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

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.

• What is your precise question ? Make what more explicit ? Commented Nov 4, 2022 at 10:17
• A joke? On the Internet? I'm not sure I believe that. Commented Nov 4, 2022 at 12:59
• Integers only have primes $p$. The sphere spectrum has primes "$p$", "really $p$", "really really $p$", "really really really $p$", ..., and "$p$ by any means". Commented Nov 4, 2022 at 14:07
• @მამუკაჯიბლაძე Can you explain (or point to an explanation of) what this thing about being “$p$”, “really $p$”, etc. means, to someone who just knows the definition of a spectrum? Commented Nov 5, 2022 at 18:41
• The comic says "natural numbers" ($\mathbb{N}$) and not "integers" ($\mathbb{Z}$) Commented Nov 5, 2022 at 21:01

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)

• What do you mean by group-like? Just that the additive structure has (homotopy) subtraction or something more than that? Commented Nov 4, 2022 at 14:20
• For $E_\infty$-algebra group-like mean "every element has an inverse", but it is enough to ask that the $\pi_0$ is a group, no higher coherence condition are needed thanks to the uniqueness of inverse (or rather all the higher conditions you may want to ask - like the existence of a map giving the inverse of an element, the existence of map specifying the homotopy $x^{-1} x = 1$, etc... will automatically be present for bifibrant objects). Commented Nov 4, 2022 at 14:28
• I really want to answer: the stable homotopy groups of sphere, but that would really fail to do justice to what is really going on. Doing arithmetic or algebraic geometry over $\mathbb{S}$ is really much much more rich and complicated than over $\mathbb{Z}$ (which is already very rich and complicated), so much so that most of it isn't fully understood yet. So, a lot is lost in many different directions, and people can probably give you a lot of examples, but probably not a general picture. Commented Nov 4, 2022 at 20:29
• In the early days of Waldhausen's work on algebraic $K$-theory, this idea was already present. In that context, the spectrum $A(X)$ for a based space $X\sim BG$ can be regarded as the $K$-theory spectrum of the group ring $S[G]$. This is much better than the $K$-theory of $\mathbb Z[\pi_0(G)]$, in that it "knows more" about geometric questions regarding $X$. Commented Nov 4, 2022 at 20:35
• This and other things (one example, related to $K$-theory, is Topological Hochschild Homology) led to the idea that "doing algebra" with $S$ rather than $\mathbb Z$ as the universal or ground ring was worthwhile. Commented Nov 4, 2022 at 20:38

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.