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Edit: It seems my question wasn't being read carefully, so I've rewritten it. I've also retagged the question.

1. Let $M$ be a smooth (pseudo-)Riemannian manifold and let $T(M)$ be the smooth tensor bundle on it. Is there a subspace $S \subset T(M)$ closed under (possibly infinitely many) finitary (possibly partially-defined) algebraic operations and finitely generated containing exactly the tensor fields on $M$ which are invariant under some fixed subgroup $G \le \mathrm{Diff}(M)$?
2. In particular, the above is an open question because e.g. the $G$-invariant tensor fields have not been classified for general $G$, how about asking that $S$ contains only isometry-invariant tensor fields and in particular contains the metric tensor, inverse metric tensor, the Riemann tensor, the Ricci tensor, and the Ricci scalar?
3. Suppose there is such an algebraic structure, and I adjoin an arbitrary tensor field to it. Will there be a non-trivial subgroup of $\mathrm{Diff}(M)$ under which the extended structure is invariant? Conversely, suppose I fix a non-trivial subgroup of $H < G$. Is there an algebraic structure of the same signature invariant under $H$ containing $S$?

Motivation

We can detect, using purely algebraic means in a certain sense, whether an arbitrary complex number $\alpha \in \mathbb{C}$ is an algebraic number: Look for the smallest subfield $\mathbb{Q}(\alpha) \le \mathbb{C}$ containing both $\mathbb{Q}$ and $\alpha$. It is a vector space over $\mathbb{Q}$, and $\alpha$ is algebraic if and only if this vector space is finite-dimensional. Moreover, we can even detect whether $\alpha$ can be expressed in terms of radicals simply by examining the automorphism group of the field extension $\mathbb{Q}(\alpha) / \mathbb{Q}$.

Here, I'm interested in whether or not something similar can be done for tensor fields on a manifold. The first question is analogous the inverse Galois problem — we have a symmetry group, and we are looking for an algebraic structure which is invariant under it. Having found such an algebraic structure, we can ask whether or not there is an analogue of the fundamental theorem of Galois theory, which establishes a bijective correspondence between subextensions of a Galois field extension and the subgroups of its automorphism group.

I'm aware of at least one result in this area, broadly interpreted as an algebraic approach to differential geometry — namely Lovelock's theorem, which classifies all the symmetric divergence-free second-order natural (0, 2)-tensors on a manifold. A corollary of this theorem, I'm told, tells us that the Einstein field equations are essentially unique (in 4 dimensions).

(Edited)

1. Let $M$ be a smooth (pseudo-)Riemannian manifold and let $T(M)$ be the smooth tensor bundle on it. Is there a subspace $S \subset T(M)$ closed under (possibly infinitely many) finitary (possibly partially-defined) algebraic operations and finitely generated containing exactly the tensor fields on $M$ which are invariant under some fixed subgroup $G \le \mathrm{Diff}(M)$?
2. In particular, the above is an open question because e.g. the $G$-invariant tensor fields have not been classified for general $G$, how about asking that $S$ contains only isometry-invariant tensor fields and in particular contains the metric tensor, inverse metric tensor, the Riemann tensor, the Ricci tensor, and the Ricci scalar?
3. Suppose there is such an algebraic structure, and I adjoin an arbitrary tensor field to it. Will there be a non-trivial subgroup of $\mathrm{Diff}(M)$ under which the extended structure is invariant? Conversely, suppose I fix a non-trivial subgroup of $H < G$. Is there an algebraic structure of the same signature invariant under $H$ containing $S$?

Motivation

We can detect, using purely algebraic means in a certain sense, whether an arbitrary complex number $\alpha \in \mathbb{C}$ is an algebraic number: Look for the smallest subfield $\mathbb{Q}(\alpha) \le \mathbb{C}$ containing both $\mathbb{Q}$ and $\alpha$. It is a vector space over $\mathbb{Q}$, and $\alpha$ is algebraic if and only if this vector space is finite-dimensional. Moreover, we can even detect whether $\alpha$ can be expressed in terms of radicals simply by examining the automorphism group of the field extension $\mathbb{Q}(\alpha) / \mathbb{Q}$.

Here, I'm interested in whether or not something similar can be done for tensor fields on a manifold. The first question is analogous the inverse Galois problem — we have a symmetry group, and we are looking for an algebraic structure which is invariant under it. Having found such an algebraic structure, we can ask whether or not there is an analogue of the fundamental theorem of Galois theory, which establishes a bijective correspondence between subextensions of a Galois field extension and the subgroups of its automorphism group.

I'm aware of at least one result in this area, broadly interpreted as an algebraic approach to differential geometry — namely Lovelock's theorem, which classifies all the symmetric divergence-free second-order natural (0, 2)-tensors on a manifold. A corollary of this theorem, I'm told, tells us that the Einstein field equations are essentially unique (in 4 dimensions).

Edit: It seems my question wasn't being read carefully, so I've rewritten it. I've also retagged the question.

1. Let $M$ be a smooth (pseudo-)Riemannian manifold and let $T(M)$ be the smooth tensor bundle on it. Is there a subspace $S \subset T(M)$ closed under (possibly infinitely many) finitary (possibly partially-defined) algebraic operations and finitely generated containing exactly the tensor fields on $M$ which are diffeomorphism-invariant?
2. If the above is an open question because e.g. the diffeomorphism-invariant tensor fields have not been classified, how about asking that $S$ contains only diffeomorphism-invariant tensor fields and in particular contains the metric tensor, inverse metric tensor, the Riemann tensor, the Ricci tensor, the Ricci scalar, and the torsion tensor?
3. Suppose there is such an algebraic structure, and I adjoin an arbitrary tensor field to it. Will there be a non-trivial subgroup of $\mathrm{Diff}(M)$ under which the extended structure is invariant? Conversely, suppose I fix a non-trivial subgroup of $G < \mathrm{Diff}(M)$. Is there an algebraic structure of the same signature invariant under $G$ containing $S$?

Motivation

We can detect, using purely algebraic means in a certain sense, whether an arbitrary complex number $\alpha \in \mathbb{C}$ is an algebraic number: Look for the smallest subfield $\mathbb{Q}(\alpha) \le \mathbb{C}$ containing both $\mathbb{Q}$ and $\alpha$. It is a vector space over $\mathbb{Q}$, and $\alpha$ is algebraic if and only if this vector space is finite-dimensional. Moreover, we can even detect whether $\alpha$ can be expressed in terms of radicals simply by examining the automorphism group of the field extension $\mathbb{Q}(\alpha) / \mathbb{Q}$.

Here, I'm interested in whether or not something similar can be done for tensor fields on a manifold. The first question is analogous the inverse Galois problem — we have a symmetry group, and we are looking for an algebraic structure which is invariant under it. Having found such an algebraic structure, we can ask whether or not there is an analogue of the fundamental theorem of Galois theory, which establishes a bijective correspondence between subextensions of a Galois field extension and the subgroups of its automorphism group.

I'm aware of at least one result in this area, broadly interpreted as an algebraic approach to differential geometry — namely Lovelock's theorem, which classifies all the symmetric divergence-free second-order natural (0, 2)-tensors on a manifold. A corollary of this theorem, I'm told, tells us that the Einstein field equations are essentially unique (in 4 dimensions).

Edit: It seems my question wasn't being read carefully, so I've rewritten it. I've also retagged the question.

1. Let $M$ be a smooth (pseudo-)Riemannian manifold and let $T(M)$ be the smooth tensor bundle on it. Is there a subspace $S \subset T(M)$ closed under (possibly infinitely many) finitary (possibly partially-defined) algebraic operations and finitely generated containing exactly the tensor fields on $M$ which are invariant under some fixed subgroup $G \le \mathrm{Diff}(M)$?
2. In particular, the above is an open question because e.g. the $G$-invariant tensor fields have not been classified for general $G$, how about asking that $S$ contains only isometry-invariant tensor fields and in particular contains the metric tensor, inverse metric tensor, the Riemann tensor, the Ricci tensor, and the Ricci scalar?
3. Suppose there is such an algebraic structure, and I adjoin an arbitrary tensor field to it. Will there be a non-trivial subgroup of $\mathrm{Diff}(M)$ under which the extended structure is invariant? Conversely, suppose I fix a non-trivial subgroup of $H < G$. Is there an algebraic structure of the same signature invariant under $H$ containing $S$?

Motivation

We can detect, using purely algebraic means in a certain sense, whether an arbitrary complex number $\alpha \in \mathbb{C}$ is an algebraic number: Look for the smallest subfield $\mathbb{Q}(\alpha) \le \mathbb{C}$ containing both $\mathbb{Q}$ and $\alpha$. It is a vector space over $\mathbb{Q}$, and $\alpha$ is algebraic if and only if this vector space is finite-dimensional. Moreover, we can even detect whether $\alpha$ can be expressed in terms of radicals simply by examining the automorphism group of the field extension $\mathbb{Q}(\alpha) / \mathbb{Q}$.

Here, I'm interested in whether or not something similar can be done for tensor fields on a manifold. The first question is analogous the inverse Galois problem — we have a symmetry group, and we are looking for an algebraic structure which is invariant under it. Having found such an algebraic structure, we can ask whether or not there is an analogue of the fundamental theorem of Galois theory, which establishes a bijective correspondence between subextensions of a Galois field extension and the subgroups of its automorphism group.

I'm aware of at least one result in this area, broadly interpreted as an algebraic approach to differential geometry — namely Lovelock's theorem, which classifies all the symmetric divergence-free second-order natural (0, 2)-tensors on a manifold. A corollary of this theorem, I'm told, tells us that the Einstein field equations are essentially unique (in 4 dimensions).

Background

Physicists have a "principle of general covariance" which basically states that physical laws (and in particular, physical quantities) can be stated in a form which is somehow coordinate-independent. The paradox is, such "coordinate-independent" quantities and equations are frequently stated in terms of (admittedly, arbitrary) coordinates! In particular, the classical characterisation of tensorial quantities goes something along the lines of "a tensor is an indexed quantity which obeys the tensor transformation law under a change of coordinates". For example, under this scheme, a vector field $V$ is something having components $V^\mu$ (w.r.t. the coordinate basis) in the coordinate system $x^\mu$ that will have components $\displaystyle V'^\nu = \frac{\partial x'^\nu}{\partial x^\mu} V^\mu$ (summing over repeated indicies; w.r.t. the coordinate basis) in the coordinate system $x'^\nu$.

It recently occurred to me that it's difficult to capture this particular form of the principle in the notation of differential geometry, let alone to formalise it — when I tried to write down the corresponding equation in the language of charts and pushforwards, I got the tautology $d\psi (V) = d(\psi \circ \phi^{-1}) \circ \phi \cdot d\phi (V)$, where $\psi$ and $\phi$ are charts with a common domain.

Examples

Let $M$ be a smooth $n$-dimensional (pseudo-)Riemannian manifold with metric $g$. Let $e_1, \ldots, e_n$ be a family of vector fields on $M$ which are linearly independent at each point in the manifold. Let $\epsilon^1, \ldots, \epsilon^n$ be the dual basis w.r.t. the metric, satisfying $\epsilon^\mu(V) = g(e_\mu, V)$ for every vector field V. Fix also an affine connection $\nabla$ on $M$.

Now, let $V$ be an arbitrary vector field. Because $\{ e_\mu \}$ forms a basis of each tangent space, there is a family of scalar fields $V^1, \ldots V^n$ so that $V = V^\mu e_\mu$ (summation convention). Conversely, if I have an arbitrary family of scalar fields $U^1, \ldots, U^n$, I can form the vector field $U = U^\mu e_\mu$. Moreover, assuming everything is nice enough, I can define a family of scalar fields $\Gamma^\sigma_{\phantom{\sigma}\mu\nu}$ which satisfies the equation $\nabla_{e_\nu} e_\mu = \Gamma^\sigma_{\phantom{\sigma}\mu\nu} e_\sigma$. And I can certainly define the (1, 2)-tensor $\Gamma = \Gamma^\sigma_{\phantom{\sigma}\mu\nu} e_\sigma \otimes \epsilon^\mu \otimes \epsilon^\nu$.

Yet, it's quite clear intuitively that $V^\mu$, $U$ and $\Gamma$ are all somehow not "genuine" scalar / vector / tensor (resp.) fields — the definitions are "unnatural" and relied on a choice of basis.

The problem

It seems to me that the problem arises because the principle of general covariance (as I understand it) is formulated intensionally rather than extensionally. But this isn't necessarily a problem — perhaps there's a way to capture some form of it extensionally, just as the concept of natural transformations captures some essence of the notion of naturality. So my question is: Given some arbitrary tensor field on a manifold, is there a way to extensionally test whether it is tensorial, in the physicists' sense?

Of course, one must first formalise what it means to be tensorial in the physicists' sense. One possibility, along the lines of algebra, is to say that a tensorial object is one built from other tensorial objects using legal operations. This coincides with the notion of "manifestly covariant". Following these lines, one could ask a preliminary question: What is the minimal set of admissible operations which will allow us to generate, from a reasonable choice of generators, a set of tensorial objects which includes all the intrinsic quantities we want, e.g. the metric tensor, the inverse metric tensor, the Riemann tensor, the Ricci tensor, the torsion tensor, etc.? Is it possible to do this without also admitting arbitrary tensors as tensorial?

At this point I must confess that I don't know much about differential geometry or mathematical physics. I've read that the principle of general covariance in general relativity is best understood as a gauge symmetry with respect to the diffeomorphism group, but the link between this and the notion of manifest covariance is not obvious to me. (Then again, who would have thought that the group of symmetries of polynomial roots have anything to do with solvability by radicals?)

Edit: It seems my question wasn't being read carefully, so I've rewritten it. I've also retagged the question.

1. Let $M$ be a smooth (pseudo-)Riemannian manifold and let $T(M)$ be the smooth tensor bundle on it. Is there a subspace $S \subset T(M)$ closed under (possibly infinitely many) finitary (possibly partially-defined) algebraic operations and finitely generated containing exactly the tensor fields on $M$ which are diffeomorphism-invariant?
2. If the above is an open question because e.g. the diffeomorphism-invariant tensor fields have not been classified, how about asking that $S$ contains only diffeomorphism-invariant tensor fields and in particular contains the metric tensor, inverse metric tensor, the Riemann tensor, the Ricci tensor, the Ricci scalar, and the torsion tensor?
3. Suppose there is such an algebraic structure, and I adjoin an arbitrary tensor field to it. Will there be a non-trivial subgroup of $\mathrm{Diff}(M)$ under which the extended structure is invariant? Conversely, suppose I fix a non-trivial subgroup of $G < \mathrm{Diff}(M)$. Is there an algebraic structure of the same signature invariant under $G$ containing $S$?

Motivation

We can detect, using purely algebraic means in a certain sense, whether an arbitrary complex number $\alpha \in \mathbb{C}$ is an algebraic number: Look for the smallest subfield $\mathbb{Q}(\alpha) \le \mathbb{C}$ containing both $\mathbb{Q}$ and $\alpha$. It is a vector space over $\mathbb{Q}$, and $\alpha$ is algebraic if and only if this vector space is finite-dimensional. Moreover, we can even detect whether $\alpha$ can be expressed in terms of radicals simply by examining the automorphism group of the field extension $\mathbb{Q}(\alpha) / \mathbb{Q}$.

Here, I'm interested in whether or not something similar can be done for tensor fields on a manifold. The first question is analogous the inverse Galois problem — we have a symmetry group, and we are looking for an algebraic structure which is invariant under it. Having found such an algebraic structure, we can ask whether or not there is an analogue of the fundamental theorem of Galois theory, which establishes a bijective correspondence between subextensions of a Galois field extension and the subgroups of its automorphism group.

I'm aware of at least one result in this area, broadly interpreted as an algebraic approach to differential geometry — namely Lovelock's theorem, which classifies all the symmetric divergence-free second-order natural (0, 2)-tensors on a manifold. A corollary of this theorem, I'm told, tells us that the Einstein field equations are essentially unique (in 4 dimensions).