Let M be a differentiable manifold. I want to define the covariant derivative $\nabla$ of tensor fields over M, as a derivation on a suitable algebra A of tensors, that is:
1) $\nabla$ is linear over tensors of the same order.
2) $\nabla$ increases the covariant order of the tensor by one.
3) $\nabla$ satisfies the Leibniz rule for arbitrary tensors of the algebra A with respect to the tensor product.
My issue is with point 3. In Physics, usually one defines the covariant derivative of an arbitrary tensor by extending the covariant derivatives of vectors and covectors, requiring that it commutes with contraction and that it satisfies the Leibniz rule for the components. However, I want to work with the tensors themselves instead of just the components. If I do that, then the Leibniz rule doesn't seem to work. For example, say S is of order (1,0) (covariant) and T is of order (0,1) (contravariant), and that the covariant derivative stacks the additional covariant component at the beginning of the tensor. Writting the tensors in terms of a basis $\partial_{j}$ and its natural cobasis $dx^{j}$:
$$ S = S_{j}dx^{j} $$
$$ T = T^{j}\partial_{j} $$
Let $\Gamma^{i}_{j,k}$ be the connection coefficients defining the covariant derivative in that particular coordinate system. Then one would have:
$$ \nabla S = \partial_{i}(S_{j}) dx^{i} \otimes dx^{j} – S_{j}\Gamma^{j}_{k,l}dx^{l} \otimes dx^{k} $$
$$ \nabla T = \partial_{i}(T^{j}) dx^{i} \otimes \partial_{j} + T^{j}\Gamma^{k}_{j,l}dx^{l} \otimes \partial_{k} $$
But then $\nabla T \otimes S$ has summands of the form $dx^{\alpha} \otimes \partial_{\beta} \otimes dx^{\gamma}$ while $ T \otimes \nabla S$ has summands of the form $\partial_{\alpha} \otimes dx^{\beta} \otimes dx^{\gamma}$, so the two cannot be summed to form $\nabla (T \otimes S )$ unless some shuffling is applied to the factors (the total number of covariant and contravariant components is the same in both).
So my question is: is there a canonical way to order the covariant and contravariant parts in a product like this, so that $\nabla$ satisfies the Leibniz rule in the way I propose?
Best Answer
I think it's much more common to let the last index position be the one introduced by differentiation, not the first. But regardless of which convention you choose, you're never going to have a Leibniz rule for total covariant derivatives of the form $\nabla (T\otimes S) = \nabla T \otimes S + T\otimes \nabla S$. (Notice, for example, that if both $S$ and $T$ have positive rank, then the differentiation index in $\nabla T \otimes S $ is never the last one, and in $T\otimes \nabla S$ it's never the first one.)
The Leibniz rule for covariant derivatives of tensor fields applies to the covariant derivative in the direction of a vector field (or vector): $$ \nabla_V(S\otimes T) = \nabla_V S \otimes T + S \otimes \nabla_V T. $$ This is true whether you put the differentiated index last or first (or somewhere else).