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分层投影块

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分层投影块

我刚刚看到这个例子（见图），我想知道如何制作类似的分层块。谢谢 在此处输入图片描述

答案1

\documentclass{beamer}
\usetheme{Warsaw}
\usecolortheme{orchid}

\usepackage{tikz}
\usetikzlibrary{shadows}
\usetikzlibrary{shapes.arrows}

\begin{document}

    \begin{frame}[plain]
        \frametitle{Conjectures on correspondence and lifting}

        \begin{block}{Theorem 3 (obtained in 1)}
            A relation of dimensions between S and S
        \end{block}

        \begin{columns}
            \begin{column}{.1\textwidth}
            \begin{tikzpicture}[>=stealth, rotate border/.style={shape border uses incircle, shape border rotate=270}]
                    \node[rotate border=-40, fill=black, minimum height=1.5cm, single arrow, single arrow head extend=.3cm, single arrow head indent=.1cm, inner sep=1.5pt] (arrow) {};
                \end{tikzpicture}
            \end{column}
            \begin{column}{.8\textwidth}
                \begin{alertblock}{}
                    \begin{itemize}
                        \item Definition
                        \item Conjection
                    \end{itemize}
                \end{alertblock}
            \end{column}
        \end{columns}

        \begin{block}{Theorem 3' (new version of Theorem 3)}
            dim S
        \end{block}

        \begin{columns}
            \begin{column}{.1\textwidth}
            \begin{tikzpicture}[>=stealth, rotate border/.style={shape border uses incircle, shape border rotate=270}]
            \node[rotate border=-40, fill=black, minimum height=1.5cm, single arrow, single arrow head extend=.3cm, single arrow head indent=.1cm, inner sep=1.5pt] (arrow) {};
          \draw[line width=5pt] (0,0) -- (1,0);
            \end{tikzpicture}
            \end{column}
            \begin{column}{.8\textwidth}
                (The RHS suggest ...)
            \end{column}
        \end{columns}       

        \begin{alertblock}{Conjectures}
            "Correspondance and Lifting"
        \end{alertblock}

    \end{frame}

\end{document}

在此处输入图片描述

答案2

@samcarter 答案的稍微简化版本：

\documentclass{beamer}
\usetheme{Boadilla}
\usecolortheme{orchid}
    \usepackage{tikz}

\begin{document}
    \begin{frame}[plain]
\frametitle{\S2 Conjectures on correspondence and lifting}

    \begin{block}{Theorem 3 (obtained in 1)}
        A relation of dimensions between $S_{k,j}^{new}(U(N))$ and $S_{k,j}^{new}(U'(N))$
    \end{block}

    \begin{columns}
        \begin{column}{.2\textwidth}
        \qquad\tikz{
                \draw[line width=1mm,-latex] (0, 0) -- (0,-2);
                \draw[line width=1mm]    (0,-1) -- (0.5,-1);
                   }%
        \end{column}%
        \hspace*{-3em}\begin{column}{.85\textwidth}
            \begin{alertblock}{}
                \begin{itemize}
                    \item Definition of paramodular newforms
                    \item Conjectural dimension formula of $S_{k,j}^{new}(U'(N))$
                \end{itemize}
            \end{alertblock}
        \end{column}
    \end{columns}

    \begin{block}{Theorem 3' (new version of Theorem 3)}
        dim $S_{k,j}^{new}(U(N))=\cdots$
    \end{block}

    \begin{columns}
        \begin{column}{.2\textwidth}
        \qquad\tikz{
                \draw[line width=1mm,-latex] (0,0)  -- (0,-1);
                   }
        \end{column}%
        \hspace*{-3em}\begin{column}{.85\textwidth}
            (The RHS suggest how we should define $S_{k,j}^{new}(U'(N))$)
        \end{column}
    \end{columns}

    \begin{alertblock}{Conjectures}
        "Correspondence and Lifting"
    \end{alertblock}
\end{frame}
    \end{document}

在此处输入图片描述

升级： 纯“TikZ”解决方案：

\documentclass{beamer}
\usetheme{Boadilla}
%\usecolortheme{orchid}
    \usepackage{tikz}
    \usetikzlibrary{backgrounds,positioning,shadows,shadows.blur,shapes.multipart}
    \tikzset{THRM/.style = {
thrm/.style = {
    shape=rectangle split, rectangle split parts=2, rounded corners,
    rectangle split part fill={blue!70!black,blue!30!gray!10},
    draw=gray, very thin,
    text width=\textwidth, align=left, inner sep=1mm
                },
blur/.style = {name=n##1,
               rounded corners, shade,
               inner sep=0pt,outer sep=0pt,
               blur shadow={shadow blur steps=7}
                },
cmnt/.style = {name=n##1,
    draw=orange, thin, fill=orange!20, rounded corners,
    text width=\textwidth-22mm, inner sep=1mm, outer sep=0mm,
    blur shadow={shadow blur steps=7}
                }
    }}

    \usepackage{amsmath}
\newcommand{\tightlist}{\vspace*{-1ex}%
  \setlength{\itemsep}{0pt}
  \setlength{\parskip}{0pt}
                        }

    \begin{document}
\begin{frame}[fragile]%,plain
\frametitle{\S2 Conjectures on correspondence and lifting}

    \hfil\begin{tikzpicture}[THRM,
    node distance = 18mm and 11mm
                    ]
%---
\node[blur=1] {\tikz\node[thrm]
    {\nodepart[text=white]{one}  Theorem 3 (obtained in \S1)
     \nodepart{two} A relation of dimensions between $S_{k,j}^{\text{new}}(U(N))$ and $S_{k,j}^{\text{new}}(U'(N))$};
     };
\node[blur=2,below=of n1] {\tikz\node[thrm]
    {\nodepart[text=white]{one}  Theorem 3' (new version of Theorem 3)
     \nodepart{two} A relation of dimensions between $S_{k,j}^{\text{new}}(U(N))$ and $S_{k,j}^{\text{new}}(U'(N))$};
     };
\node[blur=3,below=of n2] {\tikz\node[thrm,
                                      rectangle split part fill= {red!70!black,red!30!gray!10}]
    {\nodepart[text=white]{one} Conjectures
     \nodepart{two} \hfil   "Correspondence and Lifting"};
     };
\coordinate[right=of n1.south west] (A);
\path[draw,line width=1mm,shorten >=1mm,shorten <=2mm,-latex]
    (A) edge node[cmnt=4,right=11mm]
                  { \begin{itemize}\tightlist
                    \item Definition of paramodular newforms
                    \item Conjectural dimension formula of $S_{k,j}^{\text{new}}(U'(N))$
                    \end{itemize}
                    }
    (A |- n2.north)
    (A |- n2.south)
       edge node[right=11mm]
              {(The RHS suggest how we should define $S_{k,j}^{\text{new}}(U'(N))$.)}
    (A |- n3.north)
    (n4) -- (n4 -| A);
    \end{tikzpicture}
\end{frame}
    \end{document}

这种方法更简单，也更“确定”，在不同块（表示为节点）之间绘制线、箭头等。结果几乎与上面的相同。

在此处输入图片描述

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