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# 6. Discussion

This paper developed techniques for analyzing the internal
structure of distributed measurements. We introduced
entanglement, which quantifies the extent to which a
measurement is indecomposable. Entanglement can be shown
to quantify context-dependence. Moreover, positive^{} entanglement
is necessary for a system to generate more information than the
sum of its subsystems. Along the way, we constructed the quale,
which geometrically represents the compositional structure of a
distributed measurement. The information-theoretic approach
developed here is dual, in a precise sense, to the algorithmic
perspective on computation. Studying duals ${\mathfrak{m}}^{\natural}$
instead of mechanisms ${\mathfrak{m}}$ shifts the focus from *what*
the algorithm does to *how* it does it: instead of
analyzing rules we analyze functional^{} dependencies.

The intuition driving the paper is that the structure presheaf^{}
${\mathcal{F}}$ is an information-theoretic analogue of a tangent
space. A particle moving in a manifold $X$ defines a vector
field – a section^{} of the tangent space to $X$, which is a
sheaf. The tangent vector at a point depends on the particle’s
location at “nearby time-points”: it is computed by taking
the limit of difference^{} in positions at $t$ and $t+h$ as
$h\rightarrow 0$. Similarly, a system performing a measurement
generates a quale, a section of the structure presheaf
consisting of “nearby counterfactuals”. The quale is computed
by applying Bayes’ rule to determine which inputs could have led
to the output.^{1}^{1}A counterfactual input is “nearby” to
an output if it causes (leads to) that output. How far this
analogy can be developed remains to be seen.

Entanglement can be loosely considered as an
information-theoretic analogue of curvature: the extent to
which interactions within a system “warp” sections of
${\mathcal{F}}$ away from a product structure. A related approach
to geometrically analyzing the complexity of interactions was
proposed in [1]. In fact, this project began as an
attempt to reformulate [2] in terms of sheaf
cohomology using ideas from [1]. We failed at the
first step since the structure presheaf is not a sheaf.
However, the failure was instructive since it is precisely
the *obstruction* to forming a sheaf that is of interest
since it is the obstruction (entanglement) that quantifies
indecomposability and context-dependence, and only systems
whose measurements are entangled are able to generate more
information than the sum of their subsystems.

# References

- 1
N Ay, E Olbrich,
N Bertschinger & J Jost
(2006):
*A unifying framework for complexity measures of finite systems*. In: Proceedings of ECCS06, European Complex Systems Society, Oxford, UK, pp. ECCS06–174. - 2
David Balduzzi & Giulio Tononi
(2009):
*Qualia: the geometry of integrated information*. PLoS Comput Biol 5(8), p. e1000462, doi:10.1371/journal.pcbi.1000462.

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## Comments

## Enlanglement as a basic modeling idea

Thanks for the references. Thomas Edison scaled down the building blocks of electricity to economically feasible units. Edison’s first scaled model was 1/3 copper, cost-wise. The scaling down of an economic system to finite units was first implemented by Middle Kingdom Egyptians, a finite unit fraction methodology that has not been fully decoded. Greeks used the finite model. Arabs and medieval scribes modified the multiplication

^{}scaled model to a subtraction model, a system that ended with Galileo’s inverse^{}proportion square root method.## Fermat's theorem in k(i) (c0ntd)

Numbers of the type 4m+1 are not prime in k(i). However their factors are prime. Example: 5 = (2+i)(2-i). If we take one of these as the base Fermat’s theorem works with respect to other co-primes of the type 4m+1. Examples: a) ((2-i)^12-1)/13 = 904 - 792i b)((2+i)^16-1)/17 = 9696 + 20832i

## Fermat's theorem in k(i) (c0ntd)

In one of my recent messages I had stated that there are four unities in k(i) viz 1, -1, i and -i. Fermat’s theorem holds true in k(i) when we keep this in mind. For example (2+I) and 3 are co-prime. Hence ((2+I)^4 + 1)/3 = -2 + 8i. I will be giving a few more examples in the next message.