# 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.

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Posted by John Armstrong | Differential Topology, Topology

## 9 Comments	»

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

Pingback by | 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 | 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 | 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 | 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 | 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 | 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 | 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 | 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 | December 2, 2011 | Reply

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