# Petries exotic circle action

In the paper "S^1-actions on homotopy complex projective spaces" by Petrie (Bulletin of the AMS, 1972), Petrie constructs a smooth circle action on $$\mathbb{CP}^{3}$$ (page 148). The fixed point set consists of $$4$$ points and the weights at these fixed points cannot be equal to any linear circle action (A linear circle action corresponds to a subgroup $$S^{1} \hookrightarrow PGL_{4}(\mathbb{C})$$ acting on projective space in the usual way).

Question: Does this circle action preserve an almost complex structure $$J$$ on $$\mathbb{CP}^{3}$$?

Remarks. 1. It is a classical result that holomorphic $$S^{1}$$-actions on complex projective spaces are linear (with the standard complex structure), so $$J$$ cannot be integrable and have a compatible Kähler metric (by Hirzebruch-Kodaira theorem).

1. It follows from Tolmans result ("A symplectic generalisation of Petries Conjecture", Transactions of the AMS, 2010), that if the circle action preserves a symplectic form then it is linear, so there is no $$S^{1}$$-invariant symplectic form.
• Regarding remark 1, does the classical result definitely apply if $J$ is an exotic complex structure? – Oliver Nash Jul 19 '19 at 9:25
• For the standard complex structure, so I should edit my comment – Nick L Jul 19 '19 at 9:41

## 1 Answer

This is only a partial answer. What follows was the result of a discussion with Silvia Sabatini, the author of this paper, which is related to the question of the OP.

The total Chern classes of possible almost complex structures on $$\mathbb{CP}^3$$ can be parametrized by $$1 + 2kx + 2(k^2-1)x^2 + 4x^3$$

for $$x$$ a generator of $$H^2(\mathbb{CP}^3)$$ (such that $$4x$$ is the first Chern class of the standard complex structure on $$\mathbb{CP}^3$$ and $$k \in \mathbb Z$$. Note that by obstruction theory we know that the homotopy class of an a.c.s. on $$6$$-manifold is determined by the first Chern class.

Now if an $$S^1$$-action preserves an a.c.s. on a compact manifold with isolated fixed points, then the Todd genus is equal to the number of fixed points with zero negative weights in the isotropy representation of the fixed point (There are more constraints which can be found in the paper linked above). Denote by $$N_i$$ the number of fixed points with $$i$$ negative weights. Then $$N_i = N_{i-3}$$, thus, given that we have $$4$$ fixed points we have only $$0,1,2$$ as possible values of $$N_0$$. We determine the possible first Chern classes of an invariant a.c.s on $$\mathbb{CP^3}$$

The case $$N_0=0$$: one obtains that $$k=-1,0,1$$. Note that $$k=-1$$ is the almost complex structure which would correspond to a hypothetical complex structure on $$S^6$$ blowed up in a point.

The case $$N_0=1$$: it seems that the only solution is $$k=2$$ which corresponds to the standard complex structure of $$\mathbb{CP^3}$$.

The case $$N_0=2$$: no solutions.

To be honest, I was to lazy to look into Petrie's paper, but if you know the weights of the action, you can compute the Chern classes through the equivariant Chern classes and you could check if they are listed above. If they don't correspond to $$k=-1,0,1$$ then your question has a negative answer. Otherwise I don't know at the moment any other possibility to rule them out.

• The weights are computed explicitly in Petries paper (page 50). It seems that there are two fixed points with all weights positive, so there shouldn't be an invariant almost complex structure by what you said. Although for a smooth $S^{1}$-action the weights are only defined up to $\pm 1$ so we cannot quite say that if there is an invariant almost complex structure then $N_{0}=2$ in your notation (for example there could be a point with all weights negative a priori). So, if you don't mind, I will postpone accepting the answer for a couple of days in case somebody has some remarks about this. – Nick L Jul 19 '19 at 14:19