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Using the following Seifert surface of the cinquefoil knot

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I get the following Seifert matrix (of linking numbers): $$ S = \begin{pmatrix}- 1 &1 &0 &0 \\ 0 &-1 &1& 0 \\ 0& 0 &-1& 1\\ 0 &0 &0& -1\end{pmatrix}$$

I compute the matrix of the corresponding bilinear form $$ I = S^T - S = \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}$$

and a corresponding symplectic basis $$e_1 = \begin{pmatrix}0 \\0 \\0 \\1 \end{pmatrix}$$ $$f_1 = \begin{pmatrix} 0 \\1 \\1 \\0\end{pmatrix}$$

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

And I get $\mathrm{Arf}(K) = e_1^T S e_1 f_1^T S f_1 + e_2^T S e_2 f_2^T S f_2 = 2 \equiv_2 0 \neq 1$

Where is my mistake? Thanks for your help.

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It's knot a symplectic basis :,( –  Matt N. Sep 10 '12 at 11:05

1 Answer 1

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For the matrices $I$ and $S$ the following is a symplectic basis

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

$$ f_1 = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$$

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

yielding

$$ \mathrm{Arf}(K) = e_1^T S e_1 f_1^T S f_1 + e_2^T S e_2 f_2^T S f_2 = 9 \equiv_2 1$$ as it should.

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