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108 changes: 78 additions & 30 deletions blueprint/src/content.tex
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Expand Up @@ -15,11 +15,6 @@ \chapter{Continuous alternating maps}
\end{itemize}
\end{definition}

We will use notation\todo{Better \LaTeX\ notation.} \(E [\bigwedge^{I}]\to L[\mathbb k] F\)
for the type of continuous alternating maps from \(E^{I}\) to \(F\).
We will also write (in the blueprint but not in Lean code)
\(E [\bigwedge^{k}]\to L[\mathbb k] F\) for \(E [\bigwedge^{\Fin k}]\to L[\mathbb k] F\).

\section{Pointwise operations}%
\label{sec:pointwise-operations}

Expand All @@ -37,6 +32,9 @@ \subsection{Arithmetic}%
\|f\| = \operatorname{Inf} \left\{C \mid C \ge 0 \wedge \forall m, \|f(m)\| \le C\prod_{i=0}^{k-1}\|m_{i}\|\right\}.
\end{equation}
\end{theorem}
\begin{proof}
The statement immediately follows from a similar statement about continuous multilinear maps.
\end{proof}

\subsection{Prod, pi}%
\label{sec:prod-pi}
Expand All @@ -49,17 +47,25 @@ \subsection{Composition}%
\begin{theorem}%
\label{thm:clm-comp}
If \(g\colon F \to G\) is a continuous linear map between topological vector spaces over \(\mathbb k\)
and \(f\colon E [\bigwedge^{I}]\to L[\mathbb k] F\) is a continuous altenrating map,
then \(g \circ f\colon E [\bigwedge^{I}]\to L[\mathbb k] G\) is a continuous alternating map.
and \(f\colon E^{I} \to F\) is a continuous alternating map,
then \(g \circ f\colon E^{I}\to G\) is a continuous alternating map.
\end{theorem}

\begin{theorem}%
\label{thm:comp-clm}
If \(g\colon F [\bigwedge^{I}]\to L[\mathbb k] G\) is a continuous altenrating map
If \(g\colon F^{I} \to G\) is a continuous alternating map
and \(f\colon E \to F\) is a continuous linear map
then \(f^{*}g \colon E [\bigwedge^{I}]\to L[\mathbb k] G\) given by \(f^{*}g(\dots, x_{j}, \dots) = g (\dots, f(x_{j}), \dots)\) is a continuous alternating map.
then \(f^{*}g \colon E^{I} \to G\) given by \(f^{*}g(\dots, x_{j}, \dots) = g (\dots, f(x_{j}), \dots)\) is a continuous alternating map.
\end{theorem}

\begin{lemma}%
\label{lem:comp-clm-zero}
\uses{thm:comp-clm}
If \(g\colon F^{I} \to G\) is a continuous alternating map with nonempty \(I\)
(e.g., \(g\colon F^{k}\to G\) and \(k > 0\)),
then \(0^{*}g = 0\), where \(0\colon E \to F\) is the zero map.
\end{lemma}

\subsection{Flipping the arguments}%
\label{sec:flipping-arguments}

Expand All @@ -69,7 +75,7 @@ \subsection{Flipping the arguments}%
\subsection{Low order}%
\label{sec:low-order}

For \(k = 0\) and \(k = 1\), there are simple descriptions of the space of continuous altenrating forms,
For \(k = 0\) and \(k = 1\), there are simple descriptions of the space of continuous alternating forms,
given by the following definitions.

\begin{definition}%
Expand All @@ -79,7 +85,7 @@ \subsection{Low order}%
then the map sending the unique element of \(E^{I}\) to a constant \(c\)
is a continuous alternating map.
Moreover, this construction provides a continuous linear equivalence \(\mkZero\)
between \(F\) and \(E [\bigwedge^{I}]\to L[\mathbb k] G\).
between \(F\) and continuous alternating maps \(E^{I} \to G\).
\end{definition}

\begin{definition}%
Expand All @@ -105,7 +111,10 @@ \subsection{Finite dimensional}%
If the domain \(E\) is a finite dimensional topological vector space with Hausdorff topology,
then any alternating map from \(E\) is continuous.
\end{theorem}

\begin{proof}
\uses{thm:cont-mult-map-findim}
The proof immediately follows from \autoref{thm:cont-mult-map-findim} and the fact that an alternating map is a multilinear map.
\end{proof}
\subsection{Basis}%
\label{sec:basis}

Expand All @@ -122,17 +131,14 @@ \subsection{Currying}%
\begin{definition}%
\label{def:cont-alt-curry-fin}
\uses{def:cont-alt-map}
If \(f\colon E [\bigwedge^{k + 1}]\to L[\mathbb k] F\) is a \((k + 1)\)-alternating map and \(v\) is a vector,
then \(f_{v}\colon E [\bigwedge^{k}]\to L[\mathbb k] F\) given by
If \(f\colon E^{k + 1} \to F\) is a continuous alternating map and \(v\) is a vector,
then \(f_{v}\colon E^{k} \to F\) given by
\[
f_{v}(x_{0}, \dots, x_{k - 1})=f(v, x_{0}, \dots, x_{k - 1})
\]
is a \(k\)-alternating map.
Moreover, this map is linear in \(f\) and \(v\), so we can define
\begin{align}
\curryOne&\colon (E [\bigwedge^{k + 1}]\to L[\mathbb k] F) \to L[\mathbb k] E \to L[\mathbb k] E [\bigwedge^{k}]\to L[\mathbb k] F\\
(\curryOne f)(v)(x) &= f(v, x).
\end{align}
Moreover, this map is linear in \(f\) and \(v\).
We will use notation \(\curryOne f\) for the map sending \(v\) to \(f_{v}\).
\end{definition}

\begin{definition}%
Expand All @@ -143,7 +149,7 @@ \subsection{Currying}%
\left(\curry_{k,l} f(x)\right)(y) = f(x, y)
\]
defines a continuous alternating map \(E^{l}\to F\).
Moreover, this function
Moreover, this function is linear in \(f\) and is multilinear in \(x\).
\end{definition}

\begin{definition}%
Expand All @@ -167,10 +173,19 @@ \subsection{Currying}%
If \(f\) is a continuous \(k\)-alternating map, then \(\altOne(\curryOne f) = (k + 1)f\).
\end{theorem}

\begin{lemma}%
\label{lem:alt-one-linear}
\uses{def:alt-one}
The map \(\altOne(f)\) defined above is continuous and linear in \(f\).
\end{lemma}
\begin{proof}
Linearity follows from the definition.
Continuity follows from the inequality \(\|\altOne(f)(x)\|\le (k + 1)\|f\|\prod_{j}\|x_{j}\|\).
\end{proof}

\begin{remark}
We do not divide the result of \autoref{def:alt-one} by \(k + 1\)
to ensure that it works if \(\mathbb k\) has a positive characteristic.%
\todo{Are there any nontrivially normed fields of positive characteristic?}
to ensure that it works if \(\mathbb k\) has a positive characteristic.
\end{remark}

\begin{theorem}%
Expand All @@ -189,7 +204,8 @@ \subsection{Currying}%
\[
\altAdd_{k,l}(f)(x_{0}, \dots, x_{k + l - 1})=\sum_{\substack{s\subseteq [0, k + l)\\|s|=k}}{(-1)}^{\sign \sigma_{s}}f(\sigma_{s}(0), \dots, \sigma_{s}(k + l - 1)),
\]
where \(\sigma_{s}\) is a permutation of \(\Fin (k + l)\) that sends \(s\) to \([0, k)\) and its complement to \([k, k + l)\).
where for each \(k\)-element set \(s\),
\(\sigma_{s}\) is a permutation of \(\Fin (k + l)\) that sends \(s\) to \([0, k)\) and its complement to \([k, k + l)\).
The exact choice of \(\sigma_{s}\) is not important, but we choose the permutation that is monotone both on \(s\) and on its complement.
\end{definition}

Expand Down Expand Up @@ -251,7 +267,7 @@ \section{Wedge product}%

\begin{theorem}
If \(a\) is a number and \(\omega\) is a continuous alternating map,
then \(\mkZero a \wedge_{f} \omega = f(a) \circ \omega\).
then \(\mkZero a \wedge \omega = a\omega\).
\end{theorem}

\begin{theorem}
Expand All @@ -276,7 +292,7 @@ \chapter{Differential forms on a normed space}%

In this chapter we study differential forms on a normed vector space.
We use notation \(\Omega^{k}(E, F)\) for the space of differential \(k\)-forms on \(E\) taking values in \(F\),
where a \(k\)-form is a map from \(E\) to \(E [\bigwedge^{I}]\to L[\mathbb k] F\).
where a \(k\)-form is a map from \(E\) to the space of continuous alternating maps \(E^{k} \to F\).

For now, we only consider real \(k\)-forms, though we may generalize the notion to any field in the future.
\todo{Generalize \(k\)-forms to any field; if a field isn't \(\mathbb R\) or \(\mathbb C\), then only analytic \(k\)-forms satisfy \(d^{2}=0\).}
Expand All @@ -286,26 +302,52 @@ \chapter{Differential forms on a normed space}%
\section{Pullback}%
\label{sec:pullback}

\begin{definition}
\begin{definition}%
\label{def:pullback-within}
\uses{def:cont-alt-map}
\uses{thm:comp-clm}
The \emph{pullback} of \(\omega\in \Omega^{k}(F, G)\) along a map \(f\colon E\to F\) within a set \(s\)
is the \(k\)-form \(f_{s}^{*}\omega\) given by \(f^{*}\omega(x) = {(d_{s}f(x))}^{*}\omega(f(x))\),
is the \(k\)-form \(f_{s}^{*}\omega\) given by \(f_{s}^{*}\omega(x) = {(d_{s}f(x))}^{*}\omega(f(x))\),
where \(d_{s}f(x)\) is the Fréchet derivative of \(f\) within \(s\) at \(x\),
and the right-hand side is defined in~\autoref{thm:comp-clm}.
\end{definition}

\begin{definition}
\begin{lemma}%
\label{lem:pullback-within-of-non-diff}
If \(f\colon E\to F\) is not differentiable within \(s\) at \(x\) and \(k > 0\),
then \(f_{s}^{*}\omega(x)=0\) for any \(k\)-form \(\omega\).
\end{lemma}
\begin{proof}
\uses{lem:comp-clm-zero}
\uses{def:pullback-within}
This fact immediately follows from \autoref{lem:comp-clm-zero} and the fact that \(d_{s}f(x)=0\).
The latter is true because of the junk values Mathlib uses for derivatives.
\end{proof}

\begin{definition}%
\label{def:pullback-space}
\uses{def:pullback-within}
The \emph{pullback} of \(\omega\in \Omega^{k}(F, G)\) along a map \(f\colon E\to F\)
is the pullback of \(\omega\) along \(f\) within the universal set in the domain.
\end{definition}

\begin{theorem}
\begin{theorem}%
\label{thm:pullback-within-as-clm}
\uses{def:pullback-within}
The pullback \(f_{s}^{*}\omega\) within a set is a continuous linear map in \(\omega\).
\end{theorem}

\begin{theorem}
\begin{theorem}%
\label{thm:pullback-as-clm}
\uses{def:pullback-space}
The pullback \(f^{*}\omega\) is a continuous linear map in \(\omega\).
\end{theorem}

\begin{proof}
\uses{thm:pullback-within-as-clm}
This fact immediately follows from \autoref{thm:pullback-within-as-clm}.
\end{proof}

\begin{theorem}
The pullback (within a set) of the \(0\)-form given by a function \(f\colon F\to G\) along a map \(g\colon E\to F\)
is the \(0\)-form given by the function \(f\circ g\).
Expand All @@ -316,6 +358,12 @@ \section{Pullback}%
\todo[inline]{Formulate a similar statement about the pullback of a \(1\)-form.}
\todo[inline]{Formulate a similar statement about the pullback of the wedge product of forms.}

\begin{lemma}%
\label{lem:pullback-within-wedge}
Given a map \(f\colon E\to F\), a set \(s \subset E\), a continuous alternating maps \(\omega\colon F^{k}\to G\), \(\eta\colon F^{l}\to G'\), and a continuous bilinear map \(g\colon G\times G' \to H\),
we have \(f_{s}^{*}(\omega \wedge_{g}\eta) = f_{s}^{*}\omega\wedge_{g}f_{s}^{*}\eta\).
\end{lemma}

\section{Exterior derivative}%
\label{sec:exterior-derivative}

Expand Down
2 changes: 2 additions & 0 deletions blueprint/src/macros/common.tex
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Expand Up @@ -13,6 +13,8 @@
\DeclareMathOperator{\mkOne}{mk_1}

\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newcommand{\lemmaname}{Lemma}
\theoremstyle{definition}
\newtheorem{definition}{Definition}
\newcommand{\definitionname}{Definition}
Expand Down

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