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Suppose that $\mathcal{C}$ is a finitary category, so for any two objects $A$ and $B$ we have that $|\mathrm{Ext}^i(A,B)| < \infty$ for $i\geq 0$, suppose $\mathcal{C}$ has finite global dimension, so $\mathrm{Ext}^i(A,B) = \{0\}$ for all sufficiently large $i$*. In such a category $\mathcal{C}$ we can define the Euler form of two objects $$ \langle A,B \rangle_m = \left( \prod_{i=0}^{\infty} \Big|\mathrm{Ext}^i(A,B)\Big|^{(-1)^i}\right)^{\frac{1}{2}}\,. $$

This is a definition I've encountered a few times while studying quiver representations and Hall algebras. It's a constant we need to introduce when giving a Hall algebra an associative algebra structure (see these lecture notes by Olivier Schiffmann for details). I've yet to get a good grasp on what exactly this is counting in terms of the objects and morphisms in the category though. Is this related to the good ol' Euler characteristic of a topological space in some non-superficial way? How should I be thinking about this form?


* Typically I've only seen this in the context of hereditary categories, where the $\mathrm{Ext}^i$ groups vanish for $i>1$, and so we only need to consider $\mathrm{Hom}(A,B)$ and $\mathrm{Ext}^1(A,B)$ for any objects $A$ and $B$.

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This answer perhaps says things that are all obvious to the OP. $\textrm{Ext}^i(A,B)$ is a vector space over the ground field, so its cardinality is $q^d$ where $d$ is the dimension of the vector space and $q$ is the cardinality of the field. The quantity in question is therefore $(q^{1/2})^{\langle A,B\rangle}$, where $\langle A,B\rangle = \sum_i (-1)^i \textrm{dim } \textrm{Ext}^i(A,B)$. This quantity depends only on the dimension vectors of $A$ and $B$, and not on the actual module structures of $A$ and $B$. I have seen the induced form on the Grothendieck group (or equivalently the space of dimension vectors) called the Euler form, or the Euler-Ringel form. It is the alternating sum of the dimensions of the homology of a chain complex (obtained by taking a resolution of one of $A$ or $B$); I think this is what is Euler-like about it.

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