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differential geometer in training here. With regards to my background, I learned differential and Riemannian geometry from O’Neill and Lee’s series. I’m working on my algebra background (which is admittedly a bit weak) and trying to think algebraically about some of the constructions I’m familiar with in differential geometry.

If $V$ is an $R$-module over a commutative ring then let $T^k(V)$ denote the $k$-th tensor power of $V$.

In differential geometry, we’re interested in the case where $V$ is the module of smooth vector fields over the ring $R = C^\infty(M)$ of smooth functions on a smooth manifold $M$. We think of elements of $T^k(V)$ as $R$-multilinear maps with $k$ arguments $V^* \times V^* \times \ldots \times V^* \to R$ and call them contravariant tensors. Likewise, elements of $T^l(V^*)$ are thought of as $R$-multilinear maps with $l$ arguments $V \times V \times \ldots \times V \to R$ and we call them covariant tensors.

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Now it doesn’t make sense physically (as far as I know) to add tensors of different ranks or variances, but it does make sense to take their tensor product. We can take a $k$-contravariant tensor and takes its tensor product with an $l$-covariant tensor to obtain a mixed tensor of rank $(k, l)$.

We also want to deal with “tensor derivations” $D$ like the covariant derivative or Lie derivative, which are $\mathbb{R}$-linear (as opposed to $C^\infty(M)$-linear) and satisfy the Leibniz rule

$D (A \otimes B) = DA \otimes B + A \otimes DB.$

My questions are:

- How would we describe the algebraic structure involved here? It seems like some sort of graded object, but I wouldn’t say it is an $R$-algebra since it doesn’t make sense to add certain elements.
- Is this construction functorial from the category of smooth manifolds to some algebraic category?
- If so, does this construction satisfy some universal property?

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The algebraic approach is simple: whether or not it makes sense to add tensors of different ranks, that’s just what you do. If you can’t simplify a sum of two terms, you just leave it as a sum of two terms.

The tensor algebra of a module is the $R$-algebra that additively is the direct sum of all tensor powers of that module, where multiplication is given by the tensor product. This is a graded algebra over the same ring as the module, where the homogeneous components can be taken to be the tensor powers. The construction of the tensor algebra of a module is functorial, meaning any homomorphism of modules induces a graded homomorphism of tensor algebras. The construction of the homomorphism is simple; simply map the module elements, which are the homogeneous elements of degree 1, via the homomorphism into the other module, then extend to the whole algebra using the product.

If $R$ is a field or if the module is free, then the tensor algebra is the free $R$-algebra on any of its bases. That is, for every set map from a basis of the vector space (or free module) into an $R$-algebra there is a unique homomorphism from the tensor algebra into the $R$-algebra extending the set map.

I basically had the same question as you. When defining the covariant derivative as a derivation, then $\nabla (S \otimes T) $ doesn’t seem to make sense in some case unless one rearranges the covariant and contravariant factors in S and T, and I don’t know if there is a canonical way to do this… Even in certain cases where no such rearrangement is necessary to make sense of the sum in the Leibniz rule, applying it directly to a given tensor gives incorrect results: for example, if T is a purely covariant tensor, say the metric expressed in terms of the natural cobasis of a given coordinate system:

$$ g = g_{ij} dx^{i} \otimes dx^{j} $$

then applying the Leibniz rule gives an incorrect form of the components. How to make sense of this?

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