\section{Introduction}
\label{sec:intro}

Let $G$ be a simply-connected, simple compact Lie group.
Principal $G$-bundles over $S^{4}$ are classified by the value
of the second Chern class, which can take any integer value. Let
\(\namedright{P_{k}}{}{S^{4}}\)
represent the equivalence class of principal $G$-bundle whose
second Chern class is $k$. Let \ensuremath{\mathcal{G}_{k}}\ be the \emph{gauge group}
of this principal $G$-bundle, which is the group of $G$-equivariant
automorphisms of $P_{k}$ which fix $S^{4}$.

Crabb and Sutherland~\cite{CS} showed that, while there are countably
many inequivalent principal $G$-bundles, the gauge groups
$\{\ensuremath{\mathcal{G}_{k}}\}_{k\in\mathbb{Z}}$ have only finitely many distinct homotopy
types. There has been a great deal of interest recently in determining
the precise number of possible homotopy types. The following
classifications are known. For two integers $a,b$, let $(a,b)$ be
their greatest common divisor. If $G=SU(2)$ then
$\ensuremath{\mathcal{G}_{k}}\simeq\ensuremath{\mathcal{G}_{k^{\prime}}}$ if and only if $(12,k)=(12,k^{\prime})$~\cite{K};
if $G=SU(3)$ then $\ensuremath{\mathcal{G}_{k}}\simeq\ensuremath{\mathcal{G}_{k^{\prime}}}$ if and only if
$(24,k)=(24,k^{\prime})$~\cite{HK}; if $G=SU(5)$ then
$\ensuremath{\mathcal{G}_{k}}\simeq\ensuremath{\mathcal{G}_{k^{\prime}}}$ when localized at any prime $p$ or
rationally if and only if $(120,k)=(120,k^{\prime})$~\cite{Th2}; and if
$G=Sp(2)$ then $\ensuremath{\mathcal{G}_{k}}\simeq\ensuremath{\mathcal{G}_{k^{\prime}}}$ when localized at any prime $p$
or rationally if and only if $(40,k)=(40,k^{\prime})$~\cite{Th1}. Partial
classifications that are potentially off by a factor of $2$ have been
worked out for $G_{2}$~\cite{KTT} and $Sp(3)$~\cite{Cu}.

The $SU(4)$ case is noticeably absent. The $SU(5)$ case was easier
since elementary bounds on the number of homotopy types matched at
the prime $2$ but did not at the prime $3$, and it is typically easier to
work out $3$-primary problems in low dimension than $2$-primary
problems. In the $SU(4)$ case the elementary bounds do not match at $2$, and
the purpose of this paper is to resolve the difference, at least after looping.

\begin{theorem}
   \label{types}
   For $G=SU(4)$, there is a homotopy equivalence
   $\Omega\ensuremath{\mathcal{G}_{k}}\simeq\Omega\ensuremath{\mathcal{G}_{k^{\prime}}}$ when localized at any prime $p$ or
   rationally if and only if $(60,k)=(60,k^{\prime})$.
\end{theorem}

Two novel features arise in the methods used, as compared to the other
known classifications. One is the use of Miller's stable splittings of Stiefel
manifolds in order to gain some control over unstable splittings, and the
other is showing that a certain ambiguity which prevents a clear
classification statement for $\ensuremath{\mathcal{G}_{k}}$ vanishes after looping. It would be
interesting to know if these ideas give access to classifications for
$SU(n)$-gauge groups for $n\geq 6$.

One motivation for studying $SU(4)$-gauge groups is their connection
to physics, in particular, to $SU(n)$-extensions of the standard model.
For instance, the group $SU(4)$ is gauged in the Pati-Salam model~\cite{PS}
and the flavour symmetry it represents there plays a role in several other
grand unified theories~\cite{BH}. The progression of results from $SU(2)$
to $SU(5)$ and possibly beyond would be of interest to physicists studying
the $SU(n)$-gauge groups in t'Hooft's large $n$ expansion~\cite{tH}.

\section{Determining homotopy types of gauge groups}
\label{sec:prelim}

We begin by describing a context in which homotopy theory can be applied
to study gauge groups. This works for any simply-connected, simple
compact Lie group $G$ and so is stated that way. Let $BG$ and $B\ensuremath{\mathcal{G}_{k}}$
be the classifying spaces of $G$ and $\ensuremath{\mathcal{G}_{k}}$ respectively.
Let $\ensuremath{\mbox{Map}}(S^{4},BG)$ and $\mapstar(S^{4},BG)$ respectively
be the spaces of freely continuous and pointed continuous maps
between~$S^{4}$ and $BG$. The components of each space are
in one-to-one correspondence with the integers, where the
integer is determined by the degree of a map
\(\namedright{S^{4}}{}{BG}\).
By~\cite{AB,G}, there is a homotopy equivalence
$B\ensuremath{\mathcal{G}_{k}}\simeq\ensuremath{\mbox{Map}}_{k}(S^{4},BG)$ between $B\ensuremath{\mathcal{G}_{k}}$ and the component
of $\mbox{Map}(S^{4},BG)$ consisting of maps of degree~$k$.
Evaluating a map at the basepoint of $S^{4}$, we obtain a map
\(ev\colon\namedright{B\ensuremath{\mathcal{G}_{k}}}{}{BG}\)
whose fibre is homotopy equivalent to $\mapstar_{k}(S^{4},BG)$.
It is well known that each component of $\mapstar(S^{4},BG)$ is
homotopy equivalent to $\Omega^{3}_{0} G$, the component of
$\Omega^{3} G$ containing the basepoint. Putting all this together,
for each $k\in\mathbb{Z}$, there is a homotopy fibration sequence
\begin{equation}
  \label{evfib}
  \namedddright{G}{\partial_{k}}{\Omega^{3}_{0} G}{}{B\ensuremath{\mathcal{G}_{k}}}{ev}{BG}
\end{equation}
where $\partial_{k}$ is the fibration connecting map.

The order of $\partial_{k}$ plays a crucial role. By~\cite{L}, the triple adjoint
\(\namedright{S^{3}\wedge G}{}{G}\)
of $\partial_{k}$ is homotopic to the Samelson product
$\langle k\cdot i,1\rangle$, where $i$ is the inclusion of
$S^{3}$ into $G$ and $1$ is the identity map on~$G$. This implies
two things. First, the order of $\partial_{k}$ is finite. For, rationally,
$G$ is homotopy equivalent to a product of Eilenberg-MacLane spaces
and the homotopy equivalence can be chosen to be one of $H$-maps.
Since Eilenberg-MacLane spaces are homotopy commutative, any Samelson
product into such a space is null homotopic. Thus, rationally, the
adjoint of $\partial_{k}$ is null homotopic, implying that
the same is true for $\partial_{k}$ and therefore the order of
$\partial_{k}$ is finite. Second, the linearity of the Samelson product
implies that $\langle k\cdot i,1\rangle\simeq k\circ\langle i,1\rangle$,
so taking adjoints we obtain $\partial_{k}\simeq k\circ\partial_{1}$.
Thus the order of~$\partial_{k}$ is determined by the order of $\partial_{1}$.
When $G=SU(n)$, Hamanaka and Kono~\cite{HK} gave the following lower bound
on the order of $\partial_{1}$ and the number of homotopy types of \ensuremath{\mathcal{G}_{k}}.

\begin{lemma}
   \label{HKlemma}
   Let $G=SU(n)$. Then the following hold:
   \begin{letterlist}
      \item the order of $\partial_{1}$ is divisible by $n(n^{2}-1)$;
      \item if $\ensuremath{\mathcal{G}_{k}}\simeq\ensuremath{\mathcal{G}_{k^{\prime}}}$ then
                    $(n(n^{2}-1),k)=(n(n^{2}-1),k^{\prime})$.
   \end{letterlist}
\end{lemma}
\vspace{-0.8cm}~$\hfill\Box$\medskip

The calculation in~\cite{HK} that established
Lemma~\ref{HKlemma} was that a composite
\(\nameddright{\Sigma^{2n-5}\mathbb{C}P^{2}}{}{SU(n)}{\partial_{1}}{\Omega_{0}^{3} SU(n)}\)
has order~$n(n^{2}-1)$. The adjoint of this map has the same order, so
we obtain the following alternative formulation.

\begin{lemma}
   \label{loopHKlemma}
   Let $G=SU(n)$. Then the following hold:
   \begin{letterlist}
      \item the order of $\Omega\partial_{1}$ is divisible by $n(n^{2}-1)$;
      \item if $\Omega\ensuremath{\mathcal{G}_{k}}\simeq\Omega\ensuremath{\mathcal{G}_{k^{\prime}}}$ then
                    $(n(n^{2}-1),k)=(n(n^{2}-1),k^{\prime})$.
   \end{letterlist}
\end{lemma}
\vspace{-0.8cm}~$\hfill\Box$\medskip

In particular, if $G=SU(4)$ then $60$ divides the order of $\Omega\partial_{1}$
and a homotopy equivalence $\Omega\ensuremath{\mathcal{G}_{k}}\simeq\Omega\ensuremath{\mathcal{G}_{k^{\prime}}}$ implies that
$(60,k)=(60,k^{\prime})$. In Section~\ref{sec:proof} we will find an upper bound
on the order of $\Omega\partial_{1}$ that matches the lower bound.

\begin{theorem}
   \label{looppartialorder}
   The map
   \(\namedright{\Omega SU(4)}{\Omega\partial_{1}}{\Omega^{4}_{0} SU(4)}\)
   has order~$60$.
\end{theorem}

Granting Theorem~\ref{looppartialorder} for now, we can prove
Theorem~\ref{types} by using the following general result from~\cite{Th1}.
If $Y$ is an $H$-space, let
\(k\colon\namedright{Y}{}{Y}\)
be the $k^{th}$-power map.

\begin{lemma}
   \label{ptypecount}
   Let $X$ be a space and $Y$ be an $H$-space with a homotopy inverse.
   Suppose there is a map
   \(\namedright{X}{f}{Y}\)
   of order $m$, where $m$ is finite. Let $F_{k}$ be the homotopy fibre
   of $k\circ f$. If $(m,k)=(m,k^{\prime})$ then $F_{k}$ and $F_{k^{\prime}}$
   are homotopy equivalent when localized rationally or at any prime.~$\hfill\Box$
\end{lemma}

\noindent
\begin{proof}[Proof of Theorem~\ref{types}]
By Theorem~\ref{looppartialorder}, the map
\(\namedright{\Omega SU(4)}{\Omega\partial_{1}}{\Omega^{4}_{0} SU(4)}\)
has order~$60$. So Lemma~\ref{ptypecount} implies that if
$(60,k)=(60,k^{\prime})$, then $\Omega\ensuremath{\mathcal{G}_{k}}\simeq\Omega\ensuremath{\mathcal{G}_{k^{\prime}}}$ when
localized at any prime $p$ or rationally. On the other hand,
by Lemma~\ref{loopHKlemma}, if $\Omega\ensuremath{\mathcal{G}_{k}}\simeq\Omega\ensuremath{\mathcal{G}_{k^{\prime}}}$ then
$(60,k)=(60,k^{\prime})$. Thus there is a homotopy equivalence
$\Omega\ensuremath{\mathcal{G}_{k}}\simeq\Omega\ensuremath{\mathcal{G}_{k^{\prime}}}$ at each prime $p$ and rationally if and
only if $(60,k)=(60,k^{\prime})$.
\end{proof}

It remains to prove Theorem~\ref{looppartialorder}. In fact, the odd primary
components of the order of $\partial_{1}$ (and hence $\Omega\partial_{1}$
by Lemma~\ref{loopHKlemma}) are obtained  as special cases of a more general
result in~\cite{Th3}.

\begin{lemma}
   \label{oddbounds}
   Localized at $p=3$, $\partial_{1}$ has order $3$; localized at $p=5$,
   $\partial_{1}$ has order $5$; and localized at~$p>5$, $\partial_{1}$ has
   order~$1$.~$\hfill\Box$
\end{lemma}

Thus to prove Theorem~\ref{looppartialorder} we are reduced to proving
the following.

\begin{theorem}
   \label{looppartialorder2}
    Localized at $2$ the map
   \(\namedright{\Omega SU(4)}{\Omega\partial_{1}}{\Omega^{4}_{0} SU(4)}\)
   has order~$4$.
\end{theorem}

\section{An initial upper bound on the $2$-primary order of $\partial_{1}$}
\label{sec:initialbound}

For the remainder of the paper all spaces and maps will be localized at $2$
and homology will be taken with mod-$2$ coefficients. Note that some statements
that follow are valid wilthout localizing, such as Theorems~\ref{Miller}
and~\ref{Millernat}, but rather than dancing back and forth between local
and non-local statements we simply localize at $2$ throughout.

In~\cite{B}, or by different means in~\cite{KKT},
it was shown that there is a homotopy commutative square
\[\diagram
       SU(n)\rto^-{\partial_{1}}\dto^{\pi} & \Omega^{3}_{0} SU(n)\ddouble \\
       SU(n)/SU(n-2)\rto^-{f} & \Omega^{3}_{0} SU(n)
  \enddiagram\]
for some map $f$, where $\pi$ is the standard quotient map.
In our case it is well known that there is a homotopy equivalence
$SU(4)/SU(2)\simeq S^{5}\times S^{7}$. Thus there is a homotopy
commutative square
\begin{equation}
  \label{su4su2}
  \diagram
     SU(4)\rto^-{\partial_{1}}\dto^{\pi} & \Omega^{3}_{0} SU(4)\ddouble \\
     S^{5}\times S^{7}\rto^-{f} & \Omega^{3}_{0} SU(4).
  \enddiagram
\end{equation}
Taking the triple adjoint of $f$, we obtain a map
\[f'\colon\nameddright{S^{8}\vee S^{10}\vee S^{15}}{\simeq}
       {\Sigma^{3}(S^{5}\times S^{7})}{}{SU(4)}.\]
Mimura and Toda~\cite{MT} calculated the homotopy groups of $SU(4)$
through a range. The $2$-primary components of $\pi_{8}(SU(4))$,
$\pi_{10}(SU(4))$ and $\pi_{15}(SU(4))$ are
$\mathbb{Z}/8\mathbb{Z}$, $\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$
and $\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$, respectively.
Consequently, the order of $f'$ is bounded above by $8$. The order of~$f$
is therefore also bounded above by $8$. The homotopy commutativity
of~(\ref{su4su2}) then implies the following.

\begin{lemma}
   \label{initialbound}
   The order the map
   \(\namedright{SU(4)}{\partial_{1}}{\Omega^{3}_{0} SU(4)}\)
   is bounded above by $8$.~$\hfill\Box$
\end{lemma}

Ideally it should be possible to reduce this upper bound by a factor of two.
The remainder of the paper aims to show that this can be done after looping.

\section{A cofibration}
\label{sec:Cprops}

The homotopy groups of spheres will play an important role. In all
cases except one we follow Toda's notation~\cite{To}. Specifically,
(i) for $n\geq 3$, $\eta_{n}=\Sigma^{n-3}\eta_{3}$ represents the
generator of $\pi_{n+1}(S^{n})\cong\mathbb{Z}/2\mathbb{Z}$;
(ii) for $n\geq 5$, $\nu_{n}=\Sigma^{n-3}\nu_{5}$ represents the
generator of $\pi_{n+3}(S^{n})\cong\mathbb{Z}/8\mathbb{Z}$;
and (iii) differing from Toda's notation, for $n\geq 3$, $\nu'_{n}=\Sigma^{n-3}\nu'_{3}$
represents the $n-3$ fold suspension of the generator~$\nu'_{3}$ of
$\pi_{6}(S^{3})\cong\mathbb{Z}/4\mathbb{Z}$. Note that for $n\geq 5$,
$\nu'_{n}=2\nu_{n}$.
\medskip

By~(\ref{su4su2}), the map
\(\namedright{SU(4)}{\partial_{1}}{\Omega^{3}_{0} SU(4)}\)
factors as a composite
\(\nameddright{SU(4)}{\pi}{S^{5}\times S^{7}}{f}{\Omega^{3}_{0} SU(4)}\).
In this section we determine properties of the map $\pi$ and its homotopy cofibre.
To prepare, first recall some properties of $SU(4)$. There is an
algebra isomorphism $\cohlgy{SU(4)}\cong\Lambda(x,y,z)$,
where the degrees of $x,y,z$ are $3,5,7$ respectively. A $\mathbb{Z}/2\mathbb{Z}$-module
basis for $\rcohlgy{SU(4)}$ is therefore $\{x,y,z,xy,xz,yz,xyz\}$
in degrees $\{3,5,7,8,10,12,15\}$ respectively, so $SU(4)$ may be given a
$CW$-structure with one cell in each of those dimensions. There is a canonical map
\(\namedright{\Sigma\ensuremath{\mathbb{C}P^{3}}}{}{SU(4)}\)
which induces a projection onto the generating set in cohomology. Notice
that $\Sigma\ensuremath{\mathbb{C}P^{3}}$ is homotopy equivalent to the $7$-skeleton of $SU(4)$,
and there is a homotopy cofibration
\begin{equation}
  \label{CP3cofib}
  \llnameddright{S^{4}\vee S^{6}}{\eta_{3}\vee\nu'_{3}}{S^{3}}{}{\Sigma\ensuremath{\mathbb{C}P^{3}}}.
\end{equation}

Miller~\cite{M} gave a stable decomposition of Stiefel manifolds which
includes the following as a special case.

\begin{theorem}
   \label{Miller}
   There is a stable homotopy equivalence
   \[SU(4)\simeq_{S}\Sigma\ensuremath{\mathbb{C}P^{3}}\vee M\vee S^{15}\]
   where $M$ is given by the homotopy cofibration
   \[\llnameddright{S^{11}}{\nu'_{8}+\eta_{10}}{S^{8}\vee S^{10}}{}{M}.\]
\end{theorem}
\vspace{-1cm}~$\hfill\Box$\medskip

Crabb~\cite{C} and Kitchloo~\cite{Kitch} refined the stable decomposition
of Stiefel manifolds and proved that it was natural for maps
\(\namedright{SU(n)}{}{SU(n)/SU(m)}\).
In our case, this gives the following.

\begin{theorem}
   \label{Millernat}
   Stably, there is a homotopy commutative diagram
   \[\diagram
         SU(4)\rto^-{\simeq_{S}}\dto^{\pi}
             & \Sigma\ensuremath{\mathbb{C}P^{3}}\vee M\vee S^{15}\dto^{\overline{\pi}} \\
         S^{5}\times S^{7}\rto^-{\simeq_{S}} & S^{5}\vee S^{7}\vee S^{12}.
     \enddiagram\]
   where $\overline{\pi}$ is the wedge sum of: (i) the map
   \(\namedright{\Sigma\ensuremath{\mathbb{C}P^{3}}}{}{S^{5}\vee S^{7}}\)
   that collapses the bottom cell, (ii) the pinch map
   \(\lnamedright{M}{}{S^{12}}\)
   to the top cell, and (iii) the trivial map
   \(\namedright{S^{15}}{}{\ast}\).~$\hfill\Box$
\end{theorem}

Now define the space $C$ and maps $j$ and $\delta$ by the homotopy cofibration
\begin{equation}
  \label{Ccofib}
  \namedddright{SU(4)}{\pi}{S^{5}\times S^{7}}{j}{C}{\delta}{\Sigma SU(4)}.
\end{equation}
Since $\pi^{\ast}$ is an inclusion onto the subalgebra $\Lambda(y,z)$
of $\Lambda(x,y,z)\cong\cohlgy{SU(4)}$, the long exact sequence
in cohomology induced by the cofibration sequence~(\ref{Ccofib}) implies that
a $\mathbb{Z}/2\mathbb{Z}$-module basis for $\rcohlgy{C}$
is given by $\{\sigma x,\sigma xy, \sigma xz, \sigma xyz\}$
in degrees $\{4,9,11,16\}$ respectively, where the elements of
$\rcohlgy{C}$ have been identified with the image of $\delta^{\ast}$.
So as a $CW$-complex, $C$ has one cell in each of the dimensions
$\{4,9,11,16\}$. The stable homotopy type of $C$ and the stable class
of the map $j$ follow immediately from Theorem~\ref{Millernat}.

\begin{proposition}
   \label{Cstable}
   Stably, there is a homotopy commutative diagram
   \[\diagram
         S^{5}\times S^{7}\rto^-{\simeq_{S}}\dto^{j}
             & S^{5}\vee S^{7}\vee S^{12}\dto^{\overline{j}} \\
         C\rto^-(0.6){\simeq_{S}} & S^{4}\vee S^{9}\vee S^{11}\vee S^{16}
     \enddiagram\]
   where $\overline{j}$ is the wedge sum of: (i)
   \(\llnamedright{S^{5}\vee S^{7}}{\eta_{4}+\nu'_{4}}{S^{4}}\),
   and (ii) \(\llnamedright{S^{12}}{\nu'_{9}+\eta_{11}}{S^{9}\vee S^{11}}\).
\end{proposition}
\vspace{-1cm}~$\hfill\Box$\medskip

The stable decomposition of $C$ will be useful but we will ultimately need
to work with unstable information in the form of the homotopy type
of $\Sigma^{3} C$ and the homotopy class of $\Sigma^{3} j$. We start
with the homotopy type of $\Sigma^{3} C$. In general, for a $CW$-complex $X$
and positive integer~$m$, let~$X_{m}$ be the $m$-skeleton of $X$. In our case,
the $CW$-structure for $C$ implies that there are homotopy cofibrations
\begin{align}
   \label{Ccofib1}
   & \nameddright{S^{8}}{g_{1}}{S^{4}}{}{C_{9}} \\
   \label{Ccofib2}
   & \nameddright{S^{10}}{g_{2}}{C_{9}}{}{C_{11}} \\
   \label{Ccofib3}
   & \nameddright{S^{15}}{g_{3}}{C_{11}}{}{C}
\end{align}

\begin{lemma}
   \label{C9decomp}
   There is a homotopy equivalence $\Sigma^{2} (C_{9})\simeq S^{6}\vee S^{11}$.
\end{lemma}

\begin{proof}
By~\cite{To}, $\pi_{10}(S^{6})=0$, so the map $\Sigma^{2} g_{1}$
in~(\ref{Ccofib1}) is null homotopic. The asserted homotopy equivalence
for $\Sigma^{2} (C_{9})$ follows immediately.
\end{proof}

\begin{lemma}
   \label{C11decomp}
   There is a homotopy equivalence
   $\Sigma^{2} (C_{11})\simeq S^{6}\vee S^{11}\vee S^{13}$.
\end{lemma}

\begin{proof}
Substituting the homotopy equivalence in Lemma~\ref{C9decomp} into the
double suspension of~(\ref{Ccofib2}) gives a homotopy cofibration
\(\nameddright{S^{12}}{\Sigma^{2} g_{2}}{S^{6}\vee S^{11}}{}{\Sigma^{2} (C_{11})}\).
By the Hilton-Milnor Theorem, $\Sigma^{2} g_{2}\simeq a+b$ where
$a$ and $b$ are obtained by composing $\Sigma^{2} g_{2}$ with the
pinch maps to $S^{6}$ and $S^{11}$ respectively. We claim that each of $a$
and $b$ is null homotopic, implying that $\Sigma^{2} g_{2}$ is null
homotopic, from which the asserted homotopy equivalence for $\Sigma^{2} (C_{11})$
follows immediately.

By Proposition~\ref{Cstable}, $C$ is stably homotopy equivalent to a
wedge of spheres, and therefore $C_{11}$ is too. Thus $g_{2}$ is stably
trivial, implying for dimensional reasons that $a$ and $b$ are as well. On
the other hand, $a$ and $b$ are represented by classes in
$\pi_{12}(S^{6})\cong\mathbb{Z}/2\mathbb{Z}$
and $\pi_{12}(S^{11})\cong\mathbb{Z}/2\mathbb{Z}$ respectively. By~\cite{To},
these groups are generated by $\nu_{6}^{2}$ and $\eta_{11}$, both of which
are stable. Thus the only way that~$a$ and $b$ can be stably trivial is if
both are already trivial. Hence $\Sigma^{2} g_{2}$ is null homotopic.
\end{proof}

\begin{lemma}
   \label{Cdecomp}
   There is a homotopy equivalence
   $\Sigma^{3} C\simeq E\vee S^{12}\vee S^{14}$
   where $E$ is given by a homotopy cofibration
   \(\llnameddright{S^{18}}{s\cdot\bar{\nu}_{7}\nu_{15}}{S^{7}}{}{E}\)
   for some $s\in\mathbb{Z}/2\mathbb{Z}$.
\end{lemma}

\begin{proof}
Substituting the homotopy equivalence in Lemma~\ref{C11decomp} into the
double suspension of~(\ref{Ccofib3}) gives a homotopy cofibration
\(\nameddright{S^{17}}{\Sigma^{2} g_{3}}{S^{6}\vee S^{11}\vee S^{13}}{}{\Sigma^{2} C}\).
By the Hilton-Milnor Theorem, $\Sigma^{2} g_{3}\simeq a+b+c+d$ where
$a$, $b$ and $c$ are obtained by composing $\Sigma^{2} g_{3}$ with the
pinch maps to $S^{6}$, $S^{11}$ and~$S^{13}$ respectively, and $d$ is a composite
\(\nameddright{S^{17}}{}{S^{16}}{w}{S^{6}\vee S^{11}\vee S^{13}}\).
Here, $w$ is the Whitehead product of the identity maps on $S^{6}$ and $S^{11}$.
As $\Sigma w$ is null homotopic, we instead consider
\[\nameddright{S^{18}}{\Sigma^{3} g_{3}}{S^{7}\vee S^{12}\vee S^{14}}
     {}{\Sigma^{3} C}\]
where $\Sigma^{3} g_{3}\simeq\Sigma a+\Sigma b+\Sigma c$.

By Proposition~\ref{Cstable}, $C$ is stably homotopy equivalent to a wedge
of spheres, so $\Sigma^{3} g_{3}\simeq\Sigma a+\Sigma b+\Sigma c$ is
stably trivial. Thus, for dimensional reasons, each of $\Sigma a$, $\Sigma b$
and $\Sigma c$ is stably trivial. Observe that both $\Sigma b$ and $\Sigma c$
are in the stable range, impling that they are null homotopic. On the other hand,
$\Sigma a$ represents a class in $\pi_{18}(S^{7})$. By~\cite{To},
$\pi_{18}(S^{7})\cong\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$
where the order $8$ generator is the stable class $\zeta_{7}$ and the
order~$2$ generator is the unstable class $\bar{\nu}_{7}\nu_{15}$. Note
too that the stable order of $\zeta_{7}$ is $8$, so the only nontrivial unstable
class in $\pi_{18}(S^{7})$ is $\bar{\nu}_{7}\nu_{15}$. As $\Sigma a$ is stably
trivial, we obtain $\Sigma a=s\cdot\bar{\nu}_{7}\nu_{15}$ for some
$s\in\mathbb{Z}/2\mathbb{Z}$. Hence $\Sigma^{2} g_{3}$ factors as the composite
\(\llnamedright{S^{18}}{s\cdot\bar{\nu}_{7}\nu_{15}}{S^{7}}\hookrightarrow
     S^{7}\vee S^{12}\vee S^{14}\),
from which the asserted homotopy decomposition of $\Sigma^{3} C$ follows.
\end{proof}

Next, we identify $\Sigma^{3} j$. Let
\[\iota\colon\namedright{S^{7}}{}{E}\]
be the inclusion of the bottom cell.

\begin{lemma}
   \label{jclass}
   There is a homotopy commutative diagram
   \[\diagram
          S^{8}\vee S^{10}\vee S^{15}\rto^-{\simeq}\dto^{a+b+c}
               & \Sigma^{3}(S^{5}\times S^{7})\dto^{\Sigma^{3} j} \\
          E\vee S^{12}\vee S^{14}\rto^-{\simeq} & \Sigma^{3} C
     \enddiagram\]
   where $a$, $b$ and $c$ respectively are the composites
   \begin{align*}
          & a\colon\nameddright{S^{8}}{\eta_{7}}{S^{7}}{\iota}{E}\hookrightarrow
                E\vee S^{12}\vee S^{14} \\
          & b\colon\nameddright{S^{10}}{\nu'_{7}}{S^{7}}{\iota}{E}\hookrightarrow
                E\vee S^{12}\vee S^{14} \\
          & c\colon\lllnameddright{S^{15}}{\psi+\nu'_{12}+\eta_{14}}{S^{7}\vee S^{12}\vee S^{14}}
                {\iota\vee 1\vee 1}{E\vee S^{12}\vee S^{14}}
   \end{align*}
   and $\psi=t\cdot\sigma'\eta_{14}$ for some $t\in\mathbb{Z}/2\mathbb{Z}$.
\end{lemma}

\begin{proof}
By Proposition~\ref{Cstable}, the diagram in the statement of the lemma
stably homotopy commutes if~$c$ is replaced by the composite
\(c'\colon\lllnameddright{S^{15}}{\ast+\nu'_{12}+\eta_{14}}{S^{7}\vee S^{12}\vee S^{14}}
                {\iota\vee 1\vee 1}{E\vee S^{12}\vee S^{14}}\).
Since $a$ and $b$ are in the stable range, the diagram in the statement of the
lemma therefore does homotopy commute when restricted to $S^{8}\vee S^{10}$.
However, $c'$ is not in the stable range. It fails to be so only by a map
\(\psi''\colon\namedright{S^{15}}{}{S^{7}}\).
Thus if $c''$ is the composite
\(c''\colon\lllnameddright{S^{15}}{\psi''+\nu'_{12}+\eta_{14}}{S^{7}\vee S^{12}\vee S^{14}}
                {\iota\vee 1\vee 1}{E\vee S^{12}\vee S^{14}}\)
then the diagram in the statement of the lemma homotopy commutes
with $c$ replaced by $c''$.

More can be said. By~\cite{To} (stated later also in~(\ref{SU4groups})),
$\pi_{15}(S^{7})\cong\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\oplus
    \mathbb{Z}/2\mathbb{Z}$
with generators $\sigma'\nu_{14}$, $\bar{\nu}_{7}$ and $\epsilon_{7}$.
Thus $\psi''=t\cdot\sigma'\nu_{14}+u\cdot\bar{\nu}_{7} + v\cdot\epsilon_{7}$
for some $t,u,v\in\mathbb{Z}/2\mathbb{Z}$. The generators $\bar{\nu}_{7}$
and $\epsilon_{7}$ are stable while $\sigma'\nu_{14}$ is unstable. So as
$c''$ stabilizes to $c$, we must have $\psi''$ stabilizing to the trivial map.
Thus $u$ and $v$ must be zero. Hence $\psi''=t\cdot\sigma'\nu_{14}$.
Now $c''$ is exactly the map $c$ described in the statement of the lemma.
\end{proof}

\section{Preliminary information on the homotopy groups of $SU(4)$}
\label{sec:htpygroups}

This section records some information on the homotopy groups
of $SU(4)$ which will be needed subsequently. There is a homotopy fibration
\[\nameddright{S^{3}}{i}{SU(4)}{q}{S^{5}\times S^{7}}.\]
This induces a long exact sequence of homotopy groups
\[\cdots\longrightarrow\namedddright{\pi_{n+1}(S^{3}\times S^{5})}{}
     {\pi_{n}(S^{3})}{i_{\ast}}{\pi_{n}(SU(4))}{q_{\ast}}{\pi_{n}(S^{5}\times S^{7})}
     \longrightarrow\cdots\]
Following~\cite{MT}, the notation $[\alpha\oplus\beta]\in\pi_{n}(SU(4))$ means that
$[\alpha\oplus\beta]$ is a generator of $\pi_{n}(SU(4))$ with the property
that $q_{\ast}([\alpha\oplus\beta])=\alpha\oplus\beta$ for
$\alpha\in\pi_{n}(S^{5})$ and $\beta\in\pi_{n}(S^{7})$. The homotopy groups
of $SU(4)$ in low dimensions were determined by Mimura and Toda~\cite{MT}.

The information presented will be split into two parts, the first corresponding
to subsequent calculations involving $\pi_{8}(SU(4))$ and $\pi_{10}(SU(4))$,
and the second corresponding to calculations involving $\pi_{15}(SU(4))$.

First, for $r\geq 1$, let
\(\underline{2}^{r}\colon\namedright{S^{7}}{}{S^{7}}\)
be the map of degree $2^{r}$. In general, the degree two map on $S^{2n+1}$
need not induce multiplication by $2$ in homotopy groups. However,
as $S^{7}$ is an $H$-space, the degree $2$ map on $S^{7}$ is homotopic
to the $2^{nd}$-power map, implying that it does in fact induce multiplication
by $2$ in homotopy groups. We record this for later use.

\begin{lemma}
   \label{S72}
   The map
   \(\namedright{S^{7}}{\underline{2}}{S^{7}}\)
    induces multiplication by $2$ in homotopy groups.~$\hfill\Box$
\end{lemma}

\subsection{Dimensions $8$ and $10$}
The relevant table of homotopy groups from~\cite{MT} is:
\begin{equation}
  \label{SU4groups1}
  \begin{tabular}{|c|c|c|c|}\hline
          & $\pi_{7}(SU(4))$ & $\pi_{8}(SU(4))$ & $\pi_{10}(SU(4))$ \\ \hline
          $2$-component & $\mathbb{Z}$ & $\mathbb{Z}/8\mathbb{Z}$
             & $\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ \\ \hline
          generators & $[\eta_{5}^{2}\oplus\underline{2}]$
             & $[\nu_{5}\oplus\eta_{7}]$
             & $[\nu_{7}]$, $[\nu_{5}\eta_{8}^{2}]$ \\ \hline
  \end{tabular}
\end{equation}
In addition, Mimura and Toda~\cite[Lemma 6.2(i)]{MT} proved that
\(\namedright{\pi_{n+1}(S^{5}\times S^{7})}{}{\pi_{n}(S^{3})}\)
is an epimorphism for $n\in\{8,10\}$, implying the following.

\begin{lemma}
   \label{pi810inj}
   The map
   \(\namedright{\pi_{n}(SU(4))}{q_{\ast}}{\pi_{n}(S^{5}\times S^{7})}\)
   is an injection for $n\in\{8,10\}$.~$\hfill\Box$
\end{lemma}

Toda~\cite{To} proved the following relations in the homotopy groups of spheres.

\begin{lemma}
   \label{Todarelns1}
   The following hold:
   \begin{letterlist}
      \item $2\nu'_{3}\simeq\eta_{3}^{3}$;
      \item $4\nu_{5}\simeq\eta_{5}^{3}$;
      \item $\eta_{3}\nu'_{3}\simeq\ast$.
   \end{letterlist}
\end{lemma}
\vspace{-1cm}~$\hfill\Box$\medskip

For convenience, let
\[d\colon\namedright{S^{7}}{}{SU(4)}\]
represent the generator $[\eta_{5}^{2}\oplus\underline{2}]$ of $\pi_{7}(SU(4))$.

\begin{lemma}
   \label{S810dgrms}
   There are homotopy commutative diagrams
   \[\diagram
          S^{8}\rto^-{[\nu_{5}\oplus\eta_{7}]}\dto^{\eta_{7}} & SU(4)\dto^{4}
              & & S^{10}\rto^-{[\nu_{7}]}\dto^{\nu'_{7}} & SU^{4}\dto^{4} \\
          S^{7}\rto^-{d} & SU(4) & & S^{7}\rto^-{d} & SU(4).
     \enddiagram\]
\end{lemma}

\begin{proof}
By Lemma~\ref{pi810inj},
\(\namedright{\pi_{n}(SU(4))}{q_{\ast}}{\pi_{n}(S^{5}\times S^{7})}\)
is an injection for $n\in\{8,10\}$. So in both cases it suffices to show that the
asserted homotopies hold after composition with
\(\namedright{SU(4)}{q}{S^{5}\times S^{7}}\).
Since the composite
\(\nameddright{S^{7}}{d}{SU(4)}{q}{S^{5}\times S^{7}}\)
is $\eta_{5}^{2}\times\underline{2}$, the two assertions will follow if we prove:

(i) $(\eta_{5}^{2}\times\underline{2})\circ\eta_{7}\simeq q\circ 4\circ[\nu_{5}\oplus\eta_{7}]$;

(ii) $(\eta_{5}^{2}\times\underline{2})\circ\nu'_{7}\simeq q\circ 4\circ[\nu_{7}]$.

By Lemma~\ref{S72}, $\underline{2}\circ\eta_{7}\simeq 2\eta_{7}$ and
$\underline{2}\circ\nu'_{7}\simeq 2\nu'_{7}$. Since $\eta_{7}$ has order~$2$
we obtain $\underline{2}\circ\eta_{7}\simeq\ast$. By Lemma~\ref{Todarelns1}~(a),
$2\nu'_{7}\simeq\eta_{7}^{3}$. Thus (i) and (ii) reduce to proving:

(i$^{\prime}$) $\eta_{5}^{3}\simeq q\circ 4\circ[\nu_{5}\oplus\eta_{7}]$;

(ii$^{\prime}$) $\eta_{7}^{3}\simeq q\circ 4\circ[\nu_{7}]$.

Consider the diagram
\[\diagram
     S^{8}\rto^-{[\nu_{5}\oplus\eta_{7}]}\dto^{\underline{4}} & SU(4)\dto^{4} \\
     S^{8}\rto^-{[\nu_{5}\oplus\eta_{7}]}\drto_{\nu_{5}\times\eta_{7}}
         & SU(4)\dto^{q} \\
     & S^{5}\times S^{7}.
  \enddiagram\]
The top square homotopy commutes since the multiplications in
$[S^{8},SU(4)]$ induced by the $H$-structure on $SU(4)$ and the
co-$H$-structure on $S^{8}$ coincide. The bottom square homotopy
commutes by definition of $[\nu_{5}\circ\eta_{7}]$. Since $\eta_{7}$
has order $2$ and, by Lemma~\ref{Todarelns1}, $4\nu_{5}\simeq\eta_{5}^{3}$,
we obtain $(\nu_{5}\times\eta_{7})\circ\underline{4}\simeq\eta_{5}^{3}$.
Therefore $q\circ 4\circ [\nu_{5}\circ\eta_{7}]\simeq\eta_{5}^{3}$, and
so (i$^{\prime}$) holds.

Next, consider the diagram
\[\diagram
     S^{10}\rto^-{[\nu_{7}]}\dto^{\underline{4}} & SU(4)\dto^{4} \\
     S^{10}\rto^-{[\nu_{7}]}\drto_{\ast\times\nu_{7}}
         & SU(4)\dto^{q} \\
     & S^{5}\times S^{7}.
  \enddiagram\]
The two squares homotopy commute as in the previous case.
By Lemma~\ref{Todarelns}, $4\nu_{7}\simeq\eta_{7}^{3}$. Therefore
$q\circ 4\circ\beta\simeq\eta_{7}^{3}$, and so (ii$^{\prime}$) holds.
\end{proof}

\subsection{Dimension $15$}

The relevant table of homotopy groups from~\cite{MT} is:
\begin{equation}
  \label{SU4groups}
  \begin{tabular}{|c|c|c|c|}\hline
          & $\pi_{12}(SU(4))$ & $\pi_{14}(SU(4))$ & $\pi_{15}(SU(4))$ \\ \hline
          $2$-component & $\mathbb{Z}/4\mathbb{Z}$
             & $\mathbb{Z}/16\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$
             & $\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ \\ \hline
          generators & $[\sigma^{'''}]$
             & $[\eta_{5}\epsilon_{6}\oplus\sigma']$, $[\nu_{5}^{2}]\circ\nu_{11}$
             & $[\nu_{5}\oplus\eta_{7}]\circ\sigma_{8}$, $[\sigma'\eta_{14}]$ \\ \hline
  \end{tabular}
\end{equation}
In addition, Mimura and Toda~\cite[Lemma 6.2(i)]{MT} proved that
\(\namedright{\pi_{16}(S^{5}\times S^{7})}{}{\pi_{15}(S^{3})}\)
is an epimorphism, implying the following.

\begin{lemma}
   \label{pi15inj}
   The map
   \(\namedright{\pi_{15}(SU(4))}{q_{\ast}}{\pi_{15}(S^{5}\times S^{7})}\)
   is an injection.~$\hfill\Box$
\end{lemma}

The next table gives some information on the $2$-primary components of
selected homotopy groups of spheres that were determined by Toda~\cite{To}:
\begin{equation}
  \label{S15groups}
  \begin{tabular}{|c|c|c|c|}\hline
          & $\pi_{15}(S^{7})$ & $\pi_{15}(S^{12})$ & $\pi_{15}(S^{14})$ \\ \hline
          $2$-component & $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}
                  \oplus\mathbb{Z}/2\mathbb{Z}$
             & $\mathbb{Z}/8\mathbb{Z}$ & $\mathbb{Z}/2\mathbb{Z}$ \\ \hline
          generators & $\sigma'\eta_{14}$, $\bar{\nu}_{7}$, $\epsilon_{7}$
             & $\nu_{12}$ & $\eta_{14}$ \\ \hline
  \end{tabular}
\end{equation}
In addition, Toda~\cite{To} proved the following relations (the proofs are scattered
through Toda's book but a summary list can be found in~\cite[Equations 1.1 and 2.1]{O}).

\begin{lemma}
   \label{Todarelns}
   The following hold:
   \begin{letterlist}
      \item $\eta_{5}\bar{\nu}_{6}=\nu_{5}^{3}$;
      \item $\eta_{3}\nu_{4}=\nu'_{3}\eta_{6}$;
      \item $\eta_{6}\sigma'=4\bar{\nu}_{6}$;
      \item $\eta_{6}\nu_{7}=\nu_{6}\eta_{9}=0$;
      \item $\sigma^{'''}\nu_{12}=4(\nu_{5}\sigma_{8})$.
   \end{letterlist}
\end{lemma}
\vspace{-1cm}~$\hfill\Box$\medskip

Lemma~\ref{Todarelns} is used to obtain two more relations.

\begin{lemma}
   \label{sphererelns}
   The following hold:
   \begin{letterlist}
      \item $\eta_{5}^{2}\bar{\nu}_{7}=0$;
      \item $\eta_{5}^{2}\sigma'=0$.
   \end{letterlist}
\end{lemma}

\begin{proof}
In what follows, we freely use the fact that the relations in Lemma~\ref{Todarelns}
imply analogous relations for their suspensions; for example,
$\eta_{5}\bar{\nu}_{6}=\nu_{5}^{3}$ implies that $\eta_{6}\bar{\nu}_{7}=\nu_{6}^{3}$.

For part~(a), the relations in Lemma~\ref{Todarelns}~(a), (b) and (d) respectively imply
the following string of equalities:
$\eta_{5}^{2}\bar{\nu}_{7}=\eta_{5}\nu_{6}^{3}=\nu'_{5}\eta_{8}\nu_{9}^{2}=0$.

For part~(b), Lemma~\ref{Todarelns}~(c) and the fact that $\eta_{5}$ has order~$2$
imply that there are equalities $\eta_{5}^{2}\sigma'=\eta_{5}(4\bar{\nu}_{6})=0$.
\end{proof}

We now determine the homotopy classes of two maps into $SU(4)$.

\begin{lemma}
   \label{SU4relns}
   The following hold:
   \begin{letterlist}
      \item the composite
                \(\nameddright{S^{15}}{\bar{\nu}_{7}}{S^{7}}{d}{SU(4)}\)
                is null homotopic;
      \item the composite
                \(\lnameddright{S^{15}}{\sigma'\eta_{14}}{S^{7}}{d}{SU(4)}\)
                is null homotopic.
   \end{letterlist}
\end{lemma}

\begin{proof}
By Lemma~\ref{pi15inj},
\(\namedright{\pi_{15}(SU(4))}{q_{\ast}}{\pi_{15}(S^{5}\times S^{7})}\)
is an injection. So in both cases it suffices to show that the assertions hold
after composition with
\(\namedright{SU(4)}{q}{S^{5}\times S^{7}}\).
Since the composite
\(\nameddright{S^{7}}{d}{SU(4)}{q}{S^{5}\times S^{7}}\)
is $\eta_{5}^{2}\times\underline{2}$, the two assertions will follow if we prove:

(a$^{\prime}$) $(\eta_{5}^{2}\times\underline{2})\circ\bar{\nu}_{7}\simeq\ast$;

(b$^{\prime}$) $(\eta_{5}^{2}\times\underline{2})\circ\sigma'\eta_{14}\simeq\ast$.

\noindent
By Lemma~\ref{S72}, the degree two map on $S^{7}$ induces multiplication
by $2$ on homotopy groups, so as both $\bar{\nu}_{7}$
and $\sigma'\eta_{14}$ have order~$2$, it suffices to prove:

(a$^{\prime\prime}$) $\eta_{5}^{2}\bar{\nu}_{7}\simeq\ast$;

(b$^{\prime\prime}$) $\eta_{5}^{2}\sigma'\eta_{14}\simeq\ast$.

\noindent
Part~(a$^{\prime\prime}$) is the statement of Lemma~\ref{sphererelns}~(a) and
part~(b$^{\prime\prime}$) is immediate from Lemma~\ref{sphererelns}~(b).
\end{proof}

One consequence of Lemma~\ref{SU4relns} is the existence of an extension
involving the space $E$ appearing in the homotopy decomposition of $\Sigma^{3} C$
in Lemma~\ref{Cdecomp}.

\begin{lemma}
   \label{Eext}
   There is an extension
   \[\diagram
          S^{7}\rto^-{d}\dto^{\iota} & SU(4) \\
          E\urto_-{e} &
     \enddiagram\]
   for some map $e$.
\end{lemma}

\begin{proof}
By Lemma~\ref{Cdecomp}, there is a homotopy cofibration
\(\llnameddright{S^{18}}{t\cdot\bar{\nu}_{7}\nu_{15}}{S^{7}}{}{E}\)
for some $t\in\mathbb{Z}/2\mathbb{Z}$. By Lemma~\ref{SU4relns}~(a),
$d\circ\bar{\nu}_{7}$ is null homotopic. Therefore $d\circ (t\cdot\bar{\nu}_{7}\nu_{15})$
is null homotopic, implying that the asserted extension exists.
\end{proof}

\section{The proof of Theorem~\ref{looppartialorder2}}
\label{sec:proof}

Recall from~(\ref{su4su2}) that
\(\namedright{SU(4)}{\partial_{1}}{\Omega^{3}_{0} SU(4)}\)
factors as the composite
\(\nameddright{SU(4)}{\pi}{S^{5}\times S^{7}}{f}{\Omega^{3}_{0} SU(4)}\).
Let
\[f'\colon\namedright{\Sigma^{3}(S^{5}\times S^{7})}{}{SU(4)}\]
be the triple adjoint of $f$. Let $f'_{1}$, $f'_{2}$ and $f'_{3}$ be the
restrictions of the composite
\[\nameddright{S^{8}\vee S^{10}\vee S^{15}}{\simeq}{\Sigma^{3}(S^{5}\vee S^{7})}
     {f'}{SU(4)}\]
to $S^{8}$, $S^{10}$ and $S^{15}$ respectively. We wish to identify
$f'_{1}$, $f'_{2}$ and $f'_{3}$ more explicitly. Let
\(t_{1}\colon\namedright{S^{5}}{}{SU(4)}\)
and
\(t_{2}\colon\namedright{S^{7}}{}{SU(4)}\)
represent generators of $\pi_{5}(SU(4))\cong\mathbb{Z}$ and
$\pi_{7}(SU(4))\cong\mathbb{Z}$ respectively. By~\cite{MT} these generators
can be chosen so that $\pi\circ t_{1}$ is homotopic to $\underline{2}\oplus\ast$
and $\pi\circ t_{2}$ is homotopic to $\eta^{2}_{5}\oplus\underline{2}$.
So there are homotopy commutative diagrams
\begin{equation}
  \label{2Bottdgrms}
  \diagram
     S^{5}\rto^-{t_{1}}\drto_{\underline{2}\oplus\ast}
        & SU(4)\rto^-{\partial_{1}}\dto^{\pi} & \Omega^{3}_{0} SU(4)\ddouble
        & & S^{7}\rto^-{t_{2}}\drto_{\eta^{2}_{5}\oplus\underline{2}}
        & SU(4)\rto^-{\partial_{1}}\dto^{\pi} & \Omega^{3}_{0} SU(4)\ddouble \\
     & S^{5}\times S^{7}\rto^-{f} & \Omega^{3}_{0} SU(4)
        & & & S^{5}\times S^{7}\rto^-{f} & \Omega^{3}_{0} SU(4).
  \enddiagram
\end{equation}
On the other hand, since the triple adjoint of $\partial_{1}$ is the Samelson
product $\langle i,1\rangle$, the triple adjoint of $\partial_{1}\circ t_{j}$
is $\langle t_{j},1\rangle$ for $j=1,2$. Bott~\cite{B} calculated that both of
these maps have order $4$. Thus the left diagram in~(\ref{2Bottdgrms})
implies that the restriction of $f$ to $S^{5}$ has order~$8$, and the right
diagram in~(\ref{2Bottdgrms}) implies that the restriction of $f$ to $S^{7}$
has order~$8$. Thus, taking triple adjoints, $f'_{1}$ and $f'_{2}$ both
have order~$8$.

The order of $f'_{3}$ is not as clear. By~(\ref{SU4groups}),
$\pi_{15}(SU(4))\cong\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$,
so $f'_{3}$ may have order~$8$. This ambiguity will be reflected in the alternative
possibilities worked out below.

Recall from Lemma~\ref{jclass} that there is a homotopy commutative diagram
\[\diagram
       S^{8}\vee S^{10}\vee S^{15}\rto^-{\simeq}\dto^{a+b+c}
            & \Sigma^{3}(S^{5}\times S^{7})\dto^{\Sigma^{3} j} \\
       E\vee S^{12}\vee S^{14}\rto^-{\simeq} & \Sigma^{3} C
  \enddiagram\]
where $a$, $b$ and $c$ respectively are the composites
\begin{align*}
       & a\colon\nameddright{S^{8}}{\eta_{7}}{S^{7}}{\iota}{E}\hookrightarrow
             E\vee S^{12}\vee S^{14} \\
       & b\colon\nameddright{S^{10}}{\nu'_{7}}{S^{7}}{\iota}{E}\hookrightarrow
             E\vee S^{12}\vee S^{14} \\
       & c\colon\lllnameddright{S^{15}}{\psi+\nu'_{12}+\eta_{14}}{S^{7}\vee S^{12}\vee S^{14}}
             {\iota\vee 1\vee 1}{E\vee S^{12}\vee S^{14}}
\end{align*}
and $\psi=t\cdot\sigma'\eta_{14}$ for some $t\in\mathbb{Z}/2\mathbb{Z}$.
Let $c'$ be the composite
\[c'\colon\lllnameddright{S^{15}}{\psi'+\nu'_{12}+\eta_{14}}{S^{7}\vee S^{12}\vee S^{14}}
     {\iota\vee 1\vee 1}{E\vee S^{12}\vee S^{14}}\]
where $\psi'=t\cdot\sigma'\eta_{14}+\eta_{7}\sigma_{8}$. Let $\xi$
be the composite
\[\xi\colon\nameddright{E\vee S^{12}\vee S^{14}}{}{E}{e}{SU(4)}\]
where the left map is the pinch onto the first wedge summand and $e$ is
the map from Lemma~\ref{Eext}.

\begin{lemma}
   \label{alternative}
   There is a homotopy commutative diagram
   \[\diagram
          S^{8}\vee S^{10}\vee S^{15}\rrto^-{f'_{1}+f'_{2}+f'_{3}}\dto^{a+b+\gamma}
             & & SU(4)\dto^{4} \\
          E\vee S^{12}\vee S^{14}\rrto^-{\xi} & & SU(4)
     \enddiagram\]
   where $\gamma$ may be chosen to be $c$ if the order of $f'_{3}$ is at most $4$
   and $\gamma$ may be chosen to be $c'$ if the order of $f'_{3}$ is $8$. Further,
   in the latter case, the composite
   \(\namedddright{S^{15}}{\eta_{7}\sigma_{8}}{S^{7}}{\iota}{E}{e}{SU(4)}\)
   represents $4[\nu_{5}\oplus\eta_{7}]\circ\sigma_{8}$.
\end{lemma}

\begin{proof}
First, consider the diagram
\begin{equation}
  \label{dgrm1}
  \diagram
       S^{8}\vee S^{10}\rto^-{f'_{1}+f'_{2}}\dto^{\eta_{7}+\nu'_{7}}
            & SU(4)\dto^{4} \\
       S^{7}\rto^-{d}\dto^{\iota} & SU(4)\ddouble \\
       E\rto^-{e} & SU(4).
  \enddiagram
\end{equation}
Since $\pi_{8}(SU(4))\cong\mathbb{Z}/8\mathbb{Z}$ is generated by
$[\nu_{5}\oplus\eta_{7}]$ and $f'_{1}$ has order~$8$, we must have
$f'_{1}=u\cdot[\nu_{5}\oplus\eta_{7}]$ for some unit $u\in\mathbb{Z}/8\mathbb{Z}$.
Thus $4f'_{1}\simeq 4[\nu_{5}\oplus\eta_{7}]$, so the restriction of the upper square
in~(\ref{dgrm1}) to~$S^{8}$ homotopy commutes by Lemma~\ref{S810dgrms}. Similarly,
since $\pi_{10}(SU(4))\cong\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$
with $[\nu_{7}]$ being the generator of order~$8$, and $f'_{2}$ has order $8$, we
must have $4f'_{2}\simeq 4[\nu_{5}]$, so the restriction of the upper square
in~(\ref{dgrm1}) to $S^{10}$ homotopy commutes by Lemma~\ref{S810dgrms}.
The lower square in~(\ref{dgrm1}) homotopy commutes by Lemma~\ref{Eext}.
Now observe that the lower direction around~(\ref{dgrm1}) is the definition of $a+b$.
Thus~(\ref{dgrm1}) implies that the diagram in the statement of the lemma
homotopy commutes when restricted to $S^{8}\vee S^{10}$.

Second, consider the diagram
\begin{equation}
  \label{dgrm2}
  \diagram
      S^{15}\dto^{(t\cdot\sigma'\eta_{14}+\theta)+\nu'_{12}+\eta_{14}}\rrto^-{f'_{3}}
           & & SU(4)\ddto^{4} \\
      S^{7}\vee S^{12}\vee S^{15}\dto^{\iota\vee 1\vee 1} & & \\
      E\vee S^{12}\vee S^{14}\rrto^-{\xi} & & SU(4)
  \enddiagram
\end{equation}
where two possibilities for $\theta$ will be considered. In the lower direction
around the diagram, by definition, $\xi$ is the composite
\(\nameddright{E\vee S^{12}\vee S^{14}}{}{E}{e}{SU(4)}\)
where the left map is the pinch onto the first wedge summand. By
Lemma~\ref{Eext}, $\iota\circ e=d$. Thus the lower direction around the
diagram is homotopic to the composite
\(\lllnameddright{S^{15}}{t\cdot\sigma'\eta_{14}+\theta}{S^{7}}{d}{SU(4)}\).
By Lemma~\ref{SU4relns}~(b), $d\circ t\cdot\sigma'\eta_{14}$ is null homotopic.
Thus the lower direction around the diagram is in fact homotopic to the composite
\(\nameddright{S^{15}}{\theta}{S^{7}}{d}{SU(4)}\).

If $f'_{3}$ has order at most $4$ then $4f'_{3}$ is null homotopic. Taking
$\theta$ to be the constant map shows that~(\ref{dgrm2}) homotopy
commutes. Observe also that with this choice of $\theta$ the left column
in~(\ref{dgrm2}) is the definition of $c$, so we obtain the diagram in the
statement of the lemma when restricted to $S^{15}$. Now combining~(\ref{dgrm1})
and~(\ref{dgrm2}) we obtain the diagram asserted by the lemma.

Suppose that $f'_{3}$ has order~$8$. Since
$\pi_{15}(SU(4))\cong\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$
with the order~$8$ generator being $[\nu_{5}\oplus\eta_{7}]\circ\sigma_{8}$,
we obtain $4f'_{3}\simeq 4[\nu_{5}\oplus\eta_{7}]\circ\sigma_{8}$. Take
$\theta=\eta_{7}\sigma_{8}$. We claim that
$d\circ\theta\simeq 4[\nu_{5}\circ\eta_{7}]\circ\sigma_{8}$. If so then~(\ref{dgrm2})
homotopy commutes with this choice of $\theta$ and, as the left column of~(\ref{dgrm2})
is the definition of $c'$, we obtain the diagram in the statement of the lemma
when restricted to $S^{15}$. Therefore combining~(\ref{dgrm1}) and~(\ref{dgrm2})
we obtain the diagram asserted by the lemma.

It remains to show that
$d\circ\eta_{7}\sigma_{8}\simeq 4[\nu_{5}\oplus\eta_{7}]\circ\sigma_{8}$.
By Lemma~\ref{pi15inj} it suffices to compose with
\(\namedright{SU(4)}{q}{S^{5}\times S^{7}}\)
and check there. On the one hand,
$q\circ d\circ\eta_{7}\sigma_{8}\simeq
    (\eta_{5}^{2}\times\underline{2})\circ\eta_{7}\sigma_{8}\simeq
    \eta_{5}^{3}\sigma_{8}$,
where the left homotopy holds by definition of $d$ and the right homotopy is
due to the fact that $\eta_{7}$ has order~$2$ and, by Lemma~\ref{S72},
$\underline{2}$ induces multiplication by $2$ on homotopy groups. On
the other hand,
$q\circ 4[\nu_{5}\oplus\eta_{7}]\circ\sigma_{8}\simeq
     4(\nu_{5}\times\eta_{7})\circ\sigma_{8}\simeq 4\nu_{5}\sigma_{8}\simeq
     \eta_{5}^{3}\sigma_{8}$.
Here, from left to right, the first homotopy holds by definition of $[\nu_{5}\oplus\eta_{7}]$,
the second holds since $\eta_{7}$ has order~$2$, and the third holds by~\cite{To}.
Thus $d\circ\eta_{7}\simeq_{8}\simeq 4[\nu_{5}\oplus\eta_{7}]\circ\sigma_{8}$,
as claimed.
\end{proof}

Now return to the map
\(\namedright{SU(4)}{\partial_{1}}{\Omega_{0}^{3} SU(4)}\).

\begin{proposition}
   \label{order4options}
   The following hold:
   \begin{letterlist}
      \item if $f'_{3}$ has order at most $4$ then $4\circ\partial_{1}$ is null homotopic;
      \item if $f'_{3}$ has order $8$ then $4\circ\partial_{1}$ is homotopic to the composite
                \(\namedddright{SU(4)}{\pi}{S^{5}\times S^{7}}{}{S^{12}}{4\chi}
                      {\Omega^{3}_{0} SU(4)}\),
                 where the middle map is the pinch map to the top cell and $\chi$
                 is the triple adjoint of the order~$8$ generator
                 $[\nu_{5}\oplus\eta_{7}]\circ\sigma_{8}$ in $\pi_{15}(SU(4))$.
   \end{letterlist}
\end{proposition}

\begin{proof}
If the order of $f'_{3}$ is at most~$4$, then in Lemma~\ref{alternative} we
may take $\gamma=c$. Doing so, observe that by using the inverse
equivalences in Lemma~\ref{jclass} we obtain a homotopy commutative diagram
\begin{equation}
  \label{Cjdgrm}
  \diagram
        \Sigma^{3}(S^{5}\times S^{7})\rto^-{f'}\dto^{\Sigma^{3} j} & SU(4)\dto^{4} \\
        \Sigma^{3} C\rto^-{\xi'} & SU(4)
  \enddiagram
\end{equation}
where $\xi'$ is the composite
\(\nameddright{\Sigma^{3} C}{\simeq}{E\vee S^{12}\vee S^{14}}{\xi}{SU(4)}\).
Now consider the diagram
\[\diagram
      SU(4)\rto^-{\partial_{1}}\dto^{\pi} & \Omega^{3}_{0} SU(4)\ddouble \\
      S^{5}\times S^{7}\rto^-{f}\dto^{j} & \Omega^{3}_{0} SU(4)\dto^{4} \\
      C\rto & \Omega^{3}_{0} SU(4)
  \enddiagram\]
The top square homotopy commutes by~(\ref{su4su2}) while the bottom
square is the triple adjoint of~(\ref{Cjdgrm}). Since the left
column consists of two consecutive maps in a homotopy cofibration sequence
it is null homotopic. The homotopy commutativity of the diagram therefore
implies that~$4\circ\partial_{1}$ is null homotopic.

If the order of $f'_{3}$ is $8$, then in Lemma~\ref{alternative} we may
take $\gamma=c'$. Doing so, since $c'=c+\eta_{7}\sigma_{8}$, instead
of~(\ref{Cjdgrm}) we obtain a homotopy commutative diagram
\begin{equation}
  \label{Cj2dgrm}
  \diagram
        \Sigma^{3}(S^{5}\times S^{7})\rto^-{f'}\dto^{\Sigma^{3} j+\ell} & SU(4)\dto^{4} \\
        \Sigma^{3} C\rto^-{\xi'} & SU(4)
  \enddiagram
\end{equation}
where $\ell$ is the composite
\(\nameddright{\Sigma^{3}(S^{5}\times S^{7})}{}{S^{15}}{\eta_{7}\sigma_{8}}
     {S^{7}}\hookrightarrow\Sigma^{3} C\).
Now consider the diagram
\[\diagram
      \Sigma^{3} SU(4)\rto^-{\partial'_{1}}\dto^{\Sigma^{3}\pi} & SU(4)\ddouble \\
      \Sigma^{3}(S^{5}\times S^{7})\rto^-{f'}\dto^{\Sigma^{3} j+\ell} & SU(4)\dto^{4} \\
      \Sigma^{3} C\rto^-{\xi'} & SU(4)
  \enddiagram\]
where $\partial'_{1}$ is the triple adjoint of $\partial$. The top square homotopy
commutes by~(\ref{su4su2}) while the bottom square homotopy commutes
by~(\ref{Cj2dgrm}). Since $\Sigma^{3} j\circ\Sigma^{3}\pi$ are consecutive
maps in a homotopy cofibration, their composite is null homotopic. Thus
this diagram implies that $4\circ\partial'_{1}$ is homotopic to the composite
\(\namedddright{\Sigma^{3} SU(4)}{\Sigma^{3}\pi}{\Sigma^{3}(S^{5}\times S^{7})}
     {}{S^{15}}{\eta_{7}\sigma_{8}}{S^{7}}\hookrightarrow\namedright{\Sigma^{3} C}
     {\xi'}{SU(4)}\).
Notice that the pinch map to the top cell
\(\namedright{\Sigma^{3}(S^{5}\times S^{7})}{}{S^{15}}\)
is a triple suspension, while by Lemma~\ref{alternative} the composite
\(\namedright{S^{15}}{\eta_{7}\sigma_{8}}{S^{7}}\hookrightarrow\namedright{\Sigma^{3} C}
        {\xi'}{SU(4)}\)
represents $4[\nu_{5}\oplus\eta_{7}]\circ\sigma_{8}$. Thus, taking triple adjoints,
$4\circ\partial_{1}$ is homotopic to the composite
\(\namedddright{SU(4)}{\pi}{S^{5}\times S^{7}}{}{S^{12}}{4\chi}{SU(4)}\),
as asserted.
\end{proof}

\begin{remark}
It can be checked that if $f'_{3}$ has order~$8$ then there is no different
choice of the map~$\xi$ which makes $\xi\circ(a+b+c)\simeq 4f'$ in
Lemma~\ref{alternative}. The argument is to check all possible cases; it is
not included as it is not needed. However, it does imply that
$4\circ\partial_{1}$ is nontrivial; for if it were trivial then
$4\circ\partial_{1}\simeq 4\circ f\circ\pi_{1}$
would have to factor through the cofibre $C$ of $\pi$, implying that there
has to be a choice of $\xi$ such that $\xi\circ(a+b+c)\simeq 4f'$.
\end{remark}

\begin{theorem}
   \label{partialorder4}
   The following hold:
   \begin{letterlist}
      \item if $f'_{3}$ has order $4$ then $\partial_{1}$ has order $4$;
      \item if $f'_{3}$ has order $8$ then $\Omega\partial_{1}$ has order $4$.
   \end{letterlist}
\end{theorem}

\begin{proof}
By Proposition~\ref{order4options}, if $f'_{3}$ has order $4$ then $4\circ\partial_{1}$
is null homotopic, implying that $\partial_{1}$ has order at most $4$. On the
other hand, by Lemma~\ref{HKlemma}, the order of $\partial_{1}$ is divisible by $4$.
Thus $\partial_{1}$ has order $4$.

Next, in general, the quotient map
\(\namedright{X\times Y}{Q}{X\wedge Y}\)
is null homotopic after looping. For if
\(i\colon\namedright{X\vee Y}{}{X\times Y}\)
is the inclusion of the wedge into the product, then $Q\circ i$ is null
homotopic, but by the Hilton-Milnor Theorem $\Omega i$ has a right
homotopy inverse. In our case, if $f'_{3}$ has order $8$ then
Proposition~\ref{order4options} states that $4\circ\partial_{1}$
factors through the quotient map
\(\namedright{S^{5}\times S^{7}}{Q}{S^{5}\wedge S^{7}\simeq S^{12}}\).
Thus $4\Omega\partial_{1}$ is null homotopic. Consequently, $\Omega\partial_{1}$
has order at most $4$. By Lemma~\ref{loopHKlemma}, the order
of~$\Omega\partial_{1}$ is divisible by $4$. Thus $\Omega\partial_{1}$ has order $4$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{looppartialorder2}]
Proposition~\ref{partialorder4} implies that in any case the
$2$-primary component of the order of $\Omega\partial_{1}$ is $4$.
\end{proof}

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