Content-Type: multipart/mixed; boundary="-------------9905071137929" This is a multi-part message in MIME format. ---------------9905071137929 Content-Type: text/plain; name="99-147.comments" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="99-147.comments" LaTeX, 18 pages, paper 99-129 with introduction amended ---------------9905071137929 Content-Type: text/plain; name="99-147.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="99-147.keywords" supersymmetry, matrix models, zero modes ---------------9905071137929 Content-Type: application/x-tex; name="gro.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="gro.tex" \documentclass[12pt]{article} \usepackage{latexsym} \usepackage{amssymb} \usepackage{amsmath} \setlength{\textwidth}{16cm} \setlength{\textheight}{24cm} \setlength{\topmargin}{-1.5cm} \addtolength{\evensidemargin}{-1.5cm} \addtolength{\oddsidemargin}{-1.5cm} \def\i{{\rm i}} \def\e{{\rm e}} \def\d{{\rm d}} \def\one{{\sf 1}\mkern-5.0mu{\rm I}} \setcounter{secnumdepth}{2} \title{Asymptotic Form of Zero Energy Wave Functions in Supersymmetric Matrix Models} \author{J. Fr\"ohlich${}^{(a)}$ , G.M. Graf${}^{(a)}$ , D. Hasler${}^{(a)}$ , J. Hoppe${}^{(b,c)}$ , S.-T. Yau${}^{(d)}$ \\ \vspace*{-0.05truein} \\ \normalsize\it ${}^{(a)}$ Theoretische Physik, ETH-H\"onggerberg, CH--8093 Z\"urich\\ \normalsize\it \hskip -0.63cm ${}^{(b)}$ Max-Planck-Institut f\"ur Gravitationsphysik, Albert-Einstein-Institut, D-14473 Potsdam \\ \normalsize\it ${}^{(c)}$ Fachbereich Mathematik, TU Berlin, D-10623 Berlin\\ \normalsize\it ${}^{(d)}$ Department of Mathematics, Harvard University, Cambridge, MA 02138} \begin{document} \maketitle \vspace{0.4cm} \begin{abstract} We derive the power law decay, and asymptotic form, of ${\rm SU}(2)\times{\rm Spin}(d)$ invariant wave--functions satisfying $Q_\beta\psi=0$ for all $s_d=2(d-1)$ supercharges of reduced $(d+1)$--dimensional supersymmetric ${\rm SU}(2)$ Yang Mills theory, resp. of the ${\rm SU}(2)$--matrix model related to supermembranes in $d+2$ dimensions. \end{abstract} %\vspace{0.4cm} \section{Introduction} \label{sec:intr} It is generally believed that supersymmetric ${\rm SU}(N)$ matrix models in $d=9$ dimensions admit exactly one normalizable zero-energy solution for each $N>1$, while they admit none for all other dimensions for which the models can be formulated, i.e., for $d=2,3,5$. For various approaches to this problem see e.g. \cite{c1a}--\cite{c1l}. In this article, we would like to summarize (and slightly modify/extend) what is known about the behaviour of ${\rm SU}(2)$ zero--energy solutions far out at infinity in (and near) the space of configurations where the bosonic potential (the trace of all commutator--squares) vanishes. Based on some early 'negative' result concerning $N=2,\, d=2$ (that used rather different techniques/arguments; see \cite{c1a, c4a}) we started our investigation of the asymptotic behaviour, in the fall of 1997, with a Hamiltonian Born--Oppenheimer analysis of that $N=2,\, d=2$ case. Some months later, we realized that the rather complicated Hamiltonian analysis (Halpern and Schwartz \cite{c1h} had, in the meantime, derived the form of the wave function for $d=9$ near $\infty$, by Hamiltonian Born--Oppenheimer methods) can be replaced by a simple first order analysis, using only the first order operators $Q$, and first order perturbation theory. One finds that asymptotically normalizable, ${\rm SU}(2)$ and ${\rm SO}(d)$ invariant, wave functions do not exist for $d=2,3$, and $5$, in contrast to $d=9$, where there is exactly one. We close these introductory words by recalling that the models discussed below arise in at least 3 somewhat different ways: As supersymmetric extensions of regulated membrane theories in $d+2$ space--time dimensions \cite{c2a, c4a}, as reductions (to $0+1$ dimension) of $d+1$ dimensional Super Yang Mills theories \cite{c3a}--\cite{c3c}, and, for $d=9$, as a description of the dynamics of D--0 branes in superstring theory, \cite{c5a, c5b}. In this physical interpretation, the existence of a normalizable zero--energy solution is an important consistency requirement. The paper is organized as follows. In Section \ref{sec:mod} we recall the definition of the models, and in Section \ref{sec:res} we state our main result about zero--modes. The proof is given in Section \ref{sec:pr} and Appendix 1. We suggest to skip Subsection \ref{sec:pr5} and Appendix 1 at a first reading. As a warm--up the reader is advised to read Appendix 2, where a simpler model is treated by the same method. \section{The models} \label{sec:mod} The configuration space of the bosonic degrees of freedom is $X=\mathbb R^{3d}$ with coordinates \begin{equation} q=(\vec{q_1},\ldots,\vec{q_d})=(q_{sA})_{\substack{s=1,\ldots, d\\A=1,2,3}}\;. \nonumber \end{equation} To describe the fermionic degrees of freedom let, as a preliminary, \begin{equation} \gamma^i=(\gamma^i_{\alpha\beta})_{\alpha,\beta=1,\ldots,s_d}\;,\qquad (i=1,\ldots,d)\;, \label{e1.1aa} \end{equation} be the {\it real} representation of smallest dimension, called $s_d$, of the Clifford algebra with $d$ generators: $\{\gamma^s,\gamma^t\}=2\delta^{st}\one$. On the representation space, ${\rm Spin}(d)$ is realized through matrices $R\in{\rm SO}(s_d)$, so that we may view \begin{equation} {\rm Spin}(d)\hookrightarrow{\rm SO}(s_d)\;, \label{e1.1a} \end{equation} as a simply connected subgroup. We recall that \begin{equation*} s_d=\cases 2^{[d/2]}\;,& d=0,1,2 \mod 8\;,\\ 2^{[d/2]+1}\;,& \hbox{otherwise}\;, \endcases \end{equation*} where $[\cdot]$ denotes the integer part. We then consider the Clifford algebra with $s_d$ generators and its irreducible representation on $\mathcal C=\mathbb C^{2^{s_d/2}}$. On $\mathcal C^{\otimes 3}$ the Clifford generators \begin{equation} (\vec{\Theta}_1,\ldots,\vec{\Theta}_{s_d})= (\Theta_{\alpha A})_{\substack{\alpha=1,\ldots,s_d\\A=1,2,3}} \nonumber \end{equation} are defined, satisfying $\{\Theta_{\alpha A} ,\Theta_{\beta B}\}=\delta_{\alpha\beta}\,\delta_{AB}$. The Hilbert space, finally, is \begin{equation} \mathcal H={\rm L}^2(X,\mathcal C^{\otimes 3})\;. \label{e1.0} \end{equation} There is a natural representation of ${\rm SU}(2)\times{\rm Spin}(d)\ni(U,R)$ on $\mathcal H$. In fact, the group acts naturally on $X$ through its representation ${\rm SO}(3)\times{\rm SO}(d)$ (which we also denote by $(U,R)$). On $\mathcal C^{\otimes 3}$ we have the representation $\mathcal R$ of ${\rm Spin}(s_d)\ni R$ \begin{equation} \mathcal R(R)^*\Theta_{\alpha A}\mathcal R(R)=\widetilde{R}_{\alpha\beta} \Theta_{\beta A}\;, \label{e1.2} \end{equation} where $\widetilde{R}=\widetilde{R}(R)$ is its ${\rm SO}(s_d)$ representation. Through ${\rm SO}(s_d)={\rm Spin}(s_d)/\mathbb Z_2$ and (\ref{e1.1a}) we have \begin{equation} {\rm Spin}(d)\hookrightarrow{\rm Spin}(s_d)\;, \label{e1.2a} \end{equation} and thus a representation $\mathcal R$ of ${\rm Spin}(d)$. The representation $\mathcal U$ of ${\rm SU}(2)\ni U$ on $\mathcal C^{\otimes 3}$ is characterized by $\mathcal U(U)^*\Theta_{\alpha A}\mathcal U(U)=U_{AB}\Theta_{\alpha B}$. We shall now restrict to $d=2,\,3,\,5,\,9$, where $s_d=2,\,4,\,8,\,16$, the reason being that in these cases \begin{equation} s_d=2(d-1)\;, \label{e1.1} \end{equation} whereas $s_d$ is strictly larger otherwise. Eq. (\ref{e1.1}) is essential for the algebra (\ref{eq2}) below \cite{c3c}. The supercharges, acting on $\mathcal H$, are given by the $s_d$ hermitian operators \begin{equation} Q_\beta= \vec{\Theta}_\alpha \cdot \bigl( -\i \gamma^t_{\alpha\beta} \vec{\nabla}_t + \frac{1}{2} \, \vec{q}_s \times \vec{q}_t\,\gamma^{st}_{\beta\alpha} \bigr)\;, \qquad(\beta=1,\ldots, s_d)\;, \nonumber \end{equation} where $\gamma^{st}=(1/2)(\gamma^s \gamma^t -\gamma^t \gamma^s)$. These supercharges transform as scalars under ${\rm SU}(2)$ transformations generated by \begin{equation} J_{AB}=-\i(q_{sA}\partial_{sB}-q_{sB}\partial_{sA})-\frac{\i}{2} (\Theta_{\alpha A}\Theta_{\alpha B}-\Theta_{\alpha B}\Theta_{\alpha A}) \equiv L_{AB}+M_{AB}\;, \nonumber \end{equation} resp. as vectors in $\mathbb R^{s_d}$ under ${\rm Spin}(d)$ transformation generated by \begin{equation} J_{st} = -\i(\vec{q}_s\cdot\vec{\nabla}_t - \vec{q}_t\cdot\vec{\nabla}_s)-\frac{\i}{4} \,\vec{\Theta}_\alpha \gamma^{st}_{\alpha\beta} \vec{\Theta}_\beta \equiv L_{st}+M_{st}\;. \nonumber \end{equation} The anticommutation relations of the supercharges are \begin{equation} \bigl\{Q_\alpha, Q_\beta\bigr\}=\delta_{\alpha\beta}H+ \gamma^t_{\alpha\beta}q_{tA}\varepsilon_{ABC}J_{BC}\;. \label{eq2} \end{equation} Here, $H$ is the Hamiltonian \begin{equation} H = - \sum_{s=1}^9 \, \vec{\nabla}_s^2 + \sum_{s0$ and $\vec{e}{\,}^2=\sum_s E_s^2 = 1$. The dimension of the manifold is $1+2+(d-1)=3d-2(d-1)$. Points in a conical neighborhood of the manifold can be expressed in terms of tubular (or ``end--point'') coordinates \cite{b2} \begin{equation} \vec{q}_s = r \vec{e} E_s + r^{-1/2}\vec{y}_s \label{e2.00} \end{equation} with \begin{equation} \vec{y}_s\cdot \vec{e} = 0\;,\qquad\vec{y}_s E_s =\vec{0}\;. \label{e2.1} \end{equation} A prefactor has been put explicitely in front of the transversal coordinates $\vec{y}_s$, so as to anticipate the length scale $r^{-1/2}$ of the ground state. The change \begin{equation} (\vec{e},E, y)\mapsto(-\vec{e},-E, y) \label{e2.0} \end{equation} does not affect $\vec{q}_s$. Rather than identifying the two coordinates for $\vec{q}_s$, we shall look for states which are even under the antipode map (\ref{e2.0}). \smallskip We can now describe the structure of a putative ground state. \smallskip\noindent {\bf Theorem} {\it Consider the equations $Q_\beta\psi=0$ for a formal power series solution near $r=\infty$ of the form \begin{equation} \psi=r^{-\kappa}\sum_{k=0}^\infty r^{-\frac{3}{2}k}\psi_k\;, \label{e2.4} \end{equation} where: $\psi_k=\psi_k(\vec{e},E,y)$ is square integrable w.r.t. $\d e\,\d E\,\d y$; \newline \hphantom{where:} $\psi_k$ is ${\rm SU}(2)\times{\rm Spin}(d)$ invariant;\newline \hphantom{where:} $\psi_0\neq 0$.\newline Then, up to linear combinations, \begin{itemize} \item{d=9}: The solution is unique, and $\kappa=6$; \item{d=5}: There are three solutions with $\kappa=-1$ and one with $\kappa=3$; \item{d=3}: There are two solutions with $\kappa=0$; \item{d=2}: There are no solutions. \end{itemize} All solutions are even under the antipode map (\ref{e2.0}), \begin{equation*} \psi_k(\vec{e},E,y)=\psi_k(-\vec{e},-E,y)\;, \end{equation*} except for the state $d=5,\,\kappa=3$, which is odd.} \medskip\noindent {\bf Remarks} 1. The equation $Q_\beta\psi=0$ can be viewed as an ordinary differential equation in $z=r^{3/2}$ for a function taking values in ${\rm L}^2(\d e\,\d E\,\d y,\mathcal C^{\otimes 3})$ (see eq. (\ref{e3.00}) below). It turns out that $z=\infty$ is a singular point of the second kind \cite{b1}. In such a situation the series (\ref{e2.4}) is typically asymptotic to a true solution, but not convergent. \par\noindent 2. The integration measure is $\d q=\d r\cdot r^2\d e\cdot r^{d-1}\d E\cdot r^{-\frac{1}{2}\cdot 2(d-1)}\d y= r^2\d r\,\d e\,\d E\,\d y$. The wave function (\ref{e2.4}) is square integrable at infinity if $\int^\infty\d r\,r^2(r^{-\kappa})^2<\infty$, i.e., if $\kappa>3/2$. The theorem is consistent with the statement according to which {\bf only} for $d=9$ a (unique) normalizable ground state for (\ref{e1.4}) (which would have to be even) is possible. \par\noindent 3. Note that the connection of matrix models with supergravity requires the zero--energy solutions to be ${\rm Spin}(d)$ singlets only for $d=9$. \bigskip The case $d=2$ can be dealt with immediately. We may assume $\gamma^2 =\sigma_3, \, \gamma^1 = \sigma_1$ (Pauli matrices), so that \begin{equation*} M_{12}=\frac{\i}{2}\Theta_{1A}\Theta_{2A}\;, \end{equation*} with commuting terms. Since, for each $A=1,2,3,\, (\Theta_{1A}\Theta_{2A})^2 =-1/4$, we see that $M_{12}$ has spectrum in $\mathbb Z/2 +1/4$. Given that $L_{12}$ has spectrum $\mathbb Z$, no state with $J_{12}\psi=0$ is possible. We mention \cite{c1a} that, more generally, for $d=2$ no normalizable ${\rm SU}(2)$ invariant ground state exists. The proof of the theorem will thus deal with $d=9,5,3$ only. \section{Proof} \label{sec:pr} We shall first derive the power series expansion of the supercharges $Q_\beta$. To this end we note that \begin{eqnarray} \frac{\partial}{\partial q_{tA}}&=& r^{1/2}(\delta_{st}-E_sE_t)(\delta_{AB}-e_Ae_B) \frac{\partial}{\partial y_{sB}} \label{e3.000}\\ &&+r^{-1}[e_AE_t(r\frac{\partial}{\partial r}+ \frac{1}{2}y_{sB}\frac{\partial}{\partial y_{sB}})+ \i e_BE_tL_{BA}+\i e_AE_sL_{st}] +{\rm O}(r^{-5/2})\;,\nonumber \end{eqnarray} with the remainder not containing derivatives w.r.t. $r$ (see Appendix 1 for derivation). This yields \begin{equation} Q_\beta=r^{1/2}Q_\beta^0+ r^{-1}(\widehat{Q}_\beta^1r\frac{\partial}{\partial r}+Q_\beta^1) +r^{-5/2}Q_\beta^2+\ldots \label{e3.00} \end{equation} with $r$--independent operators \begin{eqnarray*} Q_\beta^0&=&-\i\Theta_{\alpha A}\gamma^t_{\alpha\beta}(\delta_{st}-E_sE_t) (\delta_{AB}-e_Ae_B)\frac{\partial}{\partial y_{sB}}+ \vec{\Theta}_\alpha \cdot(\vec{e}\times\vec{y}_t)E_s\gamma^{st}_{\beta\alpha}\;,\\ \widehat{Q}_\beta^1&=& -\i (\vec{\Theta}_\alpha\cdot\vec{e}\,)\gamma^t_{\alpha\beta}E_t\;,\\ Q_\beta^1&=&\Theta_{\alpha A}\gamma^t_{\alpha\beta} \bigl(e_BE_tL_{BA}+ e_AE_sL_{st}-\frac{\i}{2}\,e_AE_ty_{sB}\frac{\partial}{\partial y_{sB}})+ \frac{1}{2}\vec{\Theta}_\alpha\cdot(\vec{y}_s\times\vec{y}_t) \gamma^{st}_{\beta\alpha}\;. \end{eqnarray*} The explicit expressions of $Q_\beta^n,\, (n\ge 2)$ will not be needed. We then equate coefficients of powers of $r^{-3/2}$ in the equation $Q_\beta\psi=0$ with the result \begin{eqnarray} Q_\beta^0\psi_n+ \bigl(-(\kappa+\frac{3}{2}(n-1))\widehat{Q}_\beta^1+Q_\beta^1\bigr) \psi_{n-1}+Q_\beta^2\psi_{n-2}+\ldots+Q_\beta^n\psi_0&=&0\;,\nonumber\\ \qquad(n&=&0,1,\ldots)\;. \label{e3.0} \end{eqnarray} \subsection{The equation at $n=0$} \label{sec:pr1} The equation at $n=0$, \begin{equation} Q_\beta^0\psi_0=0\;, \label{e3.1} \end{equation} admits precisely the (not necessarily ${\rm SU}(2)\times{\rm Spin}(d)$ invariant) solutions \begin{equation} \psi_0(\vec{e},E, y)=\e^{-\sum_s\vec{y}_s{\,}^2/2}|F(E,\vec{e})\rangle\;, \label{e3.2} \end{equation} (with $\vec{y}$ restricted to (\ref{e2.1})), where the fermionic states $|F(E,\vec{e})\rangle$ can be described as follows: Let $\vec{n}_\pm$ be two complex vectors satisfying $\vec{n}_+\cdot\vec{n}_-=1,\,\vec{e}\times \vec{n}_\pm=\mp\i\vec{n}_\pm$ (and hence $\vec{n}_\pm\cdot\vec{n}_\pm=0$, $\vec{n}_+\times\vec{n}_-=-\i\vec{e}$\,). For any vector $v\in\mathbb R^{s_d}$ we may introduce $\vec{\Theta}(v)=\vec{\Theta}_\alpha v_\alpha$, as well as fermionic operators $\vec{\Theta}(v)\cdot\vec{n}_\pm$ satisfying canonical anticommutation relations: \begin{equation} \bigl\{\vec{\Theta}(u)\cdot\vec{n}_+,\vec{\Theta}(v)\cdot\vec{n}_-\bigr\} =u_\alpha v_\alpha\;,\qquad \bigl\{\vec{\Theta}(u)\cdot\vec{n}_\pm,\vec{\Theta}(v)\cdot\vec{n}_\pm\bigr\} =0\;. \nonumber \end{equation} Then, $|F(E,\vec{e})\rangle$ is required to obey \begin{equation} \vec{\Theta}(v)\cdot\vec{n}_\pm|F(E,\vec{e})\rangle=0 \qquad\hbox{for}\qquad E_s\gamma^sv=\pm v\;. \label{e3.4} \end{equation} To prove the above, let us note that \begin{eqnarray} &&\hskip 1cm\bigl\{Q_\alpha^0,Q_\beta^0\bigr\}= \delta_{\alpha\beta}H^0+ \gamma^t_{\alpha\beta}E_t\varepsilon_{ABC}M_{AB}e_C\;, \label{e3.4a}\\ H^0&=&\bigl[-(\delta_{st}-E_sE_t)(\delta_{AB}-e_Ae_B) \frac{\partial}{\partial y_{sA}}\frac{\partial}{\partial y_{tB}} +\sum_s\vec{y}_s^2\bigr]+ \i E_s\gamma^s_{\alpha\beta}\vec{e}\cdot \bigl(\vec{\Theta}_\alpha \times \vec{\Theta}_\beta\bigr)\nonumber\\ &\equiv& H^0_B+H^0_F\;.\nonumber \end{eqnarray} By contracting eq. (\ref{e3.4a}) against $\delta_{\alpha\beta}$, resp. $\gamma^t_{\alpha\beta}E_t$ we see that the equations (\ref{e3.1}) are equivalent to the pair of equations \begin{equation} H^0\psi_0=0\;,\qquad \varepsilon_{ABC}M_{AB}e_C\psi_0=0\;. \label{e3.4b} \end{equation} Here, $H^0_B$ is a harmonic oscillator in $2(d-1)$ degrees of freedom, with orbital ground state wave function $\e^{-\sum_s\vec{y}_s^2/2}$ and energy $2(d-1)$. On the other hand, \begin{eqnarray} H^0_F&=&-E_s\gamma^s_{\alpha\beta} \bigl((\vec{\Theta}_\alpha\cdot\vec{n}_+)(\vec{\Theta}_\beta\cdot\vec{n}_-)- (\vec{\Theta}_\alpha\cdot\vec{n}_-)(\vec{\Theta}_\beta\cdot\vec{n}_+)\bigr) \nonumber\\ &=&-s_d+ 2P^+_{\alpha\beta}(\vec{\Theta}_\alpha\cdot\vec{n}_-) (\vec{\Theta}_\beta\cdot\vec{n}_+) +2P^-_{\alpha\beta}(\vec{\Theta}_\alpha\cdot\vec{n}_+) (\vec{\Theta}_\beta\cdot\vec{n}_-)\;, \label{e3.5} \end{eqnarray} where we used the spectral decomposition $E_s\gamma^s=P^+-P^-$. In view of (\ref{e1.1}), the equation $H^0\psi_0=0$ is fulfilled iff the fermionic state is annihilated by the last two positive terms in (\ref{e3.5}), i.e., if (\ref{e3.4}) holds. The second equation (\ref{e3.4b}) is now also satisfied, since \begin{eqnarray} \frac{1}{2}\varepsilon_{ABC}M_{AB}e_C&=& -\frac{\i}{2}\vec{e}\cdot \bigl(\vec{\Theta}_\alpha \times \vec{\Theta}_\alpha \bigr)\nonumber\\ &=&\frac{1}{2} \bigl((\vec{\Theta}_\alpha\cdot\vec{n}_+)(\vec{\Theta}_\alpha\cdot\vec{n}_-)- (\vec{\Theta}_\alpha\cdot\vec{n}_-)(\vec{\Theta}_\alpha\cdot\vec{n}_+)\bigr) \nonumber\\ &=&P^-_{\alpha\beta}(\vec{\Theta}_\alpha\cdot\vec{n}_+) (\vec{\Theta}_\beta\cdot\vec{n}_-)- P^+_{\alpha\beta}(\vec{\Theta}_\alpha\cdot\vec{n}_-) (\vec{\Theta}_\beta\cdot\vec{n}_+) \label{e3.5c} \end{eqnarray} annihilates $|F(E,\vec{e})\rangle$. \par \subsection{${\rm SU}(2)\times{\rm Spin}(d)$ invariant states} \label{sec:pr2} We recall that the representation $\mathcal R[\cdot]$ of ${\rm Spin}(d)$ on $\mathcal H$ is $(\mathcal R[R]\psi)(q)=\mathcal R(R)(\psi(R^{-1}q))$, where $\mathcal R(R)$ acts on $\mathcal C^{\otimes 3}$. Similarly for ${\rm SU}(2)$. The invariant solutions among (\ref{e3.2}) are thus those which satisfy \begin{equation} \mathcal U(U)|F(E,\vec{e})\rangle =|F(E,U\vec{e})\rangle\;,\qquad \mathcal R(R)|F(E,\vec{e})\rangle =|F(RE,\vec{e})\rangle\;, \label{e3.5a} \end{equation} for $(U,R)\in{\rm SU}(2)\times{\rm Spin}(d)$. These states are in bijective correspondence to states invariant under the `little group' $(U,R)\in{\rm U}(1)\times{\rm Spin}(d-1)$, i.e., to states $|F(E,\vec{e})\rangle$ satisfying \begin{equation} \mathcal U(U)|F(E,\vec{e})\rangle =|F(E,\vec{e})\rangle\;,\qquad \mathcal R(R)|F(E,\vec{e})\rangle =|F(E,\vec{e})\rangle\;, \label{e3.5b} \end{equation} for some arbitrary but fixed $(E,\vec{e})$ and all $U,\,R$ with $U\vec{e}=\vec{e},\,RE=E$. The first relation holds on all of (\ref{e3.4}). In fact the generator (\ref{e3.5c}) of the group $\mathcal U(U)$ of rotations $U$ about $\vec{e}$ annihilates $|F(E,\vec{e})\rangle$, as we just saw. To discuss the second relation (\ref{e3.5b}) we note that the generators of ${\rm Spin}(d-1)$ (i.e., of the fermionic rotations about $E$), are $M_{st}U_sV_t$ with $U_sE_s=V_sE_s=0$. We write $M_{st}=M_{st}^{\perp}+M_{st}^{\parallel}$, where \begin{equation} M_{st}^{\perp}=-(\i/2)(\vec{\Theta}_\alpha\cdot\vec{n}_+) \gamma^{st}_{\alpha\beta}(\vec{\Theta}_\beta\cdot\vec{n}_-)\;,\qquad M_{st}^{\parallel}=-(\i/4)(\vec{\Theta}_\alpha\cdot\vec{e}\,) \gamma^{st}_{\alpha\beta}(\vec{\Theta}_\beta\cdot\vec{e}\,)\;, \label{e3.8b} \end{equation} and remark that, by a computation similar to (\ref{e3.5c}), $M_{st}^{\perp}U_sV_t$ annihilates $|F(E,\vec{e})\rangle$. As a result, we may study the representation $\mathcal R$ of the group ${\rm Spin}(d-1)$ through its embedding in the Clifford algebra generated by the $\vec{\Theta}_\alpha\cdot\vec{e}$. \par The operators $\vec{\Theta}_\alpha\cdot\vec{e}$ leave the space (\ref{e3.4}) invariant and act irreducibly on it. That space is thus isomorphic to $\mathcal C$, and ${\rm Spin}(s_d)$ acts according to (\ref{e1.2}) (with $\Theta_{\alpha A}$ replaced by $\vec{\Theta}_\alpha\cdot\vec{e}$). This representation decomposes (see e.g. \cite{b3}) as \begin{equation} \mathcal C=(2^{(s_d/2)-1})_+\oplus (2^{(s_d/2)-1})_- \label{e3.2a} \end{equation} w.r.t. the subspaces where $\Theta\equiv 2^{s_d/2}\prod_{\alpha=1}^{s_d}\vec{\Theta}_\alpha\cdot\vec{e}=+1$, resp. $-1$. The embedding (\ref{e1.2a}) and the corresponding branching of the representation (but not the statement of the theorem!) depend on the choice of the $\gamma$--matrices. In order to select a definite embedding, let \begin{equation} \gamma^d =\left( \begin{array}{cc} \one & \ 0 \\ 0 & - \one \end{array} \right) \; , \quad \gamma^{d-1} = \left( \begin{array}{cc} \ 0 &\ \one \\ \ \one & \ 0\end{array} \right) \; , \quad \gamma^j =\left( \begin{array}{cc} 0 & \i \Gamma^j\\ -\i \Gamma^j& 0 \end{array} \right) \; \label{e3.3} \end{equation} with $\Gamma^j,\,(j= 1,\ldots, d-2)$ purely imaginary, antisymmetric, and $\{ \Gamma^j, \Gamma^k \} = 2 \delta_{jk} \one_{s_d/2}$. Then (\ref{e3.2a}) branches as (see \cite{b4}, resp. \cite{c1k, c1l}) \begin{equation} \mathcal C=\cases (44\oplus 84)\oplus 128\;,&\qquad (d=9)\;,\\ (5\oplus 1\oplus 1\oplus 1)\oplus (4 \oplus 4)\;,&\qquad (d=5)\;,\\ 2\oplus (1\oplus 1) \;,&\qquad (d=3)\;, \endcases \label{e3.6} \end{equation} when viewed as a representation of ${\rm Spin}(d)$. (The choice $\widetilde{\gamma}^i_{\alpha\beta}= \widetilde{R}_{\alpha'\alpha}\gamma^i_{\alpha'\beta'} \widetilde{R}_{\beta'\beta}$ with $\widetilde{R}\in {\rm O}(s_d),\,\det\widetilde{R}=-1$ would have inverted the branching of the representations on the r.h.s. of (\ref{e3.2a})). The case $d=3$ deserves a remark, as there are additional inequivalent embeddings ${\rm Spin}(d=3)\hookrightarrow{\rm Spin}(s_d=4)$, and one has to consider the one appropriate to (\ref{e1.2a}). In fact $R\in{\rm Spin}(3)={\rm SU}(2)$ acts in the fundamental representation on $\mathbb C^2$, the irreducible representation space of the complex Clifford algebra with $3$ generators. The real representation (\ref{e3.3}) is obtained by joining two complex representations, followed by an appropriate change $T$ of basis. The embedding (\ref{e1.2a}) is thus realized through $R \mapsto T^{-1}(R\otimes\one_2)T$ and the embedding ${\rm su(2)}_{\mathbb C}\hookrightarrow{\rm so(4)}_{\mathbb C} ={\rm su(2)}_{\mathbb C}\oplus{\rm su(2)}_{\mathbb C}$ is equivalent to $u\mapsto(u,0)$. \par The further branching ${\rm Spin}(d)\hookleftarrow{\rm Spin}(d-1)$ yields \begin{equation} \mathcal C=\cases (1\oplus 8_{\rm v}\oplus 35_{\rm v}) \oplus(28\oplus 56_{\rm v}) \oplus(8_{\rm s}\oplus 8_{\rm c}\oplus 56_{\rm s}\oplus 56_{\rm c}) \;,&\qquad (d-1=8)\;,\\ 1\oplus 1\oplus 1\oplus (1\oplus 4)\oplus (2_+\oplus 2_-)\oplus(2_+\oplus 2_-) \;,&\qquad (d-1=4)\;,\\ (1_1\oplus 1_{-1})\oplus 1_0\oplus 1_0 \;,&\qquad (d-1=2)\;. \endcases \label{e3.7} \end{equation} The content of invariant states stated in the theorem is now manifest. One should notice that for $d=3$ the little group ${\rm U}(1)$ is abelian and the singlets $1_{\pm 1}$ do not correspond to invariant states. For later use we also retain the fermionic ${\rm Spin}(d)$ representation to which the remaining singlets are associated, \begin{equation} 44\quad(d=9)\;;\qquad 1, 1, 1, 5\quad(d=5)\;;\qquad 1,1\quad(d=3)\;, \label{e3.8} \end{equation} together with the corresponding eigenvalue of $\Theta$: \begin{equation} \Theta=\quad 1\quad(d=9)\;;\qquad 1, 1, 1, 1\quad(d=5)\;; \qquad -1,-1\quad(d=3)\;. \label{e3.8a} \end{equation} \subsection{Even states} \label{sec:pr3} It remains to check which of these states satisfy $|F(-E,-\vec{e})\rangle =|F(E,\vec{e})\rangle$. Let us begin by noting that by (\ref{e3.5a}) \begin{equation*} |F(-E,-\vec{e})\rangle=\e^{\i M_{AB}e_Au_B\pi}\e^{\i M_{st}E_sU_t\pi} |F(E,\vec{e})\rangle\;, \end{equation*} where $\vec{u}\in\mathbb R^3$, resp. $U\in\mathbb R^d$ are unit vectors orthogonal to $\vec{e}$, resp. $E$. The ${\rm Spin}(d)$ rotation can be factorized as $\e^{\i M_{st}E_sU_t\pi}= \e^{\i M_{st}^{\perp}E_sU_t\pi}\e^{\i M_{st}^{\parallel}E_sU_t\pi}$. We claim that $\e^{\i M_{st}^{\parallel}E_sU_t\pi}$ $|F(E,\vec{e})\rangle= \sigma|F(E,\vec{e})\rangle$ with \begin{equation} \sigma=\quad 1\quad(d=9)\;;\qquad 1, 1, 1, -1\quad(d=5)\;; \qquad 1,1\quad(d=3)\;. \label{e3.9} \end{equation} The operator represents a rotation $R\in{\rm Spin}(d)$ with $RE=-E$ in the representation (\ref{e3.8}). For $d=9$ the latter can be realized on symmetric traceless tensors $T_{ij},\,(i,j=1,\ldots, 9)$, where the ${\rm Spin}(8)$--singlet is $E_iE_j-(1/9)\delta_{ij}$, implying $\sigma=1$. For $d=5$, the last representation (\ref{e3.8}) is just the vector representation, where $\sigma=-1$. As the remaining cases are evident, eq. (\ref{e3.9}) is proven. A computation using (\ref{e3.3}) and, without loss $E=(0,\ldots,0,1),\,U=(0,\ldots,1,0)$ shows \begin{eqnarray*} \e^{\i M_{d, d-1}^{\perp}\pi}|F(E,\vec{e})\rangle&=& \prod_{\alpha=1}^{s_d/2}\e^{[ (\vec{\Theta}_\alpha\cdot\vec{n}_+) (\vec{\Theta}_{\alpha+s_d/2}\cdot\vec{n}_-)- (\vec{\Theta}_{\alpha+s_d/2}\cdot\vec{n}_+) (\vec{\Theta}_\alpha\cdot\vec{n}_-)]\pi/2}|F(E,\vec{e})\rangle\\ &=&\prod_{\alpha=1}^{s_d/2} (\vec{\Theta}_{\alpha+s_d/2}\cdot\vec{n}_+) (\vec{\Theta}_\alpha\cdot\vec{n}_-)|F(E,\vec{e})\rangle\equiv |\overline{F}(E,\vec{e})\rangle\;,\\ \e^{\i M_{AB}e_Au_B\pi}|\overline{F}(E,\vec{e})\rangle&=& \prod_{\alpha=1}^{s_d}\e^{ (\vec{\Theta}_\alpha\cdot\vec{e})(\vec{\Theta}_\alpha\cdot\vec{u})\pi} |\overline{F}(E,\vec{e})\rangle\\ &=&(-1)^{s_d/4}\Theta\prod_{\alpha=1}^{s_d/2} (\vec{\Theta}_\alpha\cdot\vec{n}_+) (\vec{\Theta}_{\alpha+s_d/2}\cdot\vec{n}_-)|\overline{F}(E,\vec{e})\rangle= |F(E,\vec{e})\rangle\;, \end{eqnarray*} where we used (\ref{e3.8a}) in the last step. Together with (\ref{e3.9}) this proves the statement of theorem concerning the invariance under (\ref{e2.0}). \subsection{The equation at $n>0$} \label{sec:pr4} We next discuss the equations (\ref{e3.0})${}_n$ with $n\ge 1$. Let $P_0$ be the orthogonal projection onto the states (\ref{e3.2}), i.e., onto the null space of $Q_\beta^0$. We replace them with an equivalent pair of equations, obtained by multiplication of (\ref{e3.0})${}_{n+1}$ with $P_0$, resp. of (\ref{e3.0})${}_n$ with $Q_\beta^0$, which is injective on the range of the complementary projection $\overline{P}_0=1-P_0$: \begin{eqnarray} P_0\bigl(-(\kappa+\frac{3}{2}n))\widehat{Q}_\beta^1+Q_\beta^1\bigr)P_0\psi_n =-P_0\bigl(Q_\beta^1\overline{P}_0\psi_n +Q_\beta^2\psi_{n-1}+\ldots+Q_\beta^{n+1}\psi_0\bigr)\;,&&\nonumber\\ \qquad (n=0,1,\ldots)\;,&& \label{e3.10}\\ (Q_\beta^0)^2\psi_n =-Q_\beta^0\Bigl( \bigl(-(\kappa+\frac{3}{2}(n-1))\widehat{Q}_\beta^1+ Q_\beta^1\bigr)\psi_{n-1} +Q_\beta^2\psi_{n-2}+\ldots+Q_\beta^n\psi_0\Bigr)\;,&&\nonumber\\ \qquad (n=1,2,\ldots)&& \label{e3.11} \end{eqnarray} (we used $P_0\widehat{Q}_\beta^1\overline{P}_0=0$). Here, and until the end of this subsection, no summation over $\beta$ is understood. The equation (\ref{e3.10}) at $n=0$ reads \begin{equation} P_0Q_\beta^1\psi_0=\kappa P_0\widehat{Q}_\beta^1\psi_0 \;(=\kappa\widehat{Q}_\beta^1\psi_0)\;. \label{e3.13} \end{equation} \par We shall verify this by explicit computation later on. Since a similar issue will show up in solving the equation (\ref{e3.10}) at $n>0$, let us also present a more general statement, whose proof is postponed to the next subsection. \smallskip\noindent {\bf Lemma} {\it Let $T_\beta$ be linear operators on the range of $P_0$, which transform as real spinors of ${\rm Spin}(d)$ and commute with the antipode map. Then, for each invariant state we have \begin{equation} T_\beta\psi_0=\kappa\widehat{Q}_\beta^1\psi_0\;, \label{e3.14} \end{equation} with $\kappa$ depending only on the associated representation (\ref{e3.8}).} \medskip\noindent We now assume having solved the equations (\ref{e3.10}, \ref{e3.11}) up to $n-1$ for ${\rm Spin}(d)$ invariant $\psi_1,\ldots\psi_{n-1}$ (which is true for $n-1=0$), and claim the same is possible for $n$. Since $Q_\beta^0$ is invertible on the range of $\overline{P}_0$, eq.~(\ref{e3.11})${}_n$ determines $\overline{P}_0\psi_n$ uniquely. The fact that the solution so obtained is independent of $\beta$ and is ${\rm Spin}(d)$ invariant may deserve a comment, because the equivalence of the equations $Q_\beta\psi=0$ and $(Q_\beta)^2\psi=0$, which holds on (\ref{e1.0}), does not apply in the sense of formal power series (\ref{e2.4}). Consider the expansion (\ref{e3.00}), i.e., \begin{equation*} Q_\beta = r^{1/2}\sum_{k=0}^\infty r^{-\frac{3}{2}k}[Q_\beta]_k\;, \qquad [Q_\beta]_k= Q_\beta^k+\delta_{1k}\widehat{Q}_\beta^1r{\partial\over\partial r}\;, \end{equation*} as well as its formal square \begin{equation*} (Q_\beta)^2 = r\sum_{k=0}^\infty r^{-\frac{3}{2}k}[(Q_\beta)^2]_k\;. \end{equation*} Notice that $(Q_\beta)^2$ is, by (\ref{eq2}), independent of $\beta$ and ${\rm Spin}(d)$ invariant as an operator on ${\rm SU}(2)$ invariant power series. Similarly, let $[Q_\beta\psi]_k$ (given by the l.h.s. of (\ref{e3.0})) and $[(Q_\beta)^2\psi]_k$ be the coefficients of the corresponding series. By induction assumption we have $[Q_\beta\psi]_k=0$ for $k=0,\ldots,\,n-1$. Since $Q_\beta(Q_\beta\psi)=(Q_\beta)^2\psi$, we obtain \begin{eqnarray*} [(Q_\beta)^2\psi]_n&=&\sum_{k=0}^n Q_\beta^k[Q_\beta\psi]_{n-k} -(\kappa+\frac{3}{2}n-2)\widehat{Q}_\beta^1[Q_\beta\psi]_{n-1} =Q_\beta^0[Q_\beta\psi]_n\;,\\{} [(Q_\beta)^2\psi]_n&=&(Q_\beta^0)^2\psi_n+\widetilde{\psi}_{n-1}\;, \end{eqnarray*} where $\widetilde{\psi}_{n-1}$ (determined by $\psi_0,\ldots\psi_{n-1}$) has the desired properties. The equation (\ref{e3.11})${}_n$, i.e., $Q_\beta^0[Q_\beta\psi]_n=0$ is thus equivalent to $(Q_\beta^0)^2\psi_n=-\widetilde{\psi}_{n-1}$, which exhibits the claim. On the other hand, invariance requires $P_0\psi_n$ to be a linear combination of invariant singlets. For the ansatz $P_0\psi_n=\lambda_n\psi_0$, eq. (\ref{e3.10})${}_n$ reads \begin{equation} \frac{3}{2}n\lambda_n\widehat{Q}_\beta^1\psi_0= -P_0\bigl(Q_\beta^1\overline{P}_0\psi_n +Q_\beta^2\psi_{n-1}+\ldots+Q_\beta^{n+1}\psi_0\bigr)\;, \nonumber \end{equation} because of (\ref{e3.13}). Again, by the lemma, this holds true for suitable $\lambda_n$. Indeed, this solution for $P_0\psi_n$ is the only one. \subsection{Proof of the lemma} \label{sec:pr5} The vectors $T_\beta\psi_0,\, (\beta=1,\ldots, s_d)$ transform under ${\rm Spin}(d)$ as real spinors, although they might be linearly dependent. By reducing matters to the little group as before, any representation of that sort is specified by the values $|F^\beta(E,\vec{e})\rangle$ of its states (see (\ref{e3.2})) at one point $(E,\vec{e})$, which are required to satisfy \begin{equation*} \widetilde{R}_{\beta\alpha}(R)|F^\alpha(E,\vec{e})\rangle= \mathcal R(R)|F^\beta(E,\vec{e})\rangle \end{equation*} for $R$ with $RE=E$. Pretending the states $|F^\beta(E,\vec{e})\rangle$ to be linearly independent, the branching ${\rm Spin}(d)\hookleftarrow{\rm Spin}(d-1)$ yields \begin{eqnarray*} 16=8_{\rm s}\oplus 8_{\rm c}\quad(d=9)\;;&&\qquad 4\oplus 4=(2_+\oplus 2_-)\oplus(2_+\oplus 2_-)\quad(d=5)\;;\\ 2\oplus 2&=&(1_1\oplus 1_{-1})\oplus(1_1\oplus 1_{-1})\quad(d=3)\;. \end{eqnarray*} For $d=9,5$ each term on the r.h.s. occurs as often as in (\ref{e3.7}), and $\psi_0$ can indeed be chosen so that the $s_d$ vectors $\widehat{Q}_\beta^1\psi_0$ are independent. Not so in the last case, where the vectors $T_\beta\psi_0$ just belong to $1_1\oplus 1_{-1}$. We continue the discussion for different values of $d$ separately. \par $\bullet\,d=9$. Any linear transformation $K$ commuting with a ${\rm Spin}(9)$ representation as above is thus of the form $K=\kappa_{\rm s}\oplus \kappa_{\rm c}$. If $K$ also commutes with the antipode map, then $\kappa_{\rm s}=\kappa_{\rm c}\equiv\kappa$. Applying this to the representation $\widehat{Q}_\beta^1\psi_0$ and to the map $K:\,\widehat{Q}_\beta^1\psi_0\mapsto T_\beta\psi_0$ yields the claim. $\bullet\,d=5$. Let us regroup $(2_+\oplus 2_-)\oplus(2_+\oplus 2_-)\cong (2_+\otimes\one_2)\oplus(2_-\otimes\one_2)$. Then any map $K$ commuting with the representation is of the form \begin{equation*} K=(\one\otimes K_+)\oplus(\one\otimes K_-)\;, \end{equation*} where $K_-$ is conjugate to $K_+$ if $K$ commutes with the antipode map. This allows for a four dimensional space of such maps $K$. To proceed further we shall again assume that $E=(0,\ldots,0,1)$ and introduce creation operators \begin{equation*} a_\alpha^*=\frac{1}{\sqrt{2}}[(\vec{\Theta}_\alpha\cdot\vec{e})+ \i(\vec{\Theta}_{\alpha+4}\cdot\vec{e})]\;,\qquad(\alpha=1,\ldots 4) \end{equation*} which then define a vacuum through $a_\alpha|0\rangle=0$. We next choose an orthonormal basis $\{\psi_0^1,\ldots, \psi_0^4\}$ for the 4-dimensional subspace of singlets in the range of $P_0$ by specifying the values of the corresponding fermionic parts (see (\ref{e3.2})) at $(E,\vec{e})$: \begin{eqnarray*} |F_0^4(E,\vec{e})\rangle&=& \frac{1}{\sqrt{2}}(|0\rangle-a_1^*a_2^*a_3^*a_4^*|0\rangle)\;,\\ |F_0^i(E,\vec{e})\rangle&=& \frac{1}{2\sqrt{2}}\widetilde{\Gamma}^i_{\alpha\beta} a_\alpha^*a_\beta^*|0\rangle= \frac{\i}{4}(\gamma^4\widetilde{\gamma}^i)_{\alpha\beta} (\vec{\Theta}_\alpha\cdot\vec{e})(\vec{\Theta}_\beta\cdot\vec{e}) |F_0^4(E,\vec{e})\rangle\;, \qquad(i=1,2,3)\;,\\ \end{eqnarray*} where \begin{equation*} \widetilde{\gamma}^i=\left( \begin{array}{cc} 0 & \i \widetilde{\Gamma}^i\\ -\i \widetilde{\Gamma}^i& 0 \end{array} \right) =\sigma^{-1}\gamma^i\sigma\;, \qquad \sigma=\left( \begin{array}{cc} \Sigma & 0 \\ 0 & \Sigma\end{array} \right) \end{equation*} with $\Sigma\in{\rm O}(4)$ and $\det\Sigma=-1$. Note that $\psi_0^4$ is the singlet belonging to the 5--dimensional fermionic representation of ${\rm Spin}(5)$. One can verify that the four maps \begin{equation*} K^i:\,\widehat{Q}_\beta^1\psi_0^1\mapsto \cases \widehat{Q}_\beta^1\psi_0^i\;,&(i=1,2,3)\;,\\ \gamma^t_{\beta\alpha}E_t \widehat{Q}_\alpha^1\psi_0^4\;,&(i=4)\;,\\ \endcases \end{equation*} besides being of the kind just discussed, are linearly independent. Therefore any map $K$ of the above form is a linear combination thereof. In particular this applies, for any $(\underline{x}, x_4)\in\mathbb R^{3+1}$, to the map $K:\,\widehat{Q}_\beta^1\psi_0^1\mapsto x_iT_\beta\psi_0^i+x_4\gamma^t_{\beta\alpha}E_t T_\alpha\psi_0^4$, hence \begin{equation*} x_iT_\beta\psi_0^i+x_4\gamma^t_{\beta\alpha}E_t T_\alpha\psi_0^4 =y_i\widehat{Q}_\beta^1\psi_0^i+ y_4\gamma^t_{\beta\alpha}E_t\widehat{Q}_\alpha^1\psi_0^4\;. \end{equation*} This defines a linear map $\kappa: (\underline{x},x_4)\mapsto(\underline{y},y_4)$ on $\mathbb R^{3+1}$. We claim that \begin{equation} \kappa: (R\underline{x},x_4)\mapsto(R\underline{y},y_4) \label{e3.13a} \end{equation} for $R\in {\rm SO}(3)$, which implies $\kappa={\rm diag}(\kappa_1=\kappa_2=\kappa_3, \kappa_4)$ and hence (\ref{e3.14}). Eq. (\ref{e3.13a}) can be proven using $R_{ij}\psi_0^i=\mathcal R\psi_0^j$ for $\mathcal R\in {\rm Spin}(8)$ projecting to $R\in{\rm Spin}(3)\subset{\rm Spin}(5) \hookrightarrow{\rm SO}(8)$. This in turn follows from (\ref{e1.2}) and from $\mathcal R\psi_0^4=\psi_0^4$. $\bullet\,d=3$. Analogously to $d=9$. \subsection{Determination of $\kappa$} \label{sec:pr6} Since $J_{AB}\psi_0=J_{st}\psi_0=0$ we may replace $Q_\beta^1$ by \begin{equation} Q_\beta^1=\Theta_{\alpha A}\gamma^t_{\alpha\beta} \bigl(-e_BE_tM_{BA}-e_AE_sM_{st} -\frac{\i}{2}\,e_AE_ty_{sB}\frac{\partial}{\partial y_{sB}})+ \frac{1}{2}\vec{\Theta}_\alpha\cdot(\vec{y}_s\times\vec{y}_t) \gamma^{st}_{\beta\alpha}\;. \label{e4.1} \end{equation} We discuss the contributions to (\ref{e3.13}) of these four terms separately. \par i) With \begin{equation} e_BM_{BA}=-\frac{\i}{2}\bigl((\vec{\Theta}_\beta\cdot\vec{e})\Theta_{\beta A}- \Theta_{\beta A}(\vec{\Theta}_\beta\cdot\vec{e})\bigr) \nonumber \end{equation} we find \begin{eqnarray*} \Theta_{\alpha A}e_BM_{BA}&=&\i\bigl( (\vec{\Theta}_\alpha \cdot\vec{n}_+)(\vec{\Theta}_\beta\cdot\vec{n}_-)+ (\vec{\Theta}_\alpha \cdot\vec{n}_-)(\vec{\Theta}_\beta\cdot\vec{n}_+) \bigr)(\vec{\Theta}_\beta\cdot\vec{e}\,)\;,\\ P_0\Theta_{\alpha A}e_BM_{BA}\psi_0&=& \i(\vec{\Theta}_\alpha\cdot\vec{e}\,)\psi_0\;, \end{eqnarray*} since only the term with $\beta=\alpha$ survives the projection $P_0$. Hence \begin{equation} -P_0\Theta_{\alpha A}\gamma^t_{\alpha\beta}e_BE_tM_{BA}\psi_0 =\widehat{Q}_\beta^1\psi_0 \label{e4.3} \end{equation} contributes 1 to $\kappa$. \par ii) Similarly, \begin{equation} -P_0(\vec{\Theta}_\alpha\cdot\vec{e}\,)\gamma^t_{\alpha\beta}E_sM_{st}\psi_0= -(\vec{\Theta}_\alpha\cdot\vec{e}\,)\gamma^t_{\alpha\beta}E_s M_{st}^{\parallel}\psi_0\;, \nonumber \end{equation} where $M_{st}^{\parallel}$ is given in (\ref{e3.8a}). For the r.h.s. we then claim \begin{equation} -(\vec{\Theta}_\alpha\cdot\vec{e}\,)\gamma^t_{\alpha\beta}E_s M_{st}^{\parallel}\psi_0=\kappa'\widehat{Q}_\beta^1\psi_0 \label{e4.3a} \end{equation} with \begin{equation} \kappa'=\cases 9\,,&\qquad(d=9)\;,\\ 0,0,0,4\,,&\qquad(d=5)\;,\\ 0,0\,,&\qquad(d=3)\;. \endcases \label{e4.3b} \end{equation} This is clear in the cases where the representation in (\ref{e3.8}) is already a singlet, i.e., when $\kappa'=0$. To prove the two remaining cases we first establish \begin{equation} -(\vec{\Theta}_\alpha\cdot\vec{e}\,)\gamma^t_{\alpha\beta}E_s M_{st}^{\parallel}\psi_0 =-\frac{\i}{2}\gamma^s_{\alpha\beta}E_s[\vec{\Theta}_\alpha\cdot\vec{e}\,, M_{ut}^{\parallel}M_{ut}^{\parallel}]\psi_0 -\i\frac{d^2-d}{8}(\vec{\Theta}_\alpha\cdot\vec{e}\,) \gamma^s_{\alpha\beta}E_s\psi_0\;, \label{e4.4} \end{equation} or the equivalent equation obtained by multiplication from the right with $E_u\gamma^u$: \begin{equation} -(\vec{\Theta}_\alpha\cdot\vec{e}\,)(\gamma^t\gamma^u)_{\alpha\beta} E_uE_sM_{st}^{\parallel}\psi_0= -\frac{\i}{2}[\vec{\Theta}_\beta\cdot\vec{e}\,, M_{ut}^{\parallel}M_{ut}^{\parallel}]\psi_0 -\i\frac{d^2-d}{8}(\vec{\Theta}_\beta\cdot\vec{e}\,)\psi_0\;. \label{e4.5} \end{equation} To this end we note that, by the invariance of $\psi_0$, its fermionic part $|F(E,\vec{e})\rangle$ at $E\in S^{d-1}$ is invariant under rotations of ${\rm Spin}(d) $ leaving $E$ fixed: $(\delta_{us}-E_uE_s)M_{sv}^{\parallel}(\delta_{vt}-E_vE_t)\psi_0=0$, i.e., \begin{equation} (M_{st}^{\parallel}E_uE_s+M_{uv}^{\parallel}E_vE_t)\psi_0= M_{ut}^{\parallel}\psi_0\;. \label{e4.5a} \end{equation} Using $\gamma^t\gamma^u=-\gamma^{ut}+\delta^{ut}\one$ and the observation just made we rewrite the l.h.s. of (\ref{e4.5}) as \begin{eqnarray*} -(\vec{\Theta}_\alpha\cdot\vec{e}\,)(\gamma^t\gamma^u)_{\alpha\beta} E_uE_sM_{st}^{\parallel}\psi_0&=& (\vec{\Theta}_\alpha\cdot\vec{e}\,) \gamma^{ut}_{\alpha\beta}E_uE_sM_{st}^{\parallel}\psi_0\cr &=&\frac{1}{2}(\vec{\Theta}_\alpha\cdot\vec{e}\,)\gamma^{ut}_{\alpha\beta} (E_uE_sM_{st}^{\parallel}-E_tE_sM_{su}^{\parallel})\psi_0\cr &=&\frac{1}{2}(\vec{\Theta}_\alpha\cdot\vec{e}\,)\gamma^{ut}_{\alpha\beta} M_{ut}^{\parallel}\psi_0\;. \end{eqnarray*} The commutation relation \begin{equation*} \i[\vec{\Theta}_\alpha\cdot\vec{e}\, ,M_{ut}^{\parallel}]= \frac{1}{2}\gamma^{ut}_{\alpha\beta}(\vec{\Theta}_\beta\cdot\vec{e}\,) \end{equation*} follows from (\ref{e1.2}) or by direct computation. It implies \begin{eqnarray*} \i[\vec{\Theta}_\alpha\cdot\vec{e}\, ,M_{ut}^{\parallel}M_{ut}^{\parallel}]&=& \frac{1}{2}\gamma^{ut}_{\alpha\beta}\{ \vec{\Theta}_\beta\cdot\vec{e}\, ,M_{ut}^{\parallel}\}= \gamma^{ut}_{\alpha\beta}(\vec{\Theta}_\beta\cdot\vec{e}\,)M_{ut}^{\parallel} -\frac{1}{2}\gamma^{ut}_{\alpha\beta} [\vec{\Theta}_\beta\cdot\vec{e}\,,M_{ut}^{\parallel}]\cr &=& \gamma^{ut}_{\alpha\beta}(\vec{\Theta}_\beta\cdot\vec{e}\,) M_{ut}^{\parallel} -\i\frac{d^2-d}{4}\,\vec{\Theta}_\alpha\cdot\vec{e}\,\;. \end{eqnarray*} Solving for the first term on the r.h.s. proves (\ref{e4.5}) and hence (\ref{e4.4}). Let us now note that for $d=9$ the fermionic part of $\psi_0$, resp. of $(\vec{\Theta}_\alpha\cdot\vec{e}\,)\psi_0$ belongs to the $44$, resp. $128$ representation of ${\rm Spin}(9)$ (see (\ref{e3.6})). Eq. (\ref{e4.4}) then implies \begin{equation*} -(\vec{\Theta}_\alpha\cdot\vec{e}\,)\gamma^t_{\alpha\beta}E_s M_{st}^{\parallel}\psi_0= (C(44)-C(128)+9)\widehat{Q}_\beta^1\psi_0=9\widehat{Q}_\beta^1\psi_0\;, \end{equation*} where we used the values \cite{b4} of the Casimir: $C(44)=C(128)=18$. In the case $d=5$ the fermionic part of $\psi_0$, resp. of $(\vec{\Theta}_\alpha\cdot\vec{e}\,)\psi_0$ belongs to the representation $5$, resp. $4\oplus 4$. We conclude that \begin{equation*} -(\vec{\Theta}_\alpha\cdot\vec{e}\,)\gamma^t_{\alpha\beta}E_s M_{st}^{\parallel}\psi_0= (C(5)-C(4)+\frac{5}{2})\widehat{Q}_\beta^1\psi_0=4\widehat{Q}_\beta^1\psi_0\;, \end{equation*} given that $C(5)=4,\,C(4)=5/2$. \par We remark that the proof of (\ref{e4.3b}) can be shortened by using the lemma, according to which (\ref{e4.3a}) holds true for some $\kappa'$. Thus, contracting with $\widehat{Q}_\beta^1\psi_0$ and summing over $\beta$, we find \begin{eqnarray*} -\kappa'(\psi_0,\widehat{Q}_\beta^1\widehat{Q}_\beta^1\psi_0)&=& -\i(\psi_0,(\vec{\Theta}_\gamma\cdot\vec{e})\gamma^u_{\gamma\beta}E_u (\vec{\Theta}_\alpha\cdot\vec{e})\gamma^t_{\alpha\beta}E_s M_{st}^{\parallel}\psi_0)\\ &=&4(\psi_0,E_uM_{ut}^{\parallel}M_{st}^{\parallel}E_s\psi_0)\\ &=&2(\psi_0,M_{ut}^{\parallel} (M_{st}^{\parallel}E_uE_s+M_{uv}^{\parallel}E_vE_t)\psi_0) =2(\psi_0,M_{ut}^{\parallel}M_{ut}^{\parallel}\psi_0)\;. \end{eqnarray*} In the step before last we relabeled indices in half the expression; in the last one we used (\ref{e4.5a}). Using $\widehat{Q}_\beta^1\widehat{Q}_\beta^1=-s_d/2$ we obtain $(s_d/2)\kappa'=2\cdot 2\cdot C$, i.e., $\kappa'=8C/s_d$, where $C$ is the Casimir in the representation (\ref{e3.8}). The above values of $C(44)\,(d=9)$ and of $C(5)\,(d=5)$ yield again (\ref{e4.3b}). \par iii) Using $\d\e^{-y^2/2}/\d y= -y\e^{-y^2/2}$ we get \begin{equation} \frac{1}{2}y_{sB}\frac{\partial}{\partial y_{sB}}\psi_0= -\frac{1}{2}y_{sB}y_{sB}\psi_0= -\frac{1}{2}\sum_{sB}(y_{sB}^2-\frac{1}{2})\psi_0- \frac{1}{4}\cdot 2(d-1)\psi_0\;, \label{e4.9} \end{equation} where the sum, consisting of second Hermite functions, is annihilated by $P_0$. \par iv) The last term in (\ref{e4.1}), when acting on $\psi_0$, is similarly annihilated by $P_0$. \smallskip Collecting terms (\ref{e4.3}, \ref{e4.3b}, \ref{e4.9}) we find \begin{equation*} \kappa=1+\kappa'-\frac{1}{2}(d-1)=\cases 6\,,&\qquad(d=9)\;,\\ -1,-1,-1,3\,,&\qquad(d=5)\;,\\ 0, 0\,,&\qquad(d=3)\;. \endcases \end{equation*} \setcounter{section}{0} \section*{Appendix 1} To prove (\ref{e3.000}) we shall compute the partial derivatives in \begin{equation} \frac{\partial}{\partial q_{tA}}= \frac{\partial r}{\partial q_{tA}}\frac{\partial}{\partial r}+ \frac{\partial e_B}{\partial q_{tA}}\frac{\partial}{\partial e_B}+ \frac{\partial E_s}{\partial q_{tA}}\frac{\partial}{\partial E_s}+ \frac{\partial y_{sB}}{\partial q_{tA}}\frac{\partial}{\partial y_{sB}}\;. \label{eq:a0} \end{equation} We regard $r,\,\vec{e},\,E,\,y$ as functions of $q$ defined by $\vec{e}{\,}^2=\sum_s E_s^2 = 1$ and (\ref{e2.00}, \ref{e2.1}) and solve for their differentials by taking different contractions of \begin{equation*} \d q_{tA}=(e_AE_t-\frac{1}{2}r^{-3/2}y_{tA})\d r+ rE_t\d e_A+re_A\d E_t+r^{-1/2}\d y_{tA}\;. \end{equation*} Using that \begin{eqnarray*} e_A\d y_{tA}+y_{tA}\d e_A=0\;,&\qquad E_t\d y_{tA}+y_{tA}\d E_t=0\;,&\\ e_A\d e_A=0\;,&\qquad E_t\d E_t=0\;,& \end{eqnarray*} the contractions are: \begin{eqnarray} e_AE_t\d q_{tA}&\!=\!&\d r\;,\nonumber\\ (\delta_{BA}-e_Be_A)E_t\d q_{tA}&\!=\!&r\d e_B-r^{-1/2}y_{tA}\d E_t\;, \label{eq:a1}\\ e_A(\delta_{st}-E_sE_t)\d q_{tA}&\!=\!&r\d E_s-r^{-1/2}y_{sA}\d e_A\;, \label{eq:a2}\\ (\delta_{BA}-e_Be_A)(\delta_{st}-E_sE_t)\d q_{tA}&\!=\!& -\frac{1}{2}r^{-3/2}y_{sB}\d r+ r^{-1/2}(\d y_{sB}+e_By_{sA}\d e_A+E_sy_{tB}\d E_t)\;. \nonumber%\label{eq:a3} \end{eqnarray} We solve (\ref{eq:a1}, \ref{eq:a2}) for $\d e_B,\, \d E_s$: \begin{eqnarray*} \d r&=&e_AE_t\d q_{tA}\;,\\ \d e_B&=&(m^{-1})_{BC}(r^{-1}(\delta_{CA}-e_Ce_A)E_t+r^{-5/2}y_{tC}e_A) \d q_{tA}\\ &=&(r^{-1}(\delta_{BA}-e_Be_A)E_t+{\rm O}(r^{-5/2}))\d q_{tA}\;,\\ \d E_s&=&(M^{-1})_{su}(r^{-1}(\delta_{ut}-E_uE_t)e_A+r^{-5/2}y_{sA}E_t) \d q_{tA}\\ &=&(r^{-1}(\delta_{st}-E_sE_t)e_A+{\rm O}(r^{-5/2}))\d q_{tA}\;,\\ \d y_{sB}&=&[r^{1/2}(\delta_{BA}-e_Be_A)(\delta_{st}-E_sE_t)+ \frac{1}{2}r^{-1}e_AE_ty_{sB}]\d q_{tA} -e_By_{sA}\d e_A-E_sy_{tB}\d E_t\;, \end{eqnarray*} where $m,\, M$ are the matrices \begin{equation*} m_{AB}=\delta_{AB}-r^{-3}y_{tA}y_{tB}\;,\qquad M_{st}=\delta_{st}-r^{-3}y_{sA}y_{tA}\;. \end{equation*} We can now read off the partial derivatives appearing in (\ref{eq:a0}) and obtain \begin{eqnarray} \frac{\partial}{\partial q_{tA}}&=& r^{1/2}(\delta_{st}-E_sE_t)(\delta_{AB}-e_Ae_B) \frac{\partial}{\partial y_{sB}} +r^{-1}[e_AE_t(r\frac{\partial}{\partial r}+ \frac{1}{2}y_{sB}\frac{\partial}{\partial y_{sB}})] \nonumber\\ &&+r^{-1}(\delta_{AC}-e_Ae_C)E_t (\delta_{CB}\frac{\partial}{\partial e_B}-e_By_{sC}\frac{\partial}{\partial y_{sB}}) \nonumber\\ &&+r^{-1}(\delta_{ut}-E_uE_t)e_A (\delta_{us}\frac{\partial}{\partial E_s}-E_sy_{uB}\frac{\partial}{\partial y_{sB}}) +{\rm O}(r^{-5/2})\;,\label{eq:a4} \end{eqnarray} with the remainder not containing derivatives w.r.t. $r$. Finally, we insert this expression into \begin{eqnarray*} \i L_{BA}&=& q_{sB}\frac{\partial}{\partial q_{sA}}-q_{sA}\frac{\partial}{\partial q_{sB}}\\ &=&[(\delta_{AC}-e_Ae_C)y_{sB}-(\delta_{BC}-e_Be_C)y_{sA}] \frac{\partial}{\partial y_{sC}}\\ &&+e_B(\delta_{AC}\frac{\partial}{\partial e_C}- e_Cy_{sA}\frac{\partial}{\partial y_{sC}}) -e_A(\delta_{BC}\frac{\partial}{\partial e_C}- e_Cy_{sB}\frac{\partial}{\partial y_{sC}})\;, \end{eqnarray*} (with no higher order corrections, as $L_{AB}$ is of exact order ${\rm O}(r^0)$) and then into \begin{equation*} \i r^{-1} e_BE_tL_{BA}=r^{-1}(\delta_{AC}-e_Ae_C)E_t (\delta_{CB}\frac{\partial}{\partial e_B}-e_By_{sC}\frac{\partial}{\partial y_{sB}})\;. \end{equation*} Similarly, we have \begin{equation*} \i r^{-1} e_AE_sL_{st}=r^{-1}(\delta_{ut}-E_uE_t)e_A (\delta_{us}\frac{\partial}{\partial E_s}-E_sy_{uB}\frac{\partial}{\partial y_{sB}})\;. \end{equation*} Together with (\ref{eq:a4}), this proves (\ref{e3.000}). \section*{Appendix 2} Consider \begin{equation} \label{xy-Hamiltonian} H = ( -{\partial_x}^2 -{\partial_y}^2 + x^2y^2 ) \one + \left( \begin{array}{cc} x & -y \\ -y & -x \end{array} \right)\;, \end{equation} which is the square of \begin{equation*} %\label{xy-supercharge} Q = \i \left( \begin{array}{cc} \partial_x & \partial_y + xy \\ \partial_y - xy & - \partial_x \end{array} \right) . \end{equation*} Just as in (\ref{e1.4}), the bosonic potential $V$ ($=x^2y^2$) is non-negative, but vanishing in regions of the configuration space that extend to infinity (causing the classical partition function to diverge). Quantum--mechanically, just as in (\ref{e1.4}), the bosonic system is stabilized by the zero point energy of fluctuations transverse to the flat directions; the fermionic matrix part in (\ref{xy-Hamiltonian}) exactly cancels this effect, causing the spectrum to cover the whole positive real axis \cite{c4b}. As simple as it is, it has remained an open question (for now more than 10 years) whether (\ref{xy-Hamiltonian}) admits a normalizable zero energy solution, or not. The argument, derived in a few lines below, gives `no' as an answer and provides the simplest illustration of our method: as $x \rightarrow + \infty, \ Q \Psi = 0$ has two approximate solutions, \begin{equation} \label{xy-solutions} \Psi_{+} = e^{-\frac{xy^2}{2}} \left( \begin{array}{c} 0 \\ 1 \end{array} \right) \quad \mathrm{and} \quad \Psi_{-} = e^{+\frac{xy^2}{2}} \left( \begin{array}{c} 1 \\ 0 \end{array} \right)\;, \end{equation} the first of which should be chosen for $\Psi_{0}$ in the asymptotic expansions \begin{equation} \label{xy-expansion} \Psi = x^{-\kappa} ( \Psi_{0} + \Psi_{1} + ... )\; . \end{equation} In this simple example, the sum $Q = \sum_{n=0}^{\infty} Q^{(n)}$ terminates after the first two terms, and \begin{equation*} 0 \stackrel{\mathrm{!}}{=} Q \Psi = \left( \left( \begin{array}{cc} 0 & \partial_y + xy \\ \partial_y -xy & 0 \end{array} \right) + \left( \begin{array}{cc} \partial_x & 0 \\ 0 & - \partial_x \end{array} \right) \right) \left( x^{- \kappa} ( \Psi_0 + \Psi_1 + ... ) \right)\;, \end{equation*} yields (as already anticipated, cp. (\ref{xy-solutions})) \begin{equation*} % \label{xy_eqn1} \left( \begin{array}{cc} 0 & \partial_y + xy \\ \partial_y -xy & 0 \end{array} \right) \Psi_0 = 0 \end{equation*} and \begin{equation} \label{xy-eqn2} \left( \begin{array}{cc} 0 & \partial_y + xy \\ \partial_y -xy & 0 \end{array} \right) \Psi_n + x^{\kappa} \left( \begin{array}{cc} \partial_x & 0 \\ 0 & - \partial_x \end{array} \right) x^{-\kappa} \Psi_{n-1} = 0\;, \qquad n = 1, 2, ... \; . \end{equation} Multiplying (\ref{xy-eqn2}) by $\Psi_{0}^{\dagger}$ and integrating over $y$ one sees that \begin{equation*} \int_{- \infty}^{+ \infty} e^{-\frac{xy^2}{2}} x^{\kappa}( 0 , - \partial_x) x^{-\kappa} \Psi_{n-1} dy \end{equation*} has to vanish, implying in particular \begin{eqnarray*} 0 & = & \int_{- \infty}^{+ \infty} \left( \frac{y^2}{2} + \frac{\kappa}{x} \right) e^{- x y^2} dy\;, \\ \nonumber \kappa & = & - \frac{1}{4} \; , \end{eqnarray*} which proves that (\ref{xy-Hamiltonian}) does not admit any square--integrable solution of the form (\ref{xy-expansion}). A different approach has recently been undertaken by Avramidi \cite{avr}. Finally note that, calculating the $\Psi_{n > 0}$ from (\ref{xy-eqn2}), yields the asymptotic expansion, $x \rightarrow + \infty $, \begin{equation*} \Psi(x,y) = x^{\frac{1}{4}} e^{-\frac{xy^2}{2}} \sum_{n=0}^{\infty} x^{- \frac{3n}{2}} \left( \begin{array}{c} \frac{y}{4x} f_n(xy^2) \\ g_n(xy^2) \end{array} \right)\;, \end{equation*} where $f_0 = 1 = g_0,\, f_1=0=g_1$, and the $f_n (s), \,g_{n}(s)$ are the (unique) polynomial solutions \begin{equation*} f_n (s) = \sum_{i = 0}^n f_{n,i} s^i\; , \qquad g_n(s) = \sum_{ i = 0}^n g_{n,i} s^i \end{equation*} of \begin{eqnarray*} 2s f'_n + ( 1 - 2s ) f_n & = & \left( 1 - 2s - 6n \right) g_n + 4s g'_n\;, \\ \nonumber 8 g'_{n+2} & = & \left( \frac{3}{4} + \frac{s}{2} + \frac{3n}{2} \right) f_n - s f'_n\; . \end{eqnarray*} \medskip \noindent {\bf Acknowledgments.\/} We thank A. Alekseev, I. Avramidi, V. Bach, F. Finster, H. Nicolai, C. Schweigert, R. Suter, P. 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