text:
\begin{document}

\renewcommand{\thefootnote}{$\star$}

\renewcommand{\PaperNumber}{087}

\FirstPageHeading

\ShortArticleName{The $N=1$ Triplet Vertex Operator Superalgebras:
Twisted Sector}

\ArticleName{The $\boldsymbol{N=1}$ Triplet Vertex Operator Superalgebras:\\
Twisted Sector\footnote{This paper is a
contribution to the Special Issue on Kac--Moody Algebras and Applications. The
full collection is available at
\href{http://www.emis.de/journals/SIGMA/Kac-Moody_algebras.html}{http://www.emis.de/journals/SIGMA/Kac-Moody{\_}algebras.html}}}

\Author{Dra\v{z}en ADAMOVI\'C~$^\dag$ and Antun MILAS~$^\ddag$}

\AuthorNameForHeading{D. Adamovi\'c  and A. Milas }

 \Address{$^\dag$~Department of Mathematics, University of Zagreb, Croatia}
\EmailD{\href{mailto:adamovic@math.hr}{adamovic@math.hr}}
\URLaddressD{\url{http://web.math.hr/~adamovic}}

\Address{$^\ddag$~Department of Mathematics and Statistics, University at
Albany (SUNY),\\
\hphantom{$^\ddag$}~Albany, NY 12222, USA}
\EmailD{\href{mailto:amilas@math.albany.edu}{amilas@math.albany.edu}}
\URLaddressD{\url{http://www.albany.edu/~am815139}}

\ArticleDates{Received August 31, 2008, in f\/inal form December 05,
2008; Published online December 13, 2008}

\Abstract{We classify irreducible $\sigma$-twisted modules for the
$N=1$ super triplet vertex operator  superalgebra $\mathcal{SW}(m)$
introduced recently [Adamovi\'{c}  D., Milas A.,  {\em Comm. Math. Phys.}, to appear,
\href{http://arxiv.org/abs/0712.0379}{arXiv:0712.0379}].
Irreducible
graded dimensions of $\sigma$-twisted modules are also determined.
These results, combined with our previous work in the untwisted
case, show that the $SL(2,\mathbb{Z})$-closure of the space spanned
by irreducible characters, irreducible supercharacters and
$\sigma$-twisted irreducible characters is $(9m+3)$-dimensional. We
present strong evidence that this is also the (full) space of
generalized characters for $\mathcal{SW}(m)$.  We are also able to
relate irreducible $\mathcal{SW}(m)$ characters to characters for
the triplet vertex algebra $\mathcal{W}(2m+1)$, studied in
[Adamovi\'{c} D., Milas A., {\em Adv. Math.} {\bf 217} (2008), 2664--2699,
\href{http://arxiv.org/abs/0707.1857}{arXiv:0707.1857}].}

\Keywords{vertex operator superalgebras; Ramond twisted
representations}

\Classification{17B69; 17B67; 17B68; 81R10}

\def \<{\langle}
\def \>{\rangle}
\def \t{\tau }
\def \a{\alpha }
\def \e{\epsilon }
\def \l{\lambda }
\def \L{ L}
\def \T{\mathcal T}
\def \b{\beta }
\def \om{\omega }
\def \o{\omega }
\def \c{\chi}
\def \ch{\chi}
\def \cg{\chi_g}
\def \ag{\alpha_g}
\def \ah{\alpha_h}
\def \ph{\psi_h}
\def \gi {\gamma}

\newcommand{\wta}{{\rm {wt} }  a }
\newcommand{\R}{\frak R}

\newcommand{\wtb}{{\rm {wt} }  b }

\newcommand{\g}{\frak g}

\def \ga{\gamma }
\newcommand{\hg}{\hat {\frak g} }
\newcommand{\hn}{\hat {\frak n} }
\newcommand{\h}{\frak h}
\newcommand{\wt}{{\rm {wt} }   }
\newcommand{\V}{\cal V}
\newcommand{\hh}{\hat {\frak h} }

\newcommand{\n}{\frak n}
\newcommand{\Z}{\Bbb Z}
\newcommand{\W}{\mathcal W}

\newcommand{\Zp}{{\Bbb Z}_{>0} }

\newcommand{\N}{{\Bbb Z}_{\ge 0} }
\newcommand{\C}{\Bbb C}
\newcommand{\Q}{\Bbb Q}
\newcommand{\vak}{\bf 1}

\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}

\newcommand{\NS}{\frak{ns} }
\newcommand{\triplet}{\mathcal{W}(p)}
\newcommand{\striplet}{\mathcal{SW}(m)}
\newcommand{\ssinglet}{\overline{{SM}(1)}}
\newcommand{\hf}{\mbox{$\frac{1}{2}$}}
\newcommand{\thf}{\mbox{$\frac{3}{2}$}}

\section{Introduction}

Many constructions and results in vertex algebra theory are easily extendable to the setup of vertex superalgebras by simply adding adjective ``super''. Still, there are results that deviate from this ``super-principle'' and new ideas are needed compared to the non-super case. For example, modular invariance for vertex operator superalgebras requires inclusion of supercharacters of (untwisted) modules, and more importantly, the characters of $\sigma$-twisted modules \cite{FFR}, where $\sigma$ is the canonical parity automorphism. Since the construction and classif\/ication of $\sigma$-twisted modules is more or less independent of the untwisted construction, many aspects of the theory need to be reworked for the twisted modules (e.g., twisted Zhu's algebra~\cite{Xu}). In fact, even for the free fermion vertex operator superalgebra, construction of $\sigma$-twisted modules is far from being trivial (see~\cite{FFR} for details).

Present work is a natural continuation of our very recent paper
\cite{AdM-str}, where we introduced a~new family of $C_2$-cof\/inite
$N=1$ vertex operator superalgebra that we call the {\em
supertriplet} family~$\mathcal{SW}(m)$, $m \geq 1$. In this
installment we focus on $\sigma$-twisted $\mathcal{SW}(m)$-modules
and their irreducible characters. The $\sigma$-twisted $\striplet$-modules are usually called modules in the {\em Ramond sector}, while untwisted (ordinary) modules are referred to as modules in the {\em Neveu--Schwarz sector}, in parallel with two distinct $N=1$ superconformal algebras.

Of course, Ramond sector has been studied for various (mostly
rational) vertex operators superalgebras (e.g.,
\cite{Li-tw,D2,M}, etc.). We should also point out
that Ramond twisted representation of certain vertex
$\mathcal{W}$-algebras were recently investigated in  \cite{Ar-t}
and \cite{KW-t}.

Here are our main results:

\begin{theorem}
The twisted Zhu's algebra $A_{\sigma}(\striplet)$ is finite-dimensional with precisely $2m\!{+}1$, non-isomorphic,  $\mathbb{Z}_2$-graded irreducible modules.
 Consequently, $\striplet$ has $2m+1$, non-iso\-mor\-phic, $\mathbb{Z}_2$-graded irreducible $\sigma$-twisted modules.
\end{theorem}

By using embedding structure of $N=1$ Feigin--Fuchs modules \cite{IK3} we can also easily compute the characters of $\sigma$-twisted $\striplet$-modules. Equipped with these formulas, results from \cite{AdM-str} about untwisted modules and supercharacters, and transformation formulas for classical Jacobi theta functions we are able to prove:
\begin{theorem}
The $SL(2,\mathbb{Z})$-closure of the space of characters, (untwisted) supercharacters and $\sigma$-twisted $\striplet$ characters is $(9m+3)$-dimensional.
\end{theorem}

Conjecturally, this is also the space of certain {\em generalized characters} studied in \cite{Miy}  (strictly speaking pseudotraces are yet-to-be def\/ined in the setup of $\sigma$-twisted modules). A closely
related conjecture is

\begin{conjecture} Let $Z(A)$ denote the center of associative algebra $A$ and $T(A)=A/[A,A]$ the trace group of $A$. Then
\[
\dim (Z(A(\striplet)))+\dim (Z(A_{\sigma}(\striplet)))=9m+3,
\]
and
\[
\dim (T(A(\striplet)))+\dim (T(A_{\sigma}(\striplet)))=9m+3,
\]
where $A(\striplet)$ is the untwisted Zhu's associative algebra of
$\striplet$ (cf.~{\rm \cite{AdM-str}}).
\end{conjecture}
The conjecture would follow if we knew more about structure of logarithmic $\striplet$-modules.

Here is a short outline of the paper.  In Section~\ref{prelim} we
recall the standard results about the $\sigma$-twisted Zhu's
algebra, the free fermion vertex superalgebra $F$, and the
construction of the $\sigma$-twisted $F$-module(s). In Section
\ref{singlet-ramond} we focus on $\sigma$-twisted modules for the
super singlet vertex algebra introduced also in \cite{AdM-str}.
Sections \ref{free-fields} and \ref{classification} deal with the
construction and classif\/ication of $\sigma$-twisted
$\striplet$-modules. Finally, in Section \ref{characters} we derive
characters formulas for irreducible $\sigma$-twisted
$\striplet$-modules, discuss modular invariance, and relate our
characters with the cha\-rac\-ters for the triplet vertex algebra
$\mathcal{W}(2m+1)$ (see for instance \cite{AdM-triplet} and \cite{FGST}).

Needless to say, this paper is largely continuation of \cite{AdM-triplet} and especially \cite{AdM-str}.  Thus the reader is strongly encouraged to consult \cite{AdM-str}, where we studied $\striplet$ in great details.

\section{Preliminaries}
\label{prelim} Let $V=V_0 \oplus V_1$ be a vertex operator
superalgebra, where as usual $V_0$ is the even and $V_1$ is the odd
subspace (cf.~\cite{FFR,HM,K}). Every vertex operator superalgebra has the canonical
parity automorphism $\sigma$, where $\sigma_{V_0}=1$ and
$\sigma_{V_1}=-1$. This leads to the notion of $\sigma$-twisted
$V$-modules, well recorded in the literature (see for example
\cite{FRW} and \cite{Xu}). As in the untwisted case, a large part of
representation theory of $\sigma$-twisted $V$-modules can be
analyzed via the $\sigma$-twisted Zhu's algebra whose construction
we recall here.

Within the same setup, consider the subspace $O(V)
\subset V$, spanned by elements of the form
\[
{\rm Res}_x \frac{(1+x)^{{\rm deg}(u)}}{x^2} Y(u,x)v,
\]
where $u \in V$ is homogeneous. It can be easily shown that
\[
{\rm Res}_x \frac{(1+x)^{{\rm deg}(u)}}{x^n} Y(u,x)v \in O(V) \qquad \mbox{for} \quad n \geq 2.
\]
Then, the vector space $A_\sigma(V)=V/O(V)$ is equipped with
an associative algebra structure via
\[
u * v={\rm Res}_x \frac{(1+x)^{\rm deg(u)}}{x} Y(u,x)v
\]
(see~\cite{Xu}). An important dif\/ference between the untwisted
associative algebra $A(V)$ and $A_\sigma(V)$ is that the latter is
$\mathbb{Z}_2$-graded, so
\[
A_\sigma(V)=A^0_\sigma(V) \oplus A^1_\sigma(V).
\]
It is also not clear that $A_{\sigma}(V) \neq 0$, while $A(V)$ is
always nontrivial. We shall often use $[a] \in A_{\sigma}(V)$ for
the image of $a \in V$ under the natural map $V \longrightarrow
A_{\sigma}(V)$.

The result we need is \cite{Xu}:

\begin{theorem} \label{zhu-corr} There is a one-to-one correspondence between irreducible
$\mathbb{Z}_{\geq 0}$-gradable $\sigma$-twisted $V$-modules and
irreducible $A_{\sigma}(V)$-modules.
\end{theorem}
In the theorem there is no reference to graded $A_{\sigma}(V)$-modules.
For practical purposes we shall need a slightly dif\/ferent version of the above theorem, because some modules are more natural if considered as $\mathbb{Z}_2$-graded modules (shorthand,  {\em graded} modules).

\begin{theorem} \label{zhu-graded-corr} There is a one-to-one correspondence between graded irreducible $\mathbb{Z}_{\geq 0}$-gradable $\sigma$-twisted $V$-module
and graded irreducible $A_{\sigma}(V)$-module.
\end{theorem}

\begin{proof}
The proof mimics the non-graded case, so we shall omit
details. As in the non-graded case,  by applying Zhu's theory, from
an irreducible graded $A_{\sigma}(V)$-module $U$ we construct a~$\sigma$-twisted graded $V$-module $L(U)$ (but of course in the
process of getting $L(U)$ will be moding out by the maximal {\em
graded} submodule). On the other hand,  if $M$ is an irreducible
graded $V$-module, then the top component $\Omega(M)$ is clearly a
graded $A_{\sigma}(V)$-module. Suppose that $\Omega(M)$ is not
graded irreducible, therefore there is a graded submodule
$\Omega(M')$. Then  we let $M'=\mathcal{U}(V[\sigma])\Omega'(M) $,
where $\mathcal{U}(V[\sigma])$ is the enveloping algebra of the Lie
algebra $V[\sigma]$.
 But $M'$ is a proper graded submodule of $M$, so we get  a contradiction.
\end{proof}

Our goals are to describe the structure of
$A_\sigma(\overline{SM(1)})$, where $\overline{SM(1)}$ is the super
singlet vertex algebra \cite{AdM-str}, and to discuss
$A_\sigma(\striplet)$ (in fact, we have a very precise conjecture
about the structure of $A_\sigma(\striplet)$). Let us recall (see
\cite{AdM-str} for details) that both $\ssinglet$ and $\striplet$
are $N=1$ superconformal vertex operator superalgebras, with the
superconformal vector $\tau$. In other words, if we let
$Y(\tau,x)=\sum\limits_{n \in \mathbb{Z}+1/2} G(n)x^{-n-3/2}$, then $G(n)$
and $L(m)$ close the $N=1$ Neveu--Schwarz superalgebra.

 It is not
hard to see that the following hold:{\samepage
\begin{gather}
 (L(-m-2)+2L(-m-1)+L(-m))v \in O(V), \qquad m \geq 2, \nonumber\\
(L(0)+L(-1))v \in O(V), \nonumber\\
 L(-m)v \equiv (-1)^m((m-1)(L(-2)+L(-1))+L(0))v \mod O(V), \qquad m \geq 2,\nonumber\\
   \sum_{n \geq 0} {\frac{3}{2} \choose n} G(-3/2-i+n) v \in O(V),
\qquad i \geq 1,   \label{ramond-relations}
 \end{gather}
  where in all formulas $v$ is a vector in $N=1$
vertex operator superalgebra $V$.}

To illustrate how to use twisted Zhu's algebra, let us classify
$\mathbb{Z}$-graded $\sigma$-twisted modules for the (neutral) free
fermion vertex operator superalgebra $F$, used in \cite{AdM-str}.
The next result is known so we will not provide its proof here.

 \begin{proposition}   We have $A_{\sigma}(F) \cong {\C}[x]/\big(x^2-\frac{1}{2}\big)$ where $x=[\phi(-1/2){\bf 1}]$,
an odd generator in the associative algebra.
\end{proposition}

 This result in particular implies there are precisely two
irreducible $\sigma$-twisted $F$-modules: $M^{\pm}$. These two
modules can be constructed explicitly. As vector spaces
\[
M^{\pm}=
\oplus_{n \geq 0} \Lambda^n (\phi({-1}),\phi({-2}),\dots),
\]
 where
$\Lambda^*$ is the exterior algebra, which is also a
$\mathbb{Z}_{\geq 0}$-graded module for the Clif\/ford algebra $K$
spanned by $\phi({n})$, $n \in \mathbb{Z}$ with anti-bracket
relations $\{ \phi(m),\phi(n) \}=\delta_{m+n,0}$. The only
dif\/ference between $M^+$ and $M^-$  is in the action of $\phi(0)$ on
the one-dimensional top subspace $M^\pm(0) $. More precisely we have
$\phi(0)|_{M^\pm(0)}=\pm \frac{1}{\sqrt{2}}$. However, notice that
$M^{\pm}$ are not $\mathbb{Z}_2$-graded thus it is more natural to examine graded
$A_{\sigma}(F)$-modules. There is a unique such module (up to parity
switch), spanned  by ${\bf 1}^{\R}$ and $\phi(0){\bf 1}^{\R}$.

 Thus
\[
M=\oplus_{n \geq 0} \Lambda^n(\phi(0), \phi({-1}),\phi({-2}),\dots) =
M^+ \oplus M^-.
\]

Then   ${\bf 1}^{\R}$ is a cyclic vector in $M$, i.e., $M= K. {\bf
1}^{\R}$. Moreover, $ M^{\pm} = K. {\bf 1}^{\pm}$,

where
\[
 {\bf 1}^{\pm} = {\bf 1} ^{\R} \pm \sqrt{2} \phi(0) {\bf 1} ^{\R}.
\]

 Next we describe the  twisted vertex operators
 \[
 \overline{Y} :  \ \ F
\otimes M \longrightarrow M[[x^{1/2},x^{-1/2}]] .
\]
 Details are
spelled out in \cite{FFR}, here we only give the explicit formula.
Def\/ine f\/irst
\[
Y\big(\phi\big({-}n-\tfrac{1}{2}\big){\bf 1},x\big)=\frac{1}{n!}\left(\frac{d}{dx}\right)^{n}\left(\sum_{m
\in \mathbb{Z}} \phi(m)x^{-m-1/2}\right),
\] acting on $M$ and use
(fermionic!) normal ordering ${\nordbullet } \ \ {\nordbullet }$ to
def\/ine
\[
Y\big(\phi\big({-}n_1-\tfrac{1}{2}\big)\cdots \phi\big({-}n_k-\tfrac{1}{2}\big){\bf 1},x\big)=\nordbullet Y\big(\phi\big({-}n_1-\tfrac{1}{2}\big),x\big) \cdots Y\big(\phi\big({-}n_k-\tfrac{1}{2}\big),x\big) \nordbullet,
\]
and extend $Y$ by linearity on all of $F$ (see \cite{FFR} and \cite{FRW} for details, especially about normal ordering).
Then, we let
\[
\overline{Y}(v,x):=Y(e^{\Delta_x}v,x),
\]
where
\begin{gather*}
  \Delta_x=\frac{1}{2} \sum_{m,n \geq 0}  C_{m,n} \phi\big(m+\tfrac{1}{2}\big)\phi\big(n+\tfrac{1}{2}\big) x^{-m-n-1}, \nonumber \\
  C_{m,n}=\frac{1}{2}\frac{m-n}{m+n+1}{-1/2 \choose m}{-1/2 \choose n}.\nonumber
\end{gather*}
Let us f\/ix the Virasoro generator
\[
\omega_s=\tfrac{1}{2} \phi\big({-}\tfrac{3}{2}\big)\phi\big({-}\tfrac{1}{2}\big){\bf 1}.
\]
Then $e^{\Delta_{x}}\omega_s=\omega_s+\frac{1}{16}x^{-2}{\bf 1}$.

The following lemma will be important in the rest of the paper.

\begin{lemma} \label{cmn} Let
\[
G(x_1,x_2)=\sum_{m=0}^\infty \sum_{n=0}^\infty C_{m,n} x_1^m x_2^n,
\]
where $C_{m,n}$ are as above. Then
\[
G(x_1,x_2)=\frac{1}{2}\left(\frac{(1+x_1)^{1/2}(1+x_2)^{-1/2}-1}{x_2-x_1}+\frac{(1+x_1)^{-1/2}(1+x_2)^{1/2}-1}{x_2-x_1}\right)  \in \mathbb{Q}[[x_1,x_2]].
\]
\end{lemma}

\begin{proof} The lemma follows directly from the identity
\[
(x_2-x_1)G(x_1,x_2)=\tfrac{1}{2}(1+x_1)^{1/2}(1+x_2)^{-1/2}+\tfrac{1}{2}(1+x_1)^{-1/2}(1+x_2)^{1/2}-1.
\]
which can be easily checked by expanding both sides as power
series in $x_1$ and $x_2$.
 \end{proof}

\section{Highest weight representations of the Ramond algebra}

 The $N=1$ Ramond  algebra $\R$ is the
inf\/inite-dimensional Lie superalgebra
\[
{\R} = \bigoplus_{n \in {\Z} } {\C} L(n) \bigoplus \bigoplus_{m
\in  {\Z} }  {\C}G(m) \bigoplus {\C}C
\]
 with commutation relations ($m,n \in {\Z} $):
\begin{gather*}
[L(m),L(n)] =(m-n)L({m+n})+\delta_{m+n,0}\frac{m^3-m}{12}C,
\nonumber  \\
[G(m),L(n)]   =
\left(m-\frac{n}{2}\right)G({m+n}),    \\
\{ G(m ),G(n) \} =
2 L(m+n)+\frac{1}{3}(m^2 -1/4)\delta_{m+n,0}C,
\\   [L(m),C] = 0, \qquad [G(m),C] = 0.
\nonumber
\end{gather*}

The representation theory of the $N=1$ Ramond algebra has been intensively
studied f\/irst in \cite{KW-85} and other papers (cf.~\cite{MR,Dor}, etc.).

Assume that $(c,h) \in {\C}^2$ such that $  24 h \ne   c $. Let
$L^{\R} (c,h)^{\pm}$ denote the irreducible  highest weight
$\R$-module generated by the highest weight vector $v_{c,h}
^{\pm}$ such that
\begin{gather*}
  G(n) v_{c,h} ^{\pm} = \pm \sqrt{h -c /24} \,\delta_{n,0} v_{c,h}
^{\pm}, \qquad
 L(n) v_{c,h} ^{\pm} = h \delta_{n,0} v_{c,h} ^{\pm} \qquad (n \ge
0). \end{gather*}
These modules can be considered as irreducible $\sigma$-twisted
modules for the Neveu--Schwarz vertex operator superalgebra (cf.~\cite{Li-tw}). Since $L^{\R} (c,h)^{\pm}$ are not ${\Z}_2$-graded
(notice that $v_{c,h}^\pm$ are eigenvectors for $G(0)$), it is more
useful to consider  graded modules. It is not hard to show that the
direct sum
\[
  L^{\R} (c,h) = L^{\R} (c,h)^{+} \oplus L^{\R} (c,h)^{-}
\]
is in fact a ${\Z}_2$-graded $\R$-module. Indeed, for
$\mathbb{Z}_2$-graded subspaces take
\[
L^{\R}(c,h)_0=U(\R)\left(v^{+}_{c,h}+\frac{1}{\sqrt{h-c/24}}v^{-}_{c,h}\right)
\]
and
\[
 L^{\R}(c,h)_1=U(\R)\left(v^{+}_{c,h}-\frac{1}{\sqrt{h-c/24}}v^{-}_{c,h}\right)
\]
 Since
$L^{\R} (c,h)$ does not contain non-trivial ${\Z}_2$-graded
submodules, we shall say that this module is ${\Z}_2$-graded
irreducible.

\begin{remark}
Details about construction of  $L^{\R} (c,h)$ and its relation to
irreducible non-graded modules can be found in~\cite{Dor} and~\cite{IK2}. In Section~\ref{free-fields} we shall present   free
f\/ields realization of modules $L^{\R} (c,h)$ and $L^{\R}
(c,h)^{\pm}$.
\end{remark}

For  $i, n, m \in {\N}$, let
\begin{gather*}
c_{2m+1,1} = \frac{3}{2} - \frac{12 m ^2 }{2m +1}, \qquad h^{1,3} = 2m +\frac{1}{2}, \\
 h^{2 i +2,2n+1}:= \frac{ (2 i + 2
 - (2n+1) (2m+1) ) ^2 - 4m^2 }{8 (2m
+1)} + \frac{1}{16}.
\end{gather*}

Let $ L(c_{2m+1,1},h)$ be the irreducible highest weight module for
the Neveu--Schwarz algebra with central charge $c$  and highest
weight~$h$. Then $L(c_{2m+1,1},0)$ is a simple vertex operator
superalgebra, $ L^{\R} (c,h)^{\pm} $ are irreducible
$\sigma$-twisted  $L(c_{2m+1,1},0)$-modules, and $ L^{\R} (c,h) $
is ${\Z}_2$-graded irreducible $\sigma$-twisted
$L(c_{2m+1,1},0)$-module.

\subsection{Intertwining operators among twisted modules}

If $V$ is a vertex operator superalgebra and $W_1$, $W_2$ and $W_3$ are three $V$-modules then we consider the space of
intertwining operators ${ W_3 \choose W_1 \ W_2}$. It is perhaps less standard to study intertwining operators between twisted
$V$-modules, so we recall the def\/inition here  (see \cite[p.~120]{Xu}).

\begin{definition} Let $W_1$, $W_2$ and $W_3$ be $\sigma_i$-twisted
$V$-modules, respectively, where $\sigma_i$ is a f\/inite order automorphism
of order $\nu_i$, with common period $T$.  An intertwining operator of type
${W_3 \choose W_1 \ \ W_2 }$ is a linear map
\[
\mathcal{Y}( \cdot, x) : \ \ W_1 \longrightarrow  {\rm End}(W_2,W_3)\{ x \},
\]
such that
\[
\mathcal{Y}(L(-1)w,x)=\frac{d}{dx}\mathcal{Y}(w,x), \qquad w \in W_1
\]
and Jacobi identity holds
\begin{gather*} \frac{1}{T x_0} \sum_{p=0}^{T-1}
\delta \left(\left(\frac{x_1-x_2}{x_0}  \right)^{-1/T} \omega_T^p
\right){Y}^{\nu_3}(\sigma_1^p v,x_1)
\mathcal{Y}(w_1,x_2) \nonumber \\
\qquad{} -(-1)^{ij}
\frac{1}{T x_0} \sum_{p=0}^{T-1}\delta \left(\left(\frac{-x_2+x_1}{x_0}
\right)^{-1/T} \omega_T^p \right) \mathcal{Y}(w_1,x_2){Y}^{\nu_2}(\sigma_1^p
v,x_1) \nonumber \\
\qquad{}=\frac{1}{T x_2} \sum_{p=0}^{T-1} \delta
\left(\left(\frac{x_1-x_0}{x_2}  \right)^{1/T} \omega_T^p
\right)\mathcal{Y}(Y^{\nu_1}(\sigma_2^p v,x_0)w_1,x_2),
\end{gather*}
 where $v \in V_i$, $w_1 \in (W_1)_j$, $i,j \in \mathbb{Z}_2$
and $\omega_T$ is a primitive $T$-th root of unity.
\end{definition}

We shall also need the following result on the fusion rules. The
proof is completely analogous to that of Proposition~4.1 of~\cite{AdM-str} (see also~\cite{IK1}).

\begin{proposition} \label{fusion-R}
For every $i=0,\dots,m-1$ and $n \geq 1$ we have: the space
\[
I   {L^{\R}(c_{2m+1,1},h)^{\pm} \choose L(c_{2m+1,1},h^{1,3}) \
L^{\R}(c_{2m+1,1},h^{2i+2,2n+1})^{\pm}}
\]
 is nontrivial only if $h
\in \{h^{2i+2,2n-1}, h^{2i+2,2n+1},h^{2i+2,2n+3} \}$, and
\[
I
{L^{\R}(c_{2m+1,1},h)^{\pm}  \choose  L(c_{2m+1,1},h^{1,3}) \
L^{\R}(c_{2m+1,1},h^{2i+2,1}) ^{\pm}}
\] is nontrivial only if
$h=h^{2i+2,3}.$

Similarly, for every $i=0,\dots,m-1$ and $n \geq 2$ we have: the space
\[
I  {L^{\R}(c_{2m+1,1},h)^{\pm} \choose L(c_{2m+1,1},h^{1,3}) \
L^{\R}(c_{2m+1,1},h^{2i+2,-2n+1})^{\pm}}
\]
 is nontrivial only if $h
\in \{ h^{2i+2,-2n-1},h^{2i+2,-2n+1}, h^{2i+2,-2n+3} \}$, and
\[
I { L^{\R}(c_{2m+1,1},h)^{\pm} \choose L(c_{2m+1,1},h^{1,3}) \
L^{\R}(c_{2m+1,1},h^{2i+2,-1})^{\pm} }
\] is nontrivial only if $h
\in \{h^{2i+2,-3},h^{2i+2,-1} \}$.

We have analogous result for fusion rules of type
\[
{L^{\R}(c_{2m+1,1},h)  \choose L(c_{2m+1,1},h^{1,3}) \
L^{\R}(c_{2m+1,1},h^{2i+2,2n+1})  }.
\]
\end{proposition}

 \begin{proof} The proof goes along the lines in~\cite{AdM-str} (cf.~\cite{M0}), with some minor modif\/ications due to
twisting. In fact, in order to avoid the  twisted version of
Frenkel--Zhu's formula we can proceed in a more straightforward
fashion. Because all modules in question are irreducible and because
all intertwining operators we are interested in are of the form
${W_3 \choose W_1 \ \ W_2}$, where $W_1$ is untwisted (so
$\nu_1=0$), the left-hand side in the Jacobi identity looks like the
ordinary Jacobi identity for intertwining operators. Thus we can use
commutator formula and the null vector conditions
\begin{gather*}
\big({-}L(-1)G\big({-}\tfrac{1}{2}\big)+(2m+1)G\big({-}\tfrac{3}{2}\big)\big)v_{1,3}=0,\\
G\big({-}\tfrac{1}{2}\big)\big({-}L(-1)G\big({-}\tfrac{1}{2}\big)+(2m+1)G\big({-}\tfrac{3}{2}\big)\big)v_{1,3}=0,
\end{gather*}
which hold in $L(c_{2m+1,1},h^{1,3})$ (here $v_{1,3}$ is the highest weight vector),
to study the matrix coef\/f\/icient
\[
f(x)=\langle w',\mathcal{Y}(v_{1,3},x) w \rangle,
\]
where $w$ and $w'$ are highest weight vectors in appropriate
modules. This leads to dif\/ferential equations for $f(x)$, which can
be solved. The general solution is a linear combination of power
functions $x^s$, where $s$ is a rational number. The rest follows by
interpreting $s$  in terms of conformal weights for the three
modules involved in~$\mathcal{Y}$.
\end{proof}

\section[$\sigma$-twisted modules for the super singlet algebra $\overline{SM(1)}$]{$\boldsymbol{\sigma}$-twisted modules for the super singlet algebra $\boldsymbol{\overline{SM(1)}}$}
\label{singlet-ramond}

In this section  $\sigma_2$ will  denote the parity automorphism of~$F$. Recall also the Heisenberg vertex operator algebra $M(1)$.
Then, as in \cite{AdM-str} we equip the space $M(1) \otimes F$
with a vertex superalgebra structure such that the total central
charge is $\frac{3}{2}(1-\frac{8m^2}{2m+1})$. We def\/ine the following superconformal and conformal
vectors:
\begin{gather*}
\tau = \frac{1}{ \sqrt{2 m +1 }} \left( \alpha(-1) {\vak}
\otimes \phi \big({-}\hf\big) {\vak} + 2 m {\vak} \otimes  \phi \big({-\thf}\big)
{\vak}
\right),  \\
\omega = \frac{1}{2 (2m+1)} ( \alpha(-1) ^{2} +  2 m \alpha(-2) )
{\vak} \otimes {\vak} +    {\vak} \otimes  {\omega}^{(s)}.
\end{gather*}

Recall also the singlet superalgebra $\overline{SM(1)}$ obtained
as the kernel of the screening operator
\[
\widetilde{Q}={\rm Res}_x Y\big(e^{\frac{-\alpha}{2m+1}} \otimes \phi\big({-}\tfrac{1}{2}\big),x\big)
\] acting from $M(1) \otimes F$ to
$M(1) \otimes e^{-\frac{\alpha}{2m+1}} \otimes F$, where $ \langle
\alpha,\alpha \rangle=2m+1$. The vertex operator superalgebra
$\overline{SM(1)}$ is generated by $\tau=G\big({-}\tfrac{3}{2}\big){\bf 1}$ and
$H=Qe^{-\alpha}$, where
\[
Q={\rm Res}_{x} Y\big(e^{\alpha} \otimes \phi\big({-}\tfrac{1}{2}\big),x\big).
\]
We will also use $\omega$ and $\widehat{H}$, where the latter is
proportional to $G\big({-}\tfrac{1}{2}\big)H$ (for details see \cite{AdM-str}).

Consider the automoprhism $\sigma= 1 \otimes \sigma_2$ of $M(1) \otimes F$, acting
nontrivially on the second tensor factor. This automorphism
plainly preserves $\overline{SM(1)}$, thus we can study
$\sigma$-twisted $\overline{SM(1)}$-modules. It is clear that
$M(1,\lambda) \otimes M$ is a graded $\sigma$-twisted
$\overline{SM(1)}$-module, and $M(1,\lambda) \otimes M^{\pm}$ are
$\sigma$-twisted $\overline{SM(1)}$-modules, where $M(1,\lambda)$ is as in
\cite{AdM-triplet} and  $\lambda \in \mathbb{C}$.

We would like to classify irreducible $\overline{SM(1)}$-modules by
virtue of Zhu's algebra. As usual, we denote by $[a] \in
A_{\sigma}(\overline{SM(1)})$ the image of $a \in \overline{SM(1)}$.

\begin{lemma} \label{formule}
Let $v_{\lambda}^{\pm} = v_{\lambda}\otimes {\bf 1}^{\pm}$ be the
highest weight vector in $M(1,\lambda) \otimes M^{\pm}$, where
$\langle \alpha,\lambda \rangle=t$. Then the twisted generators
act as
\begin{gather*}
G(0) \cdot v_{\lambda}^{\pm}=\pm \frac{t-m}{\sqrt{2(2m+1)}} v_{\lambda} ^{\pm},\\
L(0) \cdot v_{\lambda}^{\pm}=\left(\frac{t(t-2m)}{2(2m+1)}+\frac{1}{16} \right) v_{\lambda} ^{\pm},\\
 H(0) \cdot v_{\lambda}^{\pm}=\pm \frac{1}{\sqrt{2}}{t-1/2 \choose 2m} v_{\lambda} ^{\pm},\\
 \widehat{H}(0) \cdot v_{\lambda} ^{\pm} = \frac{m-t}{2m+1} {t-1/2 \choose 2m} v_{\lambda} ^{\pm}.
 \end{gather*}
\end{lemma}

\noindent \begin{proof} Recall from \cite{AdM-str} the formulas
\begin{gather*}
H=S_{2m}(\alpha) \otimes \phi\big({-}\tfrac{1}{2}\big) +S_{2m-1} \otimes \phi\big({-}\tfrac{3}{2}\big) +  \cdots + 1 \otimes \phi\big({-}2m-\tfrac{1}{2}\big),\\
 \widehat{H}=\phi(1/2)S_{2m+1}(\alpha)\phi\big({-}\tfrac{1}{2}\big) + w \\
 \hphantom{\widehat{H}}{} =S_{2m+1}(\alpha) + S_{2m -1} (\a) \otimes \phi\big({-}\tfrac{3}{2}\big) \phi\big({-}\tfrac{1}{2}\big) + \cdots + {\bf 1} \otimes \phi\big({-}2m -\tfrac{1}{2}\big) \phi\big({-}\tfrac{1}{2}\big).
 \end{gather*}
As in \cite{AdM-str} we have
\[
S_{r}(\alpha)(0) \cdot v_{\lambda} ^{\pm}={t \choose r} v_{\lambda} ^{\pm}.
\]
Then we get
\[
H(0) \cdot v_{\lambda} ^{\pm}=\pm \frac{1}{\sqrt{2}}\sum_{n=0}^{2m} {t \choose n}{-1/2 \choose 2m-n} v_{\lambda} ^{\pm}={\pm} \frac{1}{\sqrt{2}}{t-1/2 \choose 2m}
v_{\lambda} ^{\pm}.
\] The last formula is proven similarly by using
$e^{\Delta_x}$ operator. \end{proof}

Here are the main result of this section.
\begin{theorem} \label{relacije-sin} The Zhu's algebra $A_{\sigma}(\overline{SM(1)})$ is an associative algebra generated by
 $[\tau]$, $[\omega]$, $[H]$ and $[\widehat{H}]$, where the following relations hold (here $[\omega]$ is central):
\begin{gather}
 [\tau] ^2 = [\omega] - \frac{c_{2m+1,1}}{24},\nonumber\\
 [\tau]*[H] =[H]*[\tau]=\frac{-\sqrt{2m+1}}{2} [\widehat{H}],\nonumber\\
 [H] * [H] = \frac{1}{2} { \sqrt{2(2m+1)} [\tau] + {m-1/2} \choose 2
m}  ^2,  \nonumber\\
\label{important}
[H]*[H]=C_m \prod_{i=0}^{m-1}\big([\omega]-h^{2i+2,1}\big)^2,
\qquad C_m = \frac{2 ^{2m-1} (2m+1) ^{2m}}{ (2m)!^2},\\
 [\widehat{H}]*[\widehat{H}]=\frac{4}{2m+1} C_m \left([\omega] - \frac{c_{2 m+1,1}}{24}\right) \prod_{i=0}^{m-1}\big([\omega]-h^{2i+2,1}\big)^2. \nonumber
\end{gather}
In particular, Zhu's algebra is  commutative.
\end{theorem}

\begin{proof} First notice (cf. \cite{AdM-str}) that
\[
G\big({-}\tfrac{1}{2}\big){F}=-\sqrt{2m+1} \phi\big({-}\tfrac{1}{2}\big)F=-\sqrt{2m+1} \widehat{F}
\] and consequently
\[
G\big({-}\tfrac{1}{2}\big)H=-\sqrt{2m+1} \widehat{H}.
\]
We also have the relation
\[
\big((2m+1)G\big({-}\tfrac{3}{2}\big)-L(-1)G\big({-}\tfrac{1}{2}\big)\big)H=0,
\]
because $H=v_{1,3}$ is a highest weight vector.
By using (\ref{ramond-relations}) we obtain
\begin{gather*}
[\tau]*[H]=\big[G\big({-}\tfrac{3}{2}\big)H\big]+\tfrac{3}{2}\big[G\big({-}\tfrac{1}{2}\big)H\big]
=\left[\frac{1}{2m+1}L(-1)G\big({-}\tfrac{1}{2}\big)H\right]+\tfrac{3}{2}\big[G\big({-}\tfrac{1}{2}\big)H\big]\\
\phantom{[\tau]*[H]}{}
=\tfrac{1}{2}\big[G\big({-}\tfrac{1}{2}\big)H\big]=-\frac{\sqrt{2m+1}}{2}[\widehat{H}].
\end{gather*}
On the other hand, by using skew-symmetry we also have
\begin{gather*}
 [H]*[\tau]=-\big[{\rm Res}_x x^{-1} e^{x L(-1)}Y(\tau,-x)(1+x)^{2m+\frac{1}{2}}H\big] \nonumber \\
\hphantom{[H]*[\tau]}{} =-\big[{\rm Res}_x x^{-1} e^{-x L(0)}Y(\tau,-x)(1+x)^{2m+\frac{1}{2}}H\big] \nonumber \\
\hphantom{[H]*[\tau]}{}=-\big(\big[G\big({-}\tfrac{3}{2}\big)H\big]-\big[\big(2m+\tfrac{1}{2}-(2m+1)\big)
G\big({-}\tfrac{1}{2}\big)H]\big) \nonumber \\
\hphantom{[H]*[\tau]}{}=-\big[G\big({-}\tfrac{3}{2}\big)H\big]-\tfrac{1}{2}\big[G\big({-}\tfrac{1}{2}\big)H\big]=
\tfrac{1}{2}\big[G\big({-}\tfrac{1}{2}\big)H\big].
\end{gather*} Combined, we obtain
\begin{gather*}
[\tau]*[H]=[H]*[\tau] .\tag*{\qed}
\end{gather*}\renewcommand{\qed}{}
\end{proof}

Notice that relation (\ref{important}) can be written as
\[
 [H] * [H] = C_m \prod_{i=0}^{m-1}\left([\tau]^2-\frac{(2i+1-2m)^2}{8(2m+1)} \right)^2,  \qquad C_m \neq 0.
 \]

Let ${\C}[a,b]$ denote the ${\Z}_2$-graded complex commutative
associative algebra generated by odd vectors $a$, $b$.

\begin{theorem} \label{structura-singlet}
The associative algebra $A_{\sigma}(\overline{SM(1)})$ is isomorphic to the
${\Z}_2$-graded commutative associative algebra
\[
{\C}[a,b] / \la
H(a,b) \ra,
\]
 where $\la H(a,b) \ra$ is (two-sided) ideal in
${\C}[a,b]$, generated by
\[
H(a,b) = b^2 - C_m \prod_{i=0}^{m-1}\left(a^2-\frac{(2i+1-2m)^2}{8(2m+1)}\right)^2.
\]
\end{theorem}
\begin{proof} The proof is similar to that of Theorem 6.1 of
\cite{A-2003}. First we notice that we have a~surjective
homomorphism
\begin{gather*}
 \Phi : \ \ {\C}[a,b]   \rightarrow  A_{\sigma}(\overline{SM(1)}), \qquad
 a \mapsto   [\tau],\qquad
b \mapsto   [H]  .
\end{gather*}

It is easy to see that $\mbox{Ker} \,  \Phi$ is a ${\Z}_2$-graded
ideal. We shall now prove that $\mbox{Ker} \,  \Phi =\la H(a,b) \ra$.

Evidently the generating element $H(a,b)$ is even, so Theorem~\ref{relacije-sin} gives $\la H(a,b) \ra \subset \mbox{Ker} \, \Phi$.

Assume now that $K(a,b) \in \mbox{Ker} \,  \Phi$.
By using division algorithm   we get
\[
K(a,b) = A(a,b) H(a,b) + R(a,b),
\]
where
$ A(a,b), R(a,b) \in {\C}[a,b]$
and  $R(a,b)$  has degree at most $1$ in $b$. Assume that $R(a,b)
\ne 0$. Then
$R(a,b) = A(a) b + B(a)$ for certain polynomials $A,  B \in
{\C}[x]$.  We also notice that $R(a,b) \in \mbox{Ker} \, \Phi$. Since
$\mbox{Ker} \, \Phi$ is ${\Z}_2$-graded ideal, we can assume that
$R(a,b)$ is homogeneous. If  $R(a,b)$ is an even element we have
that $A$ has odd degree and $B$ has even degree, and therefore
\begin{gather}
 \deg (B) - \deg(A) \qquad \mbox{is an odd natural number.}
\label{odd-r}
\end{gather}
The case when $R(a,b)$ is odd element again leads to formula~(\ref{odd-r}).

 As in \cite{A-2003} we now shall evaluate $R(a,b)$ on
$A_{\sigma}(\overline{SM(1)})$-modules and get
\[
   A \left( \frac{t-m}{\sqrt{2(2m+1)}} \right) {t-1/2 \choose
2m} +    \sqrt{2} B \left(\frac{t-m}{\sqrt{2(2m+1)}}\right) = 0
\qquad \forall \ t \in {\C}.
\]
This implies that $\deg(B) - \deg(A) = 2m$. Contradiction.
  Therefore $R(a,b) = 0$ and $K(a,b) \in \la H(a,b)
\ra$. The proof follows. \end{proof}

\begin{remark}
By using the same arguments as  in  \cite{A-2003} and
\cite{AdM-str}, we can conclude that every irre\-ducible
$\mathbb{Z}_{\geq 0}$-gradable $\sigma$-twisted
$\overline{SM(1)}$-module  is isomorphic to an irreducible
subquotient of~\mbox{$M(1,\l) \otimes M^{\pm}$}. By using the  structure of
twisted Zhu's algebra $A_{\sigma}(\overline{SM(1)})$ and the methods
developed in~\cite{AdM-2007}, we can also construct logarithmic
$\sigma$-twisted $\overline{SM(1)}$-modules.
\end{remark}

\section[The $N=1$ Ramond module structure of twisted $V_L \otimes F$-modules]{The $\boldsymbol{N=1}$ Ramond module structure\\ of twisted $\boldsymbol{V_L \otimes F}$-modules}
\label{free-fields}

In this section we shall assume that the reader is familiar with
basic results on twisted representations of lattice vertex
superalgebras. Details can be found in \cite{BK,D2,FLM} and \cite{Xu}.

 We shall use the same notation as in
\cite{AdM-str}. Let $L= {\Z}{\a}$ be a rank one lattice with
nondegenerate form given by
$ \la \a , \a \ra = 2m+1 $, where $m \in {\Zp}$. Let $V_L$ be the
corresponding vertex  superalgebra.

For $i \in {\Z}$, we  set
\begin{gather*}
 \ga_i = \frac{i}{2m+1} \a, \qquad \ga^{R}_i = \frac{\a}{2
(2m+1)} + \frac{i}{2m+1} \a .
\end{gather*}

Then  $\sigma_1 = \exp[ \tfrac{\pi i} {2m+1}\a (0)]$ is a
canonical automorphism of order two of $V_L$. The set
$\{ V_{\ga^{R}_i + L}$,   $i=0, \dots, 2m \} $
provides all the irreducible $\sigma_1$-twisted $V_L$-modules.

\begin{remark}It is important to  notice that $\sigma_1$-twisted
$V_L$-module $(V_{\ga^{R}_i + L}, Y_{\ga^{R}_i + L}) $ can be
constructed from untwisted module $(V_{\ga_i + L}, Y_{\ga_i + L})$
as follows (cf.~\cite{Xu}):
\begin{gather*}
 V_{\ga^{R}_i + L}:= V_{\ga_i + L} \qquad \mbox{as vector space;}
\\
Y_{\ga^{R}_i + L}(\cdot,z):=Y_{\ga_i + L}\big(\Delta\big(\tfrac{\a
}{2(2m+1)},z\big)  \cdot,z\big),
\end{gather*}

where
\[
\Delta(h,z):= z^{h(0)} \exp \left( \sum_{n=1} ^{\infty} \frac{h(n)
}{-n}(-z) ^{-n} \right).
\]
\end{remark}

 Let $F$ be the fermionic vertex operator superalgebra with
central charge $1/2$ and $\sigma_2$ its parity map. Let $M$ be the
  $\sigma_2$ twisted $F$-module (cf.~\cite{FRW}).

Then $\sigma= \sigma_1 \otimes \sigma_2$ is the parity  automorphism
of order two of the vertex superalgebra $V_L \otimes F$ and
\[
  V_{\ga^{R}_i + L} \otimes M , \qquad i=0, \dots, 2m
  \]
are  $\sigma$-twisted $V_L \otimes F$-modules.

Let $(W,Y_W)$ be any  $\sigma$-twisted $V_L \otimes F$-module.
From the Jacobi identity for $\sigma$-twisted modules it follows
that the coef\/f\/icients of
\[
Y_W(\tau,z) = \sum_{n \in {\Z}} G(n) z^{-n-\tfrac{3}{2}} \qquad
\mbox{and} \qquad Y_W(\omega,z) = \sum_{n \in {\Z}} L(n) z^{-n-2}
\]
def\/ine a representation of the $N=1$ Ramond algebra.

Recall that
\[
h^{2i+2,2n+1}= \frac{ ( (2i+2) - (2n+1) (2m+1) ) ^2 -
4m^2 }{8 (2m +1)} + \frac{1}{16}.
\]

\begin{proposition}
Assume that $n \in {\Z}$. Then $e^{{\ga}^{R}_i - n \alpha} \otimes
{\bf 1} ^{\R}$ is a singular vector for the $N=1$ Ramond algebra
$\R$ and
\begin{gather*}
U(\R) \big(e^{{\ga}^{R}_i - n \alpha} \otimes
{\bf 1} ^{\R}\big) \cong L^{\R} \big(c_{2m+1,1}, h^{2i+2,2n+1}\big)\\
\hphantom{U(\R) \big(e^{{\ga}^{R}_i - n \alpha} \otimes {\bf 1} ^{\R}\big)}{}  = L^{\R}
\big(c_{2m+1,1}, h^{2i+2,2n+1}\big)^+ \oplus L^{\R} (c_{2m+1,1},
h^{2i+2,2n+1}) ^- .
\end{gather*}
Moreover,
\[
U(\R) \big(e^{{\ga}^{R}_i - n \alpha} \otimes
{\bf 1} ^{\pm}\big) \cong  L^{\R} \big(c_{2m+1,1}, h^{2i+2,2n+1}\big)^{\pm} .
\]
\end{proposition}

As in \cite{AdM-str} (see also  \cite{IK3} and \cite{MR}) we have
the following result.
\begin{lemma} The (screening)  operator
\[
Q = \mbox{\rm Res}_x Y_W\big(e^{\a}\otimes \phi\big({-}\tfrac{1}{2}\big),x\big)
\]
commutes with the action of $\R$.
\end{lemma}

\begin{remark}
By using generalized  (lattice) vertex algebras and their twisted
representations one can also def\/ine the second screening operator
$\widetilde{Q}$ acting between certain $\sigma$-twisted $V_L
\otimes F$-modules such that
\[
[\widetilde{Q}, {\R}] = 0, \qquad [Q, \widetilde{Q} ]  = 0
\] (for
details and some applications see \cite{AdM-lattice}).
\end{remark}

We shall f\/irst present results on the structure of
$\sigma$-twisted $V_L \otimes F$-modules, viewed as modules for the
$N=1$ Ramond  algebra. Each $V_{L+{\ga}^{R}_i}\otimes M$ is a
direct sum of super Feigin--Fuchs modules via
\[
V_{L+{\ga}^{R}_i} \otimes M=\bigoplus_{n \in \mathbb{Z}} \big(M(1) \otimes
e^{{\ga}^{R}_i +n \alpha}\big) \otimes M.
\]

Since operators $Q ^{j}$, $j \in {\Zp}$, commute with the action
of the Ramond algebra, they  are  actually (Lie superalgebra) intertwiners between
super Feigin--Fuchs modules inside  $V_{L+\ga^{R}_i} \otimes M$.

To simplify the notation, we shall identify $e^{\beta}$ with $e^{\beta} \otimes {\bf 1}^{\R}$ for
every $\beta \in L+ \ga^{R}_i$.

Assume that $0 \le i \le m-1$. If $Q^{j} e^{\ga^{R}_i-n \alpha}$
is nontrivial, it is a singular vector of weight
\[
\mbox{wt}\big(Q^{j} e^{\ga^R_i -n \alpha}\big) = \mbox{wt} \big(e^{\ga^R_i -n \alpha}\big) =h ^{2i+2,2n+1}.
\]

Since $\mbox{wt} \big( e^{ \ga^{R}_i + (j-n) \alpha} \big) > \mbox{wt} \big(
e^{\ga^{R}_i -n \alpha} \big)$ if $j  > 2n   $, we conclude that
\begin{gather*}
 Q^ {j} e^{\ga^{R}_i-n \alpha} = 0 \qquad \mbox{for} \quad j > 2 n.
\end{gather*} One can similarly see that for $ m \le i \le 2m$:
\begin{gather*}

 Q^ {j} e^{\ga^{R}_i-n \alpha} = 0 \qquad \mbox{for} \quad j > 2 n
+1.
\end{gather*}

The following lemma is useful for constructing singular vectors in
$V_{L+\ga^{R}_i} \otimes M$:
\begin{lemma} \label{non-triv-Ramond} \quad {}
\begin{enumerate}\itemsep=0pt
\item[$(1)$]  $ Q^{2n} e^{\ga^R_i - n\a} \ne 0$ \  for  \  $0 \le i
\le m$.

\item[$(2)$]$ Q^{2n+1} e^{\ga^R_i - n\a} \ne 0$ \ for \  $m+1 \le i
\le 2m$.
\end{enumerate}
\end{lemma}

\begin{proof} The proof uses the results on fusion rules from
Proposition~\ref{fusion-R} and is completely ana\-logous to that of
Lemma~6.1 in~\cite{AdM-str}.
\end{proof}

As in the Virasoro algebra case the $N=1$ Feigin--Fuchs modules are
classif\/ied according to their embedding structure.

For the  purposes of our paper we shall focus only on modules of
certain types (Type 4 and 5 in \cite{IK3}). These modules are either
semisimple (Type 5) or they become semisimple after quotienting with
the maximal semisimple submodule (Type 4).

The following result follows directly from Lemma
\ref{non-triv-Ramond} and the structure theory of super Feigin--Fuchs
modules, after some minor adjustments of parameters (cf. Type 4
embedding structure in \cite{IK3}).
\begin{theorem} \label{str-fock-R-1}

Assume that $i \in \{0, \dots, m-1\}$.

\begin{enumerate}\itemsep=0pt
 \item[$(i)$] As an $\R$-module, $V_{L+\ga^{R}_i} \otimes M$  is generated by the family of
singular and cosingular vectors $ \widetilde{\rm Sing}_{i}  \bigcup
\widetilde{\rm CSing}_{i}$, where
\begin{gather*}
  \widetilde{\rm Sing}_{i} =  \big\{ u_i ^{(j, n)} \ \vert \ j, n \in {\N}, \ 0 \le j \le 2n \big\}; \\
  \widetilde{\rm CSing}_{i} =  \big\{ w_i ^ {(j, n)} \ \vert \ n \in {\Zp}, 0 \le j \le 2n-1 \big\}.
  \end{gather*}
These vectors satisfy the following relations:
\[  u_i ^{(j, n)}= Q ^{j} e ^{\ga^{R}_i -n \alpha} , \qquad Q ^{j}
w_i ^{(j, n)} = e ^{\ga^{R}_i + n \a}.
\]
 The submodule generated by singular vectors
$\widetilde{\rm Sing}_i$, denoted by $R\Lambda(i+1)$, is isomorphic to
\[
\bigoplus_{n =0 } ^{\infty} (2n +1) L^{\R}\big(c_{2m+1,1}, h^{2i+2,
2n+1}\big).
\]

\item[$(ii)$] Let $R\Lambda(i+1) ^{\pm}$ denote the submodule of
$V_{L+\ga^{R}_i} \otimes M^{\pm}$ generated by singular vectors
\[
Q ^{j} \big(e ^{\ga^{R}_i -n \alpha} \otimes {\bf 1} ^{\pm}\big), \qquad n \in {\N}, \qquad 0
\le j \le 2n.
\]
Then
\[
 R\Lambda(i+1) ^{\pm } =\bigoplus_{n =0 } ^{\infty} (2n +1) L^{\R}\big(c_{2m+1,1}, h^{2i+2,
2n+1}\big) ^{\pm},
\]

and
\[
 R\Lambda(i+1) = R\Lambda(i+1) ^{+} \oplus
R\Lambda(i+1)^{-}.
\]

\item[$(iii)$] For the quotient module we have
\[
 R \Pi(m-i):=(V_{L+\ga^{R}_i} \otimes M) / R\Lambda(i+1) \cong
  \bigoplus_{n =1 } ^{\infty}
(2 n ) L^{\R}\big(c_{2m+1,1}, h^{2i+2, -2n +1}\big).
\]

Moreover, we have
\[
 R\Pi(m-i) = R\Pi(m-i) ^{+} \oplus R\Pi(m-i)^{-},
 \]
where
\[
 R\Pi(m-i) ^{\pm } = (V_{L+\ga^{R}_i} \otimes M) / R\Lambda(i+1)^{\pm} = \bigoplus_{n =1 } ^{\infty}
(2 n ) L^{\R}\big(c_{2m+1,1}, h^{2i+2, -2n +1}\big) ^{\pm}.
\]
\end{enumerate}
\end{theorem}

\begin{theorem} \label{str-fock-R-2}
Assume that $i \in \{0, \dots, m-1 \}$.

\begin{enumerate}\itemsep=0pt
 \item[$(i)$] As an  $\R$-module, $V_{L+\ga^{R}_{m+i}} \otimes M$  is generated by the family of
singular and cosingular vectors $ \widetilde{\rm Sing}'_{i}  \bigcup
\widetilde{\rm CSing}'_{i}$, where
\begin{gather*}
  \widetilde{\rm Sing}'_{i} = \big\{ u_i^ {'(j, n)} \ \vert \ n \in {\Zp}, 0 \le j \le 2n-1
\big\}; \\
 \widetilde{\rm CSing}'_{i} = \big\{ w_i^{'(j, n)} \ \vert \ j, n \in
{\N}, \ 0 \le j \le 2n \big\}.
\end{gather*}
These vectors satisfy the following relations:
\[
 u_i ^{'(j, n)}= Q ^{j} e ^{\ga^{R}_{m+i} -n \alpha} , \qquad Q
^{j} w_i ^{'(j, n)} = e ^{\ga^{R}_{m+i} + n \a}.
\]
The submodule generated by singular vectors $\widetilde{\rm Sing}_i$
is isomorphic to
\[
 R\Pi(i+1)\cong \bigoplus_{n =1 } ^{\infty} (2 n ) L^{\R}\big(c_{2m+1,1}, h^{2m -2i,
-2n +1}\big).
\]

\item[(ii)] For the quotient module we have
\[
R \Lambda(m -i)\cong (V_{L+\ga^{R}_i} \otimes M) / R \Pi(i+1) \cong
  \bigoplus_{n =0 } ^{\infty}
(2 n+1 )  L^{\R}\big(c_{2m+1,1}, h^{2m -2i, 2n+1}\big).
\]
\end{enumerate}
\end{theorem}

\newpage

\begin{theorem} \label{str-fock-R-3}\quad
\begin{enumerate}\itemsep=0pt
\item[$(i)$] As an  $\R$-module, $V_{L +
\ga^{R}_{2m}} \otimes M$ is completely reducible and  generated by
the family of singular vectors
\[
  \widetilde{\rm Sing}_{2 m} =  \big\{ u_{2 m} ^{(j, n)} : = Q ^{j} e ^{\ga^{R}_{2m} -n \alpha}  \ \vert \ n \in {\Zp}, \  j \in {\N}, \ 0 \le j \le 2n-1 \big\};
\]
 and it is isomorphic to
\[
R\Pi(m+1):=
 V_{L + \ga^{R}_{2m}} \otimes M \cong  \bigoplus_{n=1} ^{\infty} (2n)L^{\R}\big(c_{2m+1,1}, h^{4m+2, -2n
+1}\big).
\]

\item[$(ii)$] Let $R\Pi(m+1) ^{\pm}$ be the submodule of $V_{L +
\ga^{R}_{2m}} \otimes M^{\pm}$ generated by the singular vectors
\[
 Q ^{j} \big(e ^{\ga^{R}_{2m} -n \alpha} \otimes {\bf 1}^{\pm} \big),  \qquad n \in {\Zp}, \qquad  j \in {\N}, \qquad 0 \le j \le
2n-1.
\]
Then
\[
 R\Pi(m+1) ^{\pm } = \bigoplus_{n=1} ^{\infty} (2n)L^{\R}\big(c_{2m+1,1}, h^{4m+2, -2n
+1}\big)^{\pm},
\]
and
\[
 R\Pi(m+1) = R\Pi(m+1) ^{+} \oplus R\Pi(m+1)^{-}.
 \]
\end{enumerate}
\end{theorem}

\begin{remark}
In this section we actually constructed explicitly all the
non-trivial intertwining operators from Proposition \ref{fusion-R}.
\end{remark}

\section[The $\sigma$-twisted $\mathcal{SW}(m)$-modules]{The $\boldsymbol{\sigma}$-twisted $\boldsymbol{\mathcal{SW}(m)}$-modules}
\label{classification}

Since   $\striplet \subset V_L \otimes F $ is $\sigma$-invariant,
then every $\sigma$-twisted $V_L \otimes F$-module is also
a $\sigma$-twisted module for the vertex operator superalgebra
$\striplet$. In this section we shall consider $\sigma$-twisted
$V_L \otimes F$-modules from Section \ref{free-fields} as
$\sigma$-twisted $\striplet$-modules. In what follows we shall
classify all the irreducible $\sigma$-twisted $\striplet$-modules
by using Zhu's algebra~$A_{\sigma}(\striplet)$.

Following  \cite{AdM-triplet} and \cite{AdM-str}, we f\/irst  notice
the following important fact:
\begin{gather}
 Q ^2 e^{-2 \alpha} \in O(\striplet) .
\label{zhu-Q2}
\end{gather}

\begin{proposition} \label{identitet}
Let $v_{\l}^{\pm}$ be the highest weight vector in   $M(1,\l)
\otimes M^{\pm}$. We have
\[
 o\big(Q ^2 e^{-2 \alpha}\big)  v_{\l}^{\pm} =    F_m (t) \,  v_{\l} ^{\pm} ,
 \]
where $t = \la \l, \a \ra$, and
\[
F_m (t) =A_m  { t + m + 1/2 \choose 3 m + 1} { t - 1/2 \choose 3m+1 }, \qquad \mbox{where} \quad A_m = (-1)^m \frac{{2m \choose m} }{{4m+1 \choose m}}.
\]
\end{proposition}

\begin{proof} First we notice that
\begin{gather*}
  Q^{2} e^{-2\a} = w_1 + w_2,
  \end{gather*}
  where
\begin{gather*}
 w_1= \sum_{i=0} ^{\infty } e^{\a}_{-i-1} e^{\a} _i
e^{-2\a}, \qquad
 w_2 =\sum_{i,j=0} ^{\infty} e^{\a}_{i} e^{\a}_j e^{-2\a}
\otimes \phi\big({-}i-\tfrac{1}{2}\big) \phi\big({-}j -\tfrac{1}{2}\big) {\bf 1}.
\end{gather*}
The proof of Proposition~8.3 from \cite{AdM-str} gives that
\begin{gather*}
o(w_1) v_{\l}^{\pm} = {\rm Res}_{x_1} {\rm
Res}_{x_2}(x_2-x_1)^{2m} (1+x_1)^t (1+x_2)^t (x_1 x_2)^{-4m-2}
v_{\l} \\
\phantom{o(w_1) v_{\l}^{\pm}}{} = A_m { t + m   \choose 3 m + 1} { t \choose
3m+1 } v_{\l}^{\pm},
\end{gather*}
 so it remains to examine
$o(w_2)$. Recall Lemma~\ref{cmn}, so that
\[
(x_2 -x_1)G(x_1,x_2)=\tfrac{1}{2} \left( (1+x_1)^{1/2}(1+x_2)^{-1/2}+(1+x_1)^{-1/2}(1+x_2)^{1/2}-2 \right).
\]
Now, we compute
\begin{gather}
  o(w_2) \cdot v_{\l}^{\pm}
=\sum_{i,j \geq 0} o \left( e^{\alpha}_i e^{\alpha}_j e^{-2 \alpha} \otimes \phi\big({-}i-\tfrac{1}{2}\big)\phi\big({-}j-\tfrac{1}{2}\big){\bf 1} \right) \cdot v_{\l} ^{\pm} \nonumber \\
\phantom{o(w_2) \cdot v_{\l}^{\pm}}{}  =\sum_{i,j \geq 0} {\rm Res}_{x_1}{\rm Res}_{x_2} x_1^i x_2^j o\big(Y(e^{\alpha},x_1)Y(e^{\alpha},x_2)e^{-2 \alpha} \big) o\big(\psi\big({-}i-\tfrac{1}{2}\big)\psi\big({-}j-\tfrac{1}{2}\big)\big) \cdot v_{\l} ^{\pm}
 \nonumber \\
\phantom{o(w_2) \cdot v_{\l}^{\pm}}{}
 ={\rm Res}_{x_1} {\rm Res}_{x_2}  \Bigg( \sum_{i,j \geq 0} x_1^i x_2^j  c_{i,j}
{\rm Res}_{x_1} {\rm Res}_{x_2} (x_2-x_1)^{2m+1}(1+x_1)^t \nonumber\\
\phantom{o(w_2) \cdot v_{\l}^{\pm}{\rm Res}_{x_1} {\rm Res}_{x_2}  \Bigg(}{}
\times(1+x_2)^t (x_1x_2)^{-4m-2} v_{\l} ^{\pm} \Bigg) \nonumber \\
\phantom{o(w_2) \cdot v_{\l}^{\pm}}{} ={\rm Res}_{x_1} {\rm Res}_{x_2}(G(x_1,x_2) (x_2-x_1)^{2m+1}(1+x_1)^t (1+x_2)^t )(x_1x_2)^{-4m-2} v_{\l} ^{\pm} \nonumber \\
 \label{-2}
 \phantom{o(w_2) \cdot v_{\l}^{\pm}}{} =-{\rm Res}_{x_1} {\rm Res}_{x_2}(x_2-x_1)^{2m} (1+x_1)^t (1+x_2)^t (x_1 x_2)^{-4m-2} v_{\l} ^{\pm} \\
 \label{-1}
 \phantom{o(w_2) \cdot v_{\l}^{\pm}=}{}+\tfrac{1}{2}{\rm Res}_{x_1} {\rm Res}_{x_2}\left((x_2-x_1)^{2m} (x_1x_2)^{-4m-2} (1+x_1)^{t+1/2} (1+x_2)^{t-1/2} \right) v_{\l} ^{\pm} \!\!\!  \\
 \label{last}
 \phantom{o(w_2) \cdot v_{\l}^{\pm}=}{}
 +\tfrac{1}{2}{\rm Res}_{x_1} {\rm
Res}_{x_2}\left((x_2-x_1)^{2m} (x_1x_2)^{-4m-2} (1+x_1)^{t-1/2}
(1+x_2)^{t+1/2} \right) v_{\l} ^{\pm}.\!\!\!
 \end{gather}
  Now, observe that the
expression in (\ref{-2}) is precisely $-o(w_1) \cdot v_{\l}
^{\pm}$, while (\ref{-1}) and (\ref{last}) are equal.
Consequently,
\begin{gather*}
o(w_1+w_2) \cdot v_{\l} ^{\pm}={\rm Res}_{x_1} {\rm Res}_{x_2}\left((x_2-x_1)^{2m} (1+x_1)^{t-1/2} (1+x_2)^{t+1/2} (x_1x_2)^{-4m-2} \right) v_{\l} ^{\pm}.
\end{gather*}
Now, we expand the generalized rational function in the last formula and obtain
\[
o\big(Q^2e^{-2 \alpha}\big) v_{\l}^{\pm}=\sum_{k=0} ^{2 m } (-1) ^k  { 2 m \choose k } { t +1/2 \choose  4 m + 1 -
k} {t -1/2 \choose  2 m + 1+k} v_{\l}^{\pm}.
\] The sum in the last
formula can be evaluated as in \cite{AdM-triplet} and
\cite{AdM-str}. We have
\[
\sum_{k=0} ^{2 m } (-1) ^k  { 2 m \choose k } { t +1/2 \choose  4 m + 1 -
k} {t -1/2 \choose  2 m + 1+k} =A_m  { t + m + 1/2 \choose 3 m + 1} { t - 1/2 \choose 3m+1 },
\]
where $A_m$ is above. The proof follows.
\end{proof}

A direct consequence of Proposition~\ref{identitet} and relation
(\ref{zhu-Q2}) is the following important result:

\begin{theorem} \label{vazna-rel-ramond}
  In Zhu's  algebra $A_{\sigma}(\striplet)$ we have the following relation
\[
f^{\R}_m([\omega]) = 0,
\]
 where
\begin{gather*}
  f^{\R}_m(x) =  \prod_{i=0}^{3 m} \big(x - h^{2i+2,1}\big) =  \left( \prod_{i=0}^{m-1} \big(x-h^{2i+2,1}\big)^{2} \right) \left(\prod_{i=2m} ^{3 m}\big(x-h^{2i+2,1}\big)\right) .
\end{gather*}
\end{theorem}

In parallel with \cite{AdM-str} we conjecture that $f^{\R}_m(x)$ is in fact the minimal polynomial of $[\omega]$ in $A_{\sigma}(\striplet)$.

We have the following irreducibility result. The proof is similar
to that  of Theorem 3.7 in \cite{AdM-triplet}.

\begin{theorem} \label{ireducibilni-moduli-ramond-nongraded} \qquad{}
\begin{enumerate}\itemsep=0pt

\item[$(1)$] For every $0 \leq i \leq m-1$,  $R\Lambda(i+1) ^{\pm}$
are $\N$-gradable irreducible  $\sigma$-twisted
$\mathcal{SW}(m)$-modules and   the top components $R\Lambda(i+1)
^{\pm} (0)$ are $1$-dimensional irreducible
  $A_{\sigma}(\mathcal{SW}(m))$-modules.

  \item[$(1')$] For every $0 \leq i \leq m-1$,  $R\Lambda(i+1)$ is a graded irreducible  $\sigma$-twisted $\mathcal{SW}(m)$-module,
and  its  top component $R\Lambda(i+1) (0)$ is a graded irreducible
  $A_{\sigma}(\mathcal{SW}(m))$-module.

\item[$(2)$]  For every $0 \leq j \leq m$ $,  R\Pi(j+1) ^{\pm}$ are
irreducible   $\N$-gradable $\sigma$-twisted
$\mathcal{SW}(m)$-modules and the top components  $S\Pi(j+1)
^{\pm} (0)$ are irreducible  $2$-dimensional
$A_{\sigma}(\mathcal{SW}(m))$-modules.

\item[$(2')$]  For every $0 \leq j \leq m$ $,  R\Pi(j+1)$ is a
  graded irreducible $\sigma$-twisted
$\mathcal{SW}(m)$-module and its  top component $S\Pi(j+1) (0)$ is
a graded irreducible  $A_{\sigma}(\mathcal{SW}(m))$-module.
\end{enumerate}
\end{theorem}

\begin{corollary} The minimal polynomial of $[\omega]$ is divisible by
  \[
  \prod_{i=0}^{m-1} (x-h^{2i+2,1}) \left(\prod_{i=2m} ^{3 m}(x-h^{2i+2,1})\right).
  \]
Moreover, both $[\omega]$ and $[\tau]$ are units in
$A_{\sigma}(\striplet)$. Consequently, $\striplet$ has no
supersymmetric sector (i.e., there is no $\sigma$-twisted
$\striplet$-modules of highest weight $\frac{c_{2m+1,1}}{24}$).
\end{corollary}

By using similar arguments as in  \cite{AdM-triplet} and \cite{AdM-str} we have the following
result on the structure of Zhu's algebra $A_{\sigma}(\striplet)$ (as in the
proof of Theorem 4.1 we see that $[\tau]$ is central).
\begin{proposition} \label{zhu-komut-r}
The associative algebra $A_{\sigma} (\striplet)$ is  generated
by $[E]$, $[H]$, $[F]$, $[\tau]$, $[\widehat{E}]$,
$[\widehat{H}]$, $[\widehat{F}]$ and $[\omega]$.  The following
relations hold:
\begin{gather*}
 [\tau] \quad \mbox{and} \quad  [\omega]  \quad
\mbox{are in the center of} \ A_{\sigma}(\striplet),   \\
 [\tau] ^2 = [\omega] - \frac{c_{2m+1,1}}{24},
 \\
 [\tau]*[X]=\frac{-\sqrt{2m+1}}{2} [\widehat{X}], \qquad \mbox{for}
\quad X \in \{E,
F, H\}, \\
[\widehat{H}]*[\widehat{F}]-[\widehat{F}]*[\widehat{H}]=-2q([\omega])[\widehat{F}], \\
 [\widehat{H}]*[\widehat{E}]-[\widehat{E}]*[\widehat{H}]=2q([\omega])[\widehat{E}], \\
[\widehat{E}]*[\widehat{F}]-[\widehat{F}]*[\widehat{E}]=-2q([\omega])[\widehat{H}],
\end{gather*}
 where $q$ is a certain polynomial.
 \end{proposition}

Equipped with all these results we are now ready to classify irreducible $\sigma$-twisted $\striplet$-modules.

\begin{remark}
By using same arguments as in Proposition 5.6 of~\cite{ABD} we have that
for $C_2$-cof\/inite SVOAs, every week (twisted) module is admissible (see also~\cite{DZ}). Thus, the classif\/ication of
irreducible $\sigma$-twisted $\striplet$-modules reduces to classif\/ication of irreducible $\mathbb{Z}_{\geq 0}$-gradable modules.
\end{remark}

\begin{theorem} \label{class-ired-zhu}\qquad{}
\begin{enumerate}\itemsep=0pt
\item[$(i)$]The set
\[
\{ R\Pi(i)^{\pm}(0),  : 1 \leq i \leq m+1  \}
\cup \{ R\Lambda(i)^{\pm}(0) : 1 \leq i \leq m\}
\] provides, up to
isomorphism, all irreducible modules for  Zhu's algebra
$A_{\sigma}(\mathcal{SW}(m))$.

\item[$(ii)$] The set
\[
\{ R\Pi(i)(0),  : 1 \leq i \leq m+1 \} \cup
\{ R\Lambda(i)(0) : 1 \leq i \leq m\}
\] provides, up to
isomorphism, all ${\Z}_2$-graded irreducible modules for  Zhu's
algebra \linebreak $A_{\sigma}(\mathcal{SW}(m))$.
\end{enumerate}
\end{theorem}

\begin{proof}
The proof  is similar to those of Theorem 3.11 in
\cite{AdM-triplet} and Theorem 10.3 in \cite{AdM-str}.
 Assume that $U$ is an irreducible $A_{\sigma}(\mathcal{SW}(m))$-module. Relation
$f_m ^{\R} ([\omega]) = 0$ in $A_{\sigma}(\mathcal{SW}(m))$
implies that
\[
L(0) \vert U = h^{2i+2,1} \, \mbox{Id}, \qquad \mbox{for} \quad i \in \{0, \dots, m-1 \} \cup \{2m, \dots, 3m\}.
\]
Moreover, since $[\tau]$ is   in the center of
$A_{\sigma}(\mathcal{SW}(m))$, we conclude that $G(0)$ also acts on
$U$ as a~scalar. Then relation $[\tau] ^2 = [\omega] - c_{2m+1,1} /
24$ implies that
\[
  G(0) \vert U = \frac{1/2 + i -m}{\sqrt{2(2m+1)}}  \, \mbox{Id} \qquad
\mbox{or} \qquad G(0) \vert U = -\frac{1/2 + i -m}{\sqrt{2(2m+1)}}
\, \mbox{Id}.
\]
Assume f\/irst that $i= 2m + j$ for $ 0 \le j \le m$. By combining
Propositions \ref{zhu-komut-r} and Theorem
\ref{ireducibilni-moduli-ramond-nongraded} we have that
$q(h^{2i+2,1}) \ne 0$. Def\/ine
\[
 e= \frac{ 1}{ \sqrt{2} q(h^{2 i+2,1})} [\widehat{E}], \qquad f= -\frac{1}{ \sqrt{2}q(h^{2i+2,1})} [\widehat{F}], \qquad h= \frac{1}{q(h^ {2i+2,1})} [\widehat{H}] .
 \]

Therefore $U$ carries the structure of an irreducible,
$\goth{sl}_2$-module with the property that $e^2 = f^2 = 0$  and
$h \ne 0$ on $U$. This easily implies that $U$ is a
$2$-dimensional irreducible $\goth{sl}_2$-module. Moreover,  as
an $A_{\sigma}(\mathcal{SW}(m))$-module $U$ is isomorphic to
either  $R\Pi(m+1-j)^{+}(0)$ or $R\Pi(m+1-j)^{-}(0)$.

In the case  $ 0 \le i \le m-1$, as in \cite{AdM-triplet} we prove
  that $U \cong R\Lambda(i+1)^{+} (0)$ or $U \cong R\Lambda(i+1)^{-} (0)$.

  Let us now prove the second assertion. Let $ N = N^{0} \oplus
  N^{1}$ be the graded irreducible
  $A_{\sigma}(\striplet)$-module. As above, we have that
\[
L(0) \vert N= h^{2i+2,1} \, \mbox{Id}, \qquad \mbox{for} \quad i \in \{0, \dots, m-1 \} \cup \{2m, \dots, 3m\}.
\]

  Then $N= N^{+} \oplus N^{-}$, where
\begin{gather*}
 N^{\pm} = \mbox{span}_{\C} \left\{ v \pm \frac{\sqrt{2(2m+1)}}{1/2 + i -m} G(0) v  \ \vert v \in
  N^{0} \right\} \qquad \!\!\mbox{and}\!\!\qquad
 G(0) \vert N^{\pm} = \pm \frac{1/2 + i -m}{\sqrt{2(2m+1)}}  \, \mbox{Id} .
\end{gather*}
By using assertion $(i)$, we easily get that $N^{\pm} =
R\Lambda(i+1)^{\pm} (0)$ (if $ 0 \le i \le m-1$)  and  $N ^{\pm} =
R\Pi(3m+1-i)^{\pm}(0) $ (if $ 2m \le i \le 3m$).
  \end{proof}

\begin{theorem} \label{class-ired}\qquad
\begin{enumerate}\itemsep=0pt
\item[$(i)$] The set
\[
\{ R\Pi(i)^{\pm}   : 1 \leq i \leq m+1  \}
\cup \{ R\Lambda(i)^{\pm} : 1 \leq i \leq m\}
\]
 provides, up to
isomorphism, all irreducible $\sigma$-twisted
$\striplet$-modules.

\item[$(ii)$] The set
\[
\{ R\Pi(i)   : 1 \leq i \leq m+1  \} \cup \{
R\Lambda(i) : 1 \leq i \leq m\}
\] provides, up to isomorphism, all
${\Z}_2$-graded irreducible $\sigma$-twisted
$\striplet$-modules.
\end{enumerate}
\end{theorem}

So the vertex operator algebra $\striplet$ contains only f\/initely
many irreducible modules. But one can easily see that modules
$V_{L+\ga^{R}_{i}} \otimes M$ and $V_{L+\ga^{R}_{m+i}} \otimes M$
($0 \le i \le m-1$) constructed in Theorems \ref{str-fock-R-1} and
\ref{str-fock-R-2} are not completely reducible. Thus we have:

\begin{corollary} The vertex operator superalgebra $\striplet$ is
not $\sigma$-rational, i.e., the category of $\sigma$-twisted
$\striplet$-modules is not semisimple.
\end{corollary}

\begin{remark}
In our    forthcoming paper \cite{AdM-lattice}   we shall prove that
$\striplet$ also contains logarithmic $\sigma$-twisted
representations.
\end{remark}

\section[Modular properties of characters of $\sigma$-twisted  $\mathcal{SW}(m)$-modules]{Modular properties of characters\\ of $\boldsymbol{\sigma}$-twisted  $\boldsymbol{\mathcal{SW}(m)}$-modules}
 \label{characters}

We f\/irst introduce some basic modular forms
needed for description of irreducible twisted $\striplet$
characters. The Dedekind $\eta$-function is usually def\/ined as the
inf\/inite product
\[
\eta(\tau)=q^{1/24} \prod_{n=1}^\infty (1-q^n),
\]
an automorphic form of weight $\frac{1}{2}$. As usual in all these
formulas $q=e^{2 \pi i \tau}$, $\tau \in \mathbb{H}$. We also
introduce
\begin{gather*}
\goth{f}(\tau) = q^{-1/48}
\prod_{n=0
}^\infty (1+q^{n+1/2}), \\
\goth{f}_1(\tau) = q^{-1/48} \prod_{n=1}^\infty (1-q^{n-1/2}),
\\
\goth{f}_2(\tau) = q^{1/24} \prod_{n=1}^\infty (1+q^n).
\end{gather*}

Let us recall Jacobi $\Theta$-function
\[
\Theta_{j,k}(\tau,z)=\sum_{n \in \mathbb{Z}} q^{k(n+\frac{j}{2k})^2} e^{2 \pi i k z (n+ \frac{j}{2k} ) },
\]
where $j,k \in \frac{1}{2}\mathbb{Z}$.
If $j \in \mathbb{Z}+\frac{1}{2}$ or $k \in \mathbb{N}+\frac{1}{2}$ it will be useful to use the formula
\begin{gather*}
\Theta_{j,k}(\tau,z)=\Theta_{2j,4k}(\tau,z)+\Theta_{2j-4k,4k}(\tau,z).
\end{gather*}
Observe that
\[
\Theta_{j+2k,k}(\tau,z)=\Theta_{j,k}(\tau,z).
\]
We also let
\[
\partial \Theta_{j,k}(\tau,z):=\frac{1}{\pi i} \frac{d}{dz} \Theta_{j,k}(\tau,z)=\sum_{n \in \mathbb{Z}}
(2kn+j)q^{(2kn+j)^2/4k}.
\] Related $\Theta$-functions needed for
supercharacters are
\[
G_{j,k}(\tau,z)=\sum_{n \in \mathbb{Z}} (-1)^n q^{k(n+\frac{j}{2k})^2} e^{2 \pi i k z (n+ \frac{j}{2k} ) },
\]
where $j \in \mathbb{Z}$ and $k \in \frac{1}{2} \mathbb{N}$. It is easy to see that
\begin{gather*}
G_{j,k}(\tau,z)=\Theta_{2j,4k}(\tau,z)-\Theta_{2j-4k,4k}(\tau,z).
\end{gather*}
Eventually, we shall let $z=0$ so before we introduce
\[
\Theta_{j,k}(\tau):=\Theta_{j,k}(\tau,0), \qquad G_{j,k}(\tau):=G_{j,k}(\tau,0).
\]
Similarly, we def\/ine $\partial \Theta_{j,k}(\tau)$ and $\partial G_{j,k}(\tau)$.

The following formulas will be useful:
\begin{gather*}
  \Theta_{j,k}\left(\frac{-1}{\tau},\frac{z}{\tau}\right)=\frac{\sqrt{-i \tau}}{\sqrt{2k}} e^{\frac{4 \pi i k z^2}{\tau}}\left(\sum_{j'=0}^{4k-1} e^{-\pi i j' j/k} \Theta_{2j',4k}(\tau,z) \right), \\
  \partial \Theta'_{j,k}\left(\frac{-1}{\tau},\frac{z}{\tau}\right)=\frac{\sqrt{-i \tau}}{\sqrt{2k}} e^{\frac{4 \pi i k z^2}{\tau}} {8 k z} \left(\sum_{j'=0}^{4k-1} e^{-\pi i j' j/k} \Theta_{2j',4k}(\tau,z) \right)\nonumber  \\
\hphantom{\partial \Theta'_{j,k}\left(\frac{-1}{\tau},\frac{z}{\tau}\right)=}{}  + \tau \frac{\sqrt{-i \tau}}{\sqrt{2k}} e^{\frac{4 \pi i k z^2}{\tau}} \left(\sum_{j'=0}^{4k-1} e^{-\pi i j' j/k} \partial \Theta_{2j',4k}(\tau,z) \right).
\end{gather*}
If we let now $z=0$, we obtain
\begin{gather} \label{theta-1}
\Theta_{j,k}\left(\frac{-1}{\tau}\right)=\frac{\sqrt{-i \tau}}{\sqrt{2k}}\sum_{j'=0}^{4k-1} e^{\frac{-\pi i j j'}{k}} \Theta_{2j',4k}(\tau,0),
\end{gather}
and
\begin{gather} \label{theta-2}
\partial \Theta'_{j,k}\left(\frac{-1}{\tau}\right)= \tau \frac{\sqrt{-i \tau}}{\sqrt{2k}}\sum_{j'=0}^{4k-1} e^{\frac{-\pi i j j'}{k}} \partial \Theta_{2j',4k}(\tau,0).
\end{gather}

For $j \in \mathbb{Z}$ and $k \in \mathbb{N}+\frac{1}{2}$ we now have:
\begin{gather*}
 \Theta_{j,k}\left(\frac{-1}{\tau}\right)=\sqrt{\frac{-i
\tau}{2k}}\sum_{j'=0}^{2k-1} e^{-i \pi j j'/k} \Theta_{j',k}(\tau), \\
 \Theta_{j,k}(\tau+1)=e^{i \pi j^2/2k}G_{j,k}(\tau), \\
 (\partial \Theta)_{j,k}(\tau+1)=e^{i \pi j^2/2k}(\partial
G)_{j,k}(\tau), \\
 (\partial \Theta)_{j,k}\left(\frac{-1}{\tau}\right)=\tau \sqrt{-i \tau/2k}
\sum_{j'=1}^{2k-1} e^{-i \pi j j'/k} (\partial
\Theta)_{j',k}(\tau),
\end{gather*}
where we used (\ref{theta-1}) and (\ref{theta-2}), together with  $\Theta_{j,k}(\tau)=\Theta_{j+2k,k}(\tau)$  and
$\partial \Theta_{j,k}(\tau)=\partial \Theta_{j+2k,k}(\tau)$.

For $j \in \mathbb{Z}+\frac{1}{2}$ and $k \in \mathbb{N}+\frac{1}{2}$ the transformation formulas are slightly
dif\/ferent:
\begin{gather*}
 \Theta_{j,k}\left(\frac{-1}{\tau}\right)=\sqrt{\frac{-i
\tau}{2k}}\sum_{j'=0}^{2k-1} e^{-i \pi j j'/k} G_{j',k}(\tau), \\
 \Theta_{j,k}(\tau+1)=e^{i \pi j^2/2k}\Theta_{j,k}(\tau), \\
 (\partial \Theta)_{j,k}(\tau+1)=e^{i \pi j^2/2k}(\partial
\Theta)_{j,k}(\tau), \\
  (\partial \Theta)_{j,k}\left(\frac{-1}{\tau}\right)=\tau \sqrt{-i \tau/2k}
\sum_{j'=1}^{2k-1} e^{-i \pi j j'/k} (\partial
G)_{j',k}(\tau).
\end{gather*}

For a (twisted) vertex operator algebra module $W$ we def\/ine its
graded-dimension or simply character
\[
\chi_{W}(\tau)={\rm tr}|_{W} q^{L(0)-c/24}.
\]
If $V=L^{\NS}(c_{2m+1,0},0)$ and
$W=L^{\R}(c_{2m+1,0},h^{2i+2,2n+1})$, then (see \cite{IK3}, for
instance)
 \begin{gather} \label{irr-r}
 \chi_{L^{\R}(c_{2m+1,1},h^{2i+2,2n+1})}(\tau)=2 q^{\frac{m^2}{2(2m+1)}-\frac{1}{16}}\frac{\goth{f}_2
 (\tau)}{\eta(\tau)}\left(q^{h^{2i+2,2n+1}}-q^{h^{2i+2,-2n-1}}\right).
\end{gather}

By combining Theorems~\ref{str-fock-R-1}, \ref{str-fock-R-2} and
\ref{str-fock-R-3}, and formula (\ref{irr-r}) we obtain

\begin{proposition} \label{char-irr-mod-r} For $i=0,\dots,m-1$
\begin{gather} \label{sl-char}   \chi_{ R\Lambda(i+1)}(\tau)=
2 \frac{\goth{f}_2(\tau)}{\eta(\tau)}\left(\frac{2i+2}{2m+1}
\Theta_{m-i-\tfrac{1}{2},\frac{2m+1}{2}}(\tau)+\frac{2}{2m+1}(\partial
\Theta)_{m-i-\tfrac{1}{2},\frac{2m+1}{2}}(\tau) \right), \\
 \label{pi-char}
 \chi_{R \Pi(m-i)}(\tau)= 2
\frac{\goth{f}_2(\tau)}{\eta(\tau)}\left(\!\frac{2m-2i-1}{2m+1}
\Theta_{m-i-\tfrac{1}{2},\frac{2m+1}{2}}(\tau)-\frac{2}{2m+1}(\partial
\Theta)_{m-i-\tfrac{1}{2},\frac{2m+1}{2}}(\tau)\!\right)\!. \!\!\!\!\!
\end{gather}
 Also, \begin{gather*}
  \chi_{R \Pi(m+1)}(\tau)=2
\frac{\goth{f}_2(\tau)}{\eta(\tau)} \Theta_{m+
\tfrac{1}{2},\frac{2m+1}{2}}(\tau).
\end{gather*}
\end{proposition}

As in \cite{AdM-str}, the characters of irreducible
$\sigma$-twisted $\striplet$-modules can be described by using
characters of irreducible modules for the triplet vertex algebra
$\triplet$, where $p= 2 m +1$. Let $\Lambda(1), \dots, \Lambda
(p)$, $\Pi(1), \dots , \Pi(p)$ be the irreducible
$\triplet$-module. We have the following result:

\begin{proposition}\qquad{}
\begin{enumerate}\itemsep=0pt
\item[$(i)$] For $0 \le i \le m-1$, we have:
\[
 \chi_{ R\Lambda(i+1)}(\tau) = 2 \frac{\chi_{ \Lambda(2
i+2)}(\tau/2)}{\goth{f}(\tau)}.
\]

\item[$(ii)$] For $0 \le i \le m$, we have:
\[
 \chi_{ R\Pi(m+1 -i)}(\tau) = 2\frac{\chi_{ \Pi(2m - 2i+1)}(\tau/2)}{\goth{f}(\tau)}.
 \]
\end{enumerate}
\end{proposition}

Now, we recall also formulas for irreducible $\striplet$ characters and supercharacters  obtained in \cite{AdM-str}.

For $i=0,\dots,m-1$
\begin{gather} \label{orsl-char}   \chi_{S
\Lambda(i+1)}(\tau)=\frac{\goth{f}(\tau)}{\eta(\tau)}\left(\frac{2i+1}{2m+1}
\Theta_{m-i,\frac{2m+1}{2}}(\tau)+\frac{2}{2m+1}(\partial
\Theta)_{m-i,\frac{2m+1}{2}}(\tau) \right), \\
  \label{orpi-char} \chi_{S
\Pi(m-i)}(\tau)=\frac{\goth{f}(\tau)}{\eta(\tau)}\left(\frac{2m-2i}{2m+1}
\Theta_{m-i,\frac{2m+1}{2}}(\tau)-\frac{2}{2m+1}(\partial
\Theta)_{m-i,\frac{2m+1}{2}}(\tau)\right).
\end{gather}
 Also, \begin{gather}
  \label{ss-char}    \chi_{S
\Lambda(m+1)}(\tau)=\frac{\goth{f}(\tau)}{\eta(\tau)}
\Theta_{0,\frac{2m+1}{2}}(\tau).
\end{gather}

For supercharacters we have: for $i=0,\dots,m-1$
\begin{gather}\label{sl-schar}   \chi^F_{S \Lambda(i+1)}(\tau)=
 \frac{\goth{f}_1(\tau)}{\eta(\tau)} \biggl(
\frac{2i+1}{2m+1}G_{m-i,\frac{2m+1}{2}}(\tau)+ \frac{2}{2m+1} (\partial G)_{m-i,\frac{2m+1}{2}}(\tau) \biggr),\\
 \label{p1-schar}   \chi^F_{S \Pi(m-i)}(\tau)=
  \frac{\goth{f}_1(\tau)}{\eta(\tau)} \biggl(
\frac{2m-2i}{2m+1}G_{m-i,\frac{2m+1}{2}}(\tau)- \frac{2}{2m+1} (\partial G)_{m-i,\frac{2m+1}{2}}(\tau) \biggr).
\end{gather}
Also, \begin{gather}
 \label{ss-schar} \chi_{S
\Lambda(m+1)}^F(\tau)=\frac{\goth{f}_1(\tau)}{\eta(\tau)}
G_{0,\frac{2m+1}{2}}(\tau).
\end{gather}

These characters and supercharacters  can be expressed by using
characters of $W(2m+1)$-modules.

\begin{proposition} \label{SWW}\qquad {}{\samepage
\begin{itemize}\itemsep=0pt
\item[$(i)$] For  $0 \leq i \leq m$, we have
\[
\chi_{S\Lambda(i+1)}(\tau)=\frac{\chi_{\Lambda(2i+1)}(\frac{\tau}{2})}{\goth{f}_2(\tau)}, \qquad
\chi_{S\Lambda(i+1)}
^F(\tau)=\frac{\chi_{\Lambda(2i+1)}(\frac{\tau+1}{2})}{\goth{f}_2(\tau)}.
\]
\item[$(ii)$] For $0 \leq i \leq m-1$, we also have
\[
\chi_{S\Pi(m-i)}(\tau)=\frac{\chi_{\Pi(2m-2i)}(\frac{\tau}{2})}{\goth{f}_2(\tau)}, \qquad
\chi_{S\Pi(m-i)}
^F(\tau)=\frac{\chi_{\Pi(2m-2i)}(\frac{\tau+1}{2})}{\goth{f}_2(\tau)}.
\]
\end{itemize}
Here $\Lambda(i)$ and $\Pi(2m+2-i)$, $i=1, \dots, 2m+1$, are
irreducible $\mathcal{W}(2m+1)$-modules~{\rm \cite{AdM-triplet}}.}
\end{proposition}

By combining   transformation formulas for $\Theta_{j,k}(\tau)$,
$\partial \Theta_{j,k}(\tau)$, formulas
(\ref{sl-char})--(\ref{ss-char}), (\ref{sl-schar})--(\ref{ss-schar})
and Proposition~\ref{char-irr-mod-r} we obtain second main result of
our paper.

\begin{theorem} \label{thm-main-2} The $SL(2,\mathbb{Z})$ closure, called $\mathcal{H}$, of the vector space determined by $\striplet$-characters, $\striplet$-supercharacters and
$\sigma$-twisted $\striplet$-characters has the following basis:
\begin{gather*}
  \frac{\goth{f}_1(\tau)}{\eta(\tau)} G_{0,\frac{2m+1}{2}}(\tau),  \qquad  \frac{\goth{f}(\tau)}{\eta(\tau)} \Theta_{0,\frac{2m+1}{2}}(\tau), \qquad   \frac{\goth{f}_2(\tau)}{\eta(\tau)}
\Theta_{m+ \tfrac{1}{2},\frac{2m+1}{2}}(\tau), \\
  \frac{\goth{f}_1(\tau)}{\eta(\tau)} G_{m-i,\frac{2m+1}{2}}(\tau),  \qquad \frac{\goth{f}(\tau)}{\eta(\tau)}\Theta_{m-i,\frac{2m+1}{2}}(\tau),  \qquad  \frac{\goth{f}_2(\tau)}{\eta(\tau)} \Theta_{m-i-\tfrac{1}{2},\frac{2m+1}{2}}(\tau) ,  \\
  \frac{\goth{f}_1(\tau)}{\eta(\tau)} \partial G_{m-i,\frac{2m+1}{2}}(\tau),  \qquad \frac{\goth{f}(\tau)}{\eta(\tau)} \partial \Theta_{m-i,\frac{2m+1}{2}}(\tau), \qquad  \frac{\goth{f}_2(\tau)}{\eta(\tau)}  \partial \Theta_{m-i-\tfrac{1}{2},\frac{2m+1}{2}}(\tau) , \\
  \tau \frac{\goth{f}_1(\tau)}{\eta(\tau)} \partial G_{m-i,\frac{2m+1}{2}}(\tau),  \qquad \tau \frac{\goth{f}(\tau)}{\eta(\tau)} \partial \Theta_{m-i,\frac{2m+1}{2}}(\tau),  \qquad \tau \frac{\goth{f}_2(\tau)}{\eta(\tau)}  \partial \Theta_{m-i-\tfrac{1}{2},\frac{2m+1}{2}}(\tau),
\end{gather*}
where $i=0,\dots,m-1$. In particular, the space is $9m+3$ dimensional.
\end{theorem}

\subsection[Modular differential equations for $\sigma$-twisted $\striplet$ characters]{Modular dif\/ferential equations for $\boldsymbol{\sigma}$-twisted $\boldsymbol{\striplet}$ characters}

Let us recall classical $SL(2,\mathbb{Z})$ Eisenstein series ($k \geq 1$):
\[
G_{2k}(\tau)=\frac{-B_{2k}}{(2k)!}+\frac{2}{(2k-1)!}\sum_{n=1}^\infty \frac{n^{2k-1}q^n}{1-q^n},
\]
and certain linear combination of level 2 Eisenstein series ($k \geq 1$):
\begin{gather*}  G_{2k,1}(\tau)=\frac{B_{2k}}{(2k)!}+ \frac{2}{(2k-1)!} \sum_{n \geq 1}
\frac{n^{2k-1}q^n}{1+q^n},   \\
  {G}_{2k,0}(\tau)=\frac{B_{2k}(1/2)}{(2k)!}+\frac{2}{(2k-1)!}\sum_{n=1}^\infty \frac{(n-1/2)^{2k-1}q^{n-1/2}}{1+q^{n-1/2}},
\end{gather*}
where $B_{2k}(x)$ are the Bernoulli polynomials, and $B_{2k}$ are the Bernoulli numbers.

A modular dif\/ferential equation is an (ordinary) dif\/ferential equation of the form:
\[
\left( q \frac{d}{dq} \right)^k y(q)+\sum_{j=0}^{k-1} H_j(q) \left(d \frac{d}{dq} \right)^i y(q)=0,
\]
where  $H_j(q)$ are polynomials in Eisenstein series $G_{2i}$, $i \geq 1$, such that the vector
space of solutions is modular invariant.

It is known (cf.~\cite{Z}) that $C_2$-cof\/initeness condition leads
to certain modular dif\/ferential equation satisf\/ied by irreducible
characters ${\rm tr}_M q^{L(0)-c/24}$. In some instances the
degree of this dif\/ferential equation is bigger than the number of
irreducible characters.  So it is not clear what $k$ should be in
general. In the case of Virasoro minimal models, the degree of the
modular dif\/ferential equation is precisely the number of (linearly
independent) irreducible characters.

If $V$ is a vertex operator superalgebra  one can also get modular dif\/ferential equation satisf\/ied by ordinary characters but with respect to the subgroup $\Gamma_{\theta} \subset SL(2,\mathbb{Z})$, where $H_i$ are polynomials in $G_{2i}$ and $G_{2i,0}$ (see~\cite{M} for the precise  statement in the case of $N=1$ minimal models).

In \cite{AdM-ptraces} we proved that the $C_2$-cof\/initeness for the super triplet  vertex algebra $\striplet$ gives rise to a dif\/ferential equation of order $3m+1$ satisf\/ied by $2m+1$ irreducible characters (additional $m$ solutions can be interpreted as certain
{\em pseudotraces}). By applying arguments similar to those in \cite{AdM-ptraces}  it is not hard to prove

\begin{theorem} \label{diff-equation} The irreducible $\sigma$-twisted $\striplet$ characters satisfy the differential equation of the form
\[
\left(d \frac{d}{dq} \right)^{3m+1} y(q)+\sum_{j=0}^{3m} \tilde{H}_{j,0}(q)\left(q \frac{d}{dq} \right)^i y(q)=0,
\]
where $\tilde{H}_{j,0}(q)$ are certain polynomials in $G_{2i}$ and $G_{2i,1}$.
\end{theorem}

As in the case of $\triplet$-modules and ordinary $\striplet$-modules,  we expect that additional $m$ linearly independent solutions in Theorem \ref{diff-equation} have interpretation
in terms of  $\sigma$-twisted pseudotraces (cf.~Conjecture \ref{conj-twist-modular} below).

\section{Conclusion}

Here we gather a few more-or-less expected conjectures (especially in view of our earlier work~\cite{AdM-triplet} and~\cite{AdM-str}).

The f\/irst one is concerned about the structure of $A_{\sigma}(V)$. As in the case of $A(\triplet)$ and $A(\striplet)$, from our analysis in Chapter 6, we can  show that in fact
\[
{\rm dim}(A_{\sigma}(\striplet)) \geq 10m+8.
\]
But this is well below the conjectural dimension, because $10m+8$-dimensional part cannot control possible  logarithmic modules. Thus, as in the case of ordinary $\striplet$-modules, we expect
to have  $2m$ non-isomorphic (non-graded) logarithmic modules with two-dimensional top component. This then leads to the following conjecture:

\begin{conjecture} \label{conj-twist} For every $m \in \mathbb{N}$,
\[
{\rm dim}(A_{\sigma}(\striplet))=12m+8.
\]
\end{conjecture}

If we assume the existence of $m$ logarithmic modules so that Conjecture \ref{conj-twist} holds true, then the following fact is expected.
\begin{conjecture} \label{conj-twist-modular}
The vector space of (suitably defined) generalized $\sigma$-twisted characters is $3m+1$-dimensional.
\end{conjecture}

Thus,  generalized $\striplet$ characters, supercharacters and $\sigma$-twisted characters together should give rise to  a $9m+3$-dimensional modular invariant space.

\subsection*{Acknowledgments}

The second author was partially supported by NSF grant DMS-0802962.

\pdfbookmark[1]{References}{ref}

config: arxiv


TITLE: Show that this space is a vector space
QUESTION [1 upvotes]: I am looking through a paper that says:
If for the space $L_p(a,b)$, $\int_a^{b} |x(t)|^p dt$ exists, this space is a vector space.
I would like to convince myself that this is true, but I have no idea why. In an undergrad course on linear algebra, I usually used to prove that something was a vector space by showing that all axioms were satisfied. I don't know if/how that can be done here.
Any help?

REPLY [2 votes]: The reason is a little bit tricky given no background, and you would see this at the beginning of a course in so-called functional analysis when the $L_p$ spaces are introduced. The result is implied by something called Hölder's inequality, which says that
$$
\int_a^b \lvert f(x) g(x) \rvert d x
\leq
\left(\int_a^b \lvert f(x) \rvert^p d x\right)^\frac{1}{p}
\left(\int_a^b \lvert g(x) \rvert^q d x\right)^\frac{1}{q}
$$
for functions $f(x)$ and $g(x)$ and $p, q \in (1, \infty)$ satisfying $\frac{1}{p} + \frac{1}{q} = 1$. This in turn can be used to prove Minkowski's inequality, which says
$$
\left(\int_a^b \lvert f(x) + g(x) \rvert^p d x\right)^\frac{1}{p}
\leq
\left(\int_a^b \lvert f(x) \rvert^p d x\right)^\frac{1}{p}
+
\left(\int_a^b \lvert g(x) \rvert^p d x\right)^\frac{1}{p}
$$
for $p \in (1, \infty)$. The fact that $L_p([a,b])$---the space of functions $f(x)$ for which $\int_a^b \lvert f(x) \rvert^p d x < \infty$---is a vector space now follows immediately from this last inequality.
(Note that here by a "function $f(x)$" we mean something called a measurable function.)


TITLE: Examples of $(\infty,1)$-topoi that are not given as sheaves on a Grothendieck topology
QUESTION [24 upvotes]: An $(\infty,1)$-topos according to Lurie is defined as (accessible) left exact localization of a presheaf $(\infty,1)$-category $\text{P}(\mathcal C)$.
Those $(\infty,1)$-topoi $\text{Sh}(\mathcal C)$ arrising from a site $C$ correspond precisely to topological left exact localizations of $\text{P}(\mathcal C)$.
What is an example of an $(\infty,1)$-topos not given as $\text{Sh}(\mathcal C)$ - i.e. an  $(\infty,1)$-topos arrising from a non-topological localization? Should I think of them as pathological or as useful to have?

REPLY [20 votes]: Marc's examples are good ones, but let me add two more (which are closely related to each other):
1) Let $\mathcal{C}$ be an accessible $\infty$-category which admits small filtered colimits, and let $\mathcal{X}$ be the $\infty$-category
of functors from $\mathcal{C}$ to $\mathcal{S}$ which preserve small filtered colimits. Then $\mathcal{X}$ is an $\infty$-topos, but I do not know how to realize $\mathcal{X}$ as an $\infty$-category of sheaves on a site.
2) Let $X$ be a locally compact topological space, and let $\mathcal{X}$ be the $\infty$-topos of sheaves on $X$. Then $\mathcal{X}$ is exponentiable in the setting of $\infty$-topoi: that is, for every $\infty$-topos $\mathcal{Y}$, there exists another $\infty$-topos $\mathcal{Y}^{\mathcal{X}}$ such that geometric morphisms from $\mathcal{Z}$ into $\mathcal{Y}^{\mathcal{X}}$ are the same as geometric morphisms from $\mathcal{Z} \times \mathcal{X}$ into $\mathcal{Y}$
(where the product is formed in the $\infty$-category of $\infty$-topoi). I do not know if $\mathcal{Y}^{\mathcal{X}}$ can be realized as an $\infty$-topos of sheaves on a site, even if it is assumed that such a description is known for $\mathcal{Y}$.


TITLE: Method of calculating the last digit of $7^{49}$
QUESTION [6 upvotes]: I would like to point out the difference between $7^{7^7}$ and $7^{49}$. typically the notation $a^{a^{a^a}}$ means $a^{(a^a)}$. Heuristically, start from the top and work your way down. e.g. $2^{2^{{2^2}}} = 2^{(2^{({2^2})})} = 2^{(2^{4})} = 2^{16}=65536$.

REPLY [2 votes]: notice that $7^2 \equiv 49 \equiv -1 \pmod {10}$
then we have
$$7^{49} \equiv 7^{32} \times 7^{16} \times 7 \equiv (7^{2})^{16} \times (7^{2})^{8} \times 7 \equiv(-1)^{16} \times (-1)^8 \times 7\equiv 7 \pmod {10}$$
thus the answer is 7.


TITLE: Find the equation of the line tangent to the curve of the function $e^{x^2}$ in the point with coordinates $x=1$,$y=e$.
QUESTION [1 upvotes]: Find  the equation of the line tangent to the curve of the function $e^{x^2}$ in the point with coordinates $x=1$,$y=e$.
I found that tis equation is $y=2ex-e$ Is this wrong?

REPLY [2 votes]: Line tangent to a curve in the point $(x_0, y_0)$:
$$y = f'(x_0) (x-x_0)+f(x_0)$$
where $f'(x)$ is the derivative of $f(x)$ and $f(x_0)=y_0$. Then, since in this case $f(x) = e^{x^2}$, we get $f'(x)= 2x e^{x^2}$. Therefore the equation of the tangent line is:
$$y=2e(x-1)+e=2ex-2e+e=2ex-e$$
So the answer is yes, you are right

REPLY [1 votes]: here dy/dx=2xy
at(1,e) , dy/dx=2e
therefore equation of tangent line at (1,e)
y-e=2e(x-1)
y=2ex-e
Thanks


TITLE: Need help to understand this solution of $2^mp^2 +1=q^5$
QUESTION [2 upvotes]: The Question-

Find all triples $(m,p,q)$ where $ m $ is a positive integer and $ p ,
> q $ are primes.
****$2^mp^2 +1=q^5$****

After trying my best to solve this problem, and progressing a little bit, I decided to look up the solution and I am unable to quite understand it.
The solution-

clearly $q,p$ are odd
Let:$A=\sum _{i=0}^4q^i$ ,
$\large 2^mp^2=q^5-1=(q-1)\left ( A \right )$
  Let:$d=\gcd(q-1,A)$$\large \Rightarrow d|A=\left (\sum _{i=0}^4(q^i-1)
> \right )+5\Leftrightarrow d|5$
If:$d=5\large \Rightarrow p=5\Rightarrow A=5$ since$ A$ is odd , But
  $A>1+1+1+1+1=5$
then $d = 1$ $\large \Rightarrow q-1=2^m,A=p^2$ , Assume that $
> m\geq3$ $\large \Rightarrow q\equiv1[8]\Rightarrow p^2=A\equiv5[8]$
Which is impossible , so $m<3$
If:$m=1 \Rightarrow q=3,p=11$ $m=2$ $\large \Rightarrow q=5\Rightarrow
> p^2=781$
Therefore the unique solution is $(1,11,3)$

I have understood upto the part where $d$ divides 5, hence $d=1$ or $5$ but I don't understand how the solution progresses after that.
Could someone give me an explanation behind this proof, and also the thought process of the solution so that I can solve questions like this on my own

REPLY [2 votes]: I'm going to add some words to the solution.

$$2^mp^2=(q-1)A\tag1$$
where $A=q^4+q^3+q^2+q+1$.

If:$d=5\large \Rightarrow p=5\Rightarrow A=5$ since$ A$ is odd , But
  $A>1+1+1+1+1=5$

If $d=5$, then the right hand of $(1)$ is divisible by $5$, so it follows that $p=5$. Since $A$ is odd, we see that $2^m$ divides $q-1$. Noting that $5$ divides both $q-1$ and $A$, we see that $A=5$. However, this is impossible since $$A=q^4+q^3+q^2+q+1\gt 1+1+1+1+1=5$$

then $d = 1$ $\large \Rightarrow q-1=2^m,A=p^2$ , Assume that $m\geq3$ $\large \Rightarrow q\equiv1[8]\Rightarrow p^2=A\equiv5[8]$ which is impossible , so $m<3$
If:$m=1 \Rightarrow q=3,p=11$ $m=2$ $\large \Rightarrow q=5\Rightarrow p^2=781$

When $d=1$, since $A$ is odd, we see that $q-1=2^m$ and $A=p^2$.
Assume that $m\ge 3$. Then, $q=2^m+1\equiv 0+1\equiv 1\pmod 8$, so $$p^2=A=q^4+q^3+q^2+q+1\equiv 1^4+1^3+1^2+1+1\equiv 5\pmod 8$$
However, there is no integer $p$ satisfying $p^2\equiv 5\pmod 8$.
So, we get $m\lt 3$.
If $m=1$, then $q=2^1+1=3$ and $2p^2=3^5-1$ implies $p=11$.
If $m=2$, them $q=2^2+1=5$ and $4p^2=5^5-1$ implies $p=\sqrt {781}$ which is not an integer.


TITLE: Differential Equation with Integration
QUESTION [2 upvotes]: Function $f(x)$ is differentiable. Given further that
\begin{eqnarray*}
f’(x)+xf’(x-1) &=& 4 \\ \int_0^1 f(xt) dt + \int_0^x f(t-1)dt &=& x^3 +x^2 +2x
\end{eqnarray*}
Determine the function $f(x)$.
$\textbf{Attempt}$
I know $\frac{d}{dx} \int_0^x f(t-1)dt = f(x-1)$, but I am confused about what’s $\frac{d}{dx} \int_0^1 f(xt) dt $ . Maybe I can do $ - \frac{d}{dx} \int_1^x f(xt) dt - \frac{d}{dx} \int_0^x f(xt) dt $ ?

REPLY [2 votes]: You can set $xt=u$.
Setting $xt=u$ gives $x=\frac{du}{dt}$, so
$$\int_0^1 f(xt) dt =\int_{0}^{x}f(u)\cdot\frac{du}{x}$$
So, differentiating the both sides of
$$\frac 1x\int_{0}^{x}f(u)du+ \int_0^x f(t-1)dt =x^3 +x^2 +2x$$
gives
$$-\frac{1}{x^2}\int_{0}^{x}f(u)du+\frac{f(x)}{x}+f(x-1)=3x^2+2x+2,$$
i.e.
$$\int_{0}^{x}f(u)du=xf(x)+x^2f(x-1)-3x^4-2x^3-2x^2$$
Now, differentiate the both sides and use $f'(x)+xf'(x-1)=4$.


text:
\section{Proof of Lemma~\ref{lem:eigbd}} \label{sec:proofeigbd}

This section builds on a result from~\citep{bhojanapalli2018smoothed}, where a similar statement was made under different assumptions. The proof follows closely the developments therein, with appropriate changes.
Using~\eqref{eq:Riem_grad}, $Y$ is an $\varepsilon_g$-FOSP of the perturbed problem if and only if $\norm{2SY}\leq \varepsilon_g$ with $S=M+W$ and
\begin{align}
 	M & = C - (\mathcal{A}^*\circ G^{\dagger}\circ\mathcal{A})\left((C+W)YY^{*}\right).
 	\label{eq:Mproof}
\end{align}
Let $Y=P\Sigma Q^{*}$ be a thin SVD of $Y$, where $P$ is $n\times k$ with
orthonormal columns (assuming without loss of generality $k \leq n$, as otherwise $\sigma_k(Y) = 0$ deterministically) and $Q$ is $k\times k$ orthogonal. Then,
\begin{align*}
\varepsilon_{g} \geq \norm{2SY}&=\norm{2(M+W)Y}\\
&\geq 2\sigma_k(Y)\norm{(M+W)P}\\
&\geq 2\sigma_k(Y)\sqrt{\sum_{i=1}^k \sigma_{n-(i-1)}(M+W)^2}.
\end{align*}
Thus, we control the smallest singular value of $Y$ in terms of $\varepsilon$ and the $k$ smallest singular values of $M+W$:
\begin{equation}\label{eq:controlsmalleig}
 \sigma_k(Y) \leq \frac{\varepsilon_g}{2\sqrt{\sum_{i=1}^k \sigma_{n-(i-1)}(M+W)^2}}.
\end{equation}

Given that $M$ is not statistically independent of $W$, we are not able to directly apply Lemma~\ref{lem:bound_eigs}. Indeed,
$M$ depends on $W$ and on $Y$, and $Y$ itself is an $\varepsilon_g$-FOSP of the perturbed problem: a feature which depends on $W$. To tackle this issue, we cover the set of possible $M$s with a net.
Lemma~\ref{lem:bound_eigs} provides a bound for each $\bar{M}$ in this net. By union bound, we can extend the lemma for all $\bar{M}$. By taking a sufficiently dense net, we then infer that $M$ is necessarily close to one of these $\bar{M}$'s and conclude.

To this end, we first control $\norm{M-C}$ using the definitions of $R$~\eqref{boundTrace} and $K$~\eqref{boundProof}:

\begin{align*}
  \norm{M-C} & =\norm{\mathcal{A}^*\circ G^{\dagger}\circ\mathcal{A}\left((C+W)YY^{*} \right)}\\
&\leq \norm{\mathcal{A}^*\circ G^{\dagger}\circ\mathcal{A}}_{\mathrm{op}}\norm{C+W}_{\mathrm{op}}\norm{YY^{*}}\\
&\leq  K (\norm{C}_{\mathrm{op}} + \norm{W}_{\mathrm{op}})R.
\end{align*}
Since $W$ is a Gaussian Wigner matrix with variance $\sigma_W^2$, it is a well known fact
(see for instance\footnote{The reference proves the statement for complex matrices with diagonal entries equal to zero. That proof can easily be adapted to the definition of Wigner matrices used in this paper, both real and complex.} Part 1 of  Appendix A in~\citep{bandeira2016tightness}) that, with probability at least $1 - e^{-\frac{n}{2}}$,
\begin{align}
	\norm{W}_{\mathrm{op}} \leq 3\sigma_{W}\sqrt{n}.
	\label{eq:opnormWbound}
\end{align}
Hence, with probability at least $1 - e^{-\frac{n}{2}}$,
\begin{align*}
	\norm{M-C}& \leq R K (\norm{C}_{\mathrm{op}}+3\sigma_{W}\sqrt{n}) \triangleq \kappa,
\end{align*}
where we recover $\kappa$ as defined in~\eqref{eq:kappa}.

As a result, $M$ lies in a ball of center $C$ and radius $\kappa$. Moreover, from~\eqref{eq:Mproof}, we remark that $M$ lives in an affine subspace of dimension
$\mathrm{rank}(\mathcal{A}^*\circ G^{\dagger}\circ\mathcal{A})=\mathrm{rank}(\mathcal{A})$. A unit ball in Frobenius norm in $d$ dimensions admits an $\varepsilon$-net of $\left(1+\frac{3}{\varepsilon}\right)^d$ points (see for instance Lemma 1.18 in~\citep{RigolletNotes}).\footnote{The lemma in the reference shows that for any $\varepsilon \in (0,1)$ the cardinality of one such $\varepsilon$-net is bounded by $\left(3/\varepsilon\right)^d$. Furthermore, for $\varepsilon \geq 1$, there is an obvious $\varepsilon$-net of cardinality one, comprising just the origin. Hence, for any $\varepsilon > 0$, it is possible so find an $\varepsilon$-net of cardinality at most $\max\left(1, (3/\varepsilon)^d \right) \leq \left(1+3/\varepsilon\right)^d$.}
Thus, we pick a $\frac{k\sigma_W}{2 \kappa \sqrt{c_0n}}$-net on the unit ball with
$\left(1 + \frac{6 \kappa \sqrt{c_0n}}{k\sigma_W} \right)^{\mathrm{rank}(\mathcal{A})}$
points.
Rescaling by a factor $\kappa$ gives a $\frac{k\sigma_W}{2 \sqrt{c_0n}}$-net of a ball of radius $\kappa$ centered at zero.
Hence, for any $M$ as in~\eqref{eq:Mproof} there necessarily exists a point $\bar{M}$ in the net satisfying:
\begin{equation}\label{eq:MMbarineq}
 \norm{\bar{M}-M}\leq \frac{k\sigma_W}{2\sqrt{c_0n}}.
\end{equation}
Let $T\colon\Sy^{n\times n}\rightarrow \R^k$ be defined by $T_q(A) = (\sigma_{n-q+1}(A),\dots,\sigma_n(A))^{\top}$, that is: $T$ extracts the $q$
smallest singular values of $A$, in order. Then, by using the result from Exercise IV.3.5. in~\citep{Bhatia} in the first inequality,\footnote{The same result can be obtained by using
	Theorem IV.2.14 in the reference. In this setting, one considers the function
	$F(A)= -\sum_{i=1}^n \sigma_i^2(A)$; then, use the subadditive property of $F$, i.e., $F(A+B)\leq F(A)+F(B)$, and define
	$A=\bar{M}+W$ and $B=-(M+W)$.} we have:

\begin{align*}
 \norm{\bar{M}-M}&=\norm{(\bar{M}+W)-(M+W)}\\
 &= \sqrt{\sum_{i=1}^n \sigma_i^2\left((\bar{M}+W)-(M+W) \right)}\\
 &\geq \sqrt{\sum_{i=1}^n \left( \sigma_i(\bar{M}+W)-\sigma_i(M+W)\right)^2}\\
 &= \norm{T_n(\bar{M}+W)-T_n(M+W)}\\
 &\geq \norm{T_k(\bar{M}+W)-{T_k}(M+W)}  \\
 &\geq \norm{T_k(\bar{M}+W)}-\norm{T_k(M+W)},
\end{align*}
where we used the triangular inequality in the last inequality. Thus, rearranging we obtain
\begin{equation}\label{eq:ineqMbar}
 \sqrt{\sum_{i=1}^k \sigma_{n-(i-1)}(M+W)^2}\geq \sqrt{\sum_{i=1}^k \sigma_{n-(i-1)}(\bar{M}+W)^2}-\norm{\bar{M}-M}.
\end{equation}
Taking a union bound for Lemma~\ref{lem:bound_eigs} over each $\bar{M}$ in the net, we get that
\begin{equation}\label{eq:unionbound}
 \sqrt{\sum_{i=1}^k \sigma_{n-(i-1)}(\bar{M}+W)^2} \geq \frac{k\sigma_W}{\sqrt{c_0 n}}
\end{equation}
holds with probability at least
\begin{align}
	1-\exp\left(-\frac{k^2}{8}\log(8\pi)+k\log(n)+\mathrm{rank}(\mathcal{A})\cdot\log\left(1 + \frac{6 \kappa \sqrt{c_0n}}{k\sigma_W} \right)\right).
	\label{eq:probabilitysuccessfirst}
\end{align}
Combining~\eqref{eq:MMbarineq}, \eqref{eq:ineqMbar} and~\eqref{eq:unionbound}, we conclude that
\begin{align}
	 \sqrt{\sum_{i=1}^k \sigma_{n-(i-1)}({M}+W)^2} \geq \frac{k\sigma_W}{2\sqrt{c_0 n}}
\end{align}
holds with probability bounded as in~\eqref{eq:probabilitysuccessfirst}. Combining with~\eqref{eq:controlsmalleig}, we obtain
\begin{align*}
	\sigma_k(Y) \leq \frac{\varepsilon_g}{\sigma_W}\frac{\sqrt{c_0n}}{k}
\end{align*}
as desired. It remains to discuss the probability of success, which we do below.

Inside the log in~\eqref{eq:probabilitysuccessfirst}, we can safely replace $k$ with $1$, as this only hurts the probability. Then, the result holds with probability at least
\[1-\exp\left(-\frac{k^2}{8}\log(8\pi)+k\log(n)+\mathrm{rank}(\mathcal{A})\cdot
\log\left(1+\frac{6 \kappa \sqrt{c_0n}}{\sigma_W} \right) \right).   \]
We would like to constrain $k$ such that the exponential part is bounded by $\delta$. In this fashion, taking a union bound with event~\eqref{eq:opnormWbound}, we will get an overall probability of success of at least $1 - \delta - e^{-\frac{n}{2}}$.

Equivalently, $k$ must satisfy the quadratic inequality
\[ -ak^2 + bk + c \leq \log(\delta), \]
with $a,b >0,$ $c\geq 0$ defined by $a = \frac{\log(8\pi)}{8}, b = \log(n), c = \mathrm{rank}(\mathcal{A})\cdot
\log\left(1+\frac{6 \kappa\sqrt{c_0n}}{\sigma_W} \right)$. This quadratic inequality can be rewritten as:
\begin{align*}
	ak^2 - bk - c' \geq 0,
\end{align*}
with $c' = c + \log(1/\delta)$. This quadratic has two distinct real roots, one positive and one negative:
\begin{align*}
	\frac{b \pm \sqrt{b^2 + 4ac'}}{2a}.
\end{align*}
Since $k$ is positive, we deduce that $k$ needs to be larger than the positive root. The latter obeys the following inequality:\footnote{We use that, for any $u, v \geq 0$, $\sqrt{u + v} \leq \sqrt{\sqrt{u}^2 + \sqrt{v}^2 + 2\sqrt{u}\sqrt{v}} = \sqrt{u} + \sqrt{v}$.}
\begin{align*}
	\frac{b + \sqrt{b^2 + 4ac'}}{2a} & \leq \frac{b + b + 2\sqrt{ac'}}{2a} = \frac{b + \sqrt{ac'}}{a} = \frac{1}{a} b + \frac{1}{\sqrt{a}} \sqrt{c'}.
\end{align*}
Since both $1/a$ and $1/\sqrt{a}$ are smaller than 3, it is sufficient to require
\begin{align*}
	 k \geq 3 \left( b + \sqrt{c + \log(1/\delta)} \right).
\end{align*}
Assuming $\delta \leq 1$, we can use the inequality in the footnote again and find that it is sufficient to have
\begin{align*}
	k \geq 3 \left( b + \sqrt{\log(1/\delta)} + \sqrt{c} \right).
\end{align*}

Plugging in the definitions of $b$ and $c$, we find the sufficient condition (with $\delta \leq 1$):
\begin{align*}
	k & \geq 3 \left[ \log(n) + \sqrt{\log(1/\delta)} + \sqrt{ \mathrm{rank}(\mathcal{A})\cdot
		\log\left(1+\frac{6 \kappa\sqrt{c_0n}}{\sigma_W} \right) } \right].
\end{align*}

Since $\mathrm{rank}(\mathcal{A})\leq m$, we obtain the desired sufficient bound on $k$.


TITLE: Why isn't it trivial that $NP=co-NP$?
QUESTION [0 upvotes]: If we have a TM $M$ to decide a language $L\in NP$, why can't we just simulate $M$ on a universal machine $M^c$ and return the opposite?
I've read on another thread here at MS that this isn't the definition of NDTM but I don't see why.
Can you clarify that for me?

REPLY [2 votes]: By definition, a non-deterministic Turing machine M accepts a language $L$ if and only if any of the final states of $M$ when run with input $L$ are 'accepting states'. Running $M^c$ -- presumably, this is the same machine as $M$, but with reversed notion of which states are considered 'accepting' -- on some input will result in all final states being accepted if and only if the input is in $L^c$. Thus, this does not prove that $\mathrm{NP = co{-}NP}$, and there seems to be no easy way around this.


TITLE: Is there a contradiction between the 't Hooft transformation and a zero radius of convergence?
QUESTION [1 upvotes]: It is generally accepted that the radius of convergence of perturbation series in quantum field theory is zero.
E.g. 't Hooft in "Quantum Field Theory for Elementary Particles":

"The only difficulty is that these expansions will at best be
  asymptotic expansions only; there is no reason to expect a finite
  radius of convergence.”

Or Jackiw in "The Unreasonable Effectiveness of Quantum Field Theory":

"Quantum field theoretic divergences arise in several ways. First of
  all, there is the lack of convergence of the perturbation series,
  which at best is an asymptotic series."

The main argument for this comes from Dyson. (See: Dyson, Divergence of perturbation theory in quantum electrodynamics,
Phys.Rev. 85 (1952) 631–632).
Now, in "Can We Make Sense Out of Quantum Chromodynamics?" t' Hooft
 introduced a transformation (now called t' Hooft transformation, or t' Hooft renormalization scheme) that replaces the coupling constant $g$ with $g_R$ such that $\beta(g_R) =a_1 g_R+ a_2 g_R^2$. In other words, the beta function for the new parameter $g_R$ has only two terms and not infinitely many as the series for $g$.
Isn't this in contradiction with the claim that the radius of convergence is zero? Or is the series for $g_R$ still divergent although it only contains two terms? If yes, where can we observe the divergence after the t' Hooft transformation?

REPLY [2 votes]: Just a small addendum to AFT's excellent answer. " 't Hooft's clever trick" here is just a very old subject in mathematics: the conjugation of dynamical systems to their linear approximation at a fixed point of more generally to some normal form. It is as old as the method of variation of constants in the solution of ODEs. There is a version of the theory for transformations or discrete dynamical systems: $X_{n+1}=F(X_n)$, and also one for flows $\frac{dX}{dt}=F(X)$.
I will only discuss the flow case.
The dependent variable $X$ is assumed to belong to a vector space $V$. Say zero is a fixed point $F(0)=0$ and the function $F$ is defined and nice (analytic) around zero. The particular situation considered by t'Hooft is one-dimensional: $V=\mathbb{R}$, $t=\log \mu$, $X=x=g_D^2$, and
$$
F(x)=-\beta_1 x^2+\beta_2 x^3+\beta_3 x^4+\cdots
$$
To solve the conjugation problem, the first order of business is to construct the new variable $g_R^2=\psi(x)$ as a formal power series in the old one. This is not entirely trivial in the presence of zero eigenvalues and resonances. Typically, one can get rid of non resonant monomials but some monomials cannot be eliminated. I suppose this is the case for $\beta_2 x^3$.
Then one needs to construct $\psi(x)$ as a function of $x$. That depends on the wanted regularity. In the continuous setting the basic result is the Hartman-Grobman Theorem. In the $C^k$ setting, the main result is Sternberg's Theorem. In the analytic setting (having to do the convergence of the formal power series) things are quite dicey because of small divisor problems. Search keywords like: Poincaré-Koenigs Theorem, Poincaré-Dulac normal forms, etc.


TITLE: Does free phonon propagator always vanish at zero momentum?
QUESTION [5 upvotes]: When discussing electron-phonon interaction, the simplest leading electron self-energy usually has only the simplest phonon bubble diagram and excludes the 'tadpole' diagram below with a phonon propagator $D(q=0)$ connected to an electron loop. The argument is $D(q=0)$ should vanish somehow because

the phonon field $\phi(x)$ is a spatial divergence, or

$q=0$ phonon is just an overall translation.

This is what many textbooks say (e.g., page 82 top of Mahan's book (3rd ed.) and page 401 top of Fetter & Walecka's book). But it is not clear to me whether this is only true for acoustic phonons. How about optical phonons? These books seem to only vaguely refer to 'electron-phonon interaction' in terms of this issue.

REPLY [1 votes]: The argument here is physical rather than mathematical: phonons, by their physical nature, are deformation waves, and vector $q$ describes how deformation is distributed within the material. $q=0$ means that the deformation is spatially independent, i.e., that the deformation either does not exist or the material is shifted as a whole.
However, let me point out the following:

Given different possible nature of the deformation (acoustic vs. optical phonons, transverse versus longitudinal polarization) one should be cautious about interpreting $q=0$ as translation. It is better to assume that we are working from the beginning in the center-of-mass reference frame and so deformation is absent.
It is implied that $q$ lies in the first Brilouin zone. If not - as @Qmechanic has correcly pointed out, propagator is periodic in respect to the reciprocal lattice translations.
Same argument with the appropriate adjustments may work for many other types of collective excitations.


TITLE: Discriminant of isogenous elliptic curves
QUESTION [8 upvotes]: Let $E$ be an elliptic curve with a rational $p$-torsion point $P$. Then $E$ is isogenous to the elliptic curve $E' := E/\langle P \rangle$ via the mod $P$ map. I know that the conductor of $E$ and $E'$ are the same, but the discriminant might not be. Is there are relation between $\Delta_{E}$ and $\Delta_{E'}$?

REPLY [4 votes]: Section 2 in the preprint Local invariants of isogenous elliptic curves by T. and V. Dokchitser might be what you are looking for. It states a result of Coates that if $p>3$ prime and $E \to E'$ is a $p$-isogeny (this is weaker than $E$ having a $p$-torsion point) then $\Delta_E^p/{\Delta_{E'}}$ is a $12$th power, and also the new result that if $p=2$ and $p=3$ then this ratio is a third and fourth power, respectively.


text:
\section{Types of sequences}

\subsection{Associating sequences to a type}
We will make some minor adjustments to the definition of the types defined in \cite{bellini&boero&castro&rodrigues&tomita} to be able to work with larger groups. Basically, we define all the types on $\mathbb W$ as in \cite{bellini&boero&castro&rodrigues&tomita}. However, the group $\mathbb W$ used in \cite{bellini&boero&castro&rodrigues&tomita} is not the same group $\mathbb W$ we are using in this article, since there $\mathbb W$ does not have the component $\mathbf U$.

In this section we define the 11 types of sequences related to a a subgroup $G$ of $\mathbb W$ and state the theorem that every sequence is related to one of them.

The name of the sequences which are of one of these 11 types of sequences for $G$ (which will be defined in the following subsections) by $\mathcal H_G$.

The main result we will state in this section is the following, which, in particular, implies that when working with the existence of accumulation points for a sequence, by passing to a subsequence it is enough to guarantee the existence of an accumulation point for the 11 types and the convergence of the sequence of $(\frac{n!}{S} \chi_n:\, n \in \omega)$ to $0$ for each positive integer $S$. The proof is the same as the proof of Theorem 3.1. of \cite{bellini&boero&castro&rodrigues&tomita} and is omitted, although the $\mathbb W$ is different.

\begin{teo}\label{prop_subseq}
Let $f : \omega \to \mathbb W$ is given by $f(n) =n!\chi_n$ for every $n \in \omega$. Let $G$ be a subgroup of $\mathbb W$ containing $(0, \chi_n)$ for every $n \in \omega$. Let $g: \omega \to \mathbb X$ with $[g] \in G^\omega$.

Then there exists $h:\omega\rightarrow \mathbb X$ such that $h\in \mathcal{H}_G$ or $[h]$ is constant and in $G$, $c\in \mathbb X$ with $[c] \in G$, $F \in [\omega]^{< \omega}$, $p_i, q_i \in \mathbb{Z}$ with $q_i \neq 0$ for every $i \in F$, $(j_{i}:i\in F)$ increasing enumerations of subsets of $\omega$  and $j: \omega \to \omega$ strictly increasing such that

\[g \circ j = h + c + \sum_{i \in F} \frac{p_i}{q_i}   f\circ j_{i}\]

with $q_i\leq j_i(n)$ for each $n \in \omega$ and $i \in F$ (which implies $[\frac{1}{q_i} f\circ j_i] \in G^\omega$ since $q_i|((j_i(n))!)$ for each $i \in F$ and $n \in \omega$).\qed
\end{teo}

\subsection{ The types} Each type uses some point of the support and the supports are divided in three groups.

\begin{defin}[The types related to $(R_1 \cup P_1)\setminus \omega$] Let $G$ be a subgroup of $\mathbb W$. We define the first three types of sequence (with respect to $G$) as follows:\label{def_tipos_P1}
Let $g : \omega \to \mathbb X$ be such that $[g(n)]\in G$, for every $n \in \omega$. \\

We say that $g$ is \emph{of type 1} if $\supp g^{1,1}(n) \setminus \cup_{m < n} \supp g^{1,1}(m) \neq \emptyset$, for every $n \in \omega$.\\

We say that $g$ is \emph{of type 2} if $q(g^{1,1}(n)) > n$, for every $n \in \omega$.\\

We say that $g$ is \emph{of type 3} if $\{q(g^{1,1}(n)) : n \in \omega\}$ is bounded and $|p(g^{1,1}(n))| > n$, for every $n \in \omega$.
\end{defin}

\begin{defin}[The types related to $\omega$]
Let $G\subseteq \mathbb W$ be a subgroup such that $\{[\chi_n]:\, n \in \omega\} \subseteq G$. Let $g : \omega \to \mathbb X$ be such that $[g(n)] \in G$, for every $n \in \omega$. Then we define types 4 to 9 (with respect to $G$) as follows: \\

We say that $g$ is \emph{of type 4} if $q(g(n)) > n$, for every $n \in \omega$.\\

We say that $g$ is \emph{of type 5} if there exists $M \in \bigcap_{n \in \omega} \supp g^{1,0}(n)$ such that $\{q(g(n)) : n \in \omega\}$ is bounded and $|p(g(n)(M))| > n$, for every $n \in \omega$.\\

To define types 6, 7 and $8$, suppose $g$ is  such that for each $n \in \omega$, there exists $M_n \in \supp g^{1,0}(n) \setminus \cup_{m < n} \supp g^{1,0}(m)$ such that \[\left( \frac{g(n)(M_n)}{M_n !} : n \in \omega \right)\]

\noindent
is a 1-1 sequence that converges to some $ u \in  (\mathbb R\setminus \mathbb Q)\cup \{-\infty, 0, \infty\}.$
\\

We say that $g$ is of \emph{type 6} if $u=0$.\\

We say that $g$ is of \emph{type 7} if $u \in \mathbb R\setminus \mathbb Q$.\\

We say that $g$ is of \emph{type 8} if $u$ is $\infty$ or $-\infty$. \\

We say that $g$ is \emph{of type 9} if $\left\{ \dfrac{g(n)(M)}{M!} : M \in \supp g^{1,0}(n), n \in \omega \right\}$ is finite and $|\supp g^{1,0}(n)| > n$ for every $n \in \omega$.
\end{defin}

\begin{defin}[The types related to $\vec{P_0} \cup R_0$]
We define types 10 and 11 (with respect to $G$) as follows:
Let $g : \omega \to \mathbb X$ be such that $\supp g(n) \subseteq R_0 \cup (P_0 \times \omega)$ and $[g(n)]\in G$ for each $n\in \omega$ \\

We say that $g$ is \emph{of type 10} if $q(g^0(n)) > n$, for every $n \in \omega$.\\

We say that $g$ is \emph{of type 11} if  the family $\{[g(n)]: n \in \omega\}$ is an independent family whose elements have a fixed order $k$, for some positive integer $k$.
\end{defin}

Notice that if $G$ is a subgroup of $\mathbf X$ then  $\mathcal H_G=G^\omega \cap \mathcal H_{\mathbf X}$. We also set $\mathcal H=\mathcal H_{\mathbf X}$.

The following lemma is easy to verify and left to the reader.

\begin{lem}
Being a sequence of one of the types is absolute for transitive models of ZFC.
\end{lem}


TITLE: How we calculate the value of $2^\pi$ without using calculator
QUESTION [1 upvotes]: How we calculate $2^\pi$? since $\pi$ is irrational how shall I calculate this? and can we write $$(2^\pi)(2^\pi)=(2^\pi)^2$$ and if yes what will be the condition, since $\pi$ is irrational no.

REPLY [1 votes]: Here will make use of calculus approximation using differentials
$$ y= 2^x \\
\ln\ y = x\ln\ 2 \\
\frac{1\ dy}{y\ dx}\ = \ln\ 2 \\
dy = y\ln\ 2\ dx \\
dy = 2^x \ln\ 2\ dx \\
$$
now use a differential approach
$$
\Delta y = 2^x\ln 2\ \Delta x
$$
Now, using  $x=3, \ \Delta x = 0.14159265359$ we see that $\Delta y=78159144782$, that added to the original value in $2^3=8$ is $8.781581$ which is close to the value, although not as close as other methods shown here.