Help with a few annoying exterior derivatives

[etotheipi]

Homework Statement

Given $\omega \in \Lambda^p(M)$, $\eta \in \Lambda^q(M)$, show that $$\begin{align*}
\mathrm{i)} \quad &d(d\omega) = 0 \quad \\
\mathrm{ii)} \quad &d(\omega \wedge \eta) = (d\omega) \wedge \eta + (-1)^p \omega \wedge d\eta \quad \\
\mathrm{iii)} \quad &d(\varphi^* \omega) = \varphi^* (d\omega)
\end{align*}
$$with $\varphi: M \rightarrow N$ a map between $M$ and $N$.

Relevant Equations

N/A

I'm not very comfortable with these computations, so go easy on me! $$
\begin{align*}
[d(d\omega)]_{\mu_1 \dots \mu_{p+2}} &= (p+2) \partial_{[\mu_1} (d\omega)_{\mu_2 \dots \mu_{p+2}]} \\
&= (p+1)(p+2) \partial_{[\mu_1} \partial_{\mu_2} \omega_{\mu_3 \dots \mu_{p+2}]}
\end{align*}$$The last part should be zero, since every term cancels with a corresponding term of opposite sign with the indices on the mixed partial derivative switched (due to the anti-symmetrisation). For the next one, I tried to write both sides explicitly$$
\begin{align*}
[d(\omega \wedge \eta)]_{\mu_1 \dots \mu_{p+q+1}} &= (p+q+1) \partial_{[\mu_1} (\omega \wedge \eta)_{\mu_2 \dots \mu_{p+q+1}]} \\
&= \frac{(p+q+1)(p+q)!}{p!q!} \partial_{[\mu_1} \omega_{\mu_2 \dots \mu_{p+1}} \eta_{\mu_{p+2} \dots \mu_{p+q+1}]}
\end{align*}$$then, since $(d\omega)_{\alpha_1 \dots \alpha_{p+1}} = (p+1) \partial_{[\alpha_1} \omega_{\alpha_2 \dots \alpha_{p+1}]}$ and $(d\eta)_{\beta_1 \dots \beta_{q+1}} = (q+1) \partial_{[\beta_1} \eta_{\beta_2 \dots \beta_{q+1}]}$, we have$$
\begin{align*}
[(d\omega) \wedge \eta]_{\mu_1 \dots \mu_{p+q+1}} &= \frac{(p+q+1)!}{(p+1)!q!} d\omega_{[\mu_1 \dots \mu_{p+1}} \eta_{\mu_{p+2} \dots \mu_{p+q+1}]} \\
&= \frac{(p+q+1)!}{p! q!} \partial_{[\mu_1} \omega_{\mu_2 \dots \mu_{p+1}} \eta_{\mu_{p+2} \dots \mu_{p+q+1}]}
\end{align*}$$and likewise$$[\omega \wedge (d\eta)]_{\mu_1 \dots \mu_{p+q+1}} = \frac{(p+q+1)!}{p!q!} \omega_{[\mu_1 \dots \mu_{p}} \partial_{\mu_{p+1}} \eta_{\mu_{p+2} \dots \mu_{p+q+1}]}$$
If we multiply the last one by $(-1)^p$ and add it to the one before, I'm not really sure how to manipulate it into the form of (ii)? Thanks for any help!

 

[Guillermo Navas]

You can consult the book "Differential forms with Applications to the Physical Sciences, by Harley Flanders, Ed. Dover, I see that it solves that exercises.

 

[nrqed]

For the next one, I tried to write both sides explicitly$$
\begin{align*}
[d(\omega \wedge \eta)]_{\mu_1 \dots \mu_{p+q+1}} &= (p+q+1) \partial_{[\mu_1} (\omega \wedge \eta)_{\mu_2 \dots \mu_{p+q+1}]} \\
&= \frac{(p+q+1)(p+q)!}{p!q!} \partial_{[\mu_1} \omega_{\mu_2 \dots \mu_{p+1}} \eta_{\mu_{p+2} \dots \mu_{p+q+1}]}
\end{align*}$$then, we have$$
\begin{align*}
[(d\omega) \wedge \eta]_{\mu_1 \dots \mu_{p+q+1}} &= \frac{(p+q+1)!}{(p+1)!q!} d\omega_{[\mu_1 \dots \mu_{p+1}} \eta_{\mu_{p+2} \dots \mu_{p+q+1}]} \\
&= \frac{(p+q+1)!}{p! q!} \partial_{[\mu_1} \omega_{\mu_2 \dots \mu_{p+1}} \eta_{\mu_{p+2} \dots \mu_{p+q+1}]}
\end{align*}$$

Watch out. Look at your two expressions which start with $\partial_{[\mu_1 \dots]}$. They are identical, which does not make sense since in the first case the derivative should act on everything whereas in the second case it should act only on the $\omega$. Do you see how to fix this?

 

[wrobel]

You can check all the formulas for monomials by a direct calculation in local coordinates

 

[romsofia]

I think you're making your life harder by interchanging between the notation of forms, and partial derivatives. I'll suggest writing $\omega = a dx^i$ and $\eta = b dx^j$ where a and b are 0-forms (fancy terminology for functions!).

Now, for ii), it becomes $d (a dx^i \wedge b dx^j) = d (ab dx^i \wedge dx^j)...$ and you can go from there [to make this proof easy, ask yourself what does $d \omega$ and $d\eta$ look like in this notation]

HINT: do you think the $(-1)^p$ comes from properties of the exterior derivative, or potentially another "operator" in the mix?

 

[etotheipi]

Watch out. Look at your two expressions which start with $\partial_{[\mu_1 \dots]}$. They are identical, which does not make sense since in the first case the derivative should act on everything whereas in the second case it should act only on the $\omega$. Do you see how to fix this?

Ahh, thanks, yeah that makes sense! That in mind I thought to try something like
$$
\begin{align*}

[(d\omega) \wedge \eta]_{\mu_1 \dots \mu_{p+q+1}} = (-1)^{(p+1)q} [\eta \wedge (d\omega)]_{\mu_1 \dots \mu_{p+q+1}} &= (-1)^{(p+1)q} \frac{(p+q+1)!}{q! (p+1)!} \eta_{[\mu_1 \dots \mu_{q}} (d\omega)_{\mu_{q+1} \dots \mu_{p+q+1}]} \\ \\

&= (-1)^{(p+1)q} \frac{(p+q+1)!}{q! p!} \eta_{[\mu_1 \dots \mu_{q}} \partial_{\mu_{q+1}} \omega_{\mu_{q+2} \dots \mu_{p+q+1}]}

\end{align*}
$$I think that looks a bit more promising, I think the rest is just the product rule and a little bit of fiddling with the algebra, but I've got to run now so I'll try and finish and write up this one and the one with the pull-back a bit later this afternoon

I think you're making your life harder by interchanging between the notation of forms, and partial derivatives. I'll suggest writing $\omega = a dx^i$ and $\eta = b dx^j$ where a and b are 0-forms (fancy terminology for functions!).

Now, for ii), it becomes $d (a dx^i \wedge b dx^j) = d (ab dx^i \wedge dx^j)...$ and you can go from there [to make this proof easy, ask yourself what does $d \omega$ and $d\eta$ look like in this notation]

HINT: do you think the $(-1)^p$ comes from properties of the exterior derivative, or potentially another "operator" in the mix?

Is this for the case where $\omega$ and $\eta$ are both one-forms, opposed to general $p$ & $q$ forms?

 

[etotheipi]

How would you suggest doing the third one? I'm not sure where to start... (well, apart from maybe from the definitions). Let $(\partial_{(1)}, \dots \partial_{(n)})$ be a basis of $T_p M$, then if $\omega$ is a $p$-form on $N$ and we've got a $\varphi: M \rightarrow N$, then$$(\varphi^* \omega)_{\mu_1 \dots \mu_p} = (\varphi^* \omega)(\partial_{(\mu_1)}, \dots, \partial_{(\mu_p)}) = \omega(\varphi_{*} \partial_{(\mu_1)}, \dots, \varphi_{*} \partial_{(\mu_p)})$$and also$$d(\varphi^* \omega)_{\mu_1 \dots \mu_{p+1}} = (p+1) \partial_{[\mu_1} (\varphi^* \omega)_{\mu_2 \dots \mu_{p+1}]}$$whilst$$
\begin{align*}
\varphi^*(d\omega) &= (p+1) \varphi^*\left[ \frac{1}{p!} \partial_{\nu} \omega_{\mu_1, \dots, \mu_p} dx^{\nu} \wedge dx^{\mu_1} \wedge \dots \wedge dx^{\mu_p} \right] \\ \\

[\varphi^*(d\omega)]_{\mu_1, \dots, \mu_{p+1}} &= (d\omega) (\varphi_{*} \partial_{(\mu_1)}, \dots, \varphi_{*} \partial_{(\mu_{p+1})})
\end{align*}
$$I know that if $y^{\alpha}$ are some coordinates on $N$ then to push-forward the tangent basis of $T_p M$ I guess we can just do$$\varphi_{*} \partial_{(\mu_i)} = (\partial_{(\mu_i)})^{\mu} \frac{\partial y^{\alpha}}{\partial x^{\mu}} \partial_{\alpha} = \delta_{\mu_i}^{\mu} \frac{\partial y^{\alpha}}{\partial x^{\mu}} \partial_{\alpha} = \frac{\partial y^{\alpha}}{\partial x^{\mu_i}} \partial_{\alpha}$$To be honest, I'm not really sure where any of this is going ...

 

[Frabjous]

It a Big Math conspiracy. You wouldn’t be having these problems taking derivatives with index notation. If you pull the veil open a little more, you will see that ict is also the way to go.

 

[etotheipi]

It a Big Math conspiracy. You wouldn’t be having these problems taking derivatives with index notation. If you pull the veil open a little more, you will see that ict is also the way to go.

Well, unfortunately ict is a bit problematic in general manifolds

 

[Frabjous]

Big Math has spent decades suppressing ict, and the reason we cannot use it today is that we have not spent the time to make it work for general manifolds. It’s a conspiracy.

 

[romsofia]

Is this for the case where $\omega$ and $\eta$ are both one-forms, opposed to general $p$ & $q$ forms?

Nope! Just general q, and p forms (or r forms)! But why you may wonder?

So let me tell you, it's actually because of the result from i).

Quick "proof": Let $\omega = a dx \wedge dy \wedge dz$ an infamous 3 form that you know and love. Now, ask yourself, what is $d \omega$? It's simply $da \wedge dx \wedge dy \wedge dz + a d(dx) \wedge dy \wedge dz +...$ [remember that you can think of $\wedge$ as multiplication] but using the result of i) it's easy to see that $d \omega = da \wedge dx \wedge dy \wedge dz$ [And note that now we are a 4 form, so the exterior dervative made us go up i.e $d: \Lambda^3(M) \rightarrow \Lambda^4(M)$] and that's why when doing proofs such as these, it's easier to just write a general form as $\omega = a dx^i$

 

[etotheipi]

Ah thanks, okay, so am I to understand the notation as$$\omega = a \mathrm{d}x^i \equiv a \mathrm{d}x^{\alpha_1} \wedge \dots \wedge \mathrm{d}x^{\alpha_p}$$in which case presumably$$\mathrm{d}\omega = \mathrm{d}a \wedge \mathrm{d}x^i \equiv \mathrm{d}a \wedge \mathrm{d}x^{\alpha_1} \wedge \dots \wedge \mathrm{d}x^{\alpha_p}$$due to the $\mathrm{d}^2 = 0$ property you mentioned. Yes I think that makes sense, the $\mathrm{d}x^i$ part was just throwing me off

 

[etotheipi]

Okay, here's my go at (ii) using @romsofia's magic method...$$\mathrm{d}\omega \wedge \eta = b \mathrm{d}a \wedge \mathrm{d}x^i \wedge \mathrm{d}x^j$$$$(-1)^p \omega \wedge \mathrm{d}\eta = (-1)^p a \mathrm{d}x^i \wedge \mathrm{d}b \wedge \mathrm{d}x^j$$We also have$$\begin{align*}

d(\omega \wedge \eta) = d(ab \mathrm{d}x^i \wedge \mathrm{d}x^j) &= a \mathrm{d}b \wedge \mathrm{d}x^i \wedge \mathrm{d}x^j + b \mathrm{d}a \wedge \mathrm{d}x^i \wedge \mathrm{d}x^j\\

&= (-1)^p a\mathrm{d}x^i \wedge \mathrm{d}b \wedge \mathrm{d}x^j + b \mathrm{d}a \wedge \mathrm{d}x^i \wedge \mathrm{d}x^j

\end{align*}$$where we picked up the $(-1)^p$ by changing the order of the $p$-form $\mathrm{d}x^i$ and the $1$-form $\mathrm{d}b$. And that should do it! It's nice, I didn't realize it would be that simple

For (iii), how's best to go about dealing with the pull-back?