#smoothmanifolds

Motivation

I am finding it hard to both study math at this level and feel like I am retaining anything. The material is brutally extensive, so I will concisely re-derive the fundamentals of tangent spaces to the best of my abilities.

Perhaps it is my imagination, but it would not hurt to try it anyway.

Goal

We know that a smooth function from \(\mathbb R^n\) to \(\mathbb R^m\) can be approximated by its total derivative. We now want to approximate smooth maps between smooth manifolds.

To this end, let us consider two concepts:

First Concept

Usually, \(\mathbb R^n\) is thought of as a vector space with the geometric picture that its vectors are centered at the origin. We take this one step further and define \(\mathbb R^n_a:=\left\{a\right\}\times\mathbb R^n\), where \(a\) is an element of \(\mathbb R^n\), with the geometric picture that its vectors are centered at \(a\). Clearly, \(\mathbb R^n_a\) and \(\mathbb R^n\) are isomorphic. In other words, all we did was translate \(\mathbb R\). Since elements of \(\mathbb R^n_a\) have the form \(\left(a,v\right)\), where \(a\) and \(v\) are elements of \(\mathbb R^n\), we will usually denote them as \(v_a\) or \(\left.v\right|_a\), depending on the context.

Second Concept

A linear map \(X:C^\infty\left(\mathbb R^n\right)\to\mathbb R\) is called a derivation of \(C^\infty\left(\mathbb R^n\right)\) at \(a\in\mathbb R^n\) if

\[X\left(fg\right)=f\left(a\right)Xg+g\left(a\right)Xf.\]

Let \(T_a\left(\mathbb R^n\right)\) be the set of all derivations of \(C^\infty\left(\mathbb R^n\right)\) at \(a\in\mathbb R^n\). Then \(T_a\left(\mathbb R^n\right)\) is clearly a vector space.

Nice Claim

We now make the nice claim that \(\mathbb R^n_a\) and \(T_a\left(\mathbb R^n\right)\) are isomorphic with \(v_a\mapsto\left.D_v\right|_a\), where \(\left.D_v\right|_a\) is the usual directional derivative from calculus.

I will not prove it since this is a conceptual blog entry.

Pushforwards

Let \(F:M\to N\) be a smooth map between smooth manifolds, and define its pushforward \(F_*:T_pM\to T_{F\left(p\right)}N\) at a point \(p\in M\) such that

\[\left(F_*X\right)\left(f\right)=X\left(f\circ F\right),\]

where \(X\in T_pM\) and \(f\in C^\infty\left(N\right)\).

To proceed, it must be noted that if \(f\) and \(g\) agree on some neighborhood \(U\) of \(p\), then \(Xf=Xg\). Moreover, \(T_pU\) and \(T_pM\) are isomorphic. In particular, \(f\) and \(g\) need not be defined on all of \(M\).

Computations in Coordinates

Concrete Case

Consider the case where \(M=\mathbb R^m\) and \(N=\mathbb R^n\). Then

\[\left(F_*\left.\frac{\partial}{\partial x^i}\right|_p\right)f=\frac{\partial f}{\partial y^j}\left(F\left(p\right)\right)\frac{\partial F^j}{\partial x^i}\left(p\right)=\left(\left.\frac{\partial F^j}{\partial x^i}\left(p\right)\frac{\partial}{\partial y^j}\right|_{F\left(p\right)}\right)f.\]

Recall that \(x^i\) and \(y^i\) are the standard coordinates of \(\mathbb R^m\) and \(\mathbb R^n\), respectively, and that

\[\left.\frac{\partial}{\partial x^1}\right|_p,\left.\frac{\partial}{\partial x^2}\right|_p,\cdots,\left.\frac{\partial}{\partial x^m}\right|_p\]

forms a basis for \(T_p\left(\mathbb R^m\right)\). Also, recall that we tend to abuse notation tremendously to simplify expressions. For example, given a chart \(\left(U,\varphi\right)\), we identify \(U\) with \(\varphi\left(U\right)=\tilde U\) and think of \(\varphi=\left(x^i\right)\) as the identity map (always keeping in mind that these are huge sins).

Abstract Case

Now, consider the case where \(M\) and \(N\) are arbitrary smooth manifolds. Then given two charts \(\left(U,\varphi\right)\) of \(M\) and \(\left(V,\psi\right)\) of \(N\) such that \(F\left(U\right)\subseteq V\), it follows that \(\hat F=\psi\circ F\circ\varphi^{-1}\) is the coordinate representation of \(F\) and

\[\begin{align*}\left(F_*\left.\frac{\partial}{\partial x^i}\right|_p\right)&=F_*\left(\left(\varphi^{-1}\right)_*\left.\frac{\partial}{\partial x^i}\right|_{\hat p}\right)\newline&=\left(\psi^{-1}\right)_*\left(\hat F_*\left.\frac{\partial}{\partial x^i}\right|_{\hat p}\right)\newline&=\left(\psi^{-1}\right)_*\left(\left.\frac{\partial\hat F^j}{\partial x^i}\left(\hat p\right)\frac{\partial}{\partial y^j}\right|_{\hat F\left(\hat p\right)}\right)\newline&=\left.\frac{\partial\hat F^j}{\partial x^i}\left(\hat p\right)\frac{\partial}{\partial y^j}\right|_{F\left(p\right)},\end{align*}\]

where \(\hat p=\varphi\left(p\right)\) is the coordinate representation of \(p\), as usual (remember the sin, no pun intended). All of these look like complicated computations, but they are pretty straightforward.

Change of Coordinates

This will be the last section of this blog entry. I am currently discussing a typo in my text with other mathematicians. I will complete this section as soon as that issue is resolved. Also, Tumblr does not like very long posts...

Let \(\left(U,\varphi\right)=\left(U,\left(x^i\right)\right)\) and \(\left(V,\psi\right)=\left(V,\left(\tilde x^i\right)\right)\) be smooth charts of a smooth manifold, let \(p\in U\cap V\), and denote \(\psi\circ\varphi^{-1}=\left(\hat x^1,\dots,\hat x^n\right)\). Then

\[\left(\psi\circ\varphi^{-1}\right)_*\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)}=\left.\frac{\partial\hat x^j}{\partial x^i}\left(\varphi\left(p\right)\right)\frac{\partial}{\partial\hat x^j}\right|_{\psi\left(p\right)}.\]

Therefore,

\[\begin{align*}\left.\frac{\partial}{\partial x^i}\right|_p&=\left.\left(\varphi^{-1} \right )_*\frac{\partial}{\partial x^i}\right|_{\varphi\left(p \right )}\newline &=\left(\psi^{-1} \right )_*\left(\left.\left(\psi\circ\varphi^{-1} \right )_*\frac{\partial}{\partial x^i}\right|_{\varphi\left(p \right )} \right )\newline &=\left(\psi^{-1} \right )_*\left(\left.\frac{\partial\hat x^j}{\partial x^i}\left(\varphi\left(p \right ) \right )\frac{\partial}{\partial\hat x^j}\right|_{\psi\left(p \right )} \right )\newline &=\left.\frac{\partial\hat x^j}{\partial x^i}\left(\hat p \right )\frac{\partial}{\partial\hat x^j}\right|_p.\end{align*}\]