+1 for not talking about ancient religions :) - One other remark that could be made is that, while in the theory $T$ we're assessing the truth of $G$, in a meta-theory (such as ZFC) the sentence is not literally $G$ but the (fully unpacked version of) of something like "$\mathbb{N}\models G$".

@Andrei Smolensky: yes, sure. But maybe Brauer Suzuki by that phrase was meaning "existence of Jordan canonical form [for all operators]"

I think (despite its etymology) the word "canonical" is not used as synonym of "decided by the canon" in mathematics, but pretty much as a synonym of "natural" (except perhaps nuances about natural transformations, in which case one uses "natural" instead of "canonical"...)

(Sorry I made a mess with editing the comment. I just meant maybe in Olsson's definition of MOD_O it's already implicit you've to only consider isos)

Strictly speaking, the whole stack MOD_O is a not stack of groupoids (in the fibers of MOD_O over a given scheme there are not just isomorphisms). (Unless this is already implicit in the definition by Olsson) So perhaps you have to take only the sub-fibered-category with only isos. Also, are you considering Hom(X,MOD_O) as morphisms of fibered categories (over Sch/S) or just morphisms of 'abstract' categories?

I realize I'm nitpicking a century late to the party :) but shouldn't the group in the example of Tom Church be $\mathrm{Aff}^{+}(\mathbb{R})$ (which is connected) instead of $\mathrm{Aff}(\mathbb{R})$?

(Or, if the topology you're putting on $X$ is not the one inherited from the plane, could you specify how you define it?)

If I understand well, $X$ would be homeomorphic to $\mathbb{R}^2$ (I assume you're considering an open strip, otherwise it's not homogeneous so it can't be a topological group). Do you have any explicit examples of such objects?

Do you have a specific definition for a group structure on $X$ or are you considering any group structure (provided it exists...)?

If I understand correctly, you defined $OG(2,7)$ as a set or a variety. What do you mean by "a representation of $OG(2,7)$"? Did you mean "a representation of $O(V;q)$" instead, or something like that?

So, when you inflate a surface curvature becomes "less dense" but "there's totally the same amount of it". This is an intuition about curvature (possibly obvious to many) that I didn't have before your question.

@Ali Taghavi: Maybe, strangely, if I saw "dVol" written instead of "ds" I wouldn't have overseen that :) , but you're totally right. I think also the following heuristic leads to cancellation. Curvature is the reciprocal of curvature radii, each of which scales as lenght $\int \sqrt{g(\gamma'(t),\gamma'(t))}dt$, $\rho_1\mapsto r^{1/2} \cdot \rho_1$ when $g\mapsto r\cdot g$ (same for $\rho_2$), so $\kappa \mapsto r^{-1} \kappa$. The volume form scales as $\sqrt{\det(g_{ij})}$ so $dVol \mapsto r \cdot dVol $. So this results in $\kappa dVol \mapsto \kappa dVol$

(p.s. of course also Timothy Chow's explanation is perfectly right, and equivalent)

@Elliott: I think it's more about domains of functions than about powers. Starting from $f:\mathbb{R}\to [0,\infty)$, $x\mapsto x^2$, restrict to $f_{+}:=f|_{[0,\infty)}$ and $f_{-}:=f|_{(-\infty,0]}$. Let the inverses be $g_{+}:=(f_{+})^{-1}$ and $g_{-}:=(f_{-})^{-1}$. Now $-1\in (-\infty,0]=\mathrm{dom}(f_{-})$ and $-1=g_{-}(f_{-}(-1))$. So far nothing strange. Then the false proof of my comment is explained by: $-1=g_{-}(f_{-}(-1))=g_{-}(1)\neq g_{+}(1)=1$. That is, the $(\ldots)^{1/2}$ in the notation is actually $g_{-}$ but $g_{+}$ is applied instead in the last passage.

(maybe I got the powers of $r$ wrong now, but the idea should stand)

I think my comment implies that, given a $\lambda\neq 0$, every other $\lambda'$ with the same sign is reachable by rescaling $g$.