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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.

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:

  1. 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.
  2. Is this construction functorial from the category of smooth manifolds to some algebraic category?
  3. If so, does this construction satisfy some universal property?
ಠ_ಠ
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1 Answers1

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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.

Matt Samuel
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  • Ok, so then let's forget about the restriction on addition when ranks don't match up. If we let $T$ be the free (unital associative) algebra functor on the category of $R$-modules then $T(V)$ is algebra containing all the contravariant tensors. Now how do we combine $T(V)$ with $T(V^)$ to get an algebra where we can take tensor products of covariant and contravariant tensors? Would this be $T(V \oplus V^)$? – ಠ_ಠ Jan 23 '15 at 01:16
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    @user59001 $T(V\oplus V^{\ast})$ would work. I don't know if it's a problem, but the order in which the tensor products were taken would matter, i.e. $V\otimes V^{\ast}\otimes V$ is different from $V^{\ast}\otimes V\otimes V$. – Matt Samuel Jan 23 '15 at 01:19