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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?

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  • $\begingroup$ Please show your efforts as this post has a few closure votes now. $\endgroup$ – Saad Sep 29 '18 at 1:24
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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 $\mathbb R$.)

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

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