Homology of a Link Complement

Question

Setup: Let $M$ be a closed orientable connected 3-manifold and $L=\bigsqcup_{i=1}^{n} L_i$ a nullhomologous link in $M$. Let $V_i\cong S^1\times D^2$ be pairwise disjoint regular neighborhoods for the link components $L_i$ in $M$. For each $i$ let $\mu_i$ denote the meridian on $\partial V_i$, that is the curve that bounds a disk in $V_i$ but not in $\partial V_i$. Let $A=M\setminus (\cup_{i=1}^{n} int(V_i))$ denote the link complement.

Question: Can one show that the homology classes $\{[\mu_i] | 1\leq i\leq n \}$ are linearly independent in $H_1(A;\mathbb{R})$?

If $M$ is a homology sphere, this can be shown using the Mayor-Vietoris Sequence with subspaces $A=M\setminus (\cup_{i=1}^{n} int(V_i))$ and $B=\cup_{i=1}^{n} V_i$, however it is unclear to me that this technique alone is enough to answer the question for more general $M$. Any hints or references would be appreciated. Thanks!

Answer 1

I am assuming "nullhomologous link" means that every component is $\mathbb{R}$-nullhomologous in $M$.

The "half lives, half dies" principle is that the image of the boundary map $H_2(A,\partial A)\to H_1(\partial A)$ has free rank equal to half the free rank of $H_1(\partial A)$ (according to Hatcher's $3$-manifold notes, Lemma 3.5). Equivalently, using the long exact sequence, that the kernel of $H_1(\partial A)\to H_1(A)$ has free rank equal to half the free rank of $H_1(\partial A)$. Since you are working with $\mathbb{R}$ coefficients, free rank is dimension (we could work with $\mathbb{Q}$ coefficients to the same effect).

Let $A$ be the complement of an open regular neighborhood $\bigcup_{i=1}^n \operatorname{int}(V_i)$ of $L$ in $M$. Since $\partial A$ is a disjoint union of $n$ tori, $H_1(\partial A;\mathbb{R})$ has dimension $2n$. By assumption, each link component $L_i$ has a well-defined longitude curve $\lambda_i$ contained within $\partial V_i$ that is nullhomologous when included into $M-\operatorname{int}(V_i)$.

Consider a properly embedded surface $S_i\subset M-\operatorname{int}(V_i)$ whose boundary is $\lambda_i$, which exists since $\lambda_i$ is nullhomologous. We may assume that $S_i$ intersects the rest of $L$ transversely, and so we get a properly embedded surface $S_i'=S_i\cap A$ in $A$ that represents $[\lambda_i]$ as a linear combination of the meridians.

Thus, $\iota_*:H_1(\partial A;\mathbb{R})\to H_1(A;\mathbb{R})$ sends all the longitudes to linear combinations of meridians. These are now the facts:

• $H_1(\partial A;\mathbb{R})$ has $[\mu_1],[\lambda_1],\dots,[\mu_n],[\lambda_n]$ as a basis.
• $[\mu_1],\dots,[\mu_n]$ span the image of the $\iota_*$.
• $\dim\ker\iota_*=n$.

It follows by some linear algebra that $[\mu_1],\dots,[\mu_n]$ are a linearly independent in $H_1(A;\mathbb{R})$.