# From the perspective of bordism categories, where does the ring structure on Thom spectra come from?

To fix ideas, let's consider the Thom spectrum of framed bordism $M$, the spectrum whose homotopy groups are the framed bordism groups. $M$ has a ring spectrum structure inducing the product of manifolds on its homotopy groups. By Pontryagin-Thom, $M$ is the sphere spectrum $S$, which is even the initial ring spectrum.

This fact in some sense lifts to the cobordism hypothesis as follows. A suitable version of the cobordism hypothesis should assert that the framed bordism $\infty$-category $\text{Bord}^{fr}$ is the free $E_{\infty}$ $\infty$-category with duals on a point. After inverting all of the morphisms in $\text{Bord}^{fr}$ suitably, we get an $E_{\infty}$ $\infty$-groupoid, or essentially an infinite loop space, which should be the underlying space of the Thom spectrum $M$.

This tells me what $M$ looks like as a spectrum. But the higher categorical story, so far as I know, doesn't immediately tell me what $M$ looks like as a ring spectrum; I don't see how to put a familiar-looking structure on $\text{Bord}^{fr}$ inducing the ring spectrum structure on $M$.

Is there such a structure?

• How about the cartesian product in Bord^fr. – Ilias A. Jul 9 '14 at 19:51
• Some technical comments: First, the version of the cobordism hypothesis you seek is one for $(\infty,\infty)$-categories. I don't think this has been developed: The usual formulation cuts off at the $(\infty,n)$-level, declaring that all higher morphisms are isotopies of cobordisms. You seem to want all higher cobordisms, because you in particular want to capture the Thom spectrum associated to framed cobordism (which see all higher framed cobordisms between $\emptyset$ and $\emptyset$, for instance). Then the ring spectrum structure you want seems to correspond to a strange property... – Hiro Lee Tanaka Jul 9 '14 at 20:15
• You can take a $k$-morphism and a $k'$-morphism to obtain a $k+k'$-morphism. In some sense this structure is already present for the $(\infty,n)$-category of cobordisms up to dimension $n$, but I don't know if there's a name for this structure. It seems like another notion for a symmetric monoidal structure on $(\infty,n)$-categories. – Hiro Lee Tanaka Jul 9 '14 at 20:18
• @Fedotov: what do you mean by that? Note that functors $\text{Bord} \times \text{Bord} \to \text{Bord}$ take as input, say, a $j$-morphism and a $j$-morphism and return another $j$-morphism, whereas the product structure we want, as Hiro said, takes a $j$-morphism and a $k$-morphism and returns a $j+k$-morphism. In particular the cartesian product can't be responsible for this structure (I don't think $\text{Bord}$ even has cartesian products), but more generally no bifunctor can. – Qiaochu Yuan Jul 10 '14 at 8:57
• Your observation seems to be correct about non existence of the cartesian product! – Ilias A. Jul 10 '14 at 15:09

Thanks to a very helpful discussion with Clark Barwick in the homotopy theory chat, I think I now understand what's going on here. In particular, the ring spectrum structure on the sphere spectrum $\mathbb{S}$ does come from a monoidal structure on $\text{Bord}$, but I was confused about how to transport this monoidal structure from the category to the spectrum.

The monoidal structure can be thought of as coming from the universal property of $\text{Bord}$: since it's the free symmetric monoidal $\infty$-category with duals on a point, the $\infty$-category of symmetric monoidal functors $\text{Bord} \to \text{Bord}$ can canonically be identified with $\text{Bord}$ itself, and hence $\text{Bord}$ naturally acquires a monoidal structure, which I'll call $\circ$, coming from composition of functors $\text{Bord} \to \text{Bord}$. This is exactly analogous to how the free abelian group $\mathbb{Z}$ on a point canonically acquires a ring structure, and compatible under group completion with how the free spectrum on a point, namely $\mathbb{S}$, canonically acquires a ring spectrum structure. Here we need to know that symmetric monoidal $\infty$-categories are enriched over themselves, but this ought to be true by analogy both with the case of abelian groups and with the case of spectra.

This is admittedly an indirect description. It's hard to attempt a more direct description because the resulting monoidal structure isn't all that interesting on objects, and trying to describe what it does on morphisms is what got us into this mess in the first place.

So let me trudge on. In the comments I explained that I thought a monoidal structure

$$\text{Bord} \times \text{Bord} \to \text{Bord}$$

couldn't induce the usual multiplication map on $\mathbb{S}$ because monoidal structures take an $n$-morphism and an $n$-morphism and return another $n$-morphism: for example, the disjoint union does provide a monoidal structure of this form (in fact it's the monoidal structure figuring in the universal property), and the induced map

$$\pi_n(\mathbb{S}) \times \pi_n(\mathbb{S}) \to \pi_n(\mathbb{S})$$

is the usual abelian group structure on $\pi_n(\mathbb{S})$.

The problem with this story as applied to $\circ$ is that, with the natural symmetric monoidal structures on both sides, $\circ : \text{Bord} \times \text{Bord} \to \text{Bord}$ is not a symmetric monoidal functor, so it does not induce a map $\mathbb{S} \times \mathbb{S} \to \mathbb{S}$ of spectra. This issue shows up already at the level of abelian groups: with the natural abelian group structures on both sides, the multiplication map $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ is not a homomorphism of abelian groups.

As suggested by the analogy to abelian groups, $\circ$ is "bilinear": it preserves disjoint unions separately in both variables. So the fix is to use a suitable notion of "tensor product" of symmetric monoidal $\infty$-categories (suitable meaning in particular that on symmetric monoidal $\infty$-groupoids, thought of as connective spectra, it reproduces the smash product) and think of $\circ$ as instead providing a symmetric monoidal functor

$$\text{Bord} \otimes \text{Bord} \to \text{Bord}$$

which reproduces the usual ring spectrum structure $\mathbb{S} \otimes \mathbb{S} \to \mathbb{S}$ (here I am using $\otimes$ for the smash product as well). Now to see how we get a multiplication map

$$\pi_n(\mathbb{S}) \times \pi_m(\mathbb{S}) \to \pi_{n+m}(\mathbb{S})$$

it suffices to recall that the smash product of $S^n$ and $S^m$ is $S^{n+m}$.

So, one way to answer the conceptual question "how, in this situation, did we start with an $n$-morphism and an $m$-morphism and get an $n+m$-morphism" is that the universal property of $\text{Bord}$ is extremely general: it naturally acts by endomorphisms on an object in any symmetric monoidal $\infty$-category with duals whatsoever, including the "loop spaces" of $\text{Bord}$ itself! The analogous statement in stable homotopy is that the stable homotopy groups naturally give rise to operations on the homotopy groups of any spectrum whatsoever, including the shifts of the sphere spectrum itself.