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?
 A: 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 $S^3$."
A: 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
\begin{equation}
\omega_1 \leq \cdots \leq \omega_p \leq \cdots
\end{equation}
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
\begin{equation}
\mathrm{area} \, \Sigma_p = \omega_p 
\quad \text{and} \quad \mathrm{index} \, \Sigma_p \leq p.
\end{equation}
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
\begin{equation}
\mathrm{index} \, \Sigma_4 = 1.
\end{equation}
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:
\begin{equation}
2 \pi^2 < \omega_9 < 8 \pi.
\end{equation}
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.
