# Tensors constructed out of metric other than the Riemann curvature tensor

Let $(M,g_{ab})$ be a Riemannian (or Pseudo-Riemannian) manifold and let us define a tensor field as 'something' that transforms in an appropriate way under coordinate transformations. (This is how tensors are defined in Physics texts, see Gravitation & Cosmology by Steven Weinberg for example). Thus $R_{abcd}$ and $R_{ab}$ defined by taking derivatives of the metric are tensors. (One considers scalars as tensors too). These involve taking the double derivative of the metric.

My question is : Can any tensors be constructed out of the metric by taking 3 or more derivatives of the metric ? If yes, what are they ? If no, is there a proof that no quantity involving derivatives higher than the double derivatives of the metric can transform (under coordinate transformations) in the way tensors do ?

I wish to mention that this question is related to (but different from) a question that I asked here.

I must admit that this question as it stands is a bit vague. I shall try to formulate it more precisely if I can.

Any relevant references are welcome.

EDIT : As a comment points out, some such tensors can be obtained simply by taking iterated covariant derivatives of the curvature tensor. This leads me to ask whether tensors obtained in this fashion are all the tensors which can be formed by taking higher derivatives of the metric.

• Just take iterated covariant derivatives of the curvature tensor. Commented Nov 23, 2013 at 19:07
• @JackLee Thanks. You are definitely right. I should have thought of this before. However I think it is still an interesting question to ask if the tensors obtained by taking covariant derivatives are all the tensors involving higher derivatives. In the view of this, I am editing this question. Commented Nov 24, 2013 at 4:36

Let me give a simplified overview of the background related to the question. The entire story can be found in the celebrated paper M. Atiyah, R. Bott, V.K. Patodi, "On the heat equation and the index theorem", which is the canonical reference for the subject.

The Riemannian curvature $R_{abcd}$ and the Ricci tensor $R_{ab}$ mentioned in the question are constructed out of the metric in the sense that in a coordinate patch $(U, x^i)$ they are given by a universal formula involving the partial derivatives of the components of the metric. Explicit expressions can be found, e.g. here. Moreover, the formulas turn out to be polynomial in the partial derivatives of $g_{a b}$ of all orders $\ge 0$ and the inverse metric $g^{a b}$. A formula like that gives components of a tensor if they transform under change of coordinates in the tensorial way. These observations give rise to the following notion.

Let $(M,g)$ be a Riemannian manifold of dimension $n = \dim M$.

Definition 1. A metric invariant (also known as a Riemannian invariant, or an invariant of the Riemannian structure) is a section $P(g)$ of a tensor bundle over $M$, such that for any diffeomorphism $\phi \colon M \to M$ the naturality property holds: $$P(\phi^* g) = g^* P(g)$$ and in any coordinate patch $(U,x^i)$ (that also gives the coordinate trivialization of the tensor bundle where $P(g)$ has values), the components of $P(g)$ are given by a universal polynomial expression in the list of formal variables $\{ g^{i j}, \partial_{k_1} \dots \partial_{k_s} g_{i j} \}$, $s \ge 0$.

Remark 1. If P has values in $\mathbb{R}$, we have a scalar valued invariant.

Remark 2. In a similar fashion we can give a definition of a metric invariant differential operator.

Examples. The metric tensor $g_{a b}$ and its inverse $g^{a b}$, the Riemann curvature tensor $R_{a b}{}^{c}{}_{d}$, the Ricci tensor $R_{a b}$ are tensor valued metric invariants. The scalar curvature $R = g^{a b} R_{a b}$ is a scalar valued tensor invariant. More scalar valued examples are mentioned in another question of the OP. The Levi-Civita connection $\nabla^g$ of the metric $g$ is a metric invariant differential operator. See e.g. Jack Lee, Riemannian Manifolds. An Introduction to Curvature.

As @Jack Lee has pointed in the comments, using the Levi-Civita connection and the Riemannian curvature one can construct many tensor-valued metric invariants, and taking the complete contractions we obtain lots of scalar valued metric invariants. This can be formalized as follows.

Definition 2. A curvature invariant is a linear combination of partial contractions of the iterated covariant derivatives (with respect to the Levi-Civita connection $\nabla$ associated to the metric $g$) of the Riemannian curvature.

More examples can be found here.

The question in the consideration can now be reformulated as follows.

Can all the metric invariants be obtained as curvature invariants?

The answer is known to be positive. This is a consequence of the First Fundamental Theorem of the classical invariant theory. The key geometric tool that is used to reduce this problem to a problem of the representation theory of the orthogonal group is the normal (or geodesic) coordinates. See the details in the aforementioned paper.

This also holds for the metric invariant differential operators.

• I am pretty surprised to hear this. Do you know off the top of your head if the same thing is true for Lorentzian manifolds and manifolds with other signatures? (If not don't worry, I can work through the proof you cited if need be.) Commented Dec 2, 2013 at 16:41
• @Christos, it is pretty much so, with the usual precautions when dealing with arbitrary signatures. All the ingredients (the Levi-Civita connection, the geodesic coordinates etc.) are there. Please refer to J. Slovák, On invariant operations on pseudo-Riemannian manifolds, e.g. here for further details and references. Commented Dec 3, 2013 at 0:26
• Thanks a lot ! This is exactly the answer I needed. Could you please add some more references for 'Classical Invariant Theory' ? I am interested in learning this subject with a viewpoint towards applications in geometry. Commented Dec 3, 2013 at 4:03
• @user90041, I am glad that my answer helps. With regards to the references please ask another question on this site, so not only I can answer. It would be nice also if you add a link to that question in the comments here. Commented Dec 3, 2013 at 6:26
• @YuriVyatkin Yes, you have a point. I was about to ask another question but I found this Commented Dec 3, 2013 at 14:05