# Define pull-back on a manifold by pull-back on the linear space.

It appears to me that pull-back on a manifold

If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

is defined by pull-back on the linear space:

Suppose $A: V \to W$ is a linear map. Then the transpose map $A^*: W^* \to V^*$ extends to the exterior algebras, $A^*: \Lambda^p(W^*) \to \Lambda^p(V^*)$ for all $p>0$. If $T \in \Lambda^p(W^*)$, just define $A^* T \in \Lambda^p(V^*)$ by $$A^*T(v_1, \dots, v_p) = T(Av_1, \dots, Av_p)$$ for all vectors $v_1, \dots, v_p \in V$.

So my observation is to put in $df^*$ as $A^*$, $\omega$ as $T$, and $f(x) = (f^1(x), \dots, f^p(x))$ as $v_1, \dots, v_p$: $$f^*\omega(x) = (df_x)^*\omega[f(x)] = \omega(df_x f^1(x), \dots, df_x f^p(x)) = \omega \circ df_x(f^1(x), \dots, f^p(x)).$$

I know it should result in $\omega \circ df_x(x_1, \dots, x_q)$, where $p$ is the dimension of $Y$, and $q$ is the dimension of $X$. But where I messed it up?

Why $\omega[f(x)]$ is $\omega \circ df_x(x_1, \dots, x_q)$, instead of $\omega \circ df_x(f^1(x), \dots, f^p(x))$?

Here's an example showing my confusion, from James S. Cook's answer to the question Intuition about pullbacks in differential geometry

So in his answer, he states:

$$\Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)=\omega (\Psi_*(\partial_1), \Psi_*(\partial_2),\Psi_*(\partial_2),\Psi_*(\partial_4)).$$

I see it reasonable. However, in my textbook, pull-back is defined to be $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

Therefore, I understand it as: $$\Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)=\omega (\Psi_*(f^1(\partial_1,\partial_2,\partial_3,\partial_4), \cdots, \Psi_*(f^4(\partial_1,\partial_2,\partial_3,\partial_4)).$$

• the last equation is incorrect, the $f$ is used to select the differential $4$ form at $f(x)$. What you write would require the number of components for the function to match the degree of the form pulled back. But we might pull-back a 4-form on a 10-dimensional space so it'll be impossible to write that formula.(your last one) – James S. Cook Aug 2 '13 at 7:09

My answer there simply omits the point-dependence of the form. Let me put in the hideous point-dependence. Let us suppose $\Psi: M \rightarrow N$ and $\omega$ is a $4$-form at $\Psi(x) \in N$. The pull-back of $\omega$ will be a $4$-form at $x \in M$. I wrote: $$\Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)=\omega (\Psi_*(\partial_1), \Psi_*(\partial_2),\Psi_*(\partial_2),\Psi_*(\partial_4)).$$ However, explicitly: $$(\Psi^*(\omega))_x(\partial_1|x,\partial_2|x,\partial_3|x,\partial_4|x)=\omega[f(x)] (d\Psi_x(\partial_1|x), d\Psi_x(\partial_2|x),d\Psi_x(\partial_3|x),d\Psi_x(\partial_4|x)).$$ Again, the notation $\omega[f(x)]$ indicated we a selecting the $4$-form $\omega[f(x)]$ at the point $f(x)$ of the 4-form-field $\omega$. Unfortunately, we use the term differential form for both the field and the object fixed at a point. This may be part of the confusion. Hope this helps.
The secondary question of what $\partial_j$ denotes requires some discussion. I'll define it by a formula which borrows the differentiation process from the parameter space, however there are several other ways. To be clear, we consider a smooth manifold $M$ with coordinate chart $(x,U)$ and we'll focus on a point $p$. This means $x: U \rightarrow V \subseteq \mathbb{R}^m$ and $x^{-1}: V \rightarrow U$. Let $f \in C^{\infty} (p)$ which means that $f \circ \gamma$ is a smooth curve in $\mathbb{R}^m$ for all curves $\gamma$ through $p \in M$ (for our purposes here this means the derivative I'm about to write exists). We define $$\frac{\partial}{\partial x^j}{\bigg|}_p (f) = \biggl[ \frac{\partial}{\partial u^j}(f \circ x^{-1}) (u^1, \dots u^m) \biggr]\bigg|_{u=x(p)}$$ where $u^1,\dots,u^m$ are cartesian coordinates of $V$. In words, you take the function $f$ near $p$ and pull it down to a function $f \circ x^{-1}$ on $\mathbb{R}^m$ near $x(p) \in \mathbb{R}^m$. Then do plain-old partial differentiation with respect to $u^j$ and once that is done, plug $u = x(p)$. Of course, this can be phrased in terms of directional derivatives as is done: Partial Derivatives on Manifolds - Is this conclusion right? .
The neat thing about the manifold partial is that the notation hides this subtlety, but is honest for the uniformed. For example, $\frac{\partial x^j}{\partial x^i} = \delta_{ij}$ and the chain-rule (supposing $y$ is another coordinate system at the point considered) is $\frac{\partial f}{\partial x^i} = \sum_{j=1}^m \frac{\partial y^j}{\partial x^i}\frac{\partial f}{\partial y^j}$. These manifold partial derivatives are more commonly called the coordinate derivations because they satisfy the Leibniz rule: $$\partial_j|p(fg) = \partial_j|p(f)g(p)+f(p)\partial_j|p(g)$$ It is also important to know these form a natural basis for $T_pM$. There is more to say, but I think this suffices for our purposes here.
• Thanks James - what are $\partial_i$s? Or, what are $p$-vectors? – WishingFish Aug 2 '13 at 7:59
• @WishingFish here $\partial_1 = \frac{\partial}{\partial x^1}|_x$ the $x^1$ coordinate derivation (at the point $x$). The same for 2,3, and 4. GP defines these on page ... well, I can't find it. I wonder, maybe you should get a copy of Tu's An Introduction to Manifolds to parallel your study of GP? I'll add the definition of the coordinate partial derivative to my answer shortly, it's too long for here to be pretty. – James S. Cook Aug 2 '13 at 14:55