Link cobordism and the intersection of slice discs
classification
🧮 math.GT
keywords
respectivelysliceboundsdiscfollowingintersectionlinkball
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It is well-known that all 2-knots are slice. Are all 2-links slice? This is an outstanding open question. In this paper we prove the following: For any 2-component 2-link (J,K)in the 4-sphere which bounds the 5-ball B^5, there is an embedded disc 2-disc D^2_J (respectively, D^2_K) in B^5 with the following properties: J (respectively K) bounds D^2_J (respectively, D^2_K). D^2_J and D^2_K intersect transversely. the intersection of D^2_J and D^2_K in D^2_J (respectively, D^2_K) is a trivial 1-knot.
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