MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there any hope of a high-level explanation of why the fraction $\frac{1}{4}$ plays such a prominent role as a sectional curvature bound in Riemannian geometry? My (dim) understanding is that the idea is that if the sectional curvature of a manifold is constrained to be close to 1, then the manifold must be topologically a (homeomorphic to the) sphere $S^n$. Conjectured by Hopf and Rauch, proved by Berger and Klingenberg, and strengthened by Brendel and Schoen to establish diffeomorphism to $S^n$. Here is the definition of "local" $\frac{1}{4}$-pinched from Brendel and Schoen's paper "Manifolds with $1/4$-pinched curvature are space forms" (J. Amer. Math. Soc., 22(1): 287-307, January 2009; PDF):

We will say that a manifold $M$ has pointwise $1/4$-pinched sectional curvatures if $M$ has positive sectional curvature and for every point $p \in M$ the ratio of the maximum to the minimum sectional curvature at that point is less than 4.

I know the "$\frac{1}{4}$" in the $\frac{1}{4}$-pinched sphere theorem is optimal, and perhaps that is the answer to my question: $\frac{1}{4}$ appears because the theorem is false otherwise—punkt! But I am wondering if there is a high-level intelligble reason for the appearance of $\frac{1}{4}$, rather than, say, $\frac{3}{8}$, or $\frac{e}{\pi^2}$ for that matter?

I am aware this is a "fishing expedition," and a fair response is: Study the Brendel-Schoen proof closely, and enlightenment will follow!

share|cite|improve this question
A funny fact is that in odd dimension, since there are no compact rank one symmetric spaces, one can sow that the differetiable sphere theorem holds for $1/4-\epsilon(n)$-pinched manifolds (globally), this is a theorem by Ni and Wilking if I remember correctly. However, no meaningful etimates on $\epsilon(n)$ are known, except in dimension 5 (work of Berger). – Thomas Richard Oct 8 '11 at 9:08

The new proofs add a layer of complication that is unnecessary for your question. From Cheeger and Ebin, Comparison Theorems in Riemannian Geometry first paragraph of Chapter 6:

The symmetric spaces of positive curvature are known to admit metrics such that $1 \geq K_M \geq > \frac{1}{4}$; see Example 3.38. In fact, we will prove that any riemannian manifold with $1 \geq K_M > \geq \frac{1}{4},$ which is not a sphere, is isometric to one of these spaces.

Back in chapter 3, page 73,

A complete classification of symmetric spaces is available (Helgason [1962]). In particular, the only simply connected symmetric spaces having positive curvature are the spheres of constant curvatures, complex and quaternionic projective spaces, and the Cayley plane. These are sometimes referred to as the rank one symmetric spaces, and except for the spheres they have canonical metrics varying between $\frac{1}{4}$ and $1.$ As an example, we will compute the curvature of complex projective space. The calculations for the other rank one spaces are similar.

Then they begin the example, which is numbered Example 3.38. The punchline is that $\alpha$ and $\beta$ are orthogonal unit vectors, $J(\alpha)$ and $\beta$ are unit real vectors, so their real inner product lies between $-1$ and $1$ by Cauchy-Schwarz, followed by an identity about the sectional curvature that I shall write as $$ K_{\alpha \beta} = \frac{1}{4} + \frac{3}{4} \langle J(\alpha), \beta \rangle^2. $$

share|cite|improve this answer
Thanks, Will, this is useful advice: To understand the $\frac{1}{4}$, study the positive-curvature symmetric spaces! – Joseph O'Rourke Oct 8 '11 at 14:29

The number $1/4$ allows them to use a lemma by Berger to show that $M \times \mathbb{R}^2$ has non-negative isotropy curvature for a pointwise quarter-pinched manifold $M$. Non-negative isotropy curvature condition is preserved by Ricci flow.

Berger's lemma says that if the sectional curvatures at a point $p$ are bounded by $\kappa_1$ and $\kappa_2$, then the Riemann curvature tensor satisfies $R(e_1, e_2, e_3, e_4) \leq \frac{2}{3}(\kappa_2 - \kappa_1)$ for all orthonormal four-frames $\{e_1, e_2, e_3, e_4\}$ at $p$. The reference for this lemma is M. Berger, Sur quelques vari\'{e}t\'{e}s riemanniennes suffisamment pinc\'{e}es, Bull. Soc. Math. France 88, 57-71(1960).

A related result is that Yau and Zheng showed that a negatively $1/4$-pinched manifold has non-positive complex sectional curvature from this lemma. See S. T. Yau and F. Zheng, Negatively $1/4$-pinched riemannian metric on a compact K\"{a}hler manifold, Invent. Math., 103, 527-535(1991).

share|cite|improve this answer
So this reduced the problem to «why is there a $\frac23$ in Berger's result?» :) – Mariano Suárez-Alvarez Oct 8 '11 at 3:41
I just want to clarify the last part of Agol's answer. I don't think I can answer Joseph's original question. – Chenxu He Oct 8 '11 at 4:00

This was too long for a comment:

Certainly 1/4 is optimal, as Will Jagy points out, for e.g. the Fubini-Study metric on $CP^n$. The totally real tangent planes have curvature 1 (tangent to a totally geodesic $RP^2$), and the complex-invariant tangent planes to a totally geodesic $CP^1$ have curvature $1/4$. This is usually computed via the Riemannian submersion $S^{2n+1}\to CP^n$, but I think one can get a feel for this geometrically, by thinking about a quotient of $S^{2n+1}/U(1)=CP^n$. If you have $S^3/U(1)$, you get $CP^1$, and the map is the Hopf fibration. $S^3$ is the 3-sphere of radius 1, and when you quotient by $U(1)$, you get a sphere of half the diameter, since the circles of the Hopf fibration are distance $\pi/2$ apart, and thus the curvature is $1/4$. On the other hand, the $RP^n\subset CP^n$ subspaces are just a 2-fold quotient of a totally real $S^n\subset S^{2n+1}$ by the antipodal map, so have curvature $1$.

I saw Brendle give a talk about their theorem. One thing to note is that they have a local pinching theorem (pointwise), which is much stronger than a global pinching theorem. I think they actually show that they get positive isotropic curvature, which has something to do with complexifying the curvature tensor (this was introduced by Micallef-Moore I think), and is important for studying the stability of minimal 2-spheres. But I don't recall the details to be able to explain where $1/4$ comes in, but it has to do something with obtaining positive eigenvalues after complexifying the curvature tensor appropriately.

share|cite|improve this answer
Thank you, Ian, especially for the reminder that pointwise pinching is much stronger than global pinching! – Joseph O'Rourke Oct 8 '11 at 14:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.