[Math] Seifert matrices — Figure 8 knot


I've just learnt about Seifert matrices and thought it might be a good idea to compute some. Can you tell me if this is right:

enter image description here

Here $x_1^+$ denotes the push off of $x_1$. I have omitted the diagram for $x_2$ since I think that it's the same as the one for $x_1$. For the linking number of $x_1$ and $x_2^+$ I seem to get $\mathrm{lk} (x_1, x_2^+) = 0 $ and similarly for $x_1^+$ and $x_2$. I think that's wrong but I'm not sure. The Seifert matrix I get is
$$ S = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} $$

I found an example on the internet that gets Seifert matrix
$$ S = \begin{pmatrix} 1 & -1 \\ 0 & -1\end{pmatrix} $$

But since a knot can have many Seifert surfaces there can be many different Seifert matrices so this doesn't tell me much.

Question 1: In general, how can I check that my Seifert matrix is correct?

Question 2: I think the process of constructing the Seifert surface is fiddly and hence error prone. Is there a neater way to compute the Seifert matrix of a knot? (I want to use the matrix to compute the Arf invariant)

Question 3: In the picture above I'm particularly unsure about picture 3. How do I know that I attached the bands correctly?

Thanks for your help.


Assuming the other Seifert matrix is correct, my sums are wrong since mine gets me Arf invariant $1$ whereas the other matrix gives me Arf invariant $0$. Nonetheless, I'd be very grateful if someone could point out where my mistake is.

Best Answer

Here is a partial answer, I can only help you with question 3, so here goes:

Basically, you messed up when you attached the bands. If you want to reduce the probability of ballsing up when constructing the Seifert surface I suggest that you try to make it look as close as possible to the original knot diagram. In the example of the figure 8 knot, this would mean that you should draw something like this:

enter image description here

Then you will see that you get $$ S = \begin{pmatrix} -1 & 0 \\ 1 & 1\end{pmatrix} $$ which gives you $\mathcal{A} (q) = 0$ which is what you want.

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