A correspondence between higher Adams differentials and higher algebraic Novikov differentials at odd primes

For odd prime $p$, the Cartan-Eilenberg spectral sequence collapses from $E_{2}$-page with no nontrivial extensions [15, Theorem 4.4.3]. Hence we have

Note the inner degrees $|v_{n}|=|t_{n}|=2(p^{n}-1)$ are all multiples of $q$. In order for $Ext^{s-i,t-i}_{P_{*}}(\mathbb{F}_{p},I^{i}/I^{i+1})$ to be nontrivial, we need $i,s-i,t-i\geq 0$, and that $t-i\equiv 0~{}~{}\text{mod}~{}q$. Hence $i$ needs to be in the set $C_{s,t}$.

∎

For notation simplicity, we let $t(-)$ denote the inner degree and let $u(-)$ denote the motivic weight of an element in $Ext^{*,*,*}_{\mathcal{A}_{*,*}^{\mathbb{C}}}(\mathbb{F}_{p}[\tau],\mathbb{F}_{p}[\tau])$. By Proposition 2.1, we have $u(\tilde{t}_{i})=\frac{1}{2}t(\tilde{t}_{i})$ for $i\geq 1$, and $u(\tilde{\tau}_{i})=\frac{1}{2}t(\tilde{\tau}_{i})-\frac{1}{2}$ for $i\geq 0$.

Since $z$ is detected by $x\in Ext^{s-k,t-k}_{P_{*}}(\mathbb{F}_{p},I^{k}/I^{k+1})$, the number of $\tau_{i}$’s in the expression of $z$ is just $k$. Hence we conclude $u(\tilde{z})=\frac{1}{2}t(\tilde{z})-\frac{1}{2}k=\frac{t-k}{2}$. ∎

(i) By assumption, $d_{i}^{mAdams}([\tilde{z}]_{i})=0$ for each $2\leq i<r$. One can inductively show that $d_{i}^{Adams}([z]_{i})=0$ (hence $[z]_{i+1}$ is well defined) for each $2\leq i<r$ by commutativity of diagram (2.5).

(ii) Assume, for the sake of contradiction, that there exists $2\leq r_{1}<r$ such that $d_{r_{1}}^{mAdams}([\tilde{z}]_{r_{1}})\neq 0$. We claim that there exists a nontrivial differential $d_{j}^{mAdams}([u]_{j})\neq 0$, where $u\in Ext^{s+b,t+b,\frac{t}{2}-a}_{\mathcal{A}_{*,*}^{\mathbb{C}}}(\mathbb{F}_{p}[\tau],\mathbb{F}_{p}[\tau])$ is non-$\tau$-divisible, $2\leq j<r_{1}$, and $a>0$.

We let $[v_{1}]_{r_{1}}$ denote $d_{r_{1}}^{mAdams}([\tilde{z}]_{r_{1}})\neq 0$. By the commutativity of diagram (2.5), we have $\phi_{r_{1}}([v_{1}]_{r_{1}})=d_{r_{1}}^{Adams}([z]_{r_{1}})=0$. So $[v_{1}]_{r_{1}}$ is $\tau$-torsion. Let $a_{1}>0$ be the smallest integer such that $\tau^{a_{1}}[v_{1}]_{r_{1}}=0$. Then there exists a nontrivial differential $d_{r_{2}}^{mAdams}([u_{1}]_{r_{2}})=\tau^{a_{1}}[v_{1}]_{r_{2}}\neq 0$ with $2\leq r_{2}<r_{1}$. The element $[u_{1}]_{r_{2}}$ is not divisible by $\tau$ on the $E_{r_{2}}$-page. Otherwise, we have $d_{r_{2}}^{mAdams}([u_{1}]_{r_{2}}/\tau)=\tau^{a_{1}-1}[v_{1}]_{r_{2}}$, which contradicts the definition of $a_{1}$.

By Lemma 3.4, we have $[\tilde{z}]_{r_{1}}\in E_{r_{1}}^{s+k,t+k,\frac{t}{2}}(S)$. This implies

Comparing the degrees, we have

If $u_{1}$ is non-$\tau$-divisible, we can take $d_{r_{2}}^{mAdams}([u_{1}]_{r_{2}})\neq 0$ as the claimed differential.

Otherwise, since $[u_{1}]_{r_{2}}$ is not divisible by $\tau$ on the $E_{r_{2}}$-page, we conclude $u_{1}/\tau$ does not lift to the $E_{r_{2}}$-page. Then there exists differential $d_{r_{3}}^{mAdams}([u_{1}/\tau]_{r_{3}})=[v_{2}]_{r_{3}}\neq 0$ with $2\leq r_{3}<r_{2}$.

By the commutativity of diagram (2.5), we have

and $d_{r_{3}}^{mAdams}([u_{1}]_{r_{3}})=0$ since $u_{1}$ can be lifted to the $E_{r_{2}}$-page. So $[v_{2}]_{r_{3}}$ is $\tau$-torsion. Let $a_{2}>0$ be the smallest integer such that $\tau^{a_{2}}[v_{2}]_{r_{3}}=0$. Then there exists a nontrivial differential $d_{r_{4}}^{mAdams}([u_{2}]_{r_{4}})=\tau^{a_{2}}[v_{2}]_{r_{4}}\neq 0$ with $2\leq r_{4}<r_{3}$. The element $[u_{2}]_{r_{4}}$ is not divisible by $\tau$ on the $E_{r_{4}}$-page.

For the degrees, we have

If $u_{2}$ is non-$\tau$-divisible, we can take $d_{r_{4}}^{mAdams}([u_{2}]_{r_{4}})\neq 0$ as the claimed differential. Otherwise, we can repeat the process and obtain $u_{3},u_{4},\cdots$. Note there are only finitely many integers between $2$ and $r_{1}$. After repeating this process several times, we will eventually obtain a desired nontrivial differential $d_{j}^{mAdams}([u]_{j})\neq 0$, where $u\in Ext^{s+b,t+b,\frac{t}{2}-a}_{\mathcal{A}_{*,*}^{\mathbb{C}}}(\mathbb{F}_{p}[\tau],\mathbb{F}_{p}[\tau])$ is non-$\tau$-divisible, $2\leq j<r_{1}$, and $a>0$.

By Proposition 2.3 and Lemma 3.1, we can write

where $u_{i}$ can be detected by some $m_{i}\in Ext^{s+b-i,t+b-i}_{P_{*}}(\mathbb{F}_{p},I^{i}/I^{i+1})$. Comparing the motivic weights using Lemma 3.4, we get $n_{i}=\frac{b-i+2a}{2}\geq 0$ for $\tilde{u}_{i}\neq 0$.

Note $u$ is non-$\tau$-divisible. This implies the $n_{i}=0$ term is nontrivial. In particular, we have $b+2a\in C_{s+b,t+b}$. This implies

Note $x\in Ext^{s,t}_{P_{*}}(\mathbb{F}_{p},I^{k}/I^{k+1})$ implies $t\equiv 0$ mod $q$. In summary, we have

Then $s\geq 2a\geq 2(p-1)$, this contradicts $s<2p-2$. Thus we have proved (ii). ∎

Our strategy is to compare the differentials via diagram (2.20). As we will see, with the given assumptions, diagram (2.20) can be specialized to the following diagram.

By Lemma 3.5, $[\tilde{z}]_{r}$ is well defined and we have $\phi_{r}([\tilde{z}]_{r})=[z]_{r}$. By Lemma 3.4, we have $[\tilde{z}]_{r}\in E_{r}^{s+k,t+k,\frac{t}{2}}(S)$. Then we can write $d_{r}^{mAdams}([\tilde{z}]_{r})=[u]_{r}$, where $u\in Ext^{s+k+r,t+k+r-1,\frac{t}{2}}_{\mathcal{A}_{*,*}^{\mathbb{C}}}(\mathbb{F}_{p}[\tau],\mathbb{F}_{p}[\tau])$. By Proposition 2.3 and Lemma 3.1, we can write

where $u_{i}$ can be detected by some $m_{i}\in Ext^{s+k+r-i,t+k+r-1-i}_{P_{*}}(\mathbb{F}_{p},I^{i}/I^{i+1})$. Comparing the motivic weights using Lemma 3.4, we get $n_{i}=\frac{k+r-1-i}{2}$ for $\tilde{u}_{i}\neq 0$.

Since $n_{i}\geq 0$, this forces $i\leq k+r-1$. By definition, $i\in C_{s+k+r,t+k+r-1}$ implies

where we denote $q=2(p-1)$. Moreover, since $x\in Ext^{s,t}_{P_{*}}(\mathbb{F}_{p},I^{k}/I^{k+1})$, we have $t\equiv 0$ mod $q$. In summary, we have

for nontrivial $\tilde{u}_{i}$.

By assumption, we have $0<k+r-1<q$. Then (3.10) forces $i=k+r-1$, the corresponding $n_{i}=0$. So we can rewrite (3.8) as $u=\tilde{w}$, where we let $w$ denote $u_{k+r-1}$. We also let $y$ denote $m_{k+r-1}$ detecting $w$.

We have $d_{r}^{mAdams}([\tilde{z}]_{r})=[\tilde{w}]_{r}$. Note $\phi_{r}([\tilde{w}]_{r})=[w]_{r}$. The commutativity of diagram (2.5) implies $d_{r}^{Adams}([z]_{r})=[w]_{r}$.

Note $[\tilde{z}]_{r}$ and $[\tilde{w}]_{r}$ are non-$\tau$-divisible, $\psi$ sends $[\tilde{z}]_{r}$ and $[\tilde{w}]_{r}$ to the corresponding elements in $E_{r}^{*,*,*}(C\tau)$ of the same form, which we abuse the notation and still denote by $[\tilde{z}]_{r}$ and $[\tilde{w}]_{r}$ respectively. The commutativity of diagram (2.10) implies $\delta_{r}^{mAdams}([\tilde{z}]_{r})=[\tilde{w}]_{r}$.

Finally, the the isomorphism $\kappa$ associates $\tilde{z}$ with $x$ and $\tilde{w}$ with $y$. Hence $\kappa_{r}([x]_{r})=[\tilde{z}]_{r}$, $\kappa_{r}([y]_{r})=[\tilde{w}]_{r}$, and $d_{r}^{alg}([x]_{r})=[y]_{r}$.

Now we have completed diagram (3.7). The results of the theorem follow directly.

∎

We can prove the following equivalent statement: suppose $r\geq 2$ is an integer such that $z$ can be lifted to the $E_{r}$-page (of the ASS). Then $x$ can also be lifted to the $E_{r}$-page (of the algNSS).

We will establish the statement using an inductive approach. By definition, $x$ can be lifted to the $E_{2}$-page. Next, assuming that $x$ can be lifted to the $E_{i}$-page with $2\leq i<r$, we need to show that $d_{i}^{alg}([x]_{i})=0$. This will imply that $x$ can be lifted to the $E_{i+1}$-page.

We study the differential $d_{i}^{alg}([x]_{i})$ via diagram (2.20). As we will see, with the given assumptions, diagram (2.20) can be specialized to the following diagram.

By Lemma 3.5, $\tilde{z}$ can be lifted to the $E_{r}$-page of the mASS of the sphere. Hence $d_{i}^{mAdams}([\tilde{z}]_{i})=0$. Note $[\tilde{z}]_{i}$ is non-$\tau$-divisible, so $\psi$ sends $[\tilde{z}]_{i}$ to the corresponding element in $E_{r}^{*,*,*}(C\tau)$ of the same form, which we abuse the notation and still denote by $[\tilde{z}]_{i}$. The commutativity of diagram (2.10) implies $\delta_{i}^{mAdams}([\tilde{z}]_{i})=0$. Finally, using the isomorphism $\kappa$, we deduce $d_{i}^{alg}([x]_{i})=0$.