Undergrad Help with Differential Forms - self wedge product terms

[thatboi]

I'm completely new to differential forms so I am having trouble following the arguments from the following post: https://physics.stackexchange.com/q...non-trivial-fiber-bundles-chern-simons-theory
Specifically:
1.) In equation (12) of the accepted answer, where did the self wedge product terms go, i.e. $F_{D} \wedge F_{D}$?
2.) Similarly, in (18), where did terms like $A_{S_{1}} \wedge dA_{S_{1}}$ go?
3.) What is the intuition behind the last statement: " the contributions from the adjacent parts of the different patches' boundaries cancel each other". How do I see this?

Any help is appreciated.

 

[Hill]

Wedge is antisymmetric, i.e., $A \wedge B = -(B \wedge A)$. This implies that $A \wedge A = -(A \wedge A)=0$.

 

[thatboi]

Wedge is antisymmetric, i.e., $A \wedge B = -(B \wedge A)$. This implies that $A \wedge A = -(A \wedge A)=0$.

I thought this only held for A and B that are odd-number forms.

 

[fresh_42]

$a\wedge a=0$ is part of the definition of a Graßmann algebra and $d\omega \wedge d\omega =0$ for differential forms. Note that exterior (Cartan) derivatives have a sign in $d(\omega_1\wedge \omega_2).$ See Cartan derivative in
https://www.physicsforums.com/insig...#C-–-Exterior-Derivative-or-Cartan-Derivative
\begin{align*}
\int_X F\wedge F&=\int_X (F_D+F_{S^2})\wedge (F_D+F_{S^2})\\
&=\int_X\underbrace{F_D\wedge F_D}_{=0}+ \int_X (F_D\wedge F_{S^2}) +\int_X (F_{S^2}\wedge F_D)+ \int_X \underbrace{F_{S^2}\wedge F_{S^2}}_{=0} \\
&=\left(\int_D F_D\right)\cdot \left(\int_{S^2}F_{S^2}\right) +\left(\int_{S^2}F_{S^2}\right)\cdot\left(\int_D F_D\right)\\
&=2\left(\int_D F_D\right)\cdot \left(\int_{S^2}F_{S^2}\right)
\end{align*}
Hence, the question is why $\int_{D\times S^2} (F_D\wedge F_{S^2}) =\left(\int_D F_D\right)\cdot \left(\int_{S^2}F_{S^2}\right)$ where the LHS seems to be oriented and the RHS is not. I assume that $F_D\wedge F_{S^2}=F_D \times F_{S^2}$ for physical reasons but I don't know enough about the field $F$ and the physical background.

 

[thatboi]

$a\wedge a=0$ is part of the definition of a Graßmann algebra and $d\omega \wedge d\omega =0$ for differential forms. Note that exterior (Cartan) derivatives have a sign in $d(\omega_1\wedge \omega_2).$ See Cartan derivative in
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/#C-–-Exterior-Derivative-or-Cartan-Derivative
\begin{align*}
\int_X F\wedge F&=\int_X (F_D+F_{S^2})\wedge (F_D+F_{S^2})\\
&=\int_X\underbrace{F_D\wedge F_D}_{=0}+ \int_X (F_D\wedge F_{S^2}) +\int_X (F_{S^2}\wedge F_D)+ \int_X \underbrace{F_{S^2}\wedge F_{S^2}}_{=0} \\
&=\left(\int_D F_D\right)\cdot \left(\int_{S^2}F_{S^2}\right) +\left(\int_{S^2}F_{S^2}\right)\cdot\left(\int_D F_D\right)\\
&=2\left(\int_D F_D\right)\cdot \left(\int_{S^2}F_{S^2}\right)
\end{align*}
Hence, the question is why $\int_{D\times S^2} (F_D\wedge F_{S^2}) =\left(\int_D F_D\right)\cdot \left(\int_{S^2}F_{S^2}\right)$ where the LHS seems to be oriented and the RHS is not. I assume that $F_D\wedge F_{S^2}=F_D \times F_{S^2}$ for physical reasons but I don't know enough about the field $F$ and the physical background.

So were you using the fact that $d\omega \wedge d\omega =0$ above? Does this formula hold for any n-form $\omega$?

 

[fresh_42]

So were you using the fact that $d\omega \wedge d\omega =0$ above? Does this formula hold for any n-form $\omega$?

Yes. It is part of the definition of the wedge product.

The Cartan derivative on the other hand is a function of differential forms and can have therefore a different definition that involves signs depending on the rank of the differential forms.

Your question still stands: Why is the integral on the LHS not orientated although the wedge product is? I assume that we have a disc and a separate sphere and the integral over both doesn't distinguish which one is first, i.e. that $\int_X F_D \wedge F_S^2=\int_{D\times S^2}F_D\times F_{S^2}.$

 

[thatboi]

Yes. It is part of the definition of the wedge product.

The Cartan derivative on the other hand is a function of differential forms and can have therefore a different definition that involves signs depending on the rank of the differential forms.

Your question still stands: Why is the integral on the LHS not orientated although the wedge product is? I assume that we have a disc and a separate sphere and the integral over both doesn't distinguish which one is first, i.e. that $\int_X F_D \wedge F_S^2=\int_{D\times S^2}F_D\times F_{S^2}.$

Thanks for the replies. I first want to clarify your point on wedge product vs Cartan derivative. I thought the definition of the wedge product was $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$ for $p,q$ the degrees of $\alpha,\beta$ respectively. Now $F$ is a 2-form, so shouldn't the product commute?

 

[fresh_42]

Forget all of the above.

I confused myself. We have $\omega \wedge \omega =0$ for differential forms of odd degrees and especially for the underlying vector space. The exterior product is graded, i.e.
$$
\alpha \wedge \beta = (-1)^{\deg \alpha \deg\beta } \beta \wedge \alpha
$$

This explains ...
\begin{align*}
(F_D +F_{S^2})\wedge (F_D +F_{S^2})&= \underbrace{F_D\wedge F_D}_{=0}+(F_D \wedge F_{S^2})+ (F_{S^2} \wedge F_{D})+\underbrace{F_{S^2}\wedge F_{S^2}}_{=0}\\
&=(F_D \wedge F_{S^2})+(-1)^{2\cdot 2}(F_D \wedge F_{S^2})\\
&=2(F_D \wedge F_{S^2})
\end{align*}
... but it leaves me clueless as of why $\int_X F_D\wedge F_D = \int_X F_{S^2}\wedge F_{S^2}=0.$

 

[thatboi]

Forget all of the above.

I confused myself. We have $\omega \wedge \omega =0$ for differential forms of odd degrees and especially for the underlying vector space. The exterior product is graded, i.e.
$$
\alpha \wedge \beta = (-1)^{\deg \alpha \deg\beta } \beta \wedge \alpha
$$

This explains ...
\begin{align*}
(F_D +F_{S^2})\wedge (F_D +F_{S^2})&= \underbrace{F_D\wedge F_D}_{=0}+(F_D \wedge F_{S^2})+ (F_{S^2} \wedge F_{D})+\underbrace{F_{S^2}\wedge F_{S^2}}_{=0}\\
&=(F_D \wedge F_{S^2})+(-1)^{2\cdot 2}(F_D \wedge F_{S^2})\\
&=2(F_D \wedge F_{S^2})
\end{align*}
... but it leaves me clueless as of why $\int_X F_D\wedge F_D = \int_X F_{S^2}\wedge F_{S^2}=0.$

Ah ok, so we have the same question. I'm wondering if it has to do with the fact that we are integrating over $X$, which is a direct product $D \times S^{2}$ but $F_{D} \wedge F_{D}$ is only defined on $D$ so I was wondering how such an integral would even work/make sense?

 

[fresh_42]

Ah ok, so we have the same question. I'm wondering if it has to do with the fact that we are integrating over $X$, which is a direct product $D \times S^{2}$ but $F_{D} \wedge F_{D}$ is only defined on $D$ so I was wondering how such an integral would even work/make sense?

That's a nice idea. We have
$$
\int_X (F_D\wedge F_{S^2})=\int_{D\times S^2} (F_D\wedge F_{S^2})=\int_D F_D \cdot \int_{S^2} F_{S^2}
$$
By that logic, we would get
$$
\int_X (F_D\wedge F_{D})=\int_{D\times S^2} (F_D\wedge F_{D})=\int_D F_D \cdot\int_{S^2} F_{D}=\int_D F_D \cdot 0=0
$$

 

[thatboi]

That's a nice idea. We have
$$
\int_X (F_D\wedge F_{S^2})=\int_{D\times S^2} (F_D\wedge F_{S^2})=\int_D F_D \cdot \int_{S^2} F_{S^2}
$$
By that logic, we would get
$$
\int_X (F_D\wedge F_{D})=\int_{D\times S^2} (F_D\wedge F_{D})=\int_D F_D \cdot\int_{S^2} F_{D}=\int_D F_D \cdot 0=0
$$

Right, but I'm guessing $\int_{D \times S^{2}} F_{D} \wedge F_{S^{2}} = \int_{D} F_{D} \cdot \int_{S^{2}}F_{S^{2}}$ is not generally a true identity (thus your initial concern)?

 

[fresh_42]

If we consider $F$ as a two-form and $D$ and $S^2$ as vector fields instead of geometric objects, i.e. the geometric object endowed with the field strength $F$ then $0=F(D,D)=:F_D\wedge F_D$ but I'm not sure whether this is a correct point of view.

(If you have Google Chrome installed then you can have a look at
https://de.wikipedia.org/wiki/Differentialform
Chrome allows a translation into English via right-click and the option "Translate to English" which I found easier to read than https://en.wikipedia.org/wiki/Differential_form.)

 

[Orodruin]

I just had a very brief look, but it appears to me $F_D$ is a 2-form on $D$, which is 2-dimensional. Therefore $F_D\wedge F_D$ would be a 4-form, but there are no 4-forms on a two-dimensional manifold. As the tangent space is 2-dimensional, there are not enough independent one-forms to create 4-forms.

 

[Antarres]

What @Orodruin said is the correct explanation, but I will expand on that, in case you're still confused about this. It is true that on an $n$-dimensional manifold the wedge product of 2-form is commutative. However, consider the specific case where a 2-form is a product of 2 basis 1-forms(thus being a basis 2-form).
Then you have:
$$\alpha = b^1 \wedge b^2$$
and so
$$\alpha \wedge \alpha = b^1 \wedge b^2 \wedge b^1 \wedge b^2 = 0$$
just by anticommuting the basis 1-forms twice($b^1 \wedge b^1 = 0$). So in this specific case the product is zero. A general product of 2-forms is a linear combination of such products of basis forms, and such product indeed isn't zero. But a product of two same basis 2-forms constructed in this way is zero.

In your case you're looking at $F_D$ as a 2-form on a 4-dimensional manifold, however this 2-form is defined on a 2-dimensional submanifold $D$, and because this submanifold is 2-dimensional, it must be proportional to the basis form $F_D = f(x)\alpha^1 \wedge \alpha^2$, where $\alpha^i$ are basis forms of this submanifold. From what we mentioned above we have to have $F_D \wedge F_D = 0$. So what is mentioned in the post above is correct, and follows from the fact that you can't have forms of bigger dimension than 2 defined from basis forms of $D$. So despite looking at $F_D$ as a 2-form on a 4-dimensional manifold, you still know that this form must be proportional to a precise basis 2-form. Same conclusion goes for $F_{S^2}$. Hope that clears it up.