# Seifert matrices and Arf invariant — Cinquefoil knot

I have computed the following Seifert matrix for the Cinquefoil knot:

$$S = \begin{pmatrix} 1 & -1 & -1 & -1 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1& 1 & 1\end{pmatrix}$$

I also found a Seifert matrix for this knot on the internet but I still don't know how to verify the correctness of a Seifert matrix. Hence:

Question 1: How can I verify that a Seifert matrix that I computed is correct?

Now, more importantly:

I thought I could compute the Arf invariant as follows (where $K$ is my knot):

$$A(K) = A(q) = q(x_1) q(x_2) + q(x_3) q(x_4) = 1 \cdot 1 + 1 \cdot 1 = 0$$

where $x_1 = \begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, x_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, x_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} , x_4 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$ and $q(x) = x^T S x$.

But this must be wrong for the following reasons: 1. Assuming the Seifert matrix I found on the internet is correct, I get the same Arf invariant. 2. The Arf invariant of the cinquefoil knot is $1$. 3. "The Arf invariant is $0$ if a majority of classes have self-linking $0$, and is $1$ if most classes have self-linking $1$.". (Excerpt taken from here). The self-linking numbers are the diagonal entries of the Seifert matrix as far as I understand.)

Question 2: What is the correct way to compute the Arf invariant of a knot from a Seifert form / Seifert matrix?

-
It's knot obvious. – Rudy the Reindeer Sep 7 '12 at 20:26

What makes you think that $x_i$ form a symplectic basis for $S$? In fact they don't. Also, you don't want a symplectic basis for $S$ but rather for $I = S^T - S$.

Your Seifert matrix must be wrong since if I compute the Arf invariant using the $S$ with a symplectic basis $$e_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\1 \end{pmatrix} e_2 = \begin{pmatrix} -1 \\ 1 \\ 0 \\1 \end{pmatrix} f_1 = \begin{pmatrix} 0 \\ 0 \\ -1 \\1 \end{pmatrix} f_2 = \begin{pmatrix} 0 \\ -1 \\ 1 \\0 \end{pmatrix}$$ I get zero when I should be getting $1$.

Using the Seifert matrix you linked, $$S = \begin{pmatrix} -1 & -1 & 0 & -1 \\ 0 & -1 & 0 & 0 \\ -1 & -1 & -1 & -1 \\ 0 & -1 & 0 & -1 \end{pmatrix}$$

and a corresponding symplectic basis for $$I = S^T - S = \begin{pmatrix} 0 & 1 & -1 & -1 \\ -1 & 0 & -1 & -1 \\ 1 & 1 & 0 & 1 \\ -1 & 1 & -1 & 0\end{pmatrix}$$

$$e_1 = \begin{pmatrix} -1 \\ 0 \\ 0 \\1 \end{pmatrix} e_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} f_1 = \begin{pmatrix} -1 \\ 0 \\ 0 \\1 \end{pmatrix} f_2 = \begin{pmatrix} -1 \\ 0 \\ 1 \\0 \end{pmatrix}$$

yields $$A(K) = e_1^T S e_1 e_2^T S e_2 + f_1^T S f_1 f_2^T S f_2 = 7 \equiv_2 1$$ as expected.

-
Computed using this online octave "terminal". – Rudy the Reindeer Sep 10 '12 at 8:45
Of course, this matrix doesn't correspond to the Seifert surface in the drawing which is very unfortunate. – Rudy the Reindeer Sep 10 '12 at 8:47

Mathematica gives as KnotData[]: KnotData[{5, 1}, "SeifertMatrix"]

{{-1, -1, 0, -1}, {0, -1, 0, 0}, {-1, -1, -1, -1}, {0, -1, 0, -1}}

Which is closer to the first "given" matrix. The KnotData gives the arf invariant for that knot:

KnotData[{5, 1}, "ArfInvariant"]

1

I know that only gives Mathematica as the authority for the answers but that is the best I can find.

-