Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was looking for a proof that there are infinitely many prime knots and one said "take your favorite (prime) knot and consider all its Whitehead double", implying that all Whitehead doubles of a (prime?) knot are prime. But why is that so?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

It's fairly easy to see that if a knot has knot genus $1$, then it is prime. Recall that if $K$ and $K'$ are knots, $K\# K'$ is their connect sum, and $g(-)%$ is the genus of a knot, then $$g(K\# K')=g(K)+ g(K').$$That is, the genus of a knot is additive under connect sum. It follows that if $g(K)=1$, then $K$ is prime.

Now, a twisted Whitehead double of any (non-trivial) knot has genus $1$ and so it is prime$^*$. As there are infinitely many twisted Whitehead doubles of any knot, and they can be distinguished by their Alexander polynomials, it follows that there are infinitely many prime knots.

$*$ There's a great explanation of why such a knot has genus $1$ given in this answer.

share|improve this answer

Your Answer

 
discard

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.