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I am trying to prove the invariant form for the exterior derivative http://en.wikipedia.org/wiki/Exterior_derivative#Invariant_formulations_of_grad.2C_curl.2C_div.2C_and_Laplacian by following these notes http://idv.sinica.edu.tw/ftliang/diff_geom/*diff_geometry(II)/3.11/exterior_derivative_2.pdf.

I am confused however, as the beginning of the proof is to show that the RHS form is "multilinear", in the sense that, for arbitrary vectors [itex]V_i[/itex]

[tex] RHS(V_1, \ldots, fV_p, \ldots, V_{k+1}) = (-1)^{p-1} f RHS(V_1, \ldots, V_p, \ldots, V_{k+1}) [/tex]

Given that the right hand side is equal to the exterior derivative of something, does this not imply that the exterior derivative of something has the same weird symmetry. I would have thought that since the exterior derivative is a tensor, it is the usual linear

[tex] d \Omega (V_1, \ldots, fV_p, \ldots, V_{k+1}) = f d \Omega(V_1, \ldots, V_p, \ldots, V_{k+1}). [/tex]

Could someone please clarify this?

As for the general proof, after proving that things are multilinear in that unusual sense, the author states that [itex] \omega = f dx ^I[/itex] for some index set [itex] I [/itex]. Is this meant to be [itex] \omega = f_I dx ^I[/itex]?

Cheers