I'm new to triangulated category and tilting theory. To illustrate, in $Q=A_4$ the module $M=kQ$ is cluster-tilting. While I know that $M$ satisfies $\mathrm{Ext}(M,M)=\mathrm{Hom}(M,M[1])=0$ by some theory, I can't read it off from the A-R quiver of cluster category, for example the following graph. How to see that $\mathrm{Hom}(M,M[1])=0$ from the arrows?
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The diagram depicted is a presentation of the category of indecomposables and therefore allows to compute the dimension of Hom spaces between any two vertices (indecomposable objects). The Hom space itself is the quotient of the vector space spanned by paths from $M$ to $N$ modulo mesh relations (here commuting squares, including squares at the boundary where one vertex is missing). One can then easily see where Hom spaces become zero.
To just compute the dimensions of Hom spaces, use the rule that $h(?) = dim Hom(M,?)$ must satisfy
$$h(X)- \sum_i h(Z_i) + h(Y) = 0$$
on every mesh, where $X$ and $Y$ are the vertices at start and end of the mesh and the $Z_i$ are in the middle.
In the special case of cluster-type $A$, the possible $N$ with $Hom(M,N)$ not zero form a tilted rectangle on the right of $M$ with two vertices on the boundary.
F. C.
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Note that here F.C. has used M for an indecomposable representation, whereas in the question, M was the sum of several indecomposables. For the kind of approach that F.C. describes, one wants to break things down indecomposable by indecomposable.
I think there is an issue with the paragraph beginning "To just compute...". This would be true if Hom(M,.) were an exact functor, but it isn't. Maybe there is a good way to specify which meshes it is true for? In any case, the explanation in the first paragraph is good, as is the explanation in the third paragraph.
– Hugh Thomas Nov 02 '22 at 14:29
\mathrm{}or\text{}or even\operatorname{}to nicely render text like "Hom" – FShrike Apr 14 '22 at 09:21