What is the distance (metric) between $0$ and $\varepsilon$ on smooth real line in SDG/SIA?

Question

As is known, infinitesimals don't have concrete value (size) in SIA/SDG.

What is the distance (metric) between $0$ and $\varepsilon$ on smooth real line in SDG/SIA?

Thanks.

Answer 1

It is not possible in SDG/SIA to define a metric function $d: R^2 \rightarrow \{x \in R\mid x \geq 0\}$ in the traditional sense, i.e. as a function obeying $d(x,y) = d(y,x)$, $d(x,y) \leq d(x,z) + d(z,y)$, and satisfying that $x = y$ precisely if $d(x,y) = 0$.

If it was possible to define such a metric $d$ in the theory of SDG/SIA, then there would be a metric function in every model of SIA. So it suffices to show that some model of SDG fails to contain such a function.

The very first model constructed by Moerdijk and Reyes suffices. In that model, the category of smooth manifolds occurs as a full subcategory, with the role of the usual real line played by $R$, and with maps $R \rightarrow R$ corresponding 1-to-1 with everywhere-defined smooth functions $\mathbb{R} \rightarrow \mathbb{R}$, right down to their ring-theoretic (equational) properties.

We can regard a metric as a map $d: R^2 \rightarrow R$ by embedding $R_{\geq 0}$ to $R$ the obvious way. This would have to satisfy the following metric space properties, namely

1. $d(x,y) \geq 0$,
2. $d(x,y) = d(y,x)$,
3. $d(x,y) \leq d(x,z) + d(z,y)$, and
4. $x = y$ precisely if $d(x,y) = 0$.

The first and fourth properties will cause trouble.

Since this function has signature $R^2 \rightarrow R$, and the category of smooth manifolds occurs as a full subcategory of the model, this function arises from some map $\delta : \mathbb{R}^2 \rightarrow \mathbb{R}$.

First, observe that this map $\delta$ cannot take any negative values: identifying real numbers with constant smooth functions $\mathbb{R} \rightarrow \mathbb{R}$ we would get from a short argument that $d(c_1,c_2) < c_3 < 0$ holds in $R$ as well for some $c_i$, contradicting the first property. But if $\delta$ does not take negative values, its derivative has to be $0$ around the origin in every direction. If it wasn't, it would be very close to linear in a small neighborhood around $0$, and a linear function with nonzero slope and zero constant term takes negative values around zero.

Since $\delta$ has zero derivative around the origin in any direction, so does $d$. Choose an arbitrary infinitesimal $\varepsilon \in D$, set $f(x) = d(0,x)$ and use the Kock-Lawvere axiom to get that $d(0,\varepsilon) = f(\varepsilon) = d(0,0) + \varepsilon f'(0) = \varepsilon f'(0) = 0$. By the fourth property, we would obtain $\varepsilon = 0$, and since $\varepsilon$ was arbitrary, we could then conclude $\forall \varepsilon \in D. \varepsilon = 0$. That's a contradiction, since we already know that $\neg \forall \varepsilon \in D. \varepsilon = 0$ holds in SDG.

I don't know whether the existence of such a metric is consistent with SDG or not (although I doubt it): by considering this model, I merely show that the nonexistence of such a metric is consistent with SDG. As such, SDG does not let you define such a metric.

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Now, let's put this result in context. As I wrote elsewhere, the best visualization of the up-close structure of SDG's smooth real line is not as a bona fide straight line, but as a chain built from tiny metallic links. You may think of these tiny links as the nilsquare infinitesimal neighborhoods of ordinary numbers.

A function $f: R \rightarrow R$ defined on the smooth real line $R$ can twist and turn the chain links, but cannot break them. When we focus on a single chain link, we see that the transformation behaved linearly: the overall shape of the link hasn't changed, only its orientation. This is the content of the Kock-Lawvere axiom.

Under this picture, the preorder relation $x \leq y$ gets the following intuitive interpretation: the link containing the point $x$ lies to the left of the link containing the point $y$, unless they lie on the same link.

This explains why $0 \leq \varepsilon$ and $\varepsilon \leq 0$ both hold for all infinitesimals $\varepsilon \in D$, and yet you cannot conclude that $\varepsilon = 0$

I'd like to emphasize that this is not a formal definition: it is merely a way of building intuition about the line of Synthetic Differential Geometry.

We can define functions such as the absolute value that are not smooth: but to do that, we have to "remove a link of the chain". The domain of the absolute value function in SDG is not $R$, but $\{x \in R \mid x > 0 \text{ or } x < 0\}$: thanks to intuitionistic logic, this domain is not equal to $R$. Using this absolute value function, we can define a metric the usual way, as $d(x,y) = |x - y|$. But this metric is not a function $R^2 \rightarrow R$! Instead, it's a function $d: \{(x,y) \in R^2 | x > y \text{ or } x < y \} \rightarrow R$. As long as two points are on different chain links, we can define their distance: when two points are so close that they occupy the same chain link, we can't.

We don't get a "metric distance" between $0$ and $\varepsilon$ this way; they are infinitesimally close, so close that we can't tell whether one lies to the left of the other, much less measure the distance between them. This has some unexpected consequences in the way geometric reasoning works, in that only those geometric constructions can be carried out which "work uniformly" even on degenerate shapes; but this is in fact very reasonable from the perspective of the philosophy surrounding SDG.