# Layered beamer blocks

I just saw this example (see image) and I would like to know how one can do similar layered blocks. . Thanks

• On this site, a question should typically revolve around an abstract issue (e.g. "How do I get a double horizontal line in a table?") rather than a concrete application (e.g. "How do I make this table?"). Questions that look like "Please do this complicated thing for me" tend to get closed because they are either "off topic", "too broad", or "unclear". Please try to make your question clear and simple by giving a minimal working example (MWE): you'll stand a greater chance of getting help. May 30 '15 at 12:51

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

\usepackage{tikz}
\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{itemize}
\item Definition
\item Conjection
\end{itemize}
\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}

"Correspondance and Lifting"

\end{frame}

\end{document}


A slightly simplified version of @samcarter's answer:

\documentclass{beamer}
\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}
\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{itemize}
\item Definition of paramodular newforms
\item Conjectural dimension formula of $S_{k,j}^{new}(U'(N))$
\end{itemize}
\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}
\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}

"Correspondence and Lifting"
\end{frame}
\end{document}


\documentclass{beamer}
%\usecolortheme{orchid}
\usepackage{tikz}
\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,
inner sep=0pt,outer sep=0pt,
},
cmnt/.style = {name=n##1,
draw=orange, thin, fill=orange!20, rounded corners,
text width=\textwidth-22mm, inner sep=1mm, outer sep=0mm,
}
}}

\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}


In this approach it is simpler, and more "deterministic", to draw the lines, arrows etc between the different blocks (which are represented as nodes). The result is almost the same as above.