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I come from a statistics background, so we often frame things in the context of distributions and the parameters that characterize those distributions. So the most common parameter is the average value of some distribution.

Now, I was wondering whether something like a torus could have an average value? I am thinking of the simple torus that resembles a donut with a hole. That is the question, can we establish a notion of an average value for something like a torus, the same way that we can estimate an average for a 3 dimensional cloud of points?

We can apply the standard statistical algorithm to find the average. The tricky thing is that if I were to project the 3-d torus into 3 separate 1-dimensional projections, take the average value of each projection, and then recombine those averages into a 3-tuple of values, I would essentially obtain a point that was inside of the donut hole, right.

So the problem then, is that in a definitional sense we are saying that the average or central tendency of random points taken from a torus would be a point that is not in the torus--but in the hole of the torus. So this is a point that could never be reached, but which is used as the central tendency.

Now I can appreciate that a point outside of a set can represent or characterize the tendency of the set--in the same way that in point-set topology the limit of a convergent sequence can lie outside of the open set. So that is fine.

But I was wondering if there are any other ideas within perhaps differential topology or even measure theory/functional analysis which find a way to reconcile this notion of an average value with that value needing to be reachable from within the set?

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  • $\begingroup$ You could look for a point in the torus that minimizes the expected square distance to other points in the torus. I believe this is the same as minimizing the distance to the mean of the torus, which is satisfied by any point on the inner circle around the hole. Not sure if that's what you're asking about, though. $\endgroup$ – Chris Culter Aug 26 '18 at 6:43
  • $\begingroup$ That is an interesting idea. In that sense, the mean is no longer unique but does fall within the torus. I was really wondering about whether there was a more general sense of what a "mean" represents outside of stats or probability. Like does the concept of a mean map to some other concepts in more pure branches of mathematics. For example in stats we deal with transformations that preserve parameters such as the mean,variance of the data. But in a field like algebraic topology the permissible set of transformation or deformations would not necessarily preserve means and variances of objects. $\endgroup$ – krishnab Aug 26 '18 at 7:08
  • $\begingroup$ I think that you should be aiming to find an 'average' of the torus itself, rather than of a particular embedding into $\mathbb{R}^3$. $\endgroup$ – Tyrone Aug 26 '18 at 8:55
  • $\begingroup$ @Tyrone haha, very good point. I think I was coming at the problem from the sense of how do we find the mean of the torus, but also what types of transformations can I do on this object that preserve that mean. In stats/probability we are always looking at mappings between random variables where the density i.e., mean and variance (in the Gaussian case) are preserved. So I think a theory of the mean also to requires a definition of what transformations will preserve that mean. $\endgroup$ – krishnab Aug 26 '18 at 9:40
  • $\begingroup$ Your question does not address a key issue, namely: average with respect to what probability distribution on the torus? This is important whenever you ask for the "average" in a statistical sense. For example, if you asked me the "average" of the interval $[0,1]$, before answering "$\frac{1}{2}$", although I am almost willing to guess that you mean "average with respect to the probability distribution $dx$", I nonetheless would want to ask first. If you asked me the "average" of the half line $[0,\infty)$, then I certainly would need to know the probability distribution first. $\endgroup$ – Lee Mosher Aug 26 '18 at 15:06

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