A paradox:

  • Goodwillie calculus considers only finitary functors.
  • $TC$ isn't finitary.
  • Yet in some sense $\partial(TC) = \partial(K) = THH$ is the crux of the Dundas-Goodwillie-McCarthy theorem.

(Here, a finitary functor is one preserving filtered colimits[1]. $\partial$ denotes the first Goodwillie derivative. $TC,K,THH$ are respectively topological cyclic homology, algebraic $K$-theory, and topological Hochschild homology, regarded as functors from $E_1$-ring spectra to spectra.)

Obviously I don't fully understand that last point, which is just a rough idea I think I've read somewhere, and that's what I want to ask about.


  1. What does it mean to say that the first Goodwillie derivative of $TC$ is $THH$?

  2. Is there a general formalism for Goodwillie calculus of non-finitary (but, say, accessible) functors? If so, how much of the usual theory goes through?

  3. If the answer to (2) is "yes", does it specialize in the case of $TC$ to recover the answer to (1)?

[1] At any rate, in Goodwillie calculus one always requires one's functor to commute with sequential colimits. Any functor that commutes with sequential colimits and $\aleph_1$-filtered colimits commutes with all filtered colimits. $TC$ is defined from $THH$ (which commutes with filtered colimits) via a countable limit and therefore $TC$ commutes with $\aleph_1$-filtered colimits. I conclude that if $TC$ doesn't commute with filtered colimits, then it already doesn't commute with sequential colimits, and so standard Goodwillie calculus doesn't apply to it.

  • 4
    $\begingroup$ I will write a proper answer later, but you are supposed to be taking the derivative in the coefficients, that is the theorem is simply saying that $THH(A;M)=\mathrm{colim}_n\Omega^n TC(A;\Sigma^nM)$ for all ring spectra $A$ and $A$-modules $M$. $\endgroup$ Aug 16, 2018 at 15:30
  • $\begingroup$ I don't know if this helps with your $TC$ question, but this paper arxiv.org/abs/math/0601221 sets up some foundations of Goodwillie calculus for non-finitary functors. $\endgroup$ Aug 17, 2018 at 7:46

2 Answers 2


This answer addresses Question 1.

Let "ring" mean associative unital ring spectrum, say in the $A^\infty$ sense. For a functor $F$ from rings to spectra (such as $TC$), differentiating $F$ at the ring $R$ means finding the best excisive approximation to the functor from rings-having-$R$-as-a-retract to spectra, $$ (R\to S\to R)\mapsto fiber (F(S)\to F(R)). $$ The universal excisive functor takes values in $R$-bimodules. (That is, a spectrum object for the category of rings over $R$ can be encoded in an $R$-bimodule.) Call it $\Omega_R$. If $S$ is $n$-connected relative to $R$ then $\Omega_R(S)$ is related to $fiber(S\to R)$ by a roughly $2n$-connected map of bimodules. The derivative of $F$ at $R$ must be $L\circ \Omega$ for some linear functor $L$ from $R$-bimodules to spectra. Thus to name the derivative of $F$ at $R$ you have to name a certain such linear functor.

If $F$ is what I call analytic (or even if it is what I call stably excisive) then this means that $fiber (F(S)\to F(R))$ is related by a roughly $2n$-connected map to $L(S\to R)$.

If $F$ is finitary then its derivative at $R$ is a finitary linear functor, and therefore the corresponding map $L$ is given by tensoring over $R\wedge R^{op}$ with a fixed bimodule. It happens that the derivative of $TC$ is finitary (even though $TC$ is not) and the fixed bimodule in question is $R$ itself. Tensoring an $R$-bimodule $M$ with $R$ gives $THH(R;M)$, so this means that in a stable range $fiber(TC(S)\to TC(R)$ looks like $THH(R;fiber (S\to R))$.

There is an unfortunate clash of terminology. When differentiating a functor of spaces at $X$ you get an excisive functor from spaces-containing-$X$-as-a-retract to spectra. In Calculus 3 I called this the "differential" of $F$ at $X$, denoting it by $D_XF$. If $F$ is finitary then $D_XF$ is as well, and then it must be given by fiberwise smash product with a fixed parametrized spectrum over $X$. I denoted the fiber of that parametrized spectrum at $x\in X$ by $\partial_xF(X)$ and called it the (partial) derivative. So the differential is given by smashing with the derivative.

In the setting of functors of rings, the "spectra" made out of objects over $R$ correspond to $R$-bimodules, whereas in the setting of functors of spaces the "spectra" made out of objects over $X$ correspond to parametrized spectra over $X$. In the latter setting "derivative" means the parametrized spectrum that you smash with to give the differential. By analogy in the former setting it ought to mean the bimodule you tensor with to give the differential. In those terms the derivative of $TC$ at $R$ is the bimodule $R$.

  • $\begingroup$ Thanks! So really the point is that finitariness allows you to apply the Eilenberg-Watts theorem to the derivative. And I suppose the reason the derivative of $TC$ is finitary is that $TC$ restricted to square-zero extensions is finitary. $\endgroup$
    – Tim Campion
    Aug 21, 2018 at 17:17

Your first premise - that Goodwillie calculus considers only finitary functors - is wrong. Goodwillie doesn't insist on this. The only point at which finitary enters the story is when one wishes to identify homogeneous functor of degree $n$ with spectra with (naive) actions of the $n$-th symmetric group. And even without the finitary condition, Goodwillie shows that degree $n$ homogeneous functors correspond to symmetric $n$--linear functors.

In my own work, I have made much use of Goodwillie calculus applied to non--finitary functors of the form $LF$, where $L$ is a non-smashing Bousfield localization (e.g. localization with respect to a Morava $K$--theory). Composites like this are also a source of instructive examples, e.g., the composition of homogeneous functors need not be again homogeneous. See my survey paper [Goodwillie towers and chromatic homotopy: an overview. Proceedings of the Nishida Fest (Kinosaki 2003), 245–279, Geom. Topol. Monogr., 10] for more about all of this, and more references.

  • $\begingroup$ Thanks! I guess I've betrayed that what Goodwillie calculus I've learned is actually from Lurie. I was worried that the $n$-excisive approximation couldn't be computed in the usual way, but I suppose it should be fine if you just iterate it transfinitely (as long as your functor is accessible). $\endgroup$
    – Tim Campion
    Aug 21, 2018 at 17:15

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