How to prove that the gradient is a covariant tensor? (make it simple and clear)

Question

I know that all vectors are contravariant because if we transform into another basis vector we find its components change inversely to its original components. But I don't know why gradient is a covariant. How the components changes same as basis vector changes?

Answer 1

To prove that the gradient is a covariant tensor, we need to demonstrate that it transforms properly under coordinate transformations. Here's a simple and clear explanation:

Let's consider a scalar function, f(x, y, z), defined in three-dimensional space.

The gradient of this scalar function is given by: ∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)

Now, suppose we perform a coordinate transformation to a new coordinate system (u, v, w). The new coordinates (u, v, w) are functions of the original coordinates (x, y, z).

Under this coordinate transformation, the scalar function f becomes a new function F(u, v, w).

The components of the gradient in the new coordinate system can be calculated using the chain rule of differentiation: (∂F/∂u, ∂F/∂v, ∂F/∂w) = (∂x/∂u, ∂y/∂u, ∂z/∂u) ∇f

• (∂x/∂v, ∂y/∂v, ∂z/∂v) ∇f
• (∂x/∂w, ∂y/∂w, ∂z/∂w) ∇f

Notice that the terms (∂x/∂u, ∂y/∂u, ∂z/∂u), (∂x/∂v, ∂y/∂v, ∂z/∂v), and (∂x/∂w, ∂y/∂w, ∂z/∂w) are the Jacobian matrices of the coordinate transformation.

The components of the gradient in the new coordinate system are linear combinations of the components of the original gradient (∇f), multiplied by the Jacobian matrices.

This transformation property satisfies the definition of a covariant tensor. The components of the gradient transform in a specific way under coordinate transformations, indicating that it is a covariant tensor of rank 1.

Therefore, we have shown that the gradient is a covariant tensor.

Note: The explanation provided here is a simplified and intuitive explanation of the proof. For a more rigorous mathematical proof, it is necessary to consider the details of tensor calculus and the transformation laws of tensors under coordinate transformations.

Answer 2

It cannot be proved, because its a definition. The differential 1-form of any scalar function $x\to f(x)$ defines the components of the "gradient" by

$df(x) = \sum_k \partial_x f(x) dx_k $

where $dx_k$ is the differential of the coordinate map

$X_k : x \to x_k$

with its primitive definition

$\int dx_k = x_k +\text{const}, \partial_{x_m} x_k = \delta_{km}$

that is, $dx_k$ is the linearized system of x=const-hyperplanes. Integrals count the number of crossings of a path over in such family of coordinate hypersurfaces.

For the deriviatives of the general vector field in the local tangent space basis of the translation of a point

$e_k(x) = \partial_{x_k} X(x)$

there exists no a priory rule whats the derivative of the basis considered as a factor in the tensor product of scalars and basis vectors.

$d \ \sum_k v(x)_k e(x)_k = \sum_k (d v(x)_k e_k(x)) = \sum_{km}\ dx_m \ (\partial_{x_m} v_k(x) ) e_k(x) + \sum_k v_k(x) \ d e_k(x) $

where the last term, corresponding to the first term, is a tensor product of a 1-form and a vector.

It follows, that the differential of a basis vector is a linear combination of the basis vectors the same point.

The linear map is called the connection

$d e_k(x) = dx_m (\Gamma(x)_m)_{kl} e_l(x)$

The connection is a derivative, even if its not apparent. Try to do the derivative of tensor products of basis vectors, it has to coincide with the sum of derivatives over the factors.

In order to fix the unique Levi-Civita connection with the Christoffel symbols as the coefficients of the connection 1-form, one has to introduce a metric g(x), a local symmetric bilinear form on the basis of the tangent space

$e_i(x) . e_k(x) = g(x)_{ik}$

For a metric space one has the identity

$ dx_a (g_{bc} e_c ) = g_{bc} dx_a(e_c) = g_{ac} = e_a . e_c$

that is by filling one slot of the map g one can switch between tangent space $\{e_i\}$ and cotangent space $\{dx_m\}$

and the metric tensor has conseqently a mixed form, the unit matrix.

Classically this is the expression, that the cotangent basis has the inverse matrix as metric.

$g_{ik}g^{km} = \delta_k^m $

Now the Levi-Civita connection on the tangent base is defined to handle the metric as a covariant constant, so that the product rule works as usual

$\nabla_k g(e_i(x),e_j(x)) = g(\nabla_k e_i(x), e_j(x)) + g(e_i(x), \nabla_k e_j(x)) = (\Gamma(x)_k)_{is} g(e_s,e_j)(x) +(\Gamma(x)_k)_{js} g(e_i,e_s)(x)$