# “Natural” constructions of tensor fields from tensor fields on a manifold

This question begins is related to this question on physics.SE Uniqueness of Riemann Curvature Tensor, which asks roughly "what tensors can we make locally out of just the metric tensor? We can clearly make a (3,1) tensor out of just the metric, but we can't seem to locally make a vector field out just the metric.

I gave an answer there on which I'd appreciate comments. What I'd like to ask here is

Can we formalize the intuition of making a tensor out just the metric tensor?

The formalization I have come to is: constructing a tensor of type $A$ out of just a tensor of type $B$ is the same as a natural transformation between the two functors that assign to each manifold the space of tensors of type $A$ & $B$ respectively.

When I say a tensor "of some type", I mean specifying the number of upper and lower indices as well as possibly antisymmetrization/symmetrization of indices and any other $GL(n)$ invariants you can slap on. So symmetric, nondegenerate (0,2) tensor is a type.

(1) Is the above definition good? Does it capture the intuitive idea of "making a tensor field out of just another one"? Or am I missing something simple? Does this have an established name (other than natural transformation)?

The transformations from a fixed type form a tensor algebra inherited from $T(V)$: we can add, multiply by constant scalars, tensor products, contractions, symmetrizations - any $GL(n)$ invariant operation. This is kind of trivial since it doesnt know anything about the manifold, it's inherited from the vector space. So we just want to talk about the generators of this algebra, which always includes at least the trivial objects (i) identity map and (ii) map everything to $1$.

(2) Have the generators of this algebra been classified, or studied?. Or is this just asking someone to "list all the geometrical invariants ever" using fancy words? Assuming it is, some bonus questions

(3)Can anyone find a counterexample to this extreme conjecture: The entire algebra is generated by (i) metric curvature object, (ii) exterior derivatives and (iii) Hodge duals all composed with tensor algebra? (As a physicist these are the only natural geometric objects I can think of. Sad physicist)

(4)Is the only nontrivial transformation of a 2-form in $d=4$ the exterior derivative?

(5)Is there any nontrivial transformation of a symmetrized tensor of type (0,3)?

Sorry if that's a lot of questions, I just find this pretty cool.

Caveats 'n such

• Everything is smooth. I really want real analytic but I will settle for $C^\infty$.
• Probably we need to work on connected manifolds to remove some issues involving "constant" scalars taking values on different components.
• The natural transformation have lots of other algebraic structure I dont know what to do with, including regular composition.
• Edited drastically from an earlier version to pointless wall of words. See the earlier version if you want more words.
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I think questions of this form are addressed in Kolar, Michor, and Slovak (amazon.com/Natural-Operations-Differential-Geometry-Kolar/dp/…), but I haven't gone through it in detail. –  Qiaochu Yuan Jun 27 '13 at 23:56
Thanks @QiaochuYuan! Assuming they are using "natural" in the same sense it looks promising. No one would ever use the same word for two different things in math, right? –  BebopButUnsteady Jun 28 '13 at 0:28
That conjecture (3) is wrong because I forgot the Lie derivative and various generalizations. Out of a generic high index object I can make lots of independent vector fields and differential forms and I can take a lot of derivatives. I guess the spirit of the question is more "are there more and convoluted things I can do with higher index objects or is there some finite set of manipulations out of which all natural geometric transformations are made?" –  BebopButUnsteady Jun 28 '13 at 1:05
Regarding (4): I'm not sure if I understand what you mean by "transformation," but in four-dimensional geometry the operation which squares a 2-form is important (e.g. symplectic geometry). EDIT: perhaps this is included in your category of trivial things like tensor products –  Adam Saltz Jun 28 '13 at 1:56
@AdamSaltz: I am not sure what you mean by "squaring". If you mean the tensor product $\omega\otimes\omega$ than that is an "operations inherited from tensor algebra" and "kind of trivial". The reason they are trivial is that they know about the tensor field only at a point, i.e. they are just tensor algebra operation lifted to the whole field. Contrast with getting the curvature from the metric. I guess the algebraic statement would be that maps from the initial object representing $T(V)$ behave universally under a "tensor algebra operation". –  BebopButUnsteady Jun 28 '13 at 2:25