3
$\begingroup$

Let $(M,g)$ be a Riemannian manifold of dimension $n$. (In the case I am interested in, $M$ is a complex symmetric domain, but I do not think that this is relevant for the question.)

Let $N$ be a submanifold of $M$. I would like to say that $N$ is "much curved". For the time being, let me just say that the second fundamental form - aka shape tensor - of $N$ has maximal rank. Does this imply that $N$ does not contain any geodesic segment? More generally, do I get some sort of lower bound on the curvature of curves contained in $N$?

The answer is likely to be negative I think, even if I do not have a countrexample. Let me ask a more vague question. Is there a notion of curvature for $N$ which prevent it from containing geodesic segments? Or, more generally, to bound the shape tensor of curves contained in $N$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

If you assume that the codimension of $N$ is large, then your embeddings are called "free" (see the Gromov's book on h-principle). In this case there are no geodesics and all the curves in $N$ have non-vanishing normal curvature.

For small codimension, the statement does not hold, say consider the paraboloid $z=xy$. (I do not see what happens in the intermediate codimension.)

Improve this answer
$\endgroup$
10
  • $\begingroup$ Thanks! So "free embedding" is the notion of curvature that you are proposing me if I understand correctly. Do you have any reference other than Gromov's partial differential relations? $\endgroup$
    – Giulio
    Commented Nov 7, 2016 at 19:15
  • 1
    $\begingroup$ It is more like "second order regularity". Say, a map is regular if its first partial derivatives are linearly independent and the map if free if its first and second partial derivatives are linearly independent. $\endgroup$
    – Anton Petrunin
    Commented Nov 7, 2016 at 22:23
  • $\begingroup$ Thanks. If I am correct, the span of the second order derivative is the so called second osculating space, which is also the image of the second fundamental form. So you are saying that if the second fundamental form is injective (as a map from $Sym^2 TN$ ), then there are no geodesic in $N$. Is this what you mean? Do you have a reference? $\endgroup$
    – Giulio
    Commented Nov 8, 2016 at 11:34
  • $\begingroup$ @Giulio, yes that is correct. Also you can say that $\tfrac12{\cdot}n{\cdot}(n+3)$ partial derivatives $\tfrac{\partial}{\partial x_i}$, $\tfrac{\partial^2}{\partial x_i\partial x_j}$ are linearly independent. $\endgroup$
    – Anton Petrunin
    Commented Nov 8, 2016 at 14:51
  • $\begingroup$ what about a reference for geodesics? Or it is just something like that if you write out the geodesic equation you can see it as a linear relation among derivatives? $\endgroup$
    – Giulio
    Commented Nov 8, 2016 at 15:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .