# Quickly: Multlinearity of exterior derivative, and proof of invariant formula

## Main Question or Discussion Point

Hi all,

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 $V_i$

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

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
$$d \Omega (V_1, \ldots, fV_p, \ldots, V_{k+1}) = f d \Omega(V_1, \ldots, V_p, \ldots, V_{k+1}).$$

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

Cheers

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quasar987
Homework Helper
Gold Member
If you read the proof, you will see that they actually prove that RHS is multilinear in the usual sense of the word, and so that (-1)^{p-1} factor is some kind of typo.

Next, the authors make the remark that since a general k-form is (locally) a sum of terms of the form $f dx ^I$ (for some multiindex I), and since both RHS and d are linear, it suffices to check that RHS = d on forms $\omega$ of the type $\omega = f dx ^I$. It is completely unnecessary to clog up the notation by giving f a multiindex, but you may do so if its absence makes you uneasy.

If you read the proof, you will see that they actually prove that RHS is multilinear in the usual sense of the word, and so that (-1)^{p-1} factor is some kind of typo.

Next, the authors make the remark that since a general k-form is (locally) a sum of terms of the form $f dx ^I$ (for some multiindex I), and since both RHS and d are linear, it suffices to check that RHS = d on forms $\omega$ of the type $\omega = f dx ^I$. It is completely unnecessary to clog up the notation by giving f a multiindex, but you may do so if its absence makes you uneasy.
Thanks. I soon figured out that it was multilinear in the usual sense, though the initial error was confusing.

As for the next point, I just wanted to clarify that they were not assuming that all the coefficients were equal, i.e. letting $f = f_{a_1\ldots a_{p}}$ for all $a_i$.

Ah, I see --- maybe they can assume that, as long as they show that the equation is linear in $\omega$. I think they state this, but do not prove it.

Cheers

quasar987
I suppose that the complete argument would be this: pick a multiindex I and a function f, then verify the equality for $\omega=fdx^I$. Then, observe that nowhere in this is verification did we explicitely use the value of I or any special property about the function f. So the equality is valid for any multiindix I and associated function f_I and hence for any k-form by linearity.