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If the Gaussian curvature of the metric $g= f^2(x,y)(dx^2+dy^2)$ is nonzero then $f$ cannot be constant. This can be expressed by stating that the (probabilistic) variance $Var(f)$ of $f$ is nonzero (in a suitable domain). If the metric $g$ has nonzero Gaussian curvature, then $f$ is nonconstant and therefore $Var(f)$ is nonzero. Can this conclusion be quantified? Namely, assume $g$ has curvature bounded away from zero on a suitable disk. Can one get a lower bound for $Var(f)$? This has immediate applications to a stronger version of Loewner's torus inequality with isosystolic defect term a la Bonnesen, see http://arxiv.org/abs/arXiv:0803.0690 and http://arxiv.org/abs/1105.0553

Note 1. The Gaussian curvature of the metric $g= f^2(x,y)(dx^2+dy^2)$ is given by $K=\frac{-1}{2f^2}\left(\frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}\right)\log f$, so that the problem involves partial differential inequalities for the Laplacian.

Note 2. To restate the curvature hypothesis more carefully: we have a metric $r$-disk (for the metric $g$) where the curvature is bounded below by an $\epsilon>0$. The problem is to obtain a lower bound for $Var(f)$ in terms of $r$ and $\epsilon$ (certainly the estimate will become weaker as $r$ gets smaller).

Note 3 (reformulation along the lines of Robert's suggestion). On a fixed domain $\Omega$ of unit area in the $xy$-plane, we consider metrics $g=f^2(dx^2+dy^2)$ with the property that $\Omega$ contains a subdomain (say, a disk) $D$ on which Gaussian curvature $K\geq C>0$ and such that the $g$-area of $D$, i.e., $\int_D f^2 dxdy$, is at least $A>0$. We want to know whether there is a lower bound for $Var_\Omega(f)$ in terms of the constants $C$ and $A$. Is there an optimal lower bound, and if so does a rotationally symmetric metric on a disk $D$ attain it? A similar question for $K$ negative and bounded away from $0$.

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Variance with respect to what probability measure? –  Liviu Nicolaescu Apr 15 '13 at 9:27
    
Say, $f$ is defined in a unit-area domain in the $x,y$ plane. In Loewner's torus inequality, this is a fundamental domain for the torus. –  katz Apr 15 '13 at 9:31
    
@katz: I'm not familiar with your notion of '$Var(f)$'. Could you write down an explicit definition or formula? Without it, I don't see how you expect to 'quantify' any conjectured relationship. Also, when you write 'unit area domain', do you mean with respect to the standard measure in the $xy$-plane or with respect to the $g$-measure? –  Robert Bryant Apr 15 '13 at 12:19
    
The variance is the minimum of $\int(f-m)^2$ over constant $m$. The minimum is attained for $m=E(f)$, the expected value of $f$. Moreover, $E(f^2)-(E(f))^2=Var(f)$. From this combined with uniformisation one immediately deduces the strengthened form of Loewner's torus inequality. –  katz Apr 15 '13 at 12:24
    
The "unit area" is with respect to the standard area form $dxdy$. –  katz Apr 15 '13 at 12:26
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1 Answer 1

As far as I understand the question, the answer is no: for any domain $\Omega$ and any $\delta>0$, denoting by $K_f$ the curvature function of the metric $g=f^2g_{eucl}$, we have

$$\inf_{K_f\geq\delta} \mathrm{Var}(f) = \inf_{K_f\leq-\delta} \mathrm{Var}(f) = 0$$

Indeed, given any $f$ such that $K_f\geq\delta$ and any $\lambda\in(0,1)$, the function $u=\lambda f$ has $\mathrm{Var}(u)=\lambda^2 \mathrm{Var}(f)$ and $K_u=K_f/\lambda^2\geq\delta/\lambda^2>\delta$.

This seems to have to do with normalization or the volume form which is used, but I do not know how one could formulate an alternative problem with a positive answer.

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Are you proposing to take the limit as $\lambda\to 0$? But then the disk where the curvature is bounded away from zero also has area tending to zero. Obviously one needs a disk of a certain size where the curvature is bounded away from zero. For rotationally invariant metrics one can actually get explicit estimates; see the second arxiv text. –  katz Apr 15 '13 at 13:49
    
Note also that scaling does not keep you in the class of probability measures, so your construction is unclear. –  katz Apr 15 '13 at 14:39
    
I guess I did not understand the question then. You are taking a domain of unit area with respect to $dx dy$, but you compute your integrals and variance according to the normalized volume of $g$? –  Benoît Kloeckner Apr 15 '13 at 14:58
    
The domain can be taken to be a parallelogram of unit area in the $x,y$ plane, say. We have a function $f$ defining a new metric $f^2(x,y)(dx^2+dy^2)$. We are interested in the properties of this new metric. This is a reasonable set-up in studying tori because by the uniformisation theorem, every torus is conformally equivalent to a flat torus (say of unit area), and the latter can be represented by a fundamental domain in the $(x,y)$ plane. The variance of $f$ is calculated with respect to the flat metric of the $(x,y)$-plane because that's the term that arises in applying the formula... –  katz Apr 15 '13 at 16:02
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@katz: Sorry, I'm still not clear on your hypotheses. Would you agree with this: On a fixed domain $\Omega$ of unit area in the $xy$-plane, we consider metrics $g = f^2(dx^2+dy^2)$ with the property that $\Omega$ contains a subdomain $D$ on which $|K|\ge C > 0$ and such that the $g$-area of $D$, i.e., $\int_D f^2\ dxdy$, is at least $A>0$ (and, maybe, $D$ contains a disk of $g$-radius $R>0$?). We want to know whether there is a lower bound for $Var_\Omega(f)$ in terms of the constants $C$ and $A$ (and $R$?). Is that your question (maybe without the $R$ condition)? –  Robert Bryant Apr 15 '13 at 17:13
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