I’ve been trying to understand the motivation behind the definition provided by O’Neil for a tensor derivation. Specifically, I can’t seem to understand why a tensor derivation needs to commute with contraction. It is given that:
$$\mathcal D(CA) = C(\mathcal DA)$$
So is it correct to say that if I had some tensor $A=A_{i}^je^i \otimes e_j$, this would mean that the derivation of:
$$\mathcal D(A^i_i) = C(\mathcal D A)$$
If so, I’m having trouble showing this, since $\mathcal D(A^i_i)$ is just an ordinary derivative, and $\mathcal D A$ would be a covariant derivative.
Have I misunderstood the definition?
Edit: Clarifying the Confusion
If I write:
$$\mathcal D(A^i_i) = C(\mathcal D A)$$
Then the right hand side is equivalent to:
$$C(\mathcal D (A_{i}^je^i \otimes e_j))$$ $$C( A_{i}^j \mathcal D (e^i \otimes e_j))$$ $$C( A_{i}^j (\mathcal D e^i \otimes e_j + e^i \otimes \mathcal D e_j))$$
Now that I'm here, I'm not really sure how I can show that this is the same as $\mathcal D(A_i^i)$. Because I can't see why this is true, the definition provided is difficult for me to understand properly.
The contraction itself is a point-wise operation that is linear and defined independently of any kind of choices. So as an operation it can/should be viewed as being "constant" in any coordinate system. Thus it is natural to requrie that you can "differentiate through it". Alternatively, you interpret $\mathcal D(\mathcal CA)-\mathcal C(\mathcal DA)$ as $(\mathcal DC)(A)$. So you can rephrase the requirement as the fact that the contaction itself is covariantly constant, which is natural in view of the above considerations.