I have some questions about Dennis trace map in algebraic K-Theory. I was wondering if there is some conceptual way to look at this map $K(-)\rightarrow THH(-)$ (natural transformation from K-Theory to Topological Hochschild Homology). As far as I understand, Tabuada constructed a stable model category $\mathcal{M}$ containing the category of small DG-categories (and small Spectral categories). Roughly speaking, $\mathcal{M}$ is a Bousfield localization of the category of spectral presheaves on the category of small spectral categories. The algebraic K-Theory is representable by the spectral category $\mathbb{S}-Mod^{c}$ spectra which are compact. The spectral enrichment $\mathbf{Sp}_{\mathcal{M}}$ represents the algebraic (Waldhausen) K-Theory as follows $\mathbf{Sp}_{\mathcal{M}}(\mathbb{S}-Mod^{c},A)\simeq K(A-Mod^{c})$ where $A$ is a dg (or spectral category). With this notations $\mathbf{Sp}_{\mathcal{M}}(\mathbb{S}-Mod^{c},THH(-))\simeq THH(\mathbb{S})$, the Dennis trace map is represented by the class $[1]: \pi_{0}\mathbf{Sp}_{\mathcal{M}}(\mathbb{S}-Mod^{c},THH(-))\simeq \mathbf{Z}.$

My question is the following: What happens if we restrict our self to commutative $(E_{\infty})-\mathbb{S}$-Algebras and noticing that in this case $THH(R)\simeq S^{1}\otimes R$ for any $E_{\infty}$-algebra $R$. Is it possible to construct the trace map $K(-)\rightarrow S^{1}\otimes-$ (of $E_{\infty}$-algebra?!) in purely categorical comprehensive way without using Tabuada's sophisticated machinery and Dennis-Bökstedt technical results ?

PS: if there is some mathematical mistakes, please do correct me. Thank you.

  • $\begingroup$ Did you find satisfactory answers? Seemingly people like to refer to the Dennis trace map as trace of monodromy. They like to view HH(R) as derived loop space of Spec(R), and given a projective R-module M and a loop in Spec(R), they claim that the image of M is the trace of the monodromy endomorphism. It appears in Toën's survey. A formalism is presented in arxiv.org/abs/1804.00879 However, I have not digested them well yet. $\endgroup$
    – user20948
    Nov 14, 2019 at 14:15

1 Answer 1


Yes, and no. There is a "categorically comprehensive" reason for this trace map to exist, but not necessarily to construct it. And to prove that this reason is valid, one does require categorical machinery. The elevator speech answer is:

THH is an algebra in some symmetric monoidal category. K is the unit in this symmetric monoidal category, so THH receives a unique algebra map from K theory.

(If you are looking for a less theoretical reason, and want some explicit constructions, you might take a look at Kantorovitz-Miller, "An explicit description of the Dennis trace map.")

Everything I write below, I learned from Blumberg-Gepner-Tabuada, "Uniqueness of the multiplicative cyclotomic trace."

First, we note that both THH and K define functors

$\infty Cat^{perf} \to Spectra$.

On the righthand side is the category of spectra. (Take any model you like, so long as it's not the homotopy category of the model. You can take Lurie's oo-categorical model, or symmetric spectra if you like.)

On the lefthand side is the category of perfect stable oo-categories. Roughly, these are the categories that look like modules over some ring spectrum. A different way you might describe this category is as follows: The category of spectrally enriched categories, localized with respect to Morita equivalence.

Note that both categories--$\infty Cat^{perf}$ and $Spectra$--have a symmetric monoidal structure. The latter has the usual smash product, while the former has the tensor product of stable $\infty$-categories. This is given by the cocompletion of the following naive tensor product: Given two categories $A$ and $B$, the objects of $A \otimes^{naive} B$ are pairs of objects $(a,b)$, and the hom spectrum between $(a,b)$ and $(a',b')$ is given by $hom(a,a') \wedge hom(b,b')$.

Moreover, we note that both THH and K satisfy the following properties:

  1. They are lax monoidal. (In fact, THH is symmetric monoidal.) This means that we have specified natural maps $K(A) \otimes K(B) \to K(A \otimes B)$, but these need not be equivalences.

  2. They are localizing: If we have a short exact sequence of categories $A \to B \to C$, we have a cofibration sequence of spectra $K(A) \to K(B) \to K(C)$, and likewise for THH. The proof of this for THH can be found in Blumberg-Mandell ("Localization theorems in topological Hochschild homology and topological cyclic homology").

Now, consider the category of all functors $\infty Cat^{perf} \to Spectra$ satisfying (2). One can construct a symmetric monoidal structure on this category. And it turns out that any functor further satisfying (1) can be made into an $E_\infty$ algebra in this category, and that K theory is in fact the unit! Since THH satsifies (1) and (2), the corresponding algebra for THH receives a unique algebra map from K theory.

When $A$ is an $E_\infty$ ring, then $THH(A)$ is an $E_\infty$ ring as well; by the algebra map from $K$ to $THH$, one obtains an $E_\infty$ ring map $K(A) \to THH(A)$.

More generally, if $A$ is an $E_n$-algebra, then $K(A) = K(AMod)$ is an $E_{n-1}$ ring. This is because the category of $A$-modules can be given an $E_{n-1}$-structure, and $K$ theory is lax monoidal. Moreover, you can also prove that $THH(A)$ has an $E_{n-1}$ structure as well (you can see this also using factorization homology, for instance). The fact that there is an algebra map $K \to THH$ implies that one also obtains an $E_{n-1}$-algebra map $K(A) \to THH(A)$.

  • $\begingroup$ Thank you Lee Tanaka, I do think that the representability theorem for K-theory says a lot at the beginning of my question... The point is that K-theory and THH are endofunctors of E-infinity algebras, and as you said and confirmed the trace map is a map of E-infinity algebras. So my question. If M is End(Alg) the infinity category of endomorphism category of E-infinity algebras (after some localization), can we say that the mapping space Map_M(K(-),THH(-)) is connected (even contractible?) that will give us the uniqueness of Dennis trace map in this case (up to contractible choice)... $\endgroup$
    – Ilias A.
    Jun 17, 2014 at 16:43
  • $\begingroup$ In some sense, you did not answer my initial question. Or maybe I did not understand. Thank you very much anyway!! $\endgroup$
    – Ilias A.
    Jun 17, 2014 at 16:52
  • $\begingroup$ Actually, in my answer, the point isn't so much that "K-Theory and THH are endofunctors of E-infinity algebras," they are rather E-infinity algebras in a certain symmetric monoidal category (of localizing functors). Since K theory is the unit of this monoidal category, the space of algebra maps from K theory to THH has to be contractible. Put another way (tracing through what I wrote in the answer above), the (unit) map K --> THH is completely determined by the lax monoidal nature of THH. Maybe I'm being slow; can you tell me how you think of K, THH as objects of End(Alg)? $\endgroup$ Jun 17, 2014 at 16:59
  • $\begingroup$ M=End(Alg) is the category of endofunctors of commutative S-Algebras i.e., the category of functors from commutative S-Algebras to commutative S-Algebras. In my initial question I was wondering if the mapping space Map_{M}(K(-),THH(-)) was connected and even contractible. $\endgroup$
    – Ilias A.
    Jun 17, 2014 at 17:16
  • 1
    $\begingroup$ Ah, I see. I don't know whether that's true; one could try to produce a non-trivial action of Aut(THH) or Aut(K) on this space, but the only action I can think of (rotating S^1) fixes the only map from K to THH I can think of! $\endgroup$ Jun 17, 2014 at 18:10

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