I'm trying to find the mistake in the following computation: let $V$ be a (complex) vector space equipped with a hermitian inner product $\langle\cdot,\cdot\rangle$, ${\rm P}V$ be the projective space of $V$ and $\mathcal{L} \to {\rm P}V$ be the tautological line bundle. As $\mathcal{L}$ is a subbundle of ${\rm P}V \times V$, it inherits a fiber metric via $\langle\cdot,\cdot\rangle$ and so we may project the standard flat connection ${\rm D}$ on the trivial bundle to a connection $\nabla$ on $\mathcal{L}$. Now, sections of $\mathcal{L}$ correspond to homogeneous functions $\mu\colon V\setminus \{0\} \to \Bbb C$ of degree $-1$ via passing $x \mapsto \mu(x)x$ to the quotient. If $X \in \mathfrak{X}({\rm P}V)$, then the derivative of this map is $$({\rm D}_X\mu)_x = {\rm d}\mu_x(Xx)x+\mu(x)Xx,$$so since $Xx \in (\Bbb Cx)^\perp$ (since $T_{{\Bbb C}x}({\rm P}V) \cong {\rm Hom}(\Bbb C x, (\Bbb C x)^\perp)$), we obtain $(\nabla_X\mu)_x = {\rm d}\mu_x(Xx)$. And $x\mapsto {\rm d}\mu_x(Xx)$ is also homogeneous of degree $-1$.
The connection $\nabla$ is not flat, because it does not admit any local parallel sections ($\nabla \mu = 0$ implies ${\rm d}\mu_x|_{(\Bbb C x)^\perp} = 0$ for all $x$, which contradicts non-integrability of the horizontal distribution of the submersion $\Bbb S^{\dim V-1}\to {\rm P}V$).
But let's try to compute $R(X,Y)\mu$. First it seems we have $$\nabla_X\nabla_Y\mu = {\rm d}^2\mu(X,Y) + {\rm d}\mu({\rm d}Y(X)),$$where ${\rm d}^2\mu$ is ${\rm d}({\rm d}\mu)\colon V\setminus \{0\} \to {\rm Hom}(V,V^*)$ regarded as a (symmetric) bilinear form. This would imply that $$R(X,Y)\mu = {\rm d}\mu({\rm d}Y(X) - {\rm d}X(Y) - [X,Y]).$$I want to say this is zero because ${\rm D}$ is torsion-free.
- What's precisely the screw-up here?
- How to fix it?
Thanks.