Positive sectional curvature does not imply positive definite curvature operator?

Question

The curvature operator on $\Lambda^2(TM)$ is defined on decomposable bivectors by $$g(\mathfrak{R}(X \wedge Y), V \wedge W) = R(X,Y,W,V)$$ and then extended by linearity to all elements of $\Lambda^2(TM)$. It is self-adjoint, so defines a symmetric bilinear form on $\Lambda^2(TM)$. If this form is positive definite, then all sectional curvatures are positive.

My question is: are there Riemannian manifolds with positive sectional curvature but with an indefinite curvature operator?

Positive sectional curvature means that $g(\mathfrak{R}(\alpha), \alpha) > 0$ for all decomposable bivectors, which does not seem to exclude the existence of an indecomposable bivector $\beta$ for which $g(\mathfrak{R}(\beta), \beta) < 0$. In dimension 3 all bivectors are decomposable, so counterexamples can exist only starting from dimension 4.

A second question: If the answer to the previous question is "yes", what pinching of the sectional curvature does imply the positivity of the curvature operator?

Answer 1

For the first question: positive curvature operator on a compact manifold implies that the manifold is diffeomorphic to a space form, i.e., a manifold of sectional curvature one. This is due to C. Boehm and B. Wilking Manifolds with positive curvature operators are space forms Annals of Mathematics, 167 (2008), 1079–1097.

Thus most known positively curved manifolds do not admit metrics with positive curvature operator. The simplest example is $CP^n$, $n>1$.

I do not know know much about how sectional curvature pinching can imply positive curvature operator but for example, you could trace references from J.-P. Bourguignon, H. Karcher, Curvature operators: pinching estimates and geometric examples in Ann. Sci. École Norm. Sup. (4) 11 (1978), no. 1, 71–92 where some estimates on the eigenvalues of the curvature operator in terms of sectional curvature pinching can be found.

Answer 2

Positive-definiteness of the curvature operator ($R>0$) is as much stronger condition than positive sectional curvature ($\sec>0$). In fact, as Igor mentions, it follows from the work of Boehm and Wilking that only spherical space forms admit metrics with $R>0$, while many (but not so many!) other manifolds admit metrics with $\sec>0$. An almost complete list of such manifolds can be found in Section 4 of this survey of Ziller.

Of course, what happens is that although $R\colon\wedge^2 T_pM\to\wedge^2 T_pM$ is symmetric (hence diagonalizable), its eigenspaces need not intersect the Grassmannian $Gr_2(T_pM)=\{\sigma\in\wedge^2 T_pM:\sigma\wedge\sigma=0,\|\sigma\|=1\}$. Hence $R$ could have some zero (or even negative) eigenvalues at the same time as the restriction $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ of its quadratic form to $Gr_2(T_pM)$ is positive. For example, in $\mathbb C P^n$ this is what happens: $R$ is positive-semidefinite (and has nontrivial kernel), but this kernel does not intersect the Grassmannian.

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Algebraically, there is an intermediate curvature condition between $\sec>0$ and $R>0$ called strongly positive curvature, that might be of interest to you. Namely, a curvature operator $R\colon\wedge^2 V\to\wedge^2 V$ has strongly positive curvature if there exists a $4$-form $\omega\in\wedge^4V$ such that $R+\omega$ is positive-definite. Here, $\omega\in\wedge^4 V$ is identified with a symmetric endomorphism $\omega\colon\wedge^2V\to\wedge^2V$ via $$\langle\omega(\alpha),\beta\rangle=\langle\omega,\alpha\wedge\beta\rangle.$$ Clearly, $R>0$ implies strongly positive curvature (take $\omega=0$). Since $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ and $\langle\omega(\sigma),\sigma\rangle=\langle\omega,\sigma\wedge\sigma\rangle=0$ if $\sigma$ is decomposable, strongly positive curvature implies $\sec>0$. Together with R. Mendes, over the last years, I have pursued a systematic study of this curvature condition; see the following for details:

Bettiol, Renato G.; Mendes, Ricardo A. E., Strongly positive curvature, Ann. Global Anal. Geom. 53, No. 3, 287-309 (2018). ZBL1395.53058. (arXiv)

Bettiol, Renato G.; Mendes, Ricardo A. E., Strongly nonnegative curvature, Math. Ann. 368, No. 3-4, 971-986 (2017). ZBL1377.53020. (arXiv)

Bettiol, Renato G.; Mendes, Ricardo A. E., Flag manifolds with strongly positive curvature, Math. Z. 280, No. 3-4, 1031-1046 (2015). ZBL1360.53056. (arXiv)

For example, $(\mathbb C P^n,g_{FS})$ has strongly positive curvature, since $R+\varepsilon (\omega_{FS}\wedge\omega_{FS})>0$ for small $\varepsilon>0$, where $\omega_{FS}$ is the Fubini-Study $2$-form.

The upshot is that almost all known examples of manifolds with $\sec>0$ actually satisfy this much stronger condition. For manifolds with $\sec\geq0$, actually all known examples satisfy the analogous "strongly nonnegative curvature" condition (which requires $R+\omega$ to be positive-semidefinite).

Answer 3

This argument is due to Burkhard Wilking. The other answers, answer to the OP question very well but I want to give a sharp example. I.e. a manifold with positive sectional curvature and curvature operator with some negative eigenvalues:

$$\exists\ (M,g)\ \text{s.t.}\ \sec_g>0\quad \text{but not}\quad \ \mathcal{R}\geq0.$$

Any positive sectional curvature metric on $\mathsf{SU(3)}/\mathsf{T^2}$ has some negative curvature operator. To see this. note that it is an irreducible simply connected Riemannian manifold of positive sectional curvature and it is not diffeomorphic to any symmetric space; it follows that it should have some negative curvature operator by classification of closed manifolds of nonnegative curvature operator. See for instance Page 270, theorem 7.34 of

Chow, Bennett; Lu, Peng; Ni, Lei, Hamilton’s Ricci flow, Graduate Studies in Mathematics 77. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-4231-5/hbk). xxxvi, 608 p. (2006). ZBL1118.53001.