I am recently reading Rolfsen's Knots and Links, 3E6 says there is no (tame) knot group which has a presentation with deficiency two, and I am wondering why.
What I know:
- Every tame knot group (in $\mathbb{R}^3$ or $S^3$) has a finite presentation of deficiency one, by Wirtinger presentation.
- The abelianization of each knot group is infinite cyclic, by either Wirtinger presentation for tame knots or abelianizing it to get $H_1$ and apply Mayor-Vietorus.
My question is pretty simple:
Intuitively with each one of the relation in a group presentation, we can get rid of one generator at most, so abelianizing a finite group presentation can reduce the number of generators by at most the number of the relations. Since this gives $\mathbb{Z}$, we have the deficiency of any presentation of the knot group at most one. How can I make this rigorous though?
Thanks in advance for helpful ideas.
Best Answer
You've got the main idea, so I'll just explain how you might make it rigorous. Suppose you have a presentation $G=\langle g_1,\dots,g_n \mid r_1,\dots,r_m\rangle$ and want to compute the abelianization. First, a group presentation is the same as writing a group as sequence $$F_m \xrightarrow{\phi} F_n \to G \to 1$$ where $F_m$ and $F_n$ are free groups of orders $m$ and $n$, respectively, with the property that $F_n$ surjects onto $G$ and the normal closure of the image of $F_m$ is the kernel of $F_n\to G$. If we write $x_1,\dots,x_m$ for the generators of $F_m$, then we define $\phi(x_i)=r_i$. Now, we want to compute the abelianization. The abelianization functor is right exact, so applying it to this sequence we get a sequence $$\mathbb{Z}^m \xrightarrow{\phi_*} \mathbb{Z}^n \to \operatorname{Ab}G \to 0$$ that one can check is exact, where if $e_i=(0,\dots,0,1,0,\dots,0)\in\mathbb{Z}^m$ with a $1$ in position $i$, then $\phi_*(e_i)$ is a vector that gives a signed count of how many times each generator appears in $\phi(x_i)$. For example, if $r_2=g_1g_2g_1g_2^{-1}$ and $n=3$, then $\phi_*(e_2)=(2,0,0)$. In particular, $\phi_*$ can be written as an $n\times m$ matrix with values in $\mathbb{Z}$, and then we get that $\operatorname{Ab}G$ is the cokernel of this matrix.
This next bit is essentially using the structure theorem for modules over a PID. A useful tool is Smith Normal Form: there exist automorphisms of $\mathbb{Z}^m$ and $\mathbb{Z}^n$ such that the matrix for $\phi_*$ is diagonal and each diagonal entry is divisible by the next. Once it is in this form, the number of all-zero rows gives you the number of $\mathbb{Z}$ factors. If $n\geq m+2$, then there must be at least two $\mathbb{Z}$ factors. As you pointed out, knot groups have only one $\mathbb{Z}$ factor. To tie these together, we need to say that the number of $\mathbb{Z}$ factors is well-defined.
One trick is to tensor everything with $\mathbb{Q}$. Tensoring is right exact, so we get a presentation for $\mathbb{Q}\otimes \operatorname{Ab}(G)$: $$\mathbb{Q}^m \xrightarrow{\phi_{**}} \mathbb{Q}^n \to \mathbb{Q}\otimes \operatorname{Ab}(G) \to 0$$ The matrix for this presentation is given by normalizing all the nonzero entries on the diagonal to $1$ (since we can divide in $\mathbb{Q}$!). Thus, the dimension of $\mathbb{Q}\otimes \operatorname{Ab}(G)$ as a vector space over $\mathbb{Q}$ is the number of nonzero rows in the matrix, and since dimension is well defined, so is the number of $\mathbb{Z}$ factors.