% beamer-slides pdfmode % This file is a solution template for: % - Giving a talk on some subject. % - The talk is between 15min and 45min long. % - Style is ornate. % Copyright 2004 by Till Tantau . % % In principle, this file can be redistributed and/or modified under % the terms of the GNU Public License, version 2. % % However, this file is supposed to be a template to be modified % for your own needs. For this reason, if you use this file as a % template and not specifically distribute it as part of a another % package/program, I grant the extra permission to freely copy and % modify this file as you see fit and even to delete this copyright % notice. \setbeamersize{text margin left=1.5em} \setbeamersize{text margin right=1.5em} \newcommand\wider[2][3em]{% \makebox[\linewidth][c]{% \begin{minipage}{\dimexpr\textwidth+#1\relax} \raggedright#2 \end{minipage}% }% } \catcode`\@=11 \def\normalframenumbering{{\number\c@framenumber/\inserttotalframenumber}} \def\detailedframenumbering{{\number\c@framenumber/\inserttotalframenumber${} ^{[\ifnum\beamer@slideinframe=\beamer@minimum \number\beamer@slideinframe\else\advance\beamer@slideinframe by -1{} \number\beamer@slideinframe\advance\beamer@slideinframe by 1{}\fi{:} \number\c@page]}$}} \newcount \c@refinit \def\biblioframenumbering{Ref.~\advance\c@page by -\c@refinit \number\c@page \advance\c@page by \c@refinit${}^{[\number\c@page]}$} \def\setbibliopages{\c@refinit=\c@page \advance \c@refinit by -1{} \let\framenumbering=\biblioframenumbering} \def\ialign{\everycr{}\tabskip\z@skip\halign} \def\eqalign#1{\null\,\vcenter{\openup\jot\m@th \ialign{\strut\hfil$\displaystyle{##}$&$\displaystyle{{}##}$\hfil \crcr#1\crcr}}\,} \def\plainmatrix#1{\null\,\vcenter{\normalbaselines\m@th \ialign{\hfil$##$\hfil&&\quad\hfil$##$\hfil\crcr \mathstrut\crcr\noalign{\kern-\baselineskip} #1\crcr\mathstrut\crcr\noalign{\kern-\baselineskip}}}\,} \catcode`\@=12 \mode % \setbeamertemplate{background canvas}[vertical shading][bottom=red!10, % top=blue!10] \usetheme{Warsaw} \usefonttheme[onlysmall]{structurebold} \usepackage[utf8]{inputenc} \usepackage{amsmath,amssymb} \usepackage{colortbl} \usepackage[english]{babel} % Or whatever. Note that the encoding and the font should match. If T1 % does not look nice, try deleting the line with the fontenc. \font\sevenrm=cmr10 at 7pt \definecolor{ColClaim}{rgb}{0,0,0.8} \definecolor{Alert}{rgb}{0.8,0,0} \definecolor{Blank}{rgb}{1,1,1} \def\claim#1{{\color{ColClaim}#1}} \def\alert#1{{\color{Alert}#1}} \def\blank#1{{\color{Blank}#1}} \def\bddK{{{}^b\kern-1pt K}} \def\bfe{{\bf e}} \def\bfk{{\bf k}} \def\Poin{{\hbox{\sevenrm Poincar\'e}}} \let\framenumbering=\normalframenumbering \title[\ \kern-190pt\rlap{\blank{J.-P.\ Demailly (Grenoble), SISSA Trieste, October 3, 2018}}\kern181pt\rlap{\blank{ Solutions of algebraic equations}}\kern181pt \llap{\blank{\framenumbering~}}\kern-10pt] % (optional, use only with long paper titles) {Analytic and arithmetic\\ solutions of algebraic equations} %% \subtitle{Presentation Subtitle} % (optional) \author[] % (optional, use only with lots of authors) {\strut\vskip-20pt Jean-Pierre Demailly} \institute[]{\strut\vskip-20pt Institut Fourier, Université Grenoble Alpes\ \ \&\ \ Académie des Sciences de Paris} % - Use the \inst command only if there are several affiliations. % - Keep it simple, no one is interested in your street address. \date[]% (optional) {\strut\vskip-20pt SISSA Colloquium\\ 40${}^{\rm th}$ anniversary of the School Foundation\\ \vskip7pt Trieste, October 3, 2018} %%\subject{Talks} % This is only inserted into the PDF information catalog. Can be left % out. % If you have a file called "university-logo-filename.xxx", where xxx % is a graphic format that can be processed by latex or pdflatex, % resp., then you can add a logo as follows: %%\def\\{\hfil\break} \ifpdf \font\eightrm=ec-lmr10 at 8pt \else \font\eightrm=cmr10 at 8pt \fi \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\Ker}{\operatorname{Ker}} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\tors}{\operatorname{torsion}} \newcommand{\rank}{\operatorname{rank}} \newcommand{\reg}{\operatorname{reg}} \newcommand{\sing}{\operatorname{sing}} \newcommand{\bB}{{\mathbb B}} \newcommand{\bC}{{\mathbb C}} \newcommand{\bD}{{\mathbb D}} \newcommand{\bG}{{\mathbb G}} \newcommand{\bK}{{\mathbb K}} \newcommand{\bN}{{\mathbb N}} \newcommand{\bP}{{\mathbb P}} \newcommand{\bQ}{{\mathbb Q}} \newcommand{\bR}{{\mathbb R}} \newcommand{\bZ}{{\mathbb Z}} \newcommand{\cA}{{\mathcal A}} \newcommand{\cC}{{\mathcal C}} \newcommand{\cD}{{\mathcal D}} \newcommand{\cE}{{\mathcal E}} 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\newcommand{\Vol}{\mathop{\rm Vol}\nolimits} \newcommand{\pr}{\mathop{\rm pr}\nolimits} \newcommand{\NS}{\mathop{\rm NS}\nolimits} \newcommand{\GG}{{\mathop{\rm GG}\nolimits}} \newcommand{\NE}{\mathop{\rm NE}\nolimits} \newcommand{\ME}{\mathop{\rm ME}\nolimits} \newcommand{\SME}{\mathop{\rm SME}\nolimits} \newcommand{\card}{\mathop{\rm card}\nolimits} \newcommand{\alg}{{\rm alg}} \newcommand{\nef}{{\rm nef}} \newcommand{\num}{\nu} \newcommand{\ssm}{\mathop{\mathbb r}} \newcommand{\smallvee}{{\scriptscriptstyle\vee}} % figures inserted as PostScript / PDF files \ifpdf \def\RGBColor#1#2{{\pdfliteral{#1 rg}#2\pdfliteral{0 g}}} \long\def\InsertPSFigure#1 #2 #3 #4\EndFig{\par\advance\psfigurecount by 1% \pdfximage{\jobname_figures/fig\number\psfigurecount.pdf}% \setbox0=\hbox{\pdfrefximage\pdflastximage}% \psfiguredx=#1mm \advance\psfiguredx by 20mm% \hbox{$\vbox to#2mm{\vfil% \hbox{$\hskip #1mm\rlap{\smash{\raise-50mm\hbox to #1mm{\strut\kern-\psfiguredx% \pdfximage 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#5 angle #6}}#8$}} \fi \long\def\LabelTeX#1 #2 #3\ELTX{\rlap{\kern#1mm\raise#2mm\hbox{#3}}} \def\ovl{\overline} \def\build#1^#2_#3{\mathrel{\mathop{\null#1}\limits^{#2}_{#3}}} \def\bibitem[#1]#2#3{\medskip{\bf[#1]} #3} \begin{document} %%\def\pause{} % Delete this, if you do not want the table of contents to pop up at % the beginning of each subsection: %%\AtBeginSubsection[] %%{ %% \begin{frame} %% \frametitle{Outline} %% \tableofcontents[currentsection,currentsubsection] %% \end{frame} %%} % If you wish to uncover everything in a step-wise fashion, uncomment % the following command: %\beamerdefaultoverlayspecification{<+->} \begin{frame} \pgfdeclareimage[height=2cm]{uga-logo}{logo_UGA} \pgfuseimage{uga-logo} \pgfdeclareimage[height=2cm]{acad-logo}{academie_logo2} \pgfuseimage{acad-logo} \vskip-12pt \titlepage \end{frame} %%\begin{frame} %% \frametitle{Outline} %% \tableofcontents %% You might wish to add the option [pausesections] %%\end{frame} % Since this a solution template for a generic talk, very little can % be said about how it should be structured. However, the talk length % of between 15min and 45min and the theme suggest that you stick to % the following rules: % - Exactly two or three sections (other than the summary). % - At *most* three subsections per section. % - Talk about 30s to 2min per frame. So there should be between about % 15 and 30 frames, all told. %% \section*{Basic concepts} \begin{frame} \frametitle{Solutions of algebraic equations (degrees 2 and 3)} \claim{Solution of quadratic equations (Babylonians)}: $ax^2+bx+c=0$\vskip4pt $\displaystyle x_k={-b\pm\sqrt{b^2-4ac}\over 2a},~~k=1,2$\vskip8pt\pause \claim{Solution of cubic equations (Niccolo Tartaglia, Gerolamo Cardano)}: \vskip3pt $x^3+px+q=0$ has 3 complex solutions $x_k=j^ku+\overline j^kv$, $k=1,2,3$\vskip6pt where $\displaystyle j=-{1\over 2}+i{\sqrt{3}\over 2}$, $\displaystyle u={}\kern4pt\raise4pt\hbox{${}^3$}\kern-2pt \!\sqrt{-{q \over 2}+\sqrt{{q^{2} \over 4}+{p^{3} \over 27}}},~~ v=-{p/3\over u}.$\vskip1.4cm \vbox{Tartaglia\break $\sim1535$}\vskip-2cm\strut\kern1.6cm \pgfdeclareimage[height=3cm]{tartaglia}{Niccolo_Tartaglia.jpg} \pgfuseimage{tartaglia}\kern3mm \pgfdeclareimage[height=3cm]{cardano}{Gerolamo_Cardano.jpg} \pgfuseimage{cardano}\vskip-2cm\strut\kern10.2cm Cardano\vskip2cm\strut \end{frame} \begin{frame} \frametitle{Solutions of algebraic equations (higher degrees)} \claim{Solution of quartic equations (Lodovico Ferrari, 1540)}:\vskip6pt The solution of $ax^4+bx^3+cx^2+dx+e=0$ can be reduced to solving consecutively one cubic and two quadratic equations.\vskip6pt\pause However, \claim{Paolo Ruffini, Niels Abel} and finally \claim{Évariste Galois ($\sim 1831$)} show the impossibility of solving equations of degree${}\geq5$ by using only radicals. \vskip1.3cm Ferrari\vskip-1.5cm\strut\kern1.7cm \pgfdeclareimage[height=3cm]{ferrari}{Lodovico_Ferrari.jpg} \pgfuseimage{ferrari}\kern2cm \pgfdeclareimage[height=3cm]{galois}{Evariste_Galois.jpg} \pgfuseimage{galois}\vskip-2cm\strut\kern9.2cm Galois\vskip1.6cm Galois's proof relies on the fact that the symmetric group ${\frak S}_n$ of permutation of roots is not solvable for $n\geq 5$. \end{frame} \begin{frame} \frametitle{Diophantine equations} \claim{Babylonian tablet ($\sim 1800$ BC) showing Pythagorean triples} \vskip2pt \pgfdeclareimage[height=3cm]{plimpton}{Plimpton} \pgfuseimage{plimpton} \vskip3pt\pause Pythagorean triples $(a,b,c)$ are triples of positive integers \hbox{such that\kern-25pt} \vskip5pt\centerline{ $\alert{a^2+b^2=c^2},$}\vskip5pt in other words, they represent rectangular triangles.\pause\vskip6pt The general solution \claim{attributed to Euclid} is \vskip5pt\centerline{ $\alert{a=k(p^2-q^2),~~b=2kpq,~~c=k(p^2+q^2)}$}\vskip5pt where $k>0,p>q>0$ are integers. \end{frame} \begin{frame} \frametitle{Interpretation of Euclid's solution} The equation $a^2+b^2=c^2$ is equivalent to \vskip5pt\centerline{ $\displaystyle\alert{ \Big({a\over c}\Big)^2+\Big({b\over c}\Big)^2=1~~ \Leftrightarrow~~x^2+y^2=1}$}\vskip5pt by putting $x={a\over c}$, $y={b\over c}$.\pause Now, the solution is given by \vskip5pt\centerline{ $\displaystyle\alert{ x={a\over c}={p^2-q^2\over p^2+q^2},~~ y={b\over c}={2pq\over p^2+q^2}}$} \vskip5pt which is equivalent to $$\displaystyle\alert{ x={1-t^2\over 1+t^2},~~ y={2t\over 1+t^2},~~~t={q\over p}.} \leqno(*)$$ \pause Clearly, any rational value of $t\in\bQ$ yields a rational point $(x,y)\in\bQ^2$, which itself corresponds to a Pythagorean triple $(a,b,c)$. The important point is that $(*)$ is a rational parametrization of the circle. \end{frame} \begin{frame} \frametitle{Rational curves} More generally, consider a polynomial equation $P(x,y)=0$ with integer coefficients, and assume that there is a parametrization $$\alert{ x={R(t)\over Q(t)} ,~~y={V(t)\over U(t)}} $$ as rational functions with integer coefficients, then we get infinitely many rational solutions by taking $t\in\bQ$.\vskip5pt\pause Hence, the important issue is whether the curve $C=\{P(x,y)=0\}$. admits a rational parametrization.\vskip5pt\pause It is convenient to work with complex numbers and consider more generally \alert{homogeneous} polynomial equations $$\alert{ P(x,y,z)=0~~\Leftrightarrow~~P(x/z,y/z,1)=0,} $$ which amounts to work in the complex projective plane of points $[x:y:z]\simeq(x/z,y/z)$, with a line $z=0$ of ``points at infinity''. \end{frame} \begin{frame} \frametitle{Conics} \pgfdeclareimage[height=6.5cm]{paramcircle}{Rational_parameterization_of_unit_circle} \pgfuseimage{paramcircle}\vskip-15pt\pause In fact, all plane conics $C=\{ax^2+bx^2+cxy+dx+ey+f=0\}$ can be parametrized in that way, and if the coefficients are rational (with $C(\bQ)\neq\emptyset$ and $C(\bR)$ not reduced to a point), then $C$ contains infinitely many rational points. \end{frame} \begin{frame} \frametitle{Elliptic curves} Elliptic curves can be defined by plane equations of the form $$\alert{y^2=x^3+ax+b}\quad\hbox{(below, $y^2=x^3-x+1$ is shown)}$$ \pgfdeclareimage[height=3.5cm]{elliptic}{Elliptic_Curve} \pgfuseimage{elliptic}\vskip4pt\pause A fundamental property is that are isomorphic to an additive group $(\bC/\Lambda,+)$ (2-torus) where $\Lambda$ is a lattice, and they can be parametrized as $(x,y)=(\wp(t),\wp'(t))$ where $\wp$ is the Weierstrass $\wp$~function (a transcendental, $\Lambda$-periodic, meromorphic function of~$t\in\bC$). \end{frame} \begin{frame} \frametitle{Classification of curves (``Riemann surfaces'')} \pgfdeclareimage[height=7.8cm]{curves}{curves} \vskip4pt \pgfuseimage{curves} \vskip-15pt \hbox{\alert{Over $\bC$, curves are surfaces!}} \vskip22pt$~$ \end{frame} \begin{frame} \frametitle{Rational points on curves} \pgfdeclareimage[height=3cm]{mordell}{Louis_Mordell} \pgfuseimage{mordell}~~$\raise1.3cm\hbox{L.~Mordell}$\kern7mm \pgfdeclareimage[height=3cm]{faltings}{Gerd_Faltings} \pgfuseimage{faltings}~~$\raise1.3cm\hbox{G.~Faltings}$ \vskip4pt \claim{Gerd Faltings}, Fields medal 1986 for the solution of a conjecture by Louis Mordell (1922): \begin{block}{Theorem (Faltings, 1983)} Let $C$ be a smooth curve of genus $g\geq 2$ in the complex projective~plane, defined as $\{P(x,y,z)=0\}$ for some homogeneous polynomial $P(x,y,z)\in \bQ[x,y,z]$. Then the curve $C$ contains \alert{finitely many} rational points $[x:y:z]$, i.e.\ $x,y,z\in\bQ$. \end{block} \end{frame} \begin{frame} \frametitle{General Diophantine equations} General diophantine equations $P_j(x_0,x_1,\ldots,x_n)=0$, $1\leq j\leq m$ are much harder.\vskip5pt\pause \pgfdeclareimage[height=3cm]{hilbert}{David_Hilbert} \pgfuseimage{hilbert}~~$\raise1.3cm\hbox{Hilbert}$\kern2cm \pgfdeclareimage[height=3cm]{matiyasevich}{Yuri_Matiyasevich} \pgfuseimage{matiyasevich}~~$\raise1.3cm\hbox{Matiyasevich}$ \vskip1pt\pause \alert{Yuri Matiyasevich} became famous for proving in 1970 the inexistence of an algorithm deciding whether arbitrary Diophantine equations have solutions (Hilbert's 10${}^{\rm th}$ problem from the 1900 Paris congress); some equations may be undecidable.\vskip5pt\pause In fact, Diophantine equations can encompass many mathematical algorithms, e.g. there exists a polynomial $P\in\bZ[X_1,\ldots,X_{26}]$ whose set of positive values is the set of prime numbers~! \end{frame} \begin{frame} \frametitle{(Non) existence of rational \&\ elliptic curves} \vskip4pt \pgfdeclareimage[height=3cm]{clemens}{Herb_Clemens} \pgfuseimage{clemens}~~$\raise1.3cm\hbox{H.~Clemens}$\kern2cm \pgfdeclareimage[height=3cm]{voisin}{Claire_Voisin} \pgfuseimage{voisin}~~$\raise1.3cm\hbox{C.~Voisin}$ \vskip5pt \begin{block}{Theorem (Herb Clemens $\sim 1986$, Claire Voisin $\sim 1996$)} Let $X=\{P(z_0,\ldots,z_{n+1})=0\}$ be a generic $n$-dimensional algebraic hypersurface in complex projective space (i.e.\ with general enough ``random'' coefficients) of degree $d\geq 2n+1$, $n\geq 2$. Then $X$ does not contain rational or elliptic curves. \end{block} In other words, there are \alert{no entire curves $f:\bC\to X$} solving the equation $P(f(t))=0$, where $f$ is either rational or a Weierstrass type transcendental function. \end{frame} \begin{frame} \frametitle{The Kobayashi conjecture} \vskip4pt \pgfdeclareimage[height=3cm]{kobayashi}{Shoshichi_Kobayashi} \pgfuseimage{kobayashi}~~$\raise1.3cm\hbox{S.~Kobayashi}$\kern2cm \begin{block}{Conjecture (Shoshichi Kobayashi $\sim 1970$)} Let $X=\{P(z_0,\ldots,z_{n+1})=0\}$ be a generic $n$-dimensional algebraic hypersurface in complex projective space (i.e.\ with general enough ``random'' coefficients) of sufficiently large degree $d\geq d_n$. Then $X$ contains \alert{no entire holomorphic curve $f:\bC\to X$}. \end{block}\pause %%\claim{Note:} One expects $d_n=2n+1$ (same as Clemens-Voisin bound).\pause \begin{block}{Definition} $X$ is said to be \alert{Kobayashi hyperbolic} if it contains no entire holomorphic curve $f:\bC\to X$. \end{block} \end{frame} \begin{frame} \frametitle{Solution of Kobayashi conjecture (Brotbek 2016)} \vskip-5pt \begin{block}{Theorem (Brotbek 2016)} The Kobayashi conjecture holds true for $d_n\gg 1$ (non explicit), i.e.\ a generic polynomial equation $P(z)=0$ of degree $d\geq d_n$ contains no entire analytic solution $z=f(t)$, $t\in\bC$. \end{block}\pause\vskip-4pt My PhD student Ya Deng found a way to make $d_n$ explicit.\pause\vskip-2pt \begin{block}{Theorem (simpler proof, Demailly 2018)} The Kobayashi conjecture holds for $d\geq d_n=\lfloor(en)^{2n+2}/3\rfloor$. \end{block}\pause\vskip-4pt The proof is based on finding certain algebraic differential operators \vskip4pt \centerline{$\displaystyle \alert{ D(f)=\sum a_{\alpha_1\ldots\alpha_k}(f)(f')^{\alpha_1}(f'')^{\alpha_2}\ldots (f^{(k)})^{\alpha_k}}$}\vskip4pt whose coefficients $a_\alpha$ vanish along some hyperplane section \hbox{$H\subset X$.\kern-10pt}\pause\vskip-2pt \begin{block}{Theorem (Green-Griffiths 1979, D-- 1995, Siu-Yeung 1996)} If such a global operator $D$ exists on $X$, then all entire curves $f:\bC\to X$ must satisfy the differential equation $D(f)\equiv 0$. \end{block} \end{frame} \begin{frame} \frametitle{Simplified proof of Kobayashi conjecture} The ``simplified proof'' consists of finding sufficiently many global differential operators $D(f)=0$ in the form of \claim{Wronskians} \alert{$$ W(s_0,\ldots,s_k)(f)=\left| \plainmatrix{ s_0(f) & s_1(f) & \ldots &s_k(f)\cr \nabla(s_0(f)) & \nabla(s_1(f)) & \ldots &\nabla(s_k(f))\cr \vdots & & &\vdots\cr \nabla^k(s_0(f)) & \nabla^k(s_1(f)) & \ldots &\nabla^k(s_k(f))\cr}\right| $$}\pause In general they do not exist, but one can show that they exist on well chosen ``Fermat type'' hypersurfaces $\sum m_j(z)^k=0$ where the $m_j(z)=z^{\beta_j}$ are suitable monomials.\pause\vskip8pt Moreover, by using the \alert{geometric structure of jet bundles} elaborated in (D--, 1995), one can show that these operators have ``deformations'' to generic hypersurfaces of high degree. \end{frame} \begin{frame} \frametitle{Green-Griffiths conjecture} \pgfdeclareimage[height=2cm]{green}{Mark_Green} \pgfuseimage{green}~~$\raise1.3cm\hbox{M.~Green}$\kern2cm \pgfdeclareimage[height=2cm]{griffiths}{Phillip_Griffiths} \pgfuseimage{griffiths}~~$\raise1.3cm\hbox{P.~Griffiths}$ \begin{block}{Conjecture (Mark Green \&\ Phillip Griffiths 1979)} Let $X=\{P(z_0,\ldots,z_{n+1})=0\}$ be a smooth $n$-dimensional algebraic hypersurface in complex projective space. Assume $d=\deg P\geq n+2$. Then $X$ possesses an \alert{algebraic subvariety $Y\,{=}\,\{Q(z)\,{=}\,0\}$} containg all entire curves $f\,{:}\,\bC\,{\to}\,X$, i.e.\ \hbox{$Q(f)\,{=}\,0$.\kern-10pt} \end{block}\pause\vskip-2pt The smallest $Y={\rm GG}(X)$ is called the Green-Griffiths \hbox{locus of $X$.\kern-20pt}\pause\vskip-2pt \begin{block}{Theorem (Demailly 2010)} Rather than an algebraic equation $Q(f)=0$, such entire curves $f:\bC\to X=\{P=0\}$, $\deg P\geq n+2$, satisfy at least one (and in fact many) algebraic differential equations \alert{$D(f)=0$}. \end{block} \end{frame} \begin{frame} \frametitle{Lang-Vojta program} \vskip4pt \pgfdeclareimage[height=3cm]{lang}{Serge_Lang} \pgfuseimage{lang}~~$\raise1.3cm\hbox{S.~Lang}$\kern1cm \pgfdeclareimage[height=3cm]{vojta}{Paul_Vojta} \pgfuseimage{vojta}~~$\raise1.3cm\hbox{P.~Vojta}$ \vskip5pt This is a fascinating program that tries to relate the algebro-geometric properties of algebraic varieties with their arithmetic properties.\pause \begin{block}{Lang's conjecture 1987 -- very optimistic ?} For $X$ projective defined over a number field $\bK_0$, the Green-Griffiths locus \alert{${\rm GG}(X)$} in GG conjecture equals \alert{${\rm Mordell}(X)={}\hbox{smallest}~Y$} such that \alert{$X(\bK)\smallsetminus Y$ is finite}, $\forall\bK$ number field${}\supset\bK_0$. \end{block} \end{frame} \begin{frame} \frametitle{The end} \strut\vskip3mm \pgfdeclareimage[height=6.5cm]{CalabiYau}{CalabiYau} \strut\kern2cm\pgfuseimage{CalabiYau} \end{frame} \end{document} % Local Variables: % TeX-command-default: "XXTeX" % End: