Intuition about pullbacks in differential geometry

I am struggling to understand the role of pullbacks after noticing that they are used when defining an integral of $k$-forms on a manifold.

Let $F:M \to N$ be a map between differentiable manifolds. The pullback of a covector $\omega \in (T_{F(p)}N)^*$ under $F$ at a point $p \in M$ is $F^*:(T_{F(p)}N)^* \to (T_pM)^*$ with $$(F^*)_p(\omega)X = \omega((F_*)_pX)$$ where $X \in T_p(M)$.

I know that the pushforward map $F_*$ takes a tangent vector at a point in $M$ to a tangent vector at a point in $N$. Am i missing something else?

The pullback takes a covector in $N$ to a covector in $M$, but I can't see why it is important and don't have any intuition in it. I don't get why the pullback is used in integration either. I'd be grateful for any help.

• Are you comfortable with the linear algebraic fact that a linear map $V \to W$ between vector spaces gives rise to a canonical linear map $W^* \to V^*$ between the duals? – Eric O. Korman Sep 11 '12 at 15:22
• @Eric Yeah I have come across that. But it doesn't give me a good picture of what the dual operation does. – soup Sep 11 '12 at 17:30

To really connect the claims I make below with the definitions given in your post takes some effort, but since you asked for intuition here it goes...

I think the best intuition for push-forwards and pull-backs is offered from the case $\mathcal{M}=\mathcal{N}$ and both are embedded in $\mathbb{R}^n$.

• the push-forward amounts to changing coordinates for a vector field (viewed as a derivation) for example: for cartesians verses polars in $\mathbb{R}^2$ $$\frac{\partial}{\partial r} = \frac{\partial x}{\partial r}\frac{\partial}{\partial x}+ \frac{\partial y}{\partial r}\frac{\partial}{\partial y} = \cos(\theta)\frac{\partial}{\partial x}+\sin(\theta)\frac{\partial}{\partial y}$$

• the pull-back amounts to taking the total differential of the coordinate formulas, or equivalently, applying the chain-rule to swap differentials in one coordinate system for another. For example, again for two-dimensional Cartesian verse polars: $$dr = d(\sqrt{x^2+y^2}) =\frac{x}{\sqrt{x^2+y^2} }dx+\frac{y}{\sqrt{x^2+y^2} }dy$$

The role of the pullback to integration is that it allows us to lift integration defined in $\mathbb{R}^n$ up to the manifold (provided we have the partition of unity to weave things together). Essentially the abstract $dx$ on the manifold pulls down to an ordinary differential on $\mathbb{R}^n$. However, it's not quite that simple because we have $dx \wedge dy = -dy \wedge dx$ on the manifold. In an integral on $\mathbb{R}$ we have $dxdy=dydx$ provided we swap the integration bounds appropriately. This apparent contradiction is reconciled by the fact that integrals of two-forms correspond to surface integrals. In a surface integral the order of the parameters determines the orientation of the surface. All things being otherwise the same $f(x,y)dx \wedge dy$ and $f(x,y)dy \wedge dx$ pull back to the same surface integral modulo a sign. Of course this sign is important, it measures outward or inward flux for the example I'm currently discussing.

All of this said, it bugs me to no end when texts choose to omit the $\wedge$ for explicit form calculations!

Per request: an explicit example Suppose we take $\phi (A) = (A_{11},A_{12},A_{21},A_{22})$ as the coordinate chart on $\mathcal{M} = \mathbb{R}^{ 2 \times 2}$. Denote the coordinate functions by $x_{ij}: \mathcal{M} \rightarrow \mathbb{R}^4$. To integrate in this 4-dimensional manifold we need to find some 4-form with which to play. Define, $$\omega = (x_{11}^2+2)dx_{11} \wedge dx_{12} \wedge dx_{21} \wedge dx_{22}$$ To help separate concepts let me use $(u_1,u_2,u_3,u_4)$ as the standard cartesian coordinates on $\mathbb{R}^4$. Let's calculate the pull-back of $\omega$ under $\phi^{-1}=\Psi$. Since $\Psi^*(\omega)$ will be a four-form at $p=(p_1,p_2,p_3,p_4)$ on $\mathbb{R}^4$ it suffices to consider (suppressing the $p$ for now) $$\Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)=\omega (\Psi_*(\partial_1), \Psi_*(\partial_2),\Psi_*(\partial_2),\Psi_*(\partial_4))$$ We need to calculate the push-forwards to continue, $$\Psi_*(\partial_1) = \partial_{11}, \ \ \Psi_*(\partial_2) = \partial_{12}, \ \ \Psi_*(\partial_3) = \partial_{21}, \ \ \Psi_*(\partial_4) = \partial_{22}$$ Thus, \begin{align} \Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4) &=\omega (\partial_{11},\partial_{12},\partial_{21},\partial_{22}) \\ &= (x_{11}^2+2)(dx_{11} \wedge dx_{12} \wedge dx_{21} \wedge dx_{22}) (\partial_{11},\partial_{12},\partial_{21},\partial_{22}) \\ &= x_{11}^2+2 \end{align} Following the $p$ through the calculation will reveal that we find: $$\Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)_p = p_1^2+2$$ Or, if you like, set $p=u$ and find $u_1^2+2$. To perform an integration we best take some compact subset of the matrix space then you can see that it will simply reduce to integrating the function $u_1^2+2$ on the corresponding parameter space in $\mathbb{R}^4$

About half way through this I realized you only wanted a covector example. For that, we simply need an abstract one-dimensional manifold and a one-form with which to play. If you wish I'll add that later.

• I should mention, my post is mostly about fields, you should fix a point to connect better with the definitions you listed. – James S. Cook Sep 11 '12 at 16:07
• Thanks for the post. So in defining the integral over a manifold, we use the $\int_M \omega = \int_{\varphi(U)} (\varphi^{-1})^*\omega$ which defines the integral where $\omega$ is a $n$-form compactly supported in a single chart $(U, \varphi)$. So is it correct that $(\varphi^{-1})^*\omega$ pulls back $\omega$ to be a function on $\mathbb{R}^n$? Do you know of a concrete example where I can see this more clearly? – soup Sep 11 '12 at 17:27
• Thank you!!! I very much appreciate it. – soup Sep 12 '12 at 10:40
• Hi James, you posts are extremely helpful for me to learn the forms! Now I found where exactly confused me. So I understand $$\Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)=\omega (\Psi_*(\partial_1), \Psi_*(\partial_2),\Psi_*(\partial_2),\Psi_*(\partial_4)).$$ However, in the textbook, $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$ Therefore, I understand it as: $$\Psi^*(\omega)(\partial_1,\partial_2,\partial_3,\partial_4)=\omega (\Psi_*(f^1(\partial_1,\partial_2,\partial_3,\partial_4), \cdots, \Psi_*(f^4(\partial_1,\partial_2,\partial_3,\partial_4)).$$ Can you find where went wrong? – WishingFish Aug 2 '13 at 4:37
• I put the same question in the comment above here: math.stackexchange.com/questions/457649/… – WishingFish Aug 2 '13 at 4:49

I think Jame's answer is really good, I just want to add one thing, regarding intuition, that may be obvious but it took myself some time to realize.

Pullbacks is really the natural way to integrate given a certain parametrization. When you are in $\mathbb{R^n}$ it is just a fancy mechanism to compute the jacobian a.k.a. to make a correct change of variables a.k.a. to reparametrize the (hyper)surface.

For example, let's compute $\int_{R}\omega$ where $R$ is the region bounded by the circle $x^2+y^2=4$ and the line $y=x$ and $\omega=(1+\frac{y^2}{x^2})dx\wedge dy$ .

So first we want to find a suitable parametrization. This turns out to be $\phi(r,\theta)=(rcos\theta,rsin\theta)$ with $0\leq r \leq2$ and $0\leq\theta\leq\frac{\pi}{4}$. So we have :

$\int_{R}\omega=\int_{[0,1]\times[0,\frac{\pi}{4}]}φ^*(ω)=\int_0^1 \int_0^\frac{\pi}{4}\frac{1}{cos^2θ}det[J_φ]dθ\wedge dr=\int_0^1 \int_0^\frac{\pi}{4}(-tanθ)'rdθ \wedge dr$ and it is easy from now on.

The important thing to note is that through all the abstract formalism needed to define forms in abstract manifolds, what you get is a way to compute the Jacobian. Try writting all the definitions in a map between surfaces in $\mathbb{R^3}$ to gain a better understanding.

The book thath helped me clear the confusion about differential forms is A geometric approach to differential forms by Bachman and it has many excerscises and examples. When you feel comfortable you can move to Tu's Introduction to Manifolds.

• Indeed, Bachman's text has been helpful to me as well. I hope more books like that are written. – James S. Cook Oct 31 '18 at 3:43