### M Theory Lesson 268

Tom Leinster's computation of the Euler characteristic of a (finite) category uses the $n \times n$ incidence matrix, $Z$, of the underlying graph. Let $u$ be the vector $(1,1, \cdots, 1)$ of length $n$. If there exists a vector $a = (a_1,a_2,a_3, \cdots, a_n)$ such that $Z a = u$, the Euler characteristic is given by

$\chi = \sum a_{i}$

Let us try to recover the cardinality of a set from this characteristic, by generalising the set to a connected groupoid on three objects. Our favourite 3 element set will do. Now the equation shows that each hom set in the groupoid must have cardinality $1/3$ for the even weighting to work. Fortunately, this is precisely the cardinality of a group. For it to work for any number of elements $n$, this group should be something like the cyclic group of order $n$. There are $n^{2}$ such hom sets in the groupoid.

Observe how the normalisation factor here has a real effect on the possibilities for hom sets. Without the $1/3$, the vector $a$ would have to be scaled, resulting in an Euler characteristic of only $1$, for any $n$. In other words, when each hom set is the trivial group the information about the cardinality of the set is lost. The simplest possible categorification of the set n therefore uses the cyclic groups.

$\chi = \sum a_{i}$

Let us try to recover the cardinality of a set from this characteristic, by generalising the set to a connected groupoid on three objects. Our favourite 3 element set will do. Now the equation shows that each hom set in the groupoid must have cardinality $1/3$ for the even weighting to work. Fortunately, this is precisely the cardinality of a group. For it to work for any number of elements $n$, this group should be something like the cyclic group of order $n$. There are $n^{2}$ such hom sets in the groupoid.

Observe how the normalisation factor here has a real effect on the possibilities for hom sets. Without the $1/3$, the vector $a$ would have to be scaled, resulting in an Euler characteristic of only $1$, for any $n$. In other words, when each hom set is the trivial group the information about the cardinality of the set is lost. The simplest possible categorification of the set n therefore uses the cyclic groups.

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