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}} $

2015-01-18

Contravariant, Covariant, Tensor

(II) Covariance

After I understood where the term contravariant comes from, I am now ready to explain covariant. As before we have a vector space $V$ over a field $K$ with two bases \begin{align*} \v{X} &= (\vx_1,\dots,\vx_n), \qquad \vx_i\in V, \\ \v{Y} &= (\vy_1,\dots,\vy_n), \qquad \vy_j\in V \end{align*} and a set of $\mA{i}{j}\in K$ that transform $\v{X}$ into $\v{Y}$ according to \begin{equation} \vy_j= \sum_{i=1}^n \mA{i}{j}\vx_i . \label{eq:vy} \end{equation} Further we look at a linear form $f:V\to K$, i.e. a function from $V$ into $K$ that assigns an element $f(\vz)$ to each $\vz\in V$ and is linear. In particular $f$ provides us with two $n$-tupels $\tx_i:=f(\vx_i) \in K$ and $\ty_j:=f(\vy_j) \in K$, one for each of the bases.

This reminds of the coordinates of a vector $\vz\in V$ which are also $n$-tupels of values depending on the selected basis, and we can ask whether and how we transform the $\tx_i$ into the $\ty_j$. But this is not difficult: \begin{align*} \ty_j &= f(\vy_j) \\ &= f\left(\sum_{i=1}^n \mA{i}{j}\vx_i\right) & &&\text{by \eqref{eq:vy}} \\ &= \sum_{i=1}^n \mA{i}{j} f(\vx_i) \\ &= \sum_{i=1}^n \mA{i}{j} \tx_i. \end{align*} We see that the $\mA{i}{j}$ transform basis vectors $\vx_i$ into $\vy_j$ (see \eqref{eq:vy}) as well as the coefficients $\tx_i$ into $\ty_j$, hence these coefficients of a linear form $f$ transform in the same direction as the bases and are therefore covariant.

To summarize, the $\mA{i}{j}$ perform for us the following transformations:

  1. basis vector $\vx_i \longrightarrow \vy_j$ (reference)
  2. vector coordinates $y_j \longrightarrow x_i$ (contravariant, opposite direction of reference)
  3. linear form coefficients $\tx_i \longrightarrow \ty_j$ (covariant, same direction of reference)

If you were reading hoping to see indexes hop up and down between subscript and superscript, you may be disappointed, but don't dispair. I think that only now that we clearly understand that there exist two different types of basis-dependent $n$-tupels, it is time to talk about superscript indexes to differentiate contravariant tupels from covariant ones.

And the rules are simple. The components of a basis-dependent $n$-tupel have their index

  1. as a subscript, like $\tx_i$ and the basis vectors $\vx_i$ itself, if the $n$-tupel transforms covariant, and
  2. as a superscript like $x^i$, if the $n$-tupel is contravariant.