The size of a finite skeletal category C in the sense of Leinster is defined as follows: Label the objects of C by integers 1,2,...,n and let aij be the number of morphisms from i to j (for i and j between 1 and n). The size (or Euler characteristic) of C is defined as the sum of the entries of the inverse of the nxn matrix A=(aij), if the inverse exists.

Let Fq be a finite field with q elements. For every natural number i, there is up to isomorphism exactly one Fq-vector space Vi of dimension i. The number of linear maps from Vi to Vj is equal to qij. We ignore the zero dimensional vector space V0. Consider the infinite matrix


where rows and and columns are indexed by positive integers 1,2,3,... From now on let us treat q as a formal parameter, don't care about convergence issues, and set v=q-1.

Is there a notion of an inverse of Q? (The entries will probably be formal power series in v.) If the answer is yes, what is a closed form for the sum of the entries of the inverse (as a formal power series in v), i.e. the size of the category of finite dimensional Fq-vector spaces?

At least every truncation Qn of Q to an upper left nxn corner has an inverse for every positive integer n, since Qn is a Vandermonde matrix. What is the limit of the sum of the entries of Qn-1 as n goes to infinity? I believe the answer is a power series in v. Is there an explicit form?

How can you interpret the answer? Is it the Euler characteristic of some moduli space? Is it equal or related to a sum of 1/Gl(Vi)? Does something interesting happen at q=1?

  • $\begingroup$ You're probably aware of this, but when John Baez says "the cardinality of the category of finite sets is e", he's talking about the groupoid obtained by throwing out all the non-invertible morphisms. $\endgroup$ – Reid Barton Oct 18 '09 at 20:37
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    $\begingroup$ Some noodling around in Mathematica suggests that the answer is something like 1/q prod (1 - 1/q^n). $\endgroup$ – Qiaochu Yuan Oct 18 '09 at 21:21
  • $\begingroup$ Reid, what does that mean? $\endgroup$ – Philipp Lampe Oct 19 '09 at 1:19
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    $\begingroup$ Here's a comment that might help somebody else: if A(x) = a<sub>1 x + a<sub>2 x^2 + ... is a generating function, then Q is the linear transformation that sends (a<sub>1, a<sub>2, ...) to (A(q), A(q^2), ...). $\endgroup$ – Qiaochu Yuan Oct 19 '09 at 3:12
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    $\begingroup$ One curious thing I noticed related to Qiaochu Yuan's first comment: For n=3, 4, 6 (which is all I checked in Mathematica), I got that the answer is 1-(1-1/q)(1-1/q^2)(1-1/q^3)...(1-1/q^n). This is exactly the proportion of nxn matrices over F_q that are singular. $\endgroup$ – Kevin P. Costello Oct 19 '09 at 21:36

(Edit: the following should be a comment following Philipp's "Reid, what does that mean?", not an Answer. Can't see how to change it now, though. Further edit: I've appended a guess at an answer.)

Baez and Dolan have a notion of the cardinality of a groupoid. You pick one object ai from each isomorphism class, and define the cardinality to be

\sumi 1/order(Aut(ai)).

For example, if E is the groupoid of finite sets and bijections then you have to choose one i-element set ai for each natural number i; then Aut(ai) = Sn and the cardinality is e. See their paper "From finite sets to Feynman diagrams".

The cardinality of a finite groupoid is the same as its size/Euler characteristic in my sense. (And you can try your luck at extending to the infinite case.) I guess Reid was pointing out that given an arbitrary category C, not necessarily a groupoid, there are two numbers associated to it: (i) the Euler characteristic of C itself, (ii) the cardinality of the underlying groupoid of C. These are in general different (e.g. consider the two-object category consisting of a single arrow). I guess the reason why Reid made his comment was your second-to-last question: "Is it equal or related to a sum of 1/Gl(Vi)?"

As for your main question, there's weak evidence that the answer is 1. Here's the argument.

It's a theorem (in the paper you cite) that whenever you have an adjunction between a pair of finite categories, and both their Euler characteristics are defined, then their Euler characteristics are equal. It's also a theorem that whenever two categories are equivalent, the EC of one is defined iff the EC of the other is, and then they're equal. Finally, it's a theorem that if a category has a terminal object then its EC, if defined, is 1.

Now, your category is equivalent to the category C of nontrivial finite-dimensional vector spaces over Fq. The usual adjunction between vector spaces and sets restricts to an adjunction between C and the category S of nonempty finite sets. Moreover, S has a terminal object. So if the three theorems described extend (in some sense) to infinite categories, then the EC of your category is 1.

I'm not at all sure that this is the right answer, and it contradicts Qiaochu Yuan's suggestion. If the EC (or more precisely the limit that you mention) is not 1, then that's interesting, because it would imply some difference in behaviour between finite and infinite categories.

Finally, you ask 'How can one interpret the answer?' I don't think I can say much about interpreting this particular case. But in general you can think of the Euler characteristic of a category C as the same thing as the Euler characteristic of its nerve NC (a simplicial set), or its classifying space BC (the geometric realization of NC). Of course, not every topological space has a well-defined Euler characteristic, so theorems of this kind are subject to some hypotheses. The result on adjunctions mentioned above is of the flavour 'if spaces are homotopy equivalent then their Euler characteristics are equal'.

  • $\begingroup$ Thanks for deciphering my comment. By the way, comments are limited to 600 characters, and I don't think they can use any formatting, so this is a lot more readable as an answer anyways. $\endgroup$ – Reid Barton Oct 19 '09 at 5:17
  • $\begingroup$ Thank you for explaining. I had something else in mind when writing <p>\sum<sub>i</sub> 1/order(Aut(a<sub>i</sub>))</p>. Vector spaces may be viewed as representation of the quiver with one node and no arrows. Representations over F<sub>q</sub> are related to Hall algebras. The size of the Hall algebra could be something like the orbifold integral of the identity function as described by Reineke on page 10 of arxiv.org/abs/0804.3214. $\endgroup$ – Philipp Lampe Oct 19 '09 at 14:30

Following the observations made in the comments one can compute the sum of the entries of Qn-1. It turns out that Kevin Costello's formula is true for every n.

Let (a1, a2, ..., an) be the the transpose of the kth column vector of Qn-1. (Of course, this vector depends on k, but we omit the index k.) Qiaochu Yuan suggested to consider the polynomial

A(x) = a1x + a2x2 + ... + anxn.

The degree of A is n, therefore A is determined by values at n+1 points. But we know that A(0)=0 and that

A(qi) = deltaik for i = 1, 2, ... , n.

By Lagrange interpolation, A(x) is equal to

x(x-q)(x-q2)...(x-qk-1)(x-qk+1)...(x-qn) / qk(qk-q)(qk-q2)...(qk-qk-1)(qk-qk+1)...(qk-qn).

The sum a1+a2+...+an is equal to A(1). Let us work with quantized integers. We use the notation [k] = (1-qk)/(1-q) = 1+q+...+qk-1. (Note that people from quantum groups sometimes use a different convention.) Furthermore, let [n choose k] be the quantized binomial coefficient. Then, A(1) is equal to

(-1)k-1 [n choose k] qk(k-1)/2-kn.

We sum A(1) over all k. A variant of the quantum binomial theorem gives that the sum of the entries of Qn-1 is equal to 1 - (1-1/q)(1-1/q2)...(1-1/qn).

  • $\begingroup$ That's really nice! Even if you take q to be an actual number (say \geq 2) rather than a formal parameter, 1 - (1-1/q)(1-1/q^2)... is not equal to 1; it's some real number strictly between 0 and 1. This suggests that the Euler characteristic of the category of nontrivial finite-dimensional vector spaces over F_q is NOT rightfully 1, even though there is an adjunction between this category and a category that should have Euler characteristic 1 (see my earlier answer). Very interesting. $\endgroup$ – Tom Leinster Oct 22 '09 at 5:23

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