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Gromov conjectured in 1985 and LLarull proved in 1998 that: If $g > g_0$ on the sphere, then there exists some point p on the sphere with $Sc(p) < Sc_0(p)$. Here $g, g_0$ are Riemannian metrics and $g_0$ is the standard metric on the sphere. $Sc$ is the scalar curvature. His proof is by contradiction and uses some non-vanishing index on the sphere.

Is the theorem also true for sectional curvature? Or is it the case that if one Riemannian metric is bigger than the other one on the sphere, then there must exist some point such that the scalar curvature at that point is smaller than the other one?

In general, is it true that the bigger the metric, the smaller the curvature (sectional, Ricci or scalar curvaure), at least in the pointwise setting?

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    $\begingroup$ Ricci curvature is scale invariant: if you replace metric $g$ by $\lambda g$ for any constant $\lambda > 0$, the Ricci curvature is the same. Sectional curvature should scale the same as the metric, and scalar curvature like the inverse metric. So trivially the answer to your questions about sectional and ricci curvatures are in the negative. $\endgroup$
    – Willie Wong
    Commented Nov 30, 2018 at 14:30
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    $\begingroup$ @WillieWong, doesn't the sectional curvature scale like the inverse metric, as you so eloquently explained here? math.stackexchange.com/a/302140/141493 This is consistent with the interpretation of the sectional curvature of a plane in a tangent space as the scalar curvature (Gaussian curvature) of the surface obtained by applying the exponential map to the plane. $\endgroup$
    – Phillip Andreae
    Commented Nov 30, 2018 at 16:36
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    $\begingroup$ @PhillipAndreae: oops; you are right, I forgot to divide by the denominator! Scratch the part of my comment about sectional. Ricci however stands. $\endgroup$
    – Willie Wong
    Commented Nov 30, 2018 at 16:39
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    $\begingroup$ The scalar curvature is an average of sectional curvatures, so the theorem of Llarull implies that if $g > g_0$, then there exists a point $p$ and at least one plane $\sigma_p \subseteq T_pM$ such that $K(\sigma_p) < K_0(\sigma_p)$. I suppose the question is: is there a $p$ for which $K(\sigma_p) < K_0(\sigma_p)$ for all planes $\sigma_p \subseteq T_p M$? (I assume this is what is meant by "smaller sectional curvature".) $\endgroup$
    – Phillip Andreae
    Commented Nov 30, 2018 at 16:52

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As it was stated by Phillip Andreae, Llarull's theorem implies that for some sectional direction sectional curvature is $< 1$ and for some unit vector Ricci curvature is smaller than $n-1$.

But if $g>g_0$ then $\mathop{\rm diam}(\mathbb{S}^n,g_1)>\pi$. Therefore the cases of Ricci curvature (as well as sectional curvature) also follow from Myers's theorem.

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  • $\begingroup$ I guess the OP wants smaller sectional curvature for all 2-planes at some point $p$ (or smaller Ricci for all unit vectors at some point $p$)... $\endgroup$
    – Mizar
    Commented Dec 1, 2018 at 13:41
  • $\begingroup$ @Mizar he did not say that; in this case the answer is "no"; analog of Berger spheres with enlarged fibers should give a counterexample. $\endgroup$
    – Anton Petrunin
    Commented Dec 1, 2018 at 19:21
  • $\begingroup$ @Anton Petrunin, I am sorry that I don't make it clear. It is for all 2-planes in my mind as Mizar and Phillip Andrease said. What I really want to understand is described in the title. Geometry is decided by the distance if we are "back to Eulid". Therefore, the negative answer of "the bigger the metric, the smaller the curvature" surprised me. $\endgroup$
    – Jialong Deng
    Commented Dec 1, 2018 at 21:16
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Lohkamp Scalar curvature and hammocks proved that it it always possible to decrease both the metric and the scalar curvature simultaneously, as pointes out by Goette and Semmelmann in the paper Scalar curvature estimates for compact symmetric spaces. So it is not true that the bigger the metric, the smaller the scalar curvature in general.

The proof of Llarull's theorem only used the area-bigger (not the length of vector bigger) metric in the round sphere. Gomove call such metric are area extremal and asked which manifolds possess such an area extremal metric in Positive curvature, macroscopic dimension, spectral gaps and higher signatures.

As far as I know, the Gromov's question partially answered by Llarull, Min-Oo, Kramer, Goette and Semmelmann, and those examples are symmetric spaces.

Thus we can not say much about the geometric meaning of one Riemannian metric bigger than the other one on an arbitrarily smooth manifold, except some symmetric spaces in above-mentioned.

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