MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\Sigma^\infty_+$.)

share|cite|improve this question
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. – Tyler Lawson Dec 1 '09 at 17:03
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? – Reid Barton Dec 1 '09 at 19:47
up vote 5 down vote accepted

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.

share|cite|improve this answer
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? – Reid Barton Dec 2 '09 at 17:17
Oops, where that error message is was just supposed to be $(\Lambda X)^C \to \Lambda X$. – Reid Barton Dec 2 '09 at 17:18
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? – Tyler Lawson Dec 2 '09 at 18:30
Yes, my thoughts exactly (that's the behavior I want). – Reid Barton Dec 2 '09 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?

share|cite|improve this answer
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? – Reid Barton Dec 1 '09 at 5:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.