Proof. We will do a series of reductions to get to the case for some appropriate coefficients , and then we will do a direct computation.

Step 1. The aim of this step is to reduce to the particular case .

Each class $\sigma _{\epsilon } \in H_{(1,\epsilon ),0}(\mathbf {X})$ is represented by a map , well-defined up to homotopy. Fixing some choices of $\sigma _\epsilon $ we get an $E_2$ -algebra map using that $\mathbf {E_2}(-)$ is left adjoint to the forgetful functor from $E_2$ -algebras to simplicial modules. By construction this map sends $\sigma _0,\sigma _1$ to the corresponding homology classes of $\mathbf {X}$ . By assumption $\sigma _1^2-\sigma _0^2=0 \in H_{(2,0),0}(\mathbf {X})$ , so picking a nullhomotopy gives an extension to an $E_2$ -algebra map .

Now we claim that for the map c we have for $d<\operatorname {rk}(x)/2$ . Indeed, by assumption $H_{x,d}^{E_2}(\mathbf {X})=0$ for $d<\operatorname {rk}(x)-1$ and by Proposition 4.1 for $d<\operatorname {rk}(x)-1$ too. Thus, by the long exact sequence it suffices to show the claim for $(\operatorname {rk}(x)=1,d=0)$ . Both $\mathbf {X}$ and are reduced, that is, $H_{0,0}(-)$ vanishes on both of them, and hence by [Reference Galatius, Kupers and Randal-Williams5, Corollary 11.12] it suffices to show that for $\operatorname {rk}(x)=1$ , which holds by our assumption about the 0th homology of $\mathbf {X}$ and Proposition 4.1.

Now let us suppose that the theorem holds for . If we apply [Reference Galatius, Kupers and Randal-Williams5, Corollary 15.10] with $\rho (x)= \operatorname {rk}(x)/2$ , $\mu (x)=(\operatorname {rk}(x)-1)/2$ and then for $d< \mu (x)$ . But, by [Reference Galatius, Kupers and Randal-Williams5, Section 12.2.4], , giving the required reduction. Note that we use the ‘adapters construction’ of [Reference Galatius, Kupers and Randal-Williams5, Section 12.2] (see Section 2.3) to view $\mathbf {M}$ as a left -module. We will also use this construction in the rest of this proof without explicit mention.

Step 2. Now we will further reduce to the case $\mathbf {A}_{\operatorname {\mathbb {F}_{\ell }}}$ , for $\ell $ a prime number or $0$ , where $\mathbb {F}_0:=\mathbb {Q}$ .

Recall from the proof of Proposition 4.1 that . Since base-change is symmetric monoidal and preserves colimits, the cofibration sequence shows that . Thus, by the universal coefficients theorem it suffices to prove the case .

We claim that the homology groups of $\mathbf {\overline {A_{\mathbb {Z}}}}$ are finitely generated. Indeed, by [Reference Galatius, Kupers and Randal-Williams5, Theorem 16.4] each entry on the first page of the spectral sequence of the proof of Proposition 4.1 is finitely generated. Observe that this is a quite general fact that holds for any cellular $E_2$ -algebra with finitely many cells by considering the analogous cell attachment filtration. We will use this again in the proof of Theorem 3.2.

Hence, by the homology long exact sequence of $\sigma _{\epsilon } \cdot -$ , the homology groups of $\mathbf {\overline {A_{\mathbb {Z}}}}/\sigma _{\epsilon }$ are also finitely generated.

Thus, it suffices to check the cases for $\ell $ a prime number or $0$ , by another application of the universal coefficients theorem and using the finite generation of the homology groups.

Step 3. We will prove the theorem for a fixed $\ell $ , and . To simplify notation we will not write the subscripts $\operatorname {\mathbb {F}_{\ell }}$ for the rest of this proof. Consider the cell attachment filtration $\mathbf {fA} \in \operatorname {Alg}_{E_2}((\operatorname {\mathsf {sMod}}_{\operatorname {\mathbb {F}_{\ell }}}^{\mathsf {H}})^{\mathbb {Z}_{\leq }})$ as in the proof of Proposition 4.1. The filtration $0$ part is given by $\mathbf {\overline {fA}}(0)=\overline {\mathbf {E_2}(S^{(1,0),0} \sigma _0 \oplus S^{(1,1),0} \sigma _1)}$ so we can lift the maps $\sigma _{\epsilon }$ to filtered maps $\sigma _{\epsilon }: S^{(1,\epsilon ),0,0} \rightarrow \mathbf {\overline {fA}}$ . Thus, using adapters we can form the left $\mathbf {\overline {fA}}$ -module $\mathbf {\overline {fA}}/\sigma _{\epsilon }$ filtering $\mathbf {\overline {A}}/\sigma _{\epsilon }$ .

Since $\operatorname {gr}(-)$ commutes with pushouts and with $\overline {(-)}$ by [Reference Galatius, Kupers and Randal-Williams5, Lemma 12.7], we get two spectral sequences

1. (i)

2. (ii)

In order to prove the theorem it suffices to show the following claim

We will need some preparation. As in the proof of Proposition 4.1, the first spectral sequence is multiplicative and its differential satisfies $d^1(\sigma _0)=0$ , $d^1(\sigma _1)=0$ and $d^1(\rho )=\sigma _1^2-\sigma _0^2$ . Moreover, by [Reference Galatius, Kupers and Randal-Williams5, Theorem 16.4, Section 16.2], its first page is given by $\Lambda (L)$ where $\Lambda (-)$ denotes the free graded-commutative algebra, and L is the $\operatorname {\mathbb {F}_{\ell }}$ -vector space with basis $Q_{\ell }^I(y)$ such that y is a basic Lie word in $\{\sigma _0, \sigma _1, \rho \}$ and I is admissible, in the sense of [Reference Galatius, Kupers and Randal-Williams5, Section 16.2]. Let us remark that tri-degrees of the elements can be read off from the description of $\mathbf {fA}$ : on $\sigma _0,\sigma _1,\rho $ the $\mathsf {H}$ -degree is the first index in the exponent of the corresponding sphere, the total degree, $p+q$ , is the second index, and the filtration degree, p, is the last one. Then, tri-degrees of L can be found using [Reference Galatius, Kupers and Randal-Williams5, Section 16.2], and then these tri-degrees are additive under products.

The second spectral sequence is a module over the first one, and we can identify $E^1=F^1/(\sigma _{\epsilon })$ because $\sigma _{\epsilon } \cdot -$ is injective in $F^1$ , by the above description of $F^1$ , and its image is the ideal $(\sigma _{\epsilon })$ . Therefore $E^1= \Lambda (L/\operatorname {\mathbb {F}_{\ell }}\{\sigma _{\epsilon }\})$ and hence the $d^1$ differential in $F^1$ completely determines the $d^1$ differential in $E^1$ , making it into a CDGA.

Proof of Claim.

$E^2$ is given by the homology of the CDGA $(E^1,d^1)$ , and to prove the result we will introduce a ‘computational filtration’ in this CDGA that has the virtue of filtering away most of the $d^1$ differential.

We let $\mathcal {F}^{\bullet } E^1$ be the filtration in which $\sigma _{1-\epsilon }$ and $\rho $ are given filtration $0$ , the remaining elements of a homogeneous basis of $L/\operatorname {\mathbb {F}_{\ell }}\{\sigma _{\epsilon }\}$ extending these are given filtration equal to their homological degree, and then we extend the filtration to $\Lambda (L/\operatorname {\mathbb {F}_{\ell }}\{\sigma _{\epsilon }\})$ multiplicatively.

Since $d^1$ preserves this filtration we get a spectral sequence converging to $E^2$ whose first page is the homology of the associated graded $\operatorname {gr}(\mathcal {F}^{\bullet } E^1)$ . Thus, it suffices to show that $H_*(\operatorname {gr}(\mathcal {F}^{\bullet } E^1))$ already has the required vanishing line.

Let us denote by D the corresponding differential on this computational spectral sequence. Since $d^1$ lowers homological degree by $1$ we can decompose $(\operatorname {gr}(\mathcal {F}^{\bullet } E^1),D)$ as a tensor product

$$ \begin{align*}(\Lambda(\operatorname{\mathbb{F}_{\ell}}\{\sigma_{1-\epsilon},\rho\}),D) \otimes_{\operatorname{\mathbb{F}_{\ell}}} (\Lambda(L/\operatorname{\mathbb{F}_{\ell}}\{\sigma_0,\sigma_1,\rho\}),0),\end{align*} $$

where D satisfies $D(\sigma _{1-\epsilon })=0$ and $D(\rho )=(-1)^{\epsilon } \sigma _{1-\epsilon }^2$ . By the Künneth theorem the homology of this tensor product is $\operatorname {\mathbb {F}_{\ell }}\{1,\sigma _{1-\epsilon }\} \otimes _{\operatorname {\mathbb {F}_{\ell }}} \Lambda (L/\operatorname {\mathbb {F}_{\ell }}\{\sigma _0,\sigma _1,\rho \})$ when $l \neq 2$ , because graded-commutativity forces $\rho ^2=0$ ; and $\mathbb {F}_2\{1,\sigma _{1-\epsilon }\} \otimes _{\mathbb {F}_2} \mathbb {F}_2[\rho ^2] \otimes _{\mathbb {F}_2} \Lambda (L/\mathbb {F}_2\{\sigma _0,\sigma _1,\rho \})$ if $l=2$ .

By the slope of an element we shall mean the ratio between its homological degree and the rank of its $\mathsf {H}$ -valued grading. Since $\rho ^2$ has slope $1/2$ and $\sigma _{1-\epsilon }$ has homological degree $0$ and rank $1$ , in order to prove the required vanishing line it suffices to show that all the elements in $L/\operatorname {\mathbb {F}_{\ell }}\{\sigma _0,\sigma _1,\rho \}$ have slope $\geq 1/2$ . Since the slope of $Q_{\ell }^I(y)$ is always larger than or equal to the one of y, and the slope of the Browder bracket of two elements is always greater than the minimum of their slopes, the only elements in L that have slope less than $1/2$ are those in the span of $\sigma _0, \sigma _1$ , giving the result.

Proof. The idea of the proof is identical to the previous one, so we will not spell out all the details, but we will focus instead in the extra complications that arise in the computations, specially in the later steps.

Step 1. We will construct a certain cellular $E_2$ -algebra and show that it suffices to prove that for $3d \leq 2 \operatorname {rk}(x)-4$ .

The assumptions of the statement imply that $[\sigma _0,\sigma _1]= \sigma _{\epsilon } \cdot y$ for some $y \in H_{(1,1-\epsilon ),1}(\mathbf {X})$ , that for some $x \in H_{(1,\epsilon ),1}(\mathbf {X})$ and some $t \in \mathbb {Z}$ , and that for some $z \in H_{(1,1-\epsilon ),1}(\mathbf {X})$ . (Note that in fact the assumptions (i) and (ii) of the theorem are stronger than what we used here, but in fact we will need the full strength of the assumptions in proving the claim below.)

Let

By proceeding as in Step 1 of the proof of Theorem 3.1 there is an $E_2$ -algebra map sending each of $\sigma _0,\sigma _1, x,y,z$ to the corresponding homology classes in $\mathbf {X}$ with the same name.

Assuming the claim, we can apply [Reference Galatius, Kupers and Randal-Williams5, Corollary 15.10] with $\rho (x)= 2\operatorname {rk}(x)/3$ , $\mu (x)=(2\operatorname {rk}(x)-3)/3$ and to obtain the required reduction. Thus, to finish this step we just need to show the claim.

Proof of Claim.

Proceeding as in the proof of Proposition 4.1 one can compute and check that for $d<\operatorname {rk}(x)-1$ . Since $\mathbf {X}$ has the same vanishing line on its $E_2$ -homology it suffices to check that for $(\operatorname {rk}(x)=1, d=0)$ and $(\operatorname {rk}(x)=2, d=1)$ .

For $(\operatorname {rk}(x)=1,d=0)$ we use [Reference Galatius, Kupers and Randal-Williams5, Corollary 11.12] to reduce it to showing that , as in Step 1 in the proof of Theorem 3.1. This holds because the 0th homology of $\mathbf {X}$ in rank $1$ is generated by $\sigma _0,\sigma _1$ , which factor through f by construction.

For $(\operatorname {rk}(x)=2,d=1)$ the argument will be more elaborate but use the same ideas. Pick sets of -module generators $\{u_a\}_{a \in A}$ for $H_{0,1}(\mathbf {X})$ and $\{v_b\}_{b \in B}$ for $H_{(1,0),1}(\mathbf {X}) \oplus H_{(1,1),1}(\mathbf {X})$ , where each $v_b$ has $\mathsf {H}$ -grading $(1,\epsilon _b)$ for some $\epsilon _b \in \{0,1\}$ . Consider .

The map factors through the canonical map in the obvious way, so we get a long exact sequence in $E_2$ -homology for the triple :

The first term vanishes by direct computation of because $\operatorname {rk}(x)=2$ , so it suffices to show that the third term vanishes too. Using [Reference Galatius, Kupers and Randal-Williams5, Corollary 11.12] it suffices to show that for $d' \leq 1$ and $\operatorname {rk}(x') \leq 2$ .

For a given $x' \in \mathsf {H}$ with $\operatorname {rk}(x') \leq 2$ we have an exact sequence

so it suffices to show that is surjective and that is an isomorphism.

The isomorphism in degree $0$ holds because as a ring: the proof is analogous to the computation of the 0th homology of in the proof of Proposition 4.1 because the extra cells that we have in are either in degree $\geq 2$ or in degree $1$ but attached trivially, so they have no effect in the homological degree $0$ part of the spectral sequence.

The surjectivity in degree $1$ holds by construction if $\operatorname {rk}(x') \leq 1$ . When $\operatorname {rk}(x')=2$ it holds by assumptions (i) and (ii) in the statement of the theorem, which say that any class in rank $2$ and homological degree $1$ can be written in terms of stabilizations of classes in rank $1$ (where the map is surjective by construction) by $\sigma _\epsilon $ plus integer multiples of . (This uses the full assumptions (i) and (ii) in the statement, and not just the weaker version that we used before to get a map .)

Step 2. Now we will further reduce it to the case $\mathbf {S_{\operatorname {\mathbb {F}_2}}}$ for $\ell $ an odd prime or $0$ .

By proceeding as in Proposition 4.1 we find that , so it suffices to consider the case by the universal coefficients theorem.

By reasoning as in Step 2 in the proof of Theorem 3.1 we get that the homology groups of $\mathbf {S}_{\mathbb {Z}[1/2]}/ \sigma _{\epsilon }$ are finitely generated ${\mathbb {Z}[1/2]}$ -modules because $\mathbf {S}_{\mathbb {Z}[1/2]}$ only has finitely many $E_2$ -cells. Thus, another application of the universal coefficients theorem allows us to reduce to the case with $\ell $ either an odd prime or $0$ .

Step 3. Since we are working with $\operatorname {\mathbb {F}_{\ell }}$ -coefficients for a fixed $\ell $ we will drop the $\ell $ and $\operatorname {\mathbb {F}_{\ell }}$ subscripts from now on. Let us begin by considering the cellular attachment filtration of $\mathbf {S}$ , see [Reference Galatius, Kupers and Randal-Williams5, Section 6.2.1] for details, where the last grading denotes the filtration. Let us remark that we use the cell attachment filtration in which the ‘free part’ of $\mathbf {S}$ is in filtration $0$ and the other cells that we attach to it, that is, $\rho , X, Y, Z$ , are in filtration $1$ , but we are not using the CW filtration by canonically viewing $\mathbf {S}$ as a CW $E_2$ -algebra (as that would put $X, Y, Z$ in filtration $2$ instead).

$$ \begin{align*} \begin{aligned} \mathbf{fS}:=\mathbf{E_2}(S^{(1,0),0,0} \sigma_0 \oplus S^{(1,1),0,0} \sigma_1 \oplus S^{(1,\epsilon),1,0} x \oplus S^{(1,1-\epsilon),1,0} y \oplus S^{(1,1-\epsilon),1,0} z) \\ \cup_{\sigma_1^2-\sigma_0^2}^{E_2}{D^{(2,0),1,1} \rho} \cup_{Q^1(\sigma_1)-\sigma_{\epsilon} \cdot x-t Q^1(\sigma_0)}^{E_2} {D^{(2,0),2,1} X} \cup_{[\sigma_0,\sigma_1]-\sigma_{\epsilon} \cdot y}^{E_2}{D^{(2,1),2,1} Y} \\ \cup_{\sigma_{1-\epsilon} \cdot Q^1(\sigma_0)- \sigma_{\epsilon}^2 \cdot z}^{E_2}{D^{(3,1-\epsilon),2,1} Z} \in \operatorname{Alg}_{E_2}((\operatorname{\mathsf{sMod}}_{\operatorname{\mathbb{F}_{\ell}}}^{\mathsf{H}})^{\mathbb{Z}_{\leq}}) \end{aligned} \end{align*} $$

This gives two spectral sequences as in Step 3 of the proof of Theorem 3.1:

1. (i) $F^1_{x,p,q}=H_{x,p+q,p}(\overline {\operatorname {gr}(\mathbf {fS})}) \Rightarrow H_{x,p+q}(\mathbf {\overline {S}})$

2. (ii) $E^1_{x,p,q}=H_{x,p+q,p}(\overline {\operatorname {gr}(\mathbf {fS})}/\sigma _{\epsilon }) \Rightarrow H_{x,p+q}(\mathbf {\overline {S}}/\sigma _{\epsilon }).$

The first spectral sequence is multiplicative, its first page is $\Lambda (L)$ where L is the $\operatorname {\mathbb {F}_{\ell }}$ -vector space with basis $Q^I(u)$ such that u a basic Lie word in $\{\sigma _0,\sigma _1,x,y,z,\rho ,X,Y,Z\}$ and I is admissible in the sense of [Reference Galatius, Kupers and Randal-Williams5, Section 16.2]; and its $d^1$ -differential satisfies $d^1(\sigma _0)=0$ , $d^1(\sigma _1)=0$ , $d^1(x)=0$ , $d^1(y)=0$ , $d^1(z)=0$ , $d^1(\rho )=\sigma _1^2-\sigma _0^2$ , $d^1(X)=Q^1(\sigma _1)-\sigma _{\epsilon } \cdot x-t Q^1(\sigma _0)$ , $d^1(Y)=[\sigma _0,\sigma _1]-\sigma _{\epsilon } \cdot y$ and $d^1(Z)=\sigma _{1-\epsilon } \cdot Q^1(\sigma _0)-\sigma _{\epsilon }^2 \cdot z$ . Moreover, as in the previous theorem, all the tri-degrees can be found out from the description of $\mathbf {fS}$ .

The second spectral sequence has the structure of a module over the first one, and its first page is $E^1= \Lambda (L/\operatorname {\mathbb {F}_{\ell }}\{\sigma _{\epsilon }\})$ , so $(E^1,d^1)$ has the structure of a CDGA.

Thus, in order to finish the proof it suffices to show that $E^2_{x,p,q}=0$ for $p+q<(2 \operatorname {rk}(x)-3)/3$ .

We will show the required vanishing line on $E^2$ by introducing a filtration on the CDGA $(E^1,d^1)$ , similar to the one in Step 3 of the proof of Theorem 3.1. We let $\mathcal {F}^{\bullet } E^1$ be the filtration in which $\sigma _{1-\epsilon }$ , x, y, z, $\rho $ , $Q^1(\sigma _0)$ , $Q^1(\sigma _1)$ , $[\sigma _0,\sigma _1]$ , X, Y, Z are given filtration $0$ , the remaining elements of a homogeneous basis of $L/\operatorname {\mathbb {F}_{\ell }}\{\sigma _{\epsilon }\}$ extending these are given filtration equal to their homological degree, and we extend the filtration to $\Lambda (L/\operatorname {\mathbb {F}_{\ell }}\{\sigma _{\epsilon }\})$ multiplicatively.

This gives a spectral sequence converging to $E^2$ whose first page is the homology of the associated graded of the filtration $\mathcal {F}^{\bullet } E^1$ . We will show the vanishing line on the first page of this spectral sequence.

Applying [Reference Galatius, Kupers and Randal-Williams5, Theorems 16.7 and 16.8] gives that $d^1([\sigma _0,\sigma _1])=0$ , $d^1(Q^1(\sigma _0))=0$ and $d^1(Q^1(\sigma _1))=0$ . This allows to split the associated graded as a tensor product

$$ \begin{align*} \begin{aligned} (\operatorname{gr}(\mathcal{F}^{\bullet} E^1),D)= (\Lambda(\operatorname{\mathbb{F}_{\ell}}\{\sigma_{1-\epsilon},\rho,Q^1(\sigma_0),Z,Q^1(\sigma_1),X\}),D) \otimes_{\operatorname{\mathbb{F}_{\ell}}} \\ (\Lambda(\operatorname{\mathbb{F}_{\ell}}\{[\sigma_0,\sigma_1],Y\}),D) \otimes_{\operatorname{\mathbb{F}_{\ell}}} (\Lambda(\operatorname{\mathbb{F}_{\ell}}\{x,y,z\}),0) \otimes_{\operatorname{\mathbb{F}_{\ell}}} \\ (\Lambda(L/\operatorname{\mathbb{F}_{\ell}}\{\sigma_0,\sigma_1,x,y,z,\rho,Q^1(\sigma_0),Q^1(\sigma_1),[\sigma_0,\sigma_1],X,Y,Z\}),0) \end{aligned} \end{align*} $$

where D is the induced differential and satisfies $D(\sigma _{1-\epsilon })=0$ , $D(\rho )=(-1)^{\epsilon } \sigma _{1-\epsilon }^2$ , $D(Q^1(\sigma _0))=0$ , $D(Z)=\sigma _{1-\epsilon } \cdot Q^1(\sigma _0)$ , $D(Q^1(\sigma _1))=0$ , $D(X)=Q^1(\sigma _1)-t Q^1(\sigma _0)$ , $D([\sigma _0,\sigma _1])=0$ , $D(Y)=[\sigma _0,\sigma _1]$ . By the Künneth theorem it suffices to compute the homology of each of the factors separately.

By direct computation we see that

• • Elements in $\Lambda (L/\operatorname {\mathbb {F}_{\ell }}\{\sigma _0,\sigma _1,x,y,z,\rho ,Q^1(\sigma _0),Q^1(\sigma _1),[\sigma _0,\sigma _1],X,Y,Z\})$ have slope $\geq 2/3$ .

• • Elements in $\Lambda (\operatorname {\mathbb {F}_{\ell }}\{x,y,z\})$ have slope $\geq 1 \geq 2/3$ .

• • Since $\ell \neq 2$ we have $[\sigma _0,\sigma _1]^2=0$ so the homology of $(\Lambda (\operatorname {\mathbb {F}_{\ell }}\{[\sigma _0,\sigma _1],Y\}),D)$ is $\operatorname {\mathbb {F}_{\ell }}[Y^{\ell }]+ [\sigma _0,\sigma _1] \cdot \operatorname {\mathbb {F}_{\ell }}\{Y^j: \ell | j+1\}$ , where the first summand means the polynomial ring in the variable $Y^\ell $ . Since $[\sigma _0,\sigma _1]$ has bidegree $(\operatorname {rk}=2,d=1)$ , Y has bidegree $(\operatorname {rk}=2,d=2)$ and $\ell \geq 3$ then all these elements have slope $\geq 5/6 \geq 2/3$ .

Thus, it suffices to check that $H_*(\Lambda (\operatorname {\mathbb {F}_{\ell }}\{\sigma _{1-\epsilon },\rho ,Q^1(\sigma _0),Z,Q^1(\sigma _1),X\}),D)$ vanishes for $3 d<2 \operatorname {rk}-3$ , where d denotes the homological degree. The remainder of the proof will be about studying this CDGA. We will separate this as an extra step because it will require some additional filtrations and work.

Step 4. We firstly claim that it suffices to consider $t=0$ : $\sigma _{1-\epsilon }$ , $\rho $ , $Q^1(\sigma _0)$ , $Q^1(\sigma _1)$ , X, Z are now just the generators of a certain CDGA. Since both $Q^1(\sigma _0)$ and $Q^1(\sigma _1)$ lie in $\ker (D)$ and have the same homological degree and rank, the change of variables $Q^1(\sigma _1) \mapsto Q^1(\sigma _1)-t Q^1(\sigma _0)$ reparameterizes $t \mapsto 0$ .

Secondly, once we are in the case $t=0$ , we can further split the CDGA as a tensor product

$$ \begin{align*}(\Lambda(\operatorname{\mathbb{F}_{\ell}}\{\sigma_{1-\epsilon},\rho,Q^1(\sigma_0),Z\}),D) \otimes_{\operatorname{\mathbb{F}_{\ell}}} (\Lambda(X,Q^1(\sigma_1)),D)\end{align*} $$

and the homology of the second factor is $\operatorname {\mathbb {F}_{\ell }}[X^{\ell }]+ Q^1(\sigma _1) \cdot \operatorname {\mathbb {F}_{\ell }}\{X^j: \ell | j+1\}$ (since $\ell \neq 2$ ), so all its elements have slope $\geq 5/6 \geq 2/3$ . Thus, it suffices to prove that $H_*(\Lambda (\operatorname {\mathbb {F}_{\ell }}\{\sigma _{1-\epsilon },\rho ,Q^1(\sigma _0),Z\}),D)$ vanishes for $3 d<2 \operatorname {rk}-3$ . For this, we will introduce an additional filtration by giving $Q^1(\sigma _0)$ filtration $0$ , $\sigma _{1-\epsilon }$ filtration $1$ and $\rho , Z$ filtration $2$ , and then extending the filtration multiplicatively to the whole CDGA.

The differential D preserves this filtration and the associated graded splits as a tensor product

$$ \begin{align*}(\Lambda(\sigma_{1-\epsilon},\rho),D(\sigma_{1-\epsilon})=0, D(\rho)=(-1)^{\epsilon} \sigma_{1-\epsilon}^2) \otimes_{\operatorname{\mathbb{F}_{\ell}}} (\Lambda(Q^1(\sigma_0),Z),0)\end{align*} $$

so, using that $\ell \neq 2$ to compute the homology of the first factor, we get a multiplicative spectral sequence of the form

$$ \begin{align*}\mathcal{E}^1= \operatorname{\mathbb{F}_{\ell}}[\sigma_{1-\epsilon}]/(\sigma_{1-\epsilon}^2) \otimes_{\operatorname{\mathbb{F}_{\ell}}} \Lambda(Q^1(\sigma_0),Z) \Rightarrow H_{*}(\Lambda(\operatorname{\mathbb{F}_{\ell}}\{\sigma_{1-\epsilon},\rho,Q^1(\sigma_0),Z\}),D)\end{align*} $$

whose first differential satisfies $D^1(Z)=\sigma _{1-\epsilon } \cdot Q^1(\sigma _0)$ , $D^1(\sigma _{1-\epsilon })=0$ and $D^1(Q^1(\sigma _0))=0$ .

To finish the proof we will establish the required vanishing range on $\mathcal {E}^2$ . To do so, we write $\mathcal {E}^1=\operatorname {\mathbb {F}_{\ell }}\{1,\sigma _{1-\epsilon },Q^1(\sigma _0),\sigma _{1-\epsilon } \cdot Q^1(\sigma _0)\} \otimes \operatorname {\mathbb {F}_{\ell }}[Z]$ as a $\operatorname {\mathbb {F}_{\ell }}$ -vector space, and then compute $\ker (D^1),\operatorname {im}(D^1)$ explicitly as $\operatorname {\mathbb {F}_{\ell }}$ -vector spaces, where $()$ denotes the ideal generated by an element:

$$ \begin{align*}\ker(D^1)=(\sigma_{1-\epsilon})+(Q^1(\sigma_0))+\operatorname{\mathbb{F}_{\ell}}[Z^{\ell}]\end{align*} $$

and

$$ \begin{align*}\operatorname{im}(D^1)=\sigma_{1-\epsilon} \cdot Q^1(\sigma_0) \cdot \operatorname{\mathbb{F}_{\ell}}\{Z^i: \ell \nmid i+1\}.\end{align*} $$

Thus, we get that $\mathcal {E}^2=\ker (D^1)/\operatorname {im}(D^1)$ is, as a $\operatorname {\mathbb {F}_{\ell }}$ -vector space, given by

$$ \begin{align*}\mathcal{E}^2=\operatorname{\mathbb{F}_{\ell}}[Z^l]+\sigma_{1-\epsilon} \cdot \operatorname{\mathbb{F}_{\ell}}[Z]+ Q^1(\sigma_0) \cdot \operatorname{\mathbb{F}_{\ell}}[Z]+ \sigma_{1-\epsilon} \cdot Q^1(\sigma_0) \cdot \operatorname{\mathbb{F}_{\ell}}\{Z^i: \ell \mid i+1\}.\end{align*} $$

Using the bidegrees of the generators we find that the first summand vanishes for $d<2\operatorname {rk}/3$ , the second vanishes for $d<2(\operatorname {rk}-1)/3$ , the third one for $d<(2\operatorname {rk}-1)/3$ , and the fourth one for $d <(2 \operatorname {rk}-3)/3$ , as required.

Claim. $\rho ^2$ survives to $F^{\infty }$ .

Proof. Since $F^1_{(4,0),2+r,1-r}=0$ for $r \geq 1$ then $\rho ^2$ cannot be a boundary of any $d^r$ -differential. Moreover, $d^r: F^r_{(4,0),2,0} \rightarrow F^r_{(4,0),2-r,r-1}$ vanishes for $r>2$ since $\mathbf {fA}$ vanishes on negative filtration. Thus, it suffices to show that both $d^1(\rho ^2)$ and $d^2(\rho ^2)$ vanish. By the Leibniz rule we have $d^1(\rho ^2)=0$ , so we only need to show that $d^2(\rho ^2)=0$ .

Figure 1 $F^1_{(4,0),p,q}$ for small values of $p,q$ , where $\bullet $ means that the corresponding position is nonzero but not relevant for the computation below.

Since $\rho ^2=Q^1(\rho )$ and $d^1(\rho )=\sigma _1^2-\sigma _0^2$ then [Reference Galatius, Kupers and Randal-Williams5, Theorem 16.8 (i)] gives that $d^2(\rho ^2)$ is represented by $Q^1(\sigma _1^2-\sigma _0^2)$ . (As a technical note let us mention that the result we just quoted is stated for $E_{\infty }$ -algebras, but the same result holds for $E_2$ -algebras as explained in [Reference Galatius, Kupers and Randal-Williams5, Page 185].) Finally, $Q^1(\sigma _1^2-\sigma _0^2)$ vanishes by the properties of $Q^1$ shown in [Reference Galatius, Kupers and Randal-Williams5, Section 16.2.2].

Proof. The proof will be very similar to that of Theorem 3.2, so we will focus on the parts that are different and skip details.

Step 1. We will construct a certain cellular $E_2$ -algebra $\mathbf {S}$ and show that it suffices to prove that $H_{x,d}(\mathbf {\overline {S}}/(\sigma _{\epsilon },\theta ))=0$ for $3d < 2 \operatorname {rk}(x)-4$ .

The assumptions of the statement imply that $[\sigma _0,\sigma _1]= \sigma _{\epsilon } \cdot y$ for some $y \in H_{(1,1-\epsilon ),1}(\mathbf {X})$ , that $Q^1(\sigma _1)= \sigma _{\epsilon } \cdot x + t Q^1(\sigma _0)$ for some $x \in H_{(1,\epsilon ),1}(\mathbf {X})$ and some $t \in \operatorname {\mathbb {F}_2}$ , and that $\sigma _{1-\epsilon } \cdot Q^1(\sigma _0)= \sigma _{\epsilon }^2 \cdot z \in H_{(3,1-\epsilon ),1}(\mathbf {X})$ for some $z \in H_{(1,1-\epsilon ),1}(\mathbf {X})$ .

Moreover, we claim that there is $u \in H_{(4,0),3}(\mathbf {X})$ such that $Q^1(\sigma _0)^3=\sigma _{\epsilon }^2 \cdot u$ .

Indeed, condition (iv) says that $\sigma _0 \cdot Q^1(\sigma _0)= \sigma _{\epsilon }^2 \cdot \tau $ for some $\tau \in H_{(1,0),1}(\mathbf {X})$ , and then we can apply $Q^2(-)$ to both sides and use the formulae in [Reference Galatius, Kupers and Randal-Williams5, Section 16.2.2] to find $Q^1(\sigma _0)^3+ \sigma _0^2 \cdot Q^2(Q^1(\sigma _0))+\sigma _0[\sigma _0,Q^1(\sigma _0)] Q^1(\sigma _0)= \sigma _{\epsilon }^2 \cdot [\sigma _{\epsilon },\sigma _{\epsilon }] \cdot Q^1(\tau )+\sigma _{\epsilon }^4 \cdot Q^2(\tau )+\sigma _{\epsilon }^2\cdot [\sigma _{\epsilon }^2,\tau ] \cdot \tau $ , hence the result as $\sigma _{\epsilon }^2=\sigma _0^2$ and as $[\sigma _0,Q^1(\sigma _0)]=[\sigma _0,[\sigma _0,\sigma _0]]=0$ (by [Reference Galatius, Kupers and Randal-Williams5, Section 16.2.2] again). Let us remark that the computations we just described use the fact that on elements of degree $1$ , the operation $Q^2(-)$ is the ‘top operation’ denoted by $\xi $ in [Reference Galatius, Kupers and Randal-Williams5, Section 16.1.2].

Let

$$ \begin{align*} \begin{aligned} \mathbf{S}:=\mathbf{E_2}(S^{(1,0),0} \sigma_0 \oplus S^{(1,1),0} \sigma_1 \oplus S^{(1,\epsilon),1} x \oplus S^{(1,1-\epsilon),1} y \oplus S^{(1,1-\epsilon),1}z \oplus S^{(4,0),3}u) \\\cup_{\sigma_1^2-\sigma_0^2}^{E_2}{D^{(2,0),1} \rho} \cup_{Q^1(\sigma_1)-\sigma_{\epsilon} \cdot x-t Q^1(\sigma_0)}^{E_2} {D^{(2,0),2} X} \cup_{[\sigma_0,\sigma_1]-\sigma_{\epsilon} \cdot y}^{E_2}{D^{(2,1),2} Y} \\ \cup_{\sigma_{1-\epsilon} \cdot Q^1(\sigma_0)- \sigma_{\epsilon}^2 \cdot z}^{E_2}{D^{(3,1-\epsilon),2} Z} \cup_{Q^1(\sigma_0)^3-\sigma_{\epsilon}^2 \cdot u}^{E_2}{D^{(6,0),4} U} \in \operatorname{Alg}_{E_2}(\operatorname{\mathsf{sMod}}_{\operatorname{\mathbb{F}_2}}^{\mathsf{H}}) \end{aligned} \end{align*} $$

By proceeding as in Step 1 of the proof of Theorem 3.1 there is an $E_2$ -algebra map $f: \mathbf {S} \rightarrow \mathbf {X}$ sending each of $\sigma _0,\sigma _1, x,y,z,u$ to the corresponding homology classes in $\mathbf {X}$ with the same name. Moreover, we can assume that f extends any given map $\mathbf {A} \rightarrow \mathbf {X}$ and hence that it sends $\theta \mapsto \theta $ .

The proof is identical to the corresponding claim in Step 1 in the proof of Theorem 3.2. The only difference now is that $\mathbf {S}$ has a cell U below the ‘critical line’ $d=\operatorname {rk}-1$ . However, it causes no trouble since it has bidegree $(\operatorname {rk}=6,d=4)$ , so it lies on the region $3d \geq 2 \operatorname {rk}$ .

Assuming the claim, we can apply [Reference Galatius, Kupers and Randal-Williams5, Corollary 15.10] with $\rho (x)= 2\operatorname {rk}(x)/3$ , $\mu (x)=(2\operatorname {rk}(x)-4)/3$ and $\mathbf {M}=\mathbf {\overline {S}}/(\sigma _{\epsilon },\theta )$ to obtain the required reduction.

Setp 2. We proceed as in Step 3 in the proof of Theorem 3.2 to get a cell attachment filtration $\mathbf {fS} \in \operatorname {Alg}_{E_2}((\operatorname {\mathsf {sMod}}_{\operatorname {\mathbb {F}_2}}^{\mathsf {H}})^{\mathbb {Z}_{\leq }})$ . The key now is to observe that $\theta \in H_{(4,0),2}(\mathbf {S})$ lifts to a filtered map $\theta : S^{(4,0),2,2} \rightarrow \mathbf {fS}$ which maps to $\rho ^2 \in H_{*,*,*}(\operatorname {gr}(\mathbf {fS}))$ .

Indeed, $\theta \in H_{(4,0),2}(\mathbf {A})=H_{(4,0),2}(\operatorname {colim}(\mathbf {fA}))=\operatorname {colim}_f(H_{(4,0),2,f}(\mathbf {fA}))$ , so it can be represented by a class $\theta \in H_{(4,0),2,f}(\mathbf {fA})$ for some f large. Now we claim that $f=2$ is the smallest possible value such that $\theta \in H_{(4,0),2,f}(\mathbf {fA})$ : this is because by construction the image of $\theta $ in the homology of the associated graded, $H_{(4,0),2,f}(\operatorname {gr}(\mathbf {fA}))$ , is $\rho ^2 \ne 0 \in H_{(4,0),2,2}(\operatorname {gr}(\mathbf {fA}))$ . Finally observe that there is a canonical map of filtered $E_2$ -algebras $\mathbf {fA} \rightarrow \mathbf {fS}$ . Thus, we get spectral sequences

1. (i) $F^1_{x,p,q}=H_{x,p+q,p}(\overline {\operatorname {gr}(\mathbf {fS})}) \Rightarrow H_{x,p+q}(\mathbf {\overline {S}})$

2. (ii) $E^1_{x,p,q}=H_{x,p+q,p}(\overline {\operatorname {gr}(\mathbf {fS})}/(\sigma _{\epsilon },\rho ^2)) \Rightarrow H_{x,p+q}(\mathbf {\overline {S}}/(\sigma _{\epsilon },\theta )).$

The first spectral sequence is multiplicative, its first page is $\operatorname {\mathbb {F}_2}[L]$ where L is the $\operatorname {\mathbb {F}_2}$ -vector space with basis $Q^I(\alpha )$ such that $\alpha $ a basic Lie word in $\{\sigma _0,\sigma _1,x,y,z,u,\rho ,X,Y,Z,U\}$ and I is admissible; and its $d^1$ -differential satisfies $d^1(\sigma _0)=0$ , $d^1(\sigma _1)=0$ , $d^1(x)=0$ , $d^1(y)=0$ , $d^1(z)=0$ , $d^1(u)=0$ , $d^1(\rho )=\sigma _1^2-\sigma _0^2$ , $d^1(X)=Q^1(\sigma _1)-\sigma _{\epsilon } \cdot x-t Q^1(\sigma _0)$ , $d^1(Y)=[\sigma _0,\sigma _1]-\sigma _{\epsilon } \cdot y$ , $d^1(Z)=\sigma _{1-\epsilon } \cdot Q^1(\sigma _0)-\sigma _{\epsilon }^2 \cdot z$ and $d^1(U)=Q^1(\sigma _0)^3-\sigma _{\epsilon }^2 \cdot u$ .

The second spectral sequence has the structure of a module over the first one, and its first page is given by $E^1= \operatorname {\mathbb {F}_2}[L/\operatorname {\mathbb {F}_2}\{\sigma _{\epsilon }\}]/(\rho ^2)$ because $\sigma _{\epsilon } \cdot -$ is injective on $\operatorname {\mathbb {F}_2}[L]$ and $\rho ^2 \cdot -$ is injective on $\operatorname {\mathbb {F}_2}[L]/(\sigma _{\epsilon })=\operatorname {\mathbb {F}_2}[L/\operatorname {\mathbb {F}_2}\{\sigma _{\epsilon }\}]$ . Thus $(E^1,d^1)$ has the structure of a CDGA.

Thus, in order to finish the proof it suffices to show that $E^2_{x,p,q}=0$ for $p+q<(2 \operatorname {rk}(x)-4)/3$ .

Step 3. Now we will introduce additional filtrations to simplify the CDGA until we get the required result. The first filtration is similar to the one of Step 3 in the proof of Theorem 3.2: we give $\sigma _{1-\epsilon }$ , x, y, z, u, $\rho $ , $Q^1(\sigma _0)$ , $Q^1(\sigma _1)$ , $[\sigma _0,\sigma _1]$ , X, Y, Z, U filtration $0$ , we give the remaining elements of a homogeneous basis of $L/\operatorname {\mathbb {F}_2}\{\sigma _{\epsilon }\}$ extending these filtration equal to their homological degree, and we extend the filtration to $\operatorname {\mathbb {F}_2}(L/\operatorname {\mathbb {F}_2}\{\sigma _{\epsilon }\})/(\rho ^2)$ multiplicatively (which we can as $\rho $ is in filtration $0$ ).

This allows us to split the associated graded as a tensor product and all the factors are concentrated in the region $3d \geq 2 \operatorname {rk}$ except possibly the one given by

$$ \begin{align*}(\operatorname{\mathbb{F}_2}[\sigma_{1-\epsilon},\rho,Q^1(\sigma_0),Q^1(\sigma_1),X,Z,U]/(\rho^2),D)\end{align*} $$

where the nonzero part of D is characterized by $D(X)=Q^1(\sigma _1)-tQ^1(\sigma _0)$ , $D(Z)=\sigma _{1-\epsilon } \cdot Q^1(\sigma _0)$ and $D(U)=Q^1(\sigma _0)^3$ . (This computation is easier than the one of the proof of Theorem 3.2 since $\ell =2$ simplifies the homology of the other factors.)

Then, we can proceed as in Step 4 in the proof of Theorem 3.2 to go to the case $t=0$ and hence split the CDGA further to simplify it to

$$ \begin{align*}(\operatorname{\mathbb{F}_2}[\sigma_{1-\epsilon},\rho,Q^1(\sigma_0),Z,U]/(\rho^2),D).\end{align*} $$

Next we introduce a new filtration by giving $\sigma _{1-\epsilon }, \rho , Q^1(\sigma _0)$ filtration $0$ , and $Z,U$ filtration $1$ and then extending multiplicatively. The associated graded of this splits as a tensor product

$$ \begin{align*}(\operatorname{\mathbb{F}_2}[\sigma_{1-\epsilon},\rho]/(\rho^2), D(\rho)=\sigma_{1-\epsilon}^2) \otimes_{\operatorname{\mathbb{F}_2}} (\operatorname{\mathbb{F}_2}[Q^1(\sigma_0),Z,U],0)\end{align*} $$

and the homology of the first factor is precisely $\operatorname {\mathbb {F}_2}[\sigma _{1-\epsilon }]/(\sigma _{1-\epsilon }^2)$ , yielding a spectral sequence of the form

whose first differential $D^1$ satisfies $D^1(Z)= \sigma _{1-\epsilon } \cdot Q^1(\sigma _0)$ and $D^1(U)=Q^1(\sigma _0)^3$ . We will establish the required vanishing line on $\mathcal {E}^2$ of this spectral sequence. For that we will introduce yet another filtration by letting $\sigma _{1-\epsilon }, Q^1(\sigma _0), U$ have filtration $0$ and Z have filtration $1$ .

The associated graded is given by

$$ \begin{align*}(\operatorname{\mathbb{F}_2}[\sigma_{1-\epsilon},Z]/(\sigma_{1-\epsilon}^2),0) \otimes_{\operatorname{\mathbb{F}_2}} (\operatorname{\mathbb{F}_2}[Q^1(\sigma_0),U],\delta(U)=Q^1(\sigma_0)^3)\end{align*} $$

where $\delta $ is the new differential. Thus, its homology is given by

$$ \begin{align*}\operatorname{\mathbb{F}_2}[\sigma_{1-\epsilon}, Q^1(\sigma_0), Z,U^2]/(\sigma_{1-\epsilon}^2,Q^1(\sigma_0)^3)\end{align*} $$

and the $\delta ^1$ -differential satisfies $\delta ^1(Z)= \sigma _{1-\epsilon } \cdot Q^1(\sigma _0)$ . Since U has slope $2/3$ itself, in order to prove the required vanishing line we can just focus on the remaining part

$$ \begin{align*}(\operatorname{\mathbb{F}_2}[\sigma_{1-\epsilon}, Q^1(\sigma_0), Z]/(\sigma_{1-\epsilon}^2,Q^1(\sigma_0)^3),\delta^1(Z)=\sigma_{1-\epsilon} \cdot Q^1(\sigma_0)).\end{align*} $$

For that we explicitly compute $\ker (\delta ^1)$ , $\operatorname {im}(\delta ^1)$ as $\operatorname {\mathbb {F}_2}$ -vector spaces (similar to the last CDGA of the proof of Theorem 3.2).

$$ \begin{align*}\ker(\delta^1)= (\sigma_{1-\epsilon})+(Q^1(\sigma_0)^2)+\operatorname{\mathbb{F}_2}\{1,Q^1(\sigma_0)\} \cdot \operatorname{\mathbb{F}_2}[Z^2]\end{align*} $$

and

$$ \begin{align*}\operatorname{im}(\delta^1)=\operatorname{\mathbb{F}_2}\{\sigma_{1-\epsilon} \cdot Q^1(\sigma_0), \sigma_{1-\epsilon} \cdot Q^1(\sigma_0)^2\} \cdot \operatorname{\mathbb{F}_2}[Z^2].\end{align*} $$

Thus we get

$$ \begin{align*} \begin{aligned} \ker(\delta^1)/\operatorname{im}(\delta^1)= \sigma_{1-\epsilon} \cdot \operatorname{\mathbb{F}_2}[Z]+ \sigma_{1-\epsilon} \cdot Q^1(\sigma_0) \cdot \operatorname{\mathbb{F}_2}\{Z^i: 2 \nmid i\}+ \\ \sigma_{1-\epsilon} \cdot Q^1(\sigma_0)^2 \cdot \operatorname{\mathbb{F}_2}\{Z^i: 2 \nmid i\}+\operatorname{\mathbb{F}_2}\{1,Q^1(\sigma_0),Q(\sigma_0)^2\} \cdot \operatorname{\mathbb{F}_2}[Z]. \end{aligned} \end{align*} $$

Using the bidegrees of the generators it is immediate that all vanish for $3d<2 \operatorname {rk} -4$ , hence the result.

Proof. By definition (using the adapters construction as in Section 4.4) there is a cofibration of left $\mathbf {\overline {X}}$ -modules

$$ \begin{align*}S^{(4,0),2} \otimes \mathbf{\overline{X}}/\sigma_{\epsilon} \xrightarrow{\theta \cdot -} \mathbf{\overline{X}}/\sigma_{\epsilon} \rightarrow \mathbf{\overline{X}}/(\sigma_{\epsilon},\theta),\end{align*} $$

and hence a corresponding long exact sequence in homology groups which implies that $\theta \cdot -: H_{x-(4,0),d-2}(\mathbf {\overline {X}}/\sigma _{\epsilon }) \rightarrow H_{x,d}(\mathbf {\overline {X}}/\sigma _{\epsilon })$ is surjective for $3d \leq 2 \operatorname {rk}(x)-5$ and an isomorphism for $3d \leq 2 \operatorname {rk}(x)-8$ .

Similarly, the cofibration of left $\mathbf {\overline {X}}$ -modules $S^{(1,\epsilon ),0} \otimes \mathbf {\overline {X}} \xrightarrow {\sigma _{\epsilon } \cdot -} \mathbf {\overline {X}}\rightarrow \mathbf {\overline {X}}/\sigma _{\epsilon }$ gives another long exact sequence in homology groups.

Proof of (i). If $\theta ^3$ does not destabilize by $\sigma _{\epsilon }$ then the second long exact sequence gives $\theta ^3 \neq 0 \in H_{(12,0),6}(\mathbf {\overline {X}}/\sigma _{\epsilon })$ . But

$$ \begin{align*}\theta \cdot -: H_{4(k-1),2(k-1)}(\mathbf{\overline{X}}/\sigma_{\epsilon}) \rightarrow H_{4k,2k}(\mathbf{\overline{X}}/\sigma_{\epsilon})\end{align*} $$

is an isomorphism for $k \geq 4$ , so $\theta ^k \neq 0 \in H_{4k,2k}(\mathbf {\overline {X}}/\sigma _{\epsilon })$ for $k \geq 4$ (hence for $k \geq 1$ ).

Proof of (ii). If $3d \leq 2 \operatorname {rk}(x)-6$ then $3(d+2k) \leq 2 (\operatorname {rk}(x)+4k)-8$ for any $k \geq 1$ and hence the map $\theta ^k \cdot -: H_{x,d}(\mathbf {\overline {X}}/\sigma _{\epsilon }) \rightarrow H_{x+(4k,0),d+2k}(\mathbf {\overline {X}}/\sigma _{\epsilon })$ is an isomorphism for any $k \geq 1$ . Thus, it suffices to find some k for which $\theta ^k \cdot -$ is the $0$ map. We will show that in fact $k=6$ works.

Since $\theta ^3$ destabilizes by either $\sigma _0$ or $\sigma _1$ then (using that $\sigma _0^2=\sigma _1^2$ ) $\theta ^6= \alpha \cdot \sigma _{\epsilon }^2$ for some $\alpha \in H_{(22,0),12}(\mathbf {X})$ . Thus, $\theta ^6 \cdot -= \alpha \cdot (\sigma _{\epsilon }^2 \cdot -)$ as (homotopy classes of) maps $S^{(24,0),12} \otimes \mathbf {\overline {X}}/\sigma _{\epsilon } \rightarrow \mathbf {\overline {X}}/\sigma _{\epsilon }$ . Thus, it suffices to show that $\sigma _{\epsilon }^2 \cdot -: S^{(2,0),0} \otimes \mathbf {\overline {X}}/\sigma _{\epsilon } \rightarrow \mathbf {\overline {X}}/\sigma _{\epsilon }$ is nullhomotopic. This is a special case of the following general fact (see M. Ramzi, Proposition 2.3, Personal note https://drive.google.com/file/d/1_ctuCFe5ficctsfkcqjNP9jwYg2ssgwZ/view) that if $\mathbf {X}$ is an object in a stable $\infty $ -category ( $\operatorname {\mathsf {sMod}}_{\operatorname {\mathbb {F}_2}}^{\mathsf {H}}$ in our case) and $f: \mathbf {X} \rightarrow \mathbf {X}$ is a self-map then (any) induced morphism $\overline {f}: \mathbf {X}/f \rightarrow \mathbf {X}/f$ on the cofibre satisfies that $\overline {f}^2$ is nullhomotopic.

Proof. By definition

$$ \begin{align*}S_{p}^{E_1,\mathsf{G}}(x)= \underset{(x_0,\cdots,x_{p+1}) \in \mathsf{G}_{\operatorname{rk}>0}^{p+2}}{\operatorname{colim}}{\mathsf{G}(x_0 \oplus \cdots \oplus x_{p+1},x)}\end{align*} $$

and

$$ \begin{align*}S_p^{E_1,\mathsf{G^q}}(q)= \underset{(q_0,\cdots,q_{p+1}) \in \mathsf{G^q}_{\operatorname{rk}^{\mathsf{q}}>0}^{p+2}}{\operatorname{colim}}{\mathsf{G^q}(q_0 \oplus^{\mathsf{q}} \cdots \oplus^{\mathsf{q}} q_{p+1},q)}\end{align*} $$

The inclusions $\mathsf {G^q}(q_0 \oplus ^{\mathsf {q}} \cdots \oplus ^{\mathsf {q}} q_{p+1},q) \subset \mathsf {G}(x_0 \oplus \cdots \oplus x_{p+1},x)$ , for each $(q_0,\cdots ,q_{p+1}) \in \mathsf {G^q}_{\operatorname {rk}^{\mathsf {q}}>0}^{p+2}$ with $q_i \in Q(x_i)$ and $q \in Q(x)$ , assemble into canonical maps

which are compatible with the face maps of both semisimplicial sets because the natural functor $\mathsf {G^q} \rightarrow \mathsf {G}$ is monoidal. Thus, it suffices to show that $S_{p}^{E_1,\mathsf {G^q}}(q) \rightarrow S_{p}^{E_1,\mathsf {G}}(x)$ is a bijection of sets for all p.

Surjectivity: any element on the right-hand side is represented by some $\phi \in \mathsf {G}(x_0 \oplus \cdots \oplus x_{p+1},x)$ which is an isomorphism since $\mathsf {G}$ is a groupoid. Since Q is strong monoidal, $Q(\phi ): Q(x) \xrightarrow {\cong } Q(x_0) \times \cdots \times Q(x_{p+1})$ is an isomorphism. Let $q_i:= \operatorname {proj}_i ( Q(\phi )(q)) \in Q(x_i)$ , then $\phi \in \mathsf {G^q}(q_0 \oplus ^{\mathsf {q}} \cdots \oplus ^{\mathsf {q}} q_{p+1},q)$ defines an element on the left-hand side mapping to the required element.

Injectivity: suppose that two elements on the left-hand side have the same image on the right-hand side. Represent them by $\phi ^i \in \mathsf {G^q}(q_0^i \oplus ^{\mathsf {q}} \cdots \oplus ^{\mathsf {q}} q_{p+1}^i,q)$ , where $Q(\phi ^i)(q)=q_0^i \oplus ^{\mathsf {q}} \cdots \oplus ^{\mathsf {q}} q_{p+1}^i$ and $i \in \{1,2\}$ is an index.

Since $\phi ^1$ and $\phi ^2$ agree in the colimit of the right-hand side then there is an element $\phi \in \mathsf {G}(x_0 \oplus \cdots +x_{p+1},x)$ and morphisms $(\psi _0^i,\cdots ,\psi _{p+1}^i): (x_0^i,\cdots ,x_{p+1}^i) \rightarrow (x_0, \cdots ,x_{p+1})$ in $\mathsf {G}_{\operatorname {rk}>0}^{p+2}$ such that $\phi ^i \circ (\psi _0^i, \cdots ,\psi _{p+1}^i)^{-1}= \phi $ for $i \in \{1,2\}$ .

Let ${q^{\prime }_a}^i:= Q({\psi _a^i}^{-1})(q_a^i) \in Q(x_a)$ , we claim that ${q^{\prime }_a}^1={q^{\prime }_a}^2$ for $i \in \{1,2\}$ : $Q(\psi _0^i \oplus \cdots \oplus \psi _{p+1}^i) ({q^{\prime }_0}^i \oplus ^{\mathsf {q}} \cdots \oplus ^{\mathsf {q}} {q^{\prime }_{p+1}}^i)= (q_0^i \oplus ^{\mathsf {q}} \cdots \oplus ^{\mathsf {q}} q_{p+1}^i)=Q(\phi ^i)(q)$ and hence ${q^{\prime }_0}^i \oplus ^{\mathsf {q}} \cdots \oplus ^{\mathsf {q}} {q^{\prime }_{p+1}}^i=Q(\phi ^i\circ (\psi _0^i \oplus \cdots \oplus \psi _{p+1}^i)^{-1})(q)=Q(\phi )(q)$ . Since $Q(\phi )(q)$ is independent of $i \in \{1,2\}$ then the claim follows by the strong monoidality Q since ${q^{\prime }_a}^1, {q^{\prime }_a}^2 \in Q(x_a)$ for all a.

Now let $q_a:={q^{\prime }_a}^1={q^{\prime }_a}^2$ , then by definition $Q(\psi _0^i \oplus \cdots \oplus \psi _{p+1}^i) ({q}_0 \oplus ^{\mathsf {q}} \cdots \oplus ^{\mathsf {q}} {q}_{p+1})= (q_0^i \oplus ^{\mathsf {q}} \cdots \oplus ^{\mathsf {q}} q_{p+1}^i)$ and hence $(\psi _0^i, \cdots ,\psi _{p+1}^i) \in \mathsf {G^q}_{\operatorname {rk}^{\mathsf {q}>0}}^{p+2}$ . Since $\phi ^i \circ (\psi _0^i, \cdots ,\psi _{p+1}^i)^{-1}= \phi $ for $i \in \{1,2\}$ by construction, then $\phi ^1$ and $\phi ^2$ agree in the left-hand side colimit, as required.

Proof. By Theorem 5.4 and the standard connectivity estimate, the reduced homology of $S^{E_1,\mathsf {G^q}}(q)$ is concentrated in degree $\operatorname {rk}^{\mathsf {q}}(q)-2$ for any $q \in \mathsf {G^q}$ . Thus, by [Reference Galatius, Kupers and Randal-Williams5, Page 189] we have $H_{x,d}^{E_1}(\mathbf {R^q})=0$ for $d<\operatorname {rk}(x)-1$ . Finally, the ‘transferring vanishing lines up’ theorem, [Reference Galatius, Kupers and Randal-Williams5, Theorem 14.4] implies the result.

Proof. The map $(\operatorname {rk},\operatorname {Arf}): \pi _0(\mathsf {Sp^q}) \rightarrow \mathsf {H}$ is clearly surjective, it is injective and well-defined by the above discussion of the Arf invariant, and it is monoidal because $\operatorname {rk}$ is monoidal and $\operatorname {Arf}$ is also monoidal by its explicit formula.

Proof. Let $P(g)$ be the poset whose elements are submodules $0 \subsetneq M \subsetneq \mathbb {Z}^{2g}$ such that $(M,\Omega _g|_{M})$ is isomorphic to the standard symplectic form $(\mathbb {Z}^{2h},\Omega _h)$ for some $0<h<g$ , ordered by inclusion. (Note that h has to equal $\operatorname {rk}(M)/2$ if such isomorphism exists by rank reasons.) Let $P_{\bullet }(g)$ be the nerve of the poset, viewed as a semisimplicial set with p-simplices strict chains $M_0 \subsetneq M_1 \subsetneq \cdots \subsetneq M_p$ in $P(g)$ , and face maps given by forgetting elements in the chain.

The first step in the proof is about comparing $P_{\bullet }(g)$ with the $E_1$ -splitting complex.

Proof. By [Reference Galatius, Kupers and Randal-Williams5, Remark 17.11] we have the following more concrete description of $S_{\bullet }^{E_1,\mathsf {Sp}}(g)$ :

$$ \begin{align*}S_{p}^{E_1,\mathsf{Sp}}(g)= \bigsqcup_{(g_0,\cdots,g_{p+1}): \; g_i>0, \; \sum_{i} g_i= g}{\frac{Sp_{2g}(\mathbb{Z})}{Sp_{2g_0}(\mathbb{Z}) \times Sp_{2g_1}(\mathbb{Z}) \times \cdots \times Sp_{2g_{p+1}}(\mathbb{Z})}}\end{align*} $$

with the obvious face maps.

For each $0<n<g$ we let $M_n:=\mathbb {Z}^{2n} \oplus 0 \subset \mathbb {Z}^{2g}$ , so that we have a chain $M_1< \cdots < M_{g-1}$ in $P(g)$ . For each tuple $(g_0,\cdots ,g_{p+1})$ with $g_i>0$ and $\sum _{i}{g_i}=g$ we have a p-simplex $\sigma _{g_0,\cdots ,g_{p+1}}:= M_{g_0}< M_{g_0+g_1} < \cdots < M_{g_0+\cdots +g_p} \in P_p(g)$ .

The group $Sp_{2g}(\mathbb {Z})$ acts simplicially on $P_{\bullet }(g)$ , and under this action the stabilizer of $\sigma _{g_0,\cdots ,g_{p+1}}$ is precisely $Sp_{2g_0}(\mathbb {Z}) \times Sp_{2g_1}(\mathbb {Z}) \times \cdots \times Sp_{2g_{p+1}}(\mathbb {Z}) \subset Sp_{2g}(\mathbb {Z})$ . Thus, we indeed get a levelwise injective map of semisimplicial sets $S_{\bullet }^{E_1,\mathsf {Sp}}(g) \rightarrow P_{\bullet }(g)$ .

To check levelwise surjectivity, consider a chain $M_0<M_1<\cdots < M_p \in P(g)$ . By definition of $P(g)$ there are isomorphisms $(M_i,\Omega _{g}|_{lM_i}) \xrightarrow {\cong } (\mathbb {Z}^{\operatorname {rk}{M_i}},\Omega _{\operatorname {rk}(M_i)/2})$ . Let $N_i$ be the orthogonal complement of $M_i$ inside $M_{i+1}$ for $0 \le i \le p-1$ and let $N_p$ be the orthogonal complement of $M_p$ in M. Since the form on $M_{i}$ is nondegenerate for $0 \le i \le p$ then there is an orthogonal splitting $M_{i+1}=M_i \oplus N_i$ , under the convention $M_{p+1}:=M$ . Since the form on $M_{i+1}$ is nondegenerate for $0 \le i \le p$ then the form on $N_i$ is also nondegenerate for $0 \le i \le p$ , so by the classification of nondegenerate skew-symmetric forms over finitely generated free $\mathbb {Z}$ -modules there are isomorphisms $(N_i, \Omega _g|_{N_i}) \cong (\mathbb {Z}^{\operatorname {rk}{N_i}},\Omega _{\operatorname {rk}(N_i)/2})$ . Thus, we can inductively on $0 \le i \le p+1$ pick isomorphisms $(M_i,\Omega _{g}|_{lM_i}) \xrightarrow {\cong } (\mathbb {Z}^{\operatorname {rk}{M_i}},\Omega _{\operatorname {rk}(M_i)/2})$ compatible with the inclusions $M_j \subset M_i$ for $j<i$ , by extending them using the corresponding $N_i$ . Looking at the case $M=M_{p+1}$ this produces an automorphism $\phi \in Sp_{2g}(\mathbb {Z})$ which by construction maps to the chain $M_0<\cdots <M_p$ when we consider $[\phi ] \in \frac {Sp_{2g}(\mathbb {Z})}{Sp_{\operatorname {rk}(M_0)}(\mathbb {Z}) \times Sp_{\operatorname {rk}(N_0)}(\mathbb {Z}) \times \cdots \times Sp_{\operatorname {rk}(N_p)}(\mathbb {Z})} \subset S_{p}^{E_1,\mathsf {Sp}}(g)$ , hence giving surjectivity.

Let us denote $L:= (\mathbb {Z}^{2g},\Omega _g)$ . The poset $P(g)$ is then the same as $\mathcal {U}(L)_{0<-<L}$ in the sense of [Reference van der Kallen and Looijenga20, Section 1]. By [Reference van der Kallen and Looijenga20, Theorem 1.1] the poset $\mathcal {U}(L)$ is Cohen-Macaulay of dimension g, and in particular the poset $\mathcal {U}(L)_{0<-<L}$ is $(g-3)$ -connected and $(g-2)$ -dimensional, giving the result.

Proof of Theorem.

Part (i). By Lemma 6.2 and Remark 5.3 we have $\mathbf {R^q} \in \operatorname {Alg}_{E_2}(\operatorname {\mathsf {sSet}}^{\mathsf {H}})$ such that $\mathbf {R^q}(x)$ is path connected for each $x \in \mathsf {H}\setminus \{0\}$ and $\mathbf {R^q}(0)=\emptyset $ . Thus, by Remark 5.3 we have $H_{0,0}(\mathbf {R^q})=0$ and $H_{*,0}(\mathbf {\overline {\mathbf {R^q}}})=\mathbb {Z}[\sigma _0,\sigma _1]/(\sigma _1^2-\sigma _0^2)$ as a ring, where $\sigma _{\epsilon }$ is generated by a point in $\mathbf {R^q}((1,\epsilon ))$ . By Proposition 6.3, $(\mathsf {Sp},\operatorname {rk})$ satisfies the standard connectivity estimate, and thus by Corollary 5.5 we get that $H_{x,d}^{E_2}(\mathbf {R^q})=0$ for $d<\operatorname {rk}(x)-1$ .

If we now consider $\mathbf {X}:= \mathbf {\mathbf {R^q}_{\mathbb {Z}}} \in \operatorname {Alg}_{E_2}(\operatorname {\mathsf {sMod}}_{\mathbb {Z}}^{\mathsf {H}})$ then it satisfies the assumptions of Theorem 3.1 by [Reference Galatius, Kupers and Randal-Williams5, Lemma 18.2] and the properties of $(-)_{\mathbb {Z}}$ explained in Section 2. Thus the claimed homological stability for $\mathbf {R^q}$ follows.

Part (ii). This time let $\mathbf {X}:= \mathbf {\mathbf {R^q}_{\mathbb {Z}[1/2]}} \in \operatorname {Alg}_{E_2}(\operatorname {\mathsf {sMod}}_{\mathbb {Z}[1/2]}^{\mathsf {H}})$ . As before, this algebra satisfies the unnumbered assumptions of Theorem 3.2. We will check that it also verifies assumptions (i),(ii) and (iii) and then the required stability will follow from Theorem 3.2. By the universal coefficients theorem, to check them it suffices to prove that $H_{x,1}(\mathbf {R^q})$ is $2$ -torsion for $\operatorname {rk}(x) \in \{2,3\}$ , which follows from Theorems 8.7 and 8.8 in Section 8.

Secondary stability. Let $\mathbf {X}:= \mathbf {\mathbf {R^q}_{\operatorname {\mathbb {F}_2}}} \in \operatorname {Alg}_{E_2}(\operatorname {\mathsf {sMod}}_{\operatorname {\mathbb {F}_2}}^{\mathsf {H}})$ . Then Theorem 3.3 applies by Theorems 8.7 and 8.8 and the universal coefficients theorem. The result then follows by the long exact sequence of the cofibration sequence $S^{(4,0),2} \otimes \mathbf {\overline {X}}/\sigma _{\epsilon } \rightarrow \mathbf {\overline {X}}/\sigma _{\epsilon } \rightarrow \mathbf {\overline {X}}/(\sigma _{\epsilon },\theta ).$

Proof. In this case the standard connectivity estimate for $(\mathsf {MCG},\operatorname {rk})$ is proven in [Reference Galatius, Kupers and Randal-Williams6, Theorem 3.4]. Thus, proceeding as in the proof of Theorem B we can apply Theorem 3.1 to $\mathbf {\mathbf {R^q}_{\mathbb {Z}}}$ to get part (i) of the Theorem.

To prove part (ii) we consider $\mathbf {X}:=\mathbf {\mathbf {R^q}_{\mathbb {Z}[1/2]}}$ and apply Theorem 3.2. To verify assumptions (i),(ii) and (iii) we use the universal coefficients theorem and Theorems 8.1, 8.2, 8.3 and 8.4.

The secondary stability part follows by considering $\mathbf {X}:=\mathbf {\mathbf {R^q}_{\operatorname {\mathbb {F}_2}}}$ and applying Theorem 3.3, where all assumptions needed hold by Theorems 8.1, 8.2, 8.3 and 8.4.

Proof. Suppose for a contradiction that it was surjective for some $k \geq 1$ , $\epsilon , \delta \in \{0,1\}$ . By the transfer, $H_{2k}(B\Gamma _{3k,1}^{1/2}[\delta ];\mathbb {Q}) \rightarrow H_{2k}(B\Gamma _{3k,1};\mathbb {Q})$ is also surjective since the spin mapping class groups are finite index subgroups of the mapping class groups.

Thus, the stabilization map $\sigma \cdot -:H_{2k}(B\Gamma _{3k-1,1};\mathbb {Q}) \rightarrow H_{2k}(B\Gamma _{3k,1};\mathbb {Q})$ (where we are using the notation of [Reference Galatius, Kupers and Randal-Williams6]) must be surjective. By the universal coefficients theorem, $H^{2k}(B\Gamma _{3k,1};\mathbb {Q}) \rightarrow H^{2k}(B\Gamma _{3k-1,1};\mathbb {Q})$ is injective; which is false by the computations in [Reference Galatius, Kupers and Randal-Williams6, Proof of Corollary 5.8].