Covariance and Contravariance

When tensors transform covariantly, they transform the same way that the basis vectors $\boldsymbol{e_{\mu}}$ transform. When tensors transform contravariantly, they transform the opposite way that the basis vectors transform.

As an example, consider the Cartesian coordinate system:

$Bases$

and we transform the bases $\boldsymbol{e_{\mu}} \to \boldsymbol{\tilde{e}_{\mu}}$ :

\boldsymbol{\tilde{e}_{\mu}} = 2 \boldsymbol{e_{\mu}}

$Bases transformed$

The vector has to stay invariant ( $x^{\mu} \boldsymbol{e_{\mu}} = \tilde{x}^{\mu} \boldsymbol{\tilde{e}_{\mu}} = 2 \tilde{x}^{\mu} \boldsymbol{e_{\mu}}$ ). Considering the vector $\boldsymbol{x}$ :

\begin{align*} x^{\mu} \boldsymbol{e_{\mu}} &= 2 \tilde{x}^{\mu} \boldsymbol{e_{\mu}}, \\ x^{\mu} &= 2 \tilde{x}^{\mu}, \\ \tilde{x}^{\mu} &= \frac{1}{2} x^{\mu}. \end{align*}

For our vector $\boldsymbol{x} = 2 \boldsymbol{e_x} + 3 \boldsymbol{e_y}$ , the transformed components would be:

\begin{align*} \tilde{x}^x &= 1, \\ \tilde{x}^y &= \frac{3}{2}, \\ \boldsymbol{x} &= \boldsymbol{\tilde{e}_x} + \frac{3}{2} \boldsymbol{\tilde{e}_y}. \end{align*}

In general when transforming the coordinates $x^{\mu} \to \tilde{x}^{\mu}$ a general tensor $T^{\alpha \beta ...}{}_{\gamma \delta ...} \to \tilde{T}^{\alpha \beta ...}{}_{\gamma \delta ...}$ is transformed as follows:

\tilde{T}^{\alpha \beta ...}{}_{\gamma \delta ...} = \frac{\partial \tilde{x}^{\alpha}}{\partial x^{a}} \frac{\partial \tilde{x}^{\beta}}{\partial x^{b}} ... \frac{\partial x^{c}}{\partial \tilde{x}^{\gamma}} \frac{\partial x^{d}}{\partial \tilde{x}^{\delta}} ... T^{a b ...}{}_{c d ...}.

Two notes: the partial derivatives have one upper and one lower index, so we are employing the Einstein summation convention. The second note: I am using latin and greek letters. I am using latin indices for space components and greek indices for spacetime components in general relativity. Here they are used for simplicity and don't have any special meaning.

We can make sense of the above relationship by considering a vector $\boldsymbol{x}$ parametrized by $\lambda$ and its tangent vector $\frac{d\boldsymbol{x}}{d\lambda}$ . By chain rule:

\frac{d \boldsymbol{x}}{d \lambda} = \frac{\partial \boldsymbol{x}}{\partial x^{\mu}} \frac{d x^{\mu}}{d\lambda} = \frac{d x^{\mu}}{d \lambda} \boldsymbol{e_{\mu}}.

Now, consider a transformation $x^{\mu} \to \tilde{x}^{\mu}$ . The tangent vector is equal to:

\frac{d \boldsymbol{x}}{d \lambda} = \frac{\partial \boldsymbol{x}}{\partial \tilde{x}^{\mu}} \frac{d \tilde{x}^{\mu}}{d\lambda} = \frac{d \tilde{x}^{\mu}}{d \lambda} \boldsymbol{\tilde{e}_{\mu}}.

We can apply chain rule on the last part:

\frac{d \tilde{x}^{\mu}}{d \lambda} \boldsymbol{\tilde{e}_{\mu}} = \frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}} \frac{d x^{\nu}}{d \lambda} \boldsymbol{\tilde{e}_{\mu}},

implying the component $\frac{d \tilde{x}^{\mu}}{d \lambda}$ in the new basis $\boldsymbol{\tilde{e}_{\mu}}$ is equal to:

\frac{d \tilde{x}^{\mu}}{d \lambda} = \frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}} \frac{d x^{\nu}}{d \lambda},

Now, consider a covector $df$ :

df = \frac{\partial f}{\partial x^{\mu}} dx^{\mu} = \frac{\partial f}{\partial x^{\mu}} \epsilon^{\mu}.

Now, consider a transformation $x^{\mu} \to \tilde{x}^{\mu}$ . The covector is now equal to:

df = \frac{\partial f}{\partial \tilde{x}^{\mu}} d\tilde{x}^{\mu} = \frac{\partial f}{\partial \tilde{x}^{\mu}} \tilde{\epsilon}^{\mu}.

And similarly, we can apply chain rule on the last part:

\frac{\partial f}{\partial \tilde{x}^{\mu}} \tilde{\epsilon}^{\mu} = \frac{\partial f}{\partial x^{\nu}} \frac{\partial x^{\nu}}{\partial \tilde{x}^{\mu}} \tilde{\epsilon}^{\mu},

implying:

\frac{\partial f}{\partial \tilde{x}^{\mu}} = \frac{\partial f}{\partial x^{\nu}} \frac{\partial x^{\nu}}{\partial \tilde{x}^{\mu}}.

For the transformation $x^{\mu} \to \tilde{x}^{\mu}$ , the partial derivatives $\frac{\partial x^{\mu}}{\partial \tilde{x}^{\mu}}$ form the Jacobian which is used for covariant transformation:

J = \frac{\partial x^{\mu}}{\partial \tilde{x}^{\nu}} = \begin{bmatrix} \frac{\partial x^1}{\partial \tilde{x}^1} & \dots & \frac{\partial x^1}{\partial \tilde{x}^n} \\ \vdots & \ddots & \vdots \\ \frac{\partial x^n}{\partial \tilde{x}^1} & \dots & \frac{\partial x^n}{\partial \tilde{x}^n} \end{bmatrix},

And the partial derivatives $\frac{\partial \tilde{x}^{\mu}}{\partial x^{\mu}}$ form the inverse jacobian used for contravariant transformations:

J^{-1} = \frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}} = \begin{bmatrix} \frac{\partial \tilde{x}^1}{\partial x^1} & \dots & \frac{\partial \tilde{x}^1}{\partial x^n} \\ \vdots & \ddots & \vdots \\ \frac{\partial \tilde{x}^n}{\partial x^1} & \dots & \frac{\partial \tilde{x}^n}{\partial x^n} \end{bmatrix}.

Unlike classic derivatives, $\frac{dx}{df} = \frac{1}{\frac{df}{dx}}$ , the inverse partial derivative:

\frac{\partial x^{\mu}}{\partial \tilde{x}^{\nu}} \neq \frac{1}{\frac{\partial \tilde{x}^{\nu}}{\partial x^{\mu}}},

however, the Jacobian satisfies the following:

\begin{align*} J^{-1}J &= \frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}} \frac{\partial x^{\nu}}{\partial \tilde{x}^{\sigma}} \\ &= \frac{\partial \tilde{x}^{\mu}}{\partial \tilde{x}^{\sigma}} \\ &= \delta^{\mu}_{\sigma} = \begin{cases} 1 & \mu &= \sigma, \\ 0 & \mu &\neq \sigma. \end{cases} \end{align*}