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I understand that a line element is not actually a differential form but a $1$-density. My question is: is the notation $ds^2 = dx^2 + dy^2$ formal in any way? Can it be interpreted as outer or tensor products? Is it just simply an informal useful way to describe an integrable object?

I think that a straightforeward interpretation of $ds^2$ as the wedge product of two vector valued differential forms is not possible, but is there any other way I can look ar this?

After looking into the other posts related to this same subject, such as Why is arc length not a differential form? and Is $ds$ a differential form? I still can't find a useful way of thinking about this particular problem.

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Ignacio
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1 Answers1

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$ds^2$ is a symmetric covariant $2$-tensor; i.e., this notation is classical shorthand for $ds^2 = dx\otimes dx + dy\otimes dy$, and we evaluate the line integral $$\displaystyle\int_C ds = \int_C \sqrt{ds^2} = \int_a^b \sqrt{g^*(dx\otimes dx+dy\otimes dy)} = \int_a^b |g'(t)|dt,$$ for any parametrization $g\colon [a,b]\to\Bbb R^2$ of $C$, as expected. Of course, $ds$ is not a $1$-form because $ds(v) = \sqrt{ds^2(v,v)}$ is not linear in $v$.

Ted Shifrin
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  • When you say that $ds^2$ is a rank 2 covariant tensor are you implying that it is a 2-form? because as I understand it forms are just covariant tensors and $ds^2$ is supposed not to be a form. – Ignacio Apr 29 '15 at 05:12
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    $k$-forms are alternating (skew-symmetric) covariant $k$-tensors. The metric is a symmetric $2$-tensor. :) – Ted Shifrin Apr 29 '15 at 12:26
  • Thanks a lot! Great answers! – Ignacio Apr 30 '15 at 00:11