\[\DeclareMathOperator\THH{THH} \DeclareMathOperator\SL{SL} \DeclareMathOperator\GL{GL} \DeclareMathOperator\tr{tr} \DeclareMathOperator\hofib{hofib} \DeclareMathOperator\fib{fib} \DeclareMathOperator\cofib{cofib} \DeclareMathOperator\colim{colim} \DeclareMathOperator\Z{\mathbb{Z}} \newcommand\del{\partial} \newcommand\xto[1]{\xrightarrow{#1}} \renewcommand\S{\mathbb{S}} \newcommand\iso{\cong} \newcommand\equiv{\simeq}\]

In this part we will show the relative trace $\tr : K(\S\to\Z) \to \THH(\S\to\Z)$ is nonzero.

Part 1.1 - Approximation to relative $K$-theory

A repeated theme when working with these objects is to approximate $K(\S\to\Z)$ with the more well-understood space $B\SL_1\S$ – it’s more well-understood in the sense that its homotopy groups are the same as the homotopy groups of spheres (shifted by one), which are arguably not so well-understood at all, but enough for our purposes.

The comparison map arises on the left in the following diagram whose rows are fiber sequences

The algerbaic $K$-theory spaces $K(\S\to\Z),K(\S),K(\Z)$ can be identified, for example, as the $+$-construction of the bottom row. By definition, the $+$-construction preserves homology, so if we care about the effect of the map $B\SL_1\S \to B\GL_\infty(\S\to\Z)$ on first $p$-torsion in homotopy, by the Hurewicz theorem, we can look at the first $p$-torsion in homology.

The homotopy groups of $\SL_1\S$ in low degrees are known – in degree $0$ it’s trivial, and in positive degrees they coincide with the stable stem. Thus the homotopy of $B\SL_1\S$ is the stable stem shifted up by one. In particular Serre’s result, that the first $p$-torsion in $\pi_*\S$ is a copy of $\Z/p$ in degree $2p-3$, implies that the first $p$-torsion of $B\SL_1\S$ is a copy of $\Z/p$ in degree $2p-2$.

Clearly if we can show that the composite $B\SL_1\S \to K(\S\to\Z) \xto{\tr} \THH(\S\to\Z)$ is nonzero on $\pi_{2p-2}$, then the factor $\tr$ in it is also nonzero.

Part 1.2 - Approximation to relative cyclic $K$-theory

The trace map $\tr : K \to \THH$ by construction factors through the cyclic $K$-theory functor $K^{cyc}$, whose definition I’ll remind in a moment. We will see here that, analogously to our approximation of $K$ by $B\SL_1$, one can also approximate $K^{cyc}$ fairly well by the so-called cyclic nerve $B^{cyc}\SL_1$ (or rather, its free loop space). This latter space will be easier to compare to $\THH$ more directly (recall that $\THH$ is also defined in terms of a certain cyclic bar construction).

It is well-known that there’s an equivalence between $B^{cyc}R$ and the free loop space of the delooping $\Lambda BR$ natural in $R$ (ranging over a category of some sufficiently-associative monoids). Under this equivalence, we have a map $BR \to \Lambda BR$ sending a point to its constant loop, and $\Lambda BR \to BR$ evaluating a loop at its basepoint. The former is clearly a section of the latter, so the horizontal maps in the following diagram are split injective:

This is a commutative diagram of functors & natural transformations on ring spectra. We can specialize this once to $\S$ and once to $H\Z$, and by naturality these diagrams will have morphisms between them induced by $\S\to H\Z$. We shall take the fibers of all these morphisms, and get a relative diagram as follows

where we’ve used again the fact that $\SL_1\S$ is the fiber of $\GL_1\S \to \GL_1\Z$. As mentioned in the proof skeleton of Part 0, we will prove that the composite $B\SL_1\S \to K(\S\to\Z) \to \THH(\S\to\Z)$ is nonzero (in fact injective) by showing it for $B\SL_1\S \to B^{cyc}\SL_1\S \to \THH(\S\to\Z)$. Thanks to the split injectivity discussed above, the top horizontal face is ineed split injective in degree $2p-2$.

Part 1.3 - Easier computations

It remains to prove that the map $B^{cyc}\SL_1\S \to \THH(\S\to\Z)$ is injective in degree $2p-2$. Our proof of this will go kind of backwards: we compute $\pi_{2p-2}$ of both spaces, and find out that they’re both $\Z/p$. Then we use circle actions to construct a section, which makes this map split surjective. But any surjection from $\Z/p$ to $\Z/p$ is injective because it’s actually an isomorphism.

The computations are as follows. Keep in mind we are implicitly $p$-localizing all groups.

Lemma: \(\pi_{2p-2} \THH(\S\to\Z) \iso \begin{cases} \Z/p & p\text{ odd} \\ \Z/2\oplus\Z/2 & p\text{ even} \end{cases}\)

Proof: We’d like to use the long exact sequence corresponding to this relative $\THH$. Here is the relevant segment of it

\[\pi_{2p-1}\THH(\S) \xto{\ell} \pi_{2p-1}\THH(\Z) \xto{\del} \pi_{2p-2}\THH(\S\to\Z) \to \pi_{2p-2}\THH(\S) \xto{\ell} \pi_{2p-2}\THH(\Z)\]

Recall there’s an equivalence $\S \equiv \THH(\S)$, essentially because the latter is defined in terms of smash products of $\S$ with itself over itself. In fact, these smash products are only the higher simplices of the cyclic bar construction defining $\THH$ (more details in Part 1.5), so we see that all higher simplices are degenerate. Consequently, the equivalence is actually induced by a map $\S \to \THH(\S)$, realizing the inclusion of $0$-simplices. Therefore the linearlization map $\ell : \THH(\S) \to \THH(\Z)$ factors through the linearization on the level of $0$-simplices, which is just $\ell : \S \to H\Z$. The latter has trivial higher homotopy groups, meaning $\ell$ induces the zero map in all positive degrees.

We deduce that the two $\ell$’s in the displayed sequence are zero maps, and so the whole thing breaks up to SESs:

\[0 \xto{} \pi_{2p-1}\THH(\Z) \xto{\del} \pi_{2p-2}\THH(\S\to\Z) \xto{} \pi_{2p-2}\THH(\S) \xto{} 0\]

The $\THH$ of $\Z$ is known; in degree $2p-1$ it’s precisely $\Z/p$. Due to $\S \equiv \THH(\S)$ we also know what the third object is:

  • If $p$ is odd, the first $p$-torsion in $\pi_*\S$ is in degree $2p-3$, but the next $p$-torsion after that only appears in degree $4p-5$ which is $>2p-2$. In particular, the right-hand object is $0$.

  • If $p$ is even we get \(\pi_2\S = \Z/2\{\eta^2\}\). Hence we are looking at an extension of $\Z/2$ by $\Z/2$. This extension is split, thanks to the vanishing of $\ell$ described earlier (which occurs on the level of the homotopy category, modulo connected components). $\square$

A very similar computation happens for the cyclic bar construction.

Lemma: \(\pi_{2p-2}B^{cyc}\SL_1\S = \pi_{2p-2}\Lambda B\SL_1\S \iso \begin{cases} \Z/p & p\text{ odd} \\ \Z/2\oplus\Z/2 & p\text{ even} \end{cases}\)

Proof: Recall we have a split fiber sequence

\[\Omega B\SL_1\S \xto{} \Lambda B\SL_1\S \xto{} B\SL_1\S\]

The splitting means its long exact sequence breaks up nicely into SESs:

\[0 \xto{} \pi_{2p-2}\Omega B\SL_1\S \xto{} \pi_{2p-2}\Lambda B\SL_1\S \xto{} \pi_{2p-2} B\SL_1\S \xto{} 0\]

By the shifting effect of loopings & deloopings on homotopy groups, the left-hand object is $\pi_{2p-2}\Omega B\SL_1\S \iso \pi_{2p-1}B\SL_1\S \iso \pi_{2p-2}\SL_1\S$ and the right-hand is $\pi_{2p-2}B\SL_1\S \iso \pi_{2p-3}\SL_1\S$. The homotopy groups of $\SL_1\S$ in positive degrees are those of $\S$, hence with our knowledge of $p$-torsion in low degrees:

  • If $p$ is odd, this is an extension of $\Z/p$ by $0$, hence just $\Z/p$.

  • If $p$ is even, \(\pi_{2p-3}\SL_1\S = \pi_1\S = \Z/2\{\eta\}\) and \(\pi_{2p-2}\SL_1\S = \pi_2\S = \Z/2\{\eta^2\}\), and thanks to the existence of a section, the extension between them splits. $\square$

As outlined, our next aim will be to prove that the map $\Lambda B\SL_1\S \to \THH(\S\to\Z)$ is surjective in degree $2p-2$. This will be achieved over the following two subsections.

Part 1.4 - Circle actions

The fundamental important property of $\THH$ is its natural circle action,

\[\lambda : S^1_+\wedge \THH(R) \xto{} \THH(R)\]

As we already briefly mentioned in the previous subsection, the definition of $\THH(R)$ involves a simplicial set whose $0$-simplices object is $M_1R$. The functor $M_1$ is just another notation for the identity functor, emphasizing how in the current context we think about it as $1\times 1$ matrices. We also denote by $\lambda$ the same action restricted to $0$-simplices

\[\lambda : S^1_+ \wedge M_1R \xto{} \THH(R)\]

Meanwhile, the free loop space also has a circle action given by rotating loops

\[\mu : S^1_+ \wedge \Lambda B\GL_1R \xto{} \Lambda B\GL_1R\]

We can also restrict this action, along the composite $\GL_1R \to \Omega B\GL_1R \hookrightarrow \Lambda\GL_1R$;

\[\mu : S^1_+\wedge \GL_1R \xto{} \Lambda B\GL_1R\]

By functoriality, naturality, etc, these induce actions also on the corresponding homotopy fibers

\[\begin{gather} S^1_+\wedge M_1(\S\to\Z) \xto{} \THH(\S\to\Z) \\ S^1_+\wedge \GL_1(\S\to\Z) \xto{} \Lambda B\GL_1(\S\to\Z) \end{gather}\]

Earlier we already used the equivalence $\GL_1(\S\to\Z) \equiv \SL_1\S$, which holds simply because $\mathrm{GL}_1\Z$ is just \(\{\pm 1\}\) and $\mathrm{GL}_1\mathbb{S}$ is just two copies of $\mathrm{SL}_1\mathbb{S}$, one over each point. To determine $M_1(\S\to\Z)$ we look at the following square (and the fibers of its horizontal faces)

I claim that the square is a pullback, and hence the fibers are equivalent (to each other, thus to $\SL_1\S$). Indeed, we see this symbolically by the definition of $\GL_1$:

\[M_1\S \times_{M_1\Z} \GL_1\Z \cong M_1\S \times_{M_1\Z} (M_1\Z\times_{\pi_0(M_1\Z)} \pi_0(M_1\Z)^\times) \iso M_1\S \times_{\pi_0(M_1\Z)} \pi_0(M_1\Z)^\times \cong \GL_1\S\]

In the last step we’ve used the fact that linearization \(\ell : M_1\S \to M_1\Z\) induces a ring isomorphism on $\pi_0$.

To recap, we are interested in the two functors $\Lambda B\GL_1$ and $\THH$, and in their $0$-simplices functors $\GL_1$ and $M_1$, all with their circle actions, evaluated on both ring spectra $\S$ and $H\Z$, and all the fibers induced by the linearization map $\ell:\S\to H\Z$. All that fits neatly into the following commutative square, expressing the equivariance of the vertical maps wrt the circle actions:

the left face is the equivalence induced on the homotopy fibers of the pullback square considered above.

After this whole digression, we go back to our original goal here: proving that the right face of the square gives a surjection (hence bijection) on $\pi_{2p-2}$. To do so, we can in fact prove it has a section (not directly on the level of spaces, but only specifically after $\pi_{2p-2}$, or as we shall see, in a somewhat larger range).

The section will simply be the composite going all the way around the square

\[\THH(\S\to\Z) \xto{\lambda^{-1}} S^1_+\wedge\SL_1\S \xto{\equiv} S^1_+\wedge\SL_1\S \xto{\mu} \Lambda B\SL_1\S\]

Of course, $\lambda$ is not invertible in general, but we can show it’s an equivalence in a sufficient range of degrees. This will require a secondary digression into the circle action on $\THH$.

Part 1.5 - Technical digression

Warning: After writing this part, I realized the latter half of it is quite unreadable. I couldn’t be bothered rewriting it so feel free to skip – it doesn’t contain a lot of conceptual insights. Alternatively take it as a challenge to decipher all my convoluted arguments.

To establish the desired connectivity of $\lambda$ we will factor it through a different object, the “$\THH$ with coefficients”. Along the way we will also need yet another variant of the construction, so now is a good opportunity to remind the definition of $\THH$ itself, or at least, one of its many definitions.

My definition of choice is as the geometric realization of the cyclic nerve of a certain cyclic set. Briefly, the cyclic nerve is to a cyclic set is as the usual nerve is to a simplicial set. To get $\THH$, we take the geometric realization of the cyclic bar construction Let $(\mathcal{C},\otimes)$ be any symmetric monoidal $\infty$-category, and $R$ n associative monoid inside it.

\[\THH(R/\mathcal{C}) := \bigg|{R \leftleftarrows R\otimes R\Lleftarrow R\otimes R\otimes R \Lleftarrow \cdots}\bigg|\]

The two maps $R^{\otimes 2} \to R$ are the multiplication in both orders. The three maps $R^{\otimes 3} \to R^{\otimes 2}$ are those that multiply the first & second factors, or the second & third factors, or the third & first factors by “rotating” the third to the front. You also see that every $R^{\otimes k}$ has an action of the cyclic group $C_k$. These cyclic groups together form a cyclic object whose geometric realization is $S^1$:

  • $C_1$ has one element, representing the unique vertex.
  • $C_2$ has two elements, an edge from the unique vertex to itself (forming the circle) and another degenerate edge.
  • $C_3$ contains only degenerace simplices.
  • and so on…

That’s where the circle actions come from. Making all of this rigorous in the context of $\infty$-categories takes some more effort, but can be found e.g. in Nikolaus & Scholze’s paper on topological cyclic theory.

If $\mathcal{C}$ is the category of spectra with its usual smash product structure, I will just write $\THH(R)$.

In the usual bar construction, if your monoid $R$ acts from the right on some module $X$, then you can form a variant of the bar construction by letting the $k$th object be $X\otimes R^{\otimes k}$ and the multiplication $X\otimes R \to X$ is given by the module action. To do the same thing for the cyclic nerve, we need to be able to act on $X$ from the left also, for the case of rotating the last factor back in front of it. Thus if $X$ is an $R-R$-bimodule, then we write $\THH(R;X)$ for the geometric realization of this cyclic object.

Suppose now you have a map $Q\to R$. Then even more generally, let $G_k(R;Q)$ denote the colimit of all the products \(\bigotimes_{i=0}^{k}{\left\{\begin{matrix} R & \text{if }i\in s \\ Q & \text{if }i\notin s \end{matrix}\right\}}\) for every proper subset \(s \subsetneq \{0,\cdots,k\}\). The maps in this diagram are those that apply the morphism $Q\to R$ on any factor. It’s not hard to see that $G_k(R;Q)$ is again a cyclic object, so write $G(R;Q)$ for its realization (this is the same notation as Böckstedt-Madsen. I don’t know why they chose the letter $G$).

We can use this construction to re-express relative $\THH$ a bit more explicitly.

Proposition: $\THH(R\to T) \equiv G(R;M_1(R\to T))$ where, we recall the relative $\THH$ is defined by $\THH(R\to T) = \fib(\THH(R)\to\THH(T))$.

Indication of proof: Until now I’ve been a bit vague about when I work with spectra and when I work in infinite loop spaces. In this case it’s crucial to work with spectra, so that fiber sequences coincide with cofiber sequences (of course once you have a fiber sequence, you can map it through $\Omega^\infty$ which is a right adjoint, and so preserves limits).

So the goal of the proof is to show $\cofib(G(R;M_1(R\to T)) \to \THH(R)) \equiv \THH(T)$. The colimit defining $G_k(R;M_1(R\to T))$ is a $(k+1)$-dimensional cube missing its terminal object. This diagram can be completed into a full cube by allowing \(s\subseteq\{0,\cdots,k\}\) to be an equality, and the map inside the $\cofib(-)$ is that induced by the universal property of the colimit. In Goodwillie’s analysis II paper he develops the theory of cubical diagrams in more detail. Our cofiber of the comparison map, in his language, is called the total cofiber of the full cube. He then shows that the total cofiber can be computed inductively: pick any opposite “hyperfaces” of the $(k+1)$-cube, each of which will be a $k$-cube, and the rest of the cube defines a morphism between those $k$-cubes. This induces a morphism between the total cofibers of these hyperfaces, and its usual cofiber is exactly the total cofiber of the original cube.

This inductive construction implies the result, using the fact that tensor products commute with colimit in each factor individually. $\square$

Note that the cyclic object of $\THH(R;M_1(R\to T))$, whose $k$th object is $M_1(R\to T)\otimes R^{\otimes k}$, appears inside the cubical diagram that defines $G_k(R;M_1(R\to T))$. For each $i=0,\cdots,k$ let $c_i$ be the $i$th element in $C_{k+1}$, and identify \((c_i)_+\wedge M_1(R\to T)\otimes R^{\otimes k}\) with the tensor product $R^{\otimes (k+1)}$ with the $i$th object replaced by $M_1(R\to T)$. The previous object we considered was the one corresponding to $c_0$. Altogether, $(C_{k+1})_+\wedge M_1(R\to T)\otimes R^{\otimes k}$ consists of $k+1$ copies of the same thing, but with the different factor placed in a different position each time. These are precisely the cases where \(s\subseteq\{0,\cdots,k\}\) is a singleton.

You may see at this point that the sub-colimit consisting just of singleton $s$’s maps into the full colimit. The geometric realization of all that finally gives us some $S^1$-action

\[\lambda_2 : S^1_+ \wedge \THH(\S;M_1(\S\to\Z)) \xto{} \THH(\S\to\Z)\]

To relate this to our $\lambda$, we observe that there’s a homotopy equivalence \(\lambda_1 : M_1(\S\to\Z) \xto{\equiv} \THH(\S;M_1(\S\to\Z))\) Indeed, the $k$th object defining the codomain is $M_1(\S\to\Z)\otimes \S^{\otimes k}$. But $\S$ is neutral wrt tensor products, so you’re only left with the one copy of $M_1(\S\to\Z)$ in each degree.

Then $\lambda$ factors as the composite \(S^1_+\wedge M_1(\S\to\Z) \xto{\lambda_1} S^1_+\wedge \THH(\S,M_1(\S\to\Z)) \xto{\lambda_2} \THH(\S\to\Z)\)

Part 1.6 - Connectivity of $\lambda$

Lemma: $\lambda$ is at least $(4p-5)$-connected.

Proof: We will use the factorization obtained in Part 1.5. The first map, $\lambda_1$, is a homotopy equivalence. Thus it’ll suffice to show $\lambda_2$ is at least $(4p-5)$-connected. This is a map between geometric realizations of simplicial spaces, so it’d suffice to show the underlying maps in each degree are at least $(4p-5)$-connected (this follows e.g. by looking at the spectral sequences induced by skeletal filtrations).

Recall that we expressed the $k$th space of $\THH(\S;M_1(\S\to\Z))$, as a colimit

\[\colim_{\emptyset\neq s\subseteq[k]}\left(\bigotimes_{i=0}^{k}{\left\{\begin{array}{l l} M_1(\S\to\Z) & \text{if }i\in s \\ \S & \text{if }i\notin s \end{array}\right\}}\right)\]

Recall also that if $X$ is $(a-1)$-connected and $Y$ is $(b-1)$-connected, then $X\otimes Y$ is $(ab-1)$-connected. (heuristically, that’s due to smashing together the bottom-most cells). $\S$ itself is $0$-connected, in fact it’s the unit for the smash product, so we disregard it. This leaves us with \(\# s\) copies of $M_1(\S\to\Z)$. One copy of $M_1(\S\to\Z) = \fib(\S\to\Z)$ is $(2p-3-1)$-connected because, as we’ve already used numerous times, $\S\to\Z$ is $(2p-3)$-connected, working locally at $p$. Consequently the object indexed by $s$ will be \((\# s(2p-3) - 1)\)-connected, and therefore the only objects contributing in degrees $\leq 4p-5$ are those where $s$ is a singleton. There are $k+1$ such objects; let us denote by $F_i$ the $i$th one. Concretely $F_i$ is a smash product of $k+1$ factors, the $i$th being $M_1(\S\to\Z)$, and the rest of them being $\S$.

Consider then the following sub-colimit

\[\colim_{s = \{i\}}\left(\bigotimes_{i=0}^{k}{\left\{\begin{array}{l l} M_1(\S\to\Z) & \text{if }i\in s \\ \S & \text{if }i\notin s \end{array}\right\}}\right) \equiv M_1(\S\to\Z)^{\oplus(k+1)} \equiv (C_{k+1})_+\wedge M_1(\S\to\Z)\]

By the above, the inclusion map of this sub-colimit into the original colimit is $(4p-5)$-connected. We regard the space \((C_{k+1})_+\wedge M_1(\S\to\Z)\) as the $k$th object in the simplicial space structure on $S^1_+\wedge M_1(\S\to\Z)$ induced by the standard one on $S^1$ and the constant one on $M_1(\S\to\Z)$. The geometric realization of the comparison map between the sub-colimit and the colimit is then precisely our map of interest, which we’ve wanted to show is $(4p-5)$-connected. $\square$

That’s everything we needed – congratulations! Don’t forget, this was the easy part of our proof skeleton… In part 2 I’ll complete the proof of the main theorem using Waldhausen’s $K$-theory of spaces.