# exterior derivative in local coordinates

Given that the exterior derivative d acting on n-form w is defined as:

$$dw(X_1, \ldots, X_{n+1}) = \sum_{i=1}^{n+1} (-1)^{i+1}X_i(w(X_1, \ldots, \hat X_i, \ldots, X_{n+1})) + \sum_{i

where the $$X_i$$ are members of the tangent space and the hat denotes omission.

My goal is to prove that dw can be written in local coordinates as

$$dw = \partial_b w_{a_1, \ldots,a_n} dx^b \wedge dx^{a_1} \wedge , \ldots, \wedge dx^{a_n}$$

I started by writing $$w = w_{a_1, \ldots,a_n} dx^{a_1} \wedge , \ldots, \wedge dx^{a_n}$$ as an expansion of w in some chosen coordinate system, which I believe we can do because since we are only working locally, the relevant topolgy is that of $$\mathbb{R}^m$$ for some m, and therefore the n-forms have a basis in the covector space.

Then I started with the first term in the def. of dw, and wrote:

$$\sum_{i=1}^{n+1} (-1)^{i+1}X_i((w_{a_1, \ldots,a_n} dx^{a_1} \wedge , \ldots, \wedge dx^{a_n})(X_1, \ldots, \hat X_i, \ldots, X_{n+1}))$$

applying product rule(since $$X_i$$ is a derivation)

$$\sum_{i=1}^{n+1} (-1)^{i+1}X_i(w_{a_1,\ldots,a_n}) (dx^{a_1}\wedge , \ldots, \wedge dx^{a_n})(X_1, \ldots, \hat X_i, \ldots, X_{n+1}) + \sum_{i=1}^{n+1} (-1)^{i+1}w_{a_1,\ldots,a_n}X_i( (dx^{a_1}\wedge , \ldots, \wedge dx^{a_n})(X_1, \ldots, \hat X_i, \ldots, X_{n+1}))$$

And from here I'm stuck, im not sure how to complete this proof as the defintion of the exterior derivative involves its action on a number of vectors, while the statement we want to prove seems to be divorced from this notion(or rather implicitly, the vectors seem to ordered without omissions)

• You'd better start by having the correct formula to prove! Once you've done that, you should also think about appropriate choices for the vectors $X_1,\dots,X_{n+1}$. Aug 6 at 5:03
• Ah of course! the rank of the n form is wrong. Aug 6 at 18:09
• Im guessing an appropriate choice for the vectors would be basis vectors? The only thing I don't quite get is that you could just expand the $X_1,\ldots, X_{n+1}$ in terms of relevant basis vectors, however the $X_i$ would still act on the coefficients, since they are functions themselves. Aug 6 at 18:12
• So you're still missing $dx^\beta$ in your formula for $dw$. Yes, you should take the vector fields $X_i$ to be coordinate vector fields. Aug 6 at 18:33

After consulting a bit more with Spivak's "Introduction to Differential Geometry" and with help from Ted I believe I have solved it.

I forgot to recongize that as an n-form, dw could also expanded in a basis as

$$dw = dw(\frac{\partial}{\partial x^{a_1}},\ldots,\frac{\partial}{\partial x^{a_{k+1}}}) dx^{a_{1}}\wedge ,\ldots,\wedge dx^{a_{k+1}}$$

Then

$$dw(\frac{\partial}{\partial x^{a_1}},\ldots,\frac{\partial}{\partial x^{a_{k+1}}}) = \sum_{i=1}^{k+1} (-1)^{i+1}\frac{\partial}{\partial x^{a_i}}(w_{b_1,\ldots,b_n})(dx^{b_1}\wedge\ldots,\wedge dx^{b_n})(\frac{\partial}{\partial x^{a_1}},\ldots,\hat{\frac{\partial}{\partial x^{a_{k+1}}}}\ldots, \frac{\partial}{\partial x^{a_{k+1}}})$$

where the Lie bracket in the second sum causing that term to go to zero since partial derivatives commute

$$= \sum_{i=1}^{k+1} (-1)^{i+1} \frac{\partial}{\partial x^{a_i}}w_{a_1,\ldots, \hat a_i,\ldots,a_n}$$

so

$$dw = \sum_{i=1}^{k+1} (-1)^{i+1} \frac{\partial}{\partial x^{a_i}}w_{a_1,\ldots, \hat a_i,\ldots,a_n} dx^{a_{1}}\wedge ,\ldots,\wedge dx^{a_{k+1}}$$

upon rearranging of indicies for each term gives

$$dw = (k+1)\frac{\partial}{\partial x^{a_{k+1}}}w_{a_1,\ldots,a_n} dx^{a_{k+1}}\wedge dx^{a_1} ,\ldots,\wedge dx^{a_{k}}$$

The only thing that seems to be wrong is the k+1 factor which should be due to convention.

• Your factor is off more likely due to your usage of Einstein summation (assuming you're using the conventions in Spivak). You wrote $w=w_A\, dx^A$, where the sum is over all multindices of length $k$. This is wrong. If you want to extend the sum over all indices then you should have a $1/k!$ term. If you don't want this factorial term, then you should write $w=\sum_{A, \uparrow}w_A\, dx^A$, where the sum extends over strictly increasing indices of length $k$. Here, $dx^A$ denotes the wedge product $dx^{a_1}\wedge\cdots\wedge dx^{a_k}$. Aug 8 at 15:58
• I don't quite understand why the 1/k! factor is needed? what is different with the ordered sums and the einstein summation? Aug 8 at 20:06
• Let's consider a $2$-form on $\Bbb{R}^3$. $\omega=\omega_A\,dx^A$ means \begin{align} &\omega_{11}\,dx^1\wedge dx^1+ \omega_{12}\,dx^1\wedge dx^2 + \omega_{13}\,dx^1\wedge dx^3\\ +&\omega_{21}\,dx^2\wedge dx^1+ \omega_{22}\,dx^2\wedge dx^2 + \omega_{23}\,dx^2\wedge dx^3\\ +&\omega_{31}\,dx^3\wedge dx^1+ \omega_{32}\,dx^3\wedge dx^2 + \omega_{33}\,dx^3\wedge dx^3 \end{align} Do you see how there are lots of redundancies? We have $\omega_{ij}=-\omega_{ji}$ (because $\omega$ is an alternating tensor, so its components are alternating), and also $dx^i\wedge dx^j=-dx^j\wedge dx^i$. Aug 8 at 20:13
• So, all the 9 terms above collapse to $2(\omega_{12}\,dx^1\wedge dx^2+\omega_{13}\,dx^1\wedge dx^3+\omega_{23}\,dx^2\wedge dx^3)$. This factor of $2$ is really $2!$. In the general case, when you sum over all indices, there's $k!$ many times of over counting, because the indices $\omega_{a_1\dots a_k}$ are alternating, and the wedge-product $dx^{a_1}\wedge\cdots \wedge dx^{a_k}$ is also alternating. Aug 8 at 20:14
• What is correct is $\omega=\sum_A\omega_{a_1\dots a_k}\,dx^{a_1}\otimes\cdots\otimes dx^{a_k}$. This is true for any $(0,k)$ tensor field, alternating or not. But if you have that this is alternating, then you can write things in terms of the wedge product (and following Spivak's conventions for the definition of wedge product), we have $\omega=\sum_{A\uparrow}\omega_A\,dx^A=\frac{1}{k!}\omega_A\,dx^A$, where the last equality is due to the overcounting on the RHS. Aug 8 at 20:18