Pullback of a $1$-form

Question

All:

I looked at the list of similar questions, but none seemed to be done explicitly-enough to be helpful; sorry for the repeat, but maybe seeing more examples will be helpful to many.

So, I have a differentiable map $f: M \rightarrow S^1 $ , and I want to pullback $d\theta$ by f. Here is what I have:

Say we use the basis $\{ \partial/\partial x^i\}, \, i=1,2,3$ for the tangent space $T_xM$

i)We calculate $Jf=[\partial f/\partial x^1 \partial f/\partial x^2 \partial f/ \partial x^3]$

ii)We use i) to calculate the pushforward of the tangent vectors by f:

$ f_* (\partial/ \partial x^i) $=$ (\partial f/\partial x^i) (\partial/ \partial(\theta$))

iii)We evaluate $d\theta$ $(\partial f/ \partial x^i) (\partial/ \partial(\theta))$=$\partial f/ \partial x^i$

iv) We conclude : $f^* (d\theta)$=$ \partial f/ \partial x^1(d\theta)+\partial f/ \partial x^2(d\theta)+\partial f/\partial x^3(d\theta))$

Is this correct? Do I have to consider only chartwise representations, or is this a global representation for $f^*(d\theta)$ ?

EDIT: I appreciate both your explanations, but I've been confused with this for so long that I was hoping someone would answer this; I know pulling back a k-form is just the multilinear equivalent of calculating $T^*$ $V^*\rightarrow W^*$ given a linear map $T:V\rightarrow W$ between finite-dimensional vector spaces, so :

I'm trying to follow the formula:

$f^*= \Sigma_I (wof)d(y^{i_1}of)\wedge d(y^{1_2}of).....\wedge d(y^{i_n}of)$ (##)

But maybe I need to express $d(\theta)$ in a different basis?

The result I got using (##) is $f^*(d\theta)=(1of)(d(\theta of)=d(\theta)of+(\theta)odf$

Which does not seem to agree with neither answer

I would appreciate more than one explanation, but, out of fairness, I will admit now that I will accept the first answer I get, unless (both) the second one comes closely after the first in time and has something substantially better. Unfortunately, at my point level, I'm not allowed to give points for a good answer. Thanks.

Thanks for your help.

Answer 1

EDIT: Down to Earth approach.

Let $f:M\rightarrow S^1$ be the $C^{\infty}$ map of $C^{\infty}$-manifolds s.t.

$$f(x_1,\dots,x_n):=\theta, (*)$$

for all $(x_1,\dots,x_n)$ local coordinates of any point $p$ on $M$. Each point on $S^1$ is identified with an angle $\theta\in [0,2\pi)$.

In other words, the angle $\theta$ is a function of the coordinates on $M$, i.e. $\theta=\theta(x_1,\dots,x_n)$, defined through $(*)$.

The pullback $f^{*}(d\theta)$ is given by the one form on $M$

$$f^{*}(d\theta)=\sum_{i=1}^n \frac{\partial f}{\partial x_i}dx_i=\sum_{i=1}^n \frac{\partial\theta}{\partial x_i}dx_i.$$

Another (equivalent) way to write the pullback is then $$d\theta=\sum_{i=1}^n \frac{\partial\theta}{\partial x_i}dx_i.$$

Answer 2

I think it may be instructive to pull-back the form $\alpha = g(\theta)d\theta$.

Notice $f^*(\alpha)$ is a one-form on $M$ hence it is determined by its action on the basis of the tangent space $\partial_1,\partial_2, \dots , \partial_m$ assuming $dim(M)=m$ and $(x,U)$ is a chart with coordinate derivations $\partial/\partial x^j = \partial_j$ defined in the usual manner. Let us consider a particular point $x_o \in M$ for which $f(x_o) = \theta_o$. Notice, to calculate the pull-back, we really just need to calculate each component of the pull-back: $$ (f^*(\alpha))_{x_o}(\partial_j|_{x_o}) = (\alpha)_{\ \theta_o}( df_{x_o}(\partial_j|_{x_o})) = g(\theta_o)d\theta \biggl( \frac{\partial (\theta \circ f)}{\partial x^j}(x_o)\frac{\partial}{\partial \theta} |_{\theta_o} \biggr) = g(f(x_o))\frac{\partial (\theta \circ f)}{\partial x^j}(x_o). $$ Note: the definition I used above is very geometrically motivated. To calculate the pull-back I take a form in the codomain and feed it vectors on the tangent space to the codomain. How to do this? The only way natural, by pushing forward the vectors on the tangent space to the domain via $df$. This is equivalent to your formula and I do recover your formula as you read on... (you just have one small oversight at the end) continuing,

Therefore, as the above is the coefficient of $f^*(\alpha)$ with respect to $dx^j$ at $x_o$ we find: $$ f^*(\alpha)_{x_o} = \sum_{j=1}^{m} g(f(x_o))\frac{\partial (\theta \circ f)}{\partial x^j}(x_o) (dx^j)_{x_o}.$$ Hence, omitting the annoying point-dependence notation, $$ f^*\alpha = \sum_{j=1}^{m} (g \circ f )\frac{\partial (\theta \circ f)}{\partial x^j} dx^j. $$ As you can clearly see, the nature of the pull-back is very much dependent on the formula for $f$.

However, in the case you consider $d\theta$ the composition with $g=1$ is not seen which gives the rather curious result that $f^*(d\theta) = d(\theta \circ f)$.

This demonstrates the interesting property of the pull-back, as $\theta \circ f = f^* \theta$ we have: $f^*(d\theta) = d(f^* \theta)$; the pull-back commutes with exterior differentiation.

---

The function $\theta \circ f: M \rightarrow S_1 \rightarrow \mathbb{R}$ hence $d( \theta \circ f) = d\theta \circ df$ where $d\theta$ maps tangent vectors to $S_1$ to tangent vectors to $\mathbb{R}$ and $df$ maps tangent vectors to $M$ to tangent vectors on the circle. Again, the journey goes, from $T_PM$ to $T_{f(p)}S_1$ via $df_p$ then from $T_{f(p)}S_1$ to $T_{\theta (f(p))}\mathbb{R}$ via $d\theta_{f(p)}$. This is the chain-rule for manifold maps. It can be proven by lifting the chain-rule from $\mathbb{R}^n$ up through the coordinate charts.

Finally, there is a point of notational danger whenever we run into a mapping into $\mathbb{R}$, it is often the case that $T_{\theta (f(p))}\mathbb{R} = \mathbb{R}$ which means that in some equations an identification between $d/dt|_{t_o}$ and $1$ is made. For example, to use the beautiful formula $dg(X)=X(g)$ for $g: M \rightarrow \mathbb{R}$ and $X$ a vector field on $M$ we make this assumption. Note $X(g)$ is a function whereas $dg(X)$ would naturally construct a vector field, an yet we have equality. I used this in calculation of $d(\theta \circ f)$ consider that $$ d(\theta \circ f)(\partial_{j}) = \partial_j(\theta \circ f) \qquad \rightarrow \qquad d(\theta \circ f) = \sum_{j=1}^{m}\partial_j(\theta \circ f) dx^j$$ Again, I'm using the fact that a one-form $\alpha$ expressed in a coordinate basis $dx^j$ as $\alpha = \sum_{j=1}^{m}\alpha_j dx^j$ has $\alpha_j = \alpha(\partial_j)$