Assume given a smooth manifold $(\mathbb{R}^n, g)$, where the metric is a scaled identity $g = e^{2f}I$. Is there a way to know if this is always a non-positive (sectional) curvature manifold?

Note this is a parametrized manifold that is locally conformally flat. Following Einstein Manifolds [Arthur L. Besse], the Ricci tensor (in coordinates), can be shown to be:

$R = -(n-2)(H_f - \nabla f \cdot \nabla f^T ) - \frac{n-2}{n}(\Delta f + \|\nabla f\|^2)I_{n\times n}$

where $H_f$ is the Hessian of $f$.

Then $(\mathbb{R}^n, g)$ is of non-positive (sectional) curvature if $R$ is negative semi-definite.

  • 2
    $\begingroup$ You might want to check your formula for $R$, since the right hand side clearly vanishes when $n=2$ even though Riemannian surfaces (which are conformally flat) are not all flat. $\endgroup$ – Robert Bryant Jan 13 '12 at 0:33

I'm not quite sure what you mean by always non-positively curved. If you are asking if this metric is non-positively curved for any $f$ then this is false. If you are asking for conditions on $f$ ensuring that the resulting metric is non-positively curved then there is a general formula:

Let $(M,g)$ be a Riemannian manifold and let $\tilde g=e^{2f}g$ be a new metric on $M$. Let $p\in M$ and let $u,v$ be orthonormal with respect to $g$ vectors in $T_pM$ and $\sigma$ the 2-plane spanned by them.

Then $e^{2f}\tilde{K}_\sigma =K_\sigma-[Hess_f(u,u)+Hess_f(v,v)+|\nabla f|^2-\langle \nabla f, u \rangle^2-\langle \nabla f, v \rangle^2]$.

This formula is in Besse btw (Theorem 1.159) but it's written slightly differently there.

In the special case you are interested in $g$ is the canonical metric on $\mathbb R^n$ and hence $\tilde K$ is nonpositive iff $H_f(u,u)+H_f(v,v)+|\nabla f|^2-\langle \nabla f, u \rangle^2-\langle \nabla f, v \rangle^2\ge 0$ for any $p$ and any orthonormal $u$ and $v$ in $T_pM$. Note that for example it's always true if $f$ is convex.

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    $\begingroup$ So, more explicitly, what one requires for nonpositive curvature is that $|\nabla f|^2 + \lambda_1 +\lambda_2 \ge 0$, where $\lambda_1$ and $\lambda_2$ are the lowest eigenvalues of the quadratic form $\textrm{Hess}(f) - (d f)^2$. $\endgroup$ – Robert Bryant Jan 13 '12 at 12:42
  • $\begingroup$ @Robert That's a nice invariant way to state it. $\endgroup$ – Vitali Kapovitch Jan 13 '12 at 14:50
  • $\begingroup$ Thank you for your reply. This is very helpful. In the original expression, it is clear that for $n=2$ the Ricci tensor is $0$. I don't see this as clearly in your expression: $H_f(u,u) + H_f(v,v) + |\nabla f|^2 - <\nabla f,u>^2 - <\nabla f,v>^2 \ge 0$ Clearly, the last three terms will vanish if $n=2$, but it seems the sum of Hessians might not. I wonder if one of the expressions may be some tweaking. $\endgroup$ – Guillermozo Jan 13 '12 at 16:21
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    $\begingroup$ @Guillermozo: The expression you wrote in your question is for the trace free part of the Ricci curvature. It is of course identically zero in dimension 2. The full Ricci curvature tensor after a conformal change is $\tilde Ric=Ric-\Delta f\cdot g -(n-2)[Hess(f)-(df)^2+|\nabla f|^2\cdot g]$ which you can get by taking the trace of the sectional curvature formula above. $\endgroup$ – Vitali Kapovitch Jan 13 '12 at 17:19

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