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I have changed
An object $D$ whose reduced reflection is the terminal object, $Red(D) \simeq *$ is an infinitesimally thickened point.
to
… whose reduced coreflection… $\Re(D)\simeq *$
I hope this is correct.
Good point, thanks.
Have expanded the idea section and the Examples section in order to highlight some points, prompted by our discussion in another thread.
(Actually I have replaced the single sentence that used to be in the Idea-section, which wasn’t much good.)
I came across Mike’s remark from when we temporarily migrated to google groups:
The coreduced objects are the ones with “no infinitesimal behavior”, and the reduced objects are the ones “whose infinitesimal behavior is determined by their non-infinitesimal behavior”. A reduced object does contain infinitesimal points; what it lacks are “purely infinitesimal directions” while a coreduced object has no infinitesimal points,
so thought it worth adding that to a section ’Contrast between reduced and coreduced objects’.
I’m still struggling with a useful imagery to think of the equivalence between spaces which are purely reduced and spaces which are purely coreduced, in the same kind of way that one can think of a set equipped with the discrete, then the codiscrete, topology as providing an equivalence between those “pure moments”, any space being poised between those pure versions.
Is there a nice visualisable example of a non-reduced, non-coreduced space with map to it from reduced version, and map from it to coreduced version?
Is there a nice visualisable example of a non-reduced, non-coreduced space with map to it from reduced version, and map from it to coreduced version?
The canonical example of a non-reduced, non-coreduced space is the formal neighbourhood of any manifold inside any other.
Or simpler even: if we write $\mathbb{D}^1 = Spec(\mathbb{R}[x]/x^2)$ then for every manifold $X$ then $X \times \mathbb{D}^1$ is neither reduced nor coreduced.
The reduction of $X \times \mathbb{D}^1$ is $X$ (remove the infinitesimal extensions). The co-reduction is $X_{dR}$, the de Rham stack of $X$, where all infinitesimal neighbour points in $X$ are identified.
The adjoint modality opposing reduced and coreduced objects is $\R \dashv \Im$.
Ok, thanks. So perhaps what I don’t quite have is what happens in the passage across the adjoints:
$(i_! \dashv i^* \dashv i_* \dashv i^!) : \mathbf{H} \stackrel{\overset{i_!}{\hookrightarrow}}{\stackrel{\overset{i^*}{\leftarrow}}{\stackrel{\overset{i_*}{\hookrightarrow}}{\underset{i^!}{\leftarrow}}}} \mathbf{H}_{th}$$i^{\ast}$ takes the infinitesimal path groupoid of $X \times \mathbb{D}^1$ which evidently connects elements with the same $X$-coordinate. But does it also connect elements differing infinitesimally in the $X$-direction? And this gives us some (non-formal) space?
Then $i_{\ast}$ just treats this non-formal space as a formal space (without infinitesimals), while $i_!$ has to fill in infinitesimals within the $X$?
I guess I’m looking for something as simple as: start with a topological space, forget the topology, then either impose discrete or codiscrete topology.
actually $i^\ast$ just forgets that spaces may be probed by infinitesimal test spaces. This makes all infinitesimal extension disappear, because infinitesimal extension can only be seen by infinitesimal test spaces, not by finite test spaces.
Now $i_!$ simply embeds spaces without infinitesimal extension canonically into the more general “synthetic differential spaces”. That’s why reduction is
$\Re = i_!\circ i^\ast$(forget infinitesimal extension, then re-embed canonically).
To see that the other re-embedding is co-reduction
$\Im = i_! \circ i^\ast$now simply use the adjunction property $\Re \dashv \Im$: to see what probing $Im X$ by some $\mathbb{R}^n \times \mathbb{D}^k$ is we use that by adjunction morphisms
$\mathbb{R}^n \times \mathbb{D}^k \longrightarrow \Im X$are equivalent to morphisms
$\Re(\mathbb{R}^n \times \mathbb{D}^k) = \mathbb{R}^n \longrightarrow X \,.$So the finite probes of $\Im X$ are the same as those of $X$, but all infinitesimal probes of $\Im X$ are simply constant: the only infinitesimal paths in $\Im X$ are constant paths.
That we say “co-reduced” for this is quite non-standard and is only motivated from the adjunction $\Re \dashv \Im$.
Thanks! So good policy as usual to look to general spaces as things consistently probed.
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