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All:

Please forgive me, I'm new and my editing/Latex needs improvement.

I'm trying to derive the formula for change of variables for the differential form $\omega=dx\wedge dy$ in standard $xy$-coordinates in $\mathbb R^2 $, into polar coordinates in $\mathbb R^2$. I know we can use the quick-and-dirty change of variables: $$ x=r\cos t, \quad y=r\sin t. $$ Then sub-in, expand, and cancel terms with repeated $dr$'s and/or $dt$'s.

But I'm trying to use the layout in J. Lee's Smooth Manifolds, pp 303-304, given by:

Definition. Given a smooth map $F\colon M \to N$ and a form $\omega$ defined on $N$, the pullback $F^*$ is given by: \begin{equation} (F^*\omega)_p(X_1, \ldots, X_n) := \omega_{F(p)} (F_*X_1,....,F_*X_n). \qquad (**) \end{equation}

Results.

a) $F^*$ is linear on the space of smooth sections

b) $F^*(\omega\wedge \eta)=F_*(\omega)\wedge F_*(\eta)$

c) In any smooth chart, and for every multi-index index $I=(i_1,i_2,\ldots,i_k)$: $$ F^*(\sum' \omega_I dy^{i_1}\wedge dy^{i_2} \wedge \ldots \wedge dy^{i_k})=\sum'(\omega_I \circ F)d(y^{i_1} \circ F)\wedge\cdots\wedge d(y^{i_k} \circ F),$$ where $\sum'$ is a sum over increasing indices.

So far, I have:

$\omega=dx\wedge dy$; $F\colon\mathbb R^2\to\mathbb {R^2}'$, and $\omega$ is defined on the target ${\mathbb R^2}'$. We pull back by the map $F(x,y)=(r\cos t,r\sin t)$. Then, by b, $F^*\omega =F^*(dx\wedge dy) =F^*(dx)\wedge F^*(dy)$.

(Moreover, Lee has not yet defined the meaning of $dx$, nor of $dy^{i_k}$.)

Now, I can only think of using $(**)$, but this does not help: $F^*(dx):=dx(\cos t, \sin t) (F_*X_1,\ldots,F_*X_n)$ where $F_*$ is given by $F_*(X)(f)=X(f\circ F)$. Any ideas?

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@Dylan Good Job, It would have taken like years for me to fix! –  user21436 Feb 2 '12 at 0:17
    
Thanks for your help editing. –  Confused Feb 2 '12 at 0:18
    
Actually, thanks for the editing, since you did at least half. –  Confused Feb 2 '12 at 0:22
    
I've made it a bit more readable... @Confused, try to use punctuation next time: it works miracles! –  Rick Feb 2 '12 at 0:33
    
@Confused For the purpose of applying the result (c), do you agree that $$ d(r\cos t) = \cos t\,dr - r\sin t\, dt\,?$$ –  Dylan Moreland Feb 2 '12 at 0:44

1 Answer 1

up vote 9 down vote accepted

I don't know how Lee explains this, but as soon as we pick local coordinates $x^i$ we have 3 related (but distinct) things:

  1. Coordinate functions $x^i$.

  2. Coordinate vector fields $\partial_i = \partial/\partial x^i$.

  3. Coordinate covector felds $dx^i$.

These satisfy a bunch of relations, e.g. $\partial_j x^i = \delta_j^i$ (Kronecker delta) and $dx^i(\partial_j) = \delta_j^i$.

If we have a map $x = \phi(y)$ given in coordinates by $x^i = \phi^i(y^1, \cdots, y^m)$, there is a notion of pushforward of a tangent vector. If we have some tangent vector (in the $y$ coordinates which is given by $$ v = \sum_i v^i \frac{\partial}{\partial y^i} $$ then we obtain a vector $\phi_\ast v$ in the $x$ coordinates by $$ \phi_\ast v = \sum_{ij} v^i \frac{\partial \phi^j}{\partial y^i} \frac{\partial}{\partial x^j}. $$ If you think of tangent vectors as being derivations of smooth functions, this is completely obvious: it's just the chain rule.

Ok, so if you understand the chain rule, then you understand pushforwards, and if you understand pushforwards you understand pullbacks, since they are just dual to pushforwards. Remember that a $p$-form is just a (skew) multilinear function. So to define a $p$-form in the $y$ coordinates you just have to say how to act on tangent vectors. But we already know how to push tangent vectors forward to the $x$ coordinates. So if $\omega$ is a $p$-form on in the $x$-coordinates, we obtain a $p$-form in the $y$-coordinates by the formula $$ (\phi^\ast \omega)_y(v_1, \cdots, v_p) = \omega_{\phi(y)}(\phi_\ast v_1, \cdots, \phi_\ast v_p) $$ as you wrote above. To emphasize: this is all just the chain rule and linear algebra.

Now let's look at your particular example. We start with coordinates $x,y$ and want to pass to coordinates $r,t$, via $$ x = r \cos t $$ $$ y = r \sin t $$ Call this map $(r,t) \mapsto (x,y)$ $\phi$. So, how do we push vectors forward? By the chain rule, we have \begin{align} \phi_\ast(\frac{\partial}{\partial r}) &= \frac{\partial x}{\partial r} \frac{\partial}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial}{\partial y} \\ &= \cos t \frac{\partial}{\partial x} + \sin t \frac{\partial}{\partial y} \end{align} Similarly, \begin{align} \phi_\ast(\frac{\partial}{\partial t}) &= \frac{\partial x}{\partial t} \frac{\partial}{\partial x} + \frac{\partial y}{\partial t} \frac{\partial}{\partial y} \\ &= -r\sin t \frac{\partial}{\partial x} + r \cos t \frac{\partial}{\partial y} \end{align}

Now, to get the pullback of $dx \wedge dy$, we just need to evaluate it on these pushforwards (now edited to include more detail): \begin{align} (dx \wedge dy)(\phi_\ast\partial_r, \phi_\ast\partial_t) &= (dx \wedge dy)(\cos t \partial_x + \sin t \partial_y, -r\sin t \partial_x + r\cos t \partial_y) \\ &= r\cos^2 t (dx\wedge dy)(\partial_x, \partial_y) -r\sin^2 t (dx\wedge dy)(\partial_y, \partial_x) \\ &= r\cos^2 t + r \sin^2 t \\ &= r \end{align} Here, in going from the first line to the second I used the bilinearity and skew-symmetry properties of forms (e.g. $(dx\wedge dx)(\partial_x, \partial_x) = 0)$) to expand it. To go from the second to the third I used the fact that $(dx\wedge dy)(\partial_x, \partial_y) = 1$, which is true because $dx$ and $dy$ are dual to $\partial_x$ and $\partial_y$.

That is, $$ \phi^\ast (dx \wedge dy)(\partial_r, \partial_t) = r $$ So we have $$ \phi^\ast (dx \wedge dy) = r dr \wedge dt. $$

Note that this is the same result you'd obtain if you just differentiated $x = r\cos t$ and $y = r\sin t$ and then wedged the results together. The reason this works is because of two facts:

  1. Pullbacks commute with the exterior derivative $d$.

  2. The exterior derivative of a coordinate function $x^i$ is exactly the coorinate covector field $dx^i$. (Hence the notation!)

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Thanks, Jonathan, that was very helpful. Just a small (similarly-simple-minded)followup: I don't follow the evaluation of $dx/\dy$ on the pushforward; my understanding is that the wedge product of maps is a scalr multiple of the tensor product of maps, so that we would evaluate dx at costDelx+sintDely and multiply by the evaluation of dy on the second component of the pushforward. Additionally, dx is usually the projection map. Is this what you did? –  Confused Feb 2 '12 at 3:34
1  
@Confused The "wedge" in $dx \wedge dy$ is \wedge in LaTeX, by the way. –  Dylan Moreland Feb 2 '12 at 15:25
    
@Confused I edited it to include a few more details of the evaluation of the pushforward. –  Jonathan Feb 4 '12 at 16:23

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