Holomorphic projective connections on compact complex threefolds

Abstract

We prove that a holomorphic projective connection on a complex projective threefold is either flat, or it is a translation invariant holomorphic projective connection on an abelian threefold. In the second case, a generic translation invariant holomorphic affine connection on the abelian variety is not projectively flat. We also prove that a simply connected compact complex threefold with trivial canonical line bundle does not admit any holomorphic projective connection.

Appendix

The aim of this Appendix is to prove the technical Lemma 13, used in the proof of Proposition 12 (ii), namely:

$\nabla ^{A,B,C,D,E}$ is projectively flat on ${{\mathbb {T}}}^3$ if and only if $C\,=\,D$.

Proof

For that we shall compute the projective Weyl curvature tensor of $\nabla ^{A,B,C,D,E}$.

We start by computing the affine curvature tensor. To simplify the notation in the computation, $\nabla ^{A,B,C,D,E}$ is denoted simply by $\nabla$. Recall from (2.5) that the affine curvature tensor of $\nabla$ is given by the formula

$$\begin{aligned} R(X,Y)Z\,=\,\nabla _X \nabla _Y Z -\nabla _Y \nabla _X Z- \nabla _{[X, Y ]}Z. \end{aligned}$$

Substituting for X, Y and Z we get the following explicit expressions:

$$\begin{aligned}{} & {} R\bigg (\frac{\partial }{\partial \tau }, \frac{\partial }{\partial z_1}\bigg ) \frac{\partial }{\partial z_1}\,=\, \nabla _{\frac{\partial }{\partial \tau }} \nabla _{\frac{\partial }{\partial z_1}} \frac{\partial }{\partial z_1}- \nabla _{\frac{\partial }{\partial z_1}} \nabla _{\frac{\partial }{\partial \tau }} \frac{\partial }{\partial z_1}\,=\, \nabla _{\frac{\partial }{\partial \tau }} \bigg (f_{z_1,z_1}^{z_1} \frac{\partial }{\partial z_1}\bigg ) \\{} & {} \qquad \qquad \qquad \qquad \qquad \qquad - \nabla _{\frac{\partial }{\partial z_1}} \bigg (f_{\tau ,,z_1}^{\tau } \frac{\partial }{\partial \tau }+ f_{\tau ,,z_1}^{z_1}\frac{\partial }{\partial z_1}\bigg ) \\{} & {} \qquad =\, f_{z_1,z_1}^{z_1}\bigg (f_{\tau ,,z_1}^{\tau } \frac{\partial }{\partial \tau } + f_{\tau ,,z_1}^{z_1} \frac{\partial }{\partial z_1}\bigg ) -f_{\tau , z_1}^{\tau } \nabla _{\frac{\partial }{\partial z_1}} \frac{\partial }{\partial \tau } -f_{\tau , z_1}^{z_1} \nabla _{\frac{\partial }{\partial z_1}} \frac{\partial }{\partial z_1} \\{} & {} \qquad =\, f_{z_1,z_1}^{z_1}\bigg (f_{\tau ,,z_1}^{\tau } \frac{\partial }{\partial \tau } + f_{\tau ,,z_1}^{z_1} \frac{\partial }{\partial z_1}\bigg )-f_{\tau , z_1}^{\tau } \bigg (f_{z_1, \tau }^{\tau } \frac{\partial }{\partial \tau }+f_{z_1, \tau }^{z_1} \frac{\partial }{\partial z_1}\bigg )-f_{\tau , z_1}^{z_1}f_{z_1,z_1}^{z_1} \frac{\partial }{\partial z_1} \\{} & {} \qquad =\,\frac{C^2}{4} \frac{\partial }{\partial \tau } -\frac{CE}{4}\frac{\partial }{\partial z_1}. \end{aligned}$$

By symmetry we get

$$\begin{aligned} R\bigg (\frac{\partial }{\partial \tau }, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial z_2}\,=\, \frac{D^2}{4} \frac{\partial }{\partial \tau } -\frac{DE}{4}\frac{\partial }{\partial z_2}, \end{aligned}$$

and

$$\begin{aligned}{} & {} R\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial z_1}\,=\, \nabla _{\frac{\partial }{\partial z_1}} \nabla _{\frac{\partial }{\partial z_2}} \frac{\partial }{\partial z_1}- \nabla _{\frac{\partial }{\partial z_2}} \nabla _{\frac{\partial }{\partial z_1 }} \frac{\partial }{\partial z_1} \,=\, \nabla _{\frac{\partial }{\partial z_1}} \bigg (f_{z_1,z_2}^{z_1} \frac{\partial }{\partial z_1} + f_{z_1,z_2}^{z_2} \frac{\partial }{\partial z_2}\bigg ) \\{} & {} \quad - \nabla _{\frac{\partial }{\partial z_2} } \bigg (f_{z_1,z_1}^{z_1} \frac{\partial }{\partial z_1}\bigg )\,=\, f_{z_1,z_2}^{z_1}\cdot f_{z_1,z_1}^{z_1} \frac{\partial }{\partial z_1} +f_{z_1,z_2}^{z_2} \nabla _{\frac{\partial }{\partial z_1}}\frac{\partial }{\partial z_2}-f_{z_1,z_1}^{z_1} \nabla _{\frac{\partial }{\partial z_2}}\frac{\partial }{\partial z_1} \\{} & {} \quad =\, \frac{C^2}{2} \frac{\partial }{\partial z_1} +( \frac{1}{2}D-C) \nabla _{\frac{\partial }{\partial z_1}} \frac{\partial }{\partial z_2}\, =\, \frac{C^2}{2} \frac{\partial }{\partial z_1} +\bigg (\frac{1}{2}D-C\bigg ) \bigg (f_{z_1,z_2}^{z_1} \frac{\partial }{\partial z_1} + f_{z_1,z_2}^{z_2} \frac{\partial }{\partial z_2}\bigg ) \\{} & {} \quad =\, \frac{C^2}{2} \frac{\partial }{\partial z_1} +\bigg (\frac{1}{2}D-C\bigg ) \bigg (\frac{C}{2} \frac{\partial }{\partial z_1} + \frac{D}{2} \frac{\partial }{\partial z_2}\bigg )\,=\, \frac{CD}{4} \frac{\partial }{\partial z_1} + \frac{D^2-2CD}{4} \frac{\partial }{\partial z_2}. \end{aligned}$$

By symmetry we get

$$\begin{aligned} R\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial z_2}\,=\, -R\bigg (\frac{\partial }{\partial z_2}, \frac{\partial }{\partial z_1}\bigg ) \frac{\partial }{\partial z_2}\,=\,- \frac{C^2-2CD}{4} \frac{\partial }{\partial z_1} - \frac{CD}{4}\frac{\partial }{\partial z_2}. \end{aligned}$$

We have

$$\begin{aligned}{} & {} R\bigg (\frac{\partial }{\partial \tau }, \frac{\partial }{\partial z_1}\bigg ) \frac{\partial }{\partial z_2}\,=\, \nabla _{\frac{\partial }{\partial \tau }} \nabla _{\frac{\partial }{\partial z_1}} \frac{\partial }{\partial z_2}- \nabla _{\frac{\partial }{\partial z_1}} \nabla _{\frac{\partial }{\partial \tau }} \frac{\partial }{\partial z_2} \\{} & {} \quad =\,\nabla _{\frac{\partial }{\partial \tau }} \bigg (f_{z_1,z_2}^{z_1} \frac{\partial }{\partial z_1} + f_{z_1,z_2}^{z_2} \frac{\partial }{\partial z_2}\bigg ) - \nabla _{\frac{\partial }{\partial z_1}} \bigg (f_{\tau ,z_2}^{\tau } \frac{\partial }{\partial \tau } + f_{\tau ,z_2}^{z_2} \frac{\partial }{\partial z_2}\bigg ) \\{} & {} \quad =\, \frac{1}{2}C \bigg (f_{\tau , z_1}^{\tau } \frac{\partial }{\partial \tau } + f_{\tau , z_1}^{z_1} \frac{\partial }{\partial z_1}\bigg )+ \frac{1}{2} D\bigg ( f_{\tau , z_2}^{\tau } \frac{\partial }{\partial \tau } + f_{\tau , z_2}^{z_2} \frac{\partial }{\partial z_2}\bigg ) - f_{z_2, \tau }^{\tau } \nabla _{\frac{\partial }{\partial z_1}}\frac{\partial }{\partial \tau }- f_{z_2, \tau }^{z_2} \nabla _{\frac{\partial }{\partial z_1}}\frac{\partial }{\partial z_2} \\{} & {} \quad =\, \frac{1}{4} C^2 \frac{\partial }{\partial \tau }+ \frac{1}{4} CE \frac{\partial }{\partial z_1} + \frac{1}{4} D^2 \frac{\partial }{\partial \tau } + \frac{1}{4} DE \frac{\partial }{\partial z_2} - \frac{1}{2} D\bigg (f_{z_1, \tau }^{\tau } \frac{\partial }{\partial \tau } + f_{z_1, \tau }^{z_1} \frac{\partial }{\partial z_1}\bigg ) \\{} & {} \quad - \frac{1}{2} E\bigg (f_{z_1, z_2}^{z_1} \frac{\partial }{\partial z_1} + f_{z_1, z_2}^{z_2} \frac{\partial }{\partial z_2}\bigg ) \,=\, \frac{1}{4} (C^2 + D^2- CD) \frac{\partial }{\partial \tau }- \frac{1}{4}DE \frac{\partial }{\partial z_1}. \end{aligned}$$

By symmetry,

$$\begin{aligned} R\bigg (\frac{\partial }{\partial \tau }, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial z_1} \,=\, \frac{1}{4} (C^2 + D^2- CD) \frac{\partial }{\partial \tau } - \frac{1}{4}CE \frac{\partial }{\partial z_2}. \end{aligned}$$

We also compute that

$$\begin{aligned} R\bigg (\frac{\partial }{\partial z_1 }, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial \tau } \,= & {} \, \nabla _{\frac{\partial }{\partial z_1}} \nabla _{\frac{\partial }{\partial z_2}} \frac{\partial }{\partial \tau }-\nabla _{\frac{\partial }{\partial z_2}} \nabla _{\frac{\partial }{\partial z_1 }} \frac{\partial }{\partial \tau } \\= & {} \, \nabla _{\frac{\partial }{\partial z_1}} \bigg (f_{\tau , z_2}^{\tau } \frac{\partial }{\partial \tau } + f_{\tau , z_2}^{z_2} \frac{\partial }{\partial z_2}\bigg ) - \nabla _{\frac{\partial }{\partial z_2}} \bigg (f_{\tau , z_1}^{\tau } \frac{\partial }{\partial \tau } + f_{\tau , z_1}^{z_1} \frac{\partial }{\partial z_1}\bigg ) \\= & {} \,f_{\tau , z_2}^{\tau } \bigg (f_{\tau , z_1}^{z_1} \frac{\partial }{\partial z_1} + f_{\tau , z_1}^{\tau } \frac{\partial }{\partial \tau }\bigg )+ f_{\tau , z_2}^{z_2} \bigg (f_{z_1,z_2}^{z_1} \frac{\partial }{\partial z_1} + f_{z_1, z_2}^{z_2} \frac{\partial }{\partial z_2}\bigg ) \\{} & {} -f_{\tau , z_1}^{\tau } \bigg (f_{\tau ,z_2}^{z_2} \frac{\partial }{\partial z_2} + f_{\tau , z_2}^{\tau } \frac{\partial }{\partial \tau }\bigg ) - f_{\tau , z_1}^{z_1} \bigg (f_{z_1,z_2}^{z_1} \frac{\partial }{\partial z_1} + f_{z_1, z_2}^{z_2} \frac{\partial }{\partial z_2}\bigg ) \\= & {} \, \frac{DE}{4} \frac{\partial }{\partial z_1} -\frac{CE}{4}\frac{\partial }{\partial z_2}, \end{aligned}$$

and

$$\begin{aligned}{} & {} R\bigg (\frac{\partial }{\partial \tau }, \frac{\partial }{\partial z_1}\bigg ) \frac{\partial }{\partial \tau } \,=\, \nabla _{\frac{\partial }{\partial \tau }} \nabla _{\frac{\partial }{\partial z_1}} \frac{\partial }{\partial \tau }-\nabla _{\frac{\partial }{\partial z_1}} \nabla _{\frac{\partial }{\partial \tau }} \frac{\partial }{\partial \tau } \\{} & {} \quad =\, \nabla _{\frac{\partial }{\partial \tau }} \bigg (f_{\tau , z_1}^{z_1} \frac{\partial }{\partial z_1} + f_{\tau , z_1}^{\tau } \frac{\partial }{\partial \tau }\bigg )- \nabla _{\frac{\partial }{\partial z_1}} \bigg ( f_{\tau , \tau }^{\tau } \frac{\partial }{\partial \tau }+ f_{\tau , \tau }^{z_1} \frac{\partial }{\partial z_1} + f_{\tau , \tau }^{z_2} \frac{\partial }{\partial z_2}\bigg ) \\{} & {} \quad =\, f_{\tau , z_1}^{z_1} \bigg (f_{\tau , z_1}^{\tau } \frac{\partial }{\partial \tau }+f_{\tau , z_1}^{z_1} \frac{\partial }{\partial z_1}\bigg )+ f_{\tau , z_1}^{\tau } \bigg (f_{\tau , \tau }^{\tau } \frac{\partial }{\partial \tau } + f_{\tau , \tau }^{z_1} \frac{\partial }{\partial z_1} + f_{\tau , \tau }^{z_2} \frac{\partial }{\partial z_2}\bigg ) \\{} & {} \qquad - f_{\tau , \tau }^{\tau } \bigg (f_{\tau , z_1}^{z_1} \frac{\partial }{\partial z_1}+f_{\tau , z_1}^{\tau } \frac{\partial }{\partial \tau }\bigg ) - f_{\tau , \tau }^{z_1} f_{z_1, z_1}^{z_1} \frac{\partial }{\partial z_1}- f_{\tau , \tau }^{z_2} \bigg (f_{z_1, z_2}^{z_1} \frac{\partial }{\partial z_1}+f_{z_1, z_2}^{z_2} \frac{\partial }{\partial z_2}\bigg ) \\{} & {} \quad =\, \frac{EC}{4} \frac{\partial }{\partial \tau }+ \bigg (-\frac{E^2}{4}-\frac{1}{2}C(A+B)\bigg ) \frac{\partial }{\partial z_1}+ \frac{1}{2}B(C-D) \frac{\partial }{\partial z_2}. \end{aligned}$$

By symmetry,

$$\begin{aligned} R\bigg (\frac{\partial }{\partial \tau }, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial \tau } \,=\, \frac{ED}{4} \frac{\partial }{\partial \tau }+ \bigg (-\frac{E^2}{4}-\frac{1}{2}D(A+B)\bigg ) \frac{\partial }{\partial z_2}+ \frac{1}{2}A(D-C) \frac{\partial }{\partial z_1}. \end{aligned}$$

Now we compute the Ricci curvature defined in (2.7):

$$\begin{aligned}{} & {} {\textrm{Ricci}} \bigg (\frac{\partial }{\partial z_1},\, \frac{\partial }{\partial z_2}\bigg ) \,=\, {\textrm{Ricci}} \bigg (\frac{\partial }{\partial z_2},\, \frac{\partial }{\partial z_1}\bigg )\,=\,\frac{1}{4} (C^2+D^2); \\{} & {} \quad {\textrm{Ricci}} \bigg (\frac{\partial }{\partial \tau }, \,\frac{\partial }{\partial z_1}\bigg )\,=\, {\textrm{Ricci}} \bigg (\frac{\partial }{\partial z_1},\, \frac{\partial }{\partial \tau }\bigg )\,=\,\frac{1}{2}CE; \\{} & {} \quad {\textrm{Ricci}} \bigg (\frac{\partial }{\partial z_2},\, \frac{\partial }{\partial \tau }\bigg )\,=\, {\textrm{Ricci}} \bigg (\frac{\partial }{\partial \tau },\, \frac{\partial }{\partial z_2}\bigg )\,=\,\frac{1}{2}DE; \\{} & {} \quad {\textrm{Ricci}} \bigg (\frac{\partial }{\partial z_1},\, \frac{\partial }{\partial z_1}\bigg )\,=\,\frac{1}{4}(C^2+2CD-D^2); \\{} & {} \quad {\textrm{Ricci}} \bigg (\frac{\partial }{\partial z_2}, \,\frac{\partial }{\partial z_2}\bigg )\,=\,\frac{1}{4}(D^2+2CD-C^2); \\{} & {} \quad {\textrm{Ricci}} \bigg (\frac{\partial }{\partial \tau },\, \frac{\partial }{\partial \tau }\bigg )\,=\, \frac{1}{2}E^2+ \frac{1}{2}(A+B)(C+D). \end{aligned}$$

Recall from (2.6) the formula for the Weyl projective tensor W in dimension three:

$$\begin{aligned} W(X,\,Y)Z= & {} R(X,\,Y)Z-\frac{1}{4} {\textrm{Tr}}R(X,\,Y)Z-\frac{1}{2} [{\textrm{Ricci}}(Y,\,Z)X- {\textrm{Ricci}}(X,\,Z)(Y)]\\{} & {} - \frac{1}{8} [{\textrm{Tr}}R(Y,\,Z)X-{\textrm{Tr}}R(X,\,Z)(Y)]. \end{aligned}$$

Also, recall that the connection $\nabla$ is projectively flat if and only if the tensor W vanishes identically. The Weyl projective tensor W is anti-symmetric in $(X,\, Y)$ and satisfies the first Bianchi identity in (2.8).

Since ${\textrm{Ricci}}$ for $\nabla$ is symmetric, it follows that ${\textrm{Tr}}R$ vanishes identically. Connections with symmetric Ricci tensor are called equiaffine. The geometrical meaning of it is that there is a parallel holomorphic volume form [47, p. 222, Appendix A.3]. The above formula for Weyl projective tensor for $\nabla$ reduces to

$$\begin{aligned} W(X,\,Y)Z\,=\, R(X,Y)Z-\frac{1}{2} [{\textrm{Ricci}}(Y,Z)X-{\textrm{Ricci}}(X,Z)(Y) ]. \end{aligned}$$

The computation for $W(\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_2}) \frac{\partial }{\partial z_2}$ is as follows:

$$\begin{aligned}{} & {} W\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial z_2}\,=\, R\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_2}\bigg )\frac{\partial }{\partial z_2}-\frac{1}{2} [{\textrm{Ricci}}\bigg (\frac{\partial }{\partial z_2}, \frac{\partial }{\partial z_2}\bigg )\frac{\partial }{\partial z_1}-{\textrm{Ricci}}\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_2})\frac{\partial }{\partial z_2}\bigg )]\\{} & {} \quad =\,\frac{1}{4}(2CD-C^2) \frac{\partial }{\partial z_1} -\frac{1}{4}CD \frac{\partial }{\partial z_2}-\frac{1}{8}(D^2+2CD-C^2) \frac{\partial }{\partial z_1} + \frac{1}{8} (C^2+D^2)\frac{\partial }{\partial z_2} \\{} & {} \quad =\,-\frac{1}{8} (C-D)^2 \frac{\partial }{\partial z_1} + \frac{1}{8} (C-D)^2 \frac{\partial }{\partial z_2}. \end{aligned}$$

Hence

$$\begin{aligned} W\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial z_2}\,=\, -\frac{1}{8} (C-D)^2 \frac{\partial }{\partial z_1} + \frac{1}{8} (C-D)^2 \frac{\partial }{\partial z_2}. \end{aligned}$$

Also,

$$\begin{aligned}{} & {} W\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial z_1}\,=\, R\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_2}\bigg )\frac{\partial }{\partial z_1}-\frac{1}{2} [{\textrm{Ricci}}\bigg (\frac{\partial }{\partial z_2}, \frac{\partial }{\partial z_1}\bigg )\frac{\partial }{\partial z_1}-{\textrm{Ricci}}\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_1})\frac{\partial }{\partial z_2}\bigg )]\\{} & {} \quad =\,\frac{1}{4}CD \frac{\partial }{\partial z_1} + \frac{1}{4}(D^2-2CD)\frac{\partial }{\partial z_2}-\frac{1}{8}(C^2+D^2) \frac{\partial }{\partial z_1} + \frac{1}{8} (C^2+2CD-D^2)\frac{\partial }{\partial z_2} \\{} & {} \quad =-\frac{1}{8}\, (C-D)^2 \frac{\partial }{\partial z_1} + \frac{1}{8} (C-D)^2 \frac{\partial }{\partial z_2}. \end{aligned}$$

In conclusion,

$$\begin{aligned} W\bigg (\frac{\partial }{\partial z_1},\, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial z_1} \,=\, -\frac{1}{8} (C-D)^2 \frac{\partial }{\partial z_1} + \frac{1}{8} (C-D)^2 \frac{\partial }{\partial z_2}. \end{aligned}$$

We get that

$$\begin{aligned} W\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial \tau }= & {} R\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial z_2}\bigg )\frac{\partial }{\partial \tau }-\frac{1}{2} [{\textrm{Ricci}}\bigg (\frac{\partial }{\partial z_2}, \frac{\partial }{\partial \tau }\bigg )\frac{\partial }{\partial z_1} \\{} & {} -{\textrm{Ricci}}\bigg (\frac{\partial }{\partial z_1}, \frac{\partial }{\partial \tau })\frac{\partial }{\partial z_2}\bigg )]\,=\, 0. \end{aligned}$$

Hence we have

$$\begin{aligned} W\bigg (\frac{\partial }{\partial z_1},\, \frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial \tau }\,=\, 0. \end{aligned}$$

By similar direct computations we get that

$$\begin{aligned} W\bigg (\frac{\partial }{\partial \tau },\, \frac{\partial }{\partial z_1}\bigg ) \frac{\partial }{\partial z_2}\,=\,\frac{1}{8} (C-D)^2 \frac{\partial }{\partial \tau } \end{aligned}$$

and

$$\begin{aligned} W\bigg (\frac{\partial }{\partial \tau }, \,\frac{\partial }{\partial z_1}\bigg ) \frac{\partial }{\partial \tau }\,=\, \frac{1}{4}(A+B)(D-C)\frac{\partial }{\partial z_1} + \frac{1}{2} B(C-D) \frac{\partial }{\partial z_2}. \end{aligned}$$

Also by direct computation:

$$\begin{aligned} W\bigg (\frac{\partial }{\partial \tau }, \,\frac{\partial }{\partial z_1}\bigg ) \frac{\partial }{\partial z_1}\,=\, \frac{1}{8} (C-D)^2 \frac{\partial }{\partial \tau } \end{aligned}$$

and

$$\begin{aligned} W\bigg (\frac{\partial }{\partial \tau }, \,\frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial z_2} \,=\, \frac{1}{8} (C-D)^2 \frac{\partial }{\partial \tau }. \end{aligned}$$

Again by a direct computation,

$$\begin{aligned} W\bigg (\frac{\partial }{\partial z_2},\, \frac{\partial }{\partial \tau }\bigg ) \frac{\partial }{\partial \tau }\,=\, \frac{1}{2}A(C-D) \frac{\partial }{\partial z_1}+ \frac{1}{4}(A+B)(D-C)\frac{\partial }{\partial z_2}. \end{aligned}$$

The other components of the Weyl tensor can be obtained using the first Bianchi identity in (2.8). Indeed, from

$$\begin{aligned} W\bigg (\frac{\partial }{\partial z_1}, \,\frac{\partial }{\partial z_2}\bigg ) \frac{\partial }{\partial \tau }+ W(\frac{\partial }{\partial z_2},\, \frac{\partial }{\partial \tau }) \frac{\partial }{\partial z_1}+ W\bigg (\frac{\partial }{\partial \tau },\, \frac{\partial }{\partial z_1}\bigg ) \frac{\partial }{\partial z_2}\,=\,0 \end{aligned}$$

we infer that

$$\begin{aligned} W\bigg (\frac{\partial }{\partial z_2}, \,\frac{\partial }{\partial \tau }\bigg ) \frac{\partial }{\partial z_1}\,=\,-\frac{1}{8}(C-D)^2 \frac{\partial }{\partial \tau }. \end{aligned}$$

(5.2)

Notice that the Weyl projective tensor W does not depend on the parameter E. This is due to the facts that W is a projective invariant and $\nabla ^{A,B,C,D,E}$ is projectively equivalent with $\nabla ^{A,B,C,D,0}$. Indeed, let $\phi _{\tau }$ be the holomorphic one-form on ${{\mathbb {T}}}^3$ defined by

$$\begin{aligned} \phi _{\tau } \bigg (\frac{\partial }{\partial \tau }\bigg )\,=\, \frac{1}{2}E\ \ \text { and }\ \ \phi _{\tau } \bigg (\frac{\partial }{\partial z_i}\bigg )\, =\, 0 \end{aligned}$$

for $i=1,2$. Then

$$\begin{aligned} \nabla ^{A,B,C,D,E}_XY -\nabla ^{A,B,C,D,0}_XY\,=\, \phi _{\tau }(X)(Y)+ \phi _{\tau }(Y)X \end{aligned}$$

(5.3)

for all holomorphic vector fields $X,\, Y$; the identity in (5.3) being tensorial it can be easily verified for any pair of vectors chosen from the basis $(\frac{\partial }{\partial \tau }, \,\frac{\partial }{\partial z_1},\, \frac{\partial }{\partial z_2})$. From (5.3) it follows immediately that $\nabla ^{A,B,C,D,E}$ and $\nabla ^{A,B,C,D,0}$ are projectively equivalent.

From (5.2) and the expression of all components of the Weyl projective tensor, it follows that W vanishes identically if and only if $C\,=\, D$. $\square$