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

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):

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) :

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*

28more comments