Mochizuki's Gaussian Integral Analogy In his latest 115-page overview, Mochizuki spends some time explaining "alien copies" by the analogue of evaluating the Gaussian integral by squaring it and introducing a second variable/dimension. In sections 1.1 to 1.7 (pages 5-10) he goes through this, and then in section 3.8 (pages 89-94) he revisits the analogy.
I have two questions about this. Firstly, one gets that $I^2=(\int_{-\infty}^\infty e^{-x^2}\,dx)^2=\pi$, and by positivity one finds $I=\sqrt\pi$. What is the analogue in the IUT setting of "positivity" here? Do the choices up to a unit not matter somehow?
Secondly, I wonder about the scope of alien copies. At the end of the day, if I read 3.8 correctly in #10 and #11 in the table on page 92 and the computations in 3.7(iv) (page 89), the main result of the analogy is to evaluate (to first order) the left side of the inequality at the top of page 87.
So my question is: are alien copies "only" useful for this Gaussian-like evaluation, or do they play a role throughout? (NB: the word "alien" does not appear in the text from pages 22-90.) Alternatively, do they also enlighten the derivation of the inequality (see 3.7(ii) starting on page 85) in the first place, or the estimation of its right side?
Since no one has answered yet, maybe I'll add a third question. In 2.2 on page 11, a distinction (quite important I gather) between étale-like and Frobenius-like structures is made, the former being "zero mass" objects, such as (1.4) the differentials "$dx\,dy$" in the Gaussian evaluation. Are there Frobenius-like structures with the Gaussian analogue, and if so, what are they?
 A: For all the similarities, there's a significant difference between both parts of the analogy.
In the IUT side of the analogy, we are trying to compare the "mutually alien copies" at each end of the $\Theta$-link
$$\{\underline{\underline{q^{j^2}}}\} \longleftrightarrow \underline{\underline{q}}\tag{1}$$
But in the case of the Gaussian integral, we are introducing the second (alien) copy in order to evaluate the first, not to compare them.
$$\bigg(\int_{-\infty}^\infty e^{-x^2}\,dx\bigg)\bigg(\int_{-\infty}^\infty e^{-x^2}\,dx\bigg)\longleftrightarrow \quad?$$
So at some point we need to reverse the process. That's not case in IUT.
To be more precise, the alien copies in question are $\Theta^{\pm\text{ell}}\text{NF}$-Hodge theaters (denoted $\bullet$ in the diagram below) in the domain/codomain of the $\Theta$-link, and the analogy is between a pair of such a Hodge theaters and one evaluation of the Gaussian integral. But there is a infinite lattice of such a pairs at work in IUT, the log-theta lattice.

This process doesn't have to be undone in any way, in order to evaluate individual elements of the lattice, the way we need to go from $I^2$ to $I$.

The "Gaussian-like evaluation" is basically the main result of IUT, so I would say that it appear throughout and at the same time it is "only" useful for this.
A different instance of the use of alien copies would be Mochizuki's theory of mono-anabelian reconstruction. In that case we have Galois theaters and $\mathfrak{log}$-Frobenius functors between them,
$$\mathfrak{log}:\mathfrak{Th}_\mathbb{T}^\bullet [Z] \longrightarrow \mathfrak{Th}_\mathbb{T}^\bullet [Z]$$
The derivation of inequality 3.7(ii) follows from $(1)$ and the Kummer isomorphisms [IUTechIII Theorem A, (ii)], which yield
$$-|\log (\underline{\underline{\Theta}})| \leq -|\log (\underline{\underline{q}})|$$
All of this involves of course the log-theta lattice (i.e. the "alien copies").

I think the analogy is supposed to be something like this:
$$\text{zero mass} \longleftrightarrow \text{étale-like} \longleftrightarrow dx\,dy$$
$$\text{positive/net mass} \longleftrightarrow \text{Frobenius-like} \longleftrightarrow \lim \sum dx\,dy$$
