2015-01-20

# Contravariant, Covariant, Tensor

## (IV) Differentiation is Covariant

Already the fourth part about tensors and such that I write down to understand this stuff myself, but others may benefit. For the notation, please read the first and second part.

Before I start, I would like to introduce a notation which I see seldom used, but which I find quite helpful. For a function $f:(x^1,\dots,x^n)\to K$ lets denote the derivative with regard to the $k$-th parameter as $\dd{k}f$. Normally this is denoted with $\d{f}/\d{x_k}$, but this gets confusing when we have something like $\d{f(2r,4s,7t)}/\d{x_2}$, where the arguments do not contain $x_2$. From the index of $x_2$ we can see that the partial derivative with regard to the second parameter is meant, but, well, I don't like it that way. Instead I will write $\dd{2}f(2r,4s,7t)$. To be more general, I define \begin{equation} \dd{k}f(x^1,\dots,x^n) := \pderiv{x_k}f(x^1,\dots,x^n) \end{equation} to denote the derivative with regard to the $k$-th argument, without applying the inner derivative. If the $k$-th argument is not just $x_k$, but some function of other parameters, we then write \begin{align*} \frac{d}{dt}f(\dots,g_k(t),\dots) &= \sum_{k=1}^n \dd{k}f(\dots,g_k(t),\dots)\cdot \frac{d}{dt}g_k(t) \end{align*}

That said, lets look at a function defined like $f$ above on
the coordinates $x^i$ or $y^j$ of a vector $\vz=x^i\vx_i=y^j\vy_j$, using
Einstein Notation again for summing over a pair of upper and lower
indexes, where the $\vx_i$ and $\vy_j$ are basis vectors of two
different bases $\v{X}$ and $\v{Y}$ of our vector space $V$. Since
the function is defined on just the coordinates of a vector, the
value of the function is typically not unique for a given vector,
but differs with the basis used, i.e. except for special cases we
have
$$f(x^1,\dots,x^n)\neq f(y^1,\dots,y^n).$$
But lets look at the derivatives of $f$ with regard to the
coordinates and remember that $x^i = \ma{i}{j} y^j$ for the basis
change matrix $\ma{i}{j}$.
\begin{align*}
\pderiv{y_j}f(x^1,\dots,x^n)
&= \pderiv{y_j} f(\ma{1}{k}y^k,\dots,\ma{n}{k}y^k) \\
&= \sum_{i=1}^n \dd{i}f(\ma{1}{k}y^k,\dots,\ma{n}{k}y^k)
\cdot \pderiv{y_j} \ma{i}{k}y^k \\
&= \sum_{i=1}^n \dd{i}f(\ma{1}{k}y^k,\dots,\ma{n}{k}y^k)
\cdot \ma{i}{j} \\
&= \sum_{i=1}^n \dd{i}f(x^1,\dots,x^n)
\cdot \ma{i}{j} \\
&= \sum_{i=1}^n \ma{i}{j}\pderiv{x_i} f(x^1,\dots,x^n)
\end{align*}
Using Einstein Notation, this is
$$ \pderiv{y_j}f(x^1,\dots,x^n)
= \ma{i}{j}\pderiv{x_i} f(x^1,\dots,x^n) $$
where I very conciously do not remove the function arguments to
let the formula look neat. Otherwise one could be tempted
to think that *this covariant transformation* not only converts
$\partial/\d{x_i}$ into
$\partial/\d{y_j}$, but also magically replaces the $x^i$ with
$y^j$ in the argument list, which it does not.

The result shows that the differential operators $\partial/\d{x_i}$ form a basis dependent $n$-tupel which transforms covariantly. Hence it is correct that the index is a subscript.