Why some operations on tensors don't give a tensor? The gradient is a tensor $\nabla f:\mathbf{V} \to \mathbf{R}$ where the partial derivatives are evaluated at some point $(x_0, y_0, z_0)$. And evaluation of this linear form at some vector $v=(v_1,v_2,v_3)$ gives
$$
(\nabla f)(\mathbf{v}) = \partial_x f v_1 + \partial_y f v_2 + \partial_z f v_3
$$
Furthermore in going to a new coordinate system these partial derivatives transform in the expected way.
But what about a function $g:\mathbf{V} \to \mathbf{R}$ which is defined only using the partial derivative of $f$ in the $x$ direction.
$$
g(\mathbf{v}) = \partial_x f v_1 + \partial_x f v_2 + \partial_x f v_3
$$
As I understand this is not considered a tensor because in moving to a new coordinate system it does not transform correctly.
This has confused me endlessly. The strict definition considers a map such as $f:\mathbf{V} \to \mathbf{R}$ a tensor if linearity holds in each parameter. The function $g$ above certainly satisfies that. It seems to me that this definition is not used and that the definition of a tensor that is actually used consists of two parts.

*

*linearity in each parameter (i.e. multilinear form),

*and the algebraic structure of the coefficients is maintained in coordinate transformation

Because once we have calculated $\partial_x f$ it is just a scalar and we just hit $(\partial_x f,\partial_x f,\partial_x f)$ with the usual transformation for a covariant vector to get the new coefficients for $g$ in the new coordinate system. That these new coefficients don't have the right algebraic structure doesn't make multilinearity of $g$ go away. Is this at all correct?
 A: A fundamental thing to remember about vectors and their basis dependent representations is that the basis is arbitrary. Anyone has the right to express any vector in a vector space in terms of any basis of their choosing. When expressed in a basis, a vector looks like an $n$-tuple of scalars, but unless it comes with that promise of that change of basis formula, an arbitrary $n$-tuple of scalars does not actual define a vector.
A tensor is the same except now the promise has to be that the change of basis formula involves multiple factors of the change of basis matrix.
The mathematician's preferred way of achieving that is to give coordinate-free definitions of things like vector spaces and tensors built out of them. But it's fine to think of vectors and tensors by their coordinate components too, as long as you remember that you don't actually have vectors and tensors unless they behave right under change of basis.
And that's just vectors and tensors from a vector space. When we consider vector fields and tensor fields on a manifold, instead of arbitrary changes of basis on the tangent space, we consider coordinate changes of the manifold, and that induces a change of basis in the tangent space and tensors out of the tangent space. Tensors are different than tensor fields, but for brevity people often call them tensors too, so let's address both.

In linear algebra, you can make tensors out of a vector space $V$, a tensor of type $(p,q)$ is an element of the tensor product of $p$-many of copies of $V$ and $q$-many of its dual space $V^*$. In your case, a map $V\to \mathbb{R}$ is a tensor of rank $(0,1)$, also known as a linear functional or dual vector on $V$. If you wanted to, for some fixed function $f(x,y,z)$ and some coordinates $(a,b,c)$, you could compute the derivatives and evaluate at those coordinates and get numbers $\frac{\partial f}{\partial x}(a,b,c),\frac{\partial f}{\partial y}(a,b,c),\frac{\partial f}{\partial z}(a,b,c)$. You could then assemble them into a triple $\left(\frac{\partial f}{\partial x}(a,b,c),\frac{\partial f}{\partial y}(a,b,c),\frac{\partial f}{\partial z}(a,b,c)\right)$ and declare it to be a linear functional on your vector space.
Since your function $f$ and your coordinates $(a,b,c)$ had nothing to do with your vector space, you can declare its components to be this in one basis. If someone wants to know what they are in another basis, simply apply the change of basis matrix. Thus $\frac{\partial f}{\partial x}(a,b,c),\frac{\partial f}{\partial y}(a,b,c),\frac{\partial f}{\partial z}(a,b,c)$ is a valid linear functional on your three dimensional vector space. It is a tensor of type $(0,1).$
It doesn't especially matter how you chose the components of your tensor in your starting basis, so you could just as well have chosen three different numbers, including $\left(\frac{\partial f}{\partial x}(a,b,c),\frac{\partial f}{\partial x}(a,b,c),\frac{\partial f}{\partial x}(a,b,c)\right).$ That is also a valid tensor of type $(0,1).$
So to sum up, as far as the linear algebra of just a single vector space is concerned, both $(\partial_xf,\partial_yf,\partial_zf)$ and $(\partial_xf,\partial_xf,\partial_xf)$ are tensors.
The issue will come when we try to consider them tensor fields on a manifold.

In differential geometry, you do the same thing, you make tensors out of a vector space. Except now the vector space is the tangent space of a manifold $M$. And it varies from point to point. A tensor is an element of the tensor product of some number of copies of the tangent space $T_xM$ to a point $x$ and some number of copies of the cotangent space $T_xM^*$.
To make a tensor field on the manifold, we must choose a tensor from the tensor product of the tangent space at each point of the manifold.
A smooth manifold is something which locally looks like $\mathbb{R}^n$, so it admits local coordinates. Expressed in those local coordinates, a tensor of type $(p,q)$ is an array of $n^pn^q$ functions on $M$. But we may choose different coordinates around any point, and we have the write to express our tensors in any coordinates. Changing coordinates also changes the basis of the tangent space and its dual space. So it's like the change of basis operation in linear algebra, as well as a change of coordinates in the coordinate functions at the same time.
This is a constraint on what a tensor can look like. The change of basis has to match the change of coordinates. You can't swap the two basis vectors without also swapping the two coordinates. If you change the coordinates by a function, then the basis for the tangent space changes by the jacobian of that function.
So this is why $(\partial_xf,\partial_xf,\partial_xf)$ is not a tensor in the sense of differential geometry. When you change coordinates, a tensor should transform the $z$-component by the $z$-partial derivative of the change of coordinate function. But your object scales every component by the $x$-derivative.
It's also why the Christoffel symbols are not a tensor. As the derivative of a tensor, they change with an additional term, rather than a pure change of basis matrix, which is required for tensors. The easiest way to see that is that a tensor which is zero in one coordinate system must be zero in every coordinate system, which the Christoffel symbols do not safisty.
There is a theorem that an array of smooth functions is a tensor if they are $C^\infty(M)$ linear, when viewed as a function on vector fields. The Christoffel symbols also fail that test.
A: First of all if we have a function $f:\Omega\to\mathbb R$, where let's say $\Omega$ is a domain in $\mathbb R^3$, then $\nabla f(p)$ is a vector field
$$\Omega\to\mathbb R^3$$
which assign an arrow $\nabla f(p)$ at $p$. But also
a map $\Omega\to L(\mathbb R^3,\mathbb R)$
which  assigns to each $p$ in $\Omega$ a linear map
$\nabla f(p):\mathbb R^3\to\mathbb R$ given by
$$\nabla f(p)\cdot \mathbf v=\mathbf v^1\frac{\partial f}{\partial x}(p)+
\mathbf v^2\frac{\partial f}{\partial y}(p)+
\mathbf v^3\frac{\partial f}{\partial z}(p),
$$
and can be understood of how $f$ varies in the direction $\mathbf v$ at $p$.
Second, now suppose that we have a change of coordinates $\Phi:\Gamma\to\Omega$, which is a differentiable map with derivative $J\Phi$ of rank three everywhere at $\Gamma\subseteq\mathbb R^3$, where we have $\Phi(q)=p$ for a unique $q$ in
$\Gamma$.
So $f\circ\Phi:\Gamma\to\mathbb R$ (is differentiable if $f$ is)
and by the chain's rule
$\nabla(f\circ\Phi)=\nabla f\cdot J\Phi$
through $\Gamma$,  and at $q$
$$\nabla(f\circ\Phi)(q)=\nabla f(p)\cdot J\Phi(q).$$
If we think that $\Phi:(u,v,w)^{\top}\mapsto(x,y,z)^{\top}$ is the
dependence of the old coordinates into the new ones,
then
$$\left[
\frac{\partial f\circ\Phi}{\partial u}(q)\quad
\frac{\partial f\circ\Phi}{\partial v}(q)\quad
\frac{\partial f\circ\Phi}{\partial w}(q)\right]
=$$
$$
\left[\frac{\partial f}{\partial x}(p)\quad
\frac{\partial f}{\partial y}(p)\quad
\frac{\partial f}{\partial z}(p)
\right]
\left[
\begin{array}{ccc}
\dfrac{\partial x}{\partial u}(q)&\dfrac{\partial x}{\partial v}(q)&\dfrac{\partial x}{\partial w}(q)\\
\dfrac{\partial y}{\partial u}(q)&\dfrac{\partial y}{\partial v}(q)&\dfrac{\partial y}{\partial w}(q)\\
\dfrac{\partial z}{\partial u}(q)&\dfrac{\partial z}{\partial v}(q)&\dfrac{\partial z}{\partial w}(q)\\
\end{array}
\right],
$$
which means that
$$\frac{\partial f\circ\Phi}{\partial u}(q)=
\frac{\partial f}{\partial x}(p)\frac{\partial x}{\partial u}(q)+
\frac{\partial f}{\partial y}(p)\frac{\partial y}{\partial u}(q)+
\frac{\partial f}{\partial z}(p)\frac{\partial z}{\partial u}(q),
$$
$$\frac{\partial f\circ\Phi}{\partial v}(q)=
\frac{\partial f}{\partial x}(p)\frac{\partial x}{\partial v}(q)+
\frac{\partial f}{\partial y}(p)\frac{\partial y}{\partial v}(q)+
\frac{\partial f}{\partial z}(p)\frac{\partial z}{\partial v}(q),
$$
and
$$\frac{\partial f\circ\Phi}{\partial w}(q)=
\frac{\partial f}{\partial x}(p)\frac{\partial x}{\partial w}(q)+
\frac{\partial f}{\partial y}(p)\frac{\partial y}{\partial w}(q)+
\frac{\partial f}{\partial z}(p)\frac{\partial z}{\partial w}(q),
$$
are the components of $\nabla(f\circ\Phi)$ at $q$ (who controls what happen at $p$
with the new coordinates). So, $\nabla f$ is a rank one tensor.
Third, for maps $\mathbf V\times\mathbf V\to\mathbb R$ which comply something similar, would be
dubbed rank two tensors, case that is not present in the gradient operation.
