Rank of the 2-cover space of a knot space

Question

It is well known that the order of the first homology group of the double branched cover of a knot $K$ in $S^3$ is the determinant of $K$, i.e. $\lvert H_1(B_2(K),\mathbb{Z}) \rvert = \det (K)$. However, I did not find a formula for its rank. Does anyone know the answer to this question?

Answer 1

Edit: Thanks to Sucharit Sarkar for pointing out I was spreading misinformation. This replaces my previous answer completely.

The first homology of the double branched cover of a knot in $S^3$ is a finite abelian group. Since it has no free part, its rank is $0$.

Presumably you meant the minimal number of generators, which is much more interesting. I'm aware of some things you can say in terms of Fitting/elementary ideals, where the Alexander polynomial is the GCD of the zeroth one.

By considering the relationships between double branched covers $B_2(K)$, two-fold covers of the knot complement $X_2(K)$ and the infinite cyclic cover of the knot complement $X_\infty(K)$, it is possible to work out with some long exact sequences that $$H_1(B_2(K)) \cong \frac{H_1(X_\infty(K))}{(t^2-1)H_1(X_\infty(K))}$$ where $\mathbb{Z}[t^{\pm 1}]$ acts on $H_1(X_\infty(K))$ through deck transformations, with $t$ being the meridian generator of $H_1(S^3-K)$.

Now, a fact about $H_1(X_\infty(K))$ is that $1-t$ acts on it invertibly (explained by Milnor in "Infinite cyclic coverings"). Thus, $$H_1(B_2(K)) \cong \frac{H_1(X_\infty(K))}{(t+1)H_1(X_\infty(K))}.$$ So, if you take a presentation matrix for $H_1(X_\infty(K))$ as a $\mathbb{Z}[t^{\pm 1}]$-module, you can substitute in $t=-1$ to get a presentation matrix for $H_1(B_2(K))$ as a $\mathbb{Z}$-module. (A Goeritz matrix is an example of such a presentation matrix.)

This is why $\lvert H_1(B_2(K))\rvert = \lvert\Delta_K(-1)\rvert$. There is a square presentation matrix for $H_1(X_\infty(K))$ from the Seifert form, and the Alexander polynomial is the determinant of this. Plugging in $-1$ gives the determinant of the corresponding square presentation matrix for $H_1(B_2(K))$. Since $\mathbb{Z}$ is a PID, the determinant of the presentation matrix is the order of the module; this is due to Smith normal form having the same determinant, up to multiplication by $-1$.

The Alexander polynomial actually comes from a sequence of polynomials $\Delta^0_K(t),\Delta^1_K(t),\dots$, each dividing the previous, where $\Delta_K(t)=\Delta^0_K(t)$ is the zeroth (sometimes first, depending on the convention) Alexander polynomial, all defined up to multiplication by $\pm t^{k}$. The polynomial $\Delta^i_K(t)$ is the GCD of the $i$th Fitting ideal of $H_1(X_\infty(K))$ (in knot theory literature, it is the $(i-1)$th elementary ideal), which is an ideal of $\mathbb{Z}[t^{\pm 1}]$. See chapter 20 of Eisenbud's "Commutative algebra" for more details about Fitting ideals. The $i$th Fitting ideal for $H_1(X_\infty(K))$ is generated by the $(m-i)\times(m-i)$ minors of the $m\times m$ presentation matrix for $H_1(X_\infty(K))$. Hence, the $i$th Fitting ideal for $H_1(B_2(K))$ (an ideal of $\mathbb{Z}$) is given by the same-sized minors of the same presentation matrix with $t=-1$.

Let's call $\beta^i_K\in\mathbb{Z}$ the GCD (i.e., since $\mathbb{Z}$ is a PID, the generator) of the $i$th Fitting ideal of $H_1(B_2(K))$. One fact about Fitting ideals is that if $H_1(B_2(K)) = \bigoplus_{i=1}^n \mathbb{Z}/m_i\mathbb{Z}$ with $m_{i-1}\mid m_i$ for all $1<i\leq n$, then $\beta^i_K=m_1m_2\dots m_{n-i}$. Hence, $\beta^0_K=\lvert H_1(B_2(K))\rvert$, and the smallest $i$ such that $\beta^i_K=1$ gives exactly the minimum number of generators for $H_1(B_2(K))$.

To be clear, GCD for $\mathbb{Z}[t^{\pm 1}]$ means to take the smallest principal ideal containing the ideal generated by the given set of polynomials. From this, one may deduce that $\Delta^i_K(-1)$ divides $\beta^i_K$ (and it may be that they are equal, but that seems unlikely). Thus, the smallest $i$ such that $\Delta^i_K(-1)=1$ gives a lower bound for the number of generators for $H_1(B_2(K))$.

Ideals of $\mathbb{Z}[t^{\pm 1}]$ are of the form $(p(t),n)$ for $p(t)\in\mathbb{Z}[t]$ and $n\in\mathbb{Z}$, possibly with $p(t)=0$ or $n=0$. If this is the $i$th Fitting ideal of $H_1(X_\infty(K))$, then the corresponding Fitting ideal for $H_1(B_2(K))$ would be $(p(-1),n)$. Then $\beta^i_K$ is the GCD of $p(-1)$ and $n$. While the gcd of $p(t)$ and $n$ might be $1$, it might be that the gcd of $p(-1)$ and $n$ is bigger than $1$.

The zeroth Fitting ideal of $H_1(X_\infty(K))$ is principal since the presentation matrix is square, so $\beta^0_K=\Delta^0_K(-1)$. This is restating that $\lvert H_1(B_2(K)) \rvert=\lvert\beta^0_K\rvert = \lvert \Delta_K(-1)\rvert$.

The Alexander polynomials of the knot $8_{20}$ are \begin{align} \Delta^0_{8_{20}}(t) &= (1-3t+t^2)(1-t+t^2)^2 \\ \Delta^1_{8_{20}}(t) &= 1-t+t^2 \\ \Delta^2_{8_{20}}(t) &= 1 \end{align} Since $\Delta^1_{8_{20}}(-1)=3$, we know that $H_1(B_2(8_{20}))$ needs at least two generators.

I happen to know that $$ H_1(X_\infty(8_{20})) = \mathbb{Z}[t^{\pm 1}]/((1-3t+t^2)(1-t+t^2)) \oplus \mathbb{Z}[t^{\pm 1}]/(1-t+t^2) $$ Hence, $H_1(B_2(8_{20})) = \mathbb{Z}/15\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$. That's two generators.

While the Alexander polynomials detected the number of generators in this case, consider $9_{46}$. It has $$ H_1(X_\infty(9_{46})) = \mathbb{Z}[t^{\pm 1}]/(2-t)\oplus \mathbb{Z}[t^{\pm 1}]/(1-2t) $$ Hence, \begin{align} \Delta^0_{9_{46}}(t) &= (2-t)(1-2t) \\ \Delta^1_{9_{46}}(t) &= 1 \\ \end{align} Since $\Delta^0_{9_{46}}(-1)=9$ and $\Delta^1_{9_{46}}(-1)=1$, we know $H_1(B_2(9_{46}))$ has at least one generator. However, $\beta^0_{9_{46}}=9$, $\beta^1_{9_{46}}=3$, and $\beta^2_{9_{46}}=1$, so it needs exactly two generators.

One more knot: $6_1$. It has the exact same Alexander polynomials as $9_{46}$. However, $\beta^1_{6_1}=1$. Therefore, $6_1$ and $9_{46}$ can be distinguished by this piece of homological information about their double branched covers.