0$, and obviously $P$ is positive since it is a sum of squares). We can then evaluate that for this solution \begin{equation} E = J \end{equation} in the limit $J\to 0$, this solution reduces to the trivial solution where all matrices are identically zero. For the second solution, we take $P=1/4$, and we find similarly that \begin{equation} E = \frac{J^2 + \left(1-16 Q^2\right)^2}{2 - 32Q^2} \end{equation} which is of similar form if we use the variable $x=1-16 Q^2$ (this is, of the form $Ax+B x^{-1}$). Again, the minimum occurs when \begin{equation} Q = \frac{\sqrt{1-J}}{4} \end{equation} and we also find that $E = J$ identically. This is the solution where we choose to take $Q>0$. There is a similar solution with $Q<0$ that is a $\BZ_2$ reflection of this solution. When $J\to 0$, this is the standard fuzzy sphere of $2\times 2$ matrices. When $J\to 1$, this matches our other solution with $Q=0$. The three solutions meet at $J=1$. Beyond $J=1$ this solution does not exist anymore. In the third branch, we have that \begin{equation} E = \frac{2 J^2 (2 P+1)}{(4 P+1) (16 P-1)}-4 P^2+9 P-\frac{9 P}{2 P+1} \end{equation} and it is easy to solve for $J$ as a function of $P$ (essentially solving $\partial_PE=0$), giving us \begin{equation} J^2=\frac{(1-P) P \left(64 P^2+12 P-1\right)^2}{(2 P+1)^2 (8 P+1)} \end{equation} So that we end up with a parametric solution $J(P)$, rather than the other way around. This can be inverted numerically. This only makes sense if $J^2\geq 0$, so that necessarily $P\leq1$, and remember also that $1/4\geq P\geq 0$ from the reality of $Q$. But moreover, $|Q|\leq P$, and this restricts $P$ to be bigger than $1/16$. Notice that when $J\to 0$, there are various values of $P$ that can arise as roots. The only one that is new corresponds to $P=1/16$. This is a fuzzy sphere at half radius. This can be easily understood if we make an ansatz of spherical symmetry for a saddle point. In this case the matrices $X,~Y,~Z$ are all proportional to the corresponding Pauli matrices with proportionality constant $r$. Because the energy for a static configuration is quartic in the matrices, and we have that $E=0$ at $r=0$ and $E=0$ at $r=1$, and always $E\geq0$, then the energy must be proportional to $r^2(1-r)^2$. This has a maximum at $r=1/2$, which is the sphere at half radius. This solution also matches the solution at $Q=0$ that we already had when we set $P=1/4$. There is similarly a reflected solution with $Q<0$, where we take the other square root branch cut of equation~\eqref{eq:Qsol}. The new solution of the fuzzy sphere at half radius migrates to higher angular momentum as we increase $P$ from $1/16$ to $1/4$ and is a saddle point. Therefore it has Morse index one. This solution and the one that is reflected by taking $Q\to -Q$ can cancel the two minima from the BPS solution as they meet at $J=1$ with the trivial solution that has $Z=0$~throughout. The full set of solutions is depicted in figure~\ref{fig:twotimestwo}. There we can see that for low angular momenta there are five critical points. Three are minima and two are saddles with Morse index one. The saddles and two of the minima are reflected into each other by the $\BZ_2$ symmetry, and one minimum if at the fixed point. The two saddles and two of the minima annihilate each other when $J=1$. Because the minima that annihilate with the saddles are BPS, the saddles need to approach the extremal limit and thus must touch the fixed point set, because one can not descend further from the saddle to the fixed point otherwise. \begin{figure} \centerline{\includegraphics[width=3.0 in]{twotimestwo1.pdf}} ]]>