## Why the position of the indices matter

A question was asked today about what the difference was between $A_{\alpha}^{\mbox{ }\beta}$ and $A^{\beta}_{\mbox{ }\alpha}$

I’ve got to admit I wondered the same, let’s do it out and see. Let $g^{\alpha \beta}$ be the components of the Minkowski metric tensor (just so we have something to work with. Let $A_{\nu\mu}$ be a second rank tensor with matrix representation:

$\left( \begin{array}{cccc}a_{00} & a_{01} & a_{02} & a_{03}\\a_{10} & a_{11} & a_{12} & a_{13}\\a_{20} & a_{21} & a_{22} & a_{23}\\a_{30} & a_{31} & a_{32} & a_{33} \end{array} \right)$

Calculate $g^{\alpha \beta}A_{\beta \gamma}=A^{\alpha}_{\mbox{ }\gamma}$ as opposed to $A_{\beta \gamma}g^{\gamma \alpha}=A_{\beta}^{\mbox{ }\alpha}$ and compare your answers (I’m having some trouble putting in matrix equations into WordPress at the moment).

But first look carefully at the above, these are two different expressions (on both sides of the equals sign). As coefficients we can invoke the commutativity of multiplication, so $A_{\beta \gamma}g^{\gamma \alpha}=g^{\gamma \alpha}A_{\beta \gamma}$ but if we want to think of this as matrix multiplication, letting G be the metric tensor and A be the matrix represented by the components above,

$g^{\alpha \beta}A_{\beta \gamma}\to GA$

BUT!

$A_{\beta \gamma}g^{\gamma \alpha}\to AG$

and we know that in general $AG\neq GA$, and this is basically what’s going on with the mixed tensor. The order of matrix multiplication depends on which indices are being contracted over and the placement/position of indices indicates this non-commutativity.
And so we see, for this example, that $A_{\alpha}^{\mbox{ }\beta}\neq A^{\beta}_{\mbox{ }\alpha}$ because of $AG\neq GA$. But this isn’t the end of the story, the above is a justification via calculations. I suspect that there are at least two other justifications. (1) A physical justification. The mixed rank tensor in question would be the Lorentz boost. (2) A geometrical justification. Anyway, hope this helps.