The Unapologetic Mathematician

Mathematics for the interested outsider

The Lie Derivative on Forms

We’ve defined the Lie derivative L_XY of one vector field Y by another, X. This worked by using the flow of X to compare nearby points, and used the derivative of the flow to translate vectors.

Well now we know how to translate k-forms by pulling back, and thus we can define another Lie derivative:

\displaystyle L_X\omega=\lim\limits_{t\to0}\frac{1}{t}\left((\Phi_t)^*(\omega)-\omega\right)

What happens if \omega is a 0-form — a function f? We check

\displaystyle\begin{aligned}\left[L_Xf\right](p)&=\lim\limits_{t\to0}\frac{1}{t}\left(\left[(\Phi_t)^*(f)\right](p)-f(p)\right)\\&=\lim\limits_{t\to0}\frac{1}{t}\left(f\left(\Phi_t(p)\right)-f(p)\right)\\&=X_p(f)=Xf(p)\end{aligned}

That is, the Lie derivative by X acts on \Omega^0(M) exactly the same as the vector field X does itself.

I also say that the Lie derivative by X is a degree-zero derivation of the algebra \Omega(M). That is, it’s a real-linear transformation, and it satisfies the Leibniz rule:

\displaystyle L_X(\alpha\wedge\beta)=\left(L_X\alpha\right)\wedge\beta+\alpha\wedge\left(L_X\beta\right)

for any k-form \alpha and l-form \beta. Linearity is straightforward, and given linearity the Leibniz rule follows if we can show

\displaystyle L_X(\alpha_1\wedge\dots\wedge\alpha_k)=\sum\limits_{i=1}^k\alpha_1\wedge\dots\wedge\left(L_X\alpha_i\right)\wedge\dots\wedge\alpha_k

for 1-forms \alpha_i. Indeed, we can write \alpha and \beta as linear combinations of such k- and l-fold wedges, and then the Leibniz rule is obvious.

So, let us calculate:

\displaystyle\begin{aligned}L_X\left(\alpha_1\wedge\dots\wedge\alpha_k\right)=&\lim\limits_{t\to0}\frac{1}{t}\left((\Phi_t)^*\left(\alpha_1\wedge\dots\wedge\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_k\right)\\=&\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_k\right)\\=&\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge\left((\Phi_t)^*\alpha_k\right)\right)\\&+\lim\limits_{t\to0}\frac{1}{t}\left(\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge\left((\Phi_t)^*\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_k\right)\\=&\lim\limits_{t\to0}\left(\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_{k-1}\right)-\alpha_1\wedge\dots\wedge\alpha_{k-1}\right)\wedge\left((\Phi_t)^*\alpha_k\right)\right)\\&+\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_k\right)-\alpha_k\right)\\=&\left(\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_{k-1}\right)-\alpha_1\wedge\dots\wedge\alpha_{k-1}\right)\right)\wedge\alpha_k\\&+\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge L_X\alpha_k\\&=L_X\left(\alpha_1\wedge\dots\wedge\alpha_{k-1}\right)\wedge\alpha_k+\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge L_X\alpha_k\end{aligned}

So we see how we can peel off one of the 1-forms. A simple induction gives us the general case.

July 13, 2011 Posted by | Differential Topology, Topology | 5 Comments

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 | 9 Comments

   

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