Let $L\subset L'\in S^3$ be two links such that $L$ has one less number of components than $L'$. Further, $L$ is hyperbolic. Under what conditions is the link $L'$ hyperbolic. To be more specific $L, L'$ are shown in here.
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$\begingroup$ better embed the image in the post? $\endgroup$– YCorCommented Dec 13, 2019 at 22:18
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$\begingroup$ The way you drew the image, I think, you can most certainly create some incompressible torus. $\endgroup$– Anubhav MukherjeeCommented Dec 13, 2019 at 22:20
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$\begingroup$ As it seems, if it is linked with the rest of the picture, then it cannot create any non-trivial sphere. So if it doesn't create any incompressible torus, then it will be a hyperbolic link. $\endgroup$– Anubhav MukherjeeCommented Dec 13, 2019 at 22:24
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1 Answer
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There is not enough information in your picture to give a definitive answer. If you have more details about the link $L$ then an answer will probably factor through Thurston's characterisation of hyperbolic links. Purcell's book is an introduction to the subject, and the second half of Theorem 8.17 therein is the precise statement you need.
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$\begingroup$ Is it possible to say L' is hyperbolic or not given L is hyperbolic and a ribbon link? Again links L and L' are related as shown in the figure in the original question. Also, I know L is hyperbolic based on the fact that it is arborescent. Thank you! $\endgroup$ Commented Dec 14, 2019 at 19:30
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1$\begingroup$ No. It is not possible to deduce hyperbolicity of L' from properties of L. For, pull a bight out of L, tie a knot in the bight (en.wikipedia.org/wiki/Bight_(knot)#In_the_bight), and then use the green component (as in your figure) to link the bight to the rest of L. Then L' will be toroidal. $\endgroup$– Sam NeadCommented Dec 16, 2019 at 8:56