# The (upper hat of) an octahedral diagram in (la)tex

I would like to draw an octahedral diagram in my paper; I would prefer to present it as the 'upper hat' + the 'lower hat' (as it is common in the texts on triangulated categories). Could anyone tell me where I can find this diagram (certainly, the 'upper hat part' is sufficient) written down in latex. Maybe, someone could just share with me his own latex realization of this diagram (so that I could replace the original names of objects and morphisms by the ones I need)?

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Come on: make a full octahedron in one picture. ;-) – Paul Balmer May 15 '10 at 15:23

Mikhail,

Here's an upper cap in xy-pic

\xymatrix{ X'\ar[rd]^{[1]}\ar[dd]^{[1]} & & Z\ar[ll] \\ & Y\ar[ru]\ar[ld] & \\ Z'\ar[rr]^{[1]} & & X\ar[lu]\ar[uu] }

at least it's enough of one to get you started. Note that I didn't construct it by hand. I have a script for building these kinds of diagrams visually:

http://www.math.purdue.edu/~dvb/scripts/arraymaker

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I think that this is an easy exercise in xy-pic. The exact diagram you want may not be in this user's guide, but the basics of xy-pic are easy enough that you should be able to create what you want.

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Alternatively, of course, you could find the diagram you want in someone's paper on the arXiv and check out the source. – Tara Holm May 15 '10 at 14:34
Thank you; I will probably do so when I will have enough time. Yet at the moment I would prefer to use a diagram manufactured by someone else. – Mikhail Bondarko May 15 '10 at 15:28
How does one use your script? – jef Jul 7 '10 at 11:32

The same example given in xy-pic done in Tikz: More examples and code are at http://www.texample.net/tikz/examples/ and the manual http://mirror.ctan.org/graphics/pgf/base/doc/generic/pgf/pgfmanual.pdf

\documentclass[a4paper]{article}
\usepackage{tikz}
\usetikzlibrary{shapes}
\begin{document}
\begin{tikzpicture}[shorten <=7pt,shorten >=7pt]
\node [name=oct,regular polygon,regular polygon sides=4,minimum size= 12em] {};
\node at (oct.corner 1) [name=octCorner1] {$Z$};
\node at (oct.corner 2) [name=octCorner2] {$X'$};
\node at (oct.corner 3) [name=octCorner3] {$Z'$};
\node at (oct.corner 4) [name=octCorner4] {$X$};
\node at (oct.center) [name=octCenter] {$Y$};
\draw [-stealth] (oct.corner 2) -- node [right] {$[1]$} (oct.corner 3);
\draw [-stealth] (oct.corner 1) --                      (oct.corner 2);
\draw [-stealth] (oct.corner 1) --                      (oct.corner 4);
\draw [-stealth] (oct.corner 3) -- node [above] {$[1]$} (oct.corner 4);
\draw [-stealth] (oct.corner 2) -- node [right] {$[1]$} (oct.center);
\draw [-stealth] (oct.corner 4) --                      (oct.center);
\draw [-stealth] (oct.center)   --                      (oct.corner 3);
\draw [-stealth] (oct.center)   --                      (oct.corner 1);
\end{tikzpicture}
\end{document}


Changed from an octagon.

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This appears to be an octagon, not an octahedron. – S. Carnahan May 15 '10 at 20:25
Nevertheless, tikz is vastly superior to xy-pic for drawing diagrams. – André Henriques May 16 '10 at 11:56
Ah, my brain just substituted it with "octagon" when I read the question. Anyway, the above code makes the desired shape if you set "regular polygon sides=4" and place a node at "(oct.center)" – Niall Murphy May 16 '10 at 13:53