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Let $M$ be a toric Kahler manifold with $\text{dim}_{\mathbb{R}} = 4$. Let $V$ be a Killing vector field associated to the action of the two-torus $\mathbb{T}^2$ on $M$. We also assume the existence of a metric $\langle, \rangle$ and denote the Killing one-form $v = \langle V, \cdot \rangle$. I'm studying the differential equation $$ \phi (dv)^+ = -2 (1+V^2) (d\phi \wedge v)^+, \quad \phi : M \to \mathbb{R}. $$ ${}^+$ denotes self-duality where conventions are such that the symplectic form $\omega$ is anti self-dual, i.e. $\omega = \omega^-$. (In detail: $\alpha^\pm = \frac{1}{2} (\alpha \pm \star \alpha)$, $\star\alpha_{ab} = -\frac{1}{2} \epsilon_{abcd} \alpha^{cd}$.)

Clearly $\phi = 0$ is a solution. I want to show that this is the only solution.

Remarks: (1) The only thing charged under $V$ in the above equation is $\phi$. If one can assume the Lie derivative to vanish: $\mathcal{L}_V \phi = \nabla_V \phi = 0$, contraction of the above with $V$ leads to $$ V^2 d\phi = \frac{1}{(1+V^2)} \phi i_V (dv)^+ $$ where $i_v$ denotes interior multiplication.

(2) I studied the equation explicitly for $\mathbb{CP}^2$ and $\mathbb{CP}^1 \times \mathbb{CP}^1$. Apart from the trivial solution one finds a singular one whose gradient $d\phi$ connects the "poles" $V^2 = 0$. On $\mathbb{CP}^2$, $dv^+ = 0$ vanishes at one of the poles and $d\phi = 0$ here. The solution diverges at the other pole however. On $\mathbb{CP}^1 \times \mathbb{CP}^1$ the non-trivial solution diverges at both poles (which in general are $\mathbb{CP}^1$s). This time, one has $dv^+ = 0$ on the equator. Note that in both examples the non-trivial solutions diverge at "poles" that are $\mathbb{CP}^1$s; moreover, in both examples the sole critical points of $\phi$ are those where $dv^+ = 0$. Finally, in both examples $\nabla_V \phi = 0$ follows from the differential equations.

(3) One can of course consider the moment map $\mu : M \to \mathfrak{t}^*$, which induces a map $\mu^V : M \to \mathbb{R}, p \mapsto \langle \mu(p), V \rangle$. $\mathfrak{t}$ is the Lie algebra of the torus, $\langle, \rangle$ denotes the usual pairing with $\mathfrak{t}^*$. One can rephrase the above via $d\mu^V = i_V \omega$.

(4) If $p \in M$ is a critical point of $\phi$, $d\phi(p) = 0$ it follows that $dv^+(p) = 0$. In the $\mathbb{CP}^2$ example, only one such point exists. I have tried to argue that there is generally only one point on $M$ where $dv^+$ vanishes, thus $\phi$ has at most one critical point, which (with additional assumptions) would imply that no such smooth solutions exists.

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  • $\begingroup$ In your first Remark, I don't understand how you conclude that you can reduce to the case $\nabla_V\phi=0$. If the flow of $V$ induces a circle action, you can't just average $\phi$ over the circle action because the average could be identically zero, even if $\phi$ is not, couldn't it? Have you looked at the case in which $M=\mathbb{CP}^1\times\mathbb{CP}^1$ with the constant curvature product metric and $V$ is the vector field that isometrically rotates one of the $\mathbb{CP}^1$-factors? $\endgroup$ Aug 18, 2014 at 13:42
  • $\begingroup$ Regarding the reduction to $\nabla_V \phi = 0$, I had - apparently mistakenly - assumed that one could expand generic $\phi$ in a complete system of eigenfunctions of $\nabla_V$ and then solve essentially order by order. Also, thanks for pointing me to $\mathbb{CP}^1 \times \mathbb{CP}^1$. I had not checked this example. By now I studied the original (not simplified) equation for both examples and the same seems to happen. The only solutions are $\phi = 0$. Other solutions have singularities and only one critical point. The difference is that on $\mathbb{CP}^1$ there are two singularities. $\endgroup$
    – jws
    Aug 19, 2014 at 10:55
  • $\begingroup$ Are you sure there's a '$2$' in your simplified equation? BTW: Your original equation is $$d\phi=\phi\ \alpha+\psi\ v,$$ where $\alpha = i_V(dv)^+/(V^2(1+V^2))$ and $\psi$ is an unknown function. Differentiating and wedging with $v$ gives $$\phi\ d\alpha\wedge v+\psi\ dv\wedge v=0.$$ If $d\alpha\wedge v$ and $dv\wedge v$ are linearly independent, both $\phi$ and $\psi$ must vanish. Otherwise, if $dv\wedge v\not=0$, $\psi$ is a fixed multiple of $\phi$, and the equation becomes $$d\phi = \phi\ \beta,$$ which only has nonzero solutions if $d\beta=0$, so having nontrivial solutions is quite rare. $\endgroup$ Aug 19, 2014 at 13:56

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