This question might be vague and not thought-through enough.

If we have a real positive number $x$, we can start to write it as a continued fraction: $x = a_0 + \frac{1}{x_1} , \ldots , x_n=a_n + \frac{1}{x_{n+1}}$ where $a_i$ are non-negative integers, $x_i$ non-nogative real numbers less than $1$. So we can write $x=[a_0,a_1,\ldots]$. But we also may write $x=[a_0,\ldots,a_{n-1},x_n]$, i.e. we might decide that our continued fraction is finite, allowing the last term to be non-integer.

If we have a module, we can start taking a projective resolution of it. Again, we can take an infinite projective resolution, or decide to truncate it at some finite level, but then the last term will not be maybe projective.

The last term will be projective if our module was "good", i.e. of small enough (in particular, finite) cohomological dimension. In the continued fraction setting, the last term will be integer if our number was "good", i.e. rational...

Is there any (wild?) relation between rational numbers among real numbers, and modules of finite cohomological dimension among modules?


  • $\begingroup$ Why is this community wiki? Did I press something? $\endgroup$ – Sasha Feb 14 '12 at 19:53
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    $\begingroup$ This is a wild question ;-) $\endgroup$ – Fernando Muro Feb 14 '12 at 20:36
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    $\begingroup$ But intriguing nonetheless... $\endgroup$ – Todd Trimble Feb 14 '12 at 21:24
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    $\begingroup$ You can say a Taylor expansion is the same sort of thing, each step being $f_{n+1}=\frac{f_n-f_n(0)}{x}$ $\endgroup$ – Vanessa Feb 14 '12 at 22:11
  • $\begingroup$ what about thinking about the rationals as seen in ZxZ/~ as a way of choosing representatives of a Z sub-module of ZxZ? Irrationals would be the lines through zero which don't touch any other point (or more graphically, those that escape from touching nontrivial submodules) $\endgroup$ – Roberto Mizzoni Feb 14 '12 at 23:18

Hi Sasha,

As I understand it your question have 2 parts:

  1. Is there some relation between continued fractions and projective resolutions?

  2. Is there some relation between rationality of a number and finite cohomological dimension of an object in an (abelian) category?

Now let me make those questions more concrete (and of course lose information):

Can one construct a natural example of an abilian category $C$ and a real invariant $f:C \to \mathbb R$, s.t.

  1. A projective resolutions of an object $X$ in $C$ will give a continued fraction representation of $f(X)$?

  2. $f$ will map finite cohomological dimension objects to rational numbers?

Here I think that the answer for 1 is no and for 2 is yes. Let me explain why:

  1. The recursive definition of projective resolution ${P_i}$ of an object $X$ is: $X_0=X$ and an exact sequence $0 \to X_{i+1} \to P_i \to X_i \to 0$ The recursive definition of continued fraction ${a_i}$ of a number $x$ is: $x_0=x$ and $x_n=a_n+1/x_{n+1} $. It is unlikely to find an invariant that maps an exact sequence $0 \to A \to B \to C \to 0$ to numbers satisfying $c=b+1/a$, one reason is that if $B=A+C$ then $0 \to C \to B \to A \to 0$ is also exact but the relation $c=b+1/a$ is not symmetric. You can try to go to derived categories this might be relevant for squark remark but I doubt that it will help here.

  2. Consider the Grotendic group $K(C)$ and it subgroup $K_0(C)$ generated by images of protective objects (it coincide with the one generated by images of finite cohomological dimension objects). In case that you category is monoidal, I believe that one can show that $K_0(C)$ is a subring. Now you just need to find an homomorphism $f:K(C) \to \mathbb R$ s.t. $f(K_0(C)) \subset \mathbb Q$. This sounds as a reasonable task. You can ask some experts in K-theory about it, you can also as this question in MO and label it as K-theory question. You can also ask Inna Zakharevich, (http://math.mit.edu/~zakh/) she is doing K-theory and she had worked with invariant that have to do with real and rational numbers. Good luck and keep us updated.

  • $\begingroup$ Thank you, this is interesting "crystalization" of the original question. $\endgroup$ – Sasha Apr 15 '12 at 12:05

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