that this notation only works for a primitive vector of index $1$. Then one checks that the Morse complex ${\mathcal{M}}(K)$ is given by:

$$\begin{tikzpicture}[scale=1]

\node[inner sep=1pt, circle, fill=black](a) at (0,2) {};
\node[inner sep=1pt, circle, fill=black](b) at (1,0) {};
\node[inner sep=1pt, circle, fill=black](c) at (-1,0) {};
\node[inner sep=1pt, circle, fill=black](d) at (0,-1) {};
\node[inner sep=1pt, circle, fill=black](e) at (1,-3) {};
\node[inner sep=1pt, circle, fill=black](f) at (-1,-3) {};


\draw[-]  (a)--(b);
\draw[-]  (a)--(c);
\draw[-] (a)--(d);
\draw [white, line width=1.5mm] (b) -- (c) node[midway, left] {};
\draw[-]  (b)--(c);

\draw[-]  (d)--(e);
\draw[-]  (d)--(f);
\draw[-]  (e)--(f);


\draw[-] (b)--(e);
\draw[-] (c)--(f);


\node[anchor = east]  at (a) {\small{$ab$}};
\node[anchor =  west]  at (b) {\small{$bc$}};
\node[anchor = east]  at (c) {\small{$ca$}};
\node[anchor = east]  at (d) {\small{$cb$}};
\node[anchor =  west]  at (e) {\small{$ac$}};
\node[anchor = east]  at (f) {\small{$ba$}};

\end{tikzpicture}$$

The Automorphism group of ${\mathcal{M}}(K)$ {#sec: The Automorphism group}
============================================

This section is devoted to computing the automorphism group of ${\mathcal{M}}(K)$. We first show in section \[Induced Maps\] that certain automorphisms of $K$ induce automorphisms of ${\mathcal{M}}(K)$. It will then follow by Proposition \[prop: autK subgroup\] that ${\mathrm{Aut}}(K)$ is isomorphic to a subgroup of ${\mathrm{Aut}}({\mathcal{M}}(K))$. We show in Proposition \[prop: aut(M(K))=aut(K)\] that ${\mathrm{Aut}}(K)\cong {\mathrm{Aut}}({\mathcal{M}}(K))$ in the case where $K\neq 