%style set-up for construction paper %September 1994 \input amstex \documentstyle{amsppt} \magnification=1200 \redefine\R{{{\bold R}}} \redefine\Z{{{\bold Z}}} \redefine\Q{{{\bold Q}}} \redefine \Cplx{{{\bold C}}} \define\ab{\allowbreak} \define\hy{{\hbox{\it -}}} \define\one{{1\hy}} \define\hyspec{{{\hbox{\it -spec}}}} \define\spec{\text{\it spec}} \define\Vol{\text{\it Vol }} \define\mm{{{\frak g}}} \define\hh{{{\frak h}}} \define\nn{{{\frak N}}} \define\cc{{{\frak C}}} \redefine\aa{{{\frak A}}} \redefine\gg{{{\frak g}}} \define\zz{{{\frak z}}} \define\Remark{{{\flushpar {\bf Remark.}\quad}}} \define\F{{F_\tau}} \define\p#1#2{{\frac{\partial}{\partial {#1}_{#2}}}} \define\ps#1#2{{\frac{\partial^2}{\partial {#1}^2_{#2}}}} \define\m#1{{{\gg}^{(#1)}}} \define\h#1{{{\hh}^{(#1)}}} \define\g#1{{{\gg}^{(#1)}}} \define\X{{\bar{X}}} \define\Y{{\bar{Y}}} \define\ZZ{{\bar{Z}}} \define\V{{\bar{V}}} \define\ggamma{{\bar{\gamma}}} \define\BX{{\bar{X}}} \define\BY{{\bar{Y}}} \define\BZ{{\bar{Z}}} \define\BE{{\bar{E}}} \define\BF{{\bar{F}}} %\define\num{{{\bf \# \ }}} \define\Tau{{{\Cal T}}} \redefine\L{{{\lambda}}} \redefine\H{{{\Cal H}}} \redefine\xi{{{\zz}}} \define\Gup#1{{G^{(#1)}}} \define\GGup#1{{\GG^{(#1)}}} \define\GGperpup#1{{\GG_\perp^{(#1)}}} \define\Gbar{{\bar{G}}} \define\M#1{{G^{(#1)}}} \define\gggbar{{\bar{\gg}}} \define\GG{{\Gamma}} \define\G#1{{{{\Gamma}_{#1}}}} \define\GGbar{{\bar{\Gamma}}} \define\GGGbar#1{{{{\bar{\Gamma}}_{#1}}}} \define\mbar{\gggbar} \define\Mbar{\Gbar} \define\Rep{{\raise 1.5pt\hbox{$\rho$}_\GG}} \define\rep#1{{\raise 1.5pt\hbox{$\rho$}_\G{#1}}} \define\Nilmfld{{\GG \backslash G}} \define\nilmfld#1{{\G{#1} \backslash G}} \define\Nilmfldbar{{\GGbar \backslash \bar{G}}} \define\nilmfldbar#1{{\GGGbar{#1} \backslash \bar{G}}} \define\gbar{{\bar{g}}} \define\Fnsp{{L^2(\Nilmfld)}} \define\fnsp#1{{L^2(\nilmfld{#1})}} \define\fnspbar#1{{L^2(\nilmfldbar{#1})}} \define\Fnspg{{L^2(\GG \backslash G)}} \define\fnspg#1{{L^2(\G{#1} \backslash G)}} \redefine\d#1#2{{\dot{#1}_{#2}(s)}}\redefine\dL#1#2{{\dot{#1}_{#2}(s+\L)}} \define\dds{{\frac{d}{ds}}} \define\ddsz{{\frac{d}{ds}}_{|_0}} \define\dsz#1#2{{{\frac{d{#1}_{#2}}{ds}}_{|_0}}} \redefine\ss#1#2{{<\sss,{#1}_{#2}>}} \redefine\b#1#2{{\overline{{#1}}_{#2}}} \define\Pfaffian{{\overline{B}_{\lambda}}} \define\endpf{\hbox{\vrule height1.5ex width.5em}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \topmatter \pagewidth{5.5 in} \pageheight{7.5 in} \hoffset=-.02 in \title %\vskip 1.25 in A New Construction of Isospectral Riemannian Nilmanifolds with Examples \endtitle \rightheadtext{A New Construction of Isospectral Nilmanifolds} \author Ruth Gornet \endauthor \address Ruth Gornet; Texas Tech University; Department of Mathematics; Lubbock, Texas \ 79409-1042; email: gornet\@math.ttu.edu \indent September 1994 \endaddress \keywords Isospectral manifolds; Spectrum of the Laplacian on forms; Nilpotent Lie groups; Cocompact, discrete subgroups; Quasi-regular representations; Square-in\-te\-gra\-ble representations. \flushpar\indent Research at MSRI supported in part by NSF grant DMS-9022140. Research at MSRI and Texas Tech supported in part by NSF grant DMS-9409209 \endkeywords \subjclass Primary 58G25, 22E27; Secondary 53C30 \endsubjclass \abstract We present a new construction for obtaining pairs of higher-step isospectral Riemannian nilmanifolds and compare several resulting new examples. In particular, we present new examples of manifolds that are isospectral on functions, but not isospectral on one-forms. \endabstract \endtopmatter \document \subheading{\S1 Introduction} \bigskip The {\it spectrum} of a closed Riemannian manifold $(M,g)$, denoted $\spec(M,g)$, is the collection of eigenvalues with multiplicities of the associated La\-place--Bel\-tra\-mi operator acting on smooth functions. Two Riemannian manifolds $(M,g)$ and $(M',g')$ are said to be {\it isospectral} if $\spec(M,g)=\spec(M',g').$ A basic question in spectral geometry is determining what geometric information is contained in the spectrum of a Riemannian manifold. \medskip Despite considerable research in the area, only a few geometric properties are known to be spectrally determined; for example, dimension, volume, and total scalar curvature. Examples of isospectral manifolds provide us with the only means for determining properties not determined by the spectrum. \medskip Most previous constructions for producing isospectral manifolds are based solely on representation theory. In fact, most known examples can be explained by Sunada's method \cite{S} or its generalizations \cite{GW1}, \ab \cite{DG2}. (See \cite{B2} for a general overview.) All manifolds constructed by these methods are {\it strongly isospectral;} that is, all natural, strongly elliptic, self-adjoint operators on the manifolds are isospectral. In particular, these manifolds share the same $p$-form spectrum. The {\it $p$-form spectrum} of a manifold is the eigenvalue spectrum of the Laplace-Beltrami operator extended to act on smooth $p$-forms for $p$ a positive integer. \medskip The primary goal of this paper is the development of a new construction for producing pairs of isospectral nilmanifolds. Two-step nilmanifolds have been a rich source of examples of isospectral manifolds. Moreover, their geometry has been studied in some detail. (See \cite{DG1}, \ab \cite{BG}, \ab \cite{O}, \ab \cite{P1}, \ab \cite{P2}, \ab \cite{P3}, \ab \cite{GW1}, \ab \cite{GW2}, \ab \cite {G1}, \ab \cite {G2}, \ab and \cite {E}.) This new construction uses techniques from Riemannian geometry, Lie groups, and representation theory to produce pairs of isospectral nilmanifolds of arbitrary step. The higher-step examples have a much richer geometry, exhibiting many properties not previously found. \medskip While representation theory is used as a tool, this construction differs from previous ones in that the resulting pairs of isospectral manifolds need not be isospectral on one-forms, and so do not fall under the traditional Sunada setup. This property was previously exhibited by pairs of isospectral Heisenberg manifolds constructed by Gordon and Wilson \cite{GW2}, \cite{G2}. And for any choice of $P$, Ikeda \cite{I2} has constructed examples of isospectral lens spaces that are isospectral on $p$-forms for $p=0,1,\cdots,P$ but not isospectral on $(P+1)$-forms. These are the only known examples. \medskip The only other examples of isospectral manifolds that do not fall under the traditional Sunada construction are bounded domains of Urakwa \cite{U}, and nonlocally isometric examples of Szabo \cite{Sz2}, Gordon \cite{G3}, \cite{G4}, and Gordon--Wilson \cite{GW3}. \medskip Recent results of Pesce \cite{P4} now explain the Ikeda and Urakawa examples in a Sunada-like setting. This setting requires a genericity assumption that excludes nilmanifolds. Moreover, the construction presented here generalizes the method used by Gordon--Wilson to construct the Heisenberg examples. \medskip Consequently, outside of the nonlocally isometric examples mentioned above, the construction below subsumes all known examples of isospectral manifolds that do not fall under a Sunada setup. \medskip After establishing notation in Section 2, the new construction for producing pairs of higher-step isospectral nilmanifolds is presented in Section 3. In Section 4, we use the new construction to produce pairs of isospectral three-step nilmanifolds with the following combinations of properties. \medskip %%%%%%%%%%%%%table of examples%%%%%%%%%%%%% \centerline{Table I: New Examples of Isospectral Manifolds} \medskip {\eightpoint \centerline{$ \vbox{\offinterlineskip \halign{\strut\vrule#&\ #\hfil\ &&\vrule#&\ \hfil#\hfil\ \cr \noalign{\hrule} &Pair of 3-Step&&$\forall p$ Same&&Rep. Equiv.&&Isomorphic&&Same&&Same&\cr &Isospectral&&$p$-form&&Fundamental&&Fundamental&&Length&&Marked Length&\cr &Nilmanifolds&&Spectrum&&Groups&&Groups&&Spectrum&&Spectrum&\cr \noalign{\hrule} &I(7 dim)&&Yes&&Yes&&No&&No&&No&\cr \noalign{\hrule} &II(5 dim)&&Yes&&Yes&&Yes&&Yes&&No&\cr \noalign{\hrule} &III$\backslash$IV(7$\backslash$5 dim)&&No&&No&&No&&No&&No&\cr \noalign{\hrule} &V(7 dim) &&No&&No&&Yes&&Yes&&Yes&\cr \noalign{\hrule}}} $} } \bigskip The properties listed in the above table are defined as follows. Two cocompact (i.e. $\Nilmfld$ compact), discrete subgroups $\Gamma_1$ and $\Gamma_2$ of a Lie group $G$ are called {\it representation equivalent} if the associated quasi-regular representations are unitarily equivalent. (See Section 2 for details.) The {\it length spectrum} of a Riemannian manifold is the set of lengths of closed geodesics, counted with multiplicity. The multiplicity of a length is defined as the number of distinct free homotopy classes in which the length occurs. (Note: other definitions of multiplicity also appear in the literature.) The pairs of isospectral manifolds above have the same lengths of closed geodesics. However, the length spectra often differ in the multiplicities that occur. The {\it marked length spectrum} takes into account not only the lengths of the closed geodesics but also the free homotopy classes in which the geodesics occur. \medskip In Section 4 we compare the quasi-regular representations and the fundamental groups of Examples I through V. We also compare the $p$-form spectrum, but the calculations are left to an Appendix. The length spectrum and marked length spectrum of these examples will be examined in \cite{Gt4}. \medskip Example I is the first example of a pair of {\it nonisomorphic,} representation equivalent, cocompact, discrete subgroups of a nilpotent Lie group. It is also the first example of a pair of representation equivalent cocompact, discrete subgroups of a solvable Lie group producing Riemannian manifolds that do not have the same length spectrum. This example has implications in representation theory on nilpotent Lie groups and motivated \cite{Gt1} and \cite{Gt2}. \medskip We prove in the Appendix that the manifolds in Examples III, IV and V are not isospectral on one-forms. Outside of the traditional Sunada setup, no general method is known for comparing the one-form spectrum of manifolds. The methods illustrated in the Appendix are new, as previously used techniques could not be applied to the higher-step examples. The only previous examples of manifolds that are isospectral on functions but not isospectral on $p$-forms for all $p$ are the lens spaces and Heisenberg manifolds mentioned above. \medskip Example V is the first example of a pair of Riemannian manifolds with the same marked length spectrum, but not the same spectrum on one-forms. This example contrasts with two-step results relating the marked length spectrum and the $p$-form spectrum \cite{E}. This example will be studied in detail in \cite{Gt4}. \medskip Most of the contents of this paper are contained in the author's thesis at Washington University in St. Louis in partial fulfillment of the requirements for the degree of Doctor of Philosophy. The author wishes to express deep gratitude to her advisor, Carolyn S. Gordon, for all of her suggestions, encouragement and support. \bigskip \bigskip %222222222222222222222222222222222222222 \bigskip \subheading{\S2 Background and Notation} \bigskip Let $G$ be a simply connected Lie group and let $\Gamma$ be a cocompact, discrete subgroup of $G.$ A Riemannian metric $g$ is {\it left invariant} if the left translations of $G$ are isometries. The left invariant metric $g$ projects to a Riemannian metric on $\Nilmfld,$ which we also denote by $g.$ Note that a left invariant metric is determined by a choice of orthonormal basis of the Lie algebra $\gg$ of $G.$ \medskip As $G$ is unimodular, the Laplace--Beltrami operator of $(\Nilmfld, g)$ may be written $$\Delta = -\sum_{i=1}^n {E_i}^2,\tag2.1$$ where $\{E_1, \cdots, E_n\}$ is an orthonormal basis of the Lie algebra $\gg$ of $G.$ \medskip The Laplace-Beltrami operator acting on smooth $p$-forms is defined by $\Delta = d \delta + \delta d.$ Here $\delta$ is the metric adjoint of $d.$ Equivalently $\delta = (-1)^{n(p+1)+1}*d*,$ where $*$ is the Hodge-$*$ operator. We denote the {\it $p$-form spectrum} of a Riemannian manifold $(M,g)$ by $p\hy\spec(M,g).$ \medskip The quasi-regular representation $\Rep$ of $G$ on $\Fnspg$ is defined as follows: \flushpar for all $x$ in $G$ and $f$ in $\Fnspg,$ $$\Rep(x)f=f\circ R_x.$$ Here $R_x$ denotes the right action of $x$ on $\GG \backslash G.$ The quasi-regular representation is known to be unitary. \medskip We say $\G1$ and $\G2$ are {\it representation equivalent} if $\rep1$ and $\rep2$ are unitarily equivalent; that is, $\G1$ and $\G2$ are representation equivalent if there exists a unitary isomorphism $T:\fnspg1 \rightarrow \fnspg2$ such that $T(\rep1(x)f)= \rep2(x)Tf$ for every $x$ in $G$ and every $f$ in $\fnspg1.$ \bigskip \proclaim{Proposition 2.2 (Gordon--Wilson \cite{GW1}) } Let $\G1$ and $\G2$ be cocompact, discrete subgroups of a simply connected Lie group $G.$ Let $g$ be a left invariant metric on $G.$ If $\G1$ and $\G2$ are representation equivalent, then $$p\hy \spec(\G1\backslash G,g) = p\hy \spec(\G2 \backslash G,g)$$ for $p=0,1, \cdots, dim(G).$ \endproclaim \bigskip \Remark Pairs of isospectral manifolds constructed using the traditional Su\-na\-da method are of the form $(\Gamma_1 \backslash M,g)$ and $(\Gamma_2 \backslash M,g)$ where $\Gamma_1$ and $\Gamma_2$ are representation equivalent, cocompact, discrete subgroups of a group $G$ acting by isometries on a Riemannian manifold $(M,g).$ \bigskip For a Lie algebra $\gg$, denote by $\g1$ the derived algebra $[\gg,\gg]$ of $\gg.$ That is, $\g1$ is the Lie subalgebra of $\mm$ generated by all elements of the form $[X,Y]$ for $X,Y$ in $\gg.$ Inductively, define $\g{k+1}=[\gg,\g{k}].$ A Lie algebra $\mm$ is said to be {\it k-step nilpotent} if $\m{k} \equiv 0$ but $\m{k-1} \not \equiv 0.$ A Lie group $G$ is called {\it $k$-step nilpotent} if its Lie algebra is. \medskip Let $\Gup{k} = exp(\m{k})$ denote the $k$th derived subgroup of $G.$ We denote the center of $G$ by $Z(G)$ and the center of $\mm$ by $\zz.$ Note that if $G$ is $k$-step nilpotent, then $\Gup{k-1} \subset Z(G).$ \medskip Let $exp$ denote the Lie algebra exponential from $\mm$ to $G.$ The Camp\-bell--Baker--Haus\-dorff formula gives us the group operation of $G$ in terms of $\mm.$ Namely, for $X,Y \in \mm:$ $$exp(X)exp(Y) = exp(X+Y+\frac1{2}[X,Y]+\frac1{12}[X,[X,Y]]+ \frac1{12}[Y,[Y,X]]+\cdots),$$ where the remaining terms are higher-order brackets. Note that for two-step nilpotent Lie groups, only the first three terms in the right-hand side are nonzero. For three-step groups, only the first five terms are nonzero. If $\mm$ is nilpotent and $G$ is simply connected, then $exp$ is a diffeomorphism from $\mm$ onto $G.$ Denote its inverse by $log.$ \medskip Let $\G1$ and $\G2$ be cocompact, discrete subgroups of nilpotent Lie groups $G_1$ and $G_2$ respectively. Any abstract group isomorphism $\Phi:\G1 \rightarrow \G2$ extends uniquely to a Lie group isomorphism $\Phi:G_1 \rightarrow G_2.$ \medskip Let $\GG$ be a cocompact, discrete subgroup of a nilpotent Lie group $G$ with left invariant metric $g.$ The locally homogeneous space $(\Nilmfld,g)$ is called a {\it Riemannian nilmanifold.} If $G$ is an abelian Lie group, then $\GG$ is merely a lattice of full rank in $G,$ and in this case $log\GG$ is also a lattice in $\mm.$ \medskip Let $\mm_\Q= span_\Q \{ log \GG \}.$ This is a {\it rational} Lie algebra; that is, there exists a basis of $\mm$ made up of elements of $log\GG$ such that the structure constants are rational. A Lie subalgebra $\hh$ of $\mm$ is called {\it rational} if $\hh$ is spanned by $\hh \cap \mm_{\bold Q}.$ Note that the notion of rational depends on $\GG.$ If $H=exp(\hh)$ is the connected Lie subgroup of $G$ with rational Lie algebra $\hh,$ then $\GG \cap H$ is a cocompact, discrete subgroup of $H.$ The $\m{k}$ are always rational Lie subalgebras of $\mm.$ \medskip The Kirillov theory of irreducible unitary representations of nilpotent groups gives us a correspondence between irreducible unitary representations of $G$ and elements of $\mm^*,$ the dual of $\mm.$ In particular, fix $\tau \in \mm^*.$ Let $\hh$ be a rational subalgebra of $\mm$ that is maximal with respect to the property that $\tau([\hh,\hh])\equiv 0.$ The subalgebra $\hh$ is called a {\it polarization} of $\tau.$ Let $H=exp(\hh)$ be the connected subgroup of $G$ with Lie algebra $\hh.$ \medskip Define a character $\overline{\tau}$ of $H$ by $${\overline{\tau}}(h)=e^{2\pi i \tau (log(h))}\tag2.3$$ for all $h$ in $H.$ Define $\pi_\tau$ to be the irreducible representation of $G$ induced by the representation $\overline{\tau}$ of $H.$ Denote by ${\H}_\tau$ the representation space of $\pi_\tau.$ Two such irreducible representations $\pi_\tau$ and $\pi_{\tau'}$ are unitarily equivalent if and only if $\tau' = \tau \circ Ad(x)$ for some $x$ in $G.$ Here $Ad(x)$ is the adjoint map from $\mm$ to $\mm.$ \medskip For $\tau$ in $\mm^*,$ the {\it coadjoint orbit\/} of $\tau$ is $$O(\tau) = \{ \tau \circ Ad(x): x \in G \}.$$ Hence $\pi_\tau$ and $\pi_{\tau'}$ are unitarily equivalent if and only if $\tau$ and $\tau'$ lie in the same coadjoint orbit of $\mm^*.$ \medskip As $G$ is nilpotent, every irreducible representation of $G$ is unitarily equivalent to $\pi_\tau$ for some $\tau \in \mm^*,$ and the quasi-regular representation $\Rep$ is completely reducible. Thus the representation space $\Fnsp$ is unitarily isomorphic to $$\Fnsp \cong {\underset{\tau \in \Tau}\to{\bigoplus}}m(\tau){\H}_\tau$$ for some $\Tau \subset \mm^*.$ Here $m(\tau)$ denotes the multiplicity of $H_\tau,$ and we assume $\Tau$ contains at most one element of each coadjoint orbit of $\mm^*.$ \medskip A good reference for representation theory on nilpotent Lie groups is \cite{CG}. \bigskip \bigskip \subheading{\S3 A New Construction of Isospectral Nilmanifolds} \bigskip Let $G$ be a simply connected, $k$-step nilpotent Lie group with Lie algebra $\mm.$ Define $\Gbar$ to be the simply connected, $(k$--$1)$-step nilpotent Lie group $G/\Gup{k-1}.$ For $\Gamma$ a cocompact, discrete subgroup of $G,$ denote by $\bar\Gamma$ the image of $\Gamma$ under the canonical projection from $G$ onto $\Gbar.$ The group $\bar\Gamma$ is then a cocompact, discrete subgroup of $\Gbar.$ For a left invariant metric $g$ on $G,$ we associate a left invariant metric $\gbar$ on $\Gbar$ by restricting $g$ to an orthogonal complement of $\g{k-1}$ in $\mm.$ \medskip We call the $(k$--$1)$-step nilmanifold $(\Nilmfldbar,\gbar)$ the {\it quotient nilmanifold of $(\Nilmfld,g).$} By using the definition of $\gbar,$ one easily sees that the projection $(\Nilmfld, g) \rightarrow (\Nilmfldbar,\gbar)$ is a Riemannian submersion. \medskip The Lie algebra $\gggbar$ of $\Gbar$ is just $\mm / \g{k-1}.$ We denote elements of $\gggbar$ by $\bar{U},$ where $\bar{U}$ is the image of $U$ under the canonical projection from $\mm$ onto $\gggbar.$ \bigskip \proclaim{Definition 3.1} Let $G$ be a simply connected nilpotent Lie group. We say $G$ is {\it strictly nonsingular} if the following property holds: for every $z$ in $Z(G)$ and every noncentral $x$ in $G$ there exists an element $a$ in $G$ such that $$[x,a] = z.$$ Here $[x,a]$ denotes the commutator $[x,a]=xax^{-1}a^{-1}.$ Equivalently, the Lie algebra $\gg$ is {\it strictly nonsingular} if for every noncentral $X$ in $\gg,$ $$\zz \subset ad(X)(\gg).$$ That is, for every $X$ in $\gg-\zz$ and every $Z$ in $\zz$ there exists a vector $Y$ in $\gg$ such that $[X,Y]=Z.$ \endproclaim \bigskip \proclaim{Theorem 3.2} Let $G$ be a simply connected, strictly nonsingular nilpotent Lie group with left invariant metric $g.$ If \ $\G1$ and $\G2$ are cocompact, discrete subgroups of $G$ such that $$\G1 \cap Z(G) = \G2 \cap Z(G) \quad \text{and} \quad \spec(\nilmfldbar1,\gbar) = \spec(\nilmfldbar2,\gbar),$$ then $$\spec(\nilmfld1,g) = \spec(\nilmfld2,g).$$ \endproclaim \bigskip \Remark The above construction is a generalization of the construction used by Gordon and Wilson to obtain pairs of isospectral Heisenberg manifolds \cite {GW2}. If we let the Lie group $G$ be a simply connected, strictly nonsingular, two-step nilpotent Lie group with a one-dimensional center, then $G=H_n$ for some $n,$ where $H_n$ denotes the $(2n+1)$-dimensional Heisenberg group. \bigskip \pagebreak \demo{Proof of Theorem 3.2} \bigskip We use the notation of Section 2. \medskip For $i=1,2,$ let $\Tau_i$ be a subset of $\gg^*$ such that $$\fnsp{i} \cong {\underset{\tau\in\Tau_i}\to\bigoplus} {m_i(\tau)}\H_\tau.$$ Recall that $m_i(\tau)$ denotes the multiplicity of $\pi_\tau$ in the quasi-regular representation of $G$ on $\fnsp{i},$ and we assume that $\Tau_i$ contains at most one element of each coadjoint orbit of $\gg^*.$ \medskip We decompose the index set $\Tau_i =\Tau'_i \cup \Tau''_i$ by letting $$\Tau'_i = \{\tau \in \Tau_i : \tau(\zz) \equiv 0\}, \quad \text{and} \quad \Tau''_i = \{\tau \in \Tau_i : \tau(\zz )\not \equiv 0\}.$$ We likewise decompose the representation space $\fnsp{i}$ by letting $$\H'_i= {\underset{\tau\in\Tau'_i}\to\bigoplus} m_i(\tau)\H_\tau \quad \text{and} \quad \H''_i= {\underset{\tau \in \Tau''_i}\to\bigoplus} m_i(\tau)\H_\tau .$$ As representation spaces $\fnsp{i}= \H'_i \oplus \H''_i.$ \medskip Clearly $spec(\nilmfld{i}, g) = spec'(\nilmfld{i}, g) \cup spec''(\nilmfld{i}, g),$ where $spec'(\nilmfld{i}, g)$ and \break $spec''(\nilmfld{i}, g)$ are defined as the spectrum of the Laplacian restricted to acting on $\H'_i,$ and $\H''_i,$ respectively. The multiplicity of an eigenvalue in $spec(\nilmfld{i}, g)$ is equal to the sum of its multiplicities in $spec'(\nilmfld{i}, g)$ and $spec''(\nilmfld{i}, g).$ \bigskip \proclaim{Lemma 3.3} The Laplacian of $(\nilmfld{i}, g)$ acting on $\H'_i$ is precisely the Laplacian of $(\nilmfldbar{i},\gbar)$ acting on $\fnspbar{i}.$ Thus $\spec(\nilmfldbar{i},\gbar) = \spec'(\nilmfld{i},g)$ for $i=1,2.$ \endproclaim \bigskip \proclaim{Lemma 3.4} The representations of $G$ on $\H''_1$ and $\H''_2$ are unitarily equivalent, hence $\spec''(\nilmfld1,g) = \spec''(\nilmfld2,g).$ \endproclaim \bigskip Theorem 3.2 now follows. \medskip The proof of Lemma 3.3 is essentially an extension of the first part of the proof used by Gordon--Wilson to construct pairs of isospectral Heisenberg manifolds. (See \cite{GW2}, Theorem 4.1) The details are included here for completeness and because of a difference in notation. \bigskip \pagebreak \demo{Proof of Lemma 3.3} \bigskip Let $\{Z_1, Z_2, \cdots, Z_T\}$ be an orthonormal basis of $\zz = \g{k-1}.$ Extend it to $\{E_1, E_2, \allowbreak \cdots, E_N,\allowbreak Z_1, Z_2, \cdots, Z_T\},$ an orthonormal basis of $\gg.$ By (2.1), the La\-place-Bel\-tra\-mi operator of $(G,g)$ is $$\Delta = - \sum_{n=1}^N{E_n}^2 - \sum_{k=1}^K{Z_k}^2.$$ \medskip View functions in $\fnsp{i}$ as left $\G{i}$-invariant functions of $G.$ The subspace $\H'_i$ is then those functions in $\fnsp{i}$ that are independent of the center, which correspond to functions in $\fnspbar{i}$ in a natural way. So when we restrict $\Delta$ to $\H'_i,$ we have $$\Delta = - \sum_{n=1}^N{E_n}^2,$$ which corresponds to the Laplacian of $(\nilmfldbar{i},\gbar).$ \medskip The Laplacian of $(\nilmfld{i}, g)$ acting on $\H'_i$ is then precisely the Laplacian of $(\nilmfldbar{i},\gbar)$ acting on $\fnspbar{i},$ so $spec(\nilmfldbar{i},\gbar) = spec'(\nilmfld{i},g),$ as desired. $\endpf$ \enddemo \bigskip Before proving Lemma 3.4, we must introduce some of the theory of square integrable representations of nilpotent Lie groups. \bigskip \proclaim{Definition 3.5} Let $G$ be a locally compact, unimodular group with center Z(G). We say that an irreducible unitary representation $\pi$ of $G$ on a Hilbert space $\H$ is {\it square integrable\/} if there are nonzero vectors $x_1$ and $x_2$ in $\H$ such that $$\int_{G/Z(G)} |(\pi(s)x_1,x_2)|^2 d\overline{\mu}(\overline{s}) < \infty.$$ \endproclaim \bigskip Here $d\overline{\mu}(\overline{s})$ denotes integration over $G/Z(G)$ with respect to a choice of Haar measure $\overline{\mu}$ on $G/Z(G).$ As the center acts trivially, the integrand may be viewed as a function of $G/Z(G).$ \medskip Let $\xi^\perp$ be the subalgebra of $\mm^*$ defined by $\xi^\perp = \{ \mu \in \mm^* : \mu(\xi) \equiv 0 \}.$ Note that $\xi^\perp \cong (\mm/\xi)^*.$ For $\tau \in \mm^*,$ let $b_\tau$ denote the skew-symmetric, bilinear form on $\mm/\xi$ defined by $b_\tau(\BX,\BY)=\tau([X,Y])$ for all $\BX,\BY$ in $\mm/\xi.$ Here $X$ and $Y$ are any elements of $\mm$ that project onto $\BX$ and $\BY$ respectively. \bigskip \pagebreak \proclaim{Theorem 3.6 (Moore--Wolf \cite {MW})} For a linear functional $\tau$ in $\mm^*$ with coadjoint orbit $O(\tau)$ and corresponding irreducible unitary representation $\pi_\tau,$ the following three conditions are equivalent: (1) $\pi_\tau$ is square integrable. (2) $O(\tau) = \tau + \xi^\perp.$ (3) $b_\tau$ is nondegenerate on $\mm/\xi.$ \endproclaim \bigskip %\pagebreak \demo{Proof of Lemma 3.4} \bigskip Fix $\tau \in \Tau''_{i}.$ Let $Z \in \xi$ be such that $\tau(Z) \neq 0.$ By strict nonsingularity, for all noncentral $X \in \mm$ there exists $Y \in \mm$ such that $[X,Y]=Z.$ Hence $b_\tau(\BX,\BY) = \tau([X,Y]) \neq 0.$ So $b_\tau$ is nondegenerate. By Theorem 3.6, $\pi_\tau$ is square integrable. Note that $b_\tau$ nondegenerate implies that $N=dim(\mm/\xi)$ is even. \medskip Recall from Section 2 that $\pi_\tau$ is independent of the choice of $\tau$ in $O(\tau).$ By Theorem 3.6, as $\pi_\tau$ is square integrable, the coadjoint orbit $O(\tau)$ is uniquely determined by the restriction of $\tau$ to the center. We may thus assume $\tau \in \xi^*.$ \medskip Let $\alpha$ be a volume form on $G/Z(G).$ That is, let $\alpha$ be a fixed, alternating, $N$-linear form over $\mbar = \mm/\xi.$ As $b_\tau$ is nondegenerate, ${b_\tau}^{(\frac1{2}N)}=b_\tau \wedge \cdots \wedge b_\tau$ is also a volume form on $G/Z(G)$ and hence a scalar multiple of $\alpha.$ Define $P_\alpha(\tau)$ by $$P_\alpha(\tau)\alpha = b_\tau \wedge \cdots \wedge b_\tau.$$ The polynomial $P_\alpha$ is homogeneous of degree $\frac1{2}N$ on $\mm^*$ and depends only on the choice of volume form $\alpha.$ \medskip Let $\xi^*$ denote the dual of $\xi.$ Let $L=\G1 \cap Z(G) = \G2\cap Z(G)$ and let ${L}^*\subset \xi^*$ be the dual lattice of $L.$ \medskip We now use the following occurrence and multiplicity condition, also due to Moore and Wolf. \medskip \proclaim{Theorem 3.7 (Moore--Wolf \cite {MW})} Let $G$ be a nilpotent Lie group and $\GG$ a cocompact, discrete subgroup of $G.$ Let $L=\GG \cap Z(G).$ Fix a volume form $\alpha_\GG$ on $\mm/\xi$ so that $\GGbar \backslash \Mbar$ has volume one. Let $\tau$ be a nonzero element of $\xi^*$ such that $\pi_\tau$ is square integrable. The representation $\pi_\tau$ occurs in the quasi-regular representation of $G$ on $\Fnsp$ if and only if $\tau \in L^*.$ Moreover, its multiplicity $m(\tau)$ is $|P_{\alpha_\GG}(\tau)|.$ \endproclaim \bigskip By Theorem 3.7, the square integrable representation $\pi_\tau$ occurs in the quasi-regular representation of $G$ on $\fnsp{i}$ if and only if $\tau$ is contained in ${L}^*.$ Thus for $i=1,2$ the coadjoint orbits represented in $\Tau''_i$ correspond to the elements of ${L}^*.$ We may assume $\Tau''_1 = \Tau''_2.$ \medskip As the Riemannian metrics of $(\nilmfldbar1,\gbar)$ and $(\nilmfldbar2,\gbar)$ arise from the same left invariant metric $\gbar$ on $\Mbar,$ we know that the Riemannian volume forms of $(\nilmfldbar1,\gbar)$ and $(\nilmfldbar2,\gbar)$ arise from the same left invariant volume form on $\Mbar.$ We will denote this volume form and its projections onto $(\nilmfldbar1,\gbar)$ and $(\nilmfldbar2,\gbar)$ by $\Omega.$ \medskip Let $\alpha_\G1$ and $\alpha_\G2$ be as in Theorem 3.7. The volume forms $\alpha_\G{i}$ are then scalar multiples of $\Omega.$ For $i=1,2,$ let $\alpha_\G{i} = p_i\Omega.$ \medskip So $$\int_{\nilmfldbar1}\alpha_\G1 = \ 1 \ = \int_{\nilmfldbar2}\alpha_\G2$$ $$p_1\int_{\nilmfldbar1}\Omega = p_2\int_{\nilmfldbar2}\Omega$$ $$p_1\Vol(\nilmfldbar1,\gbar)= p_2\Vol(\nilmfldbar2,\gbar)$$ \medskip By hypothesis, $\spec(\nilmfldbar1,\gbar) = \spec(\nilmfldbar2,\gbar),$ and the spectrum of the La\-place-Bel\-tra\-mi operator is known to determine the volume of a closed manifold. Thus $\Vol(\nilmfldbar1,\gbar)=\Vol(\nilmfldbar2,\gbar),$ which implies $p_1 = p_2,$ and so $\alpha_\G1 = \alpha_\G2.$ \medskip As the definition of $P_{\alpha_\G{i}}$ depends only on the volume form $\alpha_\G{i},$ we must have $$P_{\alpha_\G1}(\tau) = P_{\alpha_\G2}(\tau)$$ for all $\tau$ in $\Tau''_1 = \Tau''_2.$ Hence $m_1(\tau) = m_2(\tau)$ for all $\tau$ in $\Tau''_1=\Tau''_2.$ \medskip Thus the representations of $G$ on $\H''_1$ and $\H''_2$ are unitarily equivalent, and by Proposition 2.2 $$\spec''(\nilmfld1,g) = \spec''(\nilmfld2,g),$$ as desired. \medskip The proofs of Lemma 3.4 and Theorem 3.2 are now complete. $\endpf$ \enddemo \enddemo \bigskip \bigskip \proclaim{Corollary 3.8} Let $G$ be a simply connected, strictly nonsingular nilpotent Lie group. Two cocompact, discrete subgroups {$\G1$ and $\G2$ are representation equivalent subgroups of $G$} if and only if {$\GGGbar1$ and $\GGGbar2$ are representation equivalent subgroups of $\Mbar$} and {$\G1 \cap Z(G) = \G2 \cap Z(G).$} \endproclaim \bigskip \demo{Proof of Corollary 3.8} \bigskip Let $\G1$ and $\G2$ be cocompact, discrete subgroups of $G.$ As in the proof of Theorem 3.2, we decompose the representation spaces of $\rep1$ and $\rep2$ as $$\fnsp{i}= \H'_i \oplus \H''_i,$$ for $i=1,2.$ \medskip If $\G1$ and $\G2$ are representation equivalent, then the square integrable representations occurring in the quasi-regular representations must correspond. We showed in the proof of Lemma 3.4 that the square integrable representations occurring in $\rep1$ and $\rep2$ are precisely the irreducible representations appearing in $\H''_i.$ Thus $\G1$ and $\G2$ are representation equivalent subgroups of $G$ if and only if the representations of $G$ on $\H'_1$ and $\H'_2$ are unitarily equivalent and the representations of $G$ on $\H''_1$ and $\H''_2$ are unitarily equivalent. \medskip For $i=1,2$ the irreducible components of the representation of $G$ on $\H'_i$ correspond to the linear functionals in $\Tau_i$ that are zero on the center of $\mm.$ As these functionals may be viewed as functionals on $\mbar,$ we may likewise view the irreducible components as representations of $\Mbar.$ Hence the representations of $G$ on $\H'_1$ and $\H'_2$ may be viewed as the quasi-regular representations of $\Mbar$ on $\fnspbar1$ and $\fnspbar2,$ respectively. Thus $\G1$ and $\G2$ are representation equivalent subgroups of $G$ if and only if $\GGGbar1$ and $\GGGbar2$ are representation equivalent subgroups of $\Mbar$ and the representations of $G$ on $\H''_1$ and $\H''_2$ are unitarily equivalent. \medskip Theorem 3.7 tells us that the irreducible representations occurring in $\H''_i$ correspond to the elements of the dual lattice of $\Gamma_i \cap Z(G).$ So if the representations of $G$ on $\H''_1$ and $\H''_2$ are unitarily equivalent, then the duals of $\Gamma_1\cap Z(G)$ and $\Gamma_2\cap Z(G)$ must coincide. Hence $\Gamma_1\cap Z(G)=\Gamma_2\cap Z(G).$ The proof of the forward direction is now complete. \medskip The reverse direction follows from Proposition 2.2 applied to the quotient nilmanifolds and Lemma 3.4. \endpf \enddemo \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \bigskip \subheading{\S4 New Examples of Isospectral Nilmanifolds} \bigskip Using Theorem 3.2, we construct and compare five new pairs of isospectral nilmanifolds. A summary of the properties of these examples is listed in Section 1, Table I. In this Section we compare the quasi-regular representations and the fundamental groups of these examples. The $p$-form spectra of these examples are also compared, but the calculations are left to the Appendix. The methods used in the Appendix are new, as previously used techniques could not be applied to compare the $p$-form spectrum of these higher-step examples. The length spectra and marked length spectra of these examples will be studied in \cite{Gt4}. \medskip With the exception of the column comparing the representation equivalence of the fundamental groups, all of the properties listed in Table I are geometric invariants. Hence a ``No'' in any one of these columns demonstrates that an Example is nontrivial. \medskip Before proceeding, we need the following. \bigskip \proclaim{Definition 4.1} Let $\Phi$ be a Lie group automorphism of $G.$ Let $\GG$ be a cocompact, discrete subgroup of $G.$ We say $\Phi$ is an {\it almost inner automorphism} if for all elements $x$ of $G$ there exists $a_x$ in $G$ such that $\Phi(x)=a_x x a_x^{-1}.$ \endproclaim \bigskip \proclaim{Theorem 4.2 (Gordon--Wilson \cite{GW1}) } Let $G$ be a nilpotent Lie group and let $\GG$ be a cocompact, discrete subgroup of $G.$ If $\Phi$ is an almost inner automorphism of $G,$ then $\GG$ and $\Phi(\GG)$ are representation equivalent. \endproclaim \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \bigskip \subheading{Example I} \bigskip Consider the simply connected, strictly nonsingular, three-step nilpotent Lie group $G$ with Lie algebra $$\mm = span_{\R}\{X_1, X_2, Y_1, Y_2, Z_1, Z_2, W \}$$ and Lie brackets $$[X_1,Y_1] = [X_2, Y_2] = Z_1$$ $$[X_1,Y_2] = Z_2$$ $$[X_1,Z_1] = [X_2, Z_2] = [Y_1, Y_2] = W$$ and all other basis brackets zero. \bigskip Let $\G1$ be the cocompact, discrete subgroup of $G$ generated by $$\{exp(2X_1),exp(2X_2),exp(Y_1),exp(Y_2),exp(Z_1),exp(Z_2),exp(W)\}.$$ Let $\G2$ be the cocompact, discrete subgroup of $G$ generated by $$\{exp(2X_1),exp(2X_2),exp(Y_1),exp(Y_2+\frac1{2}Z_2),exp(Z_1),exp(Z_2),exp(W)\}.$$ Note that $\G1 \cap Z(G) = \G2 \cap Z(G) = \{exp(jW):j\in\Z \}.$ \bigskip Now $\GGGbar2 = \Phi(\GGGbar1)$ where $\Phi$ is the almost inner automorphism of $\Mbar$ given on the Lie algebra level by $$\align \BX_1&\rightarrow \BX_1,\\ \BX_2 &\rightarrow \BX_2,\\ \BY_1 &\rightarrow \BY_1,\\ \BY_2 &\rightarrow \BY_2 + \frac1{2} \BZ_2,\\ \BZ_1 &\rightarrow \BZ_1,\\ \BZ_2 &\rightarrow \BZ_2.\endalign$$ \bigskip By Theorem 4.2, \ $\GGGbar1$ and $\GGGbar2$ are representation equivalent subgroups of $\Mbar.$ By Corollary 3.8, \ $\G1$ and $\G2$ are representation equivalent subgroups of $G.$ \medskip By Proposition 2.2, for any choice of left invariant metric $g$ of $G,$ !x we have \break $p\hy spec(\nilmfld1, g) = p\hy spec(\nilmfld2, g),$ for $p=0,1,\cdots,7.$ \bigskip \proclaim{Proposition 4.3 (\cite{Gt1})} The subgroups $\Gamma_1$ and $\Gamma_2$ are not isomorphic as groups. \endproclaim \bigskip \Remark The author previously established the representation equivalence of $\G1$ and $\G2$ in \cite{Gt1} by using a direct calculation. This example was presented in \cite{Gt1} as the first example of a pair of nonisomorphic, representation equivalent subgroups of a solvable Lie group. Note that a nilpotent Lie group is necessarily solvable. Also, this example was presented in \cite{Gt2} as the first example of a pair of representation equivalent subgroups of a solvable Lie group producing nilmanifolds with unequal length spectra. Contrast this example with what must happen in the two-step case. \bigskip \proclaim{Definition 4.4} Let $G$ be a two-step nilpotent Lie group and let $\GG$ be a cocompact, discrete subgroup of $G.$ We call the automorphism $\Phi$ of $G$ a $\GG${\it -equivalence} if for all $\gamma$ in $\GG$ there exists $a_\gamma$ in $G$ and $\gamma'_\gamma$ in $\GG \cap \Gup1$ such that $\Phi(\gamma)=a_\gamma \gamma a_\gamma^{-1} \gamma'_\gamma.$ \endproclaim \bigskip \proclaim{Theorem 4.5 (\cite{Gt1}, \cite{Gt2})} Let $G$ be a two-step nilpotent Lie group. Let $\G1$ and $\G2$ be cocompact, discrete subgroups of $G$. The subgroups $\G1$ and $\G2$ are representation equivalent if and only if there exists $\Phi,$ a $\G1$-equivalence of $G,$ such that $\Phi(\G1)=\G2.$ Thus if $\G1$ and $\G2$ are representation equivalent, they are necessarily isomorphic. In addition, if $\G1$ and $\G2$ are representation equivalent, then $(\nilmfld1,g)$ and $(\nilmfld2,g)$ have the same length spectrum for any choice of left invariant metric $g$ of $G.$ \endproclaim \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \bigskip \subheading{Example II} \bigskip Consider the simply connected, strictly nonsingular, three-step nilpotent Lie group $G$ with Lie algebra $$\mm = span_{\R}\{X_1,Y_1, Y_2, Z, W \}$$ and Lie brackets $$[X_1,Y_1] = Z$$ $$[X_1,Z] = [Y_1, Y_2] = W$$ and all other basis brackets zero. \bigskip Let $\G1$ be the cocompact, discrete subgroup of $G$ generated by $$\{exp(2X_1),exp(Y_1),exp(Y_2),exp(Z),exp(W)\}.$$ Let $\G2$ be the cocompact, discrete subgroup of $G$ generated by $$\{exp(2X_1),exp(Y_1+\frac1{2}Z),exp(Y_2),exp(Z),exp(W)\}.$$ Note that $\G1 \cap Z(G) = \G2 \cap Z(G) = \{exp(jW):j\in\Z \}.$ \bigskip Now $\GGGbar2 = \Phi(\GGGbar1)$ where $\Phi$ is the inner automorphism of $\Mbar$ given on the Lie algebra level by $$\align \BX_1 &\rightarrow \BX_1, \\ \BY_1 &\rightarrow \BY_1 + \frac1{2} \BZ,\\ \BY_2 &\rightarrow \BY_2, \\ \BZ &\rightarrow \BZ.\endalign$$ Note that an inner automorphism is necessarily almost inner. \medskip By Theorem 4.2, \ $\GGGbar1$ and $\GGGbar2$ are representation equivalent subgroups of $\Mbar.$ By Corollary 3.8, \ $\G1$ and $\G2$ are representation equivalent subgroups of $G.$ \bigskip By Proposition 2.2, $p\hyspec(\nilmfld1, g) = p\hyspec(\nilmfld2, g)$ for $p=0, 1, \cdots, 5,$ for any choice of left invariant metric $g$ of $G.$ \bigskip Here $\G1$ and $\G2$ are isomorphic. Indeed, a simple calculation shows that the isomorphism $\Psi$ given on the Lie algebra level by $$\align X_1 &\rightarrow X_1 + \frac1{2}Y_2,\\ Y_1 &\rightarrow Y_1 + \frac1{2}Z, \\ Y_2 &\rightarrow Y_2,\\ Z &\rightarrow Z,\\ W &\rightarrow W, \endalign$$ is an isomorphism of $G$ such that $\Psi(\G1) = \G2.$ \bigskip \proclaim{Proposition 4.6} No isomorphism between $\G1$ and $\G2$ will project to a $\GGGbar1$-equivalence of $\Mbar.$ \endproclaim \bigskip \Remark Example I illustrates that in the higher-step case, the representation equivalence of $\G1$ and $\G2$ and the isomorphism class of $\G1$ and $\G2$ need not be related. Example II shows that, in contrast to Theorem 4.5, even in the case where $\G1$ and $\G2$ are isomorphic, knowing the isomorphisms between $\G1$ and $\G2$ is not enough to use Corollary 3.8 to establish whether or not $\G1$ and $\G2$ are representation equivalent. \bigskip \demo{Proof of Proposition 4.6} \bigskip Let $\Psi$ be an isomorphism from $\G1$ to $\G2.$ Extend it to the Lie group isomorphism $\Psi:G \rightarrow G$ such that $\Psi(\G1)=\G2.$ \medskip On the Lie algebra level, any such isomorphism must preserve the following ideals of $\mm:$ (1) $\m2 = span_\R\{W\},$ (2) $\m1 = span_\R\{Z,W\},$ (3) $\aa = span_\R\{Y_2,Z,W\},$ and (4) $\cc = span_\R\{Y_1,Y_2,Z,W\},$ the centralizer of $\m1$ in $\mm.$ \flushpar To see (3), note that $ad(U)(\mm) \subset \m2$ if and only if $U \in \aa.$ \medskip Note that the generators of $\G1$ and $\G2$ presented above are canonical in the sense that every element of $\G1$ may be expressed uniquely as $$exp(2n_1X_1)exp(m_1Y_1)exp(m_2Y_2)exp(kZ)exp(jW),$$ for integers $n_1,m_1,m_2,k,j.$ Similarly for $\G2.$ \medskip As $\Psi(\G1) = \G2,$ generators of $\G1$ must go to generators of $\G2,$ and these generators must be expressible in terms of the canonical generators of $\G2,$ given above. \medskip Combining this fact with properties (1) through (4), we obtain: \medskip $\Psi_*(W) = \pm W,$ \ by (1). $\Psi_*(Z) = \pm Z + h_0W$ \ using (2). $\Psi_*(Y_2) = \pm Y_2$ mod $\m1,$ \ using (3). $\Psi_*(Y_1) = \pm (Y_1 + \frac1{2}Z) + h_1 Y_2 + h_2Z$ mod $\m2$ using (4). $\Psi_*(X_1) = \pm X_1 + \frac1{2}h_3Y_1 + \frac1{2}h_4Y_2$ mod $\m1,$ \flushpar where $h_0, \ h_1,\ h_2, \ h_3$ and $h_4$ are integers. \bigskip Finally, we use the fact that $\Psi_*$ is a Lie algebra isomorphism. By examining the $W$ coefficient of $\Psi_*([X_1,Y_1]) = [\Psi_*(X_1),\Psi_*(Y_1)]$ we have the equation $$h_0 = \pm\frac1{2} + \frac1{2}h_1h_3 \pm h_2 \pm\frac1{2}h_4.$$ Thus either $h_3 \neq 0$ or $h_4 \neq 0.$ \medskip As $\BY_1$ and $\BY_2$ are {not} in $[\BX_1,\mbar]$ and {not} in $\mbar^{(1)},$ we see that the projection of $\Psi$ cannot possibly be a $\GGGbar1$-equivalence. \endpf \enddemo \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \bigskip \subheading{Example III} \bigskip Consider again the seven-dimensional Lie group $G$ presented in Example I. \bigskip We again let $\G1$ be the cocompact, discrete subgroup of $G$ generated by $$\{exp(2X_1),exp(2X_2),exp(Y_1),exp(Y_2),exp(Z_1),exp(Z_2),exp(W)\},$$ and let $\G2$ be the cocompact, discrete subgroup of $G$ generated by $$\{exp(X_1),exp(X_2), exp(2Y_1),exp(2Y_2), exp(Z_1),exp(Z_2),exp(W)\}.$$ Note that $\G1 \cap Z(G) = \G2 \cap Z(G) = \{exp(jW):j\in\Z\}.$ \bigskip Let $g$ be the left invariant metric on $G$ defined by letting $$\{X_1, X_2, Y_1, Y_2, Z_1, Z_2, W\}$$ be an orthonormal basis of $\mm.$ \bigskip Now $\GGGbar2 = \Phi(\GGGbar1)$ where $\Phi$ is the automorphism of $\Mbar$ given on the Lie algebra level by $$\align \BX_1 &\rightarrow \BY_2, \\ \BX_2 &\rightarrow \BY_1, \\ \BY_1 &\rightarrow \BX_2, \\ \BY_2 &\rightarrow \BX_1, \\ \BZ_1 &\rightarrow -\BZ_1, \\ \BZ_2 &\rightarrow -\BZ_2.\endalign$$ \medskip The automorphism $\Phi$ is also an isometry of $(\Mbar,\gbar),$ and an isometry must preserve the spectrum. Thus $\spec(\nilmfldbar1, \gbar) = \spec(\nilmfldbar2, \gbar).$ By Theorem 3.2, $\spec(\nilmfld1, g) = \spec(\nilmfld2, g).$ \bigskip \proclaim{Proposition 4.7} The manifolds $(\nilmfld1, g)$ and $(\nilmfld2, g)$ are not isospectral on one-forms. \endproclaim \bigskip The proof of Proposition 4.7 is left to the Appendix. Note that Proposition 4.7 and Proposition 2.2 together imply that $\G1$ and $\G2$ are not representation equivalent subgroups of $G.$ \bigskip \proclaim{Proposition 4.8} The subgroups $\G1$ and $\G2$ are not isomorphic as groups. \endproclaim \bigskip \demo{Proof of Proposition 4.8} \medskip If there existed a group isomorphism between $\G1$ and $\G2,$ it would extend to a Lie group automorphism $\Psi$ of $G$ such that $\Psi(\G1)=\G2.$ \medskip Each of the following ideals of $\mm$ must be preserved by the Lie algebra automorphism $\Psi_*:$ (1) $\m2 = \xi = span_{\R}\{ W \},$ (2) $\m1 = span_{\R}\{ Z_1, Z_2, W \},$ (3) $\cc = span_{\R}\{ Y_1, Y_2, Z_1, Z_2, W \},$ the centralizer of $\m1$ in $\mm,$ and (4) $\aa = span_{\R}\{ X_2, Y_1, Z_1, Z_2, W \}.$ \flushpar To see (4), note that the image of $ad(U)$ has dimension less than three if and only if $U \in \aa.$ \medskip Now $\Psi(\G1) = \G2.$ Consequently, generators of $\G1$ must go to generators of $\G2,$ and these generators must be expressible in terms of the canonical generators of $\G2,$ given above. Combining this fact with properties (1) through (4), we obtain: \medskip $\Psi_*(Y_1) = \pm 2Y_1$ mod $\m1$ by (3) and (4). $\Psi_*(Y_2) = \pm 2Y_2$ mod $span_\R\{Y_1,Z_1,Z_2,W\}$ by (3). $\Psi_*(W) = \pm W$ by (1). \flushpar But then $\Psi_*([Y_1,Y_2]) \neq [\Psi_*(Y_1),\Psi_*(Y_2)].$ \endpf \enddemo \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \bigskip \subheading{Example IV} \bigskip Consider again the five-dimensional Lie group $G$ presented in Example II. \bigskip We again let $\G1$ be the cocompact, discrete subgroup of $G$ generated by $$\{exp(2X_1),exp(Y_1),exp(Y_2),exp(Z),exp(W)\},$$ and let $\G2$ be the cocompact, discrete subgroup of $G$ generated by $$\{exp(X_1), exp(2Y_1), exp(Y_2),exp(Z),exp(W)\}.$$ Note that $\G1 \cap Z(G) = \G2 \cap Z(G) = \{exp(jW):j\in\Z\}.$ \bigskip Let $g$ be the left invariant metric on $G$ defined by letting $$\{X_1, Y_1, Y_2, Z, W\}$$ be an orthonormal basis of $\mm.$ \bigskip Now $\GGGbar2 = \Phi(\GGGbar1)$ where $\Phi$ is the automorphism of $\Mbar$ given on the Lie alegebra level by $$\align \BX_1 &\rightarrow \BY_1,\\ \BY_1 &\rightarrow \BX_1,\\ \BY_2 &\rightarrow \BY_2, \\ \BZ &\rightarrow -\BZ.\endalign$$ \medskip The automorphism $\Phi$ is clearly an isometry of $(\Mbar,\gbar),$ and an isometry preserves the spectrum. Thus $\spec(\nilmfldbar1, \gbar) = \spec(\nilmfldbar2, \gbar).$ By Theorem 3.2, $\spec(\nilmfld1, g) = \spec(\nilmfld2, g).$ \bigskip \proclaim{Proposition 4.9} The manifolds $(\nilmfld1, g)$ and $(\nilmfld2, g)$ are not isospectral on one-forms. \endproclaim \bigskip The proof of Proposition 4.9 is left to the Appendix. \bigskip \proclaim{Proposition 4.10} The subgroups $\G1$ and $\G2$ are not isomorphic as groups. \endproclaim \bigskip \Remark The combination of properties exhibited by Examples III and IV are similar to properties exhibited by pairs of isospectral Heisenberg manifolds constructed by Gordon and Wilson \cite{GW2, G2}. \bigskip \demo{Proof of Proposition 4.10} \medskip If there existed a group isomorphism between $\G1$ and $\G2,$ it would extend to a Lie group automorphism $\Psi$ of $G$ such that $\Psi(\G1)=\G2.$ \medskip As before, generators of $\G1$ must go to generators of $\G2,$ and these generators must be expressible in terms of the canonical generators of $\G2,$ given above. Combining this with properties (1) through (4) from the proof of Proposition 4.6, we have $\Psi_*(Y_1) = \pm 2Y_1$ mod $span_\R\{Y_2,Z,W\}.$ $\Psi_*(Y_2) = \pm Y_2$ mod $\m1.$ $\Psi_*(W) = \pm W.$ \flushpar But then $\Psi_*([Y_1,Y_2]) \neq [\Psi_*(Y_1),\Psi_*(Y_2)].$ \endpf \enddemo \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \bigskip \subheading{Example V} \bigskip Consider again the seven-dimensional Lie group $G$ presented in Example I. \bigskip We fix a left invariant metric on $G$ by letting $\{E_1,E_2,E_3,E_4,E_5,E_6,E_7\}$ be an orthonormal basis of $\gg$ where $$ \align E_1&= X_1-\frac{1}{2}X_2-\frac{1}{4}Y_2,\\ E_2&= X_2-\frac{1}{4}Y_1,\\ E_3&= Y_1,\\ E_4&=Y_1+Y_2,\\ E_5&=Z_1,\\ E_6&=\frac{1}{2}Z_1+Z_2,\\ E_7&=W. \endalign $$ Let $\Phi$ be the automorphism of $G$ defined on the Lie algebra level by $$ \align X_1&\rightarrow -X_1+X_2+\frac{1}{4}Y_1+\frac{1}{2}Y_2,\\ X_2&\rightarrow X_2-\frac{1}{2}Y_1+\frac{1}{4}Z_1,\\ Y_1&\rightarrow -Y_1,\\ Y_2&\rightarrow 2Y_1+Y_2+Z_2,\\ Z_1&\rightarrow Z_1+\frac1{2}W,\\ Z_2&\rightarrow -Z_1-Z_2+\frac1{4}W,\\ W&\rightarrow -W. \endalign $$ A straightforward calculation shows that $\Phi_*([U,V])=[\Phi_*(U), \Phi_*(V)]$ for all $U,V$ in $\gg.$ Thus $\Phi$ is indeed a Lie group automorphism. \bigskip Let $\G1$ be the cocompact, discrete subgroup of $G$ generated by $$\{exp(2X_1),exp(2X_2),exp(Y_1),exp(Y_2),exp(Z_1),exp(Z_2),exp(W)\},$$ and let $\G2=\Phi(\G1).$ Note that $\G1 \cap Z(G) = \G2 \cap Z(G) = \{exp(jW):j\in\Z\}.$ \bigskip Let $\bar\Phi$ be the projection of $\Phi$ onto $\Gbar.$ Then $\bar\Phi$ factors as $\bar\Phi = \Psi_1 \circ \Psi_2$ where $\Psi_1$ is the automorphism of $\Mbar$ given on the Lie algebra level by $$\align \BX_1 &\rightarrow -\BX_1+\BX_2+\frac{1}{4}\BY_1+\frac{1}{2}\BY_2, \\ \BX_2 &\rightarrow \BX_2-\frac{1}{2}\BY_1,\\ \BY_1 &\rightarrow -\BY_1,\\ \BY_2 &\rightarrow 2\BY_1+\BY_2,\\ \BZ_1 &\rightarrow \BZ_1,\\ \BZ_2 &\rightarrow -\BZ_1-\BZ_2, \endalign$$ and $\Psi_2$ is the automorphism of $\Mbar$ given on the Lie algebra level by $$\align \BX_1 &\rightarrow \BX_1,\\ \BX_2 &\rightarrow \BX_2+\frac{1}{4}\BZ_1,\\ \BY_1 &\rightarrow \BY_1,\\ \BY_2 &\rightarrow \BY_2-\BZ_1-\BZ_2,\\ \BZ_1 &\rightarrow \BZ_1,\\ \BZ_2 &\rightarrow \BZ_2. \endalign$$ \medskip By rewriting $\Psi_1$ in terms of the orthonormal basis $\{\BE_1, \BE_2, \BE_3, \BE_4, \BE_5, \BE_6 \}$ of $\gggbar,$ one easily sees that $\Psi_1(\BE_i)=\pm\BE_i$ for $i=1, \dots, 6.$ Thus the automorphism $\Psi_1$ is also an isometry of $\Gbar$ and must preserve the spectrum. A simple calculation shows that $\Psi_2$ is an almost inner automorphism of $\Gbar,$ which by Theorem 4.2 also preserves the spectrum. Thus $\spec(\nilmfldbar1, \gbar) = \spec(\nilmfldbar2, \gbar).$ By Theorem 3.2, $\spec(\nilmfld1, g) = \spec(\nilmfld2, g).$ \bigskip \proclaim{Proposition 4.11} The manifolds $(\nilmfld1, g)$ and $(\nilmfld2, g)$ are not isospectral on one-forms. \endproclaim \bigskip The proof of Proposition 4.11 is left to the Appendix. \bigskip \Remark We will show in \cite{Gt4} that the automorphism $\Phi$ marks the length spectrum of these examples. This is the first example of a pair of manifolds with the same marked length spectrum but not the same spectrum on one-forms. \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \bigskip \subheading{Appendix: Comparing the $p$-form spectrum of nilmanifolds} \bigskip In this appendix, we show that the pairs of isospectral manifolds in Examples III, IV, and V are not isospectral on one-forms. \bigskip Recall that on smooth $p$-forms, the Laplace-Beltrami operator is defined as $$\Delta = d \delta + \delta d.$$ Here $\delta$ is the metric adjoint of $d.$ Equivalently, $\delta = (-1)^{n(p+1)+1}*d*$ where $*$ is the Hodge-$*$ operator. Let $E^p(M)$ denote the exterior algebra of smooth differential $p$-forms on $M.$ Then for $f \in C^\infty(M)$ and $\tau \in E^p(M),$ Gordon and Wilson \cite {GW1} showed that $$\Delta(f \tau) = (\Delta f)\tau + f(\Delta \tau) - 2\nabla_{\text{\it grad } f}\tau.\tag{A.1}$$ \bigskip For $G$ a simply connected Lie group with cocompact, discrete subgroup $\GG,$ view $E^p(\GG \backslash G)$ as $$E^p(\GG \backslash G) = C^\infty(\GG \backslash G) \otimes \Lambda^p(\gg^*).$$ Here elements of $\Lambda^p(\gg^*)$ are viewed both as left invariant $p$-forms of $G$ and also as elements of $E^p(\GG \backslash G).$ \bigskip \proclaim{Proposition 4.7} The nilmanifolds $(\nilmfld1, g)$ and $(\nilmfld2,g)$ as presented in Example III are not isospectral on one-forms. \endproclaim \bigskip \demo{Outline of Proof} \medskip Step 1: We decompose $\one \spec(\nilmfld{i}, g)$ into four components: $$\one \spec^{I}(\nilmfld{i}, g) \cup \one \spec^{II}(\nilmfld{i}, g) \cup \one \spec^{III}(\nilmfld{i}, g) \cup \one \spec^{IV}(\nilmfld{i},g).$$ The multiplicity of an eigenvalue in $\one \spec(\nilmfld{i},g)$ is the sum of its multiplicities in each of the four components. \bigskip Step 2: Using representation theory, we show $$\one \spec^{IV}(\nilmfld1, g)=\one \spec^{IV}(\nilmfld2, g)$$ and $$\one \spec^{III}(\nilmfld1,g) = \one \spec^{III}(\nilmfld2,g).$$ \bigskip Step 3: We show that the multiplicity of every eigenvalue in $\one \spec^{II}(\nilmfld{i}, g)$ is congruent to $0$ modulo $4.$ \bigskip Step 4: Finally, we show that the eigenvalue $\pi^2 + 1$ does not occur in $\one \spec^I(\nilmfld1, g)$ but occurs with multiplicity 2 in $\one \spec^I(\nilmfld2, g).$ \bigskip The result now follows. $\endpf$ \enddemo \bigskip \demo{Proof of Proposition 4.7} \demo{Step 1} \bigskip Using the notation of Section 2, for $i=1,2,$ let $\Tau_{i}$ be a subset of $\mm^*$ such that $$\fnsp{i} \cong {\underset{\tau\in\Tau_{i}}\to\bigoplus} {m_i(\tau)}\H_\tau.$$ \medskip Let $\Tau_i = \Tau^{I}_i \cup \Tau^{II}_i \cup \Tau^{III}_i \cup \Tau^{IV}_i$ where $$\align \Tau^{I}_i &= \{\tau \in \Tau_i : \tau(Z_1) = \tau(Z_2) = \tau(W) = 0\}, \\ \Tau^{II}_i &= \{\tau \in \Tau_i : \tau(Z_2) \neq 0, \tau(Z_1) = \tau(W) = 0 \},\\ \Tau^{III}_i &= \{\tau \in \Tau_i : \tau(Z_1) \neq 0, \tau(W) = 0\},\\ \Tau^{IV}_i &= \{\tau \in \Tau_i : \tau(W) \neq 0 \}. \endalign$$ \medskip Let $$\align \H^{I}_i ={\underset{\tau\in\Tau^{I}_{i}}\to\bigoplus}{m_i(\tau)}\H_\tau , &\qquad \H^{II}_i ={\underset{\tau\in\Tau^{II}_{i}}\to\bigoplus}{m_i(\tau)}\H_\tau,\\ \H^{III}_i ={\underset{\tau\in\Tau^{III}_{i}}\to\bigoplus}{m_i(\tau)}\H_\tau, &\qquad \H^{IV}_i ={\underset{\tau\in\Tau^{IV}_{i}}\to\bigoplus}{m_i(\tau)}\H_\tau. \endalign$$ \medskip As representation spaces, $$\fnsp{i} = \H^{I}_i \oplus \H^{II}_i \oplus \H^{III}_i \oplus \H^{IV}_i.$$ \medskip Decompose $\one \spec(\nilmfld{i},g)$ as $$\one \spec^{I}(\nilmfld{i}, g) \cup \one \spec^{II}(\nilmfld{i}, g) \cup \one \spec^{III}(\nilmfld{i}, g) \cup \one \spec^{IV}(\nilmfld{i},g),$$ where $\one \spec^I(\nilmfld{i}, g)$ is defined as the spectrum of the Laplacian %\break acting on $\H^I_i \otimes \Lambda^1(\mm^*).$ Define $\one \spec^{II}(\nilmfld{i},g), \ \one \spec^{III}(\nilmfld{i},g),$ and %\break $\one \spec^{IV}(\nilmfld{i},g)$ similarly. The multiplicity of an eigenvalue in $\one \spec(\nilmfld{i}, g)$ is equal to the sum of its multiplicities in each of the four components. \enddemo%{Step One} \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \demo{Step 2} \medskip By Lemma 3.4, the representations of $G$ on $\H^{IV}_1$ and $\H^{IV}_2$ are unitarily equivalent and $$\one\spec^{IV}(\nilmfld1, g) = \one\spec^{IV}(\nilmfld2, g).$$ \medskip We now show that the representations of $G$ on $\H^{III}_1$ and $\H^{III}_2$ are unitarily equivalent. \medskip The irreducible representations of $G$ corresponding to elements of $\mm^*$ that are zero on the center may be viewed as irreducible representations of $\Mbar = G / Z(G).$ It is easy to see that such representations of $G$ are unitarily equivalent if and only if the corresponding representations of $\Mbar$ are unitarily equivalent. \medskip For all $\tau \in \Tau^{III}_i, \tau(\xi)=0.$ We may thus use the following Proposition to calculate the representations of $G$ on $\H^{III}_1$ and $\H^{III}_2.$ \bigskip \proclaim{Proposition A.2 (see Pesce \cite{P2})} Let $N$ be a simply connected, two-step nilpotent Lie group with $\GG$ a cocompact, discrete subgroup of $N.$ Let $\tau \in \nn^*.$ Define $\nn_\tau = \{ Y \in \nn : \tau([Y,\nn]) \equiv 0 \}.$ Then the irreducible representation $\pi_\tau$ appears in the quasi-regular representation of $N$ on $L^2(\GG \backslash N)$ if and only if $\tau(log \GG \cap \nn_\tau) \subset \Z.$ The multiplicity of $\pi_\tau$ is one if $\tau(\nn^{(1)}) \equiv 0.$ Define a nondegenerate, skew-symmetric bilinear form on $\nn / \nn_\tau$ by $B_\tau(U,V)=\tau([U,V])$ for all $U,V \in \nn / \nn_\tau.$ Then if $\tau(\nn^{(1)}) \not \equiv 0,$ the multiplicity of $\pi_\tau$ is equal to $\sqrt{det B_\tau},$ where the determinant is calculated with respect to any basis of ${\Cal L}_\tau = log \GG /( log \GG \cap \nn_\tau ).$ \endproclaim \bigskip \Remark The occurrence condition above actually follows directly from a more general occurrence and multiplicity theorem due independently to Richardson \cite {R} and Howe \cite{H}. \bigskip Let $\{\alpha_1, \alpha_2, \beta_1, \beta_2, \zeta_1, \zeta_2, \omega\}$ be the dual basis to the orthonormal basis $\{X_1, X_2, Y_1, Y_2, \allowbreak Z_1, \allowbreak Z_2, W\}$ of $\mm.$ \medskip If $\tau \in \Tau^{III}_i,$ then $$\tau = A_1\alpha_1 + A_2 \alpha_2 + B_1 \beta_1 + B_2 \beta_2 + C_1 \zeta_1 + C_2 \zeta_2$$ for some $A_1, A_2, B_1, B_2, C_1, C_2 \in \R$ with $C_1 \neq 0.$ \medskip Now $\mbar_\tau = span_\R\{\BZ_1,\BZ_2\}.$ Hence $$\split log\GGGbar1 \cap \mbar_\tau &= log\GGGbar2 \cap \mbar_\tau \\ & = log(exp(\Z \BZ_1)exp(\Z \BZ_2))\\ & = span_\Z\{\BZ_1,\BZ_2\}. \endsplit$$ So $\tau(log\GGGbar{i} \cap \mbar_\tau)\subset \Z$ if and only if $C_1 \in \Z$ and $C_2 \in \Z.$ By Proposition A.2, we see that $\tau \in \Tau^{III}_i$ if and only if $C_1 \in \Z$ and $C_2 \in \Z.$ Moreover, distinct values of $C_1$ and $C_2$ determine distinct coadjoint orbits of $\mm^*.$ As these conditions are the same for both $\Tau^{III}_1$ and $\Tau^{III}_2,$ we may assume $\Tau^{III}_1 = \Tau^{III}_2.$ \medskip We now calculate multiplicities. \medskip A basis for ${\Cal L_1}_\tau = log\GGGbar1 / (log\GGGbar1 \cap \mbar_\tau)$ is $\{2\BX_1, 2\BX_2, \BY_1, \BY_2\}.$ A basis for \break ${\Cal L_2}_\tau = log\GGGbar2 / (log\GGGbar2 \cap \mbar_\tau)$ is $\{\BX_1, \BX_2, 2\BY_1, 2\BY_2\}.$ In both cases, $\sqrt{det B_\tau}= 4{C_1}^2.$ Thus for $\tau \in \Tau^{III}_1 = \Tau^{III}_2,$ the multiplicities $m_1(\tau)$ and $m_2(\tau)$ are equal. Hence the representations of $\Mbar$ are unitarily equivalent, so the representations of $G$ on $\H^{III}_1$ and $\H^{III}_2$ are unitarily equivalent. \bigskip By Proposition 2.2, $$\one spec^{III}(\nilmfld1, g) = \one spec^{III}(\nilmfld2, g),$$ as desired. \enddemo \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \demo{Step 3} \medskip We now show that for any eigenvalue in $\one spec^{II}(\nilmfld{i}, g),$ its multiplicity in \break $\one spec^{II}(\nilmfld{i}, g)$ is always congruent to 0 modulo 4. \medskip We first compute the multiplicity of the irreducible representations occurring here, using the same technique as in Step 2. \medskip For all $\tau \in \Tau^{II}_i,$ $\tau = A_1\alpha_1 + A_2 \alpha_2 + B_1 \beta_1 + B_2 \beta_2 + C_2 \zeta_2$ for some $A_1, \ab A_2, \ab B_1, \ab B_2, \ab C_2\in\R$ with $C_2 \neq 0.$ We may again use Proposition A.2. \medskip Now $\mbar_\tau = span_\R\{\BX_2,\BY_1, \BZ_1,\BZ_2\}.$ Hence $$\split log\GGGbar1 \cap \mbar_\tau &= log(exp(2\Z \BX_2)exp(\Z \BY_1)exp(\Z \BZ_1)exp(\Z \BZ_2))\\ & = span_\Z\{2\BX_2, \BY_1, \BZ_1,\BZ_2\}. \endsplit$$ So $\tau(log\GGGbar1 \cap \mbar_\tau)\subset \Z$ if and only if $A_2 \in \frac1{2}\Z, \ B_1 \in \Z$ and $C_2 \in \Z.$ \medskip However, $$\split log\GGGbar2 \cap \mbar_\tau &= log(exp(\Z \BX_2)exp(2\Z \BY_1)exp(\Z \BZ_1)exp(\Z \BZ_2))\\ & = span_\Z\{\BX_2, 2\BY_1, \BZ_1,\BZ_2\}. \endsplit$$ So $\tau(log\GGGbar2 \cap \mbar_\tau)\subset \Z$ if and only if $A_2 \in \Z, \ B_1 \in \frac1{2}\Z$ and $C_2 \in \Z.$ \medskip We now calculate $m_i(\tau).$ A basis for ${\Cal L_1}_\tau = log\GGGbar1 / (log\GGGbar1 \cap \mbar_\tau)$ is $\{2\BX_1, \BY_2\}.$ A basis for ${\Cal L_2}_\tau = log\GGGbar2 / (log\GGGbar2 \cap \mbar_\tau)$ is $\{\BX_1, 2\BY_2\}.$ In both cases $\sqrt{det B_\tau}= 2|C_2|.$ \medskip Thus if $\tau$ is in $\Tau^{II}_i,$ the irreducible representation $\pi_\tau$ occurs in the representation of $G$ on $\H^{II}_i$ with multiplicity $2|C_2|.$ As the integer $C_2 \neq 0,$ the multiplicity must be even. Hence any eigenvalue of $\Delta$ acting on $\H^{II}_i \otimes \Lambda^1(\mm^*)$ must occur in $\one spec^{II}(\nilmfld{i},g)$ with multiplicity congruent to 0 modulo $2.$ \medskip We now use the following. \medskip \proclaim{Proposition A.3 (Gordon--Wilson, \cite{GW1})} Let $G$ be a simply connected Lie group with left invariant metric $g,$ and let $\G1$ and $\G2$ be cocompact, discrete subgroups of $G.$ Let $\H_1$ and $\H_2$ be invariant subspaces of $\rep1$ and $\rep2,$ respectively. Denote by $p\hy\spec'(\nilmfld{i},g)$ the spectrum of $\Delta$ restricted to acting on $\H_i\otimes\Lambda^p(\gg^*).$ Assume there exists an automorphism $\Phi$ of $G$ such that \flushpar (1) $\Phi$ is also an isometry of $(G,g),$ \ and \flushpar (2) $\rep1$ restricted to $\H_1$ and $\rep2\circ\Phi$ restricted to $\H_2$ are unitarily equivalent, then $$p\hy\spec'(\nilmfld1,g)= p\hy\spec'(\nilmfld2,g).$$ Here $\rep2\circ\Phi$ is defined by $((\rep2\circ\Phi)(x))f=f\circ R_{\Phi(x)}$ for $x$ in $G$ and $f$ in $\fnsp1.$ \endproclaim \bigskip Let $\Phi$ be the Lie group automorphism of $G$ defined on the Lie algebra level by $$\align X_1 & \rightarrow -X_1\\ X_2 & \rightarrow X_2\\ Y_1 & \rightarrow -Y_1\\ Y_2 & \rightarrow Y_2\\ Z_1 & \rightarrow Z_1\\ Z_2 & \rightarrow -Z_2\\ W & \rightarrow -W\endalign$$ \medskip The automorphism $\Phi$ is also an isometry of $(G,g).$ Note that $\tau \circ \Phi_* = -A_1\alpha_1 + A_2\alpha_2 - B_1\beta_1 + B_2\beta_2 -C_2\zeta_2.$ Clearly, if $\tau$ satisfies Condition ($*$) or ($**$), then so does $\tau \circ \Phi_*.$ A straightforward calculation shows that since $C_2 \neq 0,$ the functionals $\tau$ and $\tau \circ \Phi_*$ are not in the same coadjoint orbit of $\mm^*,$ so $\pi_\tau$ and $\pi_{\tau \circ \Phi_*}$ are not unitarily equivalent. Thus if $\pi_\tau$ occurs in $\H^{II}_i$ with multiplicity $2|C_2|$ then so does $\pi_{\tau \circ \Phi_*},$ also with multiplicity $2|C_2|.$ \medskip Note that by (2.3) $\pi_{\tau \circ \Phi_*} = \pi_\tau \circ \Phi.$ Using Proposition A.3, any eigenvalue of $\Delta$ acting on $\H_\tau \otimes \Lambda^1(\mm^*)$ must also occur as an eigenvalue of $\Delta $ acting on $\H_{\tau \circ \Phi_*} \otimes \Lambda^1(\mm^*).$ And each of the representation spaces $\H_\tau$ and $\H_{\tau\circ\Phi_*}$ occurs in $\H^{II}_i$ with multiplicity $2|C_2|.$ Consequently, the multiplicity of any eigenvalue in $spec^{II}(\nilmfld{i},g)$ is a multiple of $4|C_2|,$ which is clearly congruent to 0 modulo 4, as desired. \enddemo \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \demo{Step 4} \medskip We now show that the eigenvalue $\pi^2 + 1$ does not occur in $\one spec^I(\nilmfld1, g)$ but occurs with multiplicity 2 in $\one spec^I(\nilmfld2, g).$ \medskip For $\tau \in \Tau^I_1$ or $\tau \in \Tau^I_2,\ \tau(\g1) \equiv 0.$ We again use Proposition A.2 to calculate the irreducible representations occurring here. We write $\tau = A_1\alpha_1 + A_2\alpha_2 + B_1\beta_1 + B_2\beta_2$ for some $A_1, A_2, B_1, B_2 \in \R.$ \medskip Now $\mbar_\tau = \mbar,$ so $\tau(log\GGGbar{i} \cap \mbar_\tau) \subset \Z$ if and only if $\tau(log \GGGbar{i}) \subset \Z.$ \medskip Thus, $\tau \in \Tau^I_1$ if and only if $$A_1,A_2 \in \frac1{2}\Z \text{ and } B_1,B_2 \in \Z,\tag{$*$}$$ and $\tau \in \Tau^I_2$ if and only if $$A_1,A_2 \in \Z \text{ and } B_1,B_2 \in \frac1{2}\Z.\tag{$**$}$$ \medskip Let $\H_\tau$ be the associated representation space of $\pi_\tau.$ Then $\H_\tau$ may be viewed as the one-dimensional subspace of $\fnsp{i}$ generated by (see Section 2): $$\split \F(exp(x_1X_1)&exp(x_2X_2)exp(y_1Y_1)exp(y_2Y_2)exp(z_1Z_1)exp(z_2Z_2)exp(wW)) \\ &= exp\{2\pi i \tau(x_1X_1 + x_2X_2 + y_1Y_1 + y_2Y_2) \}.\endsplit$$ That is, $\H_\tau = \Cplx \F.$ \medskip We now calculate $\Delta$ acting on $\H_\tau \otimes \Lambda^1(\mm^*).$ Note that if we let $\Cplx \Lambda^1(\mm^*)$ denote $\Lambda^1(\mm^*)$ with complex coefficients, then $$\H_\tau \otimes \Lambda^1(\mm^*) = \F \otimes \Cplx \Lambda^1(\mm^*).$$ \medskip Let $\F \otimes \mu \in \F \otimes \Cplx \Lambda^1(\mm^*).$ Then $\mu = a_1\alpha_1 + a_2\alpha_2 + b_1\beta_1 + b_2\beta_2 + z_1\zeta_1 + z_2\zeta_2 + w\omega$ for some $a_1, a_2, b_1, b_2, z_1, z_2, w \in \Cplx.$ \medskip As $\F$ is independent of $z_1, z_2,$ and $w$ we have $$ \align \Delta \F &= - X_1^2\F - X_2^2\F - Y_1^2\F - Y_2^2\F\\ &= 4\pi^2(A_1^2 + A_2^2 + B_1^2 + B_2^2)\F\\ &= 4\pi^2S^2\F \endalign$$ where $S^2 = A_1^2 + A_2^2 + B_1^2 + B_2^2.$ \medskip Let $*$ denote the Hodge-$*$ operator. One easily sees that $d*\mu = 0$ for all $\mu \in \mm^*.$ Hence $\delta\mu = \pm*d*\mu = 0$ for all $\mu \in \mm^*.$ Consequently $\Delta = \delta d$ on $\Lambda^1(\mm^*).$ \medskip For $\mu\in \Lambda^1(\mm^*),$ $d\mu(U,V)=-\mu([U,V])$ for all $U,V \in \mm.$ Using this fact together with the definition of $\delta$ as the metric adjoint of $d,$ one easily computes that $\Delta\alpha_1 =\Delta\alpha_2 = \Delta\beta_1 = \Delta\beta_2 =0,$ $\Delta \zeta_1 = 2\zeta_1, \Delta\zeta_2 = \zeta_2,$ and $\Delta \omega = 3\omega.$ \medskip To calculate $\Delta (\F \otimes \mu),$ it remains to calculate $$\nabla_{grad\F}\mu = 2\pi i\F(A_1\nabla_{X_1}\mu + A_2\nabla_{X_2}\mu + B_1\nabla_{Y_1}\mu + B_2\nabla_{Y_2}\mu).$$ \medskip For Lie algebras with a left invariant metric, the covariant derivatives can be calculated via the following equation: $$<\nabla_UV,U'> =\frac1{2} <[U',U],V> + \frac1{2}<[U',V],U> + \frac1{2}<[U,V],U'>,$$ for $U,V,U'$ left invariant vector fields of $\mm.$ A simple calculation shows that $\nabla_U(V^\flat) = (\nabla_UV)^\flat,$ where $V^\flat$ denotes the dual of $V$ in $\mm^*$ with respect to our choice of orthonormal basis; that is, $V^\flat(U) = $ for all $U$ in $\mm^*.$ \medskip We thus obtain the following chart: {\eightpoint $$ \vbox{\offinterlineskip \halign{\strut\vrule#&\quad#\hfil\quad&&\vrule#&\quad\hfil#\hfil\quad\cr \noalign{\hrule} &&&&&&&&&&&&&&&&\cr &$\nabla_U\mu$&&$\alpha_1$&&$\alpha_2$&&$\beta_1$&&$\beta_2$&&$\zeta_1$&&$\zeta_2$&&$\omega$&\cr &&&&&&&&&&&&&&&&\cr \noalign{\hrule} \noalign{\hrule} &&&&&&&&&&&&&&&&\cr &$X_1$&&0&&0&&$\frac1{2}\zeta_1$&&$\frac1{2}\zeta_2$&&$-\frac1{2}\beta_1+\frac1{2}\omega$&&$-\frac1{2}\beta_2$&&$-\frac1{2}\zeta_1$&\cr &&&&&&&&&&&&&&&&\cr \noalign{\hrule} &&&&&&&&&&&&&&&&\cr &$X_2$&&0&&0&&0&&$\frac1{2}\zeta_1$&&$-\frac1{2}\beta_2$&&$\frac1{2}\omega$&&$-\frac1{2}\zeta_2$&\cr &&&&&&&&&&&&&&&&\cr \noalign{\hrule} &&&&&&&&&&&&&&&&\cr &$Y_1$&&$-\frac1{2}\zeta_1$&&0&&0&&$\frac1{2}\omega$&&$\frac1{2}\alpha_1$&&0&&$-\frac1{2}\beta_2$&\cr &&&&&&&&&&&&&&&&\cr \noalign{\hrule} &&&&&&&&&&&&&&&&\cr &$Y_2$&&$-\frac1{2}\zeta_2$&&$-\frac1{2}\zeta_1$&&$-\frac1{2}\omega$&&0&&$\frac1{2}\alpha_2$&&$\frac1{2}\alpha_1$&&$\frac1{2}\beta_1$&\cr &&&&&&&&&&&&&&&&\cr \noalign{\hrule} }} $$ } \bigskip Using (A.1) and the above information, a straightforward calculation shows that if we let $E_\tau =$ $$\left(\matrix 4\pi^2S^2 & 0 & 0 & 0 & -2\pi i B_1 & -2\pi i B_2 & 0\\ 0 & 4 \pi^2S^2 & 0 & 0 & -2 \pi i B_2 & 0 & 0\\ 0 & 0 & 4 \pi^2S^2 & 0 & 2 \pi i A_1 & 0 & -2 \pi i B_2\\ 0 & 0 & 0 & 4 \pi^2S^2 & 2 \pi i A_2& 2 \pi i A_1 & 2\pi i B_1\\ 2\pi i B_1 &2 \pi i B_2 & -2 \pi i A_1 & -2 \pi i A_2 & 4 \pi ^2 S^2 + 2 &0 & 2 \pi i A_1 \\ 2\pi i B_2 & 0 & 0 & -2 \pi i A_1 & 0 & 4 \pi ^2 S^2 + 1 & 2 \pi i A_2 \\ 0 & 0 & 2 \pi i B_2 & -2 \pi i B_1 & -2 \pi i A_1 & -2 \pi i A_2 & 4 \pi^2S^2 + 3 \endmatrix \right),$$ then $\Delta(\F \otimes \mu) = \lambda(\F \otimes \mu)$ if and only if $\lambda$ is an eigenvalue of the matrix $E_\tau.$ \medskip We now calculate necessary conditions on $\tau=A_1\alpha_1+A_2\alpha_2+B_1\beta_1+B_2\beta_2$ so that $\pi^2 + 1$ is an eigenvalue of $E_\tau.$ As $\tau \in \Tau^I_1$ or $\tau \in \Tau^I_2,$ we know $A_1, A_2,B_1, B_2 \in \Q.$ If $det(E_\tau - (\pi^2+1)I_7) = 0,$ then $\pi$ is the root of a polynomial with rational coefficients. However $\pi$ is transcendental. Thus the coefficients of the powers of $\pi$ must be zero. \medskip A straightforward calculation shows that $\pi^{14}$ is the highest power of $\pi$ occurring in the polynomial, and the coefficient of $\pi^{14}$ is equal to $(4S^2-1)^7.$ Thus if $\pi^2+1$ is an eigenvalue of $E_\tau,$ then $S^2 = \frac1{4}.$ Recall $S^2 = A_1^2 + A_2^2 + B_1^2 + B_2^2.$ \medskip For $\tau$ in $\Tau^I_1,$ \ $S^2=\frac{1}{4}$ if and only if (see ($*$)) $\tau = \pm \frac1{2}\alpha_1$ or $\tau = \pm \frac1{2}\alpha_2.$ And for $\tau$ in $\Tau^I_2,$ \ $S^2=\frac{1}{4}$ if and only if (see ($**$)) $\tau = \pm \frac1{2}\beta_1$ or $\tau = \pm \frac1{2}\beta_2.$ \medskip For $\tau = \pm\frac1{2}\alpha_1, \tau =\pm\frac1{2}\alpha_2,$ or $\tau = \pm\frac1{2}\beta_2,$ a simple calculation shows that $det(E_\tau - (\pi^2+1)I_7) \neq 0.$ Thus, the eigenvalue $\pi^2 + 1$ does not arise from the Laplacian acting on $\H_{\pm\frac1{2}\alpha_1} \otimes \Lambda^1(\mm^*),$ $\H_{\pm\frac1{2}\alpha_2} \otimes \Lambda^1(\mm^*),$ or $\H_{\pm\frac1{2}\beta_2} \otimes \Lambda^1(\mm^*),$ and $$\pi^2 + 1 \not \in \one spec^I(\nilmfld1,g).$$ \medskip However for $\tau = \pm\frac1{2}\beta_1,$ $det(E_\tau - (\pi^2+1)I_7)=0.$ Thus $\pi^2+1$ is an eigenvalue for the Laplacian acting on $\H_{\pm\frac1{2}\beta_1} \otimes \Lambda^1(\mm^*).$ Indeed, the eigenspace of $\pi^2+1$ in $\H^I\otimes\Lambda^1(\mm^*)$ is $$span_\Cplx\{F_{\frac{1}{2}\beta_1}\otimes\zeta_2\ , \ F_{-\frac{1}{2}\beta_1}\otimes\zeta_2 \},$$ which has dimension 2. \medskip Thus $\pi^2 + 1 \in \one spec^I(\nilmfld2,g)$ with multiplicity 2, as desired. \enddemo%{Step 4} \bigskip The proof of Proposition 4.7 is now complete. \endpf \enddemo%{Proof of Example III} \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \proclaim{Proposition 4.9} The nilmanifolds $(\nilmfld1, g)$ and $(\nilmfld2,g)$ as presented in Example IV are not isospectral on one-forms. \endproclaim \bigskip \pagebreak \demo{Proof of Proposition 4.9} \medskip Again using the notation of Section 2, for $i=1,2,$ let $\Tau_{i}$ be a subset of $\mm^*$ such that $$\fnsp{i} \cong {\underset{\tau\in\Tau_{i}}\to\bigoplus} {m_i(\tau)}\H_\tau.$$ \medskip Let $\Tau_i = \Tau'_i \cup \Tau''_i \cup \Tau'''_i,$ where $$\align \Tau'_{i} &= \{\tau \in \Tau_{i} : \tau(\m1) \equiv 0\}, \\ \Tau''_{i} &= \{\tau \in \Tau_{i} : \tau(\m2) \equiv 0, \tau(\m1) \not \equiv 0\},\\ \Tau'''_{i} &= \{\tau \in \Tau_{i} : \tau(\m2) \not \equiv 0\}. \endalign$$ As in the proof of Proposition 4.7, decompose the representation spaces and spectrum accordingly. \medskip By Lemma 3.4, the representations of $G$ on $\H'''_1$ and $\H'''_2$ are unitarily equivalent and $$\one \spec'''(\nilmfld1, g) = \one \spec'''(\nilmfld2, g).$$ \medskip A calculation almost identical to Step 2 in the proof of Proposition 4.7 shows that the representations of $G$ on $\H''_1$ and $\H''_2$ are unitarily equivalent. Thus $$\one \spec''(\nilmfld1, g) = \one \spec''(\nilmfld2,g).$$ \medskip It remains to show that $\one \spec'(\nilmfld1,g) \neq \one \spec'(\nilmfld2,g).$ The proof of this corresponds to Step 4 in the proof of Proposition 4.7. \medskip Let $\{\alpha_1, \beta_1, \beta_2, \gamma , \omega \}$ be the dual to the orthonormal basis $\{X_1,Y_1, Y_2, Z, W\}$ given in Example IV. For $\tau \in \Tau'_i, \ \tau = A_1\alpha_1 + B_1\beta_1 + B_2\beta_2$ for some $A_1, B_1, B_2 \in \R.$ We again use Proposition A.2. \medskip Now $\mbar_\tau = \mbar.$ Hence $\tau(log\GGGbar{i} \cap \mbar_\tau) \subset \Z$ if and only if $\tau(log \GGGbar{i}) \subset \Z.$ \medskip Thus, $\tau \in \Tau^I_1$ if and only if $$A_1 \in \frac1{2}\Z \quad \text{ and } \quad B_1,B_2 \in \Z,\tag{$*$}$$ and, $\tau \in \Tau^I_2$ if and only if $$A_1,B_2 \in \Z \quad \text{ and } \quad B_1 \in \frac1{2}\Z.\tag{$**$}$$ \medskip Let $\H_\tau$ be the associated representation space of $\pi_\tau.$ As in the proof of Proposition 4.7, $\H_\tau \otimes \Lambda^1(\mm^*) = \F \otimes \Cplx \Lambda^1(\mm^*)$ where $$\split \F(exp(x_1X_1)exp(y_1&Y_1)exp(y_2Y_2)exp(zZ)exp(wW))\\ & = exp\{2\pi i \tau(x_1X_1 + y_1Y_1 + y_2Y_2) \}.\endsplit$$ \medskip Let $\F \otimes \mu \in \F \otimes \Cplx \Lambda^1(\mm^*)$ with $\mu = a_1\alpha_1 + b_1\beta_1 + b_2\beta_2 + z\zeta + w\omega$ for some $a_1, b_1, b_2, z, w \in \Cplx.$ \medskip Then $\Delta \F = 4\pi^2S^2\F$ where $S^2 = A_1^2 + B_1^2 + B_2^2.$ And $\Delta\alpha_1 = \Delta\beta_1 = \Delta\beta_2 = 0,$ $\Delta\zeta = \zeta$ and $\Delta\omega=2\omega.$ We also obtain the following chart: \medskip {\eightpoint $$ \vbox{\offinterlineskip \halign{\strut\vrule#&\quad#\hfil\quad&&\vrule#&\quad\hfil#\hfil\quad\cr \noalign{\hrule} &&&&&&&&&&&&\cr &$\nabla_U\mu$&&$\alpha_1$&&$\beta_1$&&$\beta_2$&&$\zeta$&&$\omega$&\cr &&&&&&&&&&&&\cr \noalign{\hrule} \noalign{\hrule} &&&&&&&&&&&&\cr &$X_1$&&0&&$\frac1{2}\zeta$&&0&&$-\frac1{2}\beta_1+\frac1{2}\omega$&&$-\frac1{2}\zeta$&\cr &&&&&&&&&&&&\cr \noalign{\hrule} &&&&&&&&&&&&\cr &$Y_1$&&$-\frac1{2}\zeta$&&0&&$\frac1{2}\omega$&&$\frac1{2}\alpha_1$&&$-\frac1{2}\beta_2$&\cr &&&&&&&&&&&&\cr \noalign{\hrule} &&&&&&&&&&&&\cr &$Y_2$&&0&&$-\frac1{2}\omega$&&0&&0&&$\frac1{2}\beta_1$&\cr &&&&&&&&&&&&\cr \noalign{\hrule} }} $$ } \medskip Using (A.1) and the above information, if we let $$E_\tau=\left(\matrix 4\pi^2S^2 & 0 & 0 & -2\pi i B_1 & 0\\ 0 & 4 \pi^2S^2 & 0 & 2\pi i A_1 & -2 \pi i B_2\\ 0 & 0 & 4 \pi^2S^2 & 0 & 2\pi i B_1\\ 2\pi i B_1 & -2 \pi i A_1 & 0 & 4 \pi^2 S^2 + 1 & 2 \pi i A_1 \\ 0 & 2 \pi i B_2 & -2 \pi i B_1 & -2 \pi i A_1 & 4 \pi^2S^2 + 2 \endmatrix \right),$$ then $\Delta(\F \otimes \mu) = \lambda(\F \otimes \mu)$ if and only if $\lambda$ is an eigenvalue of the matrix $E_\tau.$ \medskip We calculate necessary conditions on $\tau=A_1\alpha_1+B_1\beta_1+B_2\beta_2$ so that $\pi^2+1$ is an eigenvalue of $E_\tau.$ As we assumed $\tau\in\Tau'_1$ or $\tau\in\Tau'_2,$ we know $A_1, B_1, B_2 \in \Q.$ If $det(E_\tau - (\pi^2+1)I_5) = 0,$ then $\pi$ is the root of a polynomial with rational coefficients. However $\pi$ is transcendental. Thus the coefficients of the powers of $\pi$ must be zero. \medskip A straightforward calculation shows that $\pi^{10}$ is the highest power of $\pi$ occurring in the polynomial, and the coefficient of $\pi^{10}$ is equal to $(4S^2-1)^5.$ Thus if $\pi^2+1$ is an eigenvalue of $E_\tau,$ then $S^2 = \frac1{4}.$ Recall $S^2 = A_1^2 + B_1^2 + B_2^2.$ \medskip For $\tau$ in $\Tau'_1,$ \ $S^2=\frac1{4}$ if and only if (see ($*$)) $\tau = \pm\frac1{2}\alpha_1.$ And for $\tau$ in $\Tau'_2,$ \ $S^2=\frac1{4}$ if and only if (see ($**$)) $\tau = \pm\frac1{2} \beta_1.$ \medskip For $\tau = \pm\frac1{2}\alpha_1,$ $det(E_\tau - (\pi^2+1)I_5) = 0.$ The eigenspace of $\pi^2+1$ in $\H'\otimes\Lambda^1(\mm^*)$ is $$span_\Cplx \{F_{\frac1{2}\alpha_1}\otimes(\pi i\beta_1 + \zeta + \pi i\omega), \ F_{-\frac{1}{2}\alpha_1}\otimes(\pi i\beta_1-\zeta+\pi i\omega)\},$$ which has dimension 2. Thus $\pi^2 + 1 \in \one spec'(\nilmfld1, g)$ with multiplicity two. \medskip However, for $\tau = \pm\frac1{2}\beta_1,$ \quad $det(E_\tau - (\pi^2+1)I_5) \neq 0.$ Thus $\pi^2 + 1$ cannot occur in $\one spec'(\nilmfld2, g)$ and $$\one spec'(\nilmfld1, g) \neq \one spec'(\nilmfld2, g),$$ as desired. \medskip The proof of Proposition 4.9 is now complete. $\endpf$ \enddemo \bigskip \bigskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \proclaim{Proposition 4.11} The nilmanifolds $(\nilmfld1, g)$ and $(\nilmfld2,g)$ as presented in Example V are not isospectral on one-forms. \endproclaim \bigskip \demo{Proof of Proposition 4.11} \bigskip Using the notation of Section 2, for $i=1,2,$ let $\Tau_{i}$ be a subset of $\mm^*$ such that $$\fnsp{i} \cong {\underset{\tau\in\Tau_{i}}\to\bigoplus} {m_i(\tau)}\H_\tau.$$ \medskip Let $\Tau_i = \Tau^{I}_i \cup \Tau^{II}_i \cup \Tau^{III}_i \cup \Tau^{IV}_i$ be defined as in the proof of Proposition 4.7, and decompose the representation spaces and spectrum accordingly. \medskip By Lemma 3.4, the representations of $G$ on $\H^{IV}_1$ and $\H^{IV}_2$ are unitarily equivalent and $$\one \spec^{IV}(\nilmfld1, g) = \one \spec^{IV}(\nilmfld2, g).$$ \medskip A calculation almost identical to that in Step 2 in the proof of Proposition 4.7 shows that the representations of $G$ on $\H^{III}_1$ and $\H^{III}_2$ are unitarily equivalent. Thus $$\one \spec^{III}(\nilmfld1, g) = \one \spec^{III}(\nilmfld2,g).$$ \medskip A calculation almost identical to that in Step 3 in the proof of Proposition 4.7 shows that the representations of $G$ on $\H^{II}_1$ and $\H^{II}_2$ are unitarily equivalent. Thus $$\one \spec^{II}(\nilmfld1, g) = \one \spec^{II}(\nilmfld2,g).$$ \medskip It remains to show that $\one spec^I(\nilmfld1, g) \neq \one spec^I(\nilmfld1, g).$ \medskip For $\tau \in \Tau^I_1$ or $\tau \in \Tau^I_2,\ \tau(\g1)\equiv 0. $ We again use Proposition A.2 to calculate the irreducible representations occurring here. Let $\{\epsilon_1, \cdots, \epsilon_7\}$ be the dual to the orthonormal basis $\{E_1, \cdots, E_7\}$ given in Example V. Then $\tau = A_1\epsilon_1 + A_2\epsilon_2 + A_3\epsilon_3 + A_4\epsilon_4$ for some $A_1, \cdots, A_4 \in \R.$ \medskip Now $\mbar_\tau = \mbar.$ Hence $\tau(log\GGGbar{i} \cap \mbar_\tau) \subset \Z$ if and only if $\tau(log \GGGbar{i}) \subset \Z.$ \medskip Thus, $\tau \in \Tau^I_1$ if and only if $$2A_1+A_2-\frac{1}{4}A_3+\frac{1}{2}A_4\in\Z, \quad 2A_2+\frac{1}{2}A_3\in\Z,\quad\text{and}\quad A_3,A_4 \in \Z.\tag{$*$}$$ And $\tau \in \Tau^I_2$ if and only if $$2A_1+A_2+\frac{1}{4}A_3+\frac{1}{2}A_4\in\Z,\quad 2A_2+\frac{1}{2}A_3\in\Z,\quad\text{and}\quad A_3,A_4 \in \Z. \tag{$**$}$$ Note that the only distinction between the two conditions is in the sign of $\frac{1}{4}A_3.$ \medskip Let $\H_\tau$ be the associated representation space of $\pi_\tau.$ As in the proof of Proposition 4.7, $\H_\tau \otimes \Lambda^1(\mm^*) = \F \otimes \Cplx \Lambda^1(\mm^*)$ where $$\split \F(exp(e_1E_1)&exp(e_2E_2)exp(e_3E_3)exp(e_4E_4)exp(e_5E_5)exp(e_6E_6)exp(e_7E_7)) \\ &= exp\{2\pi i \tau(e_1E_1 + e_2E_2 + e_3E_3 + e_4E_4) \}.\endsplit$$ \medskip Let $\F \otimes \mu \in \F \otimes \Cplx \Lambda^1(\mm^*)$ with $\mu = a_1\epsilon_1 + a_2\epsilon_2 + a_3\epsilon_3 + a_4\epsilon_4 + a_5\epsilon_5 + a_6\epsilon_6 + a_7\epsilon_7$ for some $a_i \in \Cplx,$ for $i=1, \dots, 7.$ As in the proof of Proposition 4.7, $\Delta \F = 4\pi^2S^2\F$ where $S^2 = A_1^2 + A_2^2 + A_3^2 + A_4^2.$ Also, $\Delta\epsilon_1 = \Delta\epsilon_2 = \Delta\epsilon_3 = \Delta\epsilon_4 = 0,$ $\Delta\epsilon_5 = 2\epsilon_5,$ $\Delta\epsilon_6=\epsilon_6+\frac{1}{4}\epsilon_7,$ and $\Delta\epsilon_7 = \frac{1}{4}\epsilon_6+(3+\frac{1}{256}+\frac{3}{16})\epsilon_7.$ And we obtain the following chart: \bigskip {\eightpoint \centerline{ \vbox{\offinterlineskip \halign{\strut\vrule#& #\hfil &&\vrule#& \hfil#\hfil \cr \noalign{\hrule} &&&&&&&&&&&&&&&&\cr &\ $\nabla_U\mu\ $&&$\epsilon_1$&&$\epsilon_2$&&$\epsilon_3$& &$\epsilon_4$&&$\epsilon_5$&&$\epsilon_6$& &$\epsilon_7$&\cr &&&&&&&&&&&&&&&&\cr \noalign{\hrule} \noalign{\hrule} &&&&&&&&&&&&&&&&\cr &\ \ $E_1$&&0&&$-\frac{1}{32}\epsilon_7$&&$\frac1{2}\epsilon_5+\frac{1}{8}\epsilon_7$& &$\frac1{2}\epsilon_6+\frac{1}{8}\epsilon_7$&&$-\frac1{2}\epsilon_3+\frac1{2}\epsilon_7$& &$-\frac1{2}\epsilon_4$& &$\frac{1}{32}\epsilon_2-\frac{1}{8}\epsilon_3-\frac{1}{8}\epsilon_4-\frac{1}{2}\epsilon_5$&\cr &&&&&&&&&&&&&&&&\cr \noalign{\hrule} &&&&&&&&&&&&&&&&\cr &\ \ $E_2$&&$\frac{1}{32}\epsilon_7$&&0&&0&&$\frac1{2}\epsilon_5-\frac{1}{8}\epsilon_7$& &$-\frac{1}{2}\epsilon_4$&&$\frac{1}{2}\epsilon_7$& &$-\frac{1}{32}\epsilon_1+\frac{1}{8}\epsilon_4-\frac1{2}\epsilon_6$&\cr &&&&&&&&&&&&&&&&\cr \noalign{\hrule} &&&&&&&&&&&&&&&&\cr &\ \ $E_3$&&$-\frac1{2}\epsilon_5-\frac{1}{8}\epsilon_7$&&0&&0&&$\frac1{2}\epsilon_7$& &$\frac1{2}\epsilon_1$&&$0$&&$\frac{1}{8}\epsilon_1-\frac1{2}\epsilon_4$&\cr &&&&&&&&&&&&&&&&\cr \noalign{\hrule} &&&&&&&&&&&&&&&&\cr &\ \ $E_4$&&$-\frac1{2}\epsilon_6-\frac{1}{8}\epsilon_7$& &$-\frac1{2}\epsilon_5+\frac{1}{8}\epsilon_7$&&$-\frac1{2}\epsilon_7$&&0& &$\frac1{2}\epsilon_2$&&$\frac1{2}\epsilon_1$& &$\frac{1}{8}\epsilon_1-\frac{1}{8}\epsilon_2+\frac1{2}\epsilon_3$&\cr &&&&&&&&&&&&&&&&\cr \noalign{\hrule} }} } } \bigskip Using (A.1) and the above information, a straightforward calculation shows that if we let $E_\tau $ be the skew-Hermitian matrix defined by {\eightpoint $$\left(\matrix 4\pi^2S^2 & 0 & 0 & 0 & 2\pi i A_3 & 2\pi i A_4 & \pi i (-\frac{1}{8}A_2+\frac{1}{2}A_3+\frac{1}{2}A_4)\\ & 4 \pi^2S^2 & 0 & 0 & 2 \pi i A_4 & 0 & \pi i (\frac{1}{8}A_1-\frac{1}{2}A_4)\\ & & 4 \pi^2S^2 & 0 & -2 \pi i A_1 & 0 & \pi i ( -\frac{1}{2}A_1 + 2 A_4)\\ & & & 4 \pi^2S^2 & -2 \pi i A_2& -2 \pi i A_1 & \pi i (-\frac{1}{2}A_1 +\frac{1}{2}A_2 - 2 A_3)\\ & & & & 4 \pi ^2 S^2 + 2 &0 & -2 \pi i A_1 \\ & & & & & 4 \pi ^2 S^2 + 1 & - 2 \pi i A_2 +\frac{1}{4}\\ & & & & & & 4 \pi^2S^2 + \frac{817}{256} \endmatrix \right)$$ } then $\Delta(\F \otimes \mu) = \lambda(\F \otimes \mu)$ if and only if $\lambda$ is an eigenvalue of the matrix $E_\tau.$ \medskip Let $$\lambda = \frac{17}{4}\pi^2+1+\sqrt{\frac{17}{4}\pi^2+1}.$$ \medskip We now calculate necessary conditions on $\tau = A_1\epsilon_1 + A_2\epsilon_2 + A_3\epsilon_3 + A_4\epsilon_4$ so that $\lambda$ is an eigenvalue of $E_\tau.$ As $\tau\in\Tau^I_1$ or $\tau\in\Tau^I_2,$ we know $A_1, A_2, A_3, A_4 \in \Q.$ \medskip By a computation using Maple or Mathematica, if $det(E_\tau - \lambda I_7) = 0,$ then $\pi$ is the root of a polynomial with rational coefficients. However $\pi$ is transcendental. Thus the coefficients of the powers of $\pi$ must be zero. \medskip A straightforward calculation shows that $\pi^{14}$ is the highest power of $\pi$ occurring in the polynomial, and the coefficient of $\pi^{14}$ is equal to $(4S^2-\frac{17}{4})^7.$ Thus if $\lambda$ is an eigenvalue of $E_\tau,$ then $S^2 = \frac{17}{16}.$ Recall that $S^2=A_1^2+A_2^2+A_3^2+A_4^2.$ \medskip For $\tau$ in $\Tau^I_1,$ \ $S^2 = \frac{17}{16}$ if and only if (see ($*$)) $\tau = \pm (\frac{1}{4}\epsilon_2+\epsilon_3)$ or $\tau = \pm \frac{1}{4}\epsilon_1 \pm \epsilon_4.$ And for $\tau$ in $\Tau^I_2,$ \ $S^2 = \frac{17}{16}$ if and only if (see ($**$)) $\tau = \pm (\frac{1}{4}\epsilon_2-\epsilon_3)$ or $\tau = \pm \frac{1}{4}\epsilon_1 \pm \epsilon_4.$ Note that the only difference is in the sign of $\epsilon_3.$ \medskip For $\tau = \pm (\frac{1}{4}\epsilon_2+\epsilon_3)$ or $\tau = \pm \frac{1}{4}\epsilon_1 \pm \epsilon_4,$ a calculation using Maple or Mathematica shows that $det(E_\tau - \lambda I_7) \neq 0.$ Thus $\frac{17}{4}\pi^2+1+\sqrt{\frac{17}{4}\pi^2+1}\not \in \one spec^I(\nilmfld1,g).$ \medskip However, for $\tau = \pm (\frac{1}{4}\epsilon_2-\epsilon_3),$ $det(E_\tau - \lambda I_7)=0.$ Thus $\frac{17}{4}\pi^2+1+\sqrt{\frac{17}{4}\pi^2+1}$ is an eigenvalue for the Laplacian acting on $\H_{\pm (\frac{1}{4}\epsilon_2-\epsilon_3)} \otimes \Lambda^1(\mm^*),$ and $\frac{17}{4}\pi^2+1+\sqrt{\frac{17}{4}\pi^2+1} \in \one spec^I(\nilmfld2,g).$ \medskip Consequently $$1\hy spec^I(\nilmfld1, g) \neq 1\hy spec^I(\nilmfld2, g),$$ as desired. \medskip The proof of Proposition 4.11 is now complete. $\endpf$ \enddemo \bigskip \bigskip %bibliography of construction paper %September 1994 %\baselineskip=14pt \Refs \bigskip \widestnumber\key{DGGW} \ref \key B1 \manyby P. 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