# 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:

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?

Edit:

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.

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I just noticed that picture 3 is actually not quite right. But the idea behind it is, as you can see from picture 4 which should be right. –  Rudy the Reindeer Sep 4 '12 at 14:45

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:

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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You're the same user as the one who asked the question. Why are you referring to yourself as "you"? –  Rahul Sep 5 '12 at 18:36
Oops, must've taken the green pill instead of the red pill again. Multiple personality disorder and colour blindness do not mix well. But it explains why the headache went away. –  Rudy the Reindeer Sep 6 '12 at 9:38

Your Seifert algorithm seems to have gone wrong: in the fourth step it seems as if the boundary of your surface is a link instead of a knot.

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Thank you for your comment. I will have to look into it when I have more time. –  Rudy the Reindeer Jan 28 '14 at 13:59