# Linear functions on Riemannian manifolds

Let $$(M,g)$$ be a Riemannian manifold with the corresponding $$LC$$ connection $$\nabla$$. For a smooth function $$f:M \to \mathbb{R}$$, the Hessian of $$f$$ is defined as a 2-linear map on the tangent space as follows $$Hess(f)(x)(V,W)=g(\nabla_V \nabla f,W)$$ where $$V,W \in T_x M$$.

A linear function is a function whose Hessian is identically zero.

What is an example of a non constant linear function on each of the following manifolds

1)On $$S^2$$ with standard metric

2)On $$\mathbb{C}P^2$$ with Fubini Study metric

Furthermore, what can be said about the dimension of the space of linear functions?

If $$M$$ is assumed to be connected and complete, then the only such function is the zero function unless $$M$$ is isometric to a Riemannian product of the form $$\mathbb R \times N$$ for some complete Riemannian manifold $$N$$. In particular, if $$M$$ is compact (for example, $$S^2$$ or $$\mathbb C\mathbb P^2$$), then there is only the zero function.

The full proof would involve too many details to write down here, but here's a sketch. Suppose $$M$$ is complete and $$f$$ is a smooth function satisfying $$\text{Hess}(f) \equiv 0$$. Since $$\nabla f$$ usually represents a $$1$$-form rather than a vector field in Riemannian geometry, I prefer to write this as $$0 = \text{Hess}(f)(V,W) = g(\nabla _V (\text{grad}f),W),$$ for any smooth vector fields $$V,W$$, where $$\text{grad} f = (\nabla f)^\sharp= (df)^\sharp$$, with $$\sharp$$ denoting the musical isomorphism from $$T^*M$$ to $$TM$$.

Then for any vector field $$V$$, we have $$\nabla_V |\text{grad} f|^2 = 2 g (\nabla_V (\text{grad}f ), \text{grad} f) = 2\text{Hess}(V,\text{grad}f) = 0,$$ so $$|\text{grad}f|^2$$ is constant. Suppose now that it is nonzero. After multiplying $$f$$ by a constant, we can assume that $$|\text{grad} f| \equiv 1$$.

Let $$X$$ denote the vector field $$\text{grad} f$$. The fact that $$X$$ is bounded implies that it's complete. Along any integral curve $$\gamma$$ of $$X$$, $$(f\circ\gamma)'(t) = df(\gamma'(t)) = g(X, \gamma'(t) ) = g(X,X) = 1,$$ so $$f(\gamma(t)) = f(\gamma(0)) + t$$. In particular, $$f\colon M\to \mathbb R$$ is surjective. Moreover, the fact that $$\nabla X \equiv 0$$ implies that $$X$$ is a Killing vector field, i.e., $$\mathscr L_X g \equiv 0$$.

Let $$N = f^{-1}(0)$$. Because $$df$$ never vanishes, this is a nonempty embedded codimension-$$1$$ submanifold. A standard computation in Riemannian geometry using $$|\text{grad} f|\equiv 1$$ shows that $$f(x)$$ is equal to the Riemannian distance from $$x$$ to $$N$$, and the integral curves of $$X$$ are geodesics that intersect $$N$$ orthogonally.

Let $$\Phi\colon \mathbb R\times M\to M$$ be the flow of $$X$$, and let $$\phi\colon \mathbb R \times N\to M$$ be the restriction of $$\Phi$$. I claim that $$\phi$$ is a diffeomorphism. Surjectivity follows from the fact that every point of $$M$$ can be connected to $$N$$ by a minimizing geodesic, which intersects $$N$$ orthogonally and therefore is the image of an integral curve of $$X$$. Injectivity is proved as follows: if $$\phi(t,x) = \phi(t',x')$$, then $$t = f(\phi(t,x)) = f(\phi(t',x')) = t'$$. Since $$x\mapsto \phi(t,x)$$ is a diffeomorphism with inverse $$x\mapsto \phi(-t,x)$$, this implies $$x=x'$$. Then it can be shown that $$d\phi$$ never vanishes, so $$\phi$$ is a bijective local diffeomorphism and hence a diffeomorphism.

By pulling back $$g$$ by the diffeomorphism $$\phi$$, we might as well think of $$g$$ as a metric on $$\mathbb R\times N$$, and think of $$X$$ as the vector field $$\partial/\partial t$$. Both $$g$$ and the product metric are invariant under the flow of $$X$$ (since $$X$$ is a Killing vector field for both metrics), and they agree along $$\{0\}\times N$$, so they are equal everywhere.

(If you omit the hypothesis that $$M$$ is complete, then the argument above can be adapted to show that $$M$$ is locally isometric to a Riemannian product with $$\mathbb R$$.)

• Dear Prof. Lee Thank you very much for this very interesting answer. – Ali Taghavi Sep 29 '18 at 7:15
• @AliTaghavi: You’re welcome! – Jack Lee Sep 29 '18 at 13:34