Infinite room, finitely represented
Abstract. Freudenthal suspension theorem, Whitney’s embedding theorem, the stabilization hypothesis. They all root in the same simple mathematical observation: infinite room can be finitely represented.
Ok, I admit it: really, the title of this article should be “infinite-dimensional room, finite-dimensionally represented”.
The context
At one point Emily Riehl told me (that a friend of her’s told her) that
As long as you never run out of money you may as well assume that you have infinite money.
This is not financial advice, but it turns out this also applies to compositional mathematics. In this short article we’ll revisit through a simple example how $n$-dimensional stuff behaves the “same” in infinite ambient dimension as in $2n + 2$ dimensions. In this way, we can indeed represent “infinite codimensions with finite codimensions”. This observation underlies a variety of well-known phenomena in mathematics: Freudenthal suspension theorem, Whitney’s embedding theorem, the stabilization hypothesis, … .
Ok, so where are the $n+2$ codimensions coming from?
The example
Let’s start with $S^0$, but this of its two points as colored in blue and red, respectively. Embed $S^0$ in $\mathbb{R}^1$. Notice that there are two different embeddings up to isotopy, and they are definitely not isotopic: one embeds the blue point before the red, the other the red before the blue. Ok, so $\mathrm{dim} = 0 + 1$ did not work: not all embeddings are “the same” (i.e. isotopic). What about $\mathrm{dim} = 0 + 2$? Indeed, now we get to make use of the first categorical isotopy (the braid) to see that all embedding into $\mathbb{R}^2$ are indeed isotopic. We win!
Ok, let’s draw these braids: the braid comprises two wires (1-dimensional manifolds) in dimension 3 (obtained as the product $\mathbb{R}^2 \times \mathbb{R}^1$; the latter component can be thought of as a “time” dimension in which the braid progresses). We notice that there are in fact two possible braids we could consider: one with the red wire in front and blue in the back, and the other with the blue wire in front and the red in the back. Are these two isotopic? The answer, unfortunately, no.
Let’s reduce dimensions for a moment, but considering the central slice of the braid, where both points of $S^0$ “cross path”. Because they cross paths, let’s in fact look at the interval $\mathbb{R}^1$ at the central height of the respective slices. We find that we are back at the start of our story: $S^0$ embedded in $\mathbb{R}^1$! Of course, we already know that this is not going to work: we cannot isotop the two embeddings. But we already now the solution to our problem: add one more dimension!
This brings us to considers braids in dimension 4, i.e. $2 \cdot 1 + 2$. Indeed, now the two braids become isotopic! But it is not the end of the story … once more there are two such isotopies which are non-isotopic. At this point, we observe the induction: whenever we add one dimension to the isotopy, we need to add two ambient dimensions to ensure everything stays isotopic. This recovers our initial formula of having to work in dimension $2n + 2$ (i.e., codimension $n + 2$) in order to see $n$-dimensional things behave as if they where embedded in infinite codimension.
Of course, this example is just meant to provide a bit of intuition for why we need exactly $n + 2$ codimensions… it all starts with the braid!