01_25.tex

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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 3}
\date{1/25/21} % Can also use \today

\begin{document}
\maketitle
\setlength\RaggedRightParindent{\parindent}
\RaggedRight

\par\noindent
Review from Friday: Given a finite-dimensional vector space $V$ (i.e. $T_{p}M$ for some $p\in{}M$), a tensor of type $(k,l)$ is an element of
\[
	T^{(k,l)}(V)\eqdef\underbrace{V\otimes\cdots\otimes{}V}_{k}\otimes\underbrace{V^{*}\otimes\cdots\otimes{}V^{*}}_{l}\cong\{\ptxt{multilinear maps }\underbrace{V^{*}\times\cdots\times{}V^{*}}_{k}\times\underbrace{V\times\cdots\times{}V}_{l}\to\R\}
\]

\ex{
	\begin{itemize}[topsep=0pt, itemsep=0pt, leftmargin=3\parindent]
		\item Type $(0,1)$ is $V^{*}$, the set of linear maps $V\to\R$.
		\item Type $(1,0)$ is $(V^{*})^{*}\cong{}V$, by $\displaystyle\isom{V}{(V^{*})}{v}{(\alpha\mapsto\alpha(v))}$ (for $\alpha\in{}V^{*}$).
		\item Type $(0,2)$ is $V^{*}\otimes{}V^{*}\cong\set{\ptxt{bilinear }V\times{}V\to\R}$.
	\end{itemize}\up\n
}

\defn{
	We can take the \u{tensor product} (or \u{outer product}) of two tensors on the same vector space: given $\tau\in{}T^{(k,l)}(V)$ and $\tau'\in{}T^{(k',l')}(V)$, we define $\tau\otimes\tau'\in{}T^{(k+k',l+l')}(V)$ by
	\[
		(\tau\otimes\tau')(\underbrace{v_{1},\ldots,v_{k+k'}}_{\in{}V},\underbrace{\alpha_{1},\cdots,\alpha_{l+l'}}_{\in{}V^{*}})\eqdef\tau(v_{1},\ldots,v_{k},\alpha_{1},\ldots,\alpha_{k})\cdot\tau'(v_{k+1},\ldots,v_{k+k'},\alpha_{l+1},\ldots,\alpha_{l+l'})
	\]
	\n
}

\par\noindent
Note: We have to check that this map is indeed multilinear.\n

\ex{
	If $\alpha,\beta\in{}T^{(0,1)}(V)=V^{*}$, then $\alpha\otimes\beta$ is the map $\map{V\times{}V}{\R}{(v_{1},v_{2})}{\alpha(v_{1})\beta(v_{2})}$.\n
}

\par\noindent
Now, we want to consider tensors on manifolds. By the lemma from last time, given $k,l$, there's a bundle\break $\pi:T^{(k,l)}(TM)\to{}M$ with fibers $T^{(k,l)}(T_{p}M)=T^{(k,l)}(\pi\inv(p))$ for $p\in{}M$. Given coordinates $(x^{1},\ldots,x^{n})$ on $U\subseteq{}M$, we have a smooth moving frame:
\[
	\set{\pd{}{x^{i_{1}}}\otimes\cdots\otimes\pd{}{x^{i_{k}}}\otimes{}dx^{j_{1}}\otimes\cdots\otimes{}dx^{j_{l}}\setmid{}i_{a},j_{b}\in\set{1,\ldots,n}}
\]
Last time, we stated that this is a basis, and any smooth section is a linear combination of these components, with $C^{\infty}$ functions as coefficients:
\[
	\sum{}A^{i_{1}\cdots{}i_{k}}_{j_{1}\cdots{}j_{k}}\pd{}{x^{i_{1}}}\otimes\cdots\otimes\pd{}{x^{i_{k}}}\otimes{}dx^{j_{1}}\otimes\cdots\otimes{}dx^{j_{l}}
\]
Why is this true?\n

\exer{
	Check that $A^{i_{1}\cdots{}i_{k}}_{j_{1}\cdots{}j_{l}}=\tau(dx^{i_{1}},\ldots,dx^{i_{k}},\pd{}{x^{j_{1}}},\ldots,pd{}{x^{j_{l}}})$.\n
}

\ex{
	Fix a smooth $1$-form $\alpha\in\Omega^{1}(U)$. Then $\alpha=\sum_{j=1}^{n}a_{j}dx^{j}$, and $a_{j}=\alpha(\pd{}{x^{j}})$.\n
}

\defn{
	Some terminology: We say $\tau\in\Gamma(T^{(k,l)}(TM))$ is a \u{tensor}, or \u{tensor field}, on $M$ of type $(k,l)$. $k$ is the \u{contravariant degree}, and $l$ is the \u{covariant degree}.\n
}

\ex{
	Vector fields are contravariant (type $(1,0)$).\n
	$1$-forms are covariant (type $(0,1)$).\n
}

\subsection*{Einstein Summation Notation}

\par\noindent
In coordinates, the covariant indices are subindices, and the contravariant indices are superindices. The convention of Einstein summation notation is if the same index appears exactly twice in a monomial -- one upper and one lower -- then summation over that index is implied.\n

\ex{
	(Linear case) $V=\R^{n}$, $\alpha\in{}V^{*}\otimes{}V\cong\Hom(V,V)$. Then we have
	\[
		a_{j}^{i}v^{j}=\sum_{j=1}^{n}a_{j}^{i}v^{j}=\brack{\begin{pmatrix}
			a_{1}^{1} & \cdots & a_{n}^{1}\\
			\vdots & & \vdots\\
			a_{1}^{n} & \cdots & a_{n}^{n}
		\end{pmatrix}\begin{pmatrix}
			v^{1}\\
			\vdots\\
			v^{n}
		\end{pmatrix}}_{i}
	\]
	Note that $(a_{j}^{i})$ is taken to be a matrix, where $j$ is the column index and $i$ the row index.\n
}

\par\noindent
Looking ahead, we'll also have a contraction operation: $a\in{}V^{*}\otimes{}V\rightsquigarrow\alpha\in\Hom(V,V)\rightsquigarrow\tr(a)=a_{i}^{i}\in\R$. \i{This doesn't require coordinates, as trace is basis-invariant.}\n

\defn{
	A \u{Riemannian metric} $g$ on a manifold $M$ is a smooth $(0,2)$ tensor, which, at every point, as a matrix, is symmetric ($g_{p}=g_{p}\tpose$) and positive definite ($\forall{}v\in{}T_{p}M$, $v\tpose{}g_{p}v\ge{}0$, with equality iff $v=0$).\n
}

\par\noindent
Observe: $\forall{}p\in{}M$, $g_{p}:T_{p}M\times{}T_{p}M\to\R$ is a bilinear map. In coordinates $(x^{1},\ldots,x^{n})$ on $U\osubseteq{}M$, $g=g_{ij}\,dx^{i}\otimes{}dx^{j}$, where $g_{ij}\in{}C^{\infty}(U)$, and $g_{ij}=g_{ji}$. In fact, because of symmetry, we can write (using Einstein summation notation)
\begin{align*}
	g & =\frac{1}{2}\brack{g_{ij}\,dx^{i}\otimes{}dx^{j}+g_{ji}\,dx^{i}\otimes{}dx^{j}}\\
	& =\frac{1}{2}\brack{g_{ij}\,dx^{i}\otimes{}dx^{j}+g_{ij}\,dx^{j}\otimes{}dx^{i}}\\
	& =g_{ij}\frac{1}{2}\underbrace{\brack{dx^{i}\otimes{}dx^{j}+dx^{j}\otimes{}dx^{i}}}_{\hspace{-100pt}\mathrlap{\eqdef{}dx^{i}dx^{j}=dx^{j}dx^{i},\ptxt{ the symmetric product}}}\\
\end{align*}
It's also standard notation to write $ds^{2}\eqdef{}dx^{i}dx^{j}$, so we can write $g=g_{ij}\,ds^{2}$.\n

\defn{
	A \u{Riemannian manifold} is a pair $(M,g)$, where $M$ is a smooth manifold, and $g$ is a Riemannian metric on $M$.\n
}

\defn{
	If $(M,g^{M})$ and $(N,g^{N})$ are Riemannian manifolds, an \u{isometry} $F:M\to{}N$ is a diffeomorphism \st{} $\forall{}p\in{}M$, $(dF)_{p}:T_{p}M\to{}T_{F(p)}N$ preserves the metric, i.e., \st{} $\forall{}u,v\in{}T_{p}M$, $g_{F(p)}^{N}(dF_{p}(u),dF_{p}(v))=g_{p}^{M}(u,v)$.\n
}

\par\noindent
Notation: Sometimes, we'll write $g_{p}(u,v)=\iprod{v,w}_{p}=\iprod{v,w}$. (We omit the subscript when the choice of metric is obvious.)\n

\ex{
	Let $M\subset\R^{N}$, a regular submanifold. $\forall{}p\in{}M$, $T_{p}M\subset\R^{N}$ can be identified with a subspace of $T_{p}R^{N}\cong\R^{N}$. Define $g_{p}=\giprod{}$ by the usual inner product in $\R^{N}$.\n
}

\ex{
	Let $M=S^{2}\subseteq\R^{3}$. We'll compute in coordinates in the upper hemisphere. Say our coordinates are $(s,t)$ (with projection to the flat unit disk), then we have our inverse parameterization
	\[
		(s,t)\mapsto(s,t,\sqrt{1-s^{2}-t^{2}})\eqdef\vec{r}(s,t)
	\]
	(with notation as in Calc III). Then under $T_{p}S^{2}\hookrightarrow\R^{3}$, we have
	\begin{align*}
		\pd{}{s} & \mapsto\pd{\vec{r}}{s}=\paren{1,0,\frac{-s}{\sqrt{1-s^{2}-t^{2}}}}\\
		\pd{}{t} & \mapsto\pd{\vec{r}}{t}=\paren{0,1,\frac{-t}{\sqrt{1-s^{2}-t^{2}}}}
	\end{align*}
	So our metric is
	\[
		(g_{ij})=\begin{pmatrix}
			\pd{\vec{r}}{s}\cdot\pd{\vec{r}}{s} & \pd{\vec{r}}{s}\cdot\pd{\vec{r}}{t}\\
			\pd{\vec{r}}{t}\cdot\pd{\vec{r}}{s} & \pd{\vec{r}}{t}\cdot\pd{\vec{r}}{t}
		\end{pmatrix}=\begin{pmatrix}
			1-t^{2} & \frac{st}{1-s^{2}-t^{2}}\\
			\frac{st}{1-s^{2}-t^{2}} & 1-s^{2}
		\end{pmatrix}
	\]
	\n
}

\par\noindent
Observe: Any submanifold $S\subset{}M$ of a Riemannian manifold $(M,g)$ inherits a Riemannian metric by $\forall{}p\in{}S$, $T_{p}S\hookrightarrow{}T_{p}M$.\n

\end{document}