- #1

- 1,384

- 2

[tex]\mathrm{d}x \, \mathrm{d}y = \mathrm{d}x \wedge \mathrm{d}y[/tex]

since this is a 2-form, being the exterior derivative of [itex]x \, \mathrm{d}y[/itex], and therefore antisymmetric, but is this equivalent to

[tex]\mathrm{d}x \otimes \mathrm{d}y \enspace ?[/tex]

And would I be right in thinking that

[tex]\mathrm{d}x^\alpha \otimes \mathrm{d}x^\beta \neq \mathrm{d}x^\alpha \wedge \mathrm{d}x^\beta \enspace[/tex]

since

[tex]\mathrm{d}x^\alpha \otimes \mathrm{d}x^\alpha \neq \mathrm{d}x^\alpha \wedge \mathrm{d}x^\alpha \enspace = 0[/tex]