I know that for $X$ a connected space, $THH(\Sigma^\infty \Omega X) = \Sigma^\infty \Lambda X$, the suspension spectrum of the free loop space of $X$. The computation can be carried out in spaces and then transferred to spectra via $\Sigma^\infty$. What is $TC(\Sigma^\infty \Omega X)$? Can it also be computed from some kind of $TC$ on the level of spaces?

Edit: Tyler answered my question, but I want to ask a followup question: Is it fair to say that $TC(\Omega X)$ in the world of spaces, after $p$-completion, is just $X$, and is there a map (not an equivalence, because we take limits to build $TC$) $\Sigma^\infty X \to $ the thing Tyler wrote down? (Note: all my $\Sigma^\infty$ are $\Sigma^\infty_+$.)

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    $\begingroup$ Can you elaborate on what you'd mean by "TC" on the level of spaces? The stable object has the transfer, which you don't see on the space level, and I assume you don't mean simply loop-infinity of TC. $\endgroup$ Dec 1, 2009 at 17:03
  • $\begingroup$ I mean: THH(ΩX) = ΛX is a cyclotomic space, so can we construct a functor from I to Spaces (where I is the category with the Fs and Rs) and take its homotopy limit? Is it X, after p-completion? What are the Fs and Rs in this case? $\endgroup$ Dec 1, 2009 at 19:47
  • $\begingroup$ @ReidBarton You may be interested in the last section of this paper, where we outline an unstable version of TC and the cyclotomic trace: arxiv.org/pdf/1710.06414.pdf $\endgroup$ Feb 13, 2019 at 18:52

2 Answers 2


The TC spectrum, at a prime $p$, of this is the homotopy pullback of a diagram

$S^1 \wedge (\Sigma^\infty_+ \Lambda X)_{hS^1} \to \Sigma^\infty_+ \Lambda X \leftarrow \Sigma^\infty_+ \Lambda X$

after $p$-completion. Here the left-hand map is the $S^1$-transfer from homotopy orbits back to the spectrum and the right-hand map is the difference between the identity and the "$p$'th power" maps on the loop space.

This is in Bökstedt-Hsiang-Madsen's original paper defining topological cyclic homology, in section 5.

ADDED LATER: This doesn't really work on the space level, because they don't have all the structure necessary. They have the $F$ maps, but not the $R$ ones which only come about from stable considerations. Spaces with a group action really only have one notion of "fixed points," namely the honest fixed points of the group action.

However, the associated equivariant spectrum of $\Lambda X$ is built out of spaces like

$$\Omega^V \Sigma^V \Lambda X = Map(S^V, S^V \wedge \Lambda X_+)$$

where $V$ ranges over representations of $S^1$. This has two "fixed-point" objects for any cyclic group $C$: there's the fixed points, which is the space

$$Map^C(S^V, S^V \wedge \Lambda X_+)$$

of equivariant maps. There is also the collection of maps-on-fixed-points

$$Map((S^V)^C, (S^V \wedge \Lambda X_+)^C)$$

which is called the "geometric" fixed point object, and it accepts a map from the ordinary fixed points. The fact that $(\Lambda X)^C \cong \Lambda X$ implies that you can interpret this as a map $(Q \Lambda X)^C \to (Q \Lambda X)$ where the latter uses an accelerated circle. These maps give rise to the $R$ maps in the definition of $TC$, and they definitely rely on the fact that you're considering the associated spectra.

  • $\begingroup$ Thanks, this is a great explanation. For my "space-level $TC$", I was imagining using the $c$th power map $(\Lambda X)^C → \Lambda X$ for $R$ (which is an equivalence of spaces), so $TR(\Omega X, p)$ would just be $\Lambda X$ with $F$ acting as the $p$th power map. Then $TC(\Omega X, p)$ would be $\mathrm{Map}(B\mathbb{Z}[1/p], X)$, I think. That maps to $\Omega^{\infty} TC(\Sigma^\infty \Omega X, p)$, right? $\endgroup$ Dec 2, 2009 at 17:17
  • $\begingroup$ Oops, where that error message is was just supposed to be $(\Lambda X)^C \to \Lambda X$. $\endgroup$ Dec 2, 2009 at 17:18
  • $\begingroup$ Right, that makes sense and I see where you're going now. I believe you're correct and that mapping object does map to $TC$. It seems, however, like the mapping object is p-equivalent to $X$ itself and $TC$ is p-local, so this map may just factor through evaluation at the basepoint? $\endgroup$ Dec 2, 2009 at 18:30
  • $\begingroup$ Yes, my thoughts exactly (that's the behavior I want). $\endgroup$ Dec 2, 2009 at 18:45

This is probably offensively naive (sorry), but blindly following the slogan "TC is the smart homotopy theorist's refinement of homotopy $S^1$ fixed points of THH" (or a non-rational refinement of cyclic homology) would guess the suspension spectrum of unparametrized loops, $\Sigma^\infty(\Lambda X/S^1)$. Since I'm hoping to get a better sense for TC, could you point out things you know about the answer that make it clear this is nonsense?

  • $\begingroup$ Alas, I'm not even enlightened enough about TC to know for sure that this is wrong! But wouldn't the really naive thing on the level of spaces be to take the homotopy S^1 fixed points of the free loop space on X, and isn't that just X? $\endgroup$ Dec 1, 2009 at 5:45

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