Created
October 22, 2012 17:48
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LaTeX example (Beamer and TikZ): Algebraic structures
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%------------------------------------------------------------------------------- | |
\newcommand{\efivebig}{\begin{tikzpicture}[xscale=.8,yscale=.8,line width=2pt] | |
\foreach \i in {1,...,5} | |
{ \path (\i,1) coordinate (T\i); \path (\i,0) coordinate (B\i); } | |
%%%%%%% | |
\filldraw[fill= black!12,draw=black!12,line width=4pt] (T1) -- (T5) -- (B5) -- (B1) -- (T1); | |
\draw[blue] (T3) .. controls +(.1,-.5) and +(-.1,-.5) .. (T4) ; | |
\draw[blue] (B3) .. controls +(.1,.3) and +(-.1,.3) .. (B4) ; | |
\draw[blue] (T1) -- (B1); | |
\draw[blue] (T2) -- (B2); | |
\draw[blue] (T5) -- (B5); | |
%%%%%%% | |
\foreach \i in {1,...,5} | |
{ \fill (T\i) circle (4pt); \fill (B\i) circle (4pt); } | |
\end{tikzpicture}} | |
%------------------------------------------------------------------------------- | |
\newcommand{\bigdiag}{\begin{tikzpicture}[xscale=.8,yscale=.8,line width=2pt] | |
\foreach \i in {1,...,5} | |
{ \path (\i,1) coordinate (T\i); \path (\i,0) coordinate (B\i); } | |
%%%%%%% | |
\filldraw[fill= black!12,draw=black!12,line width=4pt] (T1) -- (T5) -- (B5) -- (B1) -- (T1); | |
\draw[blue] (T1) -- (B5); | |
\draw[blue] (T2) .. controls +(.1,-.4) and +(-.1,-.5) .. (T5); | |
\draw[blue] (T3) .. controls +(.1,-.3) and +(-.1,-.3) .. (T4) ; | |
\draw[blue] (B1) .. controls +(-.1,.5) and +(.1,.5) .. (B2) ; | |
\draw[blue] (B3) .. controls +(.1,.3) and +(-.1,.3) .. (B4) ; | |
%%%%%%% | |
\foreach \i in {1,...,5} | |
{ \fill (T\i) circle (4pt); \fill (B\i) circle (4pt); } | |
\end{tikzpicture}} | |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% | |
\begin{frame}[c]{Similar elements} | |
For any $d$, a $TL_k$ diagram created by the generators, we can always find a generator $e_i$ such that when $e_i$ is multiplied with $d$, the result is $x\cdot d$. | |
$$e_i\cdot d= | |
\begin{array}{c} | |
\efivebig \\ | |
\bigdiag \\ | |
\end{array} =x\cdot \begin{array}{c} | |
\bigdiag \\ | |
\end{array} =x\cdot d | |
$$ | |
This means, $\chi(d)=\chi(\frac{1}{x}e_i\cdot d)=\frac{1}{x}\chi(e_i\cdot d)=\frac{1}{x}\chi(d\cdot e_i)$ | |
\end{frame} | |
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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