Why the sphere spectrum is more correct than $\mathbb{Z}$?

Question

One may argue that $\mathbb{S}$ is more correct than $\mathbb{Z}$. Can anyone make it more explicitly? For example, what information will be lost if we work in $\mathbb{Z}$ instead of $\mathbb{S}$？

Chromatic homotopy theory may give a partial answer. I know that by chromatic homotopy theory, the sphere spectrum has more information than integer which is known as $v_n$ self map. It turns out that sphere spectrum has infinity Krull dimension. But I am a beginner of chromatic homotopy theory so I cannot tell the whole and true story. Any explanation of the advantages of $\mathbb{S}$ by chromatic homotopy theory will be appreicated.

BTW, There is a joke on the internet, shown in the picture.

Answer 1

For this to work, it is best to identify connective spectrum with spaces equipped with a group-like $E_\infty$-algebra structure (these are equivalent).

From this point of view:

• $\mathbb{Z}$ is the free abelian group on one generator.

• The sphere spectrum $\mathbb{S}$ is the free group-like $E_{\infty}$-space on one generator.

Similarly:

• $\mathbb{Z}$ is the initial ring, so the initial 0-truncated (connective) ring spectrum.

• $\mathbb{S}$ is the initial (connective) ring spectrum.

So, if you (like many people working in homotopy theory and/or higher category theory and/or homotopy type theory) think that spaces are the real fundamental objects and sets are just the reflective subcategories of 0-truncated space, then the role usually played by $\mathbb{Z}$ in traditional set-based mathematics is now played by $\mathbb{S}$, and $\mathbb{Z}$ only appear as the $0$-truncation of $\mathbb{S}$.

For example, some people have argued that one way to do algebra and geometry "below Spec $\mathbb{Z}$" (in the spirit of "the field with one element") was to do algebra and geometry over the sphere spectrum (see here or here for an example of this - but this is a fairly common idea)

Answer 2

An elementary answer to the first part of your question: Finite sets are more fundamental than their cardinalities.

Consider the category of finite sets and bijective functions. Its geometric realization (= nerve, or classifying space) has the homotopy type of $\coprod_{n\ge0} B\Sigma_n$. Forgetting the choices of bijective functions, and only remembering their existence, gives a map to the set $\mathbb{N}_0 = \{n \ge 0\}$ of non-negative integers. This way the symmetries of a finite set are disregarded: you can either fix or transpose the two elements in $\{a, b\}$, whereas the number $2$ does not intrinsically come with such structure.

Soon you want to add and multiply finite sets, using disjoint union and cartesian product, and these operations on $\coprod_{n\ge0} B\Sigma_n$ induce the usual sum and product in $\mathbb{N}_0$. We now have a map of semirings.

Shortly thereafter you want to solve equations, and need to subtract. For this, you need to ring complete the addition in $\coprod_{n\ge0} B\Sigma_n$, while preserving the multiplication (see Remark A), and the resulting ring space has the homotopy type of $\text{colim}_k \, \Omega^k S^k = \Omega^\infty \mathbb{S}$, with the ring space structure coming from the ring spectrum $\mathbb{S}$, the sphere spectrum. This now maps to the ring completion of $\mathbb{N}_0$, namely the integers $\mathbb{Z}$.

The first thing that is lost under $\mathbb{S} \to \mathbb{Z}$ is the image of the transposition of $a$ and $b$, i.e., the Mobius strip/line bundle over the circle, which corresponds to the loop in $\Omega^2 S^2$ given by the Hopf fibration $\eta \colon S^3 \to S^2$.

Remark A: One way to resolve the caveat raised in

Thomason, R. W.
Beware the phony multiplication on Quillen's A−1A.
Proc. Amer. Math. Soc. 80 (1980), no. 4, 569–573. 

is given in

Baas, Nils A.; Dundas, Bjørn Ian; Richter, Birgit; Rognes, John
Ring completion of rig categories.
J. Reine Angew. Math. 674 (2013), 43–80.