In this note I am going to introduce the notion of double forms on a Riemannian manifold and use it to define the so called -curvatures; finally I will prove an almost-Schur type theorem involving -curvatures. This extends one of the results in a previous post.

**1. Preliminaries for double forms **

** 1.1. Algebra structures **

Let be a smooth Riemannian manifold. Let be the ring of differential forms on . Considering the tensor product over the space of smooth functions, we define . The ring of double forms is then defined as .

We define a multiplication on as follows. For and , we define

and extend it linearly.

Naturally, can be regarded as a multilinear form which is skew symmetric in the first arguments and also in the last arguments. Under this identification, we have

Here is the sign of the permutation and is the permutation group of elements. Whenever possible we will omit the dot and write as .

Naturally we can regard the Riemannian metric and regard the Riemannian curvature tensor . For example, if is an orthonormal coframe for , then .

The contraction operator is defined as follows. If , then if or , otherwise is defined by

where is a local orthonormal frame on . We then define inductively.

We can give an inner product on by declaring if and elements of the form to be an orthonormal basis for .

Remark 1It should be noted that the inner product on is generally different from the inner product on the space of -tensors by a multiplicative constant. For example, using local orthonormal frame, where is the norm in the tensor algebra . However, these two inner products are the same on for . The only place where we will use is the proof of Theorem 6, and since in that proof the two inner products coincide, there will be no confusion.

Finally, the Hodge star operator is defined by

(the on the RHS is just the ordinary Hodge star operator) and extending it linearly.

The following basic but important identity will be used repeatedly ([Labbi1] Theorem 3.1):

Lemma 1Suppose , then we have

*Proof:* Equation (3) is a direct consequence of [Labbi1] Lemma 2.1 by substituting there. By the same lemma again we have

Equation (2) is clearly true for , otherwise for , by applying (1) and the above equation we have

Lemma 2If is trace-free, i.e. , then

*Proof:* By [Labbi1] Lemma 2.1, we have . Thus

** 1.2. Curvature structures **

Let denotes the symmetric elements of . For , we define its sectional curvature as follows. Let be a -dimensional plane in , then we define

where is an orthonormal basis of .

Definition 3For and , the -curvature is defined as the sectional curvature of the following -curvature tensor:

Equivalently, is the sectional curvature of on the orthogonal complement of .

Remark 2For , is called the -curvature. In particular, is half of the scalar curvature and is the sectional curvature of .Up to a constant, is the Killing-Lipschitz curvature of , which are also called the -sectional curvatures as defined by Thorpe [Thorpe].

For , we have ([Labbi1] Theorem 4.1)

which are scalar functions generalizing the scalar curvature. They are called the -scalar curvature functions, which we will denote by , i.e.

We define the -Einstein tensor by

For , are then the curvatures of the generalized Einstein tensors (See also here). By [Labbi1] Theorem 4.1, we have

The following result can be found, for example, in [Labbi2] p.179.

Lemma 4We have

Proposition 5Let be the traceless part of the generalized Einstein tensor . Then we have

*Proof:*

By Lemma 4 and (5), it is easy to compute

By iterating (1), we have for any

In particular,

Combining this with (6) and Lemma 1, we have

We conclude that

**2. Main result **

Theorem 6Let () be a closed oriented Riemannian manifold with , and , then

*Proof:* We follow the ideas in [Lellis-Topping] and [Cheng-Zhou]. We assume is not vanishing everywhere, otherwise there is nothing to prove. Let be the solution to

The solution exists because . As and by Lemma 4, we have

Then (with Remark 1 in mind)

Thus

Here we have used the fact that the first eigenvalue . So (11) becomes

Substitute this into (10), we obtain inequality (i) of (8):

The inequality (ii) of (8) can be obtained by applying Proposition 5.

As , we have

Therefore (8) can be rephrased as