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Say we have a 2-component link $L$ with components $L_1$ and $L_2$. Are there known conditions that will ensure that there exists a Seifert surface $S$ of $L_1$ with curves $\alpha_1,\beta_1,...,\alpha_g,\beta_g$ representing a basis for $H_1(S)$ such that $lk(\alpha_i,L_2)=0=lk(\beta_i,L_2)$ for all $i=1,...,g$.

(eg. Vanishing linking number, Milnor invariants, etc. on $L$?)

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I'm presuming that you mean an orientable surface $\Sigma$ in $S^3$ whose boundary is $L$. Since a basis of $H_1(\Sigma)$ has trivial linking number with $L_2$, we have $L_1$ and $L_2$ have trivial linking number.

One case in which this holds is if $L_1$ and $L_2$ bound disjoint Seifert surfaces $\Sigma_i$ (i.e. $L$ is a boundary link), then one can take $\Sigma$ to be $\Sigma_1\cup \Sigma_2$ (add a little tube if you'd like it to be connected).

If one takes the infinite cyclic cover over $S^3-L_2$, then your condition implies that the preimages of $L_1$ in this cover have pairwise trivial linking numbers. It's possible that this condition is equivalent to your condition, but it is certainly necessary.

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  • $\begingroup$ Thanks! I edited the question to make it clearer. I meant $S$ has boundary $L_1$ only. $\endgroup$ – Anthony Bosman Jan 17 '17 at 2:07
  • $\begingroup$ @AnthonyBosman: Okay, well a boundary link still works. I guess with this formulation, the preimage of $L_1$ in the infinite cyclic cover of $S^3-L_2$ should be a (infinite) boundary link. I think this might be equivalent to your condition, but not sure if it's what you're looking for. $\endgroup$ – Ian Agol Jan 17 '17 at 2:24
  • $\begingroup$ Great! I absolutely agree that $L$ a boundary link is a sufficient condition. It seems $lk(L_1,L_2)=0$ is not a sufficient condition. I'd really like to know if the Milnor invariants of $L$ vanishing is a sufficient condition (note, this is weaker than $L$ being a boundary link). $\endgroup$ – Anthony Bosman Jan 17 '17 at 2:59

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