I'm following Arone's lectures on Goodwillie calculus from Munster 2015. There he left an exercise:

Find $\partial_kF$ for $F: \text{Top}_* \to \text{Sp}$ given by $F(X) = \Sigma^\infty X^{\wedge n}/\Sigma_n$ where $X^{\wedge n} = X \wedge \cdots \wedge X$ and $\Sigma_n$ is the symmetric group, which acts on $X^{\wedge n}$ by permuting factors.

In the lectures he solves the case $n = 2$, that gives $\partial_1 F \simeq \Sigma^\infty\Sigma\mathbb{RP}^\infty$, $\partial_2F \simeq \Sigma^\infty$ and $\partial_nF \simeq *$.

I do not know how to work out the general case. I know that $\partial_kF \simeq \text{hocolim }\Omega^{km}\text{cr}_kF(S^m, \dots, S^m)$.

For $k = 1$ then $\partial_1F \simeq \text{hocolim }\Omega^mF(S^m)$. My idea is to write $F(S^m) = S^m \wedge G(S^m)$ for some $G$ in order to cancel $\Omega^m$. This can be done by choosing a $\phi: (S^m)^{\wedge n} \to (S^m)^{\wedge (n-1)}$ such that $(x_1, \dots, x_n) \mapsto (x_1 + \cdots + x_m, \phi(x_1, \dots, x_m))$ is a homemorphism. In this case $(S^m)^{\wedge n}/\Sigma_n \cong S^m \wedge (S^m)^{\wedge (n-1)}/\Sigma_n$ where the $\Sigma_n$-action on $(S^m)^{\wedge (n-1)}$ depends on $\phi$. Unfortunately, I was not able to find a $\phi$ such that this action looks nice or familiar as in the case $n = 2$.

Alternatively, I tought $(S^m)\text{^}^n/\Sigma_n \cong \text{subset}_{\leq n}(\mathbb{R^m})^+ \cong S^m \wedge \text{subset}_{\leq n, 0}(\mathbb{R^m})^+$ where $(\cdots)^+$ means compactification, $\text{subset}_{\leq n}(\cdots)$ is the space of subsets of cardinality $\leq n$ and $\text{subset}_{\leq n, 0}(\mathbb{R^m})$ is the space of those with barycenter 0. Then $\partial_1F \simeq \Sigma^\infty\text{subset}_{\leq n, 0}(\mathbb{R^\infty})$. This agrees with the previous result for $n = 2$ since any two unordered points can be identified by the line joining them ($\mathbb{RP}^\infty$) and the distance from 0 ($\Sigma$, after compactification). But

  1. It does not seem to be the "right solution".
  2. What about $\partial_nF$ for $n > 1$? I do not know how to proceed.

Thanks in advance!

[original post]


Your solution is correct as far as it goes. Let $\rho^{n-1}\cong {\mathbb R}^{n-1}$ denote the reduced standard representation of $\Sigma_n$. One way to define $\rho^{n-1}$ explicitly is to say that it is the orthogonal complement of the diagonal in $\mathbb R^n$. Let's write it out: $$ \rho^{n-1}=\{(x_1, \ldots, x_n)\in\mathbb R^n\mid x_1+\cdots+x_n=0\} $$ Let $\hat S^{n-1}$ denote the one-point compactification of $\rho^{n-1}$. This is a sphere of dimension $n-1$ with a specified action of $\Sigma_n$. More generally, for any $m\ge 0$, let us write $\hat S^{(n-1)m}$ for $(\hat S^{n-1})^{\wedge m}$.

Furthermore, $\hat S^{(n-1)m}$ is the one-point compactification of the representation $(\rho^{n-1})^m$. It follows that $\hat S^{(n-1)m}$ is homeomorphic to the unreduced suspension of the unit sphere in $(\rho^{n-1})^m$. Let me denote the unit sphere by $\tilde S^{(n-1)m-1}$. We have a $\Sigma_n$-equivariant homeomorphism $$\hat S^{(n-1)m}\cong \Sigma \tilde S^{(n-1)m-1},$$ where $\Sigma$ indicates unreduced suspension.

Now let us take colimit as $m\to \infty$. We obtain the following equivalence, where both sides are just notation for the appropriate colimits $$\hat S^{(n-1)\infty}\cong \Sigma \tilde S^{(n-1)\infty-1}.$$

Let $F_n(X)=\Sigma^\infty X^{\wedge n}_{\Sigma_n}$. By your calculation, $$\partial_1(F_n)\simeq \Sigma^\infty \Sigma {\tilde S^{(n-1)\infty-1}}/_{\Sigma_n}.$$

To analyze this further, let us calculate the fixed point set ${\tilde S^{(n-1)\infty-1}}^H$, where $H\subset \Sigma_n$. The following lemma is a quite easy exercise

Lemma: Let $H$ be a subgroup of $\Sigma_n$. Define $o(H)$ to be the number of orbits of the action of $H$ on $\{1, \ldots, n\}$. There is a homeomorphism $${\tilde S^{(n-1)-1}}^H\cong S^{o(H)-2}.$$ Note that if $H$ is a transitive subgroup of $\Sigma_n$, then $o(H)=1$ and the right hand side of this equation is $S^{-1}$, which is taken to be the empty set. More generally for $m\ge 1$, there is a homeomorphism $${\tilde S^{(n-1)m-1}}^H\cong S^{(o(H)-1)m-1}.$$ Taking limit as $m\to \infty$ we obtain the following

Corollary $(\tilde S^{(n-1)\infty-1})^H$ is contractible if $H$ is non-transitive and is empty if $H$ is transitive.

This means that $\tilde S^{(n-1)\infty-1}$ us a universal space for the family of non-transitive subgroups of $\Sigma_n$. Let us therefore introduce the notation $E_{\mathrm{nt}}\Sigma_n=\tilde S^{(n-1)\infty-1}$ and $B_{\mathrm{nt}}\Sigma_n={E_{\mathrm{nt}}\Sigma_n}/_{\Sigma_n}$. $B_{\mathrm{nt}}\Sigma_n$ is sometimes called the classifying space for the familty of non-transitive subgroups of $\Sigma_n$. We have proved the following

$$ \partial_1 F_n \simeq \Sigma^\infty \Sigma B_{\mathrm{nt}}\Sigma_n $$

Note that when $n=2$, the family of non-transitive subgroups of $\Sigma_2$ consists of just the trivial group. Because of this, $B_{\mathrm{nt}}\Sigma_2=B\Sigma_2=\mathbb R P^\infty$. So we get the expected answer for $n=2$.

But there is another interesting way to think of the spectra $\partial_1 F_n$. Let $Sp^n(X)=X^n/_{\Sigma_n}$, where $X^n$ denotes $n$-fold cartesian product of $X$ with itself. It follows that for a pointed space $X$, $X^{\wedge n}/_{\Sigma_n}\cong Sp^n(X)/Sp^{n-1}(X)$. The construction $Sp$ extends from spaces to spectra. Let $\mathtt S$ denote the sphere spectrum. There is a sequence of spectra $$ \mathtt S=Sp^1(\mathtt S)\to\cdots \to Sp^n(\mathtt S)\to \cdots Sp^\infty(\mathtt S)=H\mathbb Z $$ Note that $Sp^\infty(\mathtt S)=H\mathbb Z$ by the Dold-Thom theorem. Let us call this sequence the symmetric powers filtration of the Eilenberg Mac Lane spectrum.

The point of this is that it is easy to see that the spectra $\partial_1 F_n$ are equivalent to the subquotients of the symmetric powers filtration.

$$ \partial_1 F_n\simeq Sp^n(\mathtt S)/Sp^{n-1}(\mathtt S) $$

Remark The equivalence $Sp^n(\mathtt S)/Sp^{n-1}(\mathtt S)\simeq \Sigma^\infty \Sigma B_{\mathrm{nt}} \Sigma_n$ was initially discovered by Kathryn Lesh.

The spectra $Sp^n(\mathtt S)/Sp^{n-1}(\mathtt S)$ have been studied quite a lot, from various perspectives. The following result summarizes some facts about them that were known for a long time.

Theorem The spectrum $Sp^n(\mathtt S)/Sp^{n-1}(\mathtt S)$ and therefore also $\partial_1 F_n$ has the following properties

  1. It is rationally trivial for $n>1$
  2. It is integrally trivial unless $n$ is a power of a prime.
  3. If $n$ is a power of a prime $p$ then it is a $p$-local spectrum.

When $n=p^k$, the mod $p$-homology of $Sp^{p^k}(\mathtt S)/Sp^{p^k-1}(\mathtt S)$ is well-understood. It was initially calculated by Nakaoka. You can learn more cool details about these spectra for example in these notes of a lecture by Mike Hopkins.

Now let's consider $\partial_k F_n$ for $k\ge 1$. You can express $\partial_k F_n$ in terms of $\partial_1 F_m$ (for varying $m$) using Goodwillie's model for $\partial_k$ as the multi-linearized cross-effect. The $k$-th cross effect of $F_n$ is the following functor of $k$ variables $$ \bigvee_{n_1+\cdots + n_k=n, n_i>0} F_{n_1}(X_1)\wedge \ldots \wedge F_{n_k}(X_k) $$ After multi-linearization, we obtain the following formula $$ \partial_k F_n\simeq \bigvee_{n_1+\cdots + n_k=n, n_i>0} \partial_1 F_{n_1}\wedge \ldots \wedge \partial_1 F_{n_k} $$ where the action of $\Sigma_k$ is self-evident. For example $$ \partial_2 F_4 \simeq \partial_1 F_3 \wedge \partial_1 F_1 \vee \partial_1 F_1 \wedge \partial_1 F_3 \vee \partial_1 F_2\wedge \partial_1 F_2 $$ Where $\Sigma_2$ acts by permuting the first two wedge factors, and also permuting the two smash factors of the last wedge factor.

  • $\begingroup$ This is a great answer, thank you. I learnt more math today than in the last month. Do you know other exercises like that? It would be nice to have a Goodwillie calculus textbook one day! However, I have three questions about your answer - if you have time, of course. $\endgroup$ – Marco Nervo Nov 13 '20 at 21:19
  • $\begingroup$ 1) the equivalence $\partial_1F_n \simeq Sp^n(\mathbb{S})/Sp^{n-1}(\mathbb{S})$ does not depend on the Dold-Thom theorem that you cited before, isn't it? $\endgroup$ – Marco Nervo Nov 13 '20 at 21:19
  • 1
    $\begingroup$ 3) I do not understand how to get your result from $\partial_kF_n \simeq \text{hocolim } \Omega^{km}\text{cr}_k(S^m, ..., S^m)$ using your calculation of the cross-effect. Moreover, I am a bit surprised because I would expect $\partial_k$ to be a wedge of smashes of just $\partial_1$'s and not of $\partial_1, ..., \partial_k$. $\endgroup$ – Marco Nervo Nov 13 '20 at 22:12
  • 1
    $\begingroup$ 1) You are right, this part does not depend on Dold-Thom. 2) This is why the wedge sum in my formula is indexed by $k$-tuples $(n_1,\ldots, n_k)$ where $n_i>0$ for all $i$. Taking the homotopy fiber as you say has the effect of knocking out the tuples where some of the $n_i$s are zero. 3) I had some confusing typos in the last two formulas. Indeed I accidentally wrote $\partial_k$ where it should be $\partial_1$. Fixed them. $\endgroup$ – Gregory Arone Nov 14 '20 at 8:41
  • 1
    $\begingroup$ 2) I have used exactly the total homotopy fiber that is in the video lecture. I skipped the details of the calculation, but if you trace through all the steps you should find that the total fiber of this diagram is what I wrote. 3) We are in a special case when the functor of $k$ variables $cr_kF(X_1, \ldots, X_k)$ is equivalent to the smash product of the form $G_1(X_1)\wedge \ldots \wedge G_k(X_k)$. In this case hocolim $\Omega^{km} F(S^m, \ldots, S^m)\simeq$ hocolim $\Omega^m G_1(S^m)\wedge \ldots \wedge \Omega^m G_k(S^m)$. If you want to continue the discussion, write me an email. $\endgroup$ – Gregory Arone Nov 14 '20 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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