number of crossings (in a knot diagram) one needs to change in order to transform one knot to the other. For the question itself, it is not even known if crossing number is additive (see arxiv.org/abs/0805.4706 for the best result so far); distance-preservation would be much harder to prove. Maybe, though, a counter-example would be easier. On the other hand, Lackenby's result suggests to look for questions on "coarse geometry" of the "knot space X" e.g., if X is coarsely Euclidean or hyperbolic. –  Misha Apr 21 '13 at 17:37