Chromatic Homotopy Theory
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1 Nilpotence Theorem
\(\MU \) is a strong invariant of spaces.
If \(R\) has a map \(R\otimes R \to R\), and \(\alpha \) is a homotopy class in \(R\) in the kernel of the Hurewicz map, then by the nilpotence theorem it is nilpotent. On the other hand let \(F \to X\) be a map as in the
above theorem. It is smashnilpotent iff the dual map \(\SP \to DF\otimes X\) is, so WLOG \(F\) is a sphere. If the map is in the kernel of the \(\MU _*\) Hurewicz map, then its nulhomotopy takes place in \(\MU \otimes
X'\) for some finite spectrum in X, so we can assume \(X\) is finite, and after suspending, it can be assumed that we have a map \(\SP ^k \to X\), where \(X\) is \(0\)connected. Replacing \(X\) with \(\oplus
X^{\otimes n}\), we see that the following version of the theorem is equivalent:
Being nilpotent in the \(E\)Hurewicz map is equivalent to \(E\otimes R[\alpha ^{1}]=0\), a condition that can be checked \(p\)locally, so we will work \(p\)locally.
To prove this, we will interpolate between the \(\MU \), and the ring spectrum \(\SP \), which obviously detects nilpotence. Namely, \(\MU \) is the Thom spectrum of the identity of \(BU\), and \(BU = \Omega SU\), so \(X(n)\),
defined as the Thom spectra of \(\Omega SU(n)\) interpolate between \(X(0)=\SP \) and \(\varinjlim X(n) = \MU \). In particular if \(\alpha \otimes 1_{\MU }\) is \(0\), then so is \(\alpha \otimes 1_{X(n)}\) for
sufficiently large \(n\). Thus it suffices to show:
Fix an \(n\). To prove this, we will again interpolate between the two as follows: There is a fibration \(SU(n) \to SU(n+1) \to S^{2n+1}\), which we can loop. Then since \(\Omega S^{2n+1} = JS^{2n})\) comes with a
natural filtration \(J_kS^{2n}\). Let \(G_k'\) be the result of pulling back the fibration along \(J_{p^{k1}}\) in this filtration, and let \(G_k\) be its Thom spectrum. The following two results will then prove the
nilpotence theorem:
Let’s prove the first one first. Since \(\Omega SU(n)\) is an \(E_2\)space, \(X(n)\) is an \(E_2\)ring. We would like to first establish:

Proposition 1.6.
\[\Ext _{X(n+1)_*X(n+1)}(X(n+1)_*,X(n+1)_*(G_k\otimes R))\]
has a vanishing line of slope tending to \(0\) as \(k \to \infty \).
The Serre spectral sequence and the Thom isomorphism give that the homology of \(X(n)\) is \(\ZZ [b_1,\dots ,b_{n1}]\), where the classes \(b_i\) come from restricting the complex orientation \(\Sigma ^{2}\CC \PP
^\infty \to \MU \) to a “truncated orientation" \(\Sigma ^{2}\CC \PP ^n \to X(n)\). The EilenbergMoore specrtal sequence shows that \(G_k\) has homology \(\ZZ [b_1,\dots ,b_{n1}]\{1,b_n,\dots
,b_n^{p^k1}\}\). The truncated orientation causes Atiyah Hirzebruch spectral sequences to degenerate, resulting in the following:

Lemma 1.7. If \(n \geq m\), \(X(n)_*X_m = X(n)_*[b_1,\dots b_{m}]\).
\(X(n+1)_*G_k = X(n+1)_*[b_1,\dots ,b_n]\{1,b_{n+1},\dots ,b_{n+1}^{p^k1}\}\).
In particular, \(X(n+1)_*X(n+1) = X(n+1)_*\otimes \ZZ [b_1,\dots ,b_n]\), so it is flat over \(X(n+1)_*\) and is in fact a split Hopf algebroid. Thus the \(E_2\) term of the \(X(n+1)\)based Adams spectral sequence for
\(G_k\otimes R\) is
\[\Ext _{X(n+1)_*X(n+1)}(X(n+1)_*,X(n+1)_*(G_k\otimes R))\]
Since \(X(n+1)_*(G_k)\) is flat, we get that this is
\[\Ext _{X(n+1)_*X(n+1)}(X(n+1)_*,X(n+1)_*(G_k) \otimes _{X(n+1)_*} X(n+1)_*R))\]
Now since the Hopf algebroid is split, this is just (the localization at \(p\) is supressed):
\[\Ext _{\ZZ [b_1,\dots ,b_n]}(\ZZ ,\ZZ [b_1,\dots ,b_{n1}]\{1,b_{n},\dots , b_n^{p^k1}\} \otimes X(n+1)_*R))\]
We can identify \(\ZZ [b_1,\dots ,b_{n1}]\{1,b_{n},\dots , b_n^{p^k1}\}\) as a left comodule with the cotensor product \(\ZZ [b_1,\dots ,b_{n}]\square _{\ZZ [b_n]}\ZZ \{1,b_n,\dots ,b_n^{p^k1}\}\), so
that
\[(\ZZ [b_1,\dots ,b_{n}]\square _{\ZZ [b_n]}\ZZ \{1,b_n,\dots ,b_n^{p^k1}\}) \otimes X(n+1)_*R\]
\[=\ZZ [b_1,\dots ,b_{n}]\square _{\ZZ [b_n]}(\ZZ \{1,b_n,\dots ,b_n^{p^k1}\} \otimes X(n+1)_*R)\]
(the isomorphism is not totally obvious)
Since the comodule is extended from \(\ZZ [b_i]\), we can identify the \(\Ext \) term with
\[\Ext _{\ZZ [b_n]}(\ZZ ,\ZZ \{1,b_{n},\dots , b_n^{p^k1}\} \otimes X(n+1)_*R)\]
This is saying that action of the group \(\Spec \ZZ [b_1,\dots ,b_n]\) is induced from the subgroup \(\Spec \ZZ [b_n]\), so the group cohomology can be computed via Shapiro’s Lemma.
We would like to extablish a vanishing line on this, and so to do so, we can filter by powers of \(p\) to get a May spectral sequence from \(\Ext _{\FF _p[b_n]}(\FF _p,\FF _p\{1,b_{n},\dots , b_n^{p^k1}\} \otimes \Gr
X(n+1)_*R)\) converging to the the Ext group. Moreover, by filtering \(\Gr X(n+1)_*R\) by degree, and taking the associated spectral sequence, it suffices to establish a vanishing line for \(\Ext _{\FF _p[b_n]}(\FF _p,\FF
_p\{1,b_{n},\dots , b_n^{p^k1}\}\). But as a coalgebra \(\FF _p[b_n] = \otimes \FF _p[b_n^{p^i}]/b_n^{p^{i+1}}\). By a change of rings this is
\[\Ext _{\FF _p[b_n^{p^k}]}(\FF _p,\FF _p \otimes \Gr X(n+1)_*R)\]
But this is a sum of copies of \(\Ext _{\FF _p[b_n^{p^k}]}(\FF _p,\sigma ^k\FF _p)\), which has a vanishing line of slope \((np^{k+1}1)^{1}\) by looking at a minimal resolution. This goes to zero as \(k \to \infty \).
Now that we have our vanishing line, we can argue for Proposition 1.4 as follows:

Proof. let \(\alpha \) be a class in homotopy that is \(X(n+1)\) nilpotent. WLOG it is actually zero. Thus it has positive \(X(n+1)\)Adams filtration, so let \((ts,s)\) be its
coordinates in the Adams spectral sequence, with \(s>0\). Choose \(k\) so that the slope of the line from the origin to \((ts,s)\) is greater than that of the vanishing line established. Then for any class \(\beta \) in
\(G_k\otimes R\), \(\beta \alpha ^l\) is eventually above the vanishing line so is \(0\), so it is \(0\). Thus \(G_k \otimes R[\alpha ^{1}] = 0\) so \(\alpha \) is \(G_k\) nilpotent. □
Now let’s start to examine Theorem 1.5.
First we will produce a \(p\)local fibre sequence \(G'_k \to G'_{k+1} \to J_{p1}S^{2np^k}\). To do this, first note that the map \(J_{p^k1}S^{2n} \to JS^{2n}\) is the homotopy fibre of the JamesHopf map
\(JS^{2n} \to JS^{2np^k}\). To see this, one can see that there is an isomorphism in mod \(p\) homology via the Serre SS. Thus we can paste together the fibre sequences below.
But then we can consider the following diagram of cartesian squares:
\(G'_{k+1}\) is indeed the homotopy pullback since the pasted pullback square on the right is the same kind (i.e same on homology) as the one from before. This creates the desired fibre sequence. There are two special
things about this fibre sequence. One is that the inclusion of the fibre is an injection on mod \(p\) homology, and the other is that the action of \(\Omega J_{p1}S^{2np^k}\) extends to an action of \(\Omega ^2 S^{2np^k+1}\).
We will show that in this situation, the inclusion of the fibre induces a Bousfield equivalence on Thom spectra.
We will first study the categories \(F_r\) where the objects have the following data: \(E\) is a space with a map \(\xi \) to BU and a map \(\pi \) to \(J_{r}S^{2m}\). For \(q \leq r\), let \(E_q\) denote the homotopy
pullback of \(\pi \) via the inclusion \(J_{q}S^{2m} \to J_{r}S^{2m}\), and note it lies in \(F_q\). In particular, \(E_0\) is the fibre of \(\pi \) (which in our case is \(G'_k\)). Let \(()^\xi \) be the Thom spectrum
with respect to \(\xi \).
We will produce a natural self map \(b: \Sigma ^{2m(r+1)2} E_0 \to E_0\) such that \(E_0\) is Bousfield equivalent to \(b^{1}E_0 \vee E\). Then we will show that under our conditions, \(b^{1}E_0 = 0\). The point is
that \(b\) induces zero on mod \(p\) cohomology because of our assumption on the inclusion of the fibre. \(b\) comes from the action of an element \(\beta \) in \(\pi _*\Omega J_rS^{2m}_+\). The action extends to \(\Omega
^2S^{2m+1}_+\), and we will show that the image of \(\beta \) is \(p\)torsion. Thus since \(\Omega ^2S^{2m+1}_+[\beta ^{1}]\) is an \(E_2\)ring with \(p=0\), by Mahowald’s theorem it is a sum of \(H\FF _p\)s. Then
since \(\beta \) induces \(0\) on \(H\FF _{p*}\) it is \(0\).
\((\pi ,1)\) gives a map \(E \to J_rS^{2m}\times E\), and the latter comes with the map \(\xi \pi _2\) to \(BU\), so this is a map of spaces over \(BU\). Moreover, there is a projection to \((J_rS^{2m})^2\) and \((\pi
,1)\) covers the diagonal map. We can equip \(J_rS^{2m}\times E\) with the product filtration, but then \((\pi ,1)\) doesn’t preserve the filtration.
However, it is canonically homotopic to a map that does in the following way. Consider the two simplicial spaces, the first where the \(n^{th}\) space is \((S^2m)^{n}\), where the maps come from the monoidal structure of the
cartesian product, and let the second be where the \(n^{th}\) space is \(\bigvee _1^n S^{2m}\), and the maps come from the \(E_1\) cogroup structure on \(S^{2m}\). Then the natural inclusion of the latter as the \(2m\)
skeleton of the first extends to a map of simplicial spaces (in the infinity category of spaces).
Thus we can replace the diagonal map from before with the composite \(J_rS^{2m} \to J_r\vee _1^2S^{2m} \to J_rS^{2m} \times J_rS^{2m}\), which is filtration preserving, and lifting to \(E\) gives an essentially canonical
homotopy to a filtration preserving map.
Passing to Thom spectra, we get a filtration preserving map \(E^\xi \to J_rS^{2m}_+ \otimes E^\xi \). We can identify \(J_rS^{2m}_+\) with a truncated free \(E_1\)ring generated by \(S^{2m}\), i.e. \(\bigvee _0^r
S^{2mi}\). Via the projection onto each factor, since the map is filtration preserving we can define natural tranformations \(\theta _{i}\) as the composite \(E_{q}^\xi \to J_rS^{2m}_+ \otimes E^\xi \to \Sigma ^{2mi}
E_{qi}^\xi \). We will use the same notation to denote the map on the cofibres \(E_q^\xi /E_{qj}^\xi \), and the unfiltered version \(E^\xi /E_0^\xi \to E\).

Proof. The map \((\pi ,1,1)\) on \(E\) factors as both \((1,(\pi ,1)) \circ (\pi ,1)\), after passing to Thom spectra and projecting giving \(\theta _i \circ \theta _j\),
and also as \((\Delta , 1) \circ (\pi ,1)\),which gives \((i,j) \theta _{i+j}\) after passing to Thom spectra and projecting since the diagonal induces multiplication by \((i,j)\) on the factors of \(J_r(S^{2m})_+\).
□
Since the associated graded of \(J_r(S^{2m})\) is just spheres, the associated graded of \(E^\xi \) is also very simple.

Proof. Consider the cartesian diagram of pairs:
The lower map is a homology equivalence so the top one is too since the square is cartesian. Via the Thom isomorphism, we get a homology equivalence between the relative Thom spectra. □

Proof. Use induction on \(j\) to see the following map of cofibre sequences is an equivalence
The map on the right is an equivalence since \(\theta _j \circ \theta _1 = (j+1)\theta _{j+1}\), and the other two are equivalences by the lemma, and \(j+1\) is invertible. □
Note in the situation of application \(r=p1\) so \(r!\) is invertible since we are working \(p\)locally.
Now we can define the natural self map \(b\) on the fibre as the following composite:
\[\Sigma ^{2m(r+1)2}E^\xi _0 \xrightarrow {\theta _r^{1}} \Sigma ^{2m2}E_r^\xi /E_{r1}^\xi \xrightarrow {\delta } \Sigma ^{2m1}E_{r1}^\xi \xrightarrow {\theta _1^{1}} \Sigma ^{1}E^\xi
_{r}/E^\xi _0 \xrightarrow {\delta } E_0^\xi \]
In the situation of application, \(b\) induces \(0\) on mod \(p\) homology since the second \(\delta \) does as the inclusion of the fibre is injective on mod \(p\) homology.

Proof. Suppose \(X\otimes E_0^\xi = 0\), then clearly \(X \otimes b^{1}E_0^\xi \) is too, and since the associated graded of \(E^\xi \) are suspensions of \(E_0\), \(X\otimes
E^\xi = 0\) too. Conversely, if \(X \otimes E^\xi = 0\), the two connecting homomorphisms in the definition of \(b\) are equivalences after tensoring with \(X\), so \(X\otimes b\) is an equivalence, and \(0 = X\otimes
b^{1}E^\xi _0 =X\otimes E^\xi _0\). □
Thus to complete the proof it will suffice to show that \(b^{1}E_0^\xi \) is \(0\) in our situation. Observe that the inclusion of the fibre \(E_0 \to E\) can be taken to be a map in \(F_r\), so that by naturality of \(b\) we get
the following diagram:
In otherwords, \(E_0^\xi \) is a module over the ring \(\Omega J_rS^{2m}_+\) and \(b\) is given by multiplication by \(\beta :=b(*)\) where \(*\) is the single point object in \(F_r\). So we want \(\beta ^{1}E_0^\xi =0\).
Let’s examine \(\beta \), i.e \(E = *\). Here, since the map to \(BU\) is trivial, the Thom spectrum is just \(\Sigma ^\infty _+\). \(\beta \) is then the composite:
\[\Sigma ^{2m(r+1)2}\Omega J_r S^{2m}_+ \xrightarrow {\theta _r^{1}} \Sigma ^{2m1}E_{r1} \xrightarrow {\delta } \Sigma ^{2m1}(E_{r1+}) \xrightarrow {\theta _1^{1}} \Omega J_r S^{2m}
\xrightarrow {\delta } \Omega J_r S^{2m}_+\]
We can identify \(\theta _1: E^\xi /E_0^\xi \to S^{2m}\otimes E^\xi \) as a map \(S^1\otimes \Omega J_r S^{2m} \to S^{2m}\) more explicitly:

Proof. Since the map to \(BU\) is trivial, the Thom spectrum is \(\Sigma ^\infty _+\). Now observe the map \(E/E_0 \to E\times J_rS^{2m}\) is the evaluation map \(\Sigma
\Omega J_rS^{2m} \to J_rS^{2m}\). Then observe that in the stable splitting of \(J_rS^{2m}\), the projection onto \(S^{2m}\) used in the definition of \(\theta _1\) is the desuspension of the evaluation map. □

Proof. \(\Omega ^2S^{2m+1}_+\) is \(\Free _{E_2}S^{2m1}\) so splits as \(\bigoplus _0^\infty D_{2,r}S^{2m1}\). \(p\)locally these terms are \(0\) except when \(r\equiv
0,1 \pmod p\). For dimension reasons, \(S^{2mp2}\) must map into \(S_0\oplus S^{2m1}\oplus D_{2,p}S^{2m1}\). But by the corollary above, it must map into \(D_{2,p}S^{2m1}\), which can be identified with the
Moore spectrum \(\Sigma ^{2mp2}M(\ZZ /p\ZZ )\), which has \(p\)torsion in its \(2mp2^{th}\) homotopy group. □
Now we can complete the argument of Proposition 1.5.

Proof. Let \(r = p1\), and let us consider the situation at hand, namely \(\beta \) induces \(0\) on mod \(p\) homology and the action of the \(\Omega J_rS^{2m}\) extends to
\(\Omega ^2 S^{2m+1}\). The point is the following: \(\beta ^{1}E^\xi \) is a module over the \(E_2\)ring \(\beta ^{1}\Omega ^2 S^{2m+1}_+\). By the proposition, \(p=0\) in the localized ring, so it is an \(H\FF _p\)
module by Mahowald’s theorem. but since \(\beta \) induces \(0\) in mod \(p\) homology on \(E^\xi \), \(H\FF _p\otimes E^\xi = 0\) so \(E^\xi = 0\). □