# Weird result on forms?

1. Sep 3, 2011

### dragonlorder

I know this may sounds silly but I am confused
consider this two form for example, by substitution, I get
$$\omega = dx \wedge dy = d(rCos\theta)\wedge d(rSin\theta) = r dr \wedge d\theta$$

also consider this smooth map $$F(x,y)=(rCos\theta,rSin\theta)$$

then $$F^{*}\omega = rdr \wedge d\theta$$

which means that $$F^{*}\omega= \omega$$!?, that's just weird.

I am reading John's Lee smooth manifold book. and I saw the substitution writing at the differential form chapter. and the pullback writing at the Covector field chapter.

2. Sep 4, 2011

### quasar987

When you compute dxdy in polar coordinates, what you are doing is you're computing the pullback of dxdy by F-1. So of course when you then apply F* to the form in polar coordinates, you fall back on dxdy since you're computing

F* o (F-1)* dxdy = (F-1 o F)* dxdy = Id* dxdy = dxdy

Review the definitions if you need to.

3. Sep 4, 2011

### ForMyThunder

In the case above, $$F:\mathbb R^2\to\mathbb R^2$$ would just be the identity function, right? Since it is just a coordinate transformation. The pullback of the identity function is the identity on forms, so the relation $$F^*\omega$$ is typical for coordinate transformations.

If $$F:(r,\theta)\mapsto (r\cos\theta,r\sin\theta),$$ which I'm suspecting he meant above, computing the form $$dx\wedge dy$$ in polar coordinates is the pullback by F itself, since the pullback is a contravariant functor. This is equal to the form itself since F is the identity but it is just expressed in a different coordinate system.

Last edited: Sep 4, 2011
4. Sep 4, 2011

### quasar987

Note that F(r,O)=(rcosO,rsinO) is actually defined (at most) on (0,+\infty) x (0,2pi) in order for F to be a diffeomorphism. But clearly F is not the identity! The identity is Id(r,O)=(r,O).

5. Sep 4, 2011

### dragonlorder

Thank you for the post
I understand that $$F^*$$ pull the form on x,y R^2 back to polar
so the form would be expressed in polar R^2 as a pullback map acting on it
$$F^* \omega$$
while this seems pretty good, I noticed Lee wrote in his later chapter, $$\omega = dx \wedge dy = r dr \wedge d \theta$$ kind of throw me off. Where is the pullback operator

or is it that he changes the coordinate chart instead of change the manifold the form is on?

Last edited: Sep 4, 2011
6. Sep 5, 2011

### quasar987

That's it... if (x^i), (y^i) are coordinate charts around a point p in an abstract manifold M, a form w can be written locally around p wrt each of the coordinate systems.

This is what Lee is doing here. The manifold is R², the form is the form w defined wrt to the global chart (x,y) by w=dxdy, and now he's saying that around almost any point, this form can be written locally wrt to polar coordinates as rdrdO.

7. Sep 6, 2011

### dragonlorder

Hmm yea, Thanks for the reply!

8. Sep 7, 2011

### ForMyThunder

Okay, now I'm confused. F sends a point to itself but does so by changing coordinates. If p:R^2->R^2 is the identity on M which gives global polar coordinates and c:R^2->R^2 is the identity on N which gives global Cartesian coordinates, then cp^(-1):R^2->R^2 is F itself since it sends (r,\theta) to (r cos\theta,r sin\theta). But this is a composition of identities so F is the identity.

(r,\theta) and (r cos\theta,r sin\theta)=(x,y) are the same point but just in different coordinate systems...

9. Sep 7, 2011

### ForMyThunder

...I see now. Nevermind.