Seifert matrix from a Seifert surface I read multiple example about how to find the Seifert matrix from a Seifert surface but I just don't see it. I have trouble with the "under" and "over" crossings when we find the different loops. Here is an example from the book "Knot Theory and its Applications" by Kunio Murasugi:
example.

Is it really easy to see the different crossings in Figure 5.3.5 just from Figure 5.3.4 (b)?
(I'm also interested in other way to find it if you think this is not the easiest one).
Thank you for the help.
 A: I think it's a bit easier to see how to get a Seifert matrix from the Seifert surface from Seifert's algorithm by introducing an intermediate object.  I do not know the name of it (I've only ever seen it on the Knot Atlas wiki in the images section), but, given the pattern so far, let's call it a Seifert graph.
First, an example:

The first step is the usual one for Seifert's algorithm.  However, observe that while the usual splicing-in of the twists at the crossings gives a Seifert surface, this surface has the structure of a ribbon graph embedded in $S^3$.  A ribbon graph is like a graph but with everything thickened up: the vertices are disks and the edges are rectangular strips.  Diagram $\mathbf{F}$ in the book gives a good representation of the Seifert surface as a ribbon graph, which has two vertices and three edges.
Ribbon graphs in $S^3$ have diagrams just like knots, except now there are vertices, and the edges might be twisted.  We can draw just the "skeleton" of the ribbon graph using the following correspondences:

So: after resolving all the crossings, we get nested circles, which give disks---these are the vertices of the ribbon graph.  The crossings correspond to attaching half-twisted edges between the vertices.  This gives the Seifert graph for the knot.
The graph $\mathrm{\Gamma}(\mathbf{D})$ in the book is from taking the underlying graph of the Seifert graph.  That is, we forget about all the twists.  In fact, $\mathrm{\Gamma}(\mathbf{D})$ is a deformation retract of the Seifert graph, and as such it has the same homology groups.  For graphs, the rank of $H_1$ is the number of edges outside of a spanning forest, which in the case of the example is $2$.
Before going into a different version of the computation, at this point you can choose your favorite basis for $H_1$ (for example $\alpha_1$ and $\alpha_2$ in the diagram) then carefully visualize these curves and their pushoffs.  One thing I noticed about 5.3.4(b) is that it actually does not portray the curves correctly!  They should intersect in only one point in the surface, as illustrated here:

I would suggest you make a paper model of the Seifert surface (take two disks, tape on three half-twisted paper strips, then draw the two curves), then you can either attempt to visualize the pushoffs, or you can take a piece of string to form a pushed-off curve (lay the string along a drawn curve and tie it up into a loop).
Anyway, we can simplify the Seifert graph's diagram by flipping over a vertex to remove all the half twists, moving edges around using Reidemeister-like moves, then contracting one of the edges, like so:

In general, by flipping over all the vertices with a given orientation (in this case, with counter-clockwise orientation, from the point of view of the knot's orientation), then doing some moves like

we can get a diagram with no twists at all, since Seifert surfaces are orientable.  Any two Seifert graphs for a given knot, when in this form, are related by the following Reidemeister-like moves:

Move RI' is a restricted version of the usual first Reidemeister move; it is writhe-preserving and part of some moves Kauffman calls regular isotopy.  The second Reidemeister move is generalized to deal with vertices of arbitrary degree.  The third Reidemeister move is the usual one.  Move C is edge contraction.
With these moves, every Seifert graph has a representative with a single vertex.  Then, there is a basis for $H_1$ that is given by all the edges in the Seifert graph, which are all loops.
So, here is the mechanical calculation of the Seifert matrix for some choice of basis for $H_1$:

For the pushoff, I used the left-hand blackboard framing convention, where the curve is pushed to the left (from the point of view of someone walking along the curve in the direction of the orientation), and at the vertex the pushoff goes over anything that goes through the vertex.
Note that the linking number of $\alpha_i$ with $\alpha_i^\sharp$ is the writhe of that curve by definition.  The linking number of $\alpha_i$ with $\alpha_j^\sharp$, when $i\neq j$, is the linking number of $\alpha_i$ and $\alpha_j$ away from the vertex plus $k\in\{-1,0,1\}$, which depends on exactly how the curves interact right at the vertex.
Just to make sure I did everything more-or-less correctly, $\det(t A-A^{T})=t^2-t+1$, which is the Alexander polynomial of a trefoil knot.
Now, it's not necessary to go through all these manipulations---I just think it's easier to deal with Seifert surfaces when their surface normal always points straight out of the paper.  There is a straightforward way to just write down the Seifert matrix from the Seifert graph that comes out of Seifert's algorithm once you get a hang of how the half twists contribute to the linking numbers, though I've probably only ever done it by hand for this exact example!
