The image should already say everything. But in case it is mute... :-)
List tangles with 2n legs downward as column header, and with 2n legs
upward as row header. (n=3 here.) In the crossing point goes the
"joining" of the tangles. You get an infinite matrix. Identify the
links at the crossing points (examples in red; o means the unlink) and
replace them with some invariant of your choice. Example: Take the Jones
polynome. Then the crossings in the sixth row can be resolved, and of
course the whole infinite matrix now has only five independent vectors
and thus rank 5.
Note due to symmetry the matrix blockdiagonalizes. (If you accept that
mutants are always equal.) It's fun to compute
the matrix rank for different invariants and tangle "width" n.
Of course the rank is directly related to the number of terms in a
skein relation (if existing!) for the invariant. For n=2 the ranks of
the symmetric and antisymmetric part are: Jones, 1+1; Kauffman/Dubrovnik
1+2/2+1; and 2+2 would be the "invariant" I'm trying to prove as such since
about 30 years :-)
The construction also works for oriented knots, but due to arrow
collisions you need several matrices.
Obvious question: Surely this construction has been done before, and surely with a load of linear algebra machinery instead of one nice self-explaining picture :-) Do you have a reference handy? Also obvious: Define X to be the invariant that has ranks 2+2 as above. Any chance you're getting somewhere? (It might be possible that 2+2 can't exist since the equations force it to reduce to 2+1 or 1+2, also it might be possible some links can't be computed after all. Probably I just rewrote the skein relation approach...)
