# 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.

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

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