2

It was pointed out that dual vectors of a manifold, and hence differential 1-forms, are not dependent on the metric (Intuition behind dual vectors ('Bongs of a bell' does not help)). But doesn't the line element include both the metric and differential 1-forms?

$$ds^2 = (dx^1)^2 + (dx^2)^2.$$
Or, $$ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu}.$$

Here, isn't $dx$ a differential 1-form? Also trying to synthesize good post here (https://math.stackexchange.com/q/483191/).

Qmechanic
  • 220,844
so860
  • 145
  • 2

1 Answers1

5

The metric tensor $$\mathbb{g}~=~g_{\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}$$ is a section in the symmetric tensor product $$T^{\ast}M\odot T^{\ast}M.$$ It has an interpretation as an infinitesimal arclength square via a generalization of the Pythagorean theorem, which explains the common notation $ds^2$.

So $\mathrm{d}x^{\mu}$ is indeed a basis for covectors/1-forms, but $\mathbb{g}$ is not a 2-form, and $ds$ is not a 1-form. For more details, see my Phys.SE answer here.

Qmechanic
  • 220,844