Some align mistake [duplicate]

Question

A question why I get this error when I use align? My code is the following

\documentclass[12pt,a4paper]{report}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[spanish]{babel}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{xargs}
\usepackage[usenames]{xcolor}
\usepackage{xr-hyper}
\usepackage[left=3.00cm, right=2.50cm, top=2.50cm, bottom=3.00cm]{geometry}
\usepackage{emptypage}
\usepackage[hyphens]{url}
\usepackage{fncychap}
\usepackage{fancyhdr}
\usepackage{amsthm}

\theoremstyle{definition}
\newtheorem{teorema}{Teorema}[chapter]
\newtheorem{definicion}{Definición}[chapter]
\newtheorem{lema}{Lema}[chapter]
\newtheorem{corolario}{Corolario}[chapter]
\newtheorem{proposicion}{Proposición}[chapter]
\newtheorem{observacion}{Observación}[chapter]
\newtheorem{ejemplo}{Ejemplo}[chapter]
\newtheorem{afirmacion}{Afirmación}[chapter]
\usepackage[colorlinks=true, pdfpagemode=None, linkcolor=blue, citecolor=blue]{hyperref}
\usepackage[shortlabels]{enumitem}



\newcommand{\sub}[2]{(#1_{#2})_{#2\in \mathbb{N}}}
\newcommand{\D}{\displaystyle}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\omp}{{\omega(p)}}
\newcommand{\orb}{\mathcal{O}}



\def\span{\mathrm{Span}}
\def\defi{\it}

\def\para#1{\parax(#1)}
\def\parax(#1){\mathbf{#1}}

\def\dfun#1{\dfunx(#1)}
\def\dfunx(#1,#2,#3,#4,#5){\begin{array}{cccl}
#1:&#2&\longrightarrow &#3\\
& #4 &\mapsto&\D #1(#4):= #5
\end{array}}

\def\dfunv#1{\dfunvx(#1)}
\def\dfunvx(#1,#2,#3,#4,#5){\begin{array}{cccl}
#1:&#2&\longrightarrow &#3\\
& #4 &\mapsto&\D #1 #4:= #5
\end{array}}

%%%%%%%%%%%%%%%%%%%comandos acoplados%%%%%%%%%%%
% COLORES personales---------------------------------------------------
    \definecolor{colortitulo}{RGB}{11,17,79} %
    \definecolor{colordominante}{RGB}{11,17,79}
    \definecolor{colordominanteA}{RGB}{243,102,25}
    \definecolor{colordominanteF}{RGB}{219,68,14}
    \definecolor{colordominanteD}{RGB}{74,0,148}



%************************************************
\newcommand{\pr}{\hspace*{5.5mm}}



%************************************************
%\numberwithin{equation}{chapter}
%\renewcommand\thesection{\arabic{section}}
%\renewcommand\thesubsection{\thesection.\arabic{subsection}}
%\setcounter{secnumdepth}{3}
%\setcounter{tocdepth}{3}
%%%%%%%%%%%-----Diseño de capítulos estilo 1. 2. 3.,...%%%%%%%%%%%

%\def\LigneVerticale{\vrule height 1.2cm depth 0.5cm\hspace{0.1cm}\relax}
%\def\LignesVerticales{%
%   \let\LV\LigneVerticale\LV\LV\LV\LV\LV\LV\LV\LV\LV\LV}
\def\GrosCarreAvecUnChiffre#1{%
    \rlap{\vrule height 0.8cm width 1cm depth 0.2cm}%
    \rlap{\hbox to 1cm{\hss\mbox{\white #1}\hss}}%
    \vrule height 0pt width 1cm depth 0pt}
\def\@makechapterhead#1{\hbox{%
        \bfseries
        \Huge
    %   \LignesVerticales
        \hspace{-0.5cm}%
        \GrosCarreAvecUnChiffre{\thechapter}
        \hspace{0.3cm}\hbox{#1}%
    }\par\vskip 2cm}
    
\begin{document}
\begin{titlepage}

\renewcommand{\contentsname}{INDICE GENERAL}
\tableofcontents
\fancyhead[LE,LO]{Contenido}
\newpage
\renewcommand{\chaptername}{}

\chapter{Preliminares variedades}

\begin{definicion}
Una {\defi variedad diferenciable de clase $C^k$ de dimensión $m$}, con $1\leq k\leq +\infty$, es un espacio topológico Hausdorff $M^m$ la cual tiene una base numerable de abiertos y está dotado de un {\defi atlas diferenciable} $\mathcal{A}=\{(U_\alpha,\para{x}_\alpha):\alpha\in\Lambda\}$, es decir, una familia de homeomorfismos $\para{x}_\alpha:U_\alpha\to \para{x}_\alpha(U_\alpha)$, $\alpha\in\Lambda$ tales que:
\begin{enumerate}[{1)}]
\item cada $U_\alpha$ es un abierto de $M^m$ mientras que cada $\para{x}_\alpha(U_\alpha)$ es un abierto de $\R^m$ y además $M^m=\D\bigcup_{\alpha\in\Lambda}U_\alpha$,
\item la aplicación $\para{x}_\beta\circ\para{x}^{-1}_\alpha:\para{x}_\alpha(U_\alpha\cap U_\beta)\to\para{x}_\beta(U_\alpha\cap U_\beta)$ es un difeomorfismo de clase $C^k$ entre abiertos de $\R^m$, para cualesquiera $\alpha$ y $\beta$ en $\Lambda$ tales que $U_\alpha\cap U_\beta\neq\emptyset$.
\item la familia $\mathcal{A}=\{(U_\alpha,\para{x}_\alpha):\alpha\in\Lambda\}$ es maximal con respecto a la condición $(2)$, es decir, si $(U,\para{x})$ es una carta local tal que $\para{x}\circ\para{x}^{-1}_\alpha$ y $\para{x}_\alpha\circ\para{x}^{-1}$ son difeomorfismos de clase $C^k$ para todo $\alpha\in\Lambda$, entonces $(U,\para{x})\in\mathcal{A}.$
\end{enumerate}
\end{definicion}

A menos que se indique lo contrario, consideraremos variedades diferenciables conexas, es decir, tales que los únicos subconjuntos abiertos y cerrados simultáneamente son el propio $M^m$ y el conjunto vacío $\emptyset$. 

Los homeomorfismos $\para{x}_\alpha$ son llamados {\defi cartas locales} o {\defi coordenadas locales}, y las funciones $\para{x}_\beta\circ\para{x}^{-1}_{\alpha}$ son llamadas {\defi cambio de coordenadas}. 

Diremos que una aplicación $f:M^m\to N^n$ entre variedades diferenciblaes es {\defi diferenciable} si 
\begin{equation}\label{fdif}
\para{y}_\beta\circ f\circ\para{x}^{-1}_\alpha:\para{x}(U_\alpha\cap f^{-1}(V_\beta))\to \para{y}_\beta(V_\beta\cap f(U_\alpha))
\end{equation}
es una aplicación diferenciable para toda carta local $\para{x}_\alpha:U_\alpha\to\para{x}(U_\alpha)$ de $M^m$ y toda carta local $\para{y}_\beta:V_\beta\to\para{x}(V_\beta)$ de $N^n$ con $f(U_\alpha)\cap V_\beta\neq\emptyset$. Además, diremos que $f$ es de clase $C^k$ si $M^m$ y $N^n$ son variedades diferenciables de clase $C^k$ y toda aplicación $\para{y}_\beta\circ f\circ\para{x}^{-1}_\alpha$ en \eqref{fdif} es de clase $C^k$. Llamaremos {\defi difeomorfismo} a toda biyección $f:M^m\to N^n$ tal que tanto $f$ como $f^{-1}$ son diferenciables, si ambas aplicaciones fueran de clase $C^k$ diremos que el difeomorfismo es de clase $C^k$.

Sea $M^m$ una variedad diferenciable. Para cada $p\in M^m$, consideremos el conjunto $C_p(M^m)$ de todas las curvas $\gamma:I\to M$, donde $I$ es un intervalo abierto que contiene a $0$, tales que $\gamma(0)=p$ y $\gamma$ es diferenciable en el punto $0$, es decir la aplicación $\para{x}_\alpha\circ\gamma$ es diferenciable en el punto $0$, para toda carta local $\para{x}_\alpha:U_\alpha\to\para{x}_\alpha(U_\alpha)$ con $p\in U_\alpha$. Diremos que dos curvas $\gamma_1,\gamma_2$ en $C_p(M^m)$ son equivalentes si $\left.\D\dfrac{d}{dt}(\para{x}_\alpha\circ\gamma_1)(t)\right\vert_{t=0}=\left.\D\dfrac{d}{dt}(\para{x}_\alpha\circ\gamma_2)(t)\right\vert_{t=0}$ para toda carta local $\para{x}_\alpha:U_\alpha\to\para{x}_\alpha(U_\alpha)$. De hecho, si la igualdad vale para una carta local entonces ella vale para cualquier otra, en efecto notemos que dada cualquier otra carta local $\para{x}_\beta:U_\beta\to\para{x}(U_\beta)$ con $p\in U_\beta$, tenemos que

\begin{align}
\left.\D\dfrac{d}{dt}(\para{x}_\beta\circ\gamma_1)(t)\right\vert_{t=0}&=\left.\D\dfrac{d}{dt}((\para{x}_\beta\circ\para{x}^{-1}_\alpha)\circ(\para{x}_\alpha\circ\gamma_1))(t)\right\vert_{t=0}\\
&=\left.\D d(\para{x}_\beta\circ\para{x}^{-1}_\alpha)(\para{x}_\alpha\circ\gamma_1(0))\dfrac{d}{dt}(\para{x}_\alpha\circ\gamma_1)(t)\right\vert_{t=0}\smallskip\\
&=\left.\D d(\para{x}_\beta\circ\para{x}^{-1}_\alpha)(\para{x}_\alpha\circ\gamma_2(0))\dfrac{d}{dt}(\para{x}_\alpha\circ\gamma_2)(t)\right\vert_{t=0}\\
&=\left.\D\dfrac{d}{dt}(\para{x}_\beta\circ\gamma_2)(t)\right\vert_{t=0}.
\end{align}

Representamos por $[\gamma]$ a la clase de equivalencia de cualquier curva $\gamma\in C_p(M^m)$. El espacio tangente a la variedad $M^m$ en el punto $p$ es el conjunto de tales clases de equivalencia, el cual será denotado por $T_pM$. Dada cualquier carta local $\para{x}_\alpha:U_\alpha\to\para{x}_\alpha(U_\alpha)$


\newpage

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\addcontentsline{toc}{chapter}{5.\;\;Bibliografía }
\fancyhead[LE,LO]{Bibliografía}
%\chapter{Bibliografía}
%\renewcommand{\bibname}{Bibliografía}
\begin{thebibliography}{20}
    %\addcontentsline{toc}{chapter}{7. Bibliografía}
    \fancyhead[LE,LO]{Bibliografía}
        
\bibitem{benazic}  Benazic, R. (2007). Tópicos de ecuaciones diferenciales ordinarias. Uni, Perú.
\bibitem{ciesielski} Ciesielski, K. (2012). The Poincaré-Bendixson theorem: from Poincaré to the XXIst century. Central European Journal of Mathematics, 10(6), 2110-2128.
\bibitem{hirsch} Hirsch, M. W., Smale, S., \& Devaney, R. L. (2012). Differential equations, dynamical systems, and an introduction to chaos. Academic press.
\bibitem{palis} Palis, J. J., \& De Melo, W. (1982). Geometric theory of dynamical systems: an introduction. Springer-Verlag.
\bibitem{perko} Perko, L. (2001). Differential equations and Dynamical systems.     
\bibitem{schwartz} Schwartz, A. J. (1963). A generalization of a Poincaré-Bendixson theorem to closed two-dimensional manifolds. American Journal of Mathematics, 453-458.      
\bibitem{sotomayor1979li} Sotomayor, J. (1979). Lic{\~o}es de equac{\~o}es diferenciais ordin{\'a}rias. Instituto de Matem{\'a}tica Pura e Aplicada, CNPq.
\end{thebibliography}
\end{document}

Any idea what to do to get rid of this error?

Answer 1

Now you know why LaTeX has \newcommand and everybody recommends using it instead of \def.

Inside amsmath.sty we find

\def\measure@#1{%
    \begingroup
        [...code that's not of a concern]
            \halign{\span\align@preamble\crcr
        [...code that's not of a concern]
    \endgroup
}

and \measure@ is a very basic function for alignment displays such as align.

You'd incur in the same problem without align, but with the kernel defined \multicolumn command.

You're redefining a primitive command of TeX, something that can be done (and LaTeX does it in some places), but you must know what you're doing. And sorry, but replacing this primitive with “print something in upright type” doesn't fall in this category.

Never use \def, unless you exactly know for sure that this will have no consequences (and such knowledge is quite rare for command names that aren't tailored in a very specific way).

If you want to use “Span” in math mode, do

\DeclareMathOperator{\Span}{Span}

and use \Span. This is safe and correctly uses LaTeX, whereas \mathrm{Span} doesn't.

If you really need \def for macros with delimited arguments, do

\newcommand{\dfunx}{}
\def\dfunx(#1,#2,#3,#4,#5){...}

so you will know whether \dfunx is used by some third party package.

Just in passing: why

\def\para#1{\parax(#1)}
\def\parax(#1){\mathbf{#1}}

instead of the much simpler

\newcommand{\para}[1]{\mathbf{#1}}

which does exactly the same?

You are also defining \D for \displaystyle and use it abundantly where it does nothing at all.

Answer 2

As commented in the first comment under the question (before you posted the code that generated the error) you should use LaTeX constructs not \def so that you are warned if you are redefining TeX constructs.

\span is a TeX primitive that underlies \multicolumn so if you redefine \span you need to redefine every construct using it to use a saved version of the original primitive. That would be, at least

\sp@n from the LaTeX format

\gather@, \align@, \measure@ from amsmath

But even then your document would be likely incompatible with any package making any kind of alignment or array construct.

In practice the solution is to replace

\def\span{\mathrm{Span}}

by

\DeclareMathOperator{\Span}{Span}

then use \Span in your document,

Answer 3

I test (using recent MiKTeX) your MWE and it works fine. However, I would not use your \def but rather \mathbf as defined in LaTeX:

\documentclass[12pt,a4paper]{report}
\usepackage[T1]{fontenc}
\usepackage{amsmath}

\begin{document}
\begin{align*}
\left.\dfrac{d}{dt}\bigl(\mathbf{x}_\beta\circ\gamma_1\bigr)(t)\right\vert_{t=0}
    & = \left.\dfrac{d}{dt}\bigl((\mathbf{x}_\beta\circ\mathbf{x}^{-1}_\alpha)
            \circ\mathbf{x}_\alpha\circ\gamma_1\bigr)(t)\right\vert_{t=0}\\
&=\left. d(\mathbf{x}_\beta\circ\mathbf{x}^{-1}_\alpha)(\mathbf{x}_\alpha\circ\gamma_1(0))\dfrac{d}{dt}(\mathbf{x}_\alpha\circ\gamma_1)(t)
\right\vert_{t=0}\\
    & = \left. d(\mathbf{x}_\beta\circ\mathbf{x}^{-1}_\alpha)
            (\mathbf{x}_\alpha\circ\gamma_2(0))
            \dfrac{d}{dt}(\mathbf{x}_\alpha\circ\gamma_2)(t)\right\vert_{t=0}\\
    & = \left.\dfrac{d}{dt}(\mathbf{x}_\beta\circ\gamma_2)(t)\right\vert_{t=0}.
\end{align*}
\end{document}