A circle-valued Morse map $f:C_K \to S^1$ on the complement of a knot $K$ is said to be regular if there is a $C^\infty$ trivialisation $\Phi : N(K)\to K\times B^2(0,\epsilon)$ of a tubular neighbourhood of $K$ such that the restriction of $f$ to $N(K)\setminus K$ satisfies $f\circ\Phi^{-1}(x,z) = \frac{z}{|z|}$.

Would anyone be able to give me a better intuitive way of understanding what exactly a regular Morse map is, why for example is it "regular"? What kind of obstructions could there be to stop a Morse function from being regular? Is it the knot itself which affects whether a Morse map is regular or not?



The condition of regularity implies that the image of a meridian through $f$ winds exactly once around $S^1$.

Every $\mathbb{R}$-valued Morse function can be turned into an $S^1$-valued Morse function by including $\mathbb{R}$ along some open arc of $S^1$. Such Morse functions are not regular.

Another thing you can do is post-compose an $S^1$-valued Morse function with the degree-$n$ map $\theta\mapsto n\theta$. This is still a Morse function (when $n\neq 0$), but it fails to be regular when $n\neq\pm 1$.

Regular Morse functions are ones that, when restricted to $S^3-N(K)\to S^1$, can then be extended to $S^3\to D^2$, with $S^1$ being the boundary of the disk $D^2$.

Recall that $H^1(S^3-K;\mathbb{Z})$ is equivalent to homotopy classes of maps $S^3-K\to S^1$. Regular Morse functions are generators for $H^1$. (So, given a curve $\gamma:S^1\to S^3-K$, the image $[\gamma]\in H_1(S^3-K;\mathbb{Z})\cong\mathbb{Z}$ is given by the degree of the composition $f\circ\gamma:S^1\to S^1$.)

Every knot has a regular Morse function. This is essentially from using smooth approximation and transversality arguments, starting from a continuous map that represents a generator for $H^1$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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