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Can someone complete the following table?

$\begin{array}{cc} \text{Topology over } \mathbb{R} & \text{Topology over } \mathbb{C} & \text{Algebraic Geometry} \\\\ \hline \mathbb{R} & \mathbb{C} & \mathbb{A}^1 \\\\ \mathbb{R} P^1 \cong S^1 & \mathbb{C} P^1 \cong S^2 & \mathbb{P}^1 \\\\ {}D^1(\mathbb{R}) \cong [0,1] & D^1(\mathbb{C}) \cong [0,1] \times [0,1] & ? \end{array}$

I agree that this question is quite vague. Possible precise interpretations would be:

Is there a scheme $X$ satisfying $X(R) = D^1(R)$ for nice metric rings $R$? Or: Does the endofunctor $(X,x_0,x_1) \mapsto (X \cup_{x_0 \simeq x'_1} X',x_0,x'_1)$ (with a copy $X'$ of $X$) of the category of $k$-schemes (or locally ringed spaces over $k$) equipped with two $k$-rational points have a terminal coalgebra, similar to Freyd's characterization of $[0,1]$?

But I don't want to limit my question to this interpretation. I also would be happy with an algebraic space or stack corresponding to $[0,1]$. Note that $[-1,1]$ "is" the generalized ring and therefore affine scheme $|\mathbb{Z}_{\infty}|$ à la Durov defined as an algebraic submonad of $\mathbb{R}$.

Background: In $\mathbb{A}^1$-homotopy theory the affine line plays the role of the unit interval, right? But when we define homotopies between morphisms of schemes $X \to Y$ in the naive way as morphisms $X \times \mathbb{A}^1 \to Y$, we won't get a transitive relation. The reason is that, although $\mathbb{A}^1$ has two distinguished rational points $0,1$, we don't get $\mathbb{A}^1$ when we glue two copies of $\mathbb{A}^1$, as in the case of the unit interval.

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    $\begingroup$ 'Most' model categories do not have an interval. The idea that motivic homotopies are defined from a product with the affine line may be helpful for some purposes, but also misleading sometimes, as you suggest. Alternatively, think that motivic homotopy theory is defined by forcing the affine line to be contractible. Contractible objects may be used as 'parameter spaces' for homotopies, but while there are many contractible objects in model categories, there is seldom something as good as the interval in topology. $\endgroup$ Jun 15, 2013 at 15:09
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    $\begingroup$ @Fernando: Thank you for the comment. @Jacob: You are right, first we should find a complex interval ... roughly my question is about how to define an interval over any commutative ring. $\endgroup$ Jun 15, 2013 at 16:01
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    $\begingroup$ Dear @Martin, What is $D^1$? $\endgroup$ Jun 15, 2013 at 20:28
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    $\begingroup$ Oh, I see. Then at least in rigid geometry over a non-Archimedean field $K$, the ring $K\langle T\rangle$ of restricted formal power series (power series whose coefficients form a zero sequence) represents, on the category of $K$-Banach algebras, the functor $A\rightsquigarrow A^\circ$, where $A^\circ$ is the set of power-bounded elements, i.e., elements $a\in A$ with $\vert a^n\vert$ bounded as $n\rightarrow\infty$. When $A$ is a finite extension of $K$, this is just the closed unit ball. In fact $\mathrm{Sp}(K\langle T\rangle)$ is called the rigid analytic closed unit ball over $K$. $\endgroup$ Jun 15, 2013 at 20:49
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    $\begingroup$ But this probably isn't what you want because it isn't really algebraic. $\endgroup$ Jun 15, 2013 at 20:50

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You can define a unit interval $I$ as a co-presheaf: the set of maps from $I$ to a connected scheme $X$ is the set of triples $(x,y,\phi)$, where $x$ and $y$ are geometric points in $X$, and $\phi$ is a natural isomorphism between the corresponding fiber functors (from the category of finite étale covers of $X$ to finite sets).

I'm pretty sure this is not co-represented by a scheme (for example it needs to have maps to the spectra of fields of arbitrarily large transcendence degree), but it does have nice composition properties.

There is a discussion of objects similar to this in section 10 of Deligne's "Le Groupe Fondamental de la Droite Projective Moins Trois Points", but I think there may be later papers that are more explicit.

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    $\begingroup$ Since non-empty schemes have lots of geometric points, there is some set-theoretic subtlety in defining $I$. One can always restrict inputs to low levels of $V$ to get a more digestable but slightly less universal object. $\endgroup$
    – S. Carnahan
    Jun 16, 2013 at 15:37
  • $\begingroup$ I understand the idea of this definition, but somehow it is cheating. It only says what $I$ should do, but doesn't present $I$ as an object of algebraic geometry in the usual sense. Also, as already noted in the comment, there are even set-theoretical obstructions for this point of view. $\endgroup$ Jul 3, 2013 at 13:22
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The spectrum of a standard affinoid algebra (in one variable)?

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  • $\begingroup$ What kind of spectrum do you use here? $\endgroup$ Jun 16, 2013 at 7:40
  • $\begingroup$ The kind that lands in the category of rigid analytic spaces. I guess that's not what you wanted: you wanted something that lands in the category of schemes. Sorry. $\endgroup$ Jun 16, 2013 at 8:38
  • $\begingroup$ It's the max spectrum, the set of maximal ideals. $\endgroup$ Jun 16, 2013 at 14:28
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    $\begingroup$ @Keenan: Well... it's kind of lame to land in the category of sets ;-) $\endgroup$ Jun 16, 2013 at 16:12
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This is really more deserving of being a comment(but I don't have enough points to make them), check out this MO post What properties make $[0,1]$ a good candidate for defining fundamental groups?

Perhaps you will find some inspiration in terms of other ways to think about the role of $[0,1]$ and what its analogue should be in scheme-land.

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