The homotopy types of $U(n)$-gauge groups over lens spaces

TL;DR

This paper investigates the homotopy types of gauge groups associated with principal U(n)-bundles over lens spaces, providing insights into their topological structure.

Contribution

It offers a detailed analysis of the homotopy types of gauge groups over lens spaces, a topic not extensively covered before.

Findings

Classification of homotopy types of gauge groups

Identification of topological invariants influencing gauge group structure

Comparison with gauge groups over other 3-manifolds

Abstract

We analyse the homotopy types of gauge groups for principal $U(n)$-bundles over lens spaces.

Equations

$${\Omega }^2{\mathcal{G}}_k(L(p,q))\simeq {\Omega }^2{\mathcal{G}}_k(P^2(p))\times {\Omega }^{5}U(n);$$

$${\mathcal{G}}_k(P^2(p))\simeq {\mathcal{G}}_{\ell }(P^2(p)).$$

$$S^2\stackrel{f}{⟶}{P}^2(p)\stackrel{i}{⟶}L(p,q)$$

$$S^1\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\overline{\pi }:L(p,q)\stackrel{h}{⟶}C\stackrel{\simeq }{⟶}{S}^2\vee S^3⟶S^2$$

$$[S^2,BU(n)]\stackrel{{\underline{p}}^{\ast }}{⟶}[S^2,BU(n)]\stackrel{{\pi }^{\ast }}{⟶}[P^2(p),BU(n)]⟶[S^1,BU(n)]$$

$$[S^3,BU(n)]\stackrel{g^{\ast }}{⟶}[L(p,q),BU(n)]\stackrel{i^{\ast }}{⟶}[P^2(p),BU(n)]\stackrel{f^{\ast }}{⟶}[S^2,BU(n)].$$

$$0⟶[L(p,q),BU(n)]\stackrel{i^{\ast }}{⟶}\mathbb{Z}/p\mathbb{Z}⟶\mathbb{Z}.$$

$$U(n)\stackrel{{\partial }_{\alpha }}{⟶}\text{mbox}{Map}_{\alpha }^{\ast }(X,BU(n))⟶\text{mbox}{Map}_{\alpha }(X,BU(n))⟶BU(n).$$

$$U(n)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\text{mbox}{Map}_{\bar{k}}^{\ast }(S^2,BU(n))\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\theta :\text{mbox}{Map}_{\bar{k}}^{\ast }(S^2,BU(n))⟶\text{mbox}{Map}_0^{\ast }(S^2,BU(n))$$

$$\theta (f):S^2\stackrel{\sigma }{⟶}{S}^2\vee S^2\stackrel{f\vee ϵ}{⟶}BU(n)\vee BU(n)\stackrel{\nabla }{⟶}BU(n)$$

$$\varphi :\text{mbox}{Map}_0^{\ast }(S^2,BU(n))⟶\text{mbox}{Map}_{\bar{k}}^{\ast }(S^2,BU(n))$$

$${\theta }^{\prime }:\text{mbox}{Map}_k^{\ast }(P^2(p),BU(n))⟶\text{mbox}{Map}_0^{\ast }(P^2(p),BU(n))$$

$${\theta }^{\prime }(f):P^2(p)\stackrel{\psi }{⟶}{P}^2(p)\vee S^2\stackrel{f^{\prime }\vee ϵ}{⟶}BU(n)\vee BU(n)\stackrel{\nabla }{⟶}BU(n)$$

$${\varphi }^{\prime }:\text{mbox}{Map}_0^{\ast }(P^2(p),BU(n))⟶\text{mbox}{Map}_k^{\ast }(P^2(p),BU(n))$$

$$P^2(p)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$U(n)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$S^1\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$${\underline{d}}^{\ast }:\text{mbox}{Map}_0^{\ast }(P^2(p),BU(n))⟶\text{mbox}{Map}_0^{\ast }(P^2(p),BU(n)).$$

$$U(n)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$${\pi }_{2n-1}({\Omega }_{0}U(n))\stackrel{p}{⟶}{\pi }_{2n-1}({\Omega }_{0}U(n))\stackrel{{\pi }^{\ast }}{⟶}{\pi }_{2n-1}(\text{mbox}{Map}_0^{\ast }(P^2(p),BU(n)))$$

$$⟶{\pi }_{2n-2}({\Omega }_{0}U(n))\stackrel{p}{⟶}{\pi }_{2n-2}({\Omega }_{0}U(n)).$$

$${\pi }_{2n-1}(U(n))\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$S^2\stackrel{f}{⟶}{P}^2(p)\stackrel{i}{⟶}L(p,q)\stackrel{g}{⟶}{S}^3.$$

$$S^2\stackrel{f}{⟶}{P}^2(p)⟶L(p,q).$$

Full text

1. Introduction

Let $G$ be a simple, compact Lie group and let $P\stackrel{{\scriptstyle}}{{\longrightarrow}}M$ be a principal $G$-bundle. The gauge group of this bundle is the group of $G$-equivariant automorphisms of $P$ that fix $M$. There has been a considerable amount of work recently in trying to understand the homotopy types of gauge groups that arise in physical or geometric contexts. Most work to date has concentrated on $M$ being a simply-connected four-manifold when $G$ is simply-connected or $M$ being an orientable surface when $G=U(n)$.

In this paper we turn our attention to the case when $M$ is a $3$-manifold. If $G$ is simply-connected then $[M,BG]\cong 0$, implying that the only principal $G$-bundle is the trivial bundle, which has the trivial gauge group. We consider instead the more topologically intricate case when $G=U(n)$ and $M$ is a lens space, for then $[M,BU(n)]\not\cong 0$.

Let $p$ and $q$ be coprime integers. The lens space $L(p,q)$ is the orbit space $S^{3}/(\mathbb{Z}/s\mathbb{Z})$, where the action of $\mathbb{Z}/p\mathbb{Z}$ on $S^{3}$ is given by $(z_{0},z_{1})\longrightarrow(e^{2\pi i/p}z_{0},e^{2\pi iq/p}z_{1})$. For $n\geq 1$ and $p\geq 2$, let $\underline{p}\colon S^{n}\stackrel{{\scriptstyle}}{{\longrightarrow}}S^{n}$ be the map of degree $p$ and let $P^{n+1}(p)$ be its homotopy cofibre. The space $P^{n+1}(p)$ is the $(n+1)$-dimensional mod-$p$ Moore space. As a $CW$-complex, $L(p,q)\simeq P^{2}(p)\cup e^{3}$.

The analysis of gauge groups of principal $U(n)$-bundles over $L(p,q)$ is necessarily delicate for two reasons. First, the isomorphism classes of principal bundles is determined by $[L(p,q),BU(n)]$, and this set is determined by $[P^{2}(p),BU(n)]$ rather than $[S^{3},BU(n)]$. This is in contrast to the case when $G$ is simply-connected and $M$ is a simply-connected four-manifold or when $G=U(n)$ and $M$ is an orientable surface; in both cases $[M,BG]$ is determined by the top cell of $M$ and this leads to certain homotopy fibrations being more easily compared. Second, typically localization techniques are used to work one prime at a time, allowing for easier progress. However, as $L(p,q)$ may not be nilpotent, localization techniques may be problematic, so we approach the problem without localization. The strategy and methods used should be applicable to other cases as well.

Our main result is the following. As will be shown, there are isomorphisms $[L(p,q),BU(n)]\cong[P^{2}(p),BU(n)]\cong\mathbb{Z}/p\mathbb{Z}$. For $k\in\mathbb{Z}/p\mathbb{Z}$, let $\mathcal{G}_{k}(L(p,q))$ and $\mathcal{G}_{k}(P^{2}(p))$ be the gauge groups of the principal $U(n)$-bundles over $L(p,q)$ and $P^{2}(p)$ respectively with first Chern class $k$. For integers $a,b$, let $(a,b)$ be their greatest common divisor.

Theorem 1.1**.**

The following hold:

• (a)

if $2\,|\,p$ then there is a homotopy equivalence

[TABLE]

• (b)

if $k\equiv u\ell\bmod{n}$ where $(u,np)=1$ then there is a homotopy equivalence

[TABLE]

Theorem 1.1 (a) implies that if $2\mid p$ then the higher homotopy groups of $\mathcal{G}_{k}(L(p,q))$ are determined by $\mathcal{G}_{k}(P^{2}(p))$ and $U(n)$, and part (b) implies that if in addition $k\equiv u\ell\bmod{n}$ for $(u,np)=1$ then $\Omega^{2}\mathcal{G}_{k}(L(p,q))\simeq\Omega^{2}\mathcal{G}_{\ell}(L(p,q))$, even though the underlying principal $U(n)$-bundles might be distinct.

2. Isomorphism classes of bundles and components of mapping spaces

As a $CW$-complex $L(p,q)\simeq P^{2}(p)\cup e^{3}$, so there is a homotopy cofibration

[TABLE]

where $f$ attaches the top cell to $L(p,q)$ and $i$ is the inclusion of the $2$-skeleton. Let $\pi\colon P^{2}(p)\stackrel{{\scriptstyle}}{{\longrightarrow}}S^{2}$ be the pinch map to the top cell. Let $j$ be the composite of inclusions $j\colon S^{1}\stackrel{{\scriptstyle}}{{\longrightarrow}}P^{2}(p)\stackrel{{\scriptstyle i}}{{\longrightarrow}}L(p,q)$. Then there is a homotopy cofibration diagram

[TABLE]

that defines the space $C$ and the maps $g$ and $h$.

Lemma 2.1**.**

The map $\pi\circ f$ is null homotopic and there is a homotopy equivalence $C\simeq S^{2}\vee S^{3}$.

Proof.

The degree of $\pi\circ f$ is detected by the Bockstein in the homology of $C$, but this Bockstein is zero since the corresponding Bockstein for $L(p,q)$ is zero. Therefore $\pi\circ f\simeq\ast$, implying that $C\simeq S^{2}\vee S^{3}$. ∎

Let $\overline{\pi}$ be the composite

[TABLE]

where the right map collapses $S^{3}$ to a point. Lemma 2.1 immediately implies the following.

Corollary 2.2**.**

The pinch map $P^{2}(p)\stackrel{{\scriptstyle\pi}}{{\longrightarrow}}S^{2}$ extends across $i$ to a map $\overline{\pi}\colon L(p,q)\stackrel{{\scriptstyle}}{{\longrightarrow}}S^{2}$. $\hfill\Box$

If $n=1$ then $BU(1)$ is the Eilenberg-Mac Lane space $K(\mathbb{Z},2)$ and for any $CW$-complex $X$ the set $[X,BU(1)]$ has a group structure. If $n>1$ then the standard inclusion $BU(n)\stackrel{{\scriptstyle}}{{\longrightarrow}}BU(\infty)$ has homotopy fibre $U(\infty)/U(n)$ which is $2n$-connected. Thus if $X$ is a $CW$-complex of dimension $\leq 2n$ then there is an isomorphism $[X,BU(n)]\cong[X,BU(\infty)]$. In particular, as $BU(\infty)$ is an infinite loop space, $[X,BU(n)]$ has a group structure. In our case, each space in the homotopy cofibration sequences $S^{1}\stackrel{{\scriptstyle}}{{\longrightarrow}}P^{2}(p)\stackrel{{\scriptstyle\pi}}{{\longrightarrow}}S^{2}\stackrel{{\scriptstyle\underline{p}}}{{\longrightarrow}}S^{2}$ and $S^{2}\stackrel{{\scriptstyle f}}{{\longrightarrow}}P^{2}(p)\stackrel{{\scriptstyle i}}{{\longrightarrow}}L(p,q)\stackrel{{\scriptstyle g}}{{\longrightarrow}}S^{3}$ has dimension $\leq 4$ so for any $n\geq 1$ we obtain exact sequence of groups

[TABLE]

and

[TABLE]

Recall that $[S^{2},BU(n)]\cong\pi_{1}(U(n))\cong\mathbb{Z}$.

Lemma 2.3**.**

Let $n\geq 1$. The following hold:

• (a)

there is a group isomorphism $[P^{2}(p),BU(n)]\cong\mathbb{Z}/p\mathbb{Z}$;

• (b)

the map $\pi^{\ast}$ is reduction mod-$p$;

• (c)

there is a group isomorphism $[L(p,q),BU(n)]\stackrel{{\scriptstyle i^{\ast}}}{{\longrightarrow}}[P^{2}(p),BU(n)]$;

• (d)

the map $\overline{\pi}^{\ast}$ is reduction mod-$p$.

Proof.

In (2), since $\underline{p}$ is the map of degree $p$, the induced map $\underline{p}^{\ast}$ is multiplication by $p$. As $\pi_{2}(BU(n))\cong\mathbb{Z}$ and $\pi_{1}(BU(n))\cong 0$, exactness in (2) immediately implies that $[P^{2}(p),BU(n)]\cong\mathbb{Z}/p\mathbb{Z}$ and $\pi^{\ast}$ is reduction mod-$p$, proving parts (a) and (b).

As $\pi_{3}(BU(n))\cong 0$, $[P^{2}(p),BU(n)]\cong\mathbb{Z}/p\mathbb{Z}$ and $[S^{2},BU(n)]\cong\mathbb{Z}$, from (3) we obtain an exact sequence of groups

[TABLE]

Any homomorphism from a finite group to $\mathbb{Z}$ is trivial so, by exactness, $i^{\ast}$ is an isomorphism, proving part (c).

Since $\pi\simeq\overline{\pi}\circ i$ by Corollay 2.2, part (d) follows from parts (b) and (c). ∎

In general, if $X$ is a pointed $CW$-complex then the isomorphism classes of principal $U(n)$-bundles over $X$ are classified by the homotopy classes in $[X,BU(n)]$. If $P$ is such a bundle, classified by a map $\alpha$, let $\mathcal{G}_{\alpha}(X)$ be its gauge group. This group has a classifying space $B\mathcal{G}_{\alpha}(X)$ and by [G, AB] there is a homotopy equivlalence $B\mathcal{G}_{\alpha}(X)\simeq\mbox{Map}_{\alpha}(X,BU(n))$, where $\mbox{Map}_{\alpha}(X,BU(n))$ is the component of the space of continuous maps from $X$ to $BU(n)$ that contains $\alpha$. The subgroup $\mathcal{G}_{\alpha}^{*}(X)$ of $G$-equivariant automorphisms of $P$ that pointwise fix the fibre at the basepoint is the pointed gauge group. There is a corresponding component of the pointed mapping space, $\mbox{Map}^{\ast}_{\alpha}(X,BU(n))$, and a homotopy equivalence $B\mathcal{G}_{\alpha}^{*}(X)\simeq\mbox{Map}^{\ast}_{\alpha}(X,BU(n))$. Evaluation of maps at the basepoint gives a homotopy fibration sequence

[TABLE]

The homotopy fibre of the connecting map $\partial_{\alpha}$ is $\mathcal{G}_{\alpha}(X)$.

In our case, we have $[S^{2},BU(n)]\cong\mathbb{Z}$ and, by Lemma 2.3, $[P^{2}(p),BU(n)]\cong[L(p,q),BU(n)]\cong\mathbb{Z}/p\mathbb{Z}$. Note that, for dimensional reasons, the principal $U(n)$-bundles over $S^{2}$, $P^{2}(p)$ and $L(p,q)$ are classified by the value of the first Chern class. For $\bar{k}\in\mathbb{Z}$, let $\mathcal{G}_{\bar{k}}(S^{2})$ be the gauge group of the isomorphism class of principal $U(n)$-bundles over $S^{2}$ whose first Chern class is $\bar{k}$. For $k\in\mathbb{Z}/p\mathbb{Z}$, let $\mathcal{G}_{k}(P^{2}(p))$ and $\mathcal{G}_{k}(L(p,q))$ be the gauge groups of the isomorphism classes of principal $U(n)$-bundles over $P^{2}(p)$ and $L(p,q)$ respectively whose first Chern class is $k$. Lemma 2.3 implies that if $\bar{k}\equiv k\bmod p$ then there is a commutative diagram of fibration sequences

[TABLE]

The homotopy fibres of $\partial_{\bar{k}}(S)$, $\partial_{k}^{L}$ and $\partial_{k}^{P}$ are $\mathcal{G}_{\bar{k}}(S^{2})$, $\mathcal{G}_{k}(L(p,q))$ and $\mathcal{G}_{k}(P^{2}(p))$ respectively.

The goal is to find information about the gauge groups $\mathcal{G}_{k}(L(p,q))$ via the middle homotopy fibration in (4). However, it is not so easy to study this fibration directly, one issue being that it is unclear whether the components $\mbox{Map}^{\ast}_{k}(L(p,q),BU(n))$ are all homotopy equivalent. A similar issue appeared in work of the first author [MS] in dealing with gauge groups for principal $G$-bundles over $S^{3}$-bundles over $S^{4}$, where $G$ is a simply-connected, simple compact Lie group. The approach in that case involved localization, which needs to be avoided here since $P^{2}(p)$ need not be nilpotent. Instead, we obtain information indirectly: by [S], information about $\mathcal{G}_{\bar{k}}(S^{2})$ via the top fibration in (4) is known and in Section 3 we will determine information about $\mathcal{G}_{k}(P^{2}(p))$ via the bottom fibration in (4). In Section 4 a splitting result is proved that lets us use the information about $\mathcal{G}_{k}(P^{2}(p))$ to deduce information about $\mathcal{G}_{k}(L(p,q))$.

3. The homotopy types of $\mathcal{G}_{k}(P^{2}(p))$

By Corollary 2.2, the pinch map $P^{2}(p)\stackrel{{\scriptstyle\pi}}{{\longrightarrow}}S^{2}$ is homotopic to the composite $P^{2}(p)\stackrel{{\scriptstyle i}}{{\longrightarrow}}L(p,q)\stackrel{{\scriptstyle\overline{\pi}}}{{\longrightarrow}}S^{2}.$ So from (4) we obtain a homotopy commutative diagram of fibration sequences

[TABLE]

First, we show that all the components $\mbox{Map}^{\ast}_{k}(P^{2}(p),BU(n))$ are homotopy equivalent, and in a way that is compatible with a similar result from [S] about the components $\mbox{Map}^{\ast}_{\bar{k}}(S^{2},BU(n))$.

Lemma 3.1**.**

For $\bar{k}\in\mathbb{Z}$ and $k\in\mathbb{Z}/p\mathbb{Z}$ with $\bar{k}\equiv k\bmod{p}$, there is a homotopy commutative diagram

[TABLE]

Proof.

This was essentially proved in [S] but not stated in this form. An argument is given for the sake of completeness. Let $\epsilon\colon S^{2}\stackrel{{\scriptstyle}}{{\longrightarrow}}BU(n)$ be a fixed map with first Chern class $-\bar{k}$. Define

[TABLE]

by sending a map $f\colon S^{2}\stackrel{{\scriptstyle}}{{\longrightarrow}}BU(n)$ with first Chern class $\bar{k}$ to the composite

[TABLE]

where $\sigma$ is the comultiplication on $S^{2}$ and $\nabla$ is the folding map. Similarly, define

[TABLE]

by sending $g$ to $\nabla\circ(g\vee(-\epsilon))\circ\sigma$. Then $\theta$ and $\phi$ are continuous and the homotopy associativity of $\sigma$ implies that $\phi\circ\theta$ and $\theta\circ\phi$ are homotopic to the identity maps.

The space $P^{2}(p)$ is not a co-$H$-space. However, as $\pi$ is a homotopy cofibration connecting map there is a coaction $\psi\colon P^{2}(p)\stackrel{{\scriptstyle}}{{\longrightarrow}}P^{2}(p)\vee S^{2}$ which, when pinched to $P^{2}(p)$ is the identity map, and when pinched to $S^{2}$ is $\pi$. Further, this coaction has a homotopy associativity property: $(\psi\vee 1)\circ\psi\simeq 1\vee\sigma$. Define

[TABLE]

by sending a map $f^{\prime}\colon P^{2}(p)\stackrel{{\scriptstyle}}{{\longrightarrow}}BU(n)$ with first Chern class $k$ to the composite

[TABLE]

and define

[TABLE]

by sending $g$ to $\nabla\circ(g\vee(-\epsilon))\circ\psi$. Then, as before, $\theta^{\prime}$ is a homotopy equivalence.

Finally, the coaction $\psi$ satisfies a homotopy commutative diagram

[TABLE]

This implies that $\theta$ and $\theta^{\prime}$, and $\phi$ and $\phi^{\prime}$, are compatible, implying the homotopy commutative diagram asserted by the lemma. ∎

Using $\partial_{\bar{k}}^{S}$ to also denote the composite $U(n)\stackrel{{\scriptstyle\partial_{\bar{k}}^{S}}}{{\longrightarrow}}\mbox{Map}^{\ast}_{\bar{k}}(S^{2},BU(n)\stackrel{{\scriptstyle\simeq}}{{\longrightarrow}}\mbox{Map}^{\ast}_{0}(S^{2},BU(n)$, and similarly for $\partial_{k}^{P}$, by Lemma 3.1 the left square in (5) may be replaced with a homotopy commutative square

[TABLE]

By (5), the homotopy fibres of $\partial_{\bar{k}}^{S}$ and $\partial_{k}^{P}$ are $\mathcal{G}_{\bar{k}}(S^{2})$ and $\mathcal{G}_{k}(P^{2}(p))$ respectively.

We next identify certain self-homotopy equivalences of $\mbox{Map}^{\ast}_{k}(P^{2}(p),BU(n))$. Since $P^{2}(p)$ is not a co-$H$-space it is not immediately clear that it has a degree $d$ map for any integer $d$. However, we may define an analogue as follows. The degree $p$ map on $S^{1}$ commutes with the degree $d$ map for any $d$, so we obtain a cofibration diagram

[TABLE]

for some map $\underline{d}$. The cofibration diagram implies that, upon taking integral homology, $(\underline{d})_{\ast}$ is multiplication by $d$ on $H_{1}(P^{2}(p);\mathbb{Z})\cong\mathbb{Z}\mathit{/p}\mathbb{Z}$. Suspending, since $\Sigma P^{n}(p)\simeq P^{n+1}(p)$, we see that for any $m\geq 1$ $P^{m+2}(p)\stackrel{{\scriptstyle\Sigma^{m}\underline{d}}}{{\longrightarrow}}P^{m+2}(p)$ induces multiplication by $d$ in $H_{m+1}(P^{t+2}(p);\mathbb{Z})\cong\mathbb{Z}\mathit{/p}\mathbb{Z}$. Consequently, we obtain the following.

Lemma 3.2**.**

If $d$ is a unit in $\mathbb{Z}\mathit{/p}\mathbb{Z}$ then for any $m\geq 1$ the map $P^{m+2}(p)\stackrel{{\scriptstyle\Sigma^{m}\underline{d}}}{{\longrightarrow}}P^{m+2}(p)$ is a homotopy equivalence.

Proof.

Since $d$ is a unit in $\mathbb{Z}\mathit{/p}\mathbb{Z}$, the map $\underline{d}$ induces an isomorphism in $H_{m+1}(P^{m+2}(p);\mathbb{Z})$ and hence induces an isomorphism on $H_{\ast}(P^{m+2}(p);\mathbb{Z})$. Since $m\geq 1$, $P^{m+2}(p)$ is simply-connected, so Whitehead’s Theorem implies that $\underline{d}$ is a homotopy equivalence. ∎

For any $d\in\mathbb{Z}$, the map $\underline{d}$ induces a map

[TABLE]

Lemma 3.3**.**

If $d$ is a unit in $\mathbb{Z}\mathit{/p}\mathbb{Z}$ then $\underline{d}^{\ast}$ is a homotopy equivalence.

Proof.

The first step is to show that $(\underline{d})^{\ast}$ induces an isomorphism on homotopy groups. Note that $\mbox{Map}^{\ast}_{0}(P^{2}(p),BU(n))$ is path-connected since it is one component of $\mbox{Map}^{\ast}(P^{2}(p),BU(n))$ and for all $m\geq 1$ we have $\pi_{m}(\mbox{Map}^{\ast}_{0}(P^{2}(p),BU(n))\cong\pi_{m}(\mbox{Map}^{\ast}(P^{2}(p),BU(n))$. By the pointed Exponential Law, $\pi_{m}(\mbox{Map}^{\ast}(P^{2}(p),BU(n))\cong\mbox{Map}^{\ast}(P^{m+2}(p),BU(n))$. The effect of $\underline{d}^{\ast}$ on $\pi_{m}$ is therefore determined by applying $\mbox{Map}^{\ast}(\ \ ,BU(n))$ to the map $P^{m+2}(p)\stackrel{{\scriptstyle\Sigma^{m}\underline{d}}}{{\longrightarrow}}P^{m+2}(p)$. Since $d$ is a unit in $\mathbb{Z}\mathit{/p}\mathbb{Z}$, by Lemma 3.2, $\Sigma^{m}\underline{d}$ is a homotopy equivalence. Thus $\underline{d}^{\ast}$ induces an isomorphism on $\pi_{m}$. As this is true for all $m\geq 1$, $\underline{d}^{\ast}$ is a weak homotopy equivalence.

Observe that $P^{2}(p)$ is a $CW$-complex and $BU(n)$ may be given the structure of a $CW$-complex. Doing so, by [Mi, Theorem 3], $\mbox{Map}^{\ast}(P^{2}(p),BU(n)$ is also a $CW$-complex, and hence so is its component $\mbox{Map}^{\ast}_{0}(P^{2}(p),BU(n))$. By Whitehead’s Theorem, a weak homotopy equivalence between $CW$-complexes is a homotopy equivalence. ∎

Recall that, by elementary number theory, if $(u,n)=1$ then $u$ is a unit mod $n$, and if $(k,n)=(\ell,n)$ then $k\equiv u\ell\bmod{n}$ for some integer $u$ satisfying $(u,n)=1$.

Lemma 3.4**.**

Suppose that $(k,n)=(\ell,n)$, implying that $k\equiv u\ell\bmod{n}$ for some integer satisfying $(u,n)=1$. Then there is a homotopy commutative diagram

[TABLE]

Proof.

In [Th], refining work in [S], it was shown that $U(n)\stackrel{{\scriptstyle\partial^{S}_{1}}}{{\longrightarrow}}\Omega_{0}U(n)$ has order $n$. Therefore, $\partial^{S}_{1}$ generates a cyclic subgroup of order $n$ in the group $[U(n),\Omega_{0}U(n)]$. By [L], $\partial^{S}_{\ell}\simeq\ell\circ\partial^{S}_{1}$ and $\partial^{S}_{k}\simeq k\circ\partial^{S}_{1}$. Therefore the hypothesis that $k\equiv u\ell\bmod{n}$ implies that $u\ell\circ\partial^{S}_{1}\simeq k\circ\partial^{S}_{1}$, that is, $u\circ\partial^{S}_{\ell}\simeq\partial^{S}_{k}$. ∎

Lemma 3.5**.**

Suppose that $(k,n)=(\ell,n)$, implying that $k\equiv u\ell\bmod{n}$ for some integer satisfying $(u,n)=1$. Suppose that $(u,p)=1$ as well. Then there is a homotopy commutative diagram

[TABLE]

where $\underline{u}^{\ast}$ is a homotopy equivalence.

Proof.

Consider the diagram

[TABLE]

The left square homotopy commutes by Lemma 3.4 and the right square homotopy commutes by applying $\mbox{Map}^{\ast}(\ \ ,BU(n))$ to the right square in (7). Observe that the composite along the top row is $\partial^{P}_{\ell}$ and the composite along the bottom row is $\partial^{P}_{k}$. Thus $\underline{u}^{\ast}\circ\partial^{P}_{\ell}\simeq\partial^{P}_{k}$. Finally, the hypothesis $(u,p)=1$ implies that $u$ is a unit mod-$p$, so by Lemma 3.3, $\underline{u}^{\ast}$ is a homotopy equivalence. ∎

Note that the condition $(u,n)=1$ and $(u,p)=1$ is equivalent to the condition $(u,np)=1$.

Proposition 3.6**.**

Suppose that $k\equiv u\ell\bmod{n}$ where $(u,np)=1$. Then $\mathcal{G}_{k}(P^{2}(p))\simeq\mathcal{G}_{\ell}(P^{2}(p))$.

Proof.

The homotopy fibres of $\partial^{P}_{k}$ and $\partial^{P}_{\ell}$ are $\mathcal{G}_{k}(P^{2}(p))$ and $\mathcal{G}_{\ell}(P^{2}(p))$ respectively. Taking homotopy fibres in the homotopy commutative diagram in the statement of Lemma 3.5 gives an induced map of fibres $\varphi\colon\mathcal{G}_{\ell}(P^{2}(p))\stackrel{{\scriptstyle}}{{\longrightarrow}}\mathcal{G}_{k}(P^{2}(p))$. The fact that $\underline{u}_{\ast}$ is a homotopy equivalence implies that $\varphi$ is a homotopy equivalence. ∎

There is a partial converse to Proposition 3.6 in a limited number of cases.

Lemma 3.7**.**

Suppose that $n=\mathfrak{p}$ where $\mathfrak{p}$ is a prime, that $\mathfrak{p}$ divides $p$ and $(p!,\mathfrak{p})\neq p/\mathfrak{p}$. If $\mathcal{G}_{k}(P^{2}(p))\simeq\mathcal{G}_{\ell}(P^{2}(p))$ then $(k,\mathfrak{p})=(\ell,\mathfrak{p})$.

Proof.

To illustrate why the restrictions on $n$ and $p$ appear, they are introduced only when appropriate. For any $n$ and $p$, the homotopy cofibration $P^{2}(p)\stackrel{{\scriptstyle\pi}}{{\longrightarrow}}S^{2}\stackrel{{\scriptstyle\underline{p}}}{{\longrightarrow}}S^{2}$ induces an exact sequence

[TABLE]

[TABLE]

By [BH] or [To], $\pi_{2n}(U(n))\cong\mathbb{Z}/n!\mathbb{Z}$, and it is well known that $\pi_{2n-1}(U(n))\cong\mathbb{Z}$. As multiplication by $p$ on $\mathbb{Z}/n!\mathbb{Z}$ sends a generator $\gamma$ to $(n!,p)\gamma$ and mutliplication by $p$ on $\mathbb{Z}$ is an injection, we obtain $\pi_{2n-1}(\mbox{Map}^{\ast}_{0}(P^{2}(p),BU(n))\cong\mathbb{Z}/(n!,p)\mathbb{Z}$ and $\pi^{\ast}$ is reduction mod $(n!,p)$.

Next, consider the commutative diagram

[TABLE]

induced by (5), and note that the bottom row is exact. The fact that $\pi_{2n-1}(BU(n))\cong 0$ implies that $\pi_{2n-1}(B\mathcal{G}_{k}(P^{2}(p)))$ is isomorphic to the cokernel of $(\partial_{k}^{P})_{\ast}$. We wish to identify this cokernel in a manner related to $(n,k)$ and then compare to the $(n,\ell)$ case.

Sutherland [S] showed that the image of $(\partial_{\bar{k}}^{S})_{\ast}$ is generated by $(n-1)!(n,\bar{k})\,\gamma$. Let $\delta=\pi^{\ast}(\gamma)$. Then the image of $(\partial_{k}^{P})_{\ast}$ is generated by $(n-1)!(n,k)\,\delta$. Since $\delta$ generates a cylic group of order $(n!,p)$, the only way that $(n-1)!(n,k)\delta$ is not going to be trivial is if $(n!,p)$ has factors of $n$ which are not factors of $(n-1)!$ or $(n,k)$. The only way $n$ has factors that are not factors of $(n-1)!$ is if $n$ is a prime $\mathfrak{p}$. So from now on assume that $n=\mathfrak{p}$. This leaves two cases: $(n,k)=(\mathfrak{p},k)$ is $1$ or $\mathfrak{p}$.

Notice that now $(n!,p)=(\mathfrak{p}!,p)$. If $(\mathfrak{p},k)=1$ then $(n-1)!(n,k)\delta=(\mathfrak{p}-1)!\delta$ and $\delta$ generates $\mathbb{Z}/(n!,s)\mathbb{Z}=\mathbb{Z}/(p!,s)\mathbb{Z}$. The only way $(\mathfrak{p}-1)!\delta$ is nontrivial is if $\mathfrak{p}$ divides $(\mathfrak{p}!,p)$, implying that $\mathfrak{p}$ divides $p$. So from now on assume that $\mathfrak{p}$ divides $p$. Now the image of $(\partial^{P}_{k})_{\ast}$ is $(\mathfrak{p}-1)!\delta$ in the group $\mathbb{Z}/(\mathfrak{p}!,p)\mathbb{Z}$, and this has cokernel $\mathbb{Z}/(p/\mathfrak{p})\mathbb{Z}$. That is, $\pi_{2n-1}(B\mathcal{G}_{k}(P^{2}(p)))\cong\mathbb{Z}/(p/\mathfrak{p})\mathbb{Z}$.

If $(\mathfrak{p},k)=\mathfrak{p}$ then $(n-1)!(n,k)\delta=(\mathfrak{p}-1)!\mathfrak{p}\delta=\mathfrak{p}!\delta$ is trivial in $\mathbb{Z}/(\mathfrak{p}!,p)\mathbb{Z}$, implying that $(\partial^{P}_{k})_{\ast}$ is trivial, implying in turn that $\pi_{2n-1}(B\mathcal{G}_{k}(P^{2}(p)))\cong\mathbb{Z}/(\mathfrak{p}!,p)\mathbb{Z}$.

Assume that $(\mathfrak{p}!,p)\neq p/\mathfrak{p}$ so the cases $(\mathfrak{p},k)=1$ and $(\mathfrak{p},k)=\mathfrak{p}$ result in different groups for $\pi_{2n-1}(B\mathcal{G}_{k}(P^{2}(p)))$. The same two options occur for $\pi_{2n-1}(B\mathcal{G}_{\ell}(P^{2}(p)))$. Therefore, if $\mathcal{G}_{k}(P^{2}(p))\simeq\mathcal{G}_{\ell}(P^{2}(p))$ then $\pi_{2n-1}(B\mathcal{G}_{k}(P^{2}(p)))\cong\pi_{2n-1}(B\mathcal{G}_{\ell}(P^{2}(p)))$, implying that $(\mathfrak{p},k)=(\mathfrak{p},\ell)$. ∎

Observe that Proposition 3.6 and Lemma 3.7 both hold when $n=p=\mathfrak{p}$, noting in Proposition 3.6 that $(u,np)=(u,\mathfrak{p}^{2})=1$ implies that $(u,\mathfrak{p})=1$. Therefore there is a complete classification of gauge groups in this case.

Proposition 3.8**.**

For a prime $\mathfrak{p}$, consider the gauge groups of principal $U(\mathfrak{p})$-bundles over $P^{2}(\mathfrak{p})$. Then $\mathcal{G}_{k}(P^{2}(\mathfrak{p}))\simeq\mathcal{G}_{\ell}(P^{2}(\mathfrak{p}))$ if and only if $(\mathfrak{p},k)=(\mathfrak{p},\ell)$. $\hfill\Box$

4. A homotopy decomposition for $\Omega^{2}\mathcal{G}_{k}(L(p,q))$

In this section we prove Theorem 1.1 (a). Consider the homotopy cofibration sequence

[TABLE]

Lemma 4.1**.**

There is a homotopy equivalence $\Sigma^{2}L(p,q)\simeq P^{4}(p)\vee S^{5}$.

Proof.

Consider the homotopy cofibration

[TABLE]

In general, any closed, orientable $3$-manifold $M$ is parallelizable so by [A] it has the property that its top cell splits off stably. In our case, as $L(p,q)$ is such a manifold, the attaching map $f$ is stably trivial. As $f$ is in the stable range after two suspensions, we have $\Sigma^{2}f$ null homotopic. This implies that there is a homotopy equivalence $\Sigma^{2}L(p,q)\simeq P^{4}(p)\vee S^{5}$. ∎

Lemma 4.1 implies that the map $\Sigma^{2}L(p,q)\stackrel{{\scriptstyle\Sigma^{2}g}}{{\longrightarrow}}S^{5}$ has a right homotopy inverse. We wish to choose this right homotopy inverse carefully; the next lemma does this given a condition on $p$. Recall from Lemma 2.1 that there is a homotopy cofibration $S^{1}\stackrel{{\scriptstyle j}}{{\longrightarrow}}L(p,q)\stackrel{{\scriptstyle h}}{{\longrightarrow}}S^{2}\vee S^{3}$ and $g$ is homotopic to $h$ composed with the map collapsing $S^{2}$ to a point.

Lemma 4.2**.**

Suppose that $2\,|\,p$. Then the map $\Sigma^{2}L(p,q)\stackrel{{\scriptstyle\Sigma^{2}g}}{{\longrightarrow}}S^{5}$ has a right homotopy inverse $r\colon S^{5}\stackrel{{\scriptstyle}}{{\longrightarrow}}{\Sigma^{2}L(p,q)}$ with the property that the composite $S^{5}\stackrel{{\scriptstyle r}}{{\longrightarrow}}\Sigma^{2}L(p,q)\stackrel{{\scriptstyle\Sigma^{2}h}}{{\longrightarrow}}S^{4}\vee S^{5}$ is homotopic to the inclusion of the right wedge summand.

Proof.

Let $F$ be the homotopy fibre of $P^{3}(p)\stackrel{{\scriptstyle\Sigma\pi}}{{\longrightarrow}}S^{3}$ and consider the homology Serre exact sequence for the homotopy fibration $\Omega S^{3}\stackrel{{\scriptstyle}}{{\longrightarrow}}F\stackrel{{\scriptstyle}}{{\longrightarrow}}P^{3}(p)$. This is an exact sequence

[TABLE]

Since $H_{i}(\Omega S^{3})\cong 0\cong H_{i}(P^{3}(p))$ for $i\in\{1,3\}$ we obtain $H_{1}(F)\cong H_{3}(F)\cong 0$. Otherwise there is an exact sequence $0\stackrel{{\scriptstyle}}{{\longrightarrow}}H_{2}(\Omega S^{3})\stackrel{{\scriptstyle}}{{\longrightarrow}}H_{2}(F)\stackrel{{\scriptstyle}}{{\longrightarrow}}H_{2}(P^{3}(p))\longrightarrow 0$. As the inclusion of the bottom cell $S^{2}\stackrel{{\scriptstyle}}{{\longrightarrow}}P^{3}(p)$ lifts to $F$ it must be the case that the sequence in $H_{2}$ is $0\stackrel{{\scriptstyle}}{{\longrightarrow}}\mathbb{Z}\stackrel{{\scriptstyle p}}{{\longrightarrow}}\mathbb{Z}\stackrel{{\scriptstyle}}{{\longrightarrow}}\mathbb{Z}/p\mathbb{Z}\longrightarrow 0$, and so $H_{2}(F)\cong\mathbb{Z}$. Thus the $3$-skeleton of $F$ is homotopy equivalent to $S^{2}$.

Now consider the homotopy pullback diagram

[TABLE]

that defines the spaces $G$ and $N$. Since $S^{3}\stackrel{{\scriptstyle\Sigma f}}{{\longrightarrow}}P^{3}(p)$ composes trivially to $\Sigma L(p,q)$, it lifts to a map $\lambda\colon S^{3}\stackrel{{\scriptstyle}}{{\longrightarrow}}N$. Let $\lambda^{\prime}$ be the composite $S^{3}\stackrel{{\scriptstyle\lambda}}{{\longrightarrow}}N\stackrel{{\scriptstyle}}{{\longrightarrow}}F$. Since the $3$-skeleton of $F$ is $S^{2}$, $\lambda^{\prime}$ factors as $S^{3}\stackrel{{\scriptstyle\lambda^{\prime\prime}}}{{\longrightarrow}}S^{2}\hookrightarrow F$ for some map $\lambda^{\prime\prime}$. Thus there is a homotopy commutative diagram

[TABLE]

Let $D$ be the homotopy cofibre of $\lambda^{\prime\prime}$. Then as the composite $N\stackrel{{\scriptstyle}}{{\longrightarrow}}F\stackrel{{\scriptstyle}}{{\longrightarrow}}G$ is null homotopic (it is two consecutive maps in a homotopy fibration), there is an extension $D\stackrel{{\scriptstyle}}{{\longrightarrow}}G$.

We claim that $\Sigma\lambda^{\prime\prime}$ is null homotopic, implying that $\Sigma D\simeq S^{3}\vee S^{5}$. If so, let $r$ be the composite $r\colon S^{5}\hookrightarrow\Sigma D\stackrel{{\scriptstyle}}{{\longrightarrow}}\Sigma G\stackrel{{\scriptstyle}}{{\longrightarrow}}\Sigma L(p,q)$. Observe that as $G\stackrel{{\scriptstyle}}{{\longrightarrow}}L(p,q)\stackrel{{\scriptstyle\Sigma\overline{\pi}}}{{\longrightarrow}}S^{3}$ is a fibration, the composite $\Sigma^{2}\overline{\pi}\circ r$ is null homotopic. Further, the Blakers-Massey Theorem implies that the homotopy fibration $N\stackrel{{\scriptstyle}}{{\longrightarrow}}P^{3}(p)\stackrel{{\scriptstyle\Sigma i}}{{\longrightarrow}}\Sigma L(p,q)$ is the same as the homotopy cofibration $S^{3}\stackrel{{\scriptstyle\Sigma f}}{{\longrightarrow}}P^{3}(p)\stackrel{{\scriptstyle\Sigma i}}{{\longrightarrow}}\Sigma L(p,q)$ in dimensions $\leq 3$, and $\lambda$ is the inclusion of the bottom cell. A homology argument then shows that the composite $S^{5}\stackrel{{\scriptstyle r}}{{\longrightarrow}}\Sigma^{2}L(p,q)\stackrel{{\scriptstyle\Sigma^{2}g}}{{\longrightarrow}}S^{5}$ is homotopic to the identity map. This proves the second assertion of the lemma.

It remains to show that $\Sigma\lambda^{\prime\prime}$ is null homotopic. Suppose instead that $\Sigma\lambda^{\prime\prime}$ is essential. Then as $\pi_{4}(S^{3})\cong\mathbb{Z}/2\mathbb{Z}$, with generator $\eta$, we have $\Sigma\lambda^{\prime\prime}\simeq\eta$. By definition of $\lambda^{\prime\prime}$, the composite $S^{3}\stackrel{{\scriptstyle\lambda^{\prime\prime}}}{{\longrightarrow}}S^{2}\stackrel{{\scriptstyle}}{{\longrightarrow}}F\stackrel{{\scriptstyle}}{{\longrightarrow}}P^{3}(p)$ is homotopic to $\Sigma f$. By Lemma 4.1, $\Sigma^{2}f$ is null homotopic, implying that there is a lift

[TABLE]

for some map $\gamma$, where $M$ is the homotopy fibre of $S^{3}\stackrel{{\scriptstyle}}{{\longrightarrow}}\Sigma P^{3}(p)$. The Serre exact sequence shows that the $4$-skeleton of $M$ is $S^{3}$, so $\gamma$ factors as $S^{4}\stackrel{{\scriptstyle t\cdot\eta}}{{\longrightarrow}}S^{3}\hookrightarrow M$ for some $t\in\mathbb{Z}/2\mathbb{Z}$. As $\Sigma\lambda^{\prime\prime}$ is assumed to be essential, it must be that $t=1$. Further, the composite $S^{3}\hookrightarrow M\stackrel{{\scriptstyle}}{{\longrightarrow}}S^{3}$ is the degree $p$ map. Therefore $\Sigma\lambda^{\prime\prime}\simeq p\cdot\eta$. By hypothesis, $2\,|\,p$. Thus $p\cdot\eta\simeq\ast$, implying that $\Sigma\lambda^{\prime\prime}\simeq\ast$, a contradiction. Hence it must be the case that $\Sigma\lambda^{\prime\prime}$ is null homotopic. ∎

The aim is to use Lemma 4.2 to prove a splitting result; some care needs to be taken with components. The homotopy cofibration sequence $S^{2}\stackrel{{\scriptstyle f}}{{\longrightarrow}}P^{2}(p)\stackrel{{\scriptstyle}}{{\longrightarrow}}L(p,q)\stackrel{{\scriptstyle g}}{{\longrightarrow}}S^{3}$ induces a homotopy fibration sequence

[TABLE]

Let $\bar{f}^{\ast}$ be the restriction of $f^{\ast}$ to the $k^{th}$ component. Then we obtain a homotopy fibration diagram

[TABLE]

that defines the map $\bar{g}^{\ast}$. The map $r$ in Lemma 4.2 induces a map

[TABLE]

with the property that $r^{\ast}\circ\Omega^{2}g$ is homotopic to the identity map. Let $\overline{r}$ be the composite

[TABLE]

Then from the left square in the diagram above we immediately obtain the following.

Lemma 4.3**.**

Suppose that $2\,|\,p$. Then the composite $\overline{r}\circ\Omega^{2}\bar{g}^{\ast}$ is homotopic to the identity map. $\hfill\Box$

By Lemma 4.3, the homotopy fibration

[TABLE]

splits to give a homotopy equivalence

[TABLE]

Recall that there are homotopy equivalences $B\mathcal{G}_{k}^{*}(L(p,q))\simeq\mbox{Map}^{\ast}_{k}(L(p,q),BU(n))$ and $B\mathcal{G}_{k}^{*}(P^{2}(p))\simeq\mbox{Map}^{\ast}_{k}(P^{2}(p),BU(n))$. Also, by Lemma 3.1 there are homotopy equivalences $\Omega\mathcal{G}_{k}^{*}(P^{2}(p))\simeq\Omega\mathcal{G}_{0}^{*}(P^{2}(p))$, for any $k\in\mathbb{Z}/p\mathbb{Z}$. Using these equivalences along with $\Theta$ we obtain homotopy decompositions of the loops on the pointed gauge group $\mathcal{G}_{k}^{*}(L(p,q))$.

Lemma 4.4**.**

Suppose that $2\,|\,p$. Then there is a homotopy equivalence

[TABLE]

Remark 4.5**.**

The condition that $2\,|\,p$ is included since the map $r$ from Lemma 4.2 is used. However, if all we cared about was finding a left homotopy inverse to $\Omega^{2}\bar{g}^{\ast}$ then the weaker splitting in Lemma 4.1 could be used and the $2\,|\,p$ condition dropped from Lemmas 4.3 and 4.4. The point of using $r$ is to obtain more control over the fibration connecting map, as seen in Lemma 4.6.

Next, we relate $\Theta$ to the twice looped boundary map $\Omega^{2}\partial^{L}_{k}$ to get information about $\Omega^{2}\mathcal{G}_{k}(L(p,q))$.

Lemma 4.6**.**

Suppose that $2\,|\,p$. Then the composite $\overline{r}\circ\Omega^{2}\partial^{L}_{k}$ is null homotopic.

Proof.

By definition, $\overline{\pi}$ is the composite $L(p,q)\stackrel{{\scriptstyle h}}{{\longrightarrow}}S^{2}\vee S^{3}\stackrel{{\scriptstyle\pi_{1}}}{{\longrightarrow}}S^{2}$ where $\pi_{1}$ collapses $S^{3}$ to a point. Since $[S^{3},BU(n)]\cong 0$ the map $\pi_{1}$ induces an isomorphism $[S^{2},BU(n)]\stackrel{{\scriptstyle\pi_{1}^{\ast}}}{{\longrightarrow}}[S^{2}\vee S^{3},BU(n)]$, implying that there is a commutative diagram of evaluation fibration sequences

[TABLE]

where $\bar{k}\in\mathbb{Z}$ reduces to $k$ mod $p$. Writing

[TABLE]

the map $\pi_{1}^{\ast}$ in (9) is the inclusion into the left factor. Thus $\partial^{\vee}_{k}\simeq\partial^{S}_{k}\times\ast$.

Now consider the diagram

[TABLE]

The left square is obtained by restricting to $k^{th}$-components. Notice that the composite along the upper direction around the diagram is $\overline{r}\circ\Omega^{2}h^{\ast}$. By Lemma 4.2, the composite along the bottom row is homotopic to the projection. Thus $\overline{r}\circ\Omega^{2}h^{\ast}$ is homotopic to the projection.

Using (9) we therefore obtain a homotopy commutative diagram

[TABLE]

As $\partial^{\vee}_{k}\simeq\partial^{S}_{k}\times\ast$, the homotopy commutativity of (10) implies that $\overline{r}\circ\Omega^{2}\partial^{L}_{k}\simeq\ast$. ∎

Theorem 4.7**.**

Suppose that $2\,|\,p$. Then there is a homotopy equivalence

[TABLE]

Proof.

Combining the homotopy equivalence

[TABLE]

in (8) with the null homotopy in Lemma 4.6 gives a homotopy commutative diagram

[TABLE]

The asserted homotopy equivalence now follows, noting that $\Omega^{2}\mbox{Map}^{\ast}(S^{3},BU(n))=\Omega^{4}U(n)$. ∎

Finally, we assemble our results to prove Theorem 1.1.

Proof of Theorem 1.1.

Part (a) is Theorem 4.7 and part (b) is Proposition 3.6. ∎

Corollary 4.8**.**

Suppose that $k\equiv u\ell\bmod{n}$ where $(u,np)=1$ and that $2\,|\,p$. Then there is a homotopy equivalence $\Omega^{2}\mathcal{G}_{k}(L(p,q))\simeq\Omega^{2}\mathcal{G}_{\ell}(L(p,q))$. $\hfill\Box$

For example, take $p=2$. Then if $(k,2n)=(\ell,2n)$ there is a homotopy equivalence $\Omega^{2}\mathcal{G}_{k}(L(2,t))\simeq\Omega^{2}\mathcal{G}_{\ell}(L(2,t))$.