# If the derivative of tensors are not generally tensors, why does vector calculus work?

There's this chart on Wikipedia Suppose I have the function

$$f(x,y) = x^2y^3$$ and I compute the gradient $$\nabla f(x,y) = (2xy^3,3x^2y^2)$$

I began with a scalar field and took the derivative. In principle, I could have chosen any vector, but I decided to use the gradient which is the directional derivative in relation to unit vectors.

This is the derivative of a "scalar" by a "vector" (i.e. cell 2,1 in the grid).

According to what I've read, the derivative of a tensor is not in general a tensor (according to Steven Weinberg).

Then why does an area of mathematics like vector calculus have many rank 1 tensors with derivatives that yield other rank 1 tensors?

The example I just gave is a vector, which is a rank 1 tensor. And there are obviously countless examples of scalar by scalar derivatives, like $$\frac{d}{dx} x^2 = 2x$$ which also yield tensors (i.e. a scalar is Rank 0).

Is it fair to say then that, for certain ranks of tensors (e.g. rank 2 or less), the derivatives in general yield tensors of rank 2 or less? But for rank 3 or higher, this does not apply?

• What is your definition of a derivative for a general Tensor? We can in fact talk about derivatives of Tensor Fields - for example, we often make use of the covariant derivative of a "tensor" of rank ${n \choose k}$ which returns a sum of tensors of rank ${n \choose k}$ – Brevan Ellefsen Apr 16 '19 at 19:09

In this interpretation, it appears that partial differentiating these functions with respect to your coordinates like one does in $$\mathbb{R}^n$$ is an extremely natural way to manipulate these objects, but what Weinberg points out is that this is misguided-- differentiating these functions with respect to your coordinates depends essentially on which coordinates you're operating in, and so it tends to destroy the crucial aspect of the above statement that "the functions associated to different choices of coordinates are related in a particular way" because what looks like a simple partial derivative in one coordinate system looks like some messy combination of partial derivatives in another. In the general context of some arbitrary manifold, you cannot typically cover your manifold with a single choice of coordinates, so there's no good choice of coordinate system within which you can take partial derivatives to obtain a new tensor on the whole space-- you're going to have to worry about how it translates to coordinates covering the rest of the manifold at some point. In this case, even if your new object can be extended smoothly to a tensor on the whole manifold, it practically can't look like a partial derivative everywhere. Notice this has no dependence at all on the rank of the tensors you're starting with or attempting to construct out of coordinate derivatives.
Notice also, though, that the problems above arise when trying to make different coordinate systems agree. If you have a choice of coordinates that cover the entire manifold, you can simply ignore these issues and define new tensors through partial differentiation in that coordinate system in any way you like, accepting that if you ever work in a different coordinate system your new functions will look like a mess (think of, say, writing the curl in $$\mathbb{R}^3$$ in terms of spherical coordinates, the formulas for which can be seen here). That is, combining partial derivatives in a global coordinate system is a completely valid way of constructing new tensors from old (this approach is still not typically desirable, though, unless one has a canonical choice of such global coordinates).
That gets at what Weinberg meant and why the tensor manipulation we do in vector calculus works from the point of view of differential geometry in spite of his comment, but in some sense this is entirely the wrong approach to take. Vector calculus on $$\mathbb{R}^n$$ makes up the building blocks of smooth manifolds-- the former is logically prior. The issue Weinberg raised emerges from trying to patch notions of vector calculus together; it simply doesn't apply to vector calculus itself.