text:
\chapter{Truncation}
\noindent
In this Chapter we formally introduce the notion of truncation and state some well known results that showcase how the notion of a truncation closed subset of a Hahn field is robust in the sense that it is preserved under several extension procedures. In Section 2.2 we introduce the notion of an \textit{infinite part of a valued field} as a first-order counterpart to truncation. It allows us to extend the notion of truncation outside of the Hahn field setting. We also
consider some cases with archimedean value group where this first-order version of truncation necessarily agrees with truncation in an ambient Hahn field $C[[\fM]]$.

\section{Truncation in Hahn Fields}
\noindent
In this section $\fM$ is a monomial group.
For $f=\sum_{\fm}f_{\fm}\fm\in \smallk[[\fM]]$
and $\fn\in \fM$, the \textbf{truncation  $f|_{\fn}$ of $f$ at $\fn$} is defined by
\[ f|_{\fn}\ :=\ \sum_{\fm\succ \fn}f_{\fm}\fm, \quad\text{an element of } \smallk[[\fM].\]
Thus for $f,g\in \smallk[[\fM]]$ we have $(f+g)|_{\fn} = f|_{_\fn} +g|_{\fn}$.
A subset $S$ of $\smallk[[\fM]]$ is said to be \textbf{truncation closed} if for all $f\in S$ and $\fn\in \fM$ we have $f|_{\fn} \in S$. For
$f,g\in \smallk[[\fM]]$ we let $f\trunceq g$ mean that $f$ is a truncation of
$g$, and let $f\truncof g$ mean that $f$ is a proper truncation of $g$,
that is, $f\trunceq g$ and $f\ne g$.

\medskip\noindent
When, as in \cite{D}, Hahn fields are given as $\smallk((t^\Gamma))$,
we adapt our notation accordingly:
the \textbf{truncation of $f =\ser{f}\in \smallk((t^\Gamma))$ at $\gamma_0\in \Gamma$} is then
\[f|_{\gamma_0}\ :=\ \sum_{\gamma<\gamma_0}f_\gamma t^\gamma.\]

\noindent
We say that $g$ is a \textbf{truncation of} $f$ if there is $\gamma \in \Gamma$ such that $g=f|_\gamma$ and write $g \trunceq f$. We say that $g$ is a \textbf{proper truncation of} $f$ if there is $\gamma \in \supp(f)$ such that $g= f|_\gamma$ and we write $g\truncof f$. Note that $g\truncof f$ if and only if $g \trunceq f$ and $g\neq f$.
We have the following.
\begin{enumerate}
	\item $\smallk((t^\Gamma))$ is truncation closed,
    \item $\emptyset$ is truncation closed,
    \item $\smallk$ is truncation closed,
    \item $\{f\in K :\ \supp(f) \text{ is finite}\}$ is truncation closed,
    \item $\smallk[t^\gamma]$ is truncation closed for any $\gamma$,
    \item $\smallk[t^{\gamma_0} +t^{\gamma}]$ is not truncation closed, for all distinct $\gamma_0, \gamma \in \Gamma^{\neq}$.
\end{enumerate}

We only justify (6), as (1)--(5) are immediate. Assume $\gamma_0, \gamma\in \Gamma^{\ne}$ and $\gamma_0<\gamma$.  To show that $\smallk[t^{\gamma_0}+t^\gamma]$ is not truncation closed, it is enough
to show that for all  $P(X)\in \smallk[X]$ we have  $P(t^{\gamma_0} + t^\gamma)\neq t^{\gamma_0}$. If $P\in \smallk$, then clearly $P(t^{\gamma_0} + t^{\gamma})=P\ne t^{\gamma_0}$. It is also clear that if $\deg(P)=1$, then $P(t^{\gamma_0}+t^\gamma)\ne t^{\gamma_0}$.  If $\deg (P)=n>0$ and $\gamma>0$, then the highest exponent of $t$ in $P(t^{\gamma_0}+t^\gamma)$ is $n\gamma\ne \gamma_0$ and thus $P(t^{\gamma_0}+t^\gamma)\neq t^{\gamma_0}$. If $\deg(P)=n>1$ and $\gamma<0$, then the lowest exponent of $t$ in $P(t^{\gamma_0}+t^\gamma)$ is $n\gamma_0\ne \gamma_0$ and thus
$P(t^{\gamma_0}+t^\gamma)\ne t^{\gamma_0}$.

\medskip\noindent
Here is an easy but useful consequence of the definition:

\begin{lemma} If $V$ is a truncation closed $\smallk$-linear subspace of $\smallk[[\fM]]$, then $\supp(V)=V\cap\fM$.
\end{lemma}

\noindent
We now list some items from \cite{D} that we are going to use:

\begin{proposition} \label{D} Let $A$ be a subset of $\smallk[[\fM]]$, $R$ a subring
of $\smallk[[\fM]]$, and $E$ a $($valued$)$ subfield of $\smallk[[\fM]]$. Then:
\begin{itemize}
	\item[(i)] The ring as well as the field generated in $\smallk[[\fM]]$ by any truncation closed
subset of $\smallk[[\fM]]$ is truncation closed: \cite[Theorem 1.1]{D};
    \item[(ii)] If $R$ is truncation closed and all truncations of all $a\in A$ lie in $R[A]$, then the ring $R[A]$ is truncation closed;
    \item[(iii)] If $E$ is truncation closed and $E\supseteq \smallk$,
then the henselization $E^{\h}$ of $E$ in
$\smallk[[\fM]]$ is truncation closed: \cite[Theorem 1.2]{D};
    \item[(iv)] If $E$ is truncation closed, henselian, and $\operatorname{char}(\smallk)=0$, then any algebraic field extension of $E$ in $\smallk[[\fM]]$ is
truncation closed: \cite[Theorem 5.1]{D}.
\end{itemize}
\end{proposition}

\noindent
Item (ii) here is a variant of \cite[Corollary 2.6]{D}.
To prove (ii), assume $R$ is truncation closed and all truncations of
all $a\in A$ lie in $R[A]$. Let
\[B\ :=\  \{f\in \smallk[[\fM]]:\ f\trunceq a \text{ for some }a\in A\}.\]
Then $B$ is truncation closed, and so is $R\cup B$, and thus the ring
$R[B]=R[A]$ is truncation closed, by (i) of the proposition above.
An often used consequence of (i) and (ii) is that if $E$ is a truncation closed subfield of $\smallk[[\fM]]$ and $A\subseteq \fM$, then the subfield $E(A)$ of $\smallk[[\fM]]$ is also truncation closed.

Elaborating on (iii) and (iv) above, suppose $\smallk$ has characteristic $0$ and $E\supseteq \smallk$ is a truncation closed subfield of $\smallk[[\fM]]$. Then
\[\fM_E\ :=\  E\cap \fM\ =\ \supp(E)\cap \fM\]
is a subgroup of $\fM$ with $E\subseteq \smallk[[\fM_E]]$. Since $\smallk[[\fM_E]]$ is henselian, we have $E^{\h}\subseteq \smallk[[\fM_E]]$. We call the subfield
\[F\ :=\ \{f\in \smallk[[\fM]]:\ f \text{ is algebraic over }E\}\]
of $\smallk[[\fM]]$  the {\em algebraic closure of $E$ in $\smallk[[\fM]]$} and we can characterize it as follows:

\begin{lemma}\label{actr} Let $\fM^{\div}_{E}:=\{\fg\in \fM:\ \fg^n\in \fM_E \text{ for some }n\ge 1\}$, the divisible closure of $\fM_E$ in $\fM$. Then $\fM^{\div}_{E}$ is a subgroup of
$\fM$, $E(\fM^{\div}_{E})$ and $F$ are truncation closed subfields of $\smallk[[\fM]]$, and
\[F\ =\ E(\fM^{\div}_{E})^{\h}.\]
\end{lemma}
\begin{proof} It follows from (i) and (ii) of Proposition~\ref{D} that $E(\fM^{\div}_E)$ is truncation closed. Also,
$E(\fM^{\div}_E)\subseteq F$, hence $F$ is an algebraic extension of the henselization $E(\fM^{\div}_{E})^{\h}$, and thus $F$ is truncation closed by (iii) and (iv) of Proposition~\ref{D}. Therefore,
$\fM_F=\fM_E^{\div}$, and so $F$ is an immediate extension of $E(\fM^{\div}_E)^{\h}$. The latter is algebraically maximal, and
thus equals $F$.
\end{proof}

\subsection*{Transcendental extensions}
The above deals with extension procedures of algebraic nature.
As in \cite{D}, we also consider possibly transcendental adjunctions of
of the following kind.
Let $\mathcal{F}= (\mathcal{F}_n)$ be a family such that for each $n$: $\mathcal{F}_n\subseteq \smallk[[X_1,\ldots,X_n]]$, and for all $F\in \mathcal{F}_n$ we have $\partial F/\partial X_i \in \mathcal{F}_n$ for $i=1,\ldots,n$.
Let $K$ be a subfield of $\smallk[[\fM]]$.
We define the $\mathcal{F}$-\textit{extension}  $K(\mathcal{F},\prec 1)$
of $K$ to be
the smallest subfield of $\smallk[[\fM]]$ that contains $K$ and the set
\[  \bigcup_n \{F(f_1,\ldots,f_n):\ F\in \mathcal{F}_n,\
f_1,\dots, f_n\in K^{\prec 1}\} .\]

\begin{lemma}\label{tL16}
Suppose  $\text{char}(\smallk)=0$ and $K$ is a truncation closed subfield of $\smallk[[\fM]]$ containing $\smallk$. Then $K(\mathcal{F},\prec 1)$ is truncation closed.
\end{lemma}

\begin{proof}
Let $E$ be the largest truncation closed subfield of $K(\mathcal{F},\prec 1)$.
Towards a contradiction, assume that $E\ne K(\mathcal{F},\prec 1)$.
Let $n$ be minimal such that $F(f_1,\ldots,f_n)\notin E$ for some  $F\in \mathcal{F}_n$ and  $f_1,\ldots, f_n \in K^{\prec 1}$.
Take $f_1,\ldots,f_n\in K^{\prec 1}$ such that $F(f_1,\ldots,f_n)\notin E$ for some $F\in \mathcal{F}_n$ with lexicographically minimal $(o(f_1),\ldots, o(f_n))$.
Taking such $F$ (so $f = F(f_1,\ldots, f_n)\notin E$),
it suffices to show that then all proper truncations of $f$ lie in $E$.
If \[o(f_1)\ =\ \ldots\ =\ o(f_n) = 0\] then $f_1=\dots=f_n=0$, so
$F(f_1,\ldots, f_n)\in \smallk$.

Next assume that $f_i\neq 0$ for some $i\in \{1,\ldots,n\}$ and
 $\phi$ is a proper truncation of $f$. Every monomial in $\supp f$ is
$\succeq \fm^k$ for some $\fm\in \bigcup_{i=1}^n \supp f_i$ and some
$k\in \N$, so we can take $i\in \{1,\ldots,n\}$, $\fm\in \supp(f_i)$, and $N=N(\fm)\in \mathbb{N}$ such that $\supp(\phi)\succ \fm^k$ for all $k>N$.
Take $h,g \in K$ such that $f_i=h+g$, $ h =f_i|_{\fm}$ and $\fd(g) = \fm$.
By Taylor expansion,
\[F(f_1,\ldots,f_n)\ =\ \sum_{k=0}^\infty \frac{\partial^k F}{\partial X_i^k}(f_1,\ldots,f_{i-1},h,f_{i+1},\ldots, f_n) \frac{g^k}{k!}.\]
Thus $\phi$ is a truncation of
\[\sum_{k=0}^{N} \frac{\partial^k F}{\partial X_i^k}(f_1,\ldots,f_{i-1},h,f_{i+1},\ldots, f_n) \frac{g^k}{k!}.\]
Given that $(o(f_1),\ldots,o(f_n))$ is minimal and $o(h) < o(f_i)$, we get $\phi \in E$.
\end{proof}

\noindent
For example, if $\text{char}(\smallk)=0$, we could take
\[\cF_1=\{(1+X_1)^{-1},\ \exp X_1,\ \log(1+X_1)\},\qquad  \cF_n=\emptyset\  \text{ for }n>1\]
where $\exp X_1:=\sum_{i=0}^\infty X_1^i/i!\ $  and $\log(1+X_1):=\sum_{i=1}^{\infty}(-1)^{i+1}X_1^i/i$.

A subfield $E$ of $\smallk[[\fM]]$ is said to be $\cF$-closed if
$f(\vec a)\in E$ for all
$f(X_1,\dots, X_n)\in \cF_n$ and $\vec a\in (E\cap\ \smallo )^{\times n}$, $n=1,2,\dots$. The $\cF$-closure of a subfield
$E$ of $\smallk[[\fM]]$ is the smallest $\cF$-closed subfield
$\cF(E)$ of $\smallk[[\fM]]$ that contains $E$. The next result is \cite[Theorem 1.3]{D}. The more informative Lemma~\ref{tL16} above is not stated explicitly in \cite{D}.

\begin{corollary}\label{ta} If $\operatorname{char}(\smallk)=0$ and
$E\supseteq \smallk$ is a truncation closed subfield of $\smallk[[\fM]]$,
then its $\cF$-closure $\cF(E)$ is also truncation closed.
\end{corollary}

\noindent
Assuming $\operatorname{char}(\smallk)=0$ we now consider the meaning of the above for the case where $\cF_1=\{\exp X_1\}$ and $\cF_n=\emptyset$ for $n>1$.
(This case is needed in Chapters 7 and 8.)
Let $E$ be a subfield of $k[[\fM]]$. We say that $E$ is {\bf closed under small exponentiation\/} if $\exp(E^{\prec 1})\subseteq E$.
We define the {\bf small exponential closure of $E$\/} to be the smallest subfield $F\supseteq E$ of $\smallk[[\fM]]$ that is closed under small exponentiation;
we denote this $F$ by
$E^{\smallexp}$. Thus by Corollary~\ref{ta}, if $E\supseteq \smallk$ is truncation closed, then $E^{\smallexp}\subseteq\smallk[[\fM_E]]$ is truncation closed.

\subsection*{Additional facts on truncation}
Besides $\Gamma$ we now consider a second ordered abelian
group $\Delta$. Below we identify $\Gamma$ and $\Delta$
in the usual way with subgroups of
the lexicographically ordered sum $\Gamma\oplus \Delta$, so that
$\Gamma+\Delta=\Gamma\oplus \Delta$, with $\gamma>\Delta$ for all $\gamma\in \Gamma^{>}$. This makes $\Delta$ a convex subgroup of $\Gamma+\Delta$.
Let $\smallk_0$ be a field and $\smallk = \smallk_0((t^\Delta))$.
Then we have a field isomorphism
\[\smallk((t^\Gamma))\longrightarrow \smallk_0((t^{\Gamma+ \Delta}))\]
that is the identity on $\smallk$, namely
\[f=\sum_{\gamma}f_{\gamma}t^\gamma\mapsto
\sum_{\gamma,\delta} f_{\gamma,\delta}t^{\gamma+\delta}\]
where $f_{\gamma}=\sum_\delta f_{\gamma,\delta}t^{\delta}$ for all $\gamma$.
Below we identify $\smallk((t^\Gamma))$ with  $\smallk_0((t^{\Gamma+ \Delta}))$
via the above isomorphism.
 For a set $S\subseteq \smallk((t^\Gamma))$
this leads to two notions of truncation:
we say that $S$ is
$\smallk$-truncation closed if it is truncation closed with
$\smallk$ viewed as the coefficient
field and $\Gamma$ as the group of exponents (that is, viewing $S$ as a subset of the Hahn field $\smallk((t^\Gamma))$ over $\smallk$), and we say that
$S$ is $\smallk_0$-truncation closed if it is truncation closed  with
$\smallk_0$ viewed as the coefficient
field and $\Gamma+\Delta$ as the group of exponents (that is, viewing
$S$ as a subset of the Hahn field $\smallk_0((t^{\Gamma+\Delta}))$
over $\smallk_0$).

\begin{lemma}\label{trtr1} If $S\subseteq\smallk((t^\Gamma))$ is $\smallk_0$-truncation closed, then $S$ is $\smallk$-truncation closed.
\end{lemma}
\begin{proof} Let $f\in \smallk((t^\Gamma))$ and $\gamma\in \Gamma$. If
$ (\gamma+\Delta^{<})\cap \supp_{\smallk_0}f=\emptyset$, then the
$\smallk$-truncation of $f$ at $\gamma$ equals the $\smallk_0$-truncation
of $f$ at
$\gamma\in \Gamma +\Delta$.

If $(\gamma+\Delta^{<})\cap \supp_{\smallk_0}f\ne
\emptyset$, then the $\smallk$-truncation of $f$ at $\gamma$ equals the $\smallk_0$-truncation of $f$ at the least element of $(\gamma+\Delta^{<})\cap \supp_{\smallk_0}f$.
\end{proof}

\begin{lemma}\label{trtr2} Let $\smallk_1$ be a truncation closed subfield of the Hahn field
$\smallk_0((t^\Delta))= \smallk$ over $\smallk_0$, and let $V$ be a
$\smallk_1$-linear subspace of $\smallk_1((t^\Gamma))\subseteq \smallk((t^\Gamma))$
such that $V\supseteq \smallk_1$ and $V$ is $\smallk$-truncation closed.
Then $V$ is $\smallk_0$-truncation closed.
\end{lemma}

\begin{proof} If $\Gamma=\{0\}$, then $\smallk_1((t^\Gamma))=\smallk_1$, so
$V=\smallk_1$  is $\smallk_0$-truncation closed. In the rest of the proof
we assume $\Gamma\ne\{0\}$. Let $\beta=\gamma+\delta\in \Gamma+\Delta$ with
$\gamma\in \Gamma$ and $\delta\in \Delta$, and $f\in V$.
Let $g$ be the truncation of $f$ at $\beta$ in the Hahn field
$\smallk_0((t^{\Gamma+\Delta}))$ over $\smallk_0$ and $h$ the truncation of $f$ at $\gamma$ in the Hahn field $\smallk((t^\Gamma))$. Then $h\in V$
and $g=h+s$ with $s=\phi t^\gamma$ and $\phi\in \smallk_0((t^\Delta))$.
If $s=0$, then $g\in V$ trivially, so assume $s\ne 0$.
Then $f$ has a $\smallk$-truncation $h+\theta t^\gamma\in V$ with
$0\ne\theta\in \smallk_1$ and so $\theta t^\gamma\in V$, $t^\gamma=\theta^{-1}(\theta t^\gamma)\in V$. Moreover,
$\phi$ is a $\smallk_0$-truncation of $\theta$, so $\phi\in \smallk_1$, hence
$s=\phi t^\gamma\in V$, and thus $g\in V$.
\end{proof}

\section{Infinite Parts}
\noindent
For the rest of this section $K$ is a valued field with monomial group $\fM$ and coefficient field $C$; the latter means that
$C$ is a subfield of $K$ that is contained in the valuation ring
of $K$ and maps (necessarily isomorphically) onto the residue field under the residue map. We let $v:K^\times\rightarrow \Gamma$ be the valuation of $K$, $\smallk$ its residue field, $\mathcal{O}$ the valuation ring of $K$, and $\smallo$ the maximal ideal of $\bigO$.
Note that $\bigO=C\oplus \smallo$.

Suppose $\cU$ is a $C$-vector subspace of $K$ such that $K=\cU\oplus \bigO$, $\cU$ is closed under multiplication in $K$, and $\cU\supseteq \fM^{\succ 1}$.
We call such $\cU$ an \textit{infinite part of} $K$.

For $a\in K$ we define its $\cU$-\textit{part} to be
the unique $u\in \cU$ such that $a= u+b$ for some $b \in \bigO$, and we denote the
$\cU$-part of $a$ by $a|_{\cU}$.

\begin{example} The Hahn field $C[[\fM]]$ with coefficient field $C$ and
monomial group $\fM$, and the usual valuation given by its valuation ring
\[C[[\fM^{\preceq 1}]]\ :=\ \{f\in C[[\fM]]:\ \supp f \subseteq \fM^{\preceq 1}\},\] has infinite part
\[C[[\fM^{\succ 1}]]\ :=\ \{f\in C[[\fM]]:\ \supp f \subseteq \fM^{\succ 1}\}.\]
We call this the \textbf{canonical infinite part} of the Hahn field $C[[\fM]]$.

More generally, suppose $K$ is a truncation closed subfield of $C[[\fM]]$ with $C\subseteq K$ and $\fM\subseteq K$.
We consider $K$ as a valued subfield of this Hahn field,
and equip $K$ with its monomial group $\fM$ and its coefficient field $C$. Then $K$ has infinite part $K\cap C[[\fM^{\succ 1}]]$.
\end{example}

\subsection*{Truncation with respect to an infinite part}
For any monomial $\fm\in \fM$ we define the \textit{truncation of} $a$ \textit{at} $\fm$ \textit{with respect to} $\cU$, as $a|^\cU_{\fm}:=\fm\cdot(\fm\inv a)|_{\cU}$.
Note that for $\fm = 1$ we get $a|^{\cU}_{1} = a|_{\cU}$.
Note also that for the canonical infinite part
$\cU$ of the Hahn field $C[[\fM]]$ we have $a|^{\cU}_{\fm}=a|_{\fm}$.

The following rules are easy consequences of the definition above:

\begin{lemma} \label{Vtruncprop}
Let $a,b\in K$,  $c\in C$, and $\fm, \fn\in \fM$. Then:
\begin{enumerate}
	\item[\rm{(i)}] $(a+b)|_{\fm}^\cU =a|_{\fm}^\cU + b|_{\fm}^\cU$ and
	$(ca)|_{\fm}^{\cU}= c\cdot a|_{\fm}^{\cU}$;
    \item[\rm{(ii)}] $a\preceq \fm$ $\iff$ $a|_{\fm}^\cU = 0$;
    \item[\rm{(iii)}] $a \succ \fm $ $\iff$ $a|_{\fm}^\cU \sim a$;
    \item[\rm{(iv)}] $a|_{\fm}^{\cU}\preceq a$, by (2) and (3);
    \item[\rm{(v)}] $a- a|_{\fm}^\cU\preceq \fm $;
    \item[\rm{(vi)}] $a|_{\fm}^\cU = b|_{\fn}^\cU=0 \Rightarrow (ab)|_{\fm\fn}^\cU=0$;
    \item[\rm{(vii)}] $a|_{\fm}^\cU\neq 0 \Rightarrow a|_{\fm}^\cU\succ \fm$;
    \item[\rm{(viii)}] $\fm\cdot a|_{\fn}^\cU = (\fm a)|_{\fm\fn}^\cU$;
    \item[\rm{(ix)}] $\fm|^{\cU}_{\fn}=\fm$ if $\fm\succ \fn$, and $\fm|^{\cU}_{\fn}=0$ if $\fm\preceq \fn$.
\end{enumerate}
\end{lemma}

\noindent
The following lemmas give a commutativity property for truncation with respect to $\cU$.

\begin{lemma}\label{trunccom1}
Let $a\in K$ and $\fm\in \fM$. Then
\[  (a|_{\cU})|_{\fm}^\cU\ =\ (a|_{\fm}^{\cU})|_{\cU}\ =\
a|^{\cU}_{\max(1,\fm)}.\]
\end{lemma}

\begin{proof}
Let $a= u+ b$ with $u\in \cU$, $b\in \bigO$. Consider first the case $\fm\preceq 1$. Then

\begin{align*}
  (a|_{\cU})|_{\fm}^\cU\ &=\  u|_{\fM}^\cU \\
							&=\ \fm\cdot (\fm\inv u)|_{\cU} \\
                            &=\ \fm\cdot(\fm\inv u)\  =\ u.
\end{align*}
On the other hand,
\begin{align*}
	   a|_{\fm}^\cU\ &=\   u|_{\fm}^\cU+ b|_{\fm}^\cU\\
    				  &=\ u+ b|_{\fm}^\cU.
\end{align*}
Now $b|_{\fm}^{\cU}\preceq b$ by (iv) of Lemma~\ref{Vtruncprop}, so
$b|_{\fm}^\cU\in \bigO$, and thus
 $(a|_{\cU})|_{\fm}^\cU = u = (a|^{\cU}_{\fm})|_{\cU}$. Since $u=a|^{\cU}_{1}$, this proves the lemma for $\fm\preceq1$.
Next, assume $\fm \succ 1$. Then by (i) and (ii) of Lemma~\ref{Vtruncprop},

\begin{align*}
a|_{\fm}^\cU\ &=\   \ u|_{\fm}^\cU+\ b|_{\fm}^\cU\\
					&=\ \ u|_{\fm}^\cU\\
                    &=\  (a|_{\cU})|_{\fm}^\cU.
\end{align*}
On the other hand, $ (a|_{\fm}^{\cU})|_{\cU} = \big(\fm\cdot (\fm\inv a)|_{\cU}\big)|_{\cU} = \fm\cdot (\fm\inv a)|_\cU=a|_{\fm}^{\cU}$, since $\cU$ is closed under products.
\end{proof}

\noindent
For the proof of Lemma \ref{trunccom}, note that for $\fm, \fn \in \fM$ and $a\in K$ we have $(\fm a)|_{\fn}^\cU = \fm\cdot  a|_{\fn/ \fm}^\cU$.

\begin{lemma}\label{trunccom}
Let $a\in K$ and $\fm\preceq \fn$ in $\fM$. Then
\[  (a|_{\fm}^{\cU})|_{\fn}^{\cU}\ =\  (a|_{\fn}^{\cU})|_{\fm}^{\cU}\  =\   a|_{\fn}^\cU.\]
\end{lemma}
\begin{proof}
	To simplify notation we omit in this proof superscripts $\cU$ in expressions like  $a|^{\cU}_{\fm}$.
    Using Lemma \ref{trunccom1},
    \[\trunc{(\trunc{a}{\fm})}{\fn}\ =\ \trunc{\big(\fm\cdot(\trunc{\fm\inv a }{\cU})\big)}{\fn}\ =\ \fm\cdot\trunc{\big(\trunc{(\fm\inv a )}{\cU} \big)}{\fn/\fm}\]
    \[\ \ \ \ \ \ \ \  =\ \fm\cdot (\fm^{-1}a)|_{\fn/\fm}\ =\ (\fm\fm\inv a)|_{\fn}\ =\ a|_{\fn}.
    \]
    In a similar way we get $\trunc{(\trunc{a}{\fn})}{\fm} = \trunc{a}{\fn}$.
\end{proof}

\subsection*{$\cU$-support} For $a\in K$ we define the $\cU$-\textit{support of} $a$  as
\[\supp_\cU(a)\ =\ \{\fm\in \fM: a - \trunc{a}{\fm}^\cU\asymp \fm  \}.\]
When $K$ is a Hahn field $C[[\fM]]$ and $\cU$ is its canonical infinite part, this agrees with the
usual support $\supp a$.  Most of the following rules are easy to verify:

\begin{lemma}\label{lemsuppU}
Let $a,b\in K$ and $\fm\in \fM$. Then:
\begin{enumerate}
	\item[\rm{(i)}] $\fm \cdot \supp_\cU(a) = \supp_\cU(\fm a)$,
	\item[\rm{(ii)}] $\supp_\cU(a + b)\subseteq \supp_\cU(a)\cup\supp_\cU(b)$ and $\supp_{\cU}ca=\supp_{\cU}a$ for $c\in C^\times$,
    \item[\rm{(iii)}] $\supp_\cU(a)\cap \supp_\cU(b) = \emptyset$ $\Rightarrow$ $\supp_\cU(a + b)= \supp_\cU(a)\cup\supp_\cU(b)$,
    \item[\rm{(iv)}] $a\asymp \fm \Rightarrow \fm \in \supp_\cU(a)$,
    \item[\rm{(v)}] $ \supp_\cU(a)\preceq a$,
    \item[\rm{(vi)}] $\supp_{\cU}( \fm)=\{\fm\}$,
    \item[\rm{(vii)}] $\supp_{\cU}(a|^{\cU}_{\fm})=\supp_{\cU}(a)^{\succ \fm}$ and $\supp_\cU(a-\trunc{a}{\fm}^\cU)= \supp_\cU(a)^{\preceq \fm}$,
   \item[\rm{(viii)}] $\supp_\cU(a)\succ \fm\ \Leftrightarrow\ a=\trunc{a}{\fm}^\cU\ \Leftrightarrow\ a\in \fm\cU$.
\end{enumerate}
\end{lemma}
\begin{proof} We only do (vii), since (i)--(vi) are  easy to verify, and (viii) is an easy consequence of (vii), using also (viii) of Lemma~\ref{Vtruncprop} for the second equivalence. We have $a = \trunc{a}{\fm}^\cU + c\fm + b$ for unique $c\in C$ and $b\prec \fm$, by (v) in Lemma \ref{Vtruncprop}. Hence by (ii), (v), and (vi),
$\supp_{\cU}(a)^{\succ \fm}\subseteq  \supp_{\cU}(a|^{\cU}_{\fm})$.
Moreover,
$\supp_{\cU}(a|^{\cU}_{\fm})\succ \fm$ by Lemma~\ref{trunccom},
and then applying (ii), (v), (vi) to $a|^{\cU}_{\fm}=-a-c\fm-b$ yields
$\supp_{\cU}(a|^{\cU}_{\fm})\subseteq \supp_{\cU}(a)$,
which gives the first equality of (vii). As to the second equality,
(ii), (v), (vi) and the first equality  give
\[\supp_{\cU}(a)^{\preceq \fm}\ \subseteq\ \supp_{\cU} (c\fm+b)\ =\ \supp_\cU(a-\trunc{a}{\fm}^\cU)\ \preceq\ \fm,\]
so by (ii) and the first equality,
$\supp_{\cU}(a-a|^{\cU}_{\fm})\subseteq \supp_{\cU}(a)^{\preceq \fm}$.
\end{proof}

\iffalse
\begin{proof}
(1) Let $\fm$ be such that $a+b - \trunc{a}{\fm} - \trunc{b}{\fm} \asymp \fm$ then
\end{proof}
\fi

\subsection*{Good Infinite Parts} As before, $\cU$ is an infinite part of $K$.  We say that $\cU$ is \textbf{good} if for all $a,b \in K$ we have $\supp_\cU(ab)\subseteq \supp_\cU(a)\supp_\cU(b)$. In view of the definability of truncation with respect to $\cU$ (in the valued field $K$ equipped with $\fM$, $C$, $\cU$) and the first-order nature of the $\cU$-support, being good is also of first-order nature. We do not know whether being good is a consequence of the definition of ``infinite part''.
Note that the canonical infinite part of a Hahn field $C[[\fM]]$ is good, and so is the infinite part $K\cap C[[\fM^{\succ 1}]]$ of $K$ when $K\supseteq C(\fM)$ is a truncation closed subfield of $C[[\fM]]$.

\medskip\noindent
Assume $\cU$ is good and $K$ is  a truncation closed valued subfield of $C[[\fM]]$, with the same $C$ and $\fM$ as for $K$,  so $C(\fM)\subseteq K$.  Here ``truncation closed'' is of course with respect to truncation in the Hahn field $C[[\fM]]$.

Thus for $a\in K$ and $\fm\in \fM$ we have in principle two truncations, namely $a|_\fm$ as an element of $C[[\fM]]$ and the truncation $a|^\cU_{\fm}$ with respect to
$\cU$. Set
\[E\ :=\  \{a\in K:\ \trunc{a}{\fm} = \trunc{a}{\fm}^\cU \text{ for all } \fm\in \fM\}.\]
Thus $\supp(a)=\supp_{\cU}(a)$ for all $a\in E$.

\begin{lemma}
$E$ is a truncation closed subring of $K$ with $C[\fM]\subseteq E$.
\end{lemma}
\begin{proof}
Clearly, $E$ is truncation closed in $C[[\fM]]$, and $E$ is a $C$-linear subspace of $K$ that contains $C$ and $\fM$.  Let $a,b\in E$; it remains to show that then $ab\in E$. We do so by induction on $(o(a),o(b))$ with respect to the lexicographic ordering. If either $o(a)=0$ or $o(b)=0$, then $a=0$ or $b=0$, in which case we are done. Let $o(a)>0$ and $o(b)>0$, and $\fd\in \fM$. We want to show $\trunc{ab}{\fd} = \trunc{ab}{\fd}^\cU$. Assume that  $\fm\in \supp(a)$ and $\fn \in \supp(b)$ are such that $\fm\fn \preceq \fd$. Then
\[a = a_1 + a_2 \text{ with } a_1 := \trunc{a}{\fm}\  \text{ and }\   b= b_1+b_2 \text{ with } b_1 = \trunc{b}{\fn},\]
and $ab= a_1b + a_2b_1 + a_2b_2$. Hence
\[\trunc{(ab)}{\fd}\ =\ \trunc{(a_1b)}{\fd} + \trunc{(a_2b_1)}{\fd}\  \text{ and }\  \trunc{(ab)}{\fd}^\cU\ =\ \trunc{(a_1b)}{\fd}^\cU + \trunc{(a_2b_1)}{\fd}^\cU.\]
By induction we have $ \trunc{(a_1b)}{\fd}=\trunc{(a_1b)}{\fd}^\cU$ and $\trunc{(a_2b_1)}{\fd} = \trunc{(a_2b_1)}{\fd}^\cU$, so $\trunc{ab}{\fd} = \trunc{ab}{\fd}^\cU$.
If for all monomials $\fm \in \supp(a)$ and $\fn \in \supp(b)$ we have $\fm\fn\succ \fd$, then $\trunc{ab}{\fd} = ab = \trunc{ab}{\fd}^\cU$ where the last equality uses (8) of Lemma~\ref{lemsuppU}
and the assumption that $\cU$ is good.
\end{proof}

\noindent
This seems as far as we can go under first-order assumptions on
$\cU$. In the next subsection we indicate some cases with archimedean $\fM$ where $a|^{\cU}_{\fm}=a|_{\fm}$ for all $a\in K$ and $\fm\in \fM$.

\subsection*{The Archimedean Case}
In this subsection we assume that $\fM$ is archimedean and $K$ is a valued subfield of the Hahn field $C[[\fM]]$ with the same $C$ and $\fM$ as for $K$.  In the next lemma we do not  assume that $\cU$ is good, but derive it from the lemma. Let $\fm$ range over $\fM$.

\begin{lemma}\label{archCM} Suppose $K=C(\fM)$. Then $\cU=\bigoplus_{\fm\succ 1} C\fm$.
\end{lemma}
\begin{proof} The key point is the easily verified observation that
\[F\ :=\ \{a\in C[[\fM]]:\ \supp(a)^{\succ \fm} \text{ is finite for all } \fm\}\]
is a subfield of $C[[\fM]]$ containing $K$.
\end{proof}

\noindent It is worth saying a bit more on this case. Since
$C\cup \fM$ is truncation closed, $K=C(\fM)$ is truncation closed in $C[[\fM]]$. By Lemma~\ref{archCM} there is only one infinite part of $K$, in particular, $\cU=K\cap C[[\fM^{\succ 1}]]$ and so $\cU$ is good and
$a|^{\cU}_{\fm}=a|_{\fm}$ for all $a\in K$ and all $\fm$. We also
note that the field $F$ defined in the proof of that lemma is the closure of $C[\fM]$  (and of $C(\fM)$) in $C[[\fM]]$ with respect to the valuation topology. Thus
$F$ is henselian as a valued subfield of $C[[\fM]]$, by \cite[Corollary 3.3.5]{ADH}.  Note also that $F$ is truncation closed in $C[[\fM]]$ and has, just like $K$, only one infinite part, which is the same as
that of $K$, namely $\bigoplus_{\fm\succ 1} C\fm$.

\medskip\noindent
In the remainder of the section we assume that $\cU$ is good. Let $E$ be the subring of $K$ defined in the previous subsection. We can say more about $E$ in the archimedean case that we are considering in this subsection:

\begin{lemma}
$E$ is a subfield of $K$.
\end{lemma}
\begin{proof}
Let $f\in E$ with $f\neq 0$, so $f= c\fm(1-\epsilon)$ with $c\in C^\times$, $\fm \in \fM$ and $\epsilon \in \smallo$. Then $\epsilon\in E$ and $f\inv = c\inv \fm\inv (1-\epsilon)\inv$. Hence $f\inv \in E$ if and only if $(1-\epsilon)\inv\in E$. Let $\fd \in \fM$ be given. Since $\fM$ is archimedean, we have $n$ such that $\epsilon^n \preceq \fd$, so $(1-\epsilon)\inv = 1 + \epsilon + \cdots + \epsilon^nb$ with $b\preceq 1$. Then
\[\trunc{(1-\epsilon)\inv}{\fd}\ =\ \trunc{(1+\epsilon + \cdots + \epsilon^{n-1})}{\fd}\ =\ \trunc{(1+\epsilon + \cdots + \epsilon^{n-1})}{\fd}^\cU\ =\ \trunc{(1-\epsilon)\inv}{\fd}^\cU. \]
Thus $(1-\epsilon)\inv \in E$.
\end{proof}

\begin{lemma}
Suppose $f\in C[[t_1,\ldots,t_n]]$ and $f(a) \in K$ for all $a\in  \smallo_K^n$. Then $f(a)\in E$ for all $a\in \smallo_E^n$.
\end{lemma}
\begin{proof}
Let $\epsilon= (\epsilon_1,\ldots,\epsilon_n)\in \smallo^n_E$. Let $\fd \in \fM$ be given. Take $m$ such that $\epsilon_i^m \preceq \fd$ for $i=1,\ldots,n$. We have $f= g + t_1^mh_1 + \cdots +t_n^m h_m$ where $g\in C[t_1,\dots, t_n]$ is of degree $<m$ in $t_i$ for every $i$, and $h_i \in C[[t_1,\ldots t_n]]$ for $i=1,\dots,n$.  Then $f(\epsilon) = g(\epsilon) + b$ with $b\preceq \fd$, so $\trunc{b}{\fd} = \trunc{b}{\fd}^\cU=0$, and thus $\trunc{f(\epsilon)}{\fd}= \trunc{f(\epsilon)}{\fd}^\cU$.
\end{proof}

\begin{corollary}
If $K$ is henselian, then $E$ is a henselian valued subfield of $K$.
\end{corollary}
\begin{proof}
Assume $K$ is henselian, and let $P(X) = 1+X + a_1X^2 + \cdots + a_n X^{n+1} \in E[X]$ with $a_i \in \smallo_E$ for $i=1,\ldots,n$.  Then $P$ has a unique zero $y\preceq 1$ in $K$. It suffices to show that $y\in E$. By the implicit function theorem we have $y = f(a_1,\ldots,a_n)$ for some power series $f\in C[[t_1,\ldots,t_n]]$. Using the previous Lemma, we conclude that $y\in E$.
\end{proof}

\begin{corollary}
Suppose $C$ has characteristic 0 and $K$ is henselian. Then $E$ is algebraically closed in $K$.
\end{corollary}
\begin{proof}
Since $E$ is henselian and of equicharacteristic zero, it is algebraically maximal by \cite[Corollary 3.3.21]{ADH}.  It remains to note that $C[[\fM]]$ is an immediate extension
of $E$ in view of $E\supseteq C(\fM)$.

\end{proof}


TITLE: Why is the set $\{x \mid f(x) \not= g(x)\}$ measurable in a topological hausdorff space?
QUESTION [4 upvotes]: assuming i have a measure space $(X, \mathscr{M}_X, \mu)$ and a measurable space $(Y,\mathscr{M}_Y)$, two measurable mappings $f,g:X \to Y$ and it holds that $Y$ is a topological hausdorff space with a countable base. (I think i am supposed to assume that $\mathscr{M}_Y$ is the Borel-$\sigma$-Algebra generated by the open/closed subsets of Y)
Now i am wonderung: Why exactly is the set $\{x \in X \mid f(x) \not= g(x)\}$ measurable? in other words why is $$\{x \in X \mid f(x) \not= g(x)\} \in \mathscr{M}_X$$
I know that the diagonal $\Delta Y = \{(y,y)\mid y \in Y\}\subset Y \times Y$ is closed in any hausdorff space.
I also know that since $\Delta Y$ is closed in $Y \times Y$, its complement $(Y \times Y)\setminus \Delta Y$ is open.
However, $\{(f(x),g(x)\mid x \in X,  f(x) \not= g(x) \}$ is only a subset of $(Y \times Y)\setminus \Delta Y$ for which i can't say whether it is open or closed.
The only idea that's left was to try this:
Pick an arbitrary point $p = (f(x),g(x))$ for which $f(x)\not= g(x)$ in $Y\times Y$. Since $Y$ is a Hausdorff space, we can find disjoint open neighbourhoods $U$ around $f(x)$ and $V$ around $g(x)$ such that $U \times V$ is an open neighbourhood of $(f(x),g(x))$ with $$U \times V \cap \Delta Y = \emptyset$$
Since $\{(f(x),g(x)\mid x \in X, f(x) \not= g(x) \}$ is a union of all such open neighbourhoods $U\times V$ of points $(f(x),g(x))$ with $f(x) \not= g(x)$ it is open in $Y\times Y$.
(is this correct?)
If the last paragraph would be correct, i'd be finished.
Thank you very much for any of your help.
EDIT: New summary of what i gathered so far with the help of the recent comments and answers:
As pointed out in the comments, i did the mistake to purely focus on $$B:= \{(f(x),g(x)\mid f(x) \not= g(x), x \in X\}$$ and tried to conclude that its preimage is measurable iff $B$ is open which was a wrong assumption in the first place.
Therefore, i do not need $B$ itself to be open. It's sufficient to show that the union of all open disjoint neighbourhoods $U\times V$ of arbitrary points $f(x),g(x)$ for which $f(x) \not= g(x)$ is an open set; because what matters is its preimage and not the set (image) itself. Let $$ B' := \bigcup\{U \times V \mid U, V \in \mathcal{B}\ ;\ U \cap V = \emptyset\}$$
$B'$ is obviously an open set, therefore its preimage is mesaurable.
For the preimage of open disjoint neighbourhoods $U,V$ of $f(x)$ and $g(x)$ respectively, it holds that
$$(f\times g)^{-1}(U\times V) = \{x \mid f(x)\in U \wedge g(x) \in V\} = f^{-1}(U)\cap g^{-1}(V)$$ (this is something i actually didnt know, it didn't know the definition of cartesian products of mappings, that's also the reason i didnt understand alex's question right away)
Since $U\times V$ is open, $(f\times g)^{-1}(U \times V)$ is measurable. Since $\mathcal{B}$ is a countable base the set
$$\{x \in X\mid f(x) \not= g(x) \} = \bigcup \{ (f\times g)^{-1}(U\times V) \mid U, V \in \mathcal{B}, U \cap V = \emptyset\} = \bigcup\{ f^{-1}(U)\cap g^{-1}(V)\mid U, V \in \mathcal{B}, U \cap V = \emptyset\}$$ is a countable union of measurable sets and therefore measurable itself.

REPLY [1 votes]: In answer to the question asked, about whether the argument in the last paragraph is correct, I don't believe that it is at it stands.
The problem is with the assertion $"\ \{(
\,f(x),g(x)\ )\mid x \in X, f(x) \not= g(x) \}\ $ is a union of all such open neighbourhoods $ \dots\ $ ."  If you take the union of all the open neighbourhoods you describe you will certainly get an open set containing $\ \{(
\,f(x),g(x)\ )\mid x \in X,\, f(x) \not= g(x) \}\ $, but I don't see why it should be equal to it, and I believe I can give a counterexample to the general statement that $\ \{(
\,f(x),g(x)\ )\mid x \in X,\, f(x) \not= g(x) \}\ $ must be open under the hypotheses given
Nevertheless, I think Alex Ravsky's proof shows that the basic idea underlying the argument was sound.
Addendum: It may be helpful for me to add one of the counterexamples I was referring to in my above comments.
Take $\ X = Y = \mathbb R\ $—the set of real numbers—, and $\ \mathscr{M}_X ,\mathscr{M}_Y\ \ $ both to be the Borel sets of $\ \mathbb R\ $.  Let $\ f\left(x\right)= 0\ \mbox{ and }\ g\left(x\right)= x\ \mbox{ for all } x\in\mathbb R\ $. Then $\ \{(
\,f(x),g(x)\ )\mid x \in X,\, f(x) \not= g(x) \}\ $ is the punctured line segment $\ \left\{\ \left(\,0,x\,\right)\ \mid\  x\ne0\ \right\}\ $ of $\ \mathbb R^2\ $, which is not an open set.
Comment on proposed new proof:
Apart from some probable typos, itemised below, the new revised proof by Zest seems to me to be correct.
The following appear likely to me to be typos:

$$B' = \bigcup \left\{ U\times V\mid U,V\in B,  U\cap V=\emptyset\right\}\ .$$
This should be
$$B' = \bigcup \left\{ U\times V\mid U,V\in{\mathcal B},  U\cap V=\emptyset\right\}$$
I'm not sure what the observation "Since $\ B\ $ is a countable base $\ \mathcal B\ \dots\ $" is trying to say.  As far as I can see, all that's necessary here is the observation that $\ \mathcal B\ $ is a countable base.
A union sign is missing from the last term on the left of the final series of equations. That is, $\ \left\{\,f^{-1}\left(U\right)\cap g^{-1}\left(V\right)\mid U,V\in{\mathcal B},  U\cap V=\emptyset\right\}\ $ should be $\ \bigcup\left\{\,f^{-1}\left(U\right)\cap g^{-1}\left(V\right)\mid U,V\in{\mathcal B},  U\cap V=\emptyset\right\}\ $

The nice observation Alex Ravsky and Zest have made here, and which I had missed, is that $\ \left\{\,x\in X \mid f\left(x\right)\neq g\left(x\right)\,\right\}\ $ is actually equal to $\ \left(f\times g\right)^{-1}\left(B'\right)\ $ rather than being a strict subset of it. Since this isn't immediately obvious—at least it wasn't to me, although the proof is fairly straightforward once the fact is realised—I'd like to see a brief proof of that fact included in any formal writeup.


TITLE: Smoothness of a power of smooth non-negative function
QUESTION [0 upvotes]: Let $f$ be a non-negative infinitely smooth function on the real line. Is it true that for any constant $\alpha$ the function $f^\alpha$ is infinitely smooth?

REPLY [4 votes]: There is an interesting variant of the question: Suppose that the funktion is smooth, positive, and flat at  all zeroes  (i.e. all derivatives vanish). Is the square root again smooth?
Glaeser proved that it is continuously differebtiable and gave an example wehre it is not twice differentiable.


text:
\begin{document}
\maketitle

\begin{abstract}
We study the higher spin Dirac operators on $3$-dimensional manifolds and show that there exist two Laplace type operators for each associated bundle. Furthermore, we give lower bound estimations for the first eigenvalues of these Laplace type operators.
\end{abstract}

\section{Introduction}\label{sec:0}
In this paper, we study the higher spin Dirac operator, which is a generalization of the Dirac operator as follows (see \cite{F}, \cite{BS}, and \cite{Bu}). Let $M$ be a $n$-dimensional spin manifold and $\mathbf{Spin}(M)$ be the principal spin bundle on $M$. The irreducible unitary representation $(\rho, V_{\rho})$ of the structure group $Spin(n)$ induces the associated (irreducible) bundle $\mathbf{S}_{\rho}(M)$,
\begin{equation}
\mathbf{S}_{\rho}(M):=\mathbf{Spin}(M)\times_{\rho}V_{\rho}.
\end{equation}
For each bundle, we have the covariant derivative $\nabla$ associated to the Levi-Civita connection or the spin connection,
\begin{equation}
\nabla:\Gamma (\mathbf{S}_{\rho}(M))\to \Gamma (\mathbf{S}_{\rho}(M)\otimes T^{\ast}(M)).
\end{equation}
Here, the cotangent bundle $T^{\ast}(M)$ is the bundle corresponding to the adjoint representation $(\mathrm{Ad},\mathbf{R}^n)$ of $Spin(n)$. So we decompose the tensor bundle $\mathbf{S}_{\rho}(M)\otimes T^{\ast}(M)$ into irreducible bundles with respect to $Spin(n)$. Let $\pi_{\rho, \nu}$ be the orthogonal projection to the irreducible bundle $\mathbf{S}_{\nu}(M)$ from $\mathbf{S}_{\rho}(M)\otimes T^{\ast}(M)\simeq \oplus_{\nu}\mathbf{S}_{\nu}(M)$. Then we define the higher spin Dirac operator $D_{\rho, \nu}$ to be the composed mapping $\pi_{\rho, \nu}\circ \nabla$,
\begin{equation}
D_{\rho, \nu}:\Gamma (\mathbf{S}_{\rho}(M))\xrightarrow{\nabla} \Gamma (\mathbf{S}_{\rho}(M)\otimes T^{\ast}(M))\xrightarrow{\pi_{\rho, \nu}}\Gamma (\mathbf{S}_{\nu}(M)).
\end{equation}
In fact, the Dirac operator is given in this way. To construct the Dirac operator, we take the spinor representation $(\Delta, V_{\Delta})$ and the associated bundle $\mathbf{S}_{\Delta}(M)$. Then the tensor bundle $\mathbf{S}_{\Delta}(M)\otimes T^{\ast}(M)$ decomposes into the direct sum of only two irreducible bundles, $\mathbf{S}_{\Delta}(M)$ and $\mathbf{S}_{T}(M)$. Then the differential operator $D:=D_{\Delta,\Delta}$ is the Dirac operator and $D_{\Delta,T}$ is the twistor operator (see \cite{BFGK} and \cite{BF}). On the other hand, we know another definition of the Dirac operator by using the Clifford algebra, that is,
\begin{equation}
D=\sum_i e_i\cdot\nabla_{e_i}.\label{eqn:0-3}
\end{equation}
From the relations
\begin{equation}
e_i e_j+e_j e_i=-2\delta_{ij}, \label{eqn:0-2}
\end{equation}
we show that the Dirac operator satisfies the Bochner type identity
\begin{equation}
D^2=\nabla^{\ast}\nabla+\frac{1}{4}\kappa, \label{eqn:0-1}
\end{equation}
where $\kappa$ is the scalar curvature of $M$.

The aim of this paper is to give the Bochner type identities for the higher spin Dirac operators on $3$-dimensional spin manifolds. As mentioned above, the relations \eqref{eqn:0-2} is necessary to give the Bochner type identity for the Dirac operator. But the Clifford action does not exist on the representation spaces of $Spin(n)$ in general. So we consider linear mappings among the representation spaces, which are called {\it the Clifford homomorphisms}. For a $3$-dimensional spin manifold $M$, the structure group of $\mathbf{Spin}(M)$ is $Spin(3)=SU(2)$. Then we use the Clebsch-Gordan formula to define the Clifford homomorphisms. By using the Clifford homomorphisms, we obtain local formulas of the higher spin Dirac operators such as \eqref{eqn:0-3} and the Bochner type identities for them. Furthermore, the identities lead us to give lower bound estimations for the first eigenvalues of these operators.

In section \ref{sec:1}, we explain the Clebsch-Gordan formula for the Lie group $SU(2)$. In section \ref{sec:2}, we define the Clifford homomorphisms on the representation spaces and obtain some relations among these homomorphisms including the usual Clifford relations \eqref{eqn:0-2}. In section \ref{sec:3}, we have formulas of the higher spin Dirac operators by using the Clifford homomorphisms and investigate the properties of these operators (ellipticity, the Bochner type identities, and so on.). The interest thing is that we obtain two Laplace type operators for each associated bundle. In section \ref{sec:4}, we have the lower bound estimations for the first eigenvalue of the Laplace type operators. This estimation is a generalization of the one for the Dirac operator given in \cite{BFGK} or the Laplace-Beltrami operator in \cite{GM} and \cite{P}. In the section \ref{sec:5}, we consider the case of the $3$-dimensional manifold of the constant curvature and show that some operators commute. In the last section, as an example, we calculate all the eigenvalues of the higher spin Dirac operators on the symmetric space $S^3$.

\section{The Clebsch-Gordan formula }\label{sec:1}
In this section we shall explain the representations of $SU(2)$ and the Clebsch-Gordan formula. Let $V_m$ be the $(m+1)$-dimensional complex vector space of polynomials of degree $\le m$ in $z_m$. The inner product on $V_m$ is set by

\begin{equation}
        \left(v_m^k,v_m^l\right)=\delta_{kl}, \label{1-1}
\end{equation}
where

\begin{equation}
v_m^k:=\frac{z^k_m}{\sqrt{k!(m-k)!}}.  \label{1-2}
\end{equation}

We define a representation $\rho_{m}$ on $V_m$ by $\rho_{m}(h)z_m^k=(b z_m+d)^{m-k}(a z_m+c)^k$ for $h=
\left(
\begin{smallmatrix}
a & b \\
c & d
\end{smallmatrix}
\right)
$ in $SU(2)$.
Then $(\rho_m, V_m)$ is a finite dimensional irreducible unitary representation of $SU(2)$ called the spin-$\frac{m}{2}$ representation and all such representations are given in this way.

We denote the infinitesimal representation of $(\rho_m, V_m)$ by the same symbol $(\rho_m, V_m)$. The Lie algebra $\mathfrak{su}(2)$ of $SU(2)$ has the following basis, that is, the Pauli matrices:

    \begin{equation}

     \sigma _1:=
         \begin{pmatrix}
                 i & 0 \\
                 0 & -i
          \end{pmatrix},
   \quad

     \sigma _2:=
         \begin{pmatrix}
               0 & 1 \\
               -1 &  0
       \end{pmatrix},
 \quad

  \sigma _3:=
       \begin{pmatrix}
            0 & i \\
            i &  0
       \end{pmatrix}.  \label{1-3}
   \end{equation}

Then we show that

\begin{align}
\rho_{m}(\frac{\sigma_1}{2})z_m^k &=i(k-\frac{m}{2})z_m^k, \nonumber \\
\rho_{m}(\frac{\sigma_2}{2}+i\frac{\sigma_3}{2})z_m^k &=-k z_m^{k-1}, \label{1-4} \\
\rho_{m}(\frac{\sigma_2}{2}-i\frac{\sigma_3}{2})z_m^k &=(m-k)z^{k+1}_m. \nonumber
\end{align}

\begin{ex}
The spin-$\frac{1}{2}$ representation $(\rho_1,V_1)$ is the spinor representation on $\mathbf{C}^2$, where we identify $Spin(3)$ with $SU(2)$.
\end{ex}
\begin{ex}
The spin-$1$ representation $(\rho_2,V_2)$ is the adjoint representation on $\mathfrak{su}(2)\otimes \mathbf{C}$ of $SU(2)$, or the adjoint representation on $\mathbf{R}^3\otimes \mathbf{C}$ of $Spin(3)$. Here, the correspondence of the bases is given as follows:

\begin{equation}
z_2^0\leftrightarrow\frac{\sigma_2+i\sigma_3}{2}\leftrightarrow e_2+i e_3,
\quad
z^1_2\leftrightarrow \frac{i\sigma_1}{2}\leftrightarrow i e_1,
\quad
z_2^2\leftrightarrow\frac{\sigma_2-i\sigma_3}{2}\leftrightarrow e_2-i e_3, \label{1-5}
\end{equation}
where $z_2^i$ is in $V_2$, $\sigma_i$ in $\mathfrak{su}(2)$, and $e_i$ in $\mathbf{R}^3$.
\end{ex}

Now, we consider the unitary representation $(\rho_m\otimes \rho_n, V_m\otimes V_n)$. Then we can decompose $\rho_m\otimes \rho_n$ into its irreducible components,

\begin{equation}
\rho_m\otimes \rho_n\simeq \rho_{m+n}\oplus \rho_{m+n-2}\oplus \cdots \oplus \rho_{|m-n|}. \label{1-6}
\end{equation}

This formula is called Clebsch-Gordan formula. We need the orthogonal projection to each irreducible component from $V_m\otimes V_n$ in the next section.

\section{The Clifford homomorphisms}\label{sec:2}
In this section we shall define the Clifford homomorphisms, which is a generalization of the Clifford action.
Let $\mathbf{C}l_3$ be the complex Clifford algebra associated to $\mathbf{R}^3$ and $\{e_i\}_{1\le i \le 3}$ be the standard basis of $\mathbf{R}^3$.
We realize $\mathbf{C}l_3$ as matrix algebra $\mathbf{C}(2)\oplus \mathbf{C}(2)$ by the mapping

\begin{equation}
\mathbf{C}l_3 \ni e_i\mapsto (\sigma_i,-\sigma_i) \in \mathbf{C}(2)\oplus \mathbf{C}(2).        \label{eqn:2-1}
\end{equation}

Then the Clifford action of $e_i$ on the spinor space $V_1\simeq \mathbf{C}^2$ is given by $e_i\cdot v=\sigma_i v$. Since we would like to generalize this Clifford action on other representation spaces, we use another definition of the Clifford action as follows: we recall the irreducible decomposition

\begin{equation}
(\rho_1, V_1)\otimes (\rho_2, V_2)\simeq (\rho_{3}, V_3)\oplus (\rho_{1}, V_1) \label{eqn:2-2}
\end{equation}

and the isomorphism

\begin{equation}
(\rho_2,V_2)\simeq (ad, \mathfrak{su}(2)\otimes \mathbf{C}) \simeq (ad, \mathbf{R}^3\otimes \mathbf{C}).  \label{eqn:2-3}
\end{equation}

For $v$ in $V_1$ and $e_i$ in $\mathbf{R}^3$, we project $v\otimes e_i$ onto $V_1$ along $V_3$ orthogonally. By calculating the Clebsch-Gordan coefficients, we show that $\mathrm{pr}(v\otimes e_i)=\sigma_i v=e_i\cdot v$.

Now, we consider the representation space $V_m$. In this case, we use the irreducible decomposition

\begin{equation}
(\rho_m, V_m)\otimes (\rho_2, V_2)\simeq (\rho_{m+2}, V_{m+2})\oplus (\rho_m, V_m) \oplus (\rho_{m-2}, V_{m-2}).  \label{eqn:2-4}
\end{equation}

For $v$ in $V_m$ and $X$ in $\mathbf{R}^3$, we decompose $v\otimes X$ as

\begin{equation}
v\otimes X =(v\otimes X)^{+}+(v\otimes X)^{0}+(v\otimes X)^{-}.  \label{eqn:2-5}\end{equation}

Here, $(v\otimes X)^{0}$ is in $V_m$ and $(v\otimes X)^{\pm}$ in $V_{m\pm 2}$. Thus, we have linear mappings from $V_m$ to $V_m$ or $V_{m\pm 2}$ for any $X$ in $\mathbf{R}^3$:

\begin{align}
\rho_m^0(X)v:=&-\frac{\sqrt{m(m+2)}}{2}(v\otimes X)^{0} \in V_{m},
                         \nonumber \\
\rho_m^{+}(X)v:=&-\frac{\sqrt{(m+1)(m+2)}}{\sqrt{2}}(v\otimes X)^{+} \in V_{m+2},           \label{eqn:2-6}\\
\rho_m^{-}(X)v:=&\frac{\sqrt{m(m+1)}}{\sqrt{2}}(v\otimes X)^{-} \in V_{m-2},   \nonumber
\end{align}

where we multiply each mapping by a constant to let the calculations easier. We call these linear mappings {\it the Clifford homomorphisms}.

Calculating the Clebsch-Gordan coefficients in the decomposition \eqref{eqn:2-4}by using {\it Mathematica} (see\cite{Mathe}), we deduce explicit formulas of the Clifford homomorphisms.

\begin{prop}\label{prop:2-1}
The Clifford homomorphisms associated to $\mathbf{R}^3$ are given as follows: for the basis $\{ z_m^k\}_{0\le k \le m}$ of $V_m$ and $\{e_i\}_{1 \le i \le 3}$ in $\mathbf{R}^3$,
\begin{enumerate}
\item $\rho_m^0(\cdot):V_m\to V_m$,

\begin{align}
\rho_{m}^0(\frac{e_1}{2})z^k_m &=i(k-\frac{m}{2})z_m^k, \nonumber \\
\rho_{m}^0(\frac{e_2}{2}+i\frac{e_3}{2})z^k_m &=-k z^{k-1}_m, \label{eqn:2-7} \\
\rho_{m}^0(\frac{e_2}{2}-i\frac{e_3}{2})z^k_m &=(m-k)z^{k+1}_m. \nonumber
\end{align}

\item
$\rho_m^{+}(\cdot):V_m\to V_{m+2}$,

\begin{align}
\rho_{m}^+(\frac{e_1}{2})z^k_m &=i z^{k+1}_{m+2}, \nonumber \\
\rho_{m}^+(\frac{e_2}{2}+i\frac{e_3}{2})z^k_m &=-z^{k}_{m+2}, \label{eqn:2-8} \\
\rho_{m}^+(\frac{e_2}{2}-i\frac{e_3}{2})z^k_m &=-z^{k+2}_{m+2}. \nonumber
\end{align}

\item
$\rho_m^{-}(\cdot):V_m\to V_{m-2}$,

\begin{align}
\rho_{m}^-(\frac{e_1}{2})z^k_m &=ik(m-k)z^{k-1}_{m-2}, \nonumber \\
\rho_{m}^-(\frac{e_2}{2}+i\frac{e_3}{2})z^k_m &=k(k-1)z^{k-2}_{m-2}, \label{eqn:2-9} \\
\rho_{m}^-(\frac{e_2}{2}-i\frac{e_3}{2})z^k_m &=(m-k)(m-k-1)z^{k}_{m-2}. \nonumber
\end{align}

\end{enumerate}
\end{prop}

We remark that $\rho_m^0$ is the representation $(\rho_m,V_m)$ of $\mathfrak{su}(2)$ under the isomorphism $\mathfrak{su}(2)\simeq \mathbf{R}^3$ and $\rho_1^0$ is the usual Clifford action on the spinor space $V_1$.

Now, we shall investigate some properties of the Clifford homomorphisms.

\begin{lem}\label{lem:2-2}
For $X$ in $\mathbf{R}^3\simeq \mathfrak{su}(2)$, we have

\begin{align}
(\rho_m^0(X))^{\ast}&=-\rho_m^0(X), \label{eqn:2-10}\\
(\rho_m^{\pm}(X))^{\ast}&=-\rho_{m\pm 2}^{\mp}(X), \label{eqn:2-11}
\end{align}
where $(\cdot)^{\ast}$ is the transposed conjugate with respect to the inner product of each $V_m$.
\end{lem}

\begin{proof}
Because $\rho_m^0$ is the representation of $\mathfrak{su}(2)$, the relation \eqref{eqn:2-10} is trivial. So we shall prove that $(\rho_m^+(X))^{\ast}=-\rho_{m+2}^-(X)$. We take the complexification of \eqref{eqn:2-11} and may prove $(\rho_m^+(X+i Y))^{\ast}=-\rho_{m+2}^-(X)+i\rho_{m+2}^-(Y)$. For example, we have

\begin{equation}
\begin{split}
(\rho_{m}^+(\frac{\sigma_2}{2}-i\frac{\sigma_3}{2})z_m^k,z_{m+2}^l)
&=-(z^{k+2}_{m+2},z^l_{m+2})   \\
&=-(k+2)!(m-k)!\delta_{k+2,l},\quad \mbox{for any $k,l$}.
\end{split}  \nonumber
\end{equation}

On the other hands,

\begin{equation}
\begin{split}
(z^k_m,-\rho_{m}^-(\frac{\sigma_2}{2}+i\frac{\sigma_3}{2})z^l_{m+2})
&=-l(l-1)(z_{m}^{k},v_{m}^{l-2})\\
&=-l(l-1)k!(m-k)!\delta_{k,l-2} \\
&=-(k+2)!(m-k)!\delta_{k+2,l}, \quad \mbox{for any $k,l$}.
\end{split} \nonumber
\end{equation}

So we have $(\rho_{m}^+(\sigma_2-i\sigma_3))^{\ast}=-\rho_{m}^-(\sigma_2)+i\rho_{m}^-(\sigma_3)$. Similarly we can prove the other cases.
\end{proof}

\begin{lem}\label{lem:2-3}
For $X$ in $\mathbf{R}^3\simeq \mathfrak{su}(2)$ and $g$ in $SU(2)$, we have

\begin{align}
\rho_m^0(gXg^{-1})&=\rho_m(g)\rho_m^0(X)\rho_m(g^{-1}) , \label{eqn:2-12}\\
\rho_m^{\pm}(gXg^{-1})&=\rho_{m\pm 2}(g)\rho_m^{\pm}(X)\rho_m(g^{-1}) .\label{eqn:2-13}
\end{align}

\end{lem}

\begin{proof}
 The equation \eqref{eqn:2-12} is trivial. So we shall prove \eqref{eqn:2-13}. For an orthonormal basis $\{v_{m+2}^{k}\}$ of $V_{m+2}$, we denote the corresponding one of the irreducible component $V_{m+2}$ in $V_m\otimes V_2$ by $\{\omega_{m+2}^k\}_k$. Since $\rho_m^{+}$ is the orthogonal projection from $V_m\otimes V_2$ to $V_{m+2}$, the homomorphism $\rho_m^{+}$ is represented as
\begin{equation}
\rho_m^{+}(X)v=\sum_k (v\otimes X, \omega^k_{m+2})v^k_{m+2}, \label{eqn:2-14}
\end{equation}
where $(\cdot,\cdot)$ is inner product on $V_m\otimes V_2$. If we use another orthonormal basis $\{ \rho_{m+2}(g)v^k_{m+2}\}_k$, then we have
\begin{equation}
\rho_m^{+}(X)v=\sum_k (v\otimes X, (\rho_m\otimes \rho_2)(g)\omega^k_{m+2})\rho_{m+2}(g)v^k_{m+2}. \nonumber
\end{equation}
It follows that

\begin{equation}
\begin{split}
\rho_m^+(gXg^{-1})v&=\sum (v\otimes gXg^{-1},\omega^k_{m+2})v^k_{m+2} \\
                   &=\sum (v\otimes \rho_2(g)X,\omega^k_{m+2})v^k_{m+2}  \\
             &=\sum ((\rho_m\otimes \rho_2)(g)(\rho_m(g^{-1})v\otimes X),\omega^k_{m+2})v^k_{m+2} \\
         &=\sum (\rho_m(g^{-1})v\otimes X,(\rho_m\otimes \rho_2)(g^{-1})\omega^k_{m+2})v^k_{m+2} \\
 &=\sum (\rho_m(g^{-1})v\otimes X,\omega^k_{m+2})\rho_{m+2}(g)v^k_{m+2}\\         &=\rho_{m+2}(g)\rho_m^+(X)\rho_m(g^{-1})v.
\end{split} \nonumber
\end{equation}
Thus we have proved the lemma.
\end{proof}
The infinitesimal version of this lemma is given as follows.

\begin{lem}\label{cor:2-4}
For $X,Y$ in $\mathbf{R}^3\simeq\mathfrak{su}(2)$Cit holds that

\begin{align}
\rho_m^0([X,Y])&=[\rho_m^0(X),\rho_m^0(Y)], \label{eqn:2-15}\\
\rho_m^{\pm}([X,Y])&=\rho_{m\pm 2}^0(X)\rho_{m}^{\pm}(Y)-\rho_m^{\pm}(Y)\rho_m^0(X), \label{eqn:2-16}
\end{align}

where $[\cdot,\cdot]$ denotes the Lie bracket in $\mathfrak{su}(2)$.
\end{lem}

Now, we know that the usual Clifford actions $\{\sigma_i\}=\{e_i\cdot\}$ satisfy the relations
\begin{equation}
\sigma_i \sigma_j+\sigma_i \sigma_j =-2\delta_{ij}\quad (0\le i, j\le 3). \label{eqn:2-17}
\end{equation}
We should find what relations the Clifford homomorphisms satisfy.

\begin{lem}\label{lem:2-5}
The Clifford homomorphisms have the following relations: for $X$, $Y$ in $\mathbf{R}^3\simeq \mathfrak{su}(2)$,

\begin{align}
\rho_{m+2}^0(X)\rho_{m}^{+}(Y)-\rho_m^{+}(X)\rho_m^0(Y)
    &=\frac{m+2}{2}\rho_m^{+}([X,Y]) ,                  \label{eqn:2-18}\\
\rho_{m-2}^0(X)\rho_{m}^{-}(Y)-\rho_m^{-}(X)\rho_m^0(Y)
    &=-\frac{m}{2}\rho_m^{-}([X,Y]),                      \label{eqn:2-19}\\
 \rho_{m}^0(X)\rho_{m}^{0}(Y)+\rho_{m-2}^{+}(X)\rho_m^-(Y)
    &=\frac{m}{2}\rho_m^{0}([X,Y])-m^2(X,Y),                 \label{eqn:2-20} \\
 \rho_{m}^0(X)\rho_{m}^{0}(Y)+\rho_{m+2}^{-}(X)\rho_m^+(Y)
    &=-\frac{m+2}{2}\rho_m^{0}([X,Y])-(m+2)^2(X,Y),           \label{eqn:2-21}
\end{align}
where $(\cdot,\cdot)$ is the inner product on $\mathbf{R}^3$.

\end{lem}

\begin{proof}
By direct calculations.
\end{proof}
We remark that, for $m=1$, the relation \eqref{eqn:2-20} is the usual Clifford relation \eqref{eqn:2-17}.

\section{The higher spin bundles and the higher spin Dirac operators}\label{sec:3}
Let $M$ be the $3$-dimensional oriented Riemannian manifold. Since such a manifold is automatically a spin manifold, we have a principal spin bundle $\mathbf{Spin}(M)$ on $M$, where the structure group is $Spin(3)=SU(2)$. Then all the associated complex vector bundles are induced from the representations of $SU(2)$. For any $m\ge 0$, we define the spin-$\frac{m}{2}$ bundle $\mathbf{S}_m$ by

\begin{equation}
\mathbf{S}_m=\mathbf{S}_m(M):=\mathbf{Spin}(M)\times_{\rho_m} V_m. \label{eqn:3-1}
\end{equation}

The inner product on $V_m$ induces the one on each fiber of $\mathbf{S}_m$ naturally, which we denote by $\langle\cdot,\cdot\rangle$ on $(\mathbf{S}_m)_x$.
For example, the spin-$0$ bundle $\mathbf{S}_0$ is the trivial rank $1$ bundle $M\times \mathbf{C}\simeq \Lambda^0(M)\otimes \mathbf{C}$, the spin-$\frac{1}{2}$ bundle $\mathbf{S}_1$ is the spinor bundle, and the spin-$1$ bundle $\mathbf{S}_2$ is $T(M)\otimes \mathbf{C}\simeq \Lambda^1(M)\otimes \mathbf{C}$.

The spinor bundle $\mathbf{S}_1$ is known as a bundle of modules over the Clifford bundle $\mathbf{C}l(M)$ and the action of $T(M)$ on $\mathbf{S}_1$ is given by

\begin{equation}
T(M)\times \mathbf{S}_1\ni ([p,e_i],[p,v])\mapsto [p,e_i\cdot v]\in \mathbf{S}_1 , \label{eqn:3-2}
\end{equation}

where $p$ is in $\mathbf{Spin}(M)$, $e_i$ in $\mathbf{R}^3$, and $v$ in $V_1$.
In the same way, we define the Clifford homomorphisms of $T(M)$ on the higher spin bundle $\mathbf{S}_m$ as follows:

\begin{align}
T(M)\times \mathbf{S}_m\ni ([p,e_i],[p,v])\mapsto [p,\rho_m^0(e_i)v]\in \mathbf{S}_m , \label{eqn:3-3} \\
T(M)\times \mathbf{S}_m\ni ([p,e_i],[p,v])\mapsto [p,\rho_m^{\pm}(e_i)v]\in \mathbf{S}_{m\pm 2} . \label{eqn:3-4}
\end{align}

We can easily check from lemma \ref{lem:2-3} that these bundle homomorphisms are well-defined.

Before considering the higher spin Dirac operators on $\Gamma (M,\mathbf{S}_m)$, we recall the definition of the Dirac operator $D$ on $\Gamma(M,\mathbf{S}_1)$. Let $\nabla$ be the covariant derivative associated to the spin connection. The Dirac operator $D$ has the following (local) formula:
\begin{equation}
D=\sum_{i=1}^3e_i\cdot\nabla_{e_i}.\label{eqn:3-5}
\end{equation}
On the other hand, we know another description of $D$ as follows: the Dirac operator $D$ is said to be the composed mapping $\mathrm{pr}\circ \nabla$,

\begin{equation}
\Gamma(M,\mathbf{S}_1)\xrightarrow{\nabla}\Gamma(M,\mathbf{S}_1\otimes T^{\ast}(M))\simeq \Gamma(M,\mathbf{S}_1\otimes T(M))\xrightarrow{\mathrm{pr}} \Gamma(M,\mathbf{S}_1),   \label{eqn:3-6}
\end{equation}

where we use $\mathbf{S}_1\otimes T(M)\simeq \mathbf{S}_1\otimes \mathbf{S}_2\simeq \mathbf{S}_3\oplus \mathbf{S}_1$.

We generalize this composed mapping to give the higher spin Dirac operator (see \cite{BS}, \cite{Bu} and \cite{F}). Since the tensor bundle $\mathbf{S}_m\otimes \mathbf{S}_2$ is isomorphic to $\mathbf{S}_{m+2}\oplus \mathbf{S}_m\oplus \mathbf{S}_{m-2}$, we have three composed mappings for each bundle:

\begin{align}
D^{0}_m:\Gamma(M,\mathbf{S}_m)\xrightarrow{\nabla}\Gamma(M,\mathbf{S}_m\otimes T^{\ast}M) &\xrightarrow{\mathrm{pr}^0} \Gamma(M,\mathbf{S}_m), \label{eqn:3-7} \\
D_m^{\pm}:\Gamma(M,\mathbf{S}_m)\xrightarrow{\nabla}\Gamma(M,\mathbf{S}_m\otimes T^{\ast}M) &\xrightarrow{\mathrm{pr}^{\pm}} \Gamma(M,\mathbf{S}_{m\pm 2}). \label{eqn:3-8}
\end{align}

We call these first order differential operators {\it the higher spin Dirac operators}. In \cite{F}, Fegan show that these operators are conformally invariant first order differential operators and all such operators are given in this way.

The Clifford homomorphisms in section \ref{sec:2} lead us to represent the higher spin Dirac operators by local formulas such as \eqref{eqn:3-5}.

\begin{prop}\label{prop:3-1}
Let $M$ be the $3$-dimensional spin manifold, $\{ e_i \}_{1\le i \le 3}$ a local orthonormal frame of $T(M)$, and $\nabla$ the covariant derivative associated to the spin connection on $\mathbf{S}_m$. Then we have the following conformally invariant first order differential operators:

\begin{align}
D_m^0&=\sum_{1\le i \le 3}\rho_m^0(e_i)\nabla_{e_i}: \Gamma (M,\mathbf{S}_m)\to \Gamma (M,\mathbf{S}_m),
                           \label{eqn:3-9}   \\
D_m^{\pm}&=\sum_{1\le i \le 3}\rho_m^{\pm}(e_i)\nabla_{e_i}: \Gamma (M,\mathbf{S}_m)\to \Gamma (M,\mathbf{S}_{m\pm 2}).
                              \label{eqn:3-10}
\end{align}

\end{prop}

\begin{ex}
Some higher spin Dirac operators are well-known differential operators.
\begin{enumerate}
\item
$D_0^{+}$ is $2d$ on $\Gamma(M,\mathbf{S}_0)=\Gamma(M,\Lambda^0(M)\otimes \mathbf{C})$
 \item
$D^0_1$ is the Dirac operator $D$ and $D_1^+$ is the twistor operator on $\Gamma(M,\mathbf{S}_1)$.
\item $D_2^0$ is $2\ast d$ and  $D_2^-$ is $2d^{\ast}$ on $\Gamma(M,\mathbf{S}_2)=\Gamma(M,\Lambda^1(M)\otimes \mathbf{C})$, where $\ast$ is the Hodge star operator from $\Lambda^1(M)$ to $\Lambda^2(M)$.
\end{enumerate}
\end{ex}

From the discussion in section \ref{sec:2}, we can derive some properties of the higher spin Dirac operators.

First, we discuss the adjointness of the operators. On $\Gamma(M,\mathbf{S}_m)$, we set the inner product by
\begin{equation}
(\phi_1,\phi_2):=\int_M\langle \phi_1(x),\phi_2(x)\rangle dx,
\end{equation}
where $dx$ denotes the volume element of $M$.

\begin{prop}\label{lem:3-2}
We denote the formal adjoint of a differential operator $A$ by $A^{\ast}$. Then we have

\begin{align}
(D_m^0)^{\ast}&=D_m^0,                       \label{eqn:3-11}\\
(D_m^{\pm})^{\ast}&=D_{m\pm 2}^{\mp}.        \label{eqn:3-12}
\end{align}
In particular, $D_m^0$ is formally self adjoint.

\end{prop}

\begin{proof}
We can easily show that the Dirac operator is formally self-adjoint (for example, see \cite{LM}). In the same way, we can prove \eqref{eqn:3-11} and \eqref{eqn:3-12} by using lemma \ref{lem:2-2},.
\end{proof}

Next, we shall discuss the commutativity among the operators. So we have to introduce some curvature homomorphisms. For vector fields $X$, $Y$, the curvature $R_m$ for $\mathbf{S}_m$ is given by
\begin{align}
R_m(X,Y)&=\nabla_X \nabla_Y-\nabla_Y \nabla_X-\nabla_{[X,Y]} \quad \in \Gamma (M,\mathrm{End}(\mathbf{S}_m))\label{eqn:3-13} \\
&=\frac{1}{4}\sum_{\sigma \in S_3}\mathrm{sgn}(\sigma)\langle R(X,Y)(e_{\sigma (1)}),e_{\sigma (2)}\rangle\rho_m^0(e_{\sigma (3)}), \label{eqn:3-14}
\end{align}
where $R(\cdot ,\cdot)$ is the curvature transformation for $T(M)$ and $\{e_i\}_{1\le i \le 3}$ is a local orthonormal frame on $T(M)$. Then we obtain the following curvature homomorphisms from $\mathbf{S}_m$ to $\mathbf{S}_m$ or $\mathbf{S}_{m\pm 2}$:
\begin{align}
R_m^0:&=\sum_{\sigma \in S_3}\mathrm{sgn}(\sigma)
             \rho_m^0(e_{\sigma (1)})R_m(e_{\sigma (2)},e_{\sigma (3)})
                   \in \Gamma (M,\mathrm{End}(\mathbf{S}_m)), \label{eqn:3-15}\\
R_m^{\pm}:&=\sum_{\sigma \in S_3}\mathrm{sgn}(\sigma)
             \rho_m^{\pm}(e_{\sigma (1)})R_m(e_{\sigma (2)},e_{\sigma (3)})
                 \in \Gamma (M,\mathrm{Hom}(\mathbf{S}_m,\mathbf{S}_{m\pm 2})) .
                                                     \label{eqn:3-16}
\end{align}
Here, we show that $(R_m^0)^{\ast}=R_m^0$ and $(R_m^{\pm})^{\ast}=R_{m\pm 2}^{\mp}$. In particular, $(R_m^0)_x$ has real eigenvalues for each $x$ in $M$.
\begin{ex}
Let $\mathrm{Ric}$ be the Ricci curvature and $\kappa$ the scalar curvature. Then we have
\begin{equation}
R_1^0=\frac{1}{2}\kappa , \quad R_2^0=4\mathrm{Ric}, \quad R_0^0=R_2^-=R_0^+=0 .
                     \label{eqn:3-17}
\end{equation}
\end{ex}
The commutativity among the higher spin Dirac operators follows from Lemma \ref{lem:2-5}. The important fact is that we have two Laplace type operators on $\Gamma(M,\mathbf{S}_m)$ for each $m\ge 1$.

\begin{thm}\label{thm:3-3}
Let $\nabla^{\ast}\nabla$ be the connection Laplacian on $\mathbf{S}_m$. Then the higher spin Dirac operators satisfy the following Bochner type identities:

\begin{align}
D_m^0D_m^0+D_{m-2}^+D_m^-&=m^2\nabla^{\ast}\nabla+\frac{m}{2}R_m^0,
                    \label{eqn:3-18}\\
D_m^0D_m^0+D_{m+2}^-D_m^+&=(m+2)^2\nabla^{\ast}\nabla-\frac{m+2}{2}R_m^0,
                \label{eqn:3-19} \\
D_{m+2}^0D_m^+-D_{m}^+D_m^0&=\frac{m+2}{2}R_m^+, \label{eqn:3-20} \\
D_{m-2}^0D_m^--D_{m}^-D_m^0&=-\frac{m}{2}R_m^-, \label{eqn:3-21}
\end{align}

\end{thm}

\begin{proof}
We shall prove \eqref{eqn:3-18}. We fix $x$ in $M$ and choose an orthonormal frame $\{e_i\}$ in a neighborhood of $x$ such that $(\nabla_{e_i}e_j)_x=0$ for all $i,j$. Hence, we have $(\nabla_{e_i}\rho_m^0(e_j))_x=0$ for all $i,j$. Then it holds from Lemma \ref{lem:2-5} that
\begin{equation}
\begin{split}
D_m^0D_m^0&+D_{m-2}^+D_m^- \\
&=\sum_{i,j}\left( \rho_m^0(e_i)\nabla_{e_i}\rho_m^0(e_j)\nabla_{e_j}+\rho_{m-2}^+(e_i)\nabla_{e_i}\rho_m^-(e_j)\nabla_{e_j} \right) \\
&=\sum_{i}\left( \rho_m^0(e_i)\rho_m^0(e_i)+\rho_{m-2}^+(e_i)\rho_m^-(e_i) \right)\nabla_{e_i}\nabla_{e_i} \\
& \quad \quad \sum_{i\neq j}\left( \rho_m^0(e_i)\rho_m^0(e_j)+\rho_{m-2}^+(e_i)\rho_m^-(e_j) \right)\nabla_{e_i}\nabla_{e_j} \\
&=-m^2\sum_i \nabla_{e_i}\nabla_{e_i}+\frac{m}{2}\sum_{i<j}\rho_m^0(e_i e_j-e_j e_i)( \nabla_{e_i}\nabla_{e_j}- \nabla_{e_j}\nabla_{e_i}) \\
&=m^2 \nabla^{\ast} \nabla+\frac{m}{2}R_m^0.
 \end{split} \nonumber
\end{equation}
\end{proof}

\begin{ex}
\begin{enumerate}
\item (the case of $m=0$) The relation \eqref{eqn:3-19} means $d^{\ast}d=\nabla^{\ast}\nabla$ and the relation \eqref{eqn:3-20} does $dd=0$.
\item (the case of $m=1$) The relation \eqref{eqn:3-18} means
\begin{equation}
D^2=\nabla^{\ast} \nabla+\frac{1}{4}\kappa. \label{eqn:3-22}
\end{equation}
\item (the case of $m=2$) The relation \eqref{eqn:3-18} means
\begin{equation}
d^{\ast}d+dd^{\ast}=\nabla^{\ast}\nabla +\mathrm{Ric}. \label{eqn:3-23}
\end{equation}
and \eqref{eqn:3-21} does $dd=0$.
\end{enumerate}
\end{ex}
Now, we denote the Laplace type operators in \eqref{eqn:3-18} and \eqref{eqn:3-19} by

\begin{align}
\Delta_m:&=D_m^0D_m^0+D_{m-2}^+D_m^{-}, \label{eqn:3-24}  \\
\widetilde{\Delta}_m:&=D_m^0D_m^0+D_{m+2}^-D_m^{+}.\label{eqn:3-25}
\end{align}

If $M$ is compact, these Laplace type operators are non-negative operators and
satisfy that

\begin{gather}
\ker \Delta_m =\ker D_m^0\cap \ker D_m^-,\label{eqn:3-26}\\
\ker \widetilde{\Delta}_m =\ker D_m^0\cap \ker D_m^+,\label{eqn:3-27} \\
\ker \nabla =\ker \Delta_m \cap \ker \widetilde{\Delta}_m=\ker D_m^0\cap \ker D_m^-\cap \ker D_m^+. \label{eqn:3-28}
\end{gather}

The following corollary is the key to give lower bounds for the first eigenvalues of $\Delta_m$ and $\widetilde{\Delta}_m$.

\begin{cor}\label{cor:3-4}
The Laplace type operators $\Delta_m$ and $\widetilde{\Delta}_m$ satisfy that

\begin{equation}
(m+2)^2\Delta_m-m^2\widetilde{\Delta}_m=m(m+1)(m+2)R_m^0  \label{eqn:3-29}
\end{equation}

\end{cor}
\begin{proof}
We eliminate the connection Laplacian $\nabla^{\ast}\nabla$ from \eqref{eqn:3-18} and \eqref{eqn:3-19}.
\end{proof}

Finally, we discuss the ellipticity of the operators. Of course, it is clear that $\Delta_m$ and $\widetilde{\Delta}_m$ are elliptic.

\begin{prop}\label{prop:3-5}
\begin{enumerate}
\item
The second order differential operator $D_{m+2}^-D_m^+=(D_m^+)^{\ast}D_m^+$ is elliptic for each $m$.\item
If $m$ is odd, then the first order differential operator $D_m^0$ is elliptic. Hence $D_m^0$ is an elliptic self adjoint operator.
\end{enumerate}
\end{prop}

\begin{proof}
We investigate the ellipticity of $D_m^0$. The principal symbol of $D_m^0$ is

\begin{equation}
\sigma_{\xi} (D_m^0)=\rho_m^0(\xi) ,  \label{eqn:3-30}
\end{equation}

where $\xi=\sum \xi_i e_i$ is in $T_x^{\ast}(M)\simeq T_x(M)$. There exists $g$ in $SU(2)$ such that

\begin{equation}
g\xi g^{-1}=(\xi_1^2+\xi_2^2+\xi_3^2)^{\frac{1}{2}}e_1.  \label{eqn:3-31}
\end{equation}

Then we have

\begin{equation}
\begin{split}
\det \sigma_{\xi} (D_m^0) &=\det \rho_m(g)\rho_m^0(\xi)\rho_m(g^{-1}) \\
     &=\det \rho_m^0(g\xi g^{-1}) \\
    &=\det \rho_m^0((\xi_1^2+\xi_2^2+\xi_3^2)^{\frac{1}{2}}e_1) \\
    &=(\xi_1^2+\xi_2^2+\xi_3^2)^{\frac{m+1}{2}}\det \rho_m^0(e_1)  \\
    &=(\xi_1^2+\xi_2^2+\xi_3^2)^{\frac{m+1}{2}}\prod_{k=0}^mi(2k-m).
\end{split}  \label{eqn:3-32}
\end{equation}

It follows that, if $m$ is odd, then $\det \sigma_{\xi} (D_m^0)$ is not zero for $\xi\neq 0$. Hence $D_m^0$ is elliptic. In the same way, we verify that $D_{m+2}^-D_m^+$ is elliptic.
\end{proof}
\begin{cor}
We assume that the spin manifold $M$ is compact. Then $\ker D_m^+$ and $\ker D_{2p+1}$ are finite dimensional vector spaces for any $m$ and $p$.
\end{cor}

\section{lower bounds for the first eigenvalues of the higher spin Dirac operators}\label{sec:4}
In this section, we assume that $M$ is a $3$-dimensional connected compact spin manifold
. From corollary \ref{cor:3-4},
we have
\begin{equation}
(m+2)^2(\Delta_m\phi,\phi)-m^2(\widetilde{\Delta}_m\phi,\phi)=m(m+1)(m+2)(R_m^0(\phi),\phi), \label{eqn:4-1}
\end{equation}
where $\phi$ is a section of $\mathbf{S}_m$ and

\begin{equation}
(R_m^0(\phi),\phi):=\int_M \langle(R_m^0)_x\phi(x),\phi(x)\rangle dx.    \label{eqn:4-2}
\end{equation}

From the above equation \eqref{eqn:4-1}, we can obtain lower bounds estimations for the eigenvalues of $\Delta_m$ and $\widetilde{\Delta}_m$ depending on the curvature transformation $R_m^0$.

First, we consider a lower bound for the first eigenvalue of the Dirac operator $D=D_1^0$. It follows from \eqref{eqn:4-1} that, for a spinor $\phi$ in $\Gamma(M,\mathbf{S}_1)$,

\begin{align}
& 9\| D\phi \|^2-(\| D\phi \|^2+\| D_1^{+}\phi \|^2)\nonumber \\
=& 8\| D\phi \|^2 -\| D_1^{+}\phi \|^2 \label{eqn:4-3} \\
=& 6(R_1^0(\phi),\phi)=3(\kappa \phi,\phi). \nonumber
\end{align}

Because of $\| D_1^{+}\phi \| \ge 0$, we have

\begin{equation}
\| D\phi \|^2\ge \frac{3}{8}(\kappa \phi,\phi).  \label{eqn:4-4}
\end{equation}

If $\phi_1$ is an eigenspinor with the first eigenvalue $\lambda_1$ of $D$,
then $(\lambda_1)^2$ has a lower bound,

\begin{equation}
(\lambda_1)^2 \ge \frac{3(\kappa \phi_1,\phi_1)}{8\|\phi_1\|^2}\ge \frac{3}{8}\kappa_-.
                   \label{eqn:4-5}
\end{equation}

where

\begin{equation}
\kappa_-:=\min_{x \in M} \kappa (x).   \label{eqn:4-6}
\end{equation}

If the equality holds in \eqref{eqn:4-5}, then $\phi_1$ is in $\ker D_1^{+}$, that is, $\phi_1$ is a twistor spinor. This inequality coincides with the ones given by Friedrich (see \cite{BF}).

Next, we investigate the case of the elliptic operator $D_3^-D_1^+=(D_1^+)^{\ast}D_1^+$. It holds that
\begin{equation}
\begin{split}
(D_3^-D_1^+\phi,\phi) &=8\|D\phi\|^2-3(\kappa \phi, \phi) \\
                 &\ge -3(\kappa \phi ,\phi).  \label{eqn:4-7}
      \end{split}
   \end{equation}
If we denote the first eigenvalue of $D_3^-D_1^+$ by $\mu_1$, then we have
\begin{equation}
\mu_1 \ge -3\kappa_+,       \label{eqn:4-8}
\end{equation}
where
\begin{equation}
\kappa_+=\max_{x \in M}\kappa (x). \label{eqn:4-9}
\end{equation}

In general case ($m\ge 2$), we have the inequalities
\begin{gather}
( \Delta_m\phi, \phi)=\|D_m^0\phi \|^2+\|D_m^-\phi \|^2
\ge \frac{m(m+1)}{m+2}(\mathrm{R}_m^0(\phi),\phi ),    \label{eqn:4-10} \\
( \widetilde{\Delta}_m\phi, \phi)=\left(\|D_m^0\phi \|^2+\|D_m^+\phi \|^2\right)\ge -\frac{(m+2)(m+1)}{m}(\mathrm{R}_m^0(\phi),\phi ) .  \label{eqn:4-11}
\end{gather}
Then we give lower bounds for the first eigenvalues of $\Delta_m$ and $\widetilde{\Delta}_m$.

\begin{thm}\label{thm:4-1} We assume that there exist constants $r_{m-}$ and $r_{m+}$ such that
\begin{equation}
r_{m-}\|\phi\|^2\le (R_m^0(\phi),\phi)\le r_{m+}\|\phi\|^2
\end{equation}
for any $\phi$ in $\Gamma(M,\mathbf{S}_m)$.
\begin{enumerate}
\item Let $\lambda_1$ be the first eigenvalue of $\Delta_m$. Then we have the inequality

\begin{equation}
\lambda_1 \ge \frac{m(m+1)}{m+2}  r_{m-}.
                             \label{eqn:4-12}
\end{equation}

If the equality holds in \eqref{eqn:4-12}, the eigenvectors with the eigenvalue $\lambda_1$  is in $\ker \widetilde{\Delta}_m$.
\item
Let $\mu_1$ be the first eigenvalue of $\widetilde{\Delta}_m$. Then we have the inequality

\begin{equation}
\mu_1 \ge -\frac{(m+2)(m+1)}{m}  r_{m+}. \label{eqn:4-14}
\end{equation}

If the equality holds in \eqref{eqn:4-14}, then the eigenvectors with the eigenvalue $\mu_1$ is in $\ker \Delta_m$.
\end{enumerate}
\end{thm}

\begin{cor}[\cite{GM}]\label{cor:4-2}
We assume that there exists a constant $\mathrm{ric}_-$ such that
\begin{equation}
(\mathrm{Ric}(\phi),\phi)\ge \mathrm{ric}_-\|\phi\|^2
\end{equation}
for any $\phi$ in $\Gamma(M,\Lambda^1(M))$.
Let $\lambda_1$ be the first eigenvalue of the Laplace-Beltrami operator $dd^{\ast}+d^{\ast}d$ on $\Gamma(M,\Lambda^1(M))$. Then we have

\begin{equation}
\lambda_1 \ge \frac{3}{2}\mathrm{ric}_-.  \label{eqn:4-16}
\end{equation}

If the equality holds in \eqref{eqn:4-16}, the eigenforms with the eigenvalue $\lambda_1$ is in $\ker \widetilde{\Delta}_2=\ker d \cap \ker D_2^+$.
\end{cor}

\section{On the $3$-dimensional manifold of constant curvature}\label{sec:5}
In this section, we shall discuss the higher spin Dirac operators on $3$-dimensional manifold of constant curvature.

\begin{lem}\label{lem:5-1}
On the $3$-dimensional spin manifold $M$ of constant curvature $c$, the curvature homomorphism $R^0_m$ is $m(m+2)c$ and $R_m^{\pm}$ is zero.

\end{lem}

\begin{proof}
Since $M$ has constant curvature, it holds that, for vector fields $X$, $Y$, and $Z$,
$$R(X,Y)Z=c\{ (Y,Z)X-(X,Z)Y\}.$$
Then we have $(R(e_i,e_j)e_k,e_l)=c(\delta_{j k}\delta_{i l}-\delta_{i k}\delta_{j l})$. Hence,
$$R^0_m=-c\sum \rho_m^0(e_i)\rho_m^0(e_i)=m(m+2)c, \quad R_m^{\pm}=0.$$
Here, we use that $-\sum \rho_m^0(\sigma_i)\rho_m^0(\sigma_i)$ is the Casimir operator on $V_m$.
\end{proof}

\begin{prop}\label{prop:5-2}
On the $3$-dimensional spin manifold of constant curvature $c$, it hold that

\begin{align}
D_m^0D_m^0+D_{m-2}^+D_m^-&=m^2\nabla^{\ast}\nabla+\frac{m^2(m+2)}{2}c,
                    \label{eqn:7-2}\\
D_m^0D_m^0+D_{m+2}^-D_m^+&=(m+2)^2\nabla^{\ast}\nabla-\frac{m(m+2)^2}{2}c,
                \label{eqn:7-3} \\
D_{m+2}^0D_m^+-D_{m}^+D_m^0&=0,\label{eqn:7-4} \\
D_{m-2}^0D_m^--D_{m}^-D_m^0&=0. \label{eqn:7-5}
\end{align}
In particular, we have
\begin{gather}
\Delta_m D_m^0=D_m^0\Delta_m, \quad \widetilde{\Delta}_mD_m^0=D_m^0\widetilde{\Delta}_m, \\
\Delta_m(D_{m-2}^+D_m^-)=(D_{m-2}^+D_m^-)\Delta_m, \quad \widetilde{\Delta}_m(D_{m+2}^-D_m^+)=(D_{m+2}^-D_m^+)\widetilde{\Delta}_m.
\end{gather}
\end{prop}
We conclude from this proposition that $\Delta_m$, $D_m^0$, and $D_{m-2}^+D_m^-$ have the simultaneous eigenspaces. As an example, we will calculate the eigenvalues of these operators on $S^3$ in the next section.

\section{The spectra of the higher spin Dirac operators on $S^3$}\label{sec:6}
In this section, we calculate all the eigenvalues of the higher spin Dirac operators on the symmetric space $S^3$ with constant curvature $1$. In \cite{H}, the author gives a method for calculating of the eigenvalues and the eigenspinors for the Dirac operator on $S^3$. We can use the method in our situation and calculate the eigenvalues. So we refer to the paper \cite{H} for details.

First, we shall explain the $3$-dimensional sphere $S^3$ as the symmetric space $Spin(4)/Spin(3)$. It is well-known that $Spin(4)$ and $Spin(3)$ are isomorphic to $SU(2)\times SU(2)$ and $SU(2)$, respectively. We realize $S^3$ as $SU(2)$,

\begin{equation}
    S^3\ni x=(x_1,x_2,x_3,x_4)\mapsto h=
            \begin{pmatrix}
                x_4+ix_1 & x_2+ix_3 \\
                     -x_2+ix_3 & x_4-ix_1
        \end{pmatrix}     \label{eqn:6-1}
     \in SU(2).
\end{equation}

Therefore, the action of $SU(2)\times SU(2)$ on $S^3$ is represented by

\begin{equation}
    (SU(2)\times SU(2))\times S^3\ni(g,h)\mapsto phq^{-1}\in S^3, \label{eqn:6-2}
             \end{equation}

where $g=(p,q)$ is in $SU(2)\times SU(2)$.
Since the isotropy subgroup of $e=(0,0,0,1)$ is the subgroup $SU(2)$ in $SU(2)\times SU(2)$, we have the symmetric space $S^3$,

\begin{equation}
S^3=Spin(4)/Spin(3)=SU(2)\times SU(2)/\mathrm{diag} SU(2). \label{eqn:6-3}
\end{equation}

Here, the map `$\mathrm{diag}$' is given by

\begin{equation}
\mathrm{diag}: SU(2)\ni h \mapsto (h,h)\in SU(2)\times SU(2).  \label{eqn:6-4}
\end{equation}

Besides, the principal spin bundle $\mathbf{Spin}(S^3)$ is the Lie group $Spin(4)$, whose projection from the total space to the base space is

\begin{equation}
\mathbf{Spin}(S^3)=Spin(4)\ni g\mapsto pq^{-1}\in S^3.  \label{eqn:6-5}
\end{equation}

This principal spin bundle induces the spin $\frac{m}{2}$ bundle $\mathbf{S}_m$ as a homogeneous vector bundle:

\begin{equation}
\mathbf{S}_m:=Spin(4)\times_{\rho_m} V_m. \label{eqn:6-6}
\end{equation}

Hence the space of sections $L^2(S^3, \mathbf{S}_m)$ is a representation space of $Spin(4)$.

Now, we trivialize the vector bundle $\mathbf{S}_m$ as follows:

\begin{equation}
   \mathbf{S}_m=Spin(4)\times_{\rho_m}
               V_m \ni [g, v] \mapsto (pq^{-1},
                         \rho_m(p)v)\in S^3\times V_m.
                            \label{eqn:6-7}
   \end{equation}

So the sections of $\mathbf{S}_m$ are represented as the $\mathbf{C}^{m+1}$-valued or the $V_m$-valued functions on $S^3$.
In this situation, we can present explicit formulas of the higher spin Dirac operators on $S^3$, where the operators acts on the $V_m$-valued functions.

\begin{prop}\label{prop:6-1}
For the trivialization \eqref{eqn:6-7}, the higher spin Dirac operators on $S^3$ are the following:

\begin{align}
D_m^0&=\frac{m(m+2)}{2}+\sum \rho_{m}^0(e_i)Z_i, \label{eqn:6-8} \\
D^{\pm}_m&=\sum \rho_{m}^{\pm}(e_i)Z_i.  \label{eqn:6-9}
\end{align}

Here $Z_i$ is the right invariant vector field on $S^3=SU(2)$ corresponding to $\sigma_i$ in $\mathfrak{su}(2)$, which is given by

\begin{gather}
  Z_1 = -x_1\frac{\partial}{\partial x_4}
             +x_4\frac{\partial}{\partial x_1}
                  -x_3\frac{\partial}{\partial x_2}
                     +x_2\frac{\partial}{\partial x_3},\nonumber \\

     Z_2 = -x_2\frac{\partial}{\partial x_4}
               +x_3\frac{\partial}{\partial x_1}
                    +x_4\frac{\partial}{\partial x_2}
                          -x_1\frac{\partial}{\partial x_3},
   \label{eqn:6-10}  \\

   Z_3 = -x_3\frac{\partial}{\partial x_4}
            -x_2\frac{\partial}{\partial x_1}
                  +x_1\frac{\partial}{\partial x_2}
                      +x_4\frac{\partial}{\partial x_3}.\nonumber
\end{gather}
\end{prop}

\begin{cor}\label{cor:6-2}
The Laplace type operators $\Delta_m$ and $\widetilde{\Delta}_m$ are represented by

\begin{align}
\Delta_m &= -m^2\sum Z_i^2+m^2D_m^0-\frac{m^2(m+2)(m-2)}{4},  \label{eqn:6-11}\\
\widetilde{\Delta}_m &= -(m+2)^2\sum Z_i^2+(m+2)^2D_m^0-\frac{m(m+2)^2(m+4)}{4}. \label{eqn:6-12}
\end{align}
\end{cor}

Since the higher spin Dirac operators on $S^3$ are homogeneous differential operators, the eigenspaces are representation spaces of $Spin(4)$. So we have to decompose $L^2(S^3,\mathbf{S}_m)$ into its irreducible components. By the Frobenius reciprocity, we have the following lemma.

\begin{lem}\label{lem:6-3}
The representation space $L^2(S^3,\mathbf{S}_{m})$ decomposes into its irreducible components as follows:
\begin{enumerate}
\item (the case of $m=2p+1$)
\begin{equation}
L^2(S^3,\mathbf{S}_{2p+1})  \simeq \bigoplus_{0\le s \le p \atop k \ge p-s} E_{k,k+2s+1}\oplus E_{k+2s+1, k}. \label{eqn:6-13}
\end{equation}
\item  (the case of $m=2p$)
\begin{equation}
L^2(S^3,\mathbf{S}_{2p})  \simeq \bigoplus_{1\le s \le p \atop k \ge p-s} E_{k,k+2s}\oplus E_{k+2s, k}\bigoplus_{k\ge p}E_{k,k}.\label{eqn:6-14}
\end{equation}
\end{enumerate}
Here $E_{k,l}$ is the representation space for the exterior tensor product representation $\rho_k\widehat{\otimes}\rho_l$ of $Spin(4)=SU(2)\times SU(2)$ and $\dim E_{k,l}=(k+1)(l+1)$.
\end{lem}
We calculate the action of the higher spin Dirac operators on $E_{k,l}$ by the method given in \cite{H}. Then we have the following propositions.

\begin{prop}\label{thm:6-4}
\begin{enumerate}
\item
The eigenvalues of the self adjoint operator $D_{m}^0$ on $S^3$ are given as follows: \begin{enumerate}
\item (the case of $m=2p+1$)

\begin{equation}
\begin{cases}
(2s+1)(k+\frac{2s+3}{2}) &  \mbox{on $E_{k+2s+1,k}$}, \\
-(2s+1)(k+\frac{2s+3}{2}) & \mbox{on $E_{k,k+2s+1}$}.
\end{cases}\label{eqn:6-15}
\end{equation}
In particular, $\ker D_{2p+1}^0$ is zero for each $p$.

\item (the case of $m=2p$)
\begin{equation}
\begin{cases}
2s(k+s+1) &  \mbox{on $E_{k+2s,k}$}, \\
-2s(k+s+1) & \mbox{on $E_{k,k+2s}$}, \\
0 & \mbox{on $E_{k,k}$}.
\end{cases}\label{eqn:6-16}
\end{equation}
\end{enumerate}
\item
The eigenvalues of the second order operator $D^{+}_{m-2}D_m^-$ on $S^3$ are given as follows:
\begin{enumerate}
\item (the case of $m=2p+1$)
\begin{multline}
4(p-s)(k+1-(p-s))(p+s+1)(k+p+s+2)\\
 \mbox{on $E_{k+2s+1,k}$ or $E_{k,k+2s+1}$}.
\end{multline}
\item (the case of $m=2p$)
\begin{multline}
4(p-s)(k+1-(p-s))(p+s)(k+p+s+1)\\
 \mbox{on $E_{k+2s,k}$ or $E_{k,k+2s}$}.
\end{multline}
\end{enumerate}
\item
The eigenvalues of the second order elliptic operator $D^{-}_{m+2}D_m^+$ on $S^3$ are given as follows:
\begin{enumerate}
\item (the case of $m=2p+1$)
\begin{multline}
4(p-s+1)(k-(p-s))(p+s+2)(k+p+s+3) \\
    \quad \mbox{on $E_{k+2s+1,k}$ or $E_{k,k+2s+1}$}.
\end{multline}
\item (the case of $m=2p$)
\begin{multline}
4(p-s+1)(k-(p-s))(p+s+1)(k+p+s+2)\\
  \mbox{on $E_{k+2s,k}$ or $E_{k,k+2s}$}.
\end{multline}
\end{enumerate}
In particular,
\begin{equation}
\dim \ker D_{m}^+=\frac{1}{6}(m+1)(m+2)(m+3).
\end{equation}
\end{enumerate}
\end{prop}
\begin{prop}
\begin{enumerate}
\item
The eigenvalues of the Laplace type operator $\Delta_m$ are as follows:
\begin{enumerate}
\item (the case of $m=2p+1$)
\begin{multline}
(2s+1)^2(k+\frac{2s+3}{2})^2\\
+4(p-s)(k+1-(p-s))(p+s+1)(k+p+s+2)\\
 \quad \mbox{on $E_{k+2s+1,k}$ or $E_{k,k+2s+1}$}.
\end{multline}
\item (the case of $m=2p$)
\begin{multline}
(2s)^2(k+s+1)^2\\
 +4(p-s)(k+1-(p-s))(p+s)(k+p+s+1)\\
  \mbox{on $E_{k+2s,k}$ or $E_{k,k+2s}$}.
\end{multline}
\end{enumerate}
In particular, $\ker \Delta_{m}$ is zero.
\item The eigenvalues of the Laplace type operator $\widetilde{\Delta}_m$ are given as follows:
\begin{enumerate}
\item (the case of $m=2p+1$)
\begin{multline}
(2s+1)^2(k+\frac{2s+3}{2})^2 \\
  +4(p-s+1)(k-(p-s))(p+s+2)(k+p+s+3) \\
 \quad \mbox{on $E_{k+2s+1,k}$ or $E_{k,k+2s+1}$}.
 \end{multline}
In particular, $\ker \widetilde{\Delta}_{2p+1}$ is zero.
\item (the case of $m=2p$)
\begin{multline}
(2s)^2(k+s+1)^2  \\
    +4(p-s+1)(k-(p-s))(p+s+1)(k+p+s+2)   \\
 \quad  \mbox{on $E_{k+2s,k}$ or $E_{k,k+2s}$}.
\end{multline}
In particular, $\dim \ker \widetilde{\Delta}_{2p}=(p+1)^2$.
\end{enumerate}
\end{enumerate}
\end{prop}
\begin{rem}
For the $3$-dimensional flat manifold, that is, $T^3=S^1\times S^1\times S^1$, we can easily calculate the dimensions of the kernels for the higher spin Dirac operators:
\begin{gather}
\dim \ker D_m^+=\dim \ker \Delta_m=\dim \ker \widetilde{\Delta}_m=m+1, \\
\dim \ker D_{2p+1}^0=2p+2.
\end{gather}
\end{rem}

\section*{Acknowledgements}
The author would like to thank T. Kori for his encouragement.

config: arxiv


TITLE: $\gcd(ac,b) \geq \gcd(b,c)$
QUESTION [1 upvotes]: How could one prove that $ \gcd(ac,b) \geq \gcd(c,b) $?
My try:-  If we multiply $c$ by $a$, gcd(a,b) would multiply the gcd(c,b) to give gcd(ac,b).
But can we think of this in more vigorous mathematical terms?
I mean how would it be if we go like the smallest positive number in the set $S = \{cx+by\mid x,y\in \Bbb Z\}$ is never greater than the smallest positive number in the set $P = \{acx+by\mid x,y\in \Bbb Z\}$

REPLY [2 votes]: We could go the way you suggested : if you were to say that $\gcd(c,b) = \min\{cx+by : x,y \in \mathbb Z\}$ then call the set $\{cx+by : x,y \in \mathbb Z\} \cap \mathbb N$ as $P$ and $\{acx + by : x,y \in \mathbb Z\} \cap \mathbb N$ as $Q$. We would have $\gcd(c,b) = \min P$ and $\gcd(ac,b) = \min Q$.
If we show that $Q \subset P$, we are done? For then the smallest element of $Q$ (which exists as $Q$ is a subset of $\mathbb N$) is also in $P$ so is bigger than the smallest element of $P$.
So show that $Q \subset P$. Your hint : If $x \in \mathbb Z$ and $a \in \mathbb Z$ then $ax \in \mathbb Z$.


TITLE: Understanding of Kinetic, Potential and Change in Energy
QUESTION [1 upvotes]: From what I understand, energy is defined as the ability to do work. What does this actually mean and how is it different from work itself?
Also, according to my teacher, absolute energy does not exist. Why is that the case?
If absolute energy does not exist, and if change in energy is the difference between the energy of an object at two points in time, then how can change in energy be defined, if it is in terms of absolute energies?
Why are terms such as "kinetic energy "and "potential energy" used sometimes instead of "change in kinetic energy" and "change in potential energy" when describing energy if absolute energy does not exist?
Additionally, what is the specific difference between potential and kinetic energy? I understand that the former is dependent on position, while the latter is dependent on velocity. However, I would like a more in depth explanation because I don't understand why the change in potential energy is equal to the change in kinetic energy of an object over two points in time.

REPLY [2 votes]: You've asked a bunch of questions, albeit all related, so I will just try to break them down.

From what I understand, energy is defined as the ability to do work

Essentially yes. But the term capacity is generally used instead of ability.

What does this actually mean and how is it different from work itself?

Work is the transfer of energy from one thing to another. When that transfer occurs one thing loses energy and the other things gains energy. The other means of energy transfer is heat.

Also, according to my teacher, absolute energy does not exist. Why is
that the case?

Generally yes, energy is a relative thing. However a zero (or near zero) energy is often assigned for reference purposes.
For example, at absolute zero temperature a thermodynamic system is generally considered to have its lowest energy as the motion of atoms and molecules of a substance comes to a stop. For an ideal gas whose internal energy is strictly kinetic, one can then assign an internal energy of zero at absolute zero temperature.
Another example is assigning a gravitational potential of zero at an infinite distance away from a gravitating body.
In steam tables for water the triple point at which solid, liquid and gas phases can all exist together is used as the zero datum for specific enthalpy, entropy, and internal energy. Then all the absolute values of these quantities given in the steam table are understood to be with respect to the triple point.
As a practical matter it is generally only the difference in energy between states that is needed to determine things like work.

If absolute energy does not exist, and if change in energy is the
difference between the energy of an object at two points in time, then
how can change in energy be defined, if it is in terms of absolute
energies?

The change in energy can be defined by assigning a state as having zero energy, as I described above. For example, the internal energy of an ideal gas is kinetic only. By assigning a value of internal energy of zero at absolute zero temperature per the equation
$$U=nC_{v}T$$
One can determine changes in internal energy between two equilibrium states by the equation.
$$\Delta U_{1-2}=nC_{V}(T_{2}-T_{1})$$

Why are terms such as "kinetic energy "and "potential energy" used
sometimes instead of "change in kinetic energy" and "change in
potential energy" when describing energy if absolute energy does not
exist?

Because whenever the term "kinetic energy" and "potential energy" is used, it has to be used with respect to a specific frame of reference.
For example, if we say a car of mass $m$ having velocity $v$ has a kinetic energy of $\frac{mv^2}{2}$ we must say what is the reference frame that $v$ is measured with respect to. Generally, that reference frame would be the road. But if two such cars were moving together side by side at the same velocity with respect to the road, the kinetic energy of each car with respect to the reference frame of the other car would be zero.
Similarly, for gravitational potential energy. If an object of mass $m$ rests on my desk a height $h$ from the floor supporting my desk, that object has a gravitational potential energy of $mgh$ with respect to the floor. But If my desk is on the second floor of my house and that floor is a height $H$ from the ground, the object's gravitational potential energy with respect to the ground is $mgH$. Once again, the energy only has meaning with respect to a particular reference frame.

Additionally, what is the specific difference between potential and
kinetic energy? I understand that the former is dependent on position,
while the latter is dependent on velocity.

Your understanding is correct. Potential energy is energy of position and kinetic energy is energy of motion.

However, I would like a more in depth explanation because I don't
understand why the change in potential energy is equal to the change
in kinetic energy of an object over two points in time.

The change in potential energy is not always equal to the change in kinetic energy between two points in time. You are probably thinking about a falling body in a gravitational field where the loss of gravitational potential energy of an object between two vertical points equals the increase in the kinetic energy of the object (neglecting air resistance).
But say I apply a net horizontal force $F$ on an object of mass $m$ initially at rest with respect to my desk over a distance $d$ the net work done is $Fd$ and the change in kinetic energy is $\frac{mv^2}{2}$ where $v$ is the velocity of object after moving a distance $d$. But since the motion of the object was horizontal, there is no change in its gravitational potential energy with respect to the floor, or the ground outside.
Hope this helps.


TITLE: Solve the Distribution equation $xT= 1$
QUESTION [2 upvotes]: My Problem is to find all solutions of the distribution equation $xT =1$.
I don't know how to solve it. The solutions should be   $pv\frac1x+c$ , $c \in \mathbb{R}$.
My idea was to create a function like for: Let $T$ a Distribution. $xT=0$ then $T=c\delta$.
But I cant construct a function for my problem and I´m not sure if it´s the right way.

REPLY [1 votes]: Distributions need something to act on, so let $\phi$ be a test function. Denote the action of $T$ on $\phi$ by $\langle T,\phi \rangle$.
Functions act on test functions by integration, so that $\langle 1,\phi \rangle = \displaystyle \int_{\mathbb R} 1 \cdot \phi(x) \, dx$.
The distribution $xT$ is defined via the rule $\langle xT,\phi \rangle = \langle T,x\phi \rangle$.
The question then becomes, what distribution $T$ satisfies
$$\langle T,x\phi \rangle = \int_{\mathbb R} \phi(x) \, dx ?$$
You can get $x \phi$ in the integrand via
$$\int_{\mathbb R} \phi(x) \, dx = \lim_{\epsilon \to 0^+} \int_{\mathbb R \setminus [-\epsilon,\epsilon]} \phi(x) \, dx = \lim_{\epsilon \to 0^+} \int_{\mathbb R \setminus [-\epsilon,\epsilon]} \frac{x\phi(x)}{x} \, dx.$$
The principal value distribution $\mathrm{pv} \frac 1x$ is defined via the formula
$$\langle \mathrm{pv} \frac 1x,\phi \rangle = \lim_{\epsilon \to 0^+} \int_{\mathbb R \setminus [-\epsilon,\epsilon]} \frac{\phi(x)}{x} \, dx$$
so that $$ \int_{\mathbb R} \phi(x) \, dx = \langle \mathrm{pv} \frac 1x,x\phi \rangle.$$
Thus $\mathrm{pv} \frac 1x$ satisfies $xT = 1$.
If $T$ is an arbitrary distribution with $xT = 1$ it follows that $x(T - \mathrm{pv} \frac 1x) = 0$, implying that $T - \mathrm{pv} \frac 1x = c \delta$ for some $c$, where $\delta$ is the delta distribution.


TITLE: Is $(\prod R_i)\circ (\prod R_i)$ the same as $\prod R_i\circ R_i$?
QUESTION [0 upvotes]: For each $i\in I$, $R_i$ is a (binary) relation (on a set $X_i$). Is $(\prod R_i)\circ (\prod R_i)$ the same as $\prod R_i\circ R_i$ as relations on $\prod_{i\in I}X_i$?

REPLY [1 votes]: This is quite easy to see from the definitions, and I think you'll be better off putting it together yourself than having a full answer provided.  I expect you know the definitions involved, but just in case you're unclear on anything:
Two elements $(x_i)_{i\in I}$ and $(y_i)_{i\in I}$ of $\prod_{i\in I} X_i$ are related by $\prod_{i\in I} \rho_i$, where each $\rho_i$ is a relation on $X_i$, if $x_i\, \rho_i\, y_i$ for all $i\in I$.
If $\rho$ and $\sigma$ are relations on a set $X$, then $x,y\in X$ are related by $\rho\circ \sigma$ if there is some $z\in X$ such that $x\,\rho\, z\,\sigma\, y$.
Now mix the two definitions together in two different ways, and check whether you get the same thing both times.


TITLE: Iteration matrix and convergence
QUESTION [1 upvotes]: Assuming $G_{JA}$ is an iteration matrix for Jacobi Algorithm:
$G_{JA}(A) = I -D^{-1}A$
$D$ is the diagonal of $A$.
The sufficient condition for convergence is spectral radius less than one($\rho(G) <1$). Now, what happen if the spectral radius is equal to 1? Is there a way to set the parameter(such as initial guess) that guarantee convergence  in this case?
I would appreciate any guidance on this matter.
Thanks in advance.

REPLY [3 votes]: $\rho(G_{JA}) < 1$ is a necessary (and sufficient) condition for the convergence.
With your notation for the iteration matrix (G), we can write a stationary, first order, linear iterative method as:
$$
\mathbf{x}^{k+1} = G \mathbf{x}^k + \mathbf{f}
$$
For an iterative method, (as Jacobi is) the theorem is:

Theorem
  An iterative method is convergent $\iff$  the spectral radius of the iterative matrix $\rho(G) < 1$

I will prove only the forward implication.
Suppose for contradiction that $\rho(G) \geq 1$, i.e. $\exists$ an eigenvalue $\lambda$ of G with $|\lambda| \geq 1$; let $\mathbf{v}$ be the corresponding
eigenvector. We choose as starting point
$$\mathbf{x}^0 = \mathbf{x} + \mathbf{v}$$
With $\mathbf{x}$ the system solution. So the error is $\mathbf{e}^0 = \mathbf{v}$.
Thus:
$$
G\mathbf{e}^0 = \lambda \mathbf{e}^0
\Rightarrow
\mathbf{e}^k = G^k \mathbf{e}^0 = \lambda^k\mathbf{e}^0
$$
From the norm properties,
$||\mathbf{e}^k|| = \lambda^k \cdot ||\mathbf{e}^0||$,
but by hypothesis $|\lambda| \geq 1$, so $\mathbf{e}^k$ cannot converge to zero. So if $\mathbf{x}^k \longrightarrow \mathbf{x}$ is necessary that $\rho(G) < 1$.

Note that the convergence is global and doesn't depend on the initial $\mathbf{x}^0$.


TITLE: Finding a cycle of fixed length
QUESTION [23 upvotes]: Is there any result about the time complexity of finding a cycle of fixed length $k$ in a general graph?
All I know is that Alon, Yuster and Zwick use a technique called "color-coding",
which has a running time of $O(M(n))$, where $n$ is the number of vertices of the input graph and $M(n)$ is the time required to multiply two $n \times n$
matrices.
Is there any better result?

REPLY [1 votes]: Basically everyone above hit the nail on its head. I want to add an algorithm of counting every cycles of length 4 in an undirected graph, which runs in $O(n^3)$ instead of $O(n^4)$
Every cycle of length four includes four nodes $a$, $b$, $c$ and $d$. Thus for every pair of nodes, $u$ and $v$, you count the number of nodes that are neighbors to both $u$ and $v$. Then the problem of counting every cycle of length four transforms to the problem of selecting 2 such nodes that are adjacent to both $u$ and $v$.