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My formula of the pullback is given by

If $f:M\longrightarrow N$, then \begin{align*} f^*:\Omega ^k(N)&\longrightarrow \Omega ^k(M)\\ \omega &\longmapsto f^*\omega \end{align*}

where $$(f^*\omega )_p(X_1,...,X_k)=\omega _{f(p)}(\mathrm d f(X_1),...,\mathrm d f(X_k)).$$

I recall that $$\Omega ^k(U)=\{\theta\in Tens_k(U)\mid \forall \sigma \in \mathfrak S_k, {}^\sigma \theta=sgn(\sigma)\theta\}=\{k-differential\ form\ on\ U\}$$ where $sgn:\mathfrak S_k\longrightarrow \{-1,1\}$ is the signature.

I don't understand how to use it... For example, the transformation to polar coordinate to cartesian coordinate, i.e. $$f(r,\theta)=(x,y)=(r\cos\theta, r\sin \theta),$$ how can I compute $f^*\mathrm d x$ using this formula ? I would have $$(f^*\mathrm d x)_{(r,\theta)}\left(\frac{\partial }{\partial r},\frac{\partial }{\partial \theta}\right)=\mathrm d x_{f(r,\theta)}\left(\mathrm d f\left(\frac{\partial }{\partial r}\right), \mathrm d f\left(\frac{\partial }{\partial \theta}\right)\right) $$

By definition, $$\mathrm d f\left(X\right)=X(f)$$ where $X$ is a derivation, and thus I get $$\mathrm d x_{f(r,\theta)}\left(\mathrm d f\left(\frac{\partial }{\partial r}\right), \mathrm d f\left(\frac{\partial }{\partial \theta}\right)\right) =\mathrm d x_{(x,y)}\left(\frac{\partial f}{\partial r},\frac{\partial f}{\partial \theta}\right)=\mathrm d x_{(x,y)}\Big((\cos\theta,\sin \theta),(-r\sin\theta, r\cos\theta)\Big)=???$$

What's next ? (For your information, I know that $$f^*\mathrm d x=\cos\theta\mathrm d r-r\sin \theta\mathrm d \theta$$ but I would like to arrive at this result by using the pullback.

user301068
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    The problem is that $dx$ is a one form, and you're using it as a two-form, unless I'm misunderstanding something. Evaluate $f^*dx$ on $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta}$ separately to determine it on a basis. – Francis Begbie Jan 11 '16 at 19:52
  • @FrancisBegbie: interesting comment... I didn't think about that. Thank you. – user301068 Jan 11 '16 at 19:54
  • For future reference, there is an easy basis formula: if $\omega$ is a one form, then $\omega = \sum_j\omega\left(\frac{\partial}{\partial x_j}\right)dx_j.$ – Francis Begbie Jan 11 '16 at 20:07

1 Answers1

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$df_{r,\theta} =\pmatrix{cos\theta & -rsin\theta\cr sin\theta & rcos\theta}$.

Let $(u,v)$ be vector tangent to $(r,\theta)$, $df_{r,\theta}(u,v)=\pmatrix{cos\theta & -rsin\theta\cr sin\theta & rcos\theta}\pmatrix{u\cr v} = (cos\theta u-rsin\theta v,sin\theta u+ rcos\theta v)$.

Thus $f{^*dx}_{r,\theta}(u,v)=dx(df_{r,\theta}(u,v))=dx(cos\theta u-rsin\theta v,sin\theta u+ rcos\theta v) =cos\theta u-rsin\theta v=cos\theta dr(u,v)-rsin\theta d\theta(u,v)$

Tsemo Aristide
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