# Why might the Lawson minimal surface $\xi_{1,2}$ have a Morse index 9?

In page 15 of the article New Applications of Min-max Theory, Andre Neves said that: a wishful thinking would suggest the Lawson minimal surface $$\xi_{1,2}$$ in the 3-sphere to have a Morse index 9, but there is no real evidence.

My question is, what suggests the index to be 9? Say, can we find a 9-parameter deformation of $$\xi_{1,2}$$ which decreases its area?

## 2 Answers

Due to a recent preprint by Kapouleas and Wiygul (arXiv:1904.05812), the index of the Lawson $$\xi_{1,g}$$ surface is $$2g+3$$. I have not looked at the paper in detail so far, but let me cite from the authors introduction:

"The ideas of our proof originate with work of NK on the approximate kernel for Scherk surfaces. Our approach requires a detailed understanding of the elementary geometry of $$S^3$$ and of the surfaces involved, especially their symmetries. The proof makes also heavy use of Alexandrov reflection in the style of Schoen’s. The Courant nodal theorem [ and an argument of Montiel- Ros play important roles as well. In ongoing work we hope to extend this result to determine the index and nullity of all Lawson surfaces desingularizing intersecting great two-spheres in the round three-sphere $$S^3$$."

• Wow! A great answer!
– JSCB
Apr 12, 2019 at 16:37

This is an old thread, but I figured I would add a short comment that may explain some of the reasoning behind the 'wishful thinking' that Neves refers to.

Let us, for all integer $$p \geq 1$$, write $$\omega_p$$ for the $$p$$-width of $$\mathbf{S}^3$$, that is the min-max value associated to a $$p$$-parameter Almgren-Pitts construction. Then $$$$\omega_1 \leq \cdots \leq \omega_p \leq \cdots$$$$ is an increasing, though not strictly increasing sequence.

Moreover, for each $$p$$ there is an embedded minimal surface $$\Sigma_p \subset \mathbf{S}^3$$ with area and Morse index satisfying $$$$\mathrm{area} \, \Sigma_p = \omega_p \quad \text{and} \quad \mathrm{index} \, \Sigma_p \leq p.$$$$ Note that the index bound is not sharp in general. For example $$\omega_1 = \cdots = \omega_4 = 4 \pi$$ are all realised by an equatorial sphere, and therefore $$$$\mathrm{index} \, \Sigma_4 = 1.$$$$

The specific values of the widths are only known for relatively small $$p$$, but a PhD student of Neves, Charles Nurser, proved some estimates for the ninth width, namely: $$$$2 \pi^2 < \omega_9 < 8 \pi.$$$$

The lower bound means that $$\Sigma_9$$ is not a Clifford torus, and the upper bound that it is not a multiplicity-two copy of a minimal sphere. It must therefore be a 'new' surface, and it is not unreasonable to conjecture that it might be the Lawson surface $$\xi_{1,2}$$.

This is just a guess, but as far as the Morse index is concerned, perhaps the thinking was that $$\omega_p > \omega_{p-1}$$ might mean that $$\mathrm{index} \, \Sigma_p = p$$. For example $$\omega_5 = 2 \pi^2 > 4 \pi = \omega_4$$, and is realised by a Clifford torus, with Morse index five. However I must emphasise this is purely speculative, and I'd be very curious to hear from someone more knowledgeable.