Consider $g_1$ and $g_2$ two Riemannian metrics on a differentiable manifold $M$ of dimension $n\ge 4$. Suppose locally $g_i=f_i\sum_{j=1}^ndx_j^2$, where $f_i:M\rightarrow \mathbb{R}$ are non negative functions. Suppose that both $g_1$ and $g_2$ have non positive sectional curvature. Define the metric $\hat g$ on $M$ as $\hat g:=max\{f_1,f_2\}\sum_{j=1}^ndx_j^2$.

Short question: Will $(M,\hat g)$ have non positive curvature?

More elaborate question: A priori $\hat g$ has just continuous coefficients: we can ask if its intrinsic induced metric $d_{\hat g}$ has non positive Alexandrov curvature (it means that every $p\in M$ has an open neighborhood $U$ such that for every $x,y,z\in U$ $d_{\hat g}(z,m)^2\le \frac 1 2(d_{\hat{g}}(z,x)^2+d_{\hat{g}}(z,y)^2)-\frac 1 4d_{\hat{g}}(x,y)^2$, where $m$ is the midpoint between $x$ and $y$, definition taken from http://people.mpim-bonn.mpg.de/hwbllmnn/archiv/NPC0606.pdf)

Clearly in the open sets $U_1:=\{p\in M|f_1(p)>f_2(p)\}$ or $U_2:=\{p\in M|f_2(p)>f_1(p)\}$ this is true, because the metrics $g_1$ and $g_2$ have negative sectional curvature. But I don't know what happens in points where $f_1=f_2$.

Say $p\in M$ is such that $f_1(p)=f_2(p)$. Then every open neighborhood $U$ of $p$ will intersect both $U_1$ and $U_2$ and evaluating directly if the preceding inequality is true will be quite difficult. Is there a more straightforward way (than direct computation) to conclude that $d_{\hat{g}}$ has non positive Alexandrov curvature?

  • $\begingroup$ Is $\hat g$ even smooth ? $\endgroup$ – Amr Apr 18 at 16:27

Dimension 2 :

The answer to your question is yes in dimension 2. I am not sure about higher dimension.

Set $f_i=e^{2u_i}$, so that $f=e^{2u}$ with $u=\max (u_1,u_2)$.

The Gauss curvature of a metric $e^{2v}(dx^2+dy^2)$ is given by $-e^{-2v}\Delta v$, so that each of the $u_i$ is subharmonic, hence $u$ it self is subharmonic as the maximum of two subharmonic functions.

There is a theory of metrics on surfaces of the form $e^{2u}(dx^2+dy^2)$ where $u$ is subharmonic (or a difference of subharmonic functions) which has been developed by Reshetnyak in the 50s.

Basically, there is notion of curvature measure, which is $-\Delta_d u$, the opposite of the distributional laplacian of $u$. Since $u$ is subharmonic, this curvature measure will be nonpositive. From there you can show that $e^{2u}(dx^2+dy^2)$ has nonpositive curvature in the sense of Alexandrov.

Reshetnyak's theory is exposed in "Two-Dimensional Manifolds of Bounded Curvature", Yu. G. Reshetnyak, in Encyclopaedia of Mathematical Sciences, Volume 70 1993, Geometry IV.

Higher Dimension :

Unfortunately the formula for the conformal change of curvature in higher dimension is not so nice and to my knowledge bothered to make this approach work in higher dimensions.

The $(4,0)$ curvature tensor of $e^{2u}g$ where $g$ is flat and $u$ is smooth is given by :

$$R_u-e^{2u}g\wedge(\nabla^2u-du\otimes du+\tfrac{1}{2}|du|^2g).$$

Nonnegativity of the sectional curvature is then equivalent to the sum of the two lowest eigenvalues of $S_u=\nabla^2u-du\otimes du+\tfrac{1}{2}|du|^2g$ being nonnegative (I'll say that $S_u$ is $2$-nonnegative).

If $u=\max(u_1,u_2)$ where $u_1$ and $u_2$ are smooth it should be possible to make sense of $S_u$ at least as a symmetric tensor with measure coefficients. A starting point would be to see if the $2$-nonnegativity of $S_{u}$ (in some weak sense) can be inferred from the $2$-nonnegativity of $S_{u_1}$ and $S_{u_2}$. If $S_u$ was just $\nabla^2u$, it would boil down to a statement whose proof would be roughly the same as "sup of convex is convex" or "sup of subharmonic is subharmonic" (this is what was going on in 2d). However the occurence of the gradient terms in a nonlinear fashion makes this not as straight forward.

Then comes the real question in some sense : does the negativity of this $R_u$ in a weak sense have anything to do with the metric geometry of $(M,e^{2u}g)$ ? Maybe smoothing could help there.

PS: I've been thinking about this kind of problems for quite some time, feel free to directly email me if you have further questions.

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  • $\begingroup$ Thank you very much, but actually I'm interested in cases $n\ge 4$, so I really need to find out what happens in dimensions greater than 2.. $\endgroup$ – user99087 Oct 7 '16 at 20:54
  • $\begingroup$ Thank you! I'll think about it and in case I'll have further questions I'll email you $\endgroup$ – user99087 Oct 17 '16 at 8:04

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