Harald Kirsch

genug Unfug.

$\def\v#1{\mathfrak{#1}} \def\vx{\v{x}} \def\vy{\v{y}} \def\vz{\v{z}} \def\mA#1#2{a_{#1 #2}} \def\ma#1#2{a^{#1}_{#2}} \def\t#1{\tilde #1} \def\tx{\t{x}} \def\ty{\t{y}} \def\d#1{\partial #1} \def\dd#1{\partial_{#1}} \def\pderiv#1{\frac{\partial}{\partial #1}} $

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 $$ \pderiv{y_i}f(x^1,\dots,x^n) = \ma{i}{j}\pderiv{x_i} f(x^1,\dots,x^n) $$ 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 listl, 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.