This is a sequel to my previous post.
It is a classical result of Schur that if the Ricci curvature on a Riemannian manifold , then the scalar curvature must be constant for . This is a simple consequence of the twice contracted Bianchi identity . It is interesting to see how differs from a constant if is close to zero. In this direction Lellis and Topping proved in [Lellis-Topping] that
Theorem 1 If is a closed Riemannian manifold with nonnegative Ricci curvature, then
Here is the average of on .
In this note, we will show an analogous result for the higher -th mean curvature for closed submanifolds in space forms (Theorem 4). In particular we will show that our result implies Theorem 1 and some other results in [Cheng-Zhou] and [Cheng1]. We will also establish another version of this type of result which involves the Riemannian curvature tensor (Theorem 11), which gives a quantitative version of another result of Schur: if has sectional curvature which depends on its base point only, then its curvature is constant.
1. Submanifolds in space forms
Let be an immersed submanifold in a space form , . The second fundamental form of in is defined by and is normal-valued. Here is the connection on . We denote by , where is a local orthonormal frame on .
We define the -th mean curvature as follows. If is even,
If is odd, the -th mean curvature is a normal vector field defined by
Here is the generalized Kronecker delta symbol. In the codimension one case, i.e. is a hypersurface, by taking the inner product with a unit normal if necessary, we can assume is scalar valued. In this case the value of is given by
where are the principal curvatures. This definition of will be used whenever is a hypersurface.
If is odd,
Again in the codimension one case, we can assume is an ordinary tensor and if are the eigenvectors of , then
This definition of will be used whenever is a hypersurface.
Lemma 2 If is an immersed submanifold in a space form , let and denotes the divergence and the trace on respectively, then
Proof: The first assertion follows from the proof of [Grosjean] Lemma 2.1. But since it has only been shown in the case where is even in that paper, let us assume the odd case, and for the sake of demonstration let . Using local orthonormal frame, the first assertion follows from
where we have used the Codazzi equation (as has constant curvature). The second assertion is straightforward.
By Lemma 2 we immediately have the following Schur-type theorem which is perhaps well-known to experts (we will write instead of for a vector valued function and a tensor ):
Theorem 3 Let be a closed (compact without boundary) immersed submanifold in a space form and . If the traceless part of vanishes, i.e. , then is parallel. (In particular it is constant in the case where it can be defined as a scalar. )
By Lemma 2, as , we have . i.e. is parallel.
Theorem 4 Let () be a closed immersed orientable submanifold in a space form , . Let . Assume that either
- is even, or
- is odd and , or
- is of codimension one, i.e. a hypersurface.
Let be the average of (which is a vector when is odd and is defined by (1) in the codimension one case) and be the traceless part of . Let be the first eigenvalue for the Laplacian on and suppose the Ricci curvature of is bounded from below by , , then for , we have
Proof: We follow the ideas in [Lellis-Topping] and [Cheng-Zhou]. We do the the case where is odd (and thus ) first. We can assume is not vanishing everywhere, otherwise there is nothing to prove. Let be the solution to
The solution exists because . As and by Lemma 2, (as a vector-valued -form) we have
By Bochner formula, we have
Here by we mean the quantity . Consider
Here we have used the fact that the first eigenvalue . So (5) becomes
As , we have
Therefore (2) can be rephrased as
For the remaining cases where is even or is a hypersurface, as is scalar valued, just replace in (3) by a scalar valued function and run through the same proof, we can get the result.
Remark 1 When and is a hypersurface, as , it is easy to see that where is the traceless part of the second fundamental form. This generalizes [Perez] Theorem 3.1 (see also [Cheng-Zhou]). In the codimension one case, this recovers [Cheng1] Theorem 1.10.
Remark 2 For a embedded hypersurface in Euclidean space, having nonnegative Ricci curvature is equivalent to (i.e. convex), see [Perez] p. 48. So when , the curvature assumptions in Theorem 4 can be replaced by being convex when it is an embedded hypersurface.
2. Relations with Lovelock curvatures
In this section, we investigate the relation between Theorem 4 and an analogous result of Ge-Wang-Xia [Ge-Wang-Xia] when is even. Following [Ge-Wang-Xia], we define the Lovelock curvatures (or the so called -dimensional Euler density in Physics) of a Riemannian manifold , , by
It can be easily seen that is the sclar curvature. We can also define the generalized Einstein -tensor by defining
We have the following analogue of Lemma 2 for :
Proof: The first assertion is a straightforward calculation. For the second assertion, for the sake of demonstration let . Then this follows from
Here we have used the Bianchi identity in the second line.
We see that is divergence free and indeed is the Einstein tensor. By Lemma 5, it is clear that we can prove the analogue of Theorem 4 in this setting. Indeed, by using the same method, Ge, Wang and Xia [Ge-Wang-Xia] proved that (they have assumed , but their result can be easily extended to the version below):
Proof: We use a local orthnormal frame for computations. We also denote the dot product in by . By Gauss equation, we have
This implies the result.
Proof: To see that Theorem 4 is equivalent to Theorem 6 when is even, we just observe that by Proposition 7, clearly Theorem 6 implies our result in the even case and when is Euclidean. On the other hand, since any manifold can be isometrically embedded into some for large enough, we see that our result also implies Theorem 6.
3. Equality case of Theorem 4
It is easy to see from the proof that if the Ricci curvature assumption in Theorem 4 is strengthened to , then (as is constant). On the other hand, it is more subtle if we omit this assumption and so far we have only got some partial results in this case. The equality case for when is an embedded hypersurface has been considered in [Cheng-Zhou], in which they prove that is a distance sphere. It seems that their proof cannot be modified in our case because in their proof it is essential that contains a point whose Ricci curvature is positive, which is not true for submanifold in higher codimension in general.
The case for .
Suppose is immersed in a space form of curvature , using the same computations as in the proof of Proposition 7, we can get that
On the other hand, it is not hard to deduce that is the Einstein tensor (one way of seeing that without computing is to observe that contains only linear term involving the curvature and is divergence free, thus, up to constant, it must be the Einstein tensor). Thus
Note also that and is equal to up to a constant which only depends on and . Thus our result in this case is reduced to Theorem 1.2 of [Cheng2] or the case of Theorem 6. By the rigidity case of [Cheng2] Theorem 1.2, we deduce that the equality holds in Theorem 4 in the case if and only if is Einstein. Therefore is constant and so is by Gauss equation. In particular, if is an embedded surface in , then by [Ros] it is a sphere.
4. Another form of almost-Schur type theorem
In this section, we derive another form of Schur-type theorem which gives a quantitative version of the following classical result of Schur: if has sectional curvature which depends on its base point only, then its curvature is constant.
We first set up some notations. Let denote the space of covariant -tensor on (e.g. ). We define a product
For our purpose, our product is different from the Kulkarni-Nomizu product (ours has only half of the terms in their product). We define . It is easy to see that is the Riemann curvature tensor of a space form with curvature (we use the convention that is the sectional curvature). In local coordinates, it is given by
Proof: The first assertion follows from
Define the contraction map by . Then . By similar calculations, . Thus
Proof: Let be an orthonormal basis such that (no summation). Then if it easy to see that
Proof: From [Cheng2], we have
Plugging this into (11), we can get the result.
If the inequality in (i) becomes an equality, then by [Cheng2], is Einstein. If the equality in (ii) holds, then from the proof we have
for some constant . By taking the trace we deduce that and
Let be an orthonormal basis such that . (In this proof repeated indices will not be summed over. ) Then in this frame we have
From this we see that () is independent of , and thus the sectional curvature only depends on the base point. We thus conclude our result using either the classical Schur’s theorem or using (10).
Corollary 12 With the same assumptions as in Theorem 11, we have
Proof: It is easy to see that
Combining this with (10), we can get the result.