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

1. Jun 4, 2012

### ianhoolihan

Hi all,

I am trying to prove the invariant form for the exterior derivative http://en.wikipedia.org/wiki/Exteri...tions_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

2. Jun 5, 2012

### quasar987

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.

3. Jun 5, 2012

### ianhoolihan

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

4. Jun 5, 2012

### 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.