Let $M$ be a 4-dimensional Riemannian manifold. Assume there is a huge number (say 100) of flat totally geodesic 2-dimensional surfaces passing through a point $p\in M$ and assume that their tangent planes at $p$ lie in general position. Further, assume that sectional curvature is $\ge 0$ (or $\le 0$) in a neighborhood of $p$.

Is it true that $p$ admits a flat neighborhood?


  • These conditions easely imply that $M$ has zero curvature tensor at $p$.
  • In dimension 3 the answer is "YES"; it follows since these surfaces have to intersect along geodesics.
  • If the dimension is ≥ 4, I do not see any reason why the answer should be "YES", but I do not see a counterexample.
  • For "YES"-answer, you may consider stronger condition, say curvature operator is $\ge 0$ (or $\le 0$) or anything like that.
  • A related question: assume you have a metric on $\mathbb R^m$ with sectional curvature $\ge 0$ (or $\le 0$). Assume that metric is standard-Euclidean outside of open set $\Omega\subset \mathbb R^m$ and inside $\Omega$ scalar curvature has no zeros. What one can say about $\Omega$? (From the sketch below, it follows that one can not touch $\Omega$ by affine 2-plane in $\mathbb R^m$; it is likely that any component of intersection of $\Omega$ with a 2-plane has to be unbounded; is it all?)

Sketch for the case $\dim M=3$.

The following observation is due to S. Buyalo. You can read bit more in our "Sweeping out sectional curvature".

Assume that the number of surfaces is 3. In general position the surfaces intersect pairwise at 3 distinct geodesics, say $\alpha_1$, $\alpha_2$ and $\alpha_3$. Consider a triangle $\triangle x_1x_2x_3$ such that $x_i\in\alpha_i$ and $x_i\approx p$. It is easy to see that in $\triangle x_1x_2x_3$ angles and their model angles are equal. The curvature condition then implies that this triangle is flat.

Such triangles sweep out a flat octahedral neighborhood of $p$.

  • $\begingroup$ A great question. Could you say more about why counterexamples can exist in higher dimensions? $\endgroup$
    – Deane Yang
    Jan 2, 2011 at 1:06
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    $\begingroup$ Can you elaborate on the argument in dimension 3? $\endgroup$ Jan 2, 2011 at 2:21
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    $\begingroup$ Anton, it seems to me that in dimension 3 two flats are in general position iff they are not equal. Now look at the Riemannian product of round $S^2$ and $S^1$. It has nonnegative curvature, no flat points, and the product of any great circle and $S^1$ is a flat. Any two such flats are in general position, and there are uncountably many of them. Why isn't this a counterexample to your claim that the answer is YES dimension 3? Is your definition of "general position" different from mine? $\endgroup$ Jan 2, 2011 at 2:51
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    $\begingroup$ @Anton, could you remind us the definition of the "general position"? $\endgroup$ Jan 2, 2011 at 15:21
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    $\begingroup$ Anton, sorry I see no formal definition in the link. Could you reproduce it in your question? As I gather from Thurston's answer, I am not the only one who thinks the issue isn't crystal clear. If the definition is negotiable, would you say so explicitly? $\endgroup$ Jan 2, 2011 at 17:48

1 Answer 1


Edited 2 January 2011 to clarify:

I don't know if there is an established meaning for general position in this context: it would be nice to have a clarification of terminology (as we've seen from the comments discussion). One possible definition is that a set of subspaces is in topological general position if for any small perturbation, there is a homeomorphism taking one configuration to the other. For 2-planes in $\mathbb R^4$, this just means that any two intersect in a point. If you intersect the subspaces with a small sphere about the origin, this gives a great circle link in $S^3$; for any $n$, there is a finite set of possible link types for these links.

Example for topological general position

For topological general position, there is a relatively simple example satisfying the question: take $n$ complex 1-dimensional subspaces of $\mathbb C^2$. The intersection with $S^3$ gives $n$ circles in a Hopf fibration. Now modify the Euclidean metric of $\mathbb C^2$ by changing the metrics of concentric spheres to homogeneous metrics on $S^3$ obtained by keeping the Hopf fibers rigid, but uniformly expanding their orthogonal planes by an increasing function of radius $r$. These metrics are invariant by $U(2)$, with nonpositive curvature. (This behaviour is closely related to the geometry of $\mathbb{CP}^2$ and of complex hyperbolic space $\mathbb {CH}^2$: the family of concentric spheres in either case has this same family of metrics on $S^3$ up to a constant that depends on $r$).

Possible stronger requirements for general position

At first it may be tempting to strengthen the condition of topological general position to require that for any small enough perturbation there is a linear map carrying one pattern to another, but this condition is unreasonably strong. The linear equivalence condition is satisfied for almost every triple of 2-dimensional subspaces of a 4-dimensional vector space, since two of the subspaces can be transformed into orthogonal coordinate planes, whereupon the third subspace is the graph of a map from one to the other; by a linear transformation, this can be arranged to map basis vectors to basis vectors. When there is a fourth subspace, the map associated with the third followed by the inverse of the map associated with the fourth is a linear map from a plane to itself, and its characteristic polynomial is an invariant that varies in any open neighborhood of configurations. Note, however, that there are open sets of quadruples of 2-dimensional subspaces for which this self-map has a pair of complex conjugate eigenvalues; when this happens, the quadruple of subspaces is linearly equivalent to a quadruple of complex 1-dimensional subspaces of $\mathbb C^2$.

There are many possible intermediate conditions that could be imposed, depending on what set of special situations you focus on. If you consider all linear maps defined by triples of planes, as above, you might want to require that the rank of any composition is constant in a neighborhood. However, for many (most) configurations of $n$ planes, when $n$ is high enough, these generate a dense set of linear maps among the planes, and small perturbations can change the rank of something. A more plausible requirement is that the dimension of the Zariski closure of the set of all compositions of these maps is constant, or perhaps that the dimension of the Zariski closure of the set of all compositions associated with triples in a subset remains constant. But this definition is special to the setting of half-dimensional subspaces of an even-dimensional space so it's not ideal. It's not hard to modify this for a more general setting, but there are too many possible choices: it's not self-evident that there is one `right' definiton.

One way around the issue is to rephrase the question: is there an open set, or a set of positive measure, or a generic set (intersection of open dense sets) of $n$ 2-dimensional subspaces of $\mathbb R^4$, such that ___ [a given property is true].

Examples for an open set of subspaces

Now make a random $C^2$-small perturbation of the Hopf fibration to be a foliation of $S^3$ by geodesics, so that the perpendicular plane field $\tau$ is still a contact plane field. We can construct new metrics in a neighborhood of the origin in $\mathbb R^4$ by modifying only in the 2-planes orthogonal to the family of 2-planes that are cones on these geodesics, scaling these planes by multiplication by a function $f(r)$ that is concave upward (to aim for negative curvature). If $f$ has $C^\infty$ contact to the constant function 1 at $r = 0$, the resulting metric is smooth. It is flat in each plane that is a cone on leaf of the foliation.

I think that for appropriately chosen $f$ these metrics have nonpositive sectional curvature, but I'll need to come back to this later to back it up (or tear it down), unless someone else will do it.

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    $\begingroup$ I do not see why $f$ should exist, so please come back :) $\endgroup$ Jan 2, 2011 at 17:29
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    $\begingroup$ @Anton Petrunin: I'm still thinking about how to get a good clear geometric description of the partial differential inequalities required for $f$, for a given foliation by great circles. It's easy to write down readily satisfied inequalities for $f$ and its derivatives so that curvature is negative in any plane tangent to the sphere of radius $r$ and nonpositive in any plane orthogonal to this sphere, but I'm still trying to clarify my understanding of the cross-terms, without resorting to writing down calculations of formulas that tend to seem obscure to nonexperts (including me). Later. $\endgroup$ Jan 2, 2011 at 21:45
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    $\begingroup$ There is a simple explicit example of Bill's first kind (topological general position): Let $p(z_1,z_2)$ be a nonzero polynomial homogenous of degree $n\ge2$ with $n$ distinct linear factors. Consider the surface $S\subset\mathbb{C}^3$ defined by $z_3=p(z_1,z_2)$ endowed with the metric $g$ induced by the standard flat Hermitian one on $\mathbb{C}^3$. Then $S$ intersects the plane $z_3=0$ in $n$ distinct complex lines through the origin that are totally $g$-geodesic surfaces in $S$ and flat. This $g$ has non-positive sectional curvature (and curvature operator), but it is not flat near $0$. $\endgroup$ Oct 28, 2021 at 22:50

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