For a Serre fibration $$ F\to E \to B , $$ with $F,E,B$ having the homotopy type of finite complexes, it is known that the Euler characteristic is multiplicative: $$ \chi(E) = \chi(F)\chi(B) . $$ However, if we more generally assume that $B$ and $F$ are finitely dominated spaces, then does multiplicativity hold as well? (Recall that a finitely dominated space is a retract of a homotopy finite one.)

If true I am looking for a reference. If false, please explain.

Added later: it's true if the base is homotopy finite since we can take the fiberwise double suspension to obtain a fibration with homotopy finite fibers having the same Euler characteristic. So we only need to consider the case when the base is a finitely dominated and the fibers are homotopy finite.

Second Addition: I can solve the problem in general if I can solve it in the following case: Let $\tilde B \to B$ be a finite, regular, $n$-sheeted covering space, where $B$ is finitely dominated. Then $\chi(\tilde B) = n\chi(B)$.

(Note that $\tilde B$ is again finitely dominated, since there is a finite covering $\tilde B\times S^1 \to B\times S^1$ and by a theorem of Mather, a space $X$ is finitely dominated if and only if $X\times S^1$ is homotopy finite. But since the base $B\times S^1$ is homotopy finite, we can put a finiteness structure on the total space as well.)

Third Addition: The case alluded to in my "Second Addition" holds, by the second answer I gave below.

  • 6
    $\begingroup$ @Z.M Yes, you are mistaken. mathoverflow.net/questions/80326/… $\endgroup$
    – John Klein
    Feb 24, 2022 at 18:38
  • 3
    $\begingroup$ @ChrisSchommer-Pries I could not extract the result from the answer to the MO question since induction on the cells will fail. I also looked at the Ponto-Shulman paper--it's written in a weird way: the title has "multiplicativity" but there is no explicit multiplicativity statement in the paper. $\endgroup$
    – John Klein
    Feb 24, 2022 at 22:06
  • 2
    $\begingroup$ If you have solved it, it would be nice if you could write the arguments in more detail. In particuler the reduction to the case of covers and the Leray-Serre spectral sequence argument - it sounds hard to get rid of the $\pi_1$-action $\endgroup$ Feb 25, 2022 at 18:26
  • 9
    $\begingroup$ I think that Pedersen-Taylor 1978 prove that the total Wall invariant satisfies $w(E)=\chi(F)w(B)$, and thus that $\chi(E)=\chi(F)\chi(B)$. $\endgroup$ Feb 26, 2022 at 22:58
  • 4
    $\begingroup$ @BenWieland : is this what they prove ? From reading it it seems like it's not $\chi(F)$, but $\chi(p)$. There is a $\pi_1$-action there. Maybe I'm missing something $\endgroup$ Feb 27, 2022 at 11:33

4 Answers 4


Note Added March 1, 2022:

I now think there is a gap in deducing multiplicativity of the Euler characteristic from the Pedersen-Taylor result on the finiteness obstruction. I think the argument I give in my other answer more-or-less fills that gap.

Ben Wieland has provided a reference which answers my question.

Pedersen, Erik Kjaer; Taylor, Lawrence R. The Wall finiteness obstruction for a fibration. Amer. J. Math. 100 (1978), no. 4, 887–896.

The authors identify the image in $K_0(\Bbb Z[\pi_1(B)])$ of the Wall finiteness obstruction $\sigma(E) \in K_0(\Bbb Z[\pi_1(E)])$. The Euler characteristic is the image of this class $K_0(\Bbb Z)$.

  • $\begingroup$ Why is the image of $\chi(p)$ in $G_0(\mathbb Z) \cong K_0(\mathbb Z)$ just $\chi(F)$ ? $\endgroup$ Feb 27, 2022 at 11:34
  • 1
    $\begingroup$ The image of $\chi(p)$ under in $G_0(\mathbb Z)$ seems to exactly be a line in the Serre spectral sequence for the fibration $\endgroup$ Feb 27, 2022 at 12:08

I have already written a lot, but Oscar Randal-Williams's comment below John Klein's second answer seems to really simplify the matter, so I decided to add an answer to that effect. I can remove any of the answers I have written if somehow they're "obfuscating the view" or anything.

This argument is just a simplification, that bypasses Pedersen-Taylor's result, and also the need for Swan's other theorem (suggested by Oscar) that $\tilde K_0(\mathbb Z[\pi])$ is finite for a finite group $\pi$ (I think the proof is harder, although I might be mistaken here). The argument is therefore due to Oscar and John in this respect, this is just my write-up of it.

Namely, as in all other answers reduce to the case of a finite covering space $p: E\to B$. Letting $F$ denote the fiber, and $Q$ the (finite) image of $\pi_1(B)\to Aut(F)$, let $f: B\to BQ$ denote the corresponding classifying map.

Define the following functor from perfect $\mathbb Z[\Omega B]$-modules to $\mathbb Q$-modules: $P\mapsto B_!( (P\otimes_\mathbb Z \mathbb Q) \otimes_\mathbb Q C_*(F;\mathbb Q))$.

The observation is that $C_*(F;\mathbb Q)$, as a $\mathbb Z[\Omega B]$-module, is of the form $f^* C_*(F;\mathbb Q)$, so this is $\mathrm{colim}_{BQ} f_!((P\otimes_\mathbb Z\mathbb Q)\otimes_\mathbb Q f^*C_*(F;\mathbb Q))\simeq \mathrm{colim}_{BQ} f_!(P\otimes_\mathbb Z\mathbb Q) \otimes C_*(F;\mathbb Q)$ (where $\mathrm{colim}_{BQ}$ is derived)

In other words, this functor factors as $Perf(\mathbb Z[\Omega B])\to Perf(\mathbb Z[Q])\to Perf(\mathbb Q[Q])\to Perf(\mathbb Q)$, but the most important part is the first part.

Indeed, $Q$ is finite, and so on $K$-theory, $K_0(\mathbb ZQ)\to K_0(\mathbb QQ)$ lands in the image of $K_0(\mathbb Q)$ (under the "induction" morphism) : this is (as far as I know) a theorem of Swan's (specifically, theorem 4.2 from his book "K-theory of finite groups and orders")

Now if $P= \mathbb Q[Q]^n$ is a free $\mathbb Q[Q]$-module, $\mathrm{colim}_{BQ}(P\otimes_\mathbb Q C_*(F;\mathbb Q)) = \mathrm{colim}_{BQ}(P\otimes_\mathbb Q C_*(F^{triv}; \mathbb Q)) = P_{hQ}\otimes_\mathbb Q C_*(F; \mathbb Q)$.

In particular, on $K$-theory, this composite has the same effect as $\mathrm{colim}_B (-)\otimes_\mathbb Q C_*(F;\mathbb Q)$, and so it sends the trivial coefficient system $\mathbb Z$ to $C_*(B;\mathbb Q)\otimes_\mathbb Q C_*(F;\mathbb Q)$, which of course is another name for $\chi(B)\chi(F) \in \mathbb Z\cong K_0(\mathbb Q)$.

But this map also sends this trivial coefficient system to $C_*(E;\mathbb Q)$, which is also another name for $\chi(E)\in \mathbb Z\cong K_0(\mathbb Q)$.

This proves the claim.

  • $\begingroup$ Is it essential to take $Q$ to be the image of $\pi_1(B)\to\operatorname{Aut}(F)$ rather than $Q:=\operatorname{Aut}(F)$? Is the reduction to coverings essential? I mean, for example, is the group $\operatorname{Aut}(F)$ still finite enough when $F$ is a compact object? $\endgroup$
    – Z. M
    Mar 2, 2022 at 21:42
  • $\begingroup$ @Z.M the $Q$ part doesn't seem to be essential; on the other hand $Aut(F)$ is definitely not finite enough for a general compact (even finite!) space $F$, so that part seems essential $\endgroup$ Mar 2, 2022 at 21:56
  • $\begingroup$ I am not sure whether I understand correctly. It seems that the finiteness is mainly used in the step of invoking Swan's result. For example, I don't know what happens if I run this argument for general fiber sequences instead of coverings. $\endgroup$
    – Z. M
    Mar 2, 2022 at 22:57
  • $\begingroup$ @Z.M yes, that's exactly where it is used. But this part was the key point - if you look at Pedersen's result, or at John's other answer (or my other answer) you'll see that this is the main problem : getting rid of the $\pi_1$-action somehow. In all cases, the argument is basically to use finiteness, so it really seems essential. If you run this argument for general fiber sequences, it doesn't work, so that is what happens :) $\endgroup$ Mar 3, 2022 at 7:21
  • $\begingroup$ Sorry, I was not clear. I wanted to say general fiber sequences with fibers being compact, so there is finiteness, just no discreteness, namely, run the argument as far as possible without the extra step to pass to coverings. $\endgroup$
    – Z. M
    Mar 3, 2022 at 11:04

Here is an argument that the Euler characteristic is multiplicative for fibrations $$ F\to E \to B $$ where $F$ and $B$ are finitely dominated and $B$ is connected. Without loss in generality, we may assume that $E$ is also connected.

Step 1. The first fact is that for a homotopy pushout $\require{AMScd}$ $$ \begin{CD} X_{01} @>>> X_1\\ @VVV @VVV \\ X_0 @>>> X \end{CD} $$ of finitely dominated spaces, the Euler characteristic is additive. This follows from the Mayer-Vietoris sequence using the Betti number definition of Euler characteristic. I won't include the proof here; I think the argument is easy.

Step 2. An immediate corollary is that $$ \chi(S_B E) = 2\chi(B) - \chi(E)\, $$ where $S_B E\to B$ is the fiberwise suspension. Iterating once more, we get $$ \chi(S^2_B E) = \chi(E)\, . $$ But the fibers of $S^2_BE\to B$ are homotopy finite with the same Euler characteristic as the fibers of $E\to B$, and so this enables us to reduce to the case when $E\to B$ has homotopy finite fibers.

Step 3. The action defines a homomorphism $$ \pi_1(B) \to \text{GL}(H_\ast(F;\Bbb Z/2)) $$ and therefore a finite covering space $\tilde B \to B$. The base change (pullback) of the fibration $E\to B$ to $\tilde B$ yields a new fibration $\tilde E\to \tilde B$ which is orientable in the sense of the Serre spectral sequence and the latter fibration has fiber $F$. Moreover, both $\tilde E$ and $\tilde B$ are again finitely dominated (since $B$ is and the fibers are). It follows from Spanier's book (p. 481) that $\chi(\tilde E) = \chi(\tilde B)\chi(F)$.

Step 4. By the previous step, it is sufficient to show that the Euler characteristic is multiplicative for $n$-sheeted covering spaces $$ E\to B $$ with $B$ finitely dominated. Here $F$ is a discrete set of cardinality $n$.

For this, I will appeal to Waldhausen's functor $X\mapsto A(X)$, where $A(X)$ is the finitely dominated flavor of the algebraic $K$-theory of spaces.

Step 5. Recall that $A(X)$ can be defined as the $K$-theory of the Waldhausen category of equivariant based spaces $$ R(\ast;H)\, , $$ where $H$ is a topological group model for the loop space of $B$, i.e., $BH \simeq B$. Objects of $R(\ast;H)$ are based spaces with $H$-action that are $H$-finitely dominated, in the sense that there are retracts of up to homotopy of finite objects (a finite object is one which is built out of finitely many free cells $D^k\wedge(H_+)$.)

Note that $$ \pi_0(A(B)) = K_0(\Bbb Z[\pi_1(B)]), $$ and the class of $S^0 \in R(\ast;H)$ is identified with the unreduced version of Wall's finiteness obstruction.

Step 6. We may identify $p:E\to B$ as induced by the homomorphism $G\to H$ of suitable topological groups. Then one has a pushforward and transfer functors $$ R(\ast;H) \overset{p^!}\to R(*;G) \overset{p_*}\to R(*;H) $$ in which $p^!(U) = U$ (i.e., restrict the action) and $p_*(Z) = Z\wedge_G (H_+)$.

It follows that $$ p_*p^!(U) \cong U\wedge (F_+) , $$ where the target is given the diagonal $H$-action.

Claim: Let $F^t$ denote the finite set $F$ but with the trivial $H$-action (made cofibrant, i.e., $F^t = |F|\times EH$). Then $F_+,F^t_+\in R(\ast;H)$ define the same $K$-theory classes after pushforward to $R(\ast,e)$.

Sketch proof of Claim: With respect to the homomorphism $H\to e$, the objects $F,F^t $ push forward to $E_+$ and $(B\times F)_+$ in $R(\ast;e) = R(\ast)$. If $B$ is a finite complex, then the Additivity Theorem and induction along the skeleta of $B$ show that $ E_+ $ and $(B\times F)_+ $ define the same $K$-theory class.

If $B$ is finitely dominated, of dimension $r$ say, then by an argument of Wall, there is a finite complex $K$ of dimension $r-1$ and an $(r-1)$-connected map $K\to B$ which we can take to be a cofibration by adjusting $B$ up to homotopy equivalence. Furthermore, One can find such a map $K\to B$ such that the relative homology $H_\ast(\tilde B,\tilde K;\Bbb Z)$ is trivial except in degree $r$, where $\tilde B\to B$ is the universal cover and $\tilde K \to K$ is the pullback to $K$, and in degree $r$, we have that $P := H_r(\tilde B,\tilde K;\Bbb Z)$ is a finitely generated projective $\Bbb Z[\pi]$-module (with $\pi = \pi_1(B)$). Note that $(-1)^rP \in K_0(\Bbb Z[\pi])$ is the Wall finiteness obstruction of $B$.

As $K$ is finite, by the above one has that $E|K$ and $K\times F$ define the same class in $\pi_0(A(\ast)) = K_0(\Bbb Z)$.

Furthermore, using the fact that $\tilde B$ is $1$-connected, there a cofiber sequence of spaces with $\pi$-action $$ \tilde E_{|K} \to \tilde E \to (\tilde B/\tilde K) \wedge F_+ $$ (since the base change of $E\to B$ along $\tilde B \to B$ is a trivializable cover). Since $K$ is finite, $E_{|K}$ and $K\times F$ define the same class in $\pi_0(A(\ast))$. By additivity, it will be enough to show that $$ H_r((\tilde B/\tilde K) \wedge_{\pi} F_+) = P\otimes_{Z[\pi]} \Bbb Z[F_+] $$ is isomorphic to $P \otimes_{\Bbb Z[\pi]} \Bbb Z[F^t]$ as a $\Bbb Z$-module.

To complete the proof of the claim, it is enough to establish the following:

Lemma: Suppose that $P$ is a finitely generated projective left $\Bbb Z[\pi]$ module. Then the finitely generated free $\Bbb Z$-modules $$ P \otimes_{\Bbb Z[\pi]} \Bbb Z[F] \quad \text{ and }\quad P \otimes_{\Bbb Z[\pi]} \Bbb Z[F^t] $$ have the same rank.

Proof of Lemma (After Oscar Randall-Williams). In the above, to talk about the tensor product, we have regarded $P$ as a right module using the involution $g\mapsto g^{-1}$ of $\pi$.

Let $\Pi = \text{iso}(F)$, the automorphisms of the set $F$. Then $\Pi$ is finite. Moreover, one has a homomorphism $\pi \to \Pi$ and we can define $$ P' := P\otimes_{\Bbb Z[\pi]} \Bbb Z[\Pi] $$ Then $P' \otimes_{\Bbb Z[\Pi]} \Bbb Z[F] = P\otimes_{\Bbb Z[\pi]} \Bbb Z[F]$ and this enables to assume at the outset that $\pi$ is finite.

So assume that $\pi$ is finite.

Consider the homomorphisms $$ \rho,\rho^t: K_0(\Bbb Z[\pi]) \to K_0(\Bbb Z) $$ defined by $\rho(P) = P\otimes_{\Bbb Z[\pi]} \Bbb Z[F]$, and $\rho^t(P) = P\otimes_{\Bbb Z[\pi]} \Bbb Z[F^t]$.

It will be enough to show that these two homomorphisms coincide.

They certainly coincide for $P$ a f.g. free module of rank one and therefore for any f.g. free module.

Using Swan's Theorem that the reduced group $\tilde K_0(\Bbb Z[\pi])$ is a torsion group (with $\pi$ finite), it follows that $\rho = \rho^t$ in general. $\square$.

Let $c: \pi_0(A(B)) \to \pi_0(A(\ast))$ be the map induced by the pushforward $B\to \ast$.

Hence, by the claim, the composition $$ \begin{CD} \pi_0(A(B)) @> p^! >> \pi_0(A(E)) @>p_\ast >> \pi_0(A(B)) @> c >> \pi_0(A(\ast)) = \Bbb Z \end{CD} $$ is identified with multiplication by $\chi(F)$ in the sense that it is given by $$ U \mapsto U_{H}\wedge F^t_+ \, . $$ where $U_H$ is the reduced Borel construction.

On the other hand, if we take the pushforward of $$ p_\ast p^!(S^0) = F_+ $$ in $R(\ast;e)$ (where $e$ = trivial group), we obtain $(F_+)\wedge_H *_+ = E_+$, and this will give the Euler characteristic $\chi(E) \in \pi_0(A(\ast)) = \Bbb Z$.

If we put the above facts together, we see that the composition $$ \begin{CD} \pi_0(A(B)) @> p_*p^!>> \pi_0(A(B)) @>c >> \pi_0(A(\ast)) = \Bbb Z \end{CD} $$ has the property that it maps the class of $S^0$ to both $\chi(E)$ and $\chi(F)c([S^0])$. But $c([S^0]) = \chi(B)$, so we are done.

  • 1
    $\begingroup$ Why is each $B_i/D_i/S_i$ finitely dominated ? Or rather, why can you choose $B'$ so as to ensure that ? (There are certainly choices of $B'$ that do not have this property) $\endgroup$ Feb 28, 2022 at 8:45
  • 1
    $\begingroup$ Isn't this a counterexample to your lemma in the case of a finite group $\pi$ ? Take $M= \mathbb Q$ with the trivial $\pi$-action (which is projective rationally) then the coinvariants of $\mathbb Q[F]$ are smaller than those of $\mathbb Q[F^t]$. More generally, the action need not be free on $M$ in general. $\endgroup$ Mar 2, 2022 at 7:15
  • 3
    $\begingroup$ I think the lemma is true if you don't try to "simplify" by working over $\mathbb{Q}$. The functions $|\pi| \cdot rk_\mathbb{Z}(- \otimes_{\mathbb{Z}[\pi]} \mathbb{Z})$ and $rk_\mathbb{Z}(-)$ induce homomorphisms $K_0(\mathbb{Z}[\pi]) \to \mathbb{Z}$ which agree on free modules, so by Swan's theorem that $\tilde{K}_0(\mathbb{Z}[\pi])$ is finite when $\pi$ is a finite group it follows that they agree on all projective $\mathbb{Z}[\pi]$-modules. $\endgroup$ Mar 2, 2022 at 13:56
  • 1
    $\begingroup$ It seems to me that with this argument one can simply then apply Pedersen's result: calling $f: K_0(\mathbb Z[\pi_1B]) \to K_0(\mathbb Z)$ the projection map, we have that $f(\chi(p)\cdot -)$ and $f(\chi(F)\cdot -)$ also agree on free modules $\endgroup$ Mar 2, 2022 at 14:36
  • 1
    $\begingroup$ It also seems like in this case the rational argument works, because the rationalization of projective $\mathbb Z[\pi]$-modules is free, by another theorem of Swan's (which I think is easier to prove) $\endgroup$ Mar 2, 2022 at 15:04

$$\newcommand{\Sph}{\mathbb S} \newcommand{\THH}{\mathrm{THH}} \newcommand{\Sp}{\mathrm{Sp}}$$

The answer is that Euler characteristics are multiplicative. I can't adress the question of a reference, though, because I don't know one.

Let me point out a few things: as you point out, I think the result does not quite follow from Pedersen-Taylor's paper, essentially for the reason that $\chi(p)$ is very different from $\chi(F)$ and there is a priori no reason you can "forget the $\pi_1(B)$-action" (or at least this reason should be added to the proof - but you seem to agree that this is a gap, so I won't comment further on this).

Your other answer has what I believe to be a gap, namely I think your lemma is wrong for finite groups.

Now, about the proof. I will outline a proof below that uses three ingredients : 1- a THH approach to traces; 2- the free-loop transfer and its comparison with THH transfer; 3- a result of Linnell's on the Bass trace conjecture (I learned it from a paper of Berrick-Hesselholt's, and one can make a simpler proof using TC and the Bökstedt-Hsiand-Madsen description of TC of spherical group rings). Of these, 1- is an artifact of my personal predilections, although I'm not sure how to get rid of it; 2- is not essential (I will use it in an essential way in the proof, but I also know a different, somewhat simpler argument that doesn't use it : the reason I don't outline this simpler one is that it relies on work in progress - if you are interested I would prefer to talk about it in private); and 3 is, I believe, essential, and essentially the core of the proof.

The argument goes roughly as follows: As you suggest, reduce to the case of finite Galois covers (you did not specify this, but your argument produces a Galois cover). Identify $\chi(E)$ in terms of $\THH$, and use the comparison between free loop Becker-Gottlieb transfer and THH-transfer to compare this to $\chi(F)$ : in principle, there should be some Lefschetz numbers of the monodromy action of $\pi_1(B)$ on $H_*(F)$ that appear rather than barely $\chi(F)$, but the Linnell-Hesselholt result allows you to get rid of these monodromies.

This proof may seem complicated, but I think it will be clear at the end that some complication is necessary, see the Note in Step 4. Of course it is not a precise argument, so maybe an easy argument can be found, but I think this is unlikely.

Step 0: Clearly $B$ is assumed to be connected, and we may assume that $E$ is connected too.

Step 1 : reduction to Galois covers. John already outlined this, but for convenience (and also because, even though the proof is the same, I make a stronger claim), let me recall the proof. The Euler characteristic of $F$ can be computed using $H_*(F;\mathbb Z/2)$ (or $\mathbb Z/p$ for any prime $p$), and it is easy to prove the result if the action of $\pi_1(B)$ on these is trivial, using the Serre spectral sequence. In particular, letting $\tilde B\to B$ be a finite cover corresponding to the kernel of $\pi_1(B)\to GL(H_*(F;\mathbb Z/2))$, and $\tilde E$ the pullback of $E$ to $\tilde B$, we find $\chi(\tilde E) = \chi(\tilde B)\chi(F)$, so because $\tilde E\to E, \tilde B\to B$ have the same number of sheets, it suffices to prove that for either one, $\chi(\tilde X) = n \tilde X$, $n$ being the number of sheets. Note that the kernel of $\pi_1(B)\to GL(H_*(F;\mathbb Z/2))$ is of course normal in $B$ so $\tilde B\to B$ is a Galois cover. $\tilde E\to E$ is pulled back from it, so it is Galois too; so we reduced to Galois covers.

Step 2 : $\chi(E)$ in terms of THH. Let $X$ be finite spectrum. I claim that applying $\THH$ to the morphism $\Sp^\omega\overset{X\otimes -}\to \Sp^\omega $ yields the morphism $\chi(X) : \Sph\to \Sph$ , where I use $\otimes$ for the smash product of spectra. I will abuse notation and write functors as $\Sp\to \Sp$ to avoid having to write $^\omega$ every time, but this means I will need to justify why my functors preserve compacts.

This is a fairly well-known statement, I can expend on it if needed, but I'll take it for granted for now.

Step 3 : For a finitely dominated space $X$, observe that the functor $X^* : \Sp\to \Sp^X$ taking a spectrum to the constant parametrized spectrum over $X$ with that value preserves compacts - this is precisely because $X$ is finitely dominated, so that homotopy limits over $X$ preserve filtered colimits.

It therefore induces a morphism on $\THH$, and by the Goodwillie-Jones isomorphism (I never know the actual name, someone should tell me what it is), we have $\THH( (\Sp^X)^\omega) = \THH(\mathrm{Perf}(\Sph[\Omega X])) = \THH(\Sph[\Omega X]) = \Sigma^\infty_+ LX$, where $L$ denotes the free loop space (I'm assuming $X$ is connected, to identify parametrized spectra over $X$ with $\Sph[\Omega X]$). This morphism is therefore a morphism $\Sph\to \Sigma^\infty_+ LX$, which corresponds to an element $f_X \in \pi_0\Sigma^\infty_+ LX \cong \bigoplus_{[\gamma]\in \pi_1(X)/conj} \mathbb Z\cdot [\gamma]$, where I use brackets $[\gamma]$ to denote the conjugacy class of $\gamma$ in $\pi_1$ (or the free homotopy class of a loop).

This $f_X$ is therefore a finite sum of the form $\sum_i n_i[\gamma_i]$, where I take pairwise distinct $[\gamma_i]$'s.

Warning : It might sound "obvious" that the only $\gamma_i$ that shows up is the trivial loop. Hopefully, this is true, but it is not obvious, and as far as I know, not known so far. As I explain below, this statement is equivalent to Bass' trace conjecture.

Step 4: Relating to Bass' trace conjecture. This $f_X$ is defined in terms of $\THH$, so one way to study it is to relate it to $K$-theory. Indeed, the trace map $K\to \THH$ is natural, so we have a commutative square $$\require{AMScd}\begin{CD}K(\Sph) @>>> K(\Sph[\Omega X]) \\ @VVV @VVV \\ \THH(\Sph) @>>> \THH(\Sph[\Omega X]) \end{CD}$$

This diagram, on $\pi_0$, becomes $$\begin{CD}\mathbb Z = K_0(\mathbb Z) @>>> K_0(\mathbb Z[\pi_1 X]) \\ @V{id}VV @VVV \\ \mathbb Z = \THH_0(\mathbb S) @>f_X>> \bigoplus_{\pi_1(X)/conj}\mathbb Z = \mathrm{HH}_0(\mathbb Z[\pi_1(X)]\end{CD}$$

In particular, as the left vertical map is surjective, $f_X$ is in the image of the trace map $K_0(\mathbb Z[\pi_1X])\to \mathrm{HH}_0(\mathbb Z[\pi_1(X)])$. This is exactly the map that Bass' trace conjecture is about, namely:

Conjecture : Let $G$ be a group. The trace map $K_0(\mathbb Z[G])\to \mathrm{HH}_0(\mathbb Z[G]) \cong \bigoplus_{G/conj}\mathbb Z$ lands entirely in the summand corresponding to the neutral element of $G$.

Note, not necessary for the proof: The map $K_0(\mathbb S)\to K_0(\mathbb S[\Omega X])$ sends $1$ to the (unreduced) Wall finiteness obstruction of $X$. Because free modules over $\Sph[\Omega X]$ are sent to the summand of the neutral element of $\pi_1(X)$, and because any element of the reduced $K_0(\mathbb Z[\pi_1(X)])$ can be realized by some finitely dominated space, Bass' trace conjecture is equivalent to the statement that only the trivial loop shows up as one of the $\gamma_i$'s for all $X$. In particular, I will not claim that this is the case. Note, however, that if $X$ is finite, i.e. Wall's finiteness obstruction vanishes (in reduced $K$-theory), then only the trivial loop shows up.

Bass' trace conjecture is open, but some things are known about it:

Theorem ([Lin, Lemma 4.1], [BH, Theorem A]): Let $G$ be a group, and $g\in G$ an element such that $K_0(\mathbb ZG)\to \mathrm{HH}_0(\mathbb ZG)$ hits the summand corresponding to $g$. There exists an integer $m\geq 1$ such that for all $s\geq 1$, $g$ and $g^{s^m}$ are conjugate.

In particular, the $\gamma_i$'s that show up in $f_X$ have this property.

The core of the proof is in :

Corollary : Let $X$ be a finitely dominated space and write $f_X = \sum_i n_i[\gamma_i]$ as above. The $\gamma_i$'s vanish in any finite quotient of $\pi_1(X)$.

This is clear, as the image of $\gamma_i$ in the finite quotient has the same property, and has finite order.

Step 5 : A big diagram. Return to our situation: $p: E\to B$ is a finite Galois cover, with fiber $F$ a finite set.

Consider the following commutative diagram (there are diagonal arrows so I cannot use AMScd unfortunately, which is why I used a picture - hopefully it's readable enough, sorry for the inconvenience):

enter image description here

I use the following notation: for a map of space $f$, $f^*$ is the restriction along $f$ for parametrized spectra, and $f_!$ its (derived) left adjoint; and if $f: X\to *$ is the projection to a point, I write $X^*$ (resp. $_!$) for $f^*$ (resp. $f_!$).

From this description, it should be clear why both triangles commute. Furthermore, all functors involved preserve compact objects : indeed, their right adjoints preserve filtered colimits, either because they are of the form $f^*$, or because they are of the form $f_*$ (the right adjoint of $f^*$) for some $f$ with finitely dominated fibers.

In particular, I can apply $\THH$ to it, and get a commutative diagram.

Step 6 : The $\THH$-diagram.

It looks like (sorry, I had the same issue, I hope it's readable) :

enter image description here

The maps $f_E,f_B, \THH(p^*)$ are there by definition. The others follow from the following general claim:

Claim : Let $f:X\to Y$ be a map of spaces. The induced functor $f_!: \Sp^X\to \Sp^Y$ induces $\Sigma^\infty_+ Lf: \Sigma^\infty_+ LX\to \Sigma^\infty_+ LY$ on $\THH$.

This is not too hard to prove, and I think fairly well-known too. If needed, I can also add details about it (if it helps: I only need this statement on $\pi_0$ anyway !).

Important observation : the middle composite is $\THH$ applied to the middle composite, i.e. $\THH$ applied to $\Sp\overset{E^*}\to \Sp^E\overset{E_!}\to \Sp$, i.e. to $\Sp\overset{E\otimes -}\to \Sp$. Therefore, the middle composite is given by $\chi(E)$, by Step 2 above.

Step 7: Understanding $\THH(p^*)$.

To state the theorem, let me just make an observation: $p : E\to B$ is a finite covering space, therefore so is $Lp: LE\to LB$. In particular, this map has a Becker-Gottlieb transfer, which I will denote by $(Lp)^!$.

Theorem ([LM, Corollary 1.5]): Let $E\to B$ be a finite covering space. With respect to the identification $\THH((\Sp^X)^\omega)\simeq \Sigma^\infty_+ LX$, we have that $\THH(p^*)$ is identified with $(Lp)^! : \Sigma^\infty_+ LB\to \Sigma^\infty_+ LE$.

Corollary : In the above diagram, the vertical composite is $\Sigma^\infty_+ Lp\circ (Lp)^! : \Sigma^\infty_+ LB\to \Sigma^\infty_+ LB$. In particular, it is given by "multiplication by $\chi($fiber)$.

Warning : it is tempting to stop here, but as I will explain below there is a small subtlety, and this is where the Linnell-Hesselholt result comes in. A first subtlety is that in general, $LB$ is disconnected, and so the multiplication by "$\chi($fiber$)$" depends on the component under consideration.

Step 8 : Putting things together.

By the Important observation in Step 6, the middle composite of the diagram is $\chi(E)$. Because the bottom triangle commutes, one can write this composite as right-down-up instead of righ-right, and then up-down-down-up, because the top triangle commutes.

Going up gives us $f_B = \sum_i n_i[\gamma_i]$, then down-down is multiplication by $\chi(\mathrm{fib}_{\gamma_i})$, where I let $\mathrm{fib}_{\gamma_i}$ denote the fiber of $LE\to LB$ at the loop $\gamma_i$.

By Step 4, each $\gamma_i$ vanishes in $\pi_1(B)/\pi_1(E)$ (which is a finite quotient of $\pi_1(B)$, as $\pi_1(E)$ is normal in $\pi_1(B)$, and $E\to B$ is a finite cover). In particular, one can compute this fiber to be exactly $\pi_1(B)/\pi_1(E)$.

Here's the argument:

Because $E\to B$ is a Galois cover, letting $H\triangleleft G$ denote the corresponding normal inclusion, we have a homotopy pullback

$$\begin{CD}E@>>> * \\ @VVV @VVV \\ B @>>> B(G/H) \end{CD}$$

which remains a pullback after applying $L$. Now taking the fiber over $\gamma_i$, as $\gamma_i$ vanishes in $L(B(G/H))$ (more precisely: the composite $\{\gamma_i\}\to LB \to L(B(G/H))$ is homotopic to $L(*\to B(G/H))$), we can just take the outer pullback, which is $\Omega (LB(G/H), triv)$. For any group $K$, $\Omega(LBK, triv) = K$.

This proves the claim, and in particular the Euler characteristic of the fiber is $|G/H| = \chi(F)$ regardless of the $\gamma_i$, so that at the end of the day, the composite up-down-down sends $1$ to $\sum_i n_i\chi(F)[\gamma_i]$, and then going up one last time sends all loops to $1$, so up-down-down-up sends $1$ to $\sum_i n_i \chi(F)$.

So $\chi(E) = \sum_i n_i \chi(F)$. But $\sum_i n_i$ only depends on $B$ (the $n_i$'s are the ones appearing in $f_B$), so we can apply this to the cover $B\to B$, and we find $\sum_i n_i = \chi(B)$.

Therefore, $\chi(E) = \chi(B)\chi(F)$, as claimed, and we are (finally!) done.

Conclusion : The point is that in Pedersen's paper, $\chi(p)$ alone cannot be related to $\chi(F)$, it's exactly by using also $B$ that one gets to $\chi(F)$. Now this fact uses strong results about Bass' trace conjecture, and it seems like an elementary proof cannot explain this kind of "cancellation".

But maybe I am wrong. At least, I think this is a proof - of course it would be better if a more elementary one existed. (I mentioned that I knew a simpler proof, but it still uses the Linnell-Hesselholt result)

References :

[BH] : Berrick-Hesselholt, Topological Hochschild homology and the Bass trace conjecture

[Lin] : Linnell, Decomposition of augmentation ideals and relation modules

[LM] : Lind-Malkiewich, The transfer map of free loop spaces

  • $\begingroup$ It will take time for me to digest this, but thanks for your write up. I am surprised that there appears to be no elementary proof. $\endgroup$
    – John Klein
    Mar 2, 2022 at 12:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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