# The Unapologetic Mathematician

## Pulling Back Forms

We’ve just seen that smooth real-valued functions are differential forms with grade zero. We also know that functions pull back along smooth maps; if $g\in\mathcal{O}_NV$ is a smooth function on an open subset $V\subseteq N$ and if $f:M\to N$ is a smooth map, then $g\circ f:f^{-1}(V)\to\mathbb{R}$ is a smooth function — $g\circ f\in\mathcal{O}_{f^{-1}(V)}M$.

It turns out that all $k$-forms pull back in a similar way. But the “value” of a $k$-form doesn’t only depend on a point, but on $k$ vectors at that point. Functions pull back because smooth maps push points forward. It turns out that vectors push forward as well, by the derivative. And so we can define the pullback of a $k$-form $\alpha$:

$\displaystyle \left[\left[f^*\alpha\right](p)\right](v_1,\dots,v_k)=\left[\alpha(f(p))\right](f_{*p}(v_1),\dots,f_{*p}(v_k))$

Here $\alpha$ is a $k$-form on a region $V\subseteq N$, $p$ is a point in $f^{-1}(V)\subseteq M$, and the $v_i$ are $k$ vectors in $\mathcal{T}_pM$. Since the differential $f_{*p}:\mathcal{T}_pM\to\mathcal{T}_{f(p)}N$ is a linear function and $\alpha(f(p))$ is a multilinear function on $\mathcal{T}_{f(p)}N^{\otimes k}$, $\left[f^*\alpha\right](p)$ is a multilinear function on $\mathcal{T}_pM^{\otimes k}$, as asserted.

This pullback $f^*:\Omega_N(V)\to\Omega_M(f^{-1}(V))$ is a homomorphism of graded algebras. Since it sends $k$-forms to $k$-forms, it has degree zero. To show that it’s a homomorphism, we must verify that it preserves addition, scalar multiplication by functions, and exterior multiplication. If $\alpha$ and $\beta$ are $k$-forms in $\Omega_N(V)$, we can check

\displaystyle\begin{aligned}\left[\left[f^*(\alpha+\beta)\right](p)\right](v_1,\dots,v_k)&=\left[[\alpha+\beta](f(p))\right](f_{*p}(v_1),\dots,f_{*p}(v_k))\\&=\left[\alpha(f(p))+\beta(f(p))\right](f_{*p}(v_1),\dots,f_{*p}(v_k))\\&=\left[\alpha(f(p))\right](f_{*p}(v_1),\dots,f_{*p}(v_k))+\left[\beta(f(p))\right](f_{*p}(v_1),\dots,f_{*p}(v_k))\\&=\left[\left[f^*\alpha\right](p)\right](v_1,\dots,v_k)+\left[\left[f^*\beta\right](p)\right](v_1,\dots,v_k)\end{aligned}

so $f^*(\alpha+\beta)=f^*\alpha+f^*\beta$. Also if $g\in\mathcal{O}(V)$ we can check

\displaystyle\begin{aligned}\left[\left[f^*(g\alpha)\right](p)\right](v_1,\dots,v_k)&=\left[[g\alpha](f(p))\right](f_{*p}(v_1),\dots,f_{*p}(v_k))\\&=\left[g(f(p))\alpha(f(p))\right](f_{*p}(v_1),\dots,f_{*p}(v_k))\\&=g(f(p))\left[\alpha(f(p))\right](f_{*p}(v_1),\dots,f_{*p}(v_k))\\&=\left[f^*g\right](p)\left[\left[f^*\alpha\right](p)\right](v_1,\dots,v_k)\end{aligned}

As for exterior multiplication, we will use the fact that we can write any $k$-form $\alpha$ as a linear combination of $k$-fold products of $1$-forms. Thus we only have to check that

\displaystyle\begin{aligned}\left[f^*(\alpha^1\wedge\dots\wedge\alpha^k)\right](v_1,\dots,v_k)&=\left[(\alpha^1\wedge\dots\wedge\alpha^k)\circ f\right](f_{*p}v_1,\dots,f_{*p}v_k)\\&=\det\left(\left[\alpha_i\circ f\right](f_{*p}v_j)\right)\\&=\det\left(\left[f^*\alpha_i\right](v_j)\right)\\&=\left[(f^*\alpha_1)\wedge\dots\wedge(f^*\alpha_k)\right](v_1,\dots,v_k)\end{aligned}

Thus $f^*$ preserves the wedge product as well, and thus gives us a degree-zero homomorphism of the exterior algebras.

Advertisements

July 13, 2011 - Posted by | Differential Topology, Topology

## 13 Comments »

1. […] now we know how to translate -forms by pulling back, and thus we can define another Lie […]

Pingback by The Lie Derivative on Forms « The Unapologetic Mathematician | July 13, 2011 | Reply

2. […] seen that if is a smooth map of manifolds that we can pull back differential forms, and that this pullback is a degree-zero homomorphism of graded algebras. But […]

Pingback by Pullbacks on Cohomology « The Unapologetic Mathematician | July 21, 2011 | Reply

3. […] How shall we define the “integral” of over ? The most natural thing in the world is to pull back the form along to get a -form on . Then we can […]

Pingback by Integration on Singular Cubes « The Unapologetic Mathematician | August 3, 2011 | Reply

4. […] find this pullback of we must work out how to push forward vectors from . That is, we must work out the derivative of […]

Pingback by An Example (part 3) « The Unapologetic Mathematician | August 24, 2011 | Reply

5. […] is orientable — we can just use to orient — and given a choice of top form on we can pull it back along to give an orientation of […]

Pingback by Compatible Orientations « The Unapologetic Mathematician | August 29, 2011 | Reply

6. […] further, let’s say we have a compactly-supported -form on . We can use to pull back from to . Then I say […]

Pingback by Integrals and Diffeomorphisms « The Unapologetic Mathematician | September 12, 2011 | Reply

7. […] manifold to another so that we can compare them, but we’ve seen one case where we can do it: pulling back differential forms. This works because differential forms are entirely made from contravariant vector fields, so we […]

Pingback by Isometries « The Unapologetic Mathematician | September 27, 2011 | Reply

8. […] take to be an orientation-preserving embedding — a singular cube of top dimension. Then the pullback for some strictly-positive function . We conclude […]

Pingback by Compact Oriented Manifolds without Boundary have Nontrivial Homology « The Unapologetic Mathematician | November 24, 2011 | Reply

9. […] forms — together with the exterior derivative — gives us a chain complex. Since pullbacks of differential forms commute with the exterior derivative, they define a chain map between two […]

Pingback by The Poincaré Lemma (setup) « The Unapologetic Mathematician | December 2, 2011 | Reply

10. Great post! Are you working with duals here? Because my professor changed the whole theory to work on duals. I mean, when you write $f_{*p} : T_{p}M \to T_{f(p)}N$ is different from $f^*_p : T^*_p M \to T^*_{f(p)}N$, right? Is there any shortcut to use everything you’re doing here, but using duals instead? Because these posts are really clear!

Comment by aaronmaroja | October 1, 2015 | Reply

• Okay, so 1-forms are dual to tangent vector fields. The tangent bundle $TM$ (with fiber over a point $T_pM$) is covariant, meaning that a map $f:M\to N$ induces a “push-forward” map $f_*:TM\to TN$ that acts as a linear transformation on each fiber $f_{*p}: T_pM\to T_{f(p)}N$. This $f_*$ is exactly the differential $df$; it takes a point $p$ and a tangent vector $v$ at $p$ and gives a tangent vector at the point $f(p)$ we write in general as $df(p;v)$. In one dimension this looks more familiar: $f'(p)v$. In higher dimensions it’s basically the chain rule.

Just like a vector field is a (smooth) choice of a vector at each point, a 1-form is a (smooth) choice of a covector at each point. The bundle of covectors is $T^*M$ is contravariant, meaning that $f:M\to N$ induces a “pull-back” map $f^*:T^*N\to T^*M$. In fact, this doesn’t just work for 1-forms; it works for differential forms of any degree. The definition for $k$-forms is the first display-set equation in the post above. Basically it defines the pull-back of a $k$-form on $N$ (which should be a $k$-form on $M$) by its value on $k$ tangent vectors on $M$. The natural thing to do is push those vectors forward to vectors on $N$, and evaluate the original $k$-form on them there. This is a pretty common construction when dealing with dualities like 1-forms and vector fields.

In general, there’s not a “natural” map $T^*M\to T^*N$ induced by a map $f: M\to N$; the natural map goes the other way, just like there’s no natural pull-back on vectors $TN\to TM$ induced by $f$. However, if you equip both bundles with a metric — and if you’re always thinking of these manifolds as sitting inside regular $n$-dimensional space you get a metric induced from that inclusion — then you can use that to change vectors into covectors and vice versa. In that case, you do get a push-forward on differential forms. The catch is, the choice of metric is a choice of how to measure lengths of tangent vectors, and angles between tangent vectors to the same point. That says a lot about the “shape” of your space, and puts you into the realm of differential geometry; everything I’m doing here is without a choice of metric, and is differential topology. That might explain the difference.

As for $\Omega_NV$ and $\mathcal{O}_NV$, I’m dealing here with the sheaf of smooth sections of the bundles of forms and functions. That is, given an open subset $V\subseteq N$ we consider the algebra of smooth real-valued functions on $V$ and call this algebra $\mathcal{O}_NV$. In a small enough $V$, you can think of this as “smooth slices” of the product space $V\times\mathbb{R}$. Similarly, $\Omega^k_NV$ is the space of smooth $k$-forms defined on $V$, which is a module over the algebra $\mathcal{O}_NV$. The important thing here is that we’re not just considering one stalk $\Lambda^k_qN$ over one point $q\in N$ at a time, but all the stalks over all the points $q\in V$, and that the choice of a point in each stalk must be “smooth” as we vary $q$.

Comment by John Armstrong | October 1, 2015 | Reply

• Thank you very much for your time.

Comment by aaronmaroja | October 2, 2015 | Reply

11. Sorry if I’m not familiar with your notation. I couldn’t find where you define those objects $\Omega_N (V)$, $\mathcal O_N V$, etc.

Comment by aaronmaroja | October 1, 2015 | Reply