Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given two knots $K$ and $L$. With Seifert matrices $M_{K}$ and $M_{L}$ respectively, then the matrix

$\begin{bmatrix}M_K & \\ &M_L\end{bmatrix}$

is a Seifert matrix of the connected sum $K+L$.

Therefore a knot is prime if and only if it has a Seifert matrix that is not S-equivalent to a matrix of this form.

Edit: This is incorrect, every knot is S-equivalent to a prime knot.

I have two questions;

1) Is what I have said correct? No

2) My understanding is that identifying whether a certain knot is prime or not was a non-trivial question, whereas identifying S-equivalence was relatively easy. Is there some hidden difficulty I am missing?

share|cite|improve this question
What do you mean by "a matrix of this form"? Beware that a too naive statement cannot be true: it is easy to find complicated matrix for the unknot, so you can have complicated blocks in the Seifert matrix of any (prime) knot... – PseudoNeo Feb 20 '12 at 23:26
I see now the logic is flawed, I was equating S-equivalence with equality for the 'only if' statement. – Dilitante Feb 21 '12 at 13:06
I am also reconsidering the statement that S-equivalence of matrices is easily identified – Dilitante Feb 21 '12 at 13:21
up vote 1 down vote accepted

For the sake of having an answer:

"In general, it is nontrivial to determine if a given knot is prime or composite (Hoste et al. 1998). However, in the case of alternating knots, Menasco (1984) showed that a reduced alternating diagram represents a prime knot iff the diagram is itself prime ("an alternating knot is prime iff it looks prime"; Hoste et al. 1998)."

Wolfram Mathworld, Prime Knot.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.