6 The Generalized Leibniz Rule and Generalized Mahowald Trick

Now, we introduce the theorem of the Generalized Leibniz Rule, a valuable tool for computing new Adams differentials.

Theorem 6.1 Generalized Leibniz Rule

Let \(f: X\to Y\) be a map between two classical spectra. Suppose that \(2\le n\le r\), \(e(f)\le m\le n-2+e(f)\), \(l\ge e(f)\), and we have

\[ x\in Z_{r-1}^{s,t}(X), \hspace{0.5cm} y\in Z_{r-1-m+e(f)}^{s+m,t+m}(Y) \]

\[ x_\infty \in Z_\infty ^{s+r,t+r-1}(X), \hspace{0.5cm} y_\infty \in Z_\infty ^{s+r+l,t+r+l-1}(Y) \]

and the following conditions hold:

\(d_{r}(x)=x_\infty \),
\(d_{m}^{f,E_n}(x)=y\),
\(d_{l}^{f,E_\infty }(x_\infty )=y_\infty \),
the differential in (1) has no crossing on the \(E_n\)-page or (2) has no crossing.
the differential in (3) has no crossing.

Then we have an Adams differential

\[ d_{r+l-m}(y)= y_\infty . \]

Proof ▶

Consider the commutative diagram of synthetic spectra

\[ \begin{CD} \end{CD} \Sigma ^{0,e(f)}\nu X/\lambda ^{n-1} @{\gt}{\hat f_{n-1}}{\gt}{\gt} \nu Y/\lambda ^{n-1} \\ @V{\delta _X}VV @VV{\delta _Y}V \\ \Sigma ^{1,-n+1+e(f)} \nu X @{\gt}{\hat f}{\gt}{\gt} \Sigma ^{1,-n+1}\nu Y \end{CD} \]

By condition (1) and Proposition 4.5, we have

\begin{equation} \label{eq:leibniz-delta-diff} d^{\delta _X}_r(x)=\lambda ^{r-n}x_\infty . \end{equation}

6.1

By conditions (2) and (3), and Definition 5.2, we have

\begin{equation} \label{eq:leibniz-fhat-diff} d^{\hat f_{n-1}}_m(x)=\lambda ^{m-e(f)} y \end{equation}

6.2

and

\begin{equation*} d^{\hat f}_l(x_\infty )=\lambda ^{l-e(f)} y_\infty , \end{equation*}

which implies

\begin{equation} \label{eq:leibniz-fhat-xinfty} d^{\hat f}_l(\lambda ^{r-n}x_\infty )=\lambda ^{r+l-n-e(f)}y_\infty . \end{equation}

6.3

Applying Proposition 4.11 and Proposition 5.7 to conditions (4) and (5), we know that the differential in (6.1) or (6.2) has no crossing, and that the differential in (6.3) has no crossing.

Therefore, we can apply Corollary 0.53 and obtain

\[ d^{\delta _Y}_{r+l-m}(\lambda ^{m-e(f)} y)=\lambda ^{r+l-n-e(f)} y_\infty . \]

By Remark 4.8, this is equivalent to

\[ d_{r+l-m}(y)= y_\infty . \]

Remark 6.2

We can further generalize Theorem 6.1 by using the conditions in Theorem 0.52 rather than those Corollary 0.53. We leave this generalization to the reader.

Theorem 6.3 Generalized Leibniz Rule, level \(k\)

Let \(f: X\to Y\) be a map between two classical spectra. Suppose that \(2\le n\le r\), \(e(f)\le m\le n-2+e(f)\), \(l\ge e(f)\), \(k\ge 0\), and we have

\[ x\in Z_{r-1}^{s,t}(X), \hspace{0.5cm} y\in Z_{r-1-m+e(f)}^{s+m,t+m}(Y) \]

\[ x_\infty \in Z_\infty ^{s+r,t+r-1}(X), \hspace{0.5cm} y_\infty \in Z_\infty ^{s+r+l,t+r+l-1}(Y) \]

and the following conditions hold:

\(d_{r}(x)=x_\infty \),
\(d_{m}^{f,E_n, k}(x)=y\),
\(d_{l}^{f,E_\infty , k}(x_\infty )=y_\infty \),
the differential in (1) has no crossing on the \(E_n\)-page or (2) has no crossing.
the differential in (3) has no crossing.

Then we have an Adams differential

\[ d_{r+l-m}(y)= y_\infty . \]

Proof ▶

Consider the commutative diagram of synthetic spectra

By condition (1) and Proposition 4.5, we have

\begin{equation} \label{eq:leibniz-k-delta-diff} d^{\delta _X}_r(x)=\lambda ^{r-n}x_\infty . \end{equation}

6.4

By conditions (2) and (3), and Definition 5.4, we have

\begin{equation} \label{eq:leibniz-k-fhat-diff} d^{\hat f_{n-1}}_m(\lambda ^k x)=\lambda ^{m-e(f)+k} y \end{equation}

6.5

and

\begin{equation*} d^{\hat f}_l(\lambda ^k x_\infty )=\lambda ^{l-e(f)+k} y_\infty , \end{equation*}

which implies

\begin{equation} \label{eq:leibniz-k-fhat-xinfty} d^{\hat f}_l(\lambda ^{r-n+k}x_\infty )=\lambda ^{r+l-n-e(f)+k}y_\infty . \end{equation}

6.6

Applying Proposition 4.11 and Proposition 5.11 to conditions (4) and (5), we know that the differential in (6.4) or (6.5) has no crossing, and that the differential in (6.6) has no crossing.

Therefore, we can apply Corollary 0.53 and obtain

\[ d^{\delta _Y}_{r+l-m}(\lambda ^{m-e(f)+k} y)=\lambda ^{r+l-n-e(f)+k} y_\infty . \]

By Remark 4.8, this is equivalent to

\[ d_{r+l-m}(y)= y_\infty . \]

Theorem 6.4 Generalized Leibniz Rule, first direction

Let \(f: X\to Y\) be a map between two classical spectra. Suppose that \(2\le n\le r\), \(e(f)\le m\le n-2+e(f)\), \(l\ge e(f)\), \(k\ge 0\), and we have

\[ x\in Z_{r-1}^{s,t}(X), \hspace{0.5cm} y\in Z_{r-1-m+e(f)}^{s+m,t+m}(Y) \]

\[ x_\infty \in Z_\infty ^{s+r,t+r-1}(X), \hspace{0.5cm} y_\infty \in Z_\infty ^{s+r+l,t+r+l-1}(Y) \]

and the following conditions hold:

\(d_{r}(x)=x_\infty \),
\(d_{m}^{f,E_n, k}(x)=y\),
\(d_{r+l-m}(y)= y_\infty \),
the differential in (1) has no crossing on the \(E_n\)-page or (2) has no crossing,
the differential in (3) has no crossing.

Then we have an \((f,E_\infty )\)-extension

\[ d_{l}^{f,E_\infty , r-n+k}(x_\infty )=y_\infty . \]

Proof ▶

Consider the commutative diagram of synthetic spectra

By condition (1) and Proposition 4.5, we have

\begin{equation} \label{eq:leibniz-fwd-delta-diff} d^{\delta _X}_r(\lambda ^k x)=\lambda ^{r-n+k}x_\infty . \end{equation}

6.7

By condition (2) and Definition 5.4, we have

\begin{equation} \label{eq:leibniz-fwd-fhat-diff} d^{\hat f_{n-1}}_m(\lambda ^k x)=\lambda ^{m-e(f)+k} y. \end{equation}

6.8

By condition (3) and Remark 4.8, we have

\begin{equation} \label{eq:leibniz-fwd-delta-y} d^{\delta _Y}_{r+l-m}(\lambda ^{m-e(f)+k} y)=\lambda ^{r+l-n-e(f)+k} y_\infty . \end{equation}

6.9

Applying Proposition 4.11 and Proposition 5.11 to conditions (4) and (5), we know that the differential in (6.7) or (6.8) has no crossing, and that the differential in (6.9) has no crossing.

Therefore, we can apply Corollary 0.53 to the transposed commutative diagram (with \(\delta _X\) as the first map, \(\hat f_{n-1}\) as the second map, and \(\delta _Y\) as the known target extension) and obtain

\[ d^{\hat f}_l(\lambda ^{r-n+k}x_\infty )=\lambda ^{r+l-n-e(f)+k}y_\infty . \]

By definition, this is equivalent to

\[ d_{l}^{f,E_\infty , r-n+k}(x_\infty )=y_\infty . \]

We emphasize that the no-crossing conditions in Theorem 6.1, the Generalized Leibniz Rule, are crucial. Without the no-crossing condition, the conclusion could be false.

Remark 6.5

A version of the Generalized Leibniz Rule without the no-crossing conditions is presented in the synthetic setting in Chua’s work [ . However, there is no doubt that this version is incorrect. For further details, see Remark 6.7.

Remark 6.6

The essential \((f, E_\infty )\)-extension:

\[ d_1^{f, E_\infty } (d_0) = h_0 d_0, \]

is a crossing for both the \((f, E_\infty )\)-extension,

\[ d_2^{f, E_\infty } (h_0 h_3^2) = 0, \]

and the inessential \((f, E_\infty )\)-extension:

\[ d_2^{f, E_\infty } (h_0 h_3^2) = h_0 d_0. \]

This indicates that if we were to disregard the no-crossing condition (5), and apply the Generalized Leibniz Rule to these two cases of the \((f, E_\infty )\)-extensions, we would arrive at two conflicting classical statements:

\[ d_3(h_0h_4) = 0, \ d_3(h_0h_4) = h_0 d_0. \]

Remark 6.7

In Chua’s work [ , it is stated that for a synthetic map \(\alpha : X\rightarrow Y\), and an element \(x \in \pi _{*,*}X/\lambda \), there exists a differential from a maximal \(\alpha \)-extension of \(x\) to a maximal \(\alpha \)-extension of \(d_r(x)\). According to [ , a maximal \(\alpha \)-extension of \(x'\) is defined as \(\alpha [x']\), where \([x']\) represents a lift of \(x'\) to the \(E_{r'}\)-page, chosen such that \(\alpha [x']\) is the most \(\lambda \)-divisible among all such lifts.

In the context of the counter-example above, let \(r=2, r'=\infty \),

\[ \alpha = [h_0]: S^{0,1} \rightarrow S^{0,0}, \]

and consider \(x = h_4\) in Ext, with \(x' = d_2(x) = h_0h_3^2\).

The \((f, E_\infty )\)-extension:

\[ d_2^{f, E_\infty } (h_0 h_3^2) = 0 \]

is equivalent to the existence of a synthetic homotopy class \([h_0 h_3^2]\) in \(\pi _{14, 17} S^{0,0}\), such that

\[ [h_0 h_3^2] \cdot [h_0] = 0 \ \text{in} \ \pi _{14,18}S^{0,0}. \]

Similarly, the inessential \((f, E_\infty )\)-extension:

\[ d_2^{f, E_\infty } (h_0 h_3^2) = h_0 d_0 \]

implies the existence of another synthetic homotopy class \([h_0 h_3^2]\) in \(\pi _{14, 17} S^{0,0}\), such that

\[ [h_0 h_3^2] \cdot [h_0] = [h_0d_0] \ \text{in} \ \pi _{14,18}S^{0,0}. \]

Between these two lifts of \(h_0h_3^2\), the first \([h_0h_3^2]\) is clearly the maximal \([h_0]\)-extension according to Chua’s definition [ . Consequently, the incorrect version of the Generalized Leibniz Rule in [ , would lead to an incorrect conclusion:

\[ d_3(h_0h_4) = 0. \]

Next, we discuss the Generalized Mahowald Trick.

In order to apply the Generalized Leibniz Rule, we need to provide a method for computing extensions on specific Adams \(E_k\)-pages. This is provided by Theorem 6.9 (the Generalized Mahowald Trick). The crux of the proof of the Generalized Mahowald Trick lies in the following lemma.

Lemma 6.8 May [

Let \(X\to Y\to Z\to \Sigma X\) and \(X'\to Y'\to Z'\to \Sigma X'\) be distinguished triangles of (synthetic) spectra. By smashing these distinguished triangles together, we obtain the following commutative diagram of cofiber sequences:

\[ \begin{CD} \end{CD} X\wedge X' @{\gt}{\gt}{\gt} Y\wedge X' @{\gt}{\gt}{\gt} Z\wedge X' \\ @VVV @VVV @VVV \\ X\wedge Y' @{\gt}{\gt}{\gt} Y\wedge Y' @{\gt}{\gt}{\gt} Z\wedge Y' \\ @VVV @VVV @VVV \\ X\wedge Z' @{\gt}{\gt}{\gt} Y\wedge Z' @{\gt}{\gt}{\gt} Z\wedge Z' \end{CD} \]

If \(a\in \pi _n(X\wedge Z')\) and \(b\in \pi _n(Y\wedge Y')\) map to the same element in \(\pi _n(Y\wedge Z')\), then there exists \(c\in \pi _n(Z\wedge X')\) such that

\(b\) and \(c\) map to the same element in \(\pi _n(Z\wedge Y')\), and
\(a\) and \(c\) map to the same element via boundary maps in \(\pi _{n-1}(X\wedge X')\).

Proof ▶

The proof follows directly from [ .

In fact, consider \(V\) in Axiom TC3 of [ and the corresponding commutative diagram. By [ we know that \(V\) is the pull back of the following diagram.

\[ \begin{CD} \end{CD} V @{\gt}{\gt}{\gt} Y\wedge Y' \\ @VVV @VVV \\ X\wedge Z' @{\gt}{\gt}{\gt} Y\wedge Z' \end{CD} \]

where the square is a pullback. Since \(a\) and \(b\) map to the same element in \(\pi _n(Y\wedge Z')\), we know that we can find \(v\in V\) such that \(v\) maps to \(a\) in \(\pi _n(X\wedge Z')\) and \(b\) in \(\pi _n(Y\wedge Y')\). Then we let \(c=j_3(v)\in \pi _n(Z\wedge X')\), where \(j_3: V \rightarrow Z \wedge X'\) is the map in the commutative diagram in Axiom TC3 of [ . This lemma follows from the commutativity of the diagram.

Theorem 6.9 Generalized Mahowald Trick

Consider a distinguished triangle of spectra

\[ X\xrightarrow {f} Y\xrightarrow {g} Z\xrightarrow {h} \Sigma X \]

with \(e(f)+e(g)+e(h)=1\). Suppose that \(r=n+m+l\), \(n_1=n-e(f)\ge 1\), \(m_1=m-e(g)\ge 0\), \(l_1=l-e(h)\ge 0\), and

\[ \begin{array}{ll} x\in Z_{n_1}^{s+l,t+l-1}(X), & y\in Z_{m_1+1}^{s+n+l,t+n+l-1}(Y),\\ \bar x\in Z_{r-1}^{s,t}(Z), & \bar y\in Z_\infty ^{s+r,t+r-1}(Z), \end{array} \]

such that

\(d^{h,E_{r'}}_{l}\bar x=x\), where \(r'=r-m_1=n_1+l_1+1\),
\(d_r \bar x=\bar y\),
the \((h,E_{r'})\)-extension in (1) has no crossing, or the Adams differential (2) has no crossing on the \(E_{r'}\)-page,
\(d^{g,E_{m_1+2}}_{m}y=\bar y\),
for \(0 \le i \le n-1\), the Adams \(E_\infty \)-page of \(Y\) vanishes at the positions that are potential targets of \((f,E_r)\)-extensions of degree \(i\) from \(x\): \(E_\infty ^{s+l+i, t+l+i-1}(Y) = 0\).

Then we have \(x\in Z_{n+m+e(h)}^{s+l,t+l-1}(X)\) and

\[ d^{f,E_{n+m+1+e(h)}}_{n}x\equiv y \mod B_{r'}^{s+n+l,t+n+l-1}(Y). \]

(See Figure 4.)

tikz diagram — Figure 4 A demonstration of Theorem 6.9

\(\begin{CD} \end{CD} \bigwedge @. \nu X @{\gt}{\hat f}{\gt}{\gt} \Sigma ^{0,-\! e(f)}\nu Y @{\gt}{\hat g}{\gt}{\gt} \Sigma ^{0,-\! e(f)\! -\! e(g)}\nu Z @{\gt}{\hat h}{\gt}{\gt} \Sigma ^{1,0}\nu X \\ S^{0,0}/\lambda ^{r} @. @. @. @. [\lambda ^{l_1}x] \\ @V{\rho }VV @. @. @. @VVV \\ S^{0,0}/\lambda ^{r'-1} @. @. [\bar x] @{\gt}{\gt}{\gt} [\lambda ^{l_1}x] \\ @V{\delta _{r'-1}}VV @. @. @VVV @. \\ S^{1,-r'+1}/\lambda ^{m_1+1} @. [y] @{\gt}{\gt}{\gt} [\lambda ^{m_1}\bar y] @. \\ @V{\lambda ^{r'-1}}VV @. @VVV @. @. \\ S^{1,0}/\lambda ^{r} @. [\lambda ^{l_1}x] @{\gt}{\quad }{\gt}{\gt} [\lambda ^{r'-1}y] @. \end{CD}\)

Figure 5 Elements in the homotopy groups of the smash products

Proof ▶

Consider the following two distinguished triangles of synthetic spectra

\begin{gather*} \nu X\xrightarrow {\hat f} \Sigma ^{0,-e(f)}\nu Y\xrightarrow {\hat g} \Sigma ^{0,-e(f)-e(g)}\nu Z\xrightarrow {\hat h} \Sigma ^{0,-1}\nu \Sigma X=\Sigma ^{1,0}\nu X\\ S^{0,0}/\lambda ^{r}\xrightarrow {\rho } S^{0,0}/\lambda ^{r'-1} \xrightarrow {\delta _{r'-1}} S^{1,-r'+1}/\lambda ^{m_1+1} \xrightarrow {\lambda ^{r'-1}} S^{1,0}/\lambda ^{r} \end{gather*}

and their smash product. See Figure 5.

By condition (1), we have

\begin{equation} \label{eq:mahowald-cond-a} d^{\hat h_{r'-1}}_l(\bar x)=\lambda ^{l_1}x \end{equation}

6.10

where \(\hat h_{r'-1}\) is the map \(\Sigma ^{0,e(h)}\nu Z/\lambda ^{r'-1}\to \nu X/\lambda ^{r'-1}\) induced by \(\hat h\). By condition (2), we have

\begin{equation} \label{eq:mahowald-cond-b} d^{\delta _{r'-1}}_r(\bar x)=\lambda ^{m_1}\bar y. \end{equation}

6.11

Applying Proposition 4.11 and Proposition 5.7 to condition (3), we know that the differential in (6.10) or the differential in (6.11) has no crossing. Hence, there exists

\[ [\bar x]\in \{ \bar x\} \subset \pi _{t-s,t}(\nu Z/\lambda ^{r'-1}) \]

such that

\[ \hat h_{r'-1}([\bar x])\in \{ \lambda ^{l_1}x\} \subset \pi _{t-s-1, t-1+e(h)}(\nu X/\lambda ^{r'-1}) \]

and

\[ \delta _{r'-1}([\bar x])\in \{ \lambda ^{m_1}\bar y\} \subset \pi _{t-s-1,t+r'-1}(\nu Z/\lambda ^{m_1+1}). \]

By condition (4), we have

\begin{equation*} d^{\hat g_{m_1+1}}_m(y)=\lambda ^{m_1}\bar y. \end{equation*}

This implies that there exists

\[ [y]\in \{ y\} \subset \pi _{t-s-1,t+n+l-1}(\nu Y/\lambda ^{m_1+1}) \]

such that

\[ \hat g_{m_1+1}([y])\in \{ \lambda ^{m_1}\bar y\} \subset \pi _{t-s-1, t+r'-1}(\nu Z/\lambda ^{m_1+1}), \]

where \(\hat g_{m_1+1}\) is the map \(\Sigma ^{0,e(g)}\nu Y/\lambda ^{m_1+1}\to \nu Z/\lambda ^{m_1+1}\) induced by \(\hat g\).

Due to degree reasons (Proposition 0.122), \(\lambda ^{m_1}\bar y\) detects a unique element in homotopy,

\[ [\lambda ^{m_1}\bar y]\in \pi _{t-s-1, t+r'-1}(\nu Z)/\lambda ^{m_1+1}. \]

Therefore, we have

\[ \delta _{r'-1}([\bar x])=\hat g_{m_1+1}([y])=[\lambda ^{m_1}\bar y], \]

By Lemma 6.8, there exists

\[ [\lambda ^{l_1}x]\in \pi _{t-s-1,t-1+e(h)}(\nu X/\lambda ^r), \]

such that

\begin{gather} \rho ([\lambda ^{l_1}x])=\hat h_{r'-1}([\bar x]),\label{eq:mahowald-rho-eq}\\ \hat f_r([\lambda ^{l_1}x])=\lambda ^{r'-1}[y],\label{eq:mahowald-fhat-eq} \end{gather}

where \(\rho \) is the map \(\nu X/\lambda ^r\to \nu X/\lambda ^{r'-1}\), and \(\hat f_r\) is the map \(\Sigma ^{0,e(f)}\nu X/\lambda ^r\to \nu Y/\lambda ^r\) induced by \(\hat f\).

The equality in (6.12) indicates that \([\lambda ^{l_1}x]\) can be lifted to the \(E_\infty \)-page of \(\nu X/\lambda ^r\), so we have

\[ x\in Z_{n+m+e(h)}^{s+l,t+l-1}(X). \]

The equation in (6.13) indicates that

\[ d^{\hat f_r}_n(\lambda ^{l_1}x)=\lambda ^{r'-1}y. \]

Therefore, by dividing \(\lambda ^{l_1}\), we have

\[ d^{\hat f_{n+m+e(h)}}_n(x)\equiv \lambda ^{n_1}y \mod B_{r'}^{s+n+l,t+n+l-1}(Y) \]

for \(x\) in the \(E_\infty \) page of \(\nu X/\lambda ^{r-l_1}=\nu X/\lambda ^{n+m+e(h)}\). By definition, this is equivalent to

\[ d^{f, E_{n+m+1+e(h)}}_nx\equiv y \mod B_{r'}^{s+n+l,t+n+l-1}(Y). \]

Remark 6.10

There is also a version of the Generalized Mahowald Trick where the assumptions involve \((g, E_{m_1+2})\)-extensions and \((h, E_{r'})\)-extensions of level \(k\). We leave this generalization to the reader.

We refer to Theorem 6.9 as the Generalized Mahowald Trick, as this approach was first utilized by Mahowald and his collaborators in various works (see, for example, [ ), particularly in the case where \(m_1 = l_1 = 0\). The synthetic setting advances this method by allowing for the consideration of cases where \(m_1{\gt}0, l_1{\gt}0\) as well.

Remark 6.11

Several versions of classical generalizations of Mahowald’s original trick appear in the literature and are often referred to as geometric boundary theorems. Notable examples include the works of Behrens [ , and Ma [ .

Remark 6.12

A synthetic generalization of the Mahowald Trick, again lacking any no-crossing conditions, is also presented in Chua’s work [ . This version is also incorrect for similar reasons.

The outcome of the Generalized Mahowald Trick, as stated in Theorem 6.9, is an \(f\)-extension on a specific Adams page. In practice, the source of an \((f, E_r)\)-extension may survive to later Adams pages, prompting interest in the \((f, E_{{\gt}r})\)-extensions. The following Propositions 6.13 and 6.14 describe the relationships between extensions across different pages.

Proposition 6.13

Suppose that we have an \((f,E_r)\)-extension \(d^{f,E_r}_n(x)=y\), where \(x\in Z_{r-1}^{s,t}(X)\) and \(y\in Z_{r-1-n+e(f)}^{s+n,t+n}(Y)\). Then for all \(2\le r'{\lt}r\), we also have

\begin{equation} \label{eq:dr'} d^{f,E_{r'}}_n(x)=y. \end{equation}

6.14

Furthermore, if \(d^{f,E_r}_n(x)=y\) is essential and \(n\le r'-2+e(f)\), then (6.14) is inessential if and only if there exists some \(0{\lt}a'\le n-e(f)\) and an element

\[ x'\in Z_{r'-1-a'}^{s+a',t+a'}(X)\backslash Z_{r-1-a'}^{s+a',t+a'}(X) \]

such that

\[ d^{\hat f_{r'-1}}_{n-a'-b}(\lambda ^{a'} x')=\lambda ^{n-b-e(f)} y' \]

for some \(b\ge 0\).

Proof ▶

Consider the following commutative diagram:

\[ \begin{CD} \end{CD} \Sigma ^{0,e(f)}\nu X/\lambda ^{r-1} @{\gt}{\hat f_{r-1}}{\gt}{\gt} \nu Y/\lambda ^{r-1} \\ @V{\rho _X}VV @VV{\rho _Y}V \\ \Sigma ^{0,e(f)}\nu X/\lambda ^{r'-1} @{\gt}{\hat f_{r'-1}}{\gt}{\gt} \nu Y/\lambda ^{r'-1} \end{CD} \]

By Corollary 0.57, \((\rho _X,\rho _Y)\) induces a map from the \(\hat f_{r-1}\)-ESS to the \(\hat f_{r'-1}\)-ESS. Therefore, by naturality,

\[ d^{\hat f_{r-1}}_n(x)=\lambda ^{n-e(f)} y\ \text{implies} \ d^{\hat f_{r'-1}}_n(x)=\lambda ^{n-e(f)} y, \]

which, by definition, is

\[ d^{f,E_{r'}}_n(x)=y. \]

Next, we prove the second part of the proposition. By Definition 5.2, the \((f,E_r)\)-extension (6.14) is inessential if and only if there exists \(0{\lt}a\le n-e(f)\) and

\[ \lambda ^a x'\in E_\infty ^{s+a,t+a,t}(\nu X/\lambda ^{r'-1})\cong Z_{r'-1-a}^{s+a,t+a}(X)/B_{1+a}^{s+a,t+a}(X), \]

such that

\begin{equation} \label{eq:inessential-crossing} d^{\hat f_{r'-1}}_{n-a}(\lambda ^ax')=\lambda ^{n-e(f)} y, \end{equation}

6.15

and this differential is not induced by \((\rho _X,\rho _Y)\), as we assume that \(d^{f,E_r}_n(x)=y\) is essential.

There are two scenarios where this differential is not induced by \((\rho _X,\rho _Y)\). The first case occurs when \(\lambda ^ax'\) is not in the image of \(\rho _X\) at all, which is equivalent to \(x'\notin Z_{r-1-a}^{s+a,t+a}(X)\). (See Figure 6.)

The second case occurs when \(\lambda ^ax'\) is in the image of \(\rho _X\), but it supports an essential \(\hat f_{r-1}\)-extension that is strictly shorter than the differential (6.15).

To further explore this scenario, we replace \(x\) with \(\lambda ^ax'\) and analyze successive essential extensions. This process is repeated iteratively until the first case is reached. Ultimately, this leads to the existence of some \(0{\lt}a'\le n-e(f)\) and an element

\[ x''\in Z_{r'-1-a'}^{s+a',t+a'}(X)\backslash Z_{r-1-a'}^{s+a',t+a'}(X) \]

such that

\[ d^{\hat f_{r'-1}}_{n-a'-b}(\lambda ^{a'} x'')=\lambda ^{n-b-e(f)} y' \]

for some \(b{\gt}0\). This represents a crossing of \(d^{\hat f_{r'-1}}_n(x)=\lambda ^{n-e(f)}y\).

The following Corollary 6.14 is the contrapositive statement of the second part of Proposition 6.13.

Corollary 6.14

Suppose \(r\ge r'\) and there exists an \((f,E_{r'})\)-extension

\[ d^{f,E_{r'}}_n(x)=y \]

for \(x\in Z_{r-1}^{s,t}(X)\) and \(y\in Z_{r'-1-n+e(f)}^{s+n,t+n}(Y)\). Assume that this extension has no crossing of the form \(d^{f,E_{r'-a}}_{n-a-b}(x')=y'\) for any \(b{\gt}0\), \(0{\lt}a\le n-e(f)\), and

\[ x'\in Z_{r'-1-a}^{s+a,t+a}(X)\backslash Z_{r-1-a}^{s+a,t+a}(X). \]

Under these conditions, we also have an \((f,E_{r})\)-extension

\[ d^{f,E_{r}}_n(x)=y. \]

Example 6.15

In Example ??, we use the Generalized Mahowald Trick to establish the \((f, E_3)\)-extension:

\[ d_2^{f, E_3} (h_0h_4^2) = h_0 p, \]

for the map \(f=\nu :S^3 \rightarrow S^0\), as discussed in Example ??(2). However, to apply the Generalized Leibniz Rule in proving the classical Adams differential

\[ d_3(h_2h_5) = h_0p \]

in Example ??, we need the \(E_\infty \)-page version of the \((f, E_3)\)-extension:

\[ d_2^{f, E_\infty } (h_0h_4^2) = h_0 p. \]

We apply Corollary 6.14 to confirm this.

As checked in Example ??(2), the \((f, E_3)\)-extension has no crossing. Consequently, by Corollary 6.14, we obtain the required \((f, E_\infty )\)-extension:

\[ d_2^{f, E_\infty } (h_0h_4^2) = h_0 p. \]