In [arXiv:math/0605371, Theorem 4 on p.8](https://arxiv.org/pdf/math/0605371v1.pdf), there is the following statement:


Let $X$ be a unit Killing vector field on a $n$-dimensional Riemannian manifold $M$. Then the Ricci curvature $\operatorname{Ric}$ of the manifold $M$ satisfies the condition $\operatorname{Ric}(X,X)\geq 0$.




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Now, let $H^3$ be a hyperbolic 3-sphere, i.e. the set of all points $x\in\mathbb{R}^{1,3}$ with $\langle x,\, x\rangle=-1$, where $\langle \cdot,\cdot\rangle$ is the scalar product in Lorentz $n$-space with metric $(\eta\_{ij})=\text{diag}(-1,1,1,1)$


Then consider the (complex valued) unit Killing vector $X=(x\_2, x\_1,-i x\_4,i x\_3)$
for which $\langle X, \, X\rangle=1$ (using standard hyper-spherical coordinates).



> 
> **Question:** Isn't this a counter-example? Are imaginary Killing fields not allowed?
> 
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Unit Killing vector fields on pseudo Riemannian manifolds
All vector fields in Riemannian geometry papers are assumed real and smooth, unless otherwise stated. All vector fields in complex algebraic geometry papers are assumed complex and holomorphic, unless otherwise stated.


