text:
\begin{document}

\title{Ansatz for $(-1)^{n-1}\nabla p_n$}

\author[1]{Soumya D Banerjee}
\author[2]{Mahir Bilen Can}
\author[3]{Adriano Garsia}
\affil[1]{{\small Tulane University; sbanerjee@tulane.edu}}
\affil[2]{{\small Tulane University, mahirbilencan@gmail.com}}
\affil[3]{{\small UCSD; garsiaadriano@gmail.com}}

\normalsize

\date{\today}
\maketitle

\begin{abstract}

We construct a special family of equivariant coherent sheaves on the Hilbert scheme on $n$-points in the affine plane.
The equivariant Euler characteristic of these sheaves are closely related to the symmetic functions $(-1)^{n-1} \nabla p_n$.
We prove a higher cohomology vanishing result of these sheaves.
It follows from the Bridgeland-King-Reid correspondence that there is an effective $S_n$ module underlying the aforementioned family of symmetric functions.

\vspace{.2cm}

\noindent
\textbf{Keywords:} Hilbert scheme of points, Atiyah-Bott-Lefschetz formula, nabla operator, symmetric functions,
Macdonald polynomials\\
\noindent
\textbf{MSC:} {14L30, 14F99, 14C05, 05E05}
\end{abstract}

\maketitle

\section{Introduction}

Let $S_n$ denote the symmetric group of permutations of the set $\{1,\dots, n\}$,
and let $\mathbf{x} \cup \mathbf{y}$ denote the set of $2n$ algebraically
independent variables defined by
$$
\mathbf{x}:=\{x_1,\dots, x_n\}\ \text{ and }\  \mathbf{y}:=\{y_1,\dots, y_n\}.
$$
We consider the diagonal action of $S_n$
on $\C[\mathbf{x},\mathbf{y}]:=\C[x_1,\dots, x_n,y_1,\dots, y_n]$ given by
\[
\sigma\cdot x_i = x_{\sigma(i)} \text{ and } \sigma\cdot y_i = y_{\sigma(i)}\qquad
\text{for $i=1,\dots, n$ and $\sigma \in S_n$.}
\]
Let $I_+$ denote the ideal generated by the $S_n$-invariant homogeneous polynomials of positive degree
in $\C[\mathbf{x},\mathbf{y}]$. The {\em ring of diagonal coinvariants} is the quotient ring
$$
R_n :=\C[\mathbf{x},\mathbf{y}] / I_+.
$$
Clearly, $R_n$ has the structure of an infinite dimensional graded $S_n$-module,
whose grading is induced by the bi-degrees of the elements of $\C[\mathbf{x},\mathbf{y}]$.
It is known for sometime that the corresponding bi-graded Frobenius
character of $R_n$ is given by the Bergeron-Garsia's ``nabla'' operator, $\nabla$
applied to the $n$-th elementary symmetric function, $e_n$.
Roughly speaking, $\nabla$ is a $\mbb{Q}$-linear operator
defined on the ring of symmetric functions, denoted by $\Lambda$, such that the
`modified Macdonald symmetric functions' are eigenfunctions with prescribed eigenvalues
(see Section~\ref{S:Prelim} for details).
We say that
a symmetric function $f \in \Lambda$ has a representation if
$f$ is the Frobenius character of an $S_n$-module $F$.
In particular, $\nabla e_n$ has the representation $F=R_n$.
It is a natural question to ask if $\nabla f$ has a representation
for any other well known symmetric function.
In general, for arbitrary $f\in \Lambda$, this question is difficult to answer.
A prerequisite to have a positive answer is that $f$ is {\em Schur-positive},
that is to say, the expansion of $f$ in the Schur
basis for $\Lambda$ has nonnegative integral coefficients.
We known from the work of Loehr and Warrington in~\cite{LoehrWarrington} that
the symmetric functions $(-1)^{n-1}\nabla p_{n}$ ($n\in \N$) are Schur-positive.
The purpose of our article is to construct, in an algebro-geometric way,
a representation for each
element of the set $\{(-1)^{n-1}\nabla p_{n}:\ n\in \N\}$.
We will use the techniques that are developed by Mark Haiman to prove
the $n!$ conjecture (see~\cite{HaimanSurvey} for an overview).
\\

The fundamental geometric object in this story is the
Hilbert scheme of $n$ points in the complex affine plane,
which is denoted by $H_{n}$.
It is a (fine) moduli space, parametrizing the ideals of colength $n$ in the coordinate ring of $\C^2$
The closed points of the scheme $H_{n}$ are given by
\begin{align}\label{A:variety}
H_n(\C) := \lbrace  I:\; I \subset \C[x,y] \text{ is an ideal such that } \dim_{\C} \C[x,y]/I = n  \rbrace.
\end{align}
Although it is not completely obvious from (\ref{A:variety}),
$H_{n}$ turns out to be a smooth irreducible $2n$-dimensional algebraic variety over $\C$.
To set-up our notation and to further motivate our discussion,
we will briefly review some important categories attached to $H_n$.

The Hilbert-Chow map,
\begin{align} \label{E:HilbertChow}
\sigma: H_n \rightarrow \C^{2n}/S_{n}
\end{align}
describes $H_{n}$ as a relative projective scheme over $\C^{2n}/S_{n}$,
and moreover, (\ref{E:HilbertChow}) turns out to be a crepant resolution of the isolated singularity.
Consequently, the Bridgeland-King-Reid theorem provides a categorical equivalence,
\begin{align}\label{E:catequivalence}
\Phi: \mathsf{D}^{b}(H_{n}) \longrightarrow \mathsf{D}^{b}_{S_{n}}(\C^{2n}),
\end{align}
between the bounded derived category of $S_{n}$-equivariant sheaves on $\C^{2n}$
and the bounded derived category of coherent sheaves on $H_{n}$.
Let $T$ denote the algebraic torus $\C^*\times \C^*$ with coordinates $(q,t)$.
The coordinatewise multiplication action of $T$ on $\C^2$ gives an action $T\times H_n\rightarrow H_n$.
Along with (\ref{E:catequivalence}), this torus action induces an isomorphism, also denoted by $\Phi$,
between the equivariant Grothendieck groups,
\begin{align} \label{E:ktheoryisom}
\Phi: K^{0}_{T}(H_{n}) \cong K^{0}_{S_{n \times T}}(\C^{2n}).
\end{align}
By (\ref{E:ktheoryisom}), the questions about $S_{n}$-modules
translate into geometric questions about the equivariant sheaves on  $H_{n}$.
More precisely, the $T$-equivariant Euler-characteristic of a sheaf
$\mcal{F}$, computed with the canonical polarization coming from (\ref{E:HilbertChow}),
equals the Frobenius character of the $S_{n}$-representation $\Phi(\mcal{F})$.
Indeed, $K^{0}_{S_{n \times T}}(\C^{2n})$ is freely generated,
as a $\Z[q,t,q^{-1},t^{-1}]$-module, by the free
$\C[\mathbf{x},\mathbf{y}]$-modules
$V^\lambda \otimes \C[\mathbf{x},\mathbf{y}]$,
where $\lambda$ runs over all partitions of $n$, and
$V^\lambda$ is the irreducible $S_n$-module
corresponding to $\lambda$.
Now, by applying the Euler-Frobenius characteristic
map, it is not difficult to see that
$K^{0}_{T}(H_{n})$ is isomorphic to the algebra
of symmetric functions $f$ with the property that
$f[(1-q)(1-t)Z]$ has coefficients in $\Z[q,t,q^{-1},t^{-1}]$.
Here, $(1-q)(1-t)Z$ is the symmetric function
$(1-q)(1-t) (z_1+z_2+\cdots )$ and the brackets in
$f[(1-q)(1-t)Z]$ indicates that we are applying the
``plethystic substitution.''

The  {\em isospectral Hilbert
scheme}, denoted by $X_n$, is the reduced fiber product
$X_n := H_n \times_{\C^{2n}/S_n} \C^{2n}$,
defined as in Figure~\ref{A:F1}.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[scale=.75]
\node at (-2.25,1.5) (a) {$X_n$};
\node at (2.25,1.5) (b) {$\C^{2n}$};
\node at (-2.25,-1) (c) {$H_n$};
\node at (2.25,-1) (d) {$\C^{2n}/S_n$};
\node at (-2.7,0) {};
\node at (2.6,0) {};
\node at (2.75,.35) {$\pi$};
\node at (0,-.5) {$\sigma$};
\node at (-2.75,.35) {$\rho$};
\draw[->,thick] (a) to (b);
\draw[->,thick] (a) to (c);
\draw[->,thick] (b) to (d);
\draw[->,thick] (c) to (d);
\end{tikzpicture}
\end{center}
\label{A:F1}
\caption{The isospectral Hilbert scheme}
\end{figure}

In \cite{HaimanJams}, Haiman showed that the coherent sheaf $\mcal{P} :=
\rho_{\ast}(\mcal{O}_{X_{n}})$ is a locally free sheaf of rank $n!$ on $H_{n}$.
The resulting vector bundle, called the {\em Procesi bundle}, turns out to be of fundamental importance.
In particular, the class of $\mcal{P}|_{Z_{n}}$ in $K^{0}_{T}(H_{n})$, where $Z_{n}$ is the reduced
fiber $\sigma^{-1}( \pi(0))$ in $H_{n}$, corresponds to the representation underlying $\nabla e_{n}$
 under the isomorphism  $\Phi$.

We are now ready explain the main result of our note.

We describe two $T$-equivariant coherent
sheaves $\mcal{F}$ and $\mcal{F}^{\prime}$ on $H_n$.
A priori, the first sheaf is an object of $\Der^{b}_{T}(H_{n}, \Z)$,
and the second sheaf lives in $\Der^{b}_{T}(H_{n}, \Z[1/n])$.
Roughly speaking, these sheaves are the subsheaves of some
standard vector bundles on $H_{n}$ twisted with suitable \'{e}tale local-systems coming from the torus $T$.
By using a version of the Atiyah-Bott-Lefschetz theorem,
we calculate the $T$-equivariant Euler-characteristic of these sheaves.
We show for the first sheaf that
the resulting expression is the Frobenius series of $(-1)^{n-1}n \nabla p_{n}$, and for the
second sheaf, it is $(-1)^{n-1}\nabla p_{n}$.
We recorded these facts as Theorem~\ref{t:main} in Section~\ref{S:Prelim}.

The structure of our paper is as follows. In Section~\ref{S:Prelim},
we set up notation to explain the strategy of the proof of our Theorem~\ref{t:main}.
Also in the same section, we briefly review some relevant results of Haiman, Loehr and Warrington.
The final Section~\ref{S:Proof} is devoted to the proof of the theorem.

\begin{comment}
An interesting facet of our construction is that the description of $\mcal{F}$ (resp. $\mcal{F}^\prime$) is somehwat opaque in the abelian category of sheaves. Our
experience suggests that finding an explicit description of this sheaf directly is probably difficult. However, in the
derived catgeory things simplify greatly and we are able to obtain a
very explicit description. This suggests that a better understanding of the derived category of $H_n$ might provide deeper
insights about symmetric functions.
\end{comment}

\subsubsection*{Acknowledgements}
The authors thank Eugene Gorsky for pointing out a crucial error in the first version of this paper.
The authors thank Al Vitter for very helpful discussions.

\section{Preliminaries}\label{S:Prelim}

A partition $\mu$ of $n$ is a nonincreasing sequence
of nonnegative integers $\mu=(\mu_1,\mu_2,\dots)$,
with $\sum_{i\geq 1} \mu_i = n$.
We will use the French notation to represent the diagram of
partitions as in Figure~\ref{F:F}. Therefore, there are
$\mu_1$ squares in the bottom row, there are $\mu_2$
squares in the second row, and so on.
The squares in the diagram of $\mu$
will be called the cells of the partition.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[scale=.75]

\node at (1.25,1.75) {$x$};

\draw[-,thick] (0,0) -- (0,5) -- (1,5) -- (1,4) -- (3,4) -- (3,1) -- (5,1) -- (5,0) -- (0,0);
\draw[dotted] (0.5,0) -- (.5,5);
\draw[dotted] (1,0) rectangle (1,4);
\draw[dotted] (1.5,0) rectangle (1.5,4);
\draw[dotted] (2,0) rectangle (2,4);
\draw[dotted] (2.5,0) rectangle (2.5,4);
\draw[dotted] (3,0) rectangle (3,1);
\draw[dotted] (3.5,0) rectangle (3.5,1);
\draw[dotted] (4,0) rectangle (4,1);
\draw[dotted] (4.5,0) rectangle (4.5,1);
\draw[dotted] (5,0) rectangle (5,1);

\draw[dotted] (0,0.5) rectangle (5,.5);
\draw[dotted] (0,1) rectangle (4,1);
\draw[dotted] (0,1.5) rectangle (3,1.5);

\draw[dotted] (0,2) rectangle (3,2);
\draw[dotted] (0,2.5) rectangle (3,2.5);
\draw[dotted] (0,3) rectangle (3,3);
\draw[dotted] (0,3.5) rectangle (3,3.5);
\draw[dotted] (0,4) rectangle (1,4);
\draw[dotted] (0,4.5) rectangle (1,4.5);

\end{tikzpicture}
\end{center}
\caption{The diagram of a partition in French notation.}
\label{F:F}
\end{figure}

Let $\mu$ be a partition of $n$ with parts $\mu = (\mu_1,\mu_2,\dots )$.
In this case, we write $\mu \vdash n$.
The partition that is conjugate to $\mu$ is denoted by
$\mu^\prime$.
We will use $n(\mu)$ to denote the following sum:
$$
n(\mu) = \sum_i (i-1) \mu_i.
$$
The dominance order on partitions,
denoted by $\leq$, is defined by
$$
(\mu_1,\mu_2,\dots) \leq (\lambda_1,\lambda_2,\dots) \iff
\mu_1+ \dots +\mu_i \leq \lambda_1+ \dots +\lambda_i \
\text{ for all } i \geq 1.
$$

Let $x$ be a cell in $\mu$.
The {\em arm, coarm, leg}, and the {\em coleg of $x$} are defined by
\begin{enumerate}
          \item $a(x) = $ arm: the number of cells that are directly above $x$;
          \item $a^{\prime}(x) = $ coarm: the number cells that are directly below $x$;
          \item $l(x) = $ leg: the number of cells to the right of $x$;
          \item $l^{\prime}(x) = $ coleg:  the number of cells to the left of $x$.
\end{enumerate}
For example, the arm, coarm, leg, and the coleg of $x$ in Figure~\ref{F:F}
are 4,3,3, and 2, respectively.
In this notation, we identify the cells $x$
in the partition $\mu$ with their ``cartesian coordinates''
$(a^\prime (x) , l^\prime (x))$.
We denote by $\square_{n}$
the partition with $n$ equal parts
all of which equals $n$,
$$
\square_{n} := \{ (r,s) \in \Z \times \Z \; | 0 \leq r,s \leq n-1 \}.
$$
In addition, we define
\begin{enumerate}
\item  $[t]_{n} := 1 + t + \ldots+ t^{n-1}$ and $[q]_{n} := 1 + q + \ldots+ q^{n-1}$;
\item $B_{\mu} := \sum_{x \in \mu} q^{a^{\prime}(x)}t^{l^{\prime}(x)}$;
\item $ T_{\mu} := q^{n(\mu^{\prime})}t^{n(\mu)}$;
\item $\Pi_{\mu} := \prod_{x \in \mu \setminus \{ 0,0 \}} (1 - q^{a^{\prime}(x)}t^{l^{\prime}(c)})$.
\end{enumerate}

Let $\Lambda$ denote the algebra of symmetric functions
in the variables $z=z_1,z_2,\dots $ with coefficients in $\Q(q,t)$.
Then $\Lambda$ is a graded algebra, graded by the degree of the symmetric functions,
$\Lambda = \bigoplus_{n\geq 0} \Lambda_n$.
It is well known that the elementary symmetric functions, power-sum symmetric functions,
and the Schur functions, are $\Q$-bases for $\Lambda$ (see~\cite[Chapter I]{Macdonald}).

The {\em modified Macdonald symmetric functions}, which form a $\Q(q,t)$-basis for $\Lambda$,
are uniquely determined by the following three properties:
\begin{enumerate}
\item $\widetilde{H}_\mu (z;q,t) \in \Q(q,t)
\{ s_\lambda \lbrack Z/(1-q) \rbrack :\ \mu \leq \lambda \}$,
\item $\widetilde{H}_\mu (z;q,t) \in \Q(q,t)
\{ s_\lambda \lbrack Z/(1-q) \rbrack :\ \mu' \leq \lambda \}$,
\item $\widetilde{H}_\mu \lbrack 1;q,t \rbrack =1$,
\end{enumerate}
where the brackets indicate the plethystic substitution,
$Z/(1-q)$ is the symmetric function
$\frac{1}{1-q} (z_1+z_2+\cdots )$, and
$s_\lambda(z)$ denotes a Schur function.
The Bergeron-Garsia operator on $\Lambda$ is explicitly given by setting
\[
\nabla ( \widetilde{H}_\mu(z;q,t) )
= t^{n(\mu)} q^{ n(\mu^\prime)} \widetilde{H}_\mu(z;q,t).
\]It turns out that $\nabla e_n$ is the Frobenius series
of the ring of diagonal coinvariants $R_n$ and
its expansion in the (modified) Macdonald basis is given by
\begin{align}\label{E:en}
\nabla e_n =    \sum_{\mu \vdash n} \left( \frac{ (1-q)(1-t) \Pi_{\mu} B_{\mu}}
{ \prod_{x \in \mu} (1 - t^{1+l(x)}q^{-a(x)})(1 - t^{-l(x)}q^{1+ a(x)})} \right) \widetilde{H}_{\mu}(z;q,t),
\end{align}
see~\cite[Proposition 3.5.26]{HaimanSurvey}.

In \cite{LoehrWarrington}, Loehr and Warrington obtained an expansion of the symmetric functions $(1)^{n-1} \nabla
p_{n}$ in the (modified) Macdonald basis,
\begin{align}\label{E:pn}
(-1)^{n-1}\nabla p_{n}= \sum_{\mu \vdash n} \left (\frac{(1-t^{n})(1- q^{n}) \Pi_{\mu} T_{\mu}}
{\prod_{x \in \mu}(q^{a(x)} - t^{1+ l(x)})(t^{l(x)} - q^{1+ a(x)}) } \right) \widetilde{H}_{\mu}(z;q,t).
\end{align}
By using the identities
\begin{align*}
  n(\mu) = \sum_{x \in \mu} l(x) \ \text{ and }\ n(\mu^{\prime}) = \sum_{x \in \mu^{\prime}} l(x) = \sum_{x \in \mu} a(x),
\end{align*}
let us rewrite \eqref{E:pn} in a form that is similar to \eqref{E:en};
\begin{align}\label{E:refinedpn}
(-1)^{n-1}\nabla p_{n}= \sum_{\mu \vdash n} \left (\frac{(1-t)(1- q) \Pi_{\mu} [t]_{n}[q]_{n}}
{\prod_{x \in \mu}(1 - t^{1+l(x)}q^{-a(x)})(1 - t^{-l(x)} q^{1+a(x)}) } \right) \widetilde{H}_{\mu}(z;q,t).
\end{align}

The purpose of this note is to prove the following theorem.

\begin{theorem}\label{t:main}
  There exists a coherent sheaf $\mathcal{F}$ (resp. $\mcal{F}^{\prime}$) on the Hilbert scheme $H_n$
  such that the Frobenius series of the $S_n$-module under the derived equivalence  $\Phi(\mcal{F} \otimes \mcal{P})$ (resp. $\Phi(\mcal{F}^\prime \otimes \mcal{P})$) is given by $(-1)^{n-1}n \nabla p_{n}$ (resp. $(-1)^{n-1}\nabla p_{n}$).
\end{theorem}

Let us summarize the major steps of the proof.
\begin{enumerate}

\item We present a construction of $\mcal{F}$ (resp. $\mcal{F}^\prime$).

\item By using the Atiyah-Bott-Lefschetz formula, we calculate the
equivariant Euler characteristics of the sheaf
$\mcal{F} \otimes \mcal{P}$ (and of $\mcal{F}^\prime \otimes \mcal{P}$), and
we show that it has the expected form as in (\ref{E:refinedpn}).

\item We show that the higher cohomology of $\mcal{F}$ (resp. of $\mcal{F}^\prime$) vanishes.
This proves that under the Bridgeland-King-Reid correspondence the resulting
  $S_n$-module, which is a priori a virtual representation,
is an honest $S_n$-representation.
\end{enumerate}

\subsection{Some useful facts about Hilbert scheme of points on the plane}

In this subsection, we will recall some basic facts about $H_{n}$ that we will use in the sequel.
For the detailed proofs of these facts, we refer to~\cite[Section 2]{HaimanDiscrete}.

The two-dimensional torus $T$ with coordinates $(q,t)$ acts on $\C^{2}$.
On the coordinate ring $\C[x,y]$ of $\C^2$, this action is given by $(q,t)\cdot x = qx$ and $(q,t)\cdot y = ty$.
The induced action on closed points of $H_{n}$ is given by the pull-back of ideals.
Explicitly, if $I = \langle p(x,y) \rangle$, then
$$
(q,t)\cdot I = \langle p(q^{-1}x, t^{-1}y) \rangle.
$$

We know from the Gotzmann-Grothendieck construction that $H_n$ has an imbedding into a
high dimensional Grassmann variety.
By pulling back the standard open covering of the Grassmann variety,
we get an open cover $\{U_{\mu}\}$ of $H_{n}$ indexed by the partitions of $n$.
Explicitly, the closed points of $U_{\mu}$ are given by
\begin{align}\label{E:standardopen}
U_\mu := \{ I \in H_n:\ \C[x,y]/I \ \text{admits a vector-space basis of monomials } x^{h}y^{k} \mtxt{ for } (h,k) \in
  \mu \}.
\end{align}
The coordinate ring $\mcal{O}_{U_{\mu}}$  is generated by functions  $c^{rs}_{hk}$.
On an ideal $I \in U_{\mu}$, these functions take the values
\[
x^{r}y^{s} = \sum_{(h,k) \in \mu} c^{rs}_{hk}(I) x^{h}y^{k} \mod (I)
\]
for all $r,s \geq 0$.
As a result, the $T$-action on $c^{rs}_{hk}$ is given by $(q,t) c^{rs}_{hk} = q^{r-h}t^{s-k}c^{rs}_{hk}$.
It follows from the definitions that the above open covering is stable under the $T$-action.
Furthermore, the $T$-fixed points of $H_{n}$ are precisely the monomial ideals
\[
I_\mu := \langle x^h y^k :\ (h,k) \notin \mu \rangle.
\]
The maximal ideal $\mfrk{m}_{\mu}$ at $I_{\mu}$ is described by
\[
\mfrk{m}_{\mu} = \{ c^{rs}_{hk} | \; (r,s) \notin \mu \}.
\]

Let $\pi: F_{n}\rightarrow H_n$ denote the universal family over $H_{n}$.
We set
$ \mcal{B}:= \pi_{\ast} \mcal{O}_{F_{n}}.$
The sheaf $\mcal{B}$ is locally free, and it has rank $n$.
The trace map
\[
\Tr: \mcal{B} \rarr \mcal{O}_{H_{n}}
\]
splits the canonical inclusion $ \mcal{O}_{H_{n}} \rightarrow  \mcal{B}$, and hence,
we obtain an equivariant splitting of vector bundles,
\[
\mcal{B} = \mcal{B}^{\prime} \oplus \mcal{O}_{H_{n}}.
\]

\begin{Remark}
Following Haiman, we will call $\mcal{B}$ the {\em tautological bundle} on $H_{n}$.
Note that the Hilbert-Chow map $\sigma: H_{n} \rarr \C^{2n}/S_{n}$ is a blow-up of $\C^{2n}/S_{n}$.
This provides $H_{n}$ with a canonical ample line bundle $\mcal{O}(1)$.
It turns out that $\wedge^{n} \mcal{B} = \mcal{O}(1)$ (see~\cite{HaimanDiscrete}).
\end{Remark}

Let $[0]$ denote the image of origin $0\in \C^{2n}$ under the canonical projection
$\C^{2n} \rarr \C^{2n}/S_{n}$.
The \emph{zero fiber}, denoted by $Z_{n}$ is the closed subscheme $ \sigma^{-1}([0])$.
By Brian\c{c}on's work~\cite{Briancon}, we know that $Z_{n}$ is an irreducible scheme of dimension
$n-1$. In~\cite{HaimanDiscrete}, Haiman showed that the structure sheaf $\mcal{O}_{Z_{n}}$ is $T$-equivariantly
isomorphic to a perfect complex in derived category of coherent sheaves in $H_{n}$. More precisely, he proved
the following result.
\begin{theorem}
    In the derived category of $T$-equivariant coherent sheaves on $H_{n}$, the sheaf $\mcal{O}_{Z_{n}}$ is isomorphic to
 \begin{align}\label{E:KoszulResolution}
   0 \rightarrow \mc{B} \otimes \wedge^{n+1} (\mc{B}^\prime \oplus \mc{O}_t \oplus \mc{O}_q)\rightarrow \dots \rightarrow
\mc{B} \otimes (\mc{B}^\prime \oplus \mc{O}_t \oplus \mcal{O}_q)
\rightarrow \mc{B} \rarr 0,
 \end{align}
where all tensor products are over $\mcal{O}_{H_{n}}$ and $\mcal{O}_{t}$ (resp. $\mcal{O}_{q}$) denote trivial  line bundles
 on $H_{n}$ with $T$-characters $t$ (resp. $q$).
\end{theorem}

\section{Proof of the Theorem} \label{S:Proof}
\subsection{Step 1.}

Let $p_1$ and $p_2$ denote the canonical projections as in Figure~\ref{D:label}.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[scale=.75]
\node at (-2.25,1.5) (a) {$H_n \times T$};
\node at (1.75,1.5) (b) {$T$};
\node at (-2,-1.65) (c) {$H_n$};

\node at (0,2) {$p_2$};
\node at (-2.85,0) {$p_1$};

\draw[->,thick] (a) to (b);
\draw[->,thick] (-2.25,1.1) to (-2.25,-1.1);

\end{tikzpicture}
\end{center}
\label{D:label}
\caption{}
\end{figure}

\begin{Lemma}
If $\mcal{F}$ is a quasi-coherent sheaf on $H_{n} \times T$,
then we have $R^{i}p_{\ast}(\mcal{F}) = 0 $ for all $i > 0$.
\end{Lemma}

\begin{proof}
    This follows directly from Corollary I.3.2 Chapter III of  \cite{EGA}.
\end{proof}

We will interchangeably use the same letter to denote a vector bundle and the associated sheaf of sections.
Let $\chi$ be a character of the torus $T$, and let $\mcal{O}_{\chi}$ denote the associated line bundle on $T$.
Then we define $L_\chi$ as follows;
\[
L_{\chi} := p_{1,\ast}\circ p_{2}^{\ast}(\mcal{O}_{\chi}).
\]

\begin{Proposition}
The quasi-coherent sheaf $L_{\chi}$ is a $T$-equivariant line bundle on $H_{n}$.
\end{Proposition}

\begin{proof}

First, we will show that $L_{\chi}$ is a line bundle.
But this is clear since for any open subscheme $U$ of $H_{n}$,
we have sections
\[
L_{\chi}(U) := \mcal{O}_{H_{n}}(U) \otimes_{\C} \C \cdot \chi.
\]
This shows that, indeed, $L_{\chi}$ is a line bundle.

To show that it is $T$-equivariant, we will construct an isomorphism
$\varphi: p_{1}^{\ast}L_{\chi} \rarr a^{\ast} L_{\chi}$, where $a:H_{n} \times T \rarr H_{n}$ is the action
map. We use the $T$-stable affine-covering of $\{ U_{\mu}\}$ of $H_{n}$ constructed as above.
Restricted to the open set $U_{\mu}$, we have the canonical identifications
$\varphi_{\mu}: p_{1}^{\ast}L_{\chi}(U_{\mu}) \cong a^{\ast} L_{\chi}(U_{\mu})$.
These sections agree on the intersections $U_{\mu} \cap U_{\lambda}$, and, hence,
they glue to a global isomorphism of sheaves. This proves our assertion.
\end{proof}

For a cell $(r,s)$ in the square grid $\square_{n}$, let us denote by $\chi_{r,s}$ the character
$\chi_{r,s} : T\rightarrow \C^{\ast}$ defined by
\[
\chi_{r,s} (q,t) := q^{r}t^{s}.
\]
\begin{Definition} \label{D:VandN}
Let
\[
\mathcal{V} := \bigoplus_{(r,s) \in \square_{n}} L_{\chi_{r,s}}
\]
denote a $T$-equivariant coherent sheaf on $H_{n}$.
\end{Definition}

\subsubsection{A variation: the construction of $\mcal{V}^{\prime}$.}

Let $\iota$ denote the following isogeny:
\begin{align}
\iota: T &\longrightarrow T \notag \\
(q,t) &\longmapsto (q^{n},t^{n}). \label{A:isogeny}
\end{align}
It fits to a commutative diagram as in Figure~\ref{F:2}.
Then, for a line bundle $\mcal{O}_{\chi}$ on $T$, we define
$$
L^{\prime}_{\chi} := p_{1, \ast} \circ \tau^{\ast} \mcal{O}_{\chi}.
$$

\begin{figure}[htbp]
\begin{center}
\begin{tikzpicture}[scale=.75]
\node at (-2.25,1.5) (a) {$H_n \times T$};
\node at (1.75,1.5) (b) {$T$};
\node at (1.75,-1.65) (c) {$T$};
\node at (-2.25,-1.65) (d) {$H_n$};
\node at (-2.75,0) {$p_1$};
\node at (0,2) {$p_2$};
\node at (-.5,-.5) {$\tau$};
\node at (2.2,0) {$\iota$};
\draw[->,thick] (a) to (c);
\draw[->,thick] (a) to (b);
\draw[->,thick] (b) to (c);
\draw[->,thick] (a) to (d);
\end{tikzpicture}
\end{center}
\caption{}
\label{F:2}
\end{figure}

By using the same arguments as before,
we can show that $L_{\chi}^{\prime}$ is a $T$-equivariant line bundle on $H_{n}$.
We define, analogous to $\mcal{V}$, a $T$-equivariant coherent sheaf
\begin{align*}
\mathcal{V}^{\prime} &:= \oplus_{(r,s) \in \square_{n}} L^{\prime}_{\chi_{r,s}}.
\end{align*}

\subsubsection{The descriptions of $\mcal{F}$ and $\mcal{F}^\prime$.}

Let $\mcal{K}^\bullet$ denote the complex
\begin{align*}
  0 \rightarrow\mcal{V}\otimes \mc{B} \otimes \wedge^{n+1} (\mc{B}^\prime \oplus \mc{O}_t \oplus \mc{O}_q)
  \stackrel{\partial_{n+1}}{\rightarrow} \dots \stackrel{\partial_{2}}{ \rightarrow}
\mcal{V} \otimes\mc{B} \otimes (\mc{B}^\prime \oplus \mc{O}_t \oplus \mcal{O}_q)
\stackrel{\partial_{1}}{\rightarrow} \mcal{V} \otimes \mc{B} \rarr 0.
\end{align*}
For any positive integer $j$, let $\partial^\prime_i$ denote the composition $i \circ \partial_i \circ \Tr$ in the diagram
\eqref{E:differential} below.
\begin{equation} \label{E:differential}
\begin{tikzcd}
    \mcal{V} \otimes \mcal{O}_{H_n} \otimes \wedge^{j+1} (\mc{B}^\prime \oplus \mc{O}_t \oplus \mc{O}_q)  \arrow[d,"i"]
    \arrow[r, dashed, "\partial^\prime_i"] & \mcal{V} \otimes \mcal{O}_{H_n} \otimes \wedge^{j} (\mc{B}^\prime \oplus \mc{O}_t \oplus \mc{O}_q) \\
    \mcal{V} \otimes \mcal{B} \otimes \wedge^{j+1} (\mc{B}^\prime \oplus \mc{O}_t \oplus \mc{O}_q) \arrow[r,"\partial_i"] & \mcal{V} \otimes \mcal{B} \otimes \wedge^{j} (\mc{B}^\prime \oplus \mc{O}_t \oplus \mc{O}_q) \arrow[u,"\Tr"]
\end{tikzcd}
\end{equation}
In \eqref{E:differential}, $i$ is induced by the canonical push-forward
$\mcal{O}_{H_{n}} \hookrightarrow \mcal{B}$, and $\Tr$ is induced by the
trace map $\Tr: \mcal{B} \rarr \mcal{O}_{H_n}$.
The following proposition is clear from straightforward computation.

\begin{Proposition}
We have $\partial^\prime_i \circ \partial^\prime_{i-1} = 0$.
By putting the maps $\partial^\prime_i$ together,
we get a bounded  complex $\mcal{S}^\bullet$ given by
  \[
  \mcal{S}^\bullet \defeq  0 \rightarrow\mcal{V}\otimes \mc{O}_{H_n} \otimes \wedge^{n+1} (\mc{B}^\prime \oplus \mc{O}_t \oplus \mc{O}_q)
\rarr \dots \rarr \mcal{V} \otimes\mc{O}_{H_n} \otimes (\mc{B}^\prime \oplus \mc{O}_t \oplus \mcal{O}_q) \rarr \mcal{V} \otimes \mc{O}_{H_n} \rarr 0.
\]
Moreover, we have the chain maps $i: \mcal{S}^\bullet \rarr \mcal{K}^\bullet$ and $\Tr: \mcal{K}^\bullet \rarr
\mcal{S}^\bullet$.  Finally, $\Tr \circ i $ is the identity map on $\mcal{S}^\bullet$.
\end{Proposition}

In fact, the following is true.
\begin{Proposition}
  In the bounded derived category of equivariant sheaves, the complex $\mcal{S}^\bullet$ is a direct summand of
  $\mcal{K}^\bullet$.
\end{Proposition}

\begin{proof}
This follows from Lemma 1.2.8 of \cite{ANeeman} and the previous proposition.
\end{proof}

The complex $\mcal{K}^\bullet$ is exact except at degree zero.
Consequently, the complex $\mcal{S}^\bullet$ is also exact except possibly at degree zero.
In fact, from the description of the resolution \eqref{E:KoszulResolution},
it follows that $\mcal{S}^\bullet$ is not exact at degree zero
(see the discussion after Proposition 2.2 in~\cite{HaimanInven}.)

\begin{Definition}
Let $\mcal{F}$ denote the image of $\mcal{S}^\bullet$ inside $\mcal{O}_{Z_n} \otimes  \mcal{V}$
under the quasi-isomorphism $\mcal{K}^\bullet \rarr \mcal{O}_{Z_n} \otimes \mcal{V}$.
In the above construction, if $\mcal{V}^\prime$ is used instead of $\mcal{V}$,
then let $\mcal{F}^\prime$ denote the corresponding image.
\end{Definition}

\subsection{Step 2.}

In this subsection, we will calculate the equivariant Euler characteristic of the sheaf $\mcal{F} \otimes \mcal{P}$
(resp. $\mcal{F}^{\prime} \otimes \mcal{P}$) by using an appropriate version of the Atiyah-Bott-Lefschetz
formula.

We know from~\cite[(3.1)]{HaimanDiscrete} that
\begin{align}\label{A:H's formula}
\sum_{i=0}^{n} (-1)^{i} \Tr_{H^{i}(H_{n}, \mcal{E})} (\tau) = \sum_{\mu \vdash n} \frac{\sum_{i=0}^{n} (-1)^{i}
  \Tr_{\Tor_{i}(k(I_{\mu}), \mcal{E})}(\tau)}{\det_{C(I_{\mu})}(1- \tau)},
\end{align}
where $\mcal{E}$ is any bounded complex of coherent sheaves,
$\tau \in T$, $k(I_\mu)$ is the residue field and $C(I_{\mu})$ is the tangent space the $T$-fixed point $I_{\mu}$.
We will evaluate the right hand side of (\ref{A:H's formula}) for the sheaf $\mcal{E} = \mcal{F} \otimes \mcal{P}$.
Note that the terms in the denominator, namely $\det_{C(I_{\mu})}(1 - \tau)$, are well known from the
work of Ellingsrud and Str\o mme (see~\cite{Ellingsrud}).
So, we will calculate the numerator.
The tor-modules $\Tor_i(k(I_\mu),\mcal{F} \otimes \mcal{P})$ are obtained as the
homology of the complex $\mcal{S}^\bullet\otimes \mcal{P} \otimes k(I_\mu)$.
By using the multiplicative property of the trace map with respect to tensor products,
we obtain
\begin{align}\label{E:CalculatingTraces}
\sum_{i} (-1)^{i}\Tr_{\Tor_{i}(k(I_{\mu}), \mcal{F} \otimes \mcal{P})}(\tau) =
\Tr_{k(I_{\mu})}(\tau) \cdot \Tr_{\mcal{P}_{I_\mu}}(\tau) \cdot \Tr_{\mcal{V}_{I_\mu}}(\tau) \cdot \left( \sum_{i}(-1)^{i}
  \Tr_{\wedge^{i}(\mcal{B}^{\prime} \oplus \mcal{O}_{t} \oplus \mcal{O}_{q})|_{I_{\mu}}} \right )
\end{align}

Next, we look at the explicit formulation of the right hand side of (\ref{E:CalculatingTraces})
for the specialization $\tau = (q,t)$.

\begin{itemize}
\item It follows from the descriptions of the coordinate rings of the affine open sets $U_{\mu}$,
and the $T$-actions on their coordinates as in (\ref{E:standardopen}),
that $\Tr_{k(I_{\mu})}(q,t) = n$.

\item By~\cite[Proposition 5.4.1]{HaimanSurvey}, we know that
$\Tr_{\mathcal{P}_{I_\mu}}(q,t)$ is the modified Macdonald polynomial.

\item
It follows immediately from the construction of $\mathcal{V}$ that
\[
\mathsf{tr}_{\mathcal{V}_{I_\mu}}(q,t) = [t]_{n}[q]_{n}.
\]

\item
The final term involving the alternating sum of traces equals
$(1-t)(1-q) \prod_{\mu}$ (see~\cite[Theorem 2]{HaimanDiscrete}).
\end{itemize}

Now, by putting these identities together on the right hand side of (\ref{E:CalculatingTraces}),
we see that the equivariant Euler characteristic of $\mcal{F}$ equals $n$ times the
Frobenius characteristic of $(-1)^{n-1} \nabla p_{n}$.

\begin{Remark}
If we use the sheaf $\mathcal{F}^{\prime}$ instead of $\mcal{F}$, then
on the right-hand side of equation (\ref{E:CalculatingTraces}) we get
$\mathsf{tr}_{\mathcal{V}_{I_\mu}^{\prime}}(q,t)$ instead of $\Tr_{\mcal{V}}(q,t)$.
This contributes the factor $[t]_{n}[q]_{n}/n$.
\end{Remark}

\subsection{Step 3.}

In the previous subsection, we calculated the equivariant Euler characteristic
\[\sum_{i} (-1)^{i}
\mathsf{tr}_{H^{i}(H_{n},\mcal{F} \otimes \mcal{P})}(\tau)
\]
at $\tau = (q,t)$.
In this section, we will prove a cohomology vanishing result, which will show that the summands
$\Tr_{H^i(H_n, \mcal{F}\otimes \mcal{P})}$ vanish for $i > 0$. In particular, this will imply that,
under the derived equivalence, $\Phi(\mcal{F} \otimes \mcal{P})$ is an honest $S_n$-representation
(and not just a virtual representation) with the expected Frobenius character series.

\begin{Proposition}\label{P:vanishing}
For all $i > 0$, the higher cohomology groups vanish,
\begin{align}\label{A:vanish}
H^{i}(H_{n}, \mcal{F} \otimes \mcal{P}) = 0.
\end{align}
Furthermore, the vanishing result still holds true if $\mcal{F}$ is replaced by $\mcal{F}^{\prime}$ in (\ref{A:vanish}).
\end{Proposition}

\begin{proof}
  Let $\Gamma$ denote the global section functor on $H_n$. We set $\mcal{G} \defeq \mcal{F}\otimes \mcal{P}$ and
  $\mcal{G}^\bullet \defeq \mcal{S}^\bullet \otimes^{L} \mcal{P}$. Passing to the derived category we have isomorphisms
  \begin{align}\label{E: cohomologie}
    \RR\Gamma(\mcal{G}) = \RR\Gamma(\mcal{G}^\bullet).
  \end{align}
  The terms of the complex $\mcal{G}^\bullet$ are direct summands of
  $(\mcal{P} \otimes \mcal{B}^{\otimes l})^{n^2}$.
  Since cohomology commutes with direct sums, it follows from~\cite[Theorem 2.1]{HaimanInven} that
  $\mcal{G}^\bullet$ is a complex of acyclic objects for $\Gamma$.
  Hence, we have
  \[
  \RR\Gamma(\mcal{G}) = \Gamma(\mcal{G^\bullet}).
  \]
  The complex $\Gamma(\mcal{G}^\bullet)$ has zero cohomology in positive degrees.
  So, we get $H^i(H_n, \mcal{G}) = 0$ for $i > 0$.
  Since $H^i(H_n, \mcal{G}) = 0 $ for $i < 0$, the proof of the first claim is complete.

 For the proof of the second claim, we apply the same arguments.

\end{proof}

\bibliography{ReferenceNabla}
\bibliographystyle{alpha}

\end{document}


text:
\begin{document}
\maketitle

\vspace{1.75cm}

\begin{abstract}

In this paper we study the geometry of the set of real square roots of $\pm I_2$. After some introductory remarks, we begin our study by deriving by quite elementary methods the forms of the real square roots of $\pm I_2$. We then discuss the interpretations of these square roots as transformations of the cartesian $(x,y)$-plane. To study the geometry of the set of square roots of $\pm I_2$ we consider a slightly more general  set of square matrices of order $2$ and show that these sets are hyperboloids of one sheet or hyperboloids of two sheets. From these general results we conclude that the set of involutory matrices of order 2 is a hyperboloid of one sheet and the set of skew-involutory matrices of order 2 is a hyperboloid of two sheets. The relations between the geometrical properties of the hyperboloids and the set of square roots of $I_2$ are also investigated. We then proceed to obtain the forms of the involutory matrices of order 2 by more advanced methods. We have considered two approaches: in the first approach we use the concept of a function of a matrix and in the second approach we use concepts of split-quaternions.
\end{abstract}

\newpage
\
\vspace{1in}
\tableofcontents
\newpage

\section{Introduction}
\subsection{Matrix square roots}

Let $P$ be a square matrix over a field $F$. A square matrix $R$ over $F$ such that $R^2=P$ is called a {\em square root of $P$}.

The term ``matrix" was coined by James Joseph Sylvester in 1850 and matrix algebra was developed by Arthur Cayley in his ``Memoir on the Theory of Matrices'' published in 1858 \cite{Cayley}. In this memoir, Cayley considered matrix square roots also.

An arbitrary  square matrix may have no, or a finite number of, or an infinite number of square roots.  For example,  it can be seen that:

\begin{itemize}
\item
$\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$ has {\em no} square root;
\item
$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ has exactly {\em two} square roots, namely, $\begin{bmatrix} \pm 1 &  \phantom{\pm} 0 \\ \phantom{\pm} 0 &  \phantom{\pm} 0 \end{bmatrix}$;
\item
$\begin{bmatrix} 1 & 0 \\  0 & 4 \end{bmatrix}$ has exactly {\em four} square roots, namely, $\begin{bmatrix} \pm 1 &  \phantom{\pm} 0 \\  \phantom{\pm} 0 & \pm 2\end{bmatrix}$;
\item
$\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$ has an {\em infinite number} of square roots some of which are given by $\begin{bmatrix}2 \cos \phi  & \phantom{-}2\sin\phi  \\2 \sin\phi  & - 2\cos\phi \end{bmatrix}$ for arbitrary values of $\phi$.
\end{itemize}

The matrix square root is one of the most  commonly occurring matrix functions,  and it arises in several contexts: for example in the
matrix sign function, the generalized eigenvalue problem, the polar decomposition, and the geometric mean problem. There are also a
variety of methods for computing the matrix square root (see, for example, \cite{Nigham}, \cite{Jean}). When the order of the matrix is $2$, the matrix becomes particularly simple and there are special formulas and techniques to compute the square roots of matrices of order $2$ (see, for example, \cite{Sullivan},\cite{Sam}).

\subsection{Involutory and skew-involutory matrices}

In this paper, we study the square roots of a very special and  a very interesting square matrix of order $2$, namely, the identity matrix $I_2= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ of order $2$. In the literature, square roots of the identity matrix $I$ is called an {\em involutory matrix} and the square roots of $-I$ are called {\em skew-involutory matrices}. Thus this paper is concerened with the geometry of the set of involutory and skew-involutory matrices of order 2.

In Section \ref{S2}, we first find {\em all} the square roots of  the identity matrix $I_2$  by elementary methods. In Section \ref{S3} we discuss the geometrical interpretations of these square roots as transformations of the cartesian $xy$-plane.

\section{The square roots of $I_2$: Elementary approach}\label{S2}

Let $R=\begin{bmatrix} a & b \\ c& d \end{bmatrix}$ be a square root of $I_2$. The defining condition $R^2 = I_2$ now implies the following relations:

\begin{align}
a^2 + bc & = 1 \label{Eq1}\\
(a+d)b & = 0 \label{Eq2}\\
(a+d)c & =0 \label{Eq3}\\
d^2 + bc & =1 \label{Eq4}
\end{align}

\subsubsection*{Case 1}

If $a+d\ne 0$ then from Eq.\eqref{Eq2} and  Eq.\eqref{Eq3} we get $b=0$ and $c=0$. Also from Eq.\eqref{Eq1} and Eq.\eqref{Eq4} we have $a=\pm 1$ and $d=\pm 1$. The condition $a+d\ne 0$ will be stisfied only when $a=d=1$ or $a=d=-1$. Thus, if $a+d\ne 0$, we have:
\begin{enumerate}[{\hspace{\parindent}\bf {1}(a)}]
\item
$R=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}=I_2$
\item
$ R=\begin{bmatrix}-1 & \phantom{-}0 \\ \phantom{-}0 & -1 \end{bmatrix}=-I_2$.
\end{enumerate}

\subsubsection*{Case 2}

Now assume that $a+d=0$ so that $d=-a$. Then Eq.\eqref{Eq2} and Eq.\eqref{Eq3} are obviously satisfied and the other two equations become equivalent to each other. We have to distinguish two cases:

\subsubsection*{Case 2(i)}

Let $a^2=1$ so that $bc=0$ and so $b=0$ or $c=0$. This case yields the following square roots of $I_2$:
\begin{enumerate}[{\bf {\hspace{\parindent} 2(i)}(a) }]
\item
$R= \begin{bmatrix}\phantom{-}1 & \phantom{-}b \\ \phantom{-}0 & -1 \end{bmatrix}$ (See Figure \ref{Fig1})
\item
$R=\begin{bmatrix}-1 & \phantom{-}b \\ \phantom{-}0 & \phantom{-}1 \end{bmatrix}$ (See Figure \ref{Fig2})
\item
$R=\begin{bmatrix}\phantom{-}1 & \phantom{-}0 \\\phantom{-} c & -1 \end{bmatrix}$ (See Figure \ref{Fig3})
\item
$R=\begin{bmatrix}-1 & \phantom{-}0 \\ \phantom{-}c & \phantom{-}1 \end{bmatrix}$ (See Figure \ref{Fig4})
\end{enumerate}

\subsubsection*{Case 2(ii)}

Let $a^2 \ne 1$ so that $b\ne 0$ and $c\ne 0$. Now from Eq.\eqref{Eq1} we have
$$
c=\frac{1-a^2}{b}.
$$
This yields the following general expression for square roots of $I_2$:

\begin{enumerate}[{\bf \hspace{\parindent} {2(ii)}(a)}]
\item
$R=\begin{bmatrix} a & \phantom{-}b \\ \dfrac{1-a^2}{b} & -a \end{bmatrix}.$
\end{enumerate}

This shows that the set of square roots of $I_2$ is a two-parameter family of matrices.

\section{Geometrical interpretations of the square roots of $I_2$}\label{S3}

\subsection{Matrices of order $2$ as transformations of a plane}

The two-dimensional vector space $\mathbb R^2$ over $\mathbb R$ can be geometrically realised as the cartesian $xy$-plane. If $P(x, y)$ is any point in this plane, we may represent it by the column vector ${\mathbf x} = \begin{bmatrix} x \\ y \end{bmatrix}$.

Let $T=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ be any square matrix of order $2$ over $\mathbb R$. Then the mapping $T:\mathbb R^2 \longrightarrow \mathbb R^2$ defined by $\mathbf x\mapsto  T\mathbf x$ defines the following   transformation of the $xy$-plane:

\begin{equation}\label{Trans}
(x,y)\mapsto (ax+by, cx+dy).
\end{equation}

Let $S$ be another square matrix of order $2$, then the transformation defined by the product $ST$ can be expressed as
$$
\mathbf x\mapsto  T\mathbf x \mapsto S(T \mathbf x).
$$
This represents successive applications of two transformations: first, the transformation represented by $T$ and then the transformation represented by $S$.

A transformation such as one given by Eq.\eqref{Trans} can be described in terms of elementary geometric transformations like, rotation, magnification, translation, point and line reflections, etc. In fact if $T=I_2$, the map $\mathbf x\mapsto  T\mathbf x$ is the identity map of  $\mathbb R^2$ onto itself.

\subsection{Square roots of $I_2$ as transformations of a plane}

In this section we describe the various square roots of $I_2$ in terms of these elementary geometric transformations.  As aids to visualisation we have also included diagrams illustrating the transformations. In the diagrams, the maps from $P$ to $Q$ and from $Q$ to $P$ are both given by $T$ so that $T^2$ is the map from $P$ to $P$ which is the identity mapping.

\subsubsection*{Case 1}

\begin{enumerate}[{\bf {\hspace{\parindent} 1}(a) }]
\item
As already mentioned, the square root $I_2$ of $I_2$ is the identity mapping of $\mathbb R^2$.
\item
The matrix $-I_2$ represents the point-reflection about the coordinate origin  (see Figure \ref{Fig8}).
\end{enumerate}

\begin{figure}[!h]
\begin{center}
\includegraphics{SquareRootOfIdentityMatrix08.png}
\caption{Geometrical interpretation the square root $-I_2$ of $I_2$ (Case 1(b))}\label{Fig8}
\end{center}
\end{figure}

\subsubsection*{Case 2(i)(a)}
Next we have
$$
\begin{bmatrix}\phantom{-}1 & \phantom{-}b \\ \phantom{-}0 & -1 \end{bmatrix} =
\begin{bmatrix}\phantom{-}1 & \phantom{-}0 \\ \phantom{-}0 & -1 \end{bmatrix}
\begin{bmatrix}\phantom{-}1 & \phantom{-}b \\ \phantom{-}0 & \phantom{-1}1 \end{bmatrix}
$$
The latter factor represents the map
$$
(x,y)\mapsto (x+by, y)
$$
which is a translation of the point $P(x,y)$ parallel to $x$-axis through a distance of $by$. The former factor represents the map
$$
(x,y)\mapsto (x, -y)
$$
which is a refection  in the $x$-axis. The transfromations, and their repeated applications, are shown in Figure~\ref{Fig1}.

\begin{figure}
\begin{center}
\includegraphics{SquareRootOfIdentityMatrix02.png}
\caption{Geometrical interpretation the square root  of $I_2$ (Case 2(i)a) }\label{Fig1}
\end{center}
\end{figure}

\subsubsection*{Case 2(i)(b)}
Next we have
$$
\begin{bmatrix} -1 & \phantom{-}b \\ \phantom{-}0 &  \phantom{-}1 \end{bmatrix} =
\begin{bmatrix}\phantom{-}1 & \phantom{-}b \\ \phantom{-}0 & \phantom{-}1 \end{bmatrix}
\begin{bmatrix}-1 & \phantom{-}0 \\ \phantom{-}0 &  \phantom{-}1 \end{bmatrix}
$$
The latter factor represents the map
$$
(x,y)\mapsto (-x, y)
$$
which is a refection  in the $y$-axis. The former factor represents the map

$$
(x,y)\mapsto (x+by,y)
$$

which is a translation of the point $P(x,y)$ parallel to $x$-axis through a distance of $by$.
 The transfromations, and their repeated applications, are shown in Figure~\ref{Fig2}.

\begin{figure}[!t]
\begin{center}
\includegraphics[width=12cm]{SquareRootOfIdentityMatrix06.png}
\caption{Geometrical interpretation the square root  of $I_2$ (Case 2(i)b)}\label{Fig2}
\end{center}
\end{figure}

\begin{figure}[!h]
\begin{center}
\includegraphics{SquareRootOfIdentityMatrix07.png}
\caption{Geometrical interpretation the square root  of $I_2$ (Case 2(i)c) }\label{Fig4}
\end{center}
\end{figure}

\begin{figure}[!h]
\begin{center}
\includegraphics{SquareRootOfIdentityMatrix03.png}
\caption{Geometrical interpretation the square root  of $I_2$ (Case 2(i)d) }\label{Fig3}
\end{center}
\end{figure}

\subsubsection*{Case 2(i)(c)}
Next we have
$$
\begin{bmatrix} \phantom{-}1 & \phantom{-}0 \\ \phantom{-}c &  -1 \end{bmatrix} =
\begin{bmatrix}\phantom{-}1 & \phantom{-}0 \\ \phantom{-}c & \phantom{-}1 \end{bmatrix}
\begin{bmatrix}\phantom{-}1 & \phantom{-}0 \\ \phantom{-}0 &  -1 \end{bmatrix}
$$
The latter factor represents the map
$$
(x,y)\mapsto (x, -y)
$$
which is a refection  in the $x$-axis. The former factor represents the map

$$
(x,y)\mapsto (x,cx+y)
$$

which is a translation of the point $P(x,y)$ parallel to $y$-axis through a distance of $cx$.
 The transfromations, and their repeated applications, are shown in Figure~\ref{Fig4}.

\subsubsection*{Case 2(i)(d)}
Next we have
$$
\begin{bmatrix} -1 & \phantom{-}0 \\ \phantom{-}c &  \phantom{-}1 \end{bmatrix} =
\begin{bmatrix} -1  & \phantom{-}0 \\ \phantom{-}0 & \phantom{-}1 \end{bmatrix}
\begin{bmatrix}\phantom{-}1 & \phantom{-}0 \\ \phantom{-}c &  \phantom{-}1 \end{bmatrix}
$$
The latter factor represents the map
$$
(x,y)\mapsto (x, cx+y)
$$
which is a translation of the point $P(x,y)$ parallel to $y$-axis through a distance of $cx$.
The former matrix represents
$$
(x,y)\mapsto (-x,y)
$$

which is a refection  in the $y$-axis. The transfromations, and their repeated applications, are shown in Figure~\ref{Fig3}.

\subsubsection*{Case 2(ii): The general case}

Now let us consider the general square root of $I_2$ given by

$$
X=\begin{bmatrix} a & b \\ \frac{1-a^2}{b} & -a \end{bmatrix}.
$$
Letting $a=\rho\cos \phi$ and $b=\rho \sin\phi$, the matrix $X$ can be written as
\begin{equation}\label{R1R2}
R=R_3R_2R_1 + R_4
\end{equation}
where
$$
R_1=\begin{bmatrix}\phantom{-}\cos \phi & \sin\phi  \\ - \sin\phi  & \cos \phi  \end{bmatrix}, \quad
R_2 = \begin{bmatrix} 1 & \phantom{-}0 \\ 0 & -1 \end{bmatrix},\quad
R_3 = \rho I_2, \quad
R_4 = \begin{bmatrix} 0 & 0 \\ \frac{1-\rho^2}{\rho \sin\phi} & 0 \end{bmatrix}.
$$
The matrices $R_1, \ldots, R_4$ have the following geometrical interpretations (see Figure \ref{Fig6}).
\begin{itemize}
\item
$R_1$ represents a rotation through the angle $\phi  $ in the clockwise direction.
\item
$R_2$ represents a reflection in the $x$-axis.
\item
 $R_3$  represents a magnification by a factor $\rho$.
\item
We have
$$
R_4\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix} 0  \\
\frac{1-\rho^2}{\rho\sin\phi} x \end{bmatrix}.
$$
Hence the term ``$+ \,R_4$'', that is, the addition of $R_4$ to $R_1R_2R_3$, in Eq.\eqref{R1R2} represents
  a translation through a distance $\frac{1-\rho^2}{r\sin\phi} x $ parallel to $y$-axis.
\end{itemize}

\begin{figure}[t]
\begin{center}
\includegraphics[height=12cm]{SquareRootOfIdentityMatrix05.png}
\caption{Geometrical interpretation of the general square root of $I_2$}\label{Fig6}
\end{center}
\end{figure}

\section{A special case of the general case}

As a special case, let us consider those square roots of $I_2$ which are symmetric. The square roots discussed in Case 1 are clearly symmetric and those  discussed in Case 2 are symmetric only if $b=0$ or $c=0$ in which case they reduce to the matrices given in Case 1.

The square roots discussed in the general case (that is, Case 3) are symmetric if
$$
\frac{a^2-1}{b} =b;
$$
that is
$$
a^2+b^2=1.
$$
Hence in  such a case we can
take $a=\cos \phi $ and $b=\sin\phi $ for some $\phi $. Then the square root of $I_2$ takes the form
\begin{equation}\label{Householder}
X=\begin{bmatrix}\cos \phi & \phantom{-}\sin \phi \\
\sin\phi  & -\cos\phi  \end{bmatrix}
\end{equation}

Choosing a Pythagorean triple $(r,s,t)$ and letting $\cos \phi =\frac{r}{t}$ and $\sin\phi = \frac{s}{t}$, we have
$$
X=\frac{1}{t}\begin{bmatrix} r & \phantom{-}s \\ s & -r \end{bmatrix}.
$$
This case has been discussed in \cite{Gazette}.

\subsection{Geometrical interpretation: Rotation followed by reflection}

This square root has an interesting and obvious geometrical interpretation.
The matrix can be written as
$$
R=R_2R_1
$$
where
$$
R_2= \begin{bmatrix}1 & \phantom{-}0  \\
0 & -1 \end{bmatrix}, \quad R_1 = \begin{bmatrix}\phantom{-}\cos \phi & \sin \phi \\
- \sin\phi & \cos\phi \end{bmatrix}.
$$

Here $R_2$ represents the reflection in the $x$-axis of points in the $xy$-cartesian plane and $R_1$ represents a rotation of points clockwise through an angle $\phi$ about the origin of the coordinate system (see Figure \ref{Fig5}).  Thus this special square root is a combination of rotation and reflection.

\begin{figure}[!h]
\begin{center}
\includegraphics{SquareRootOfIdentityMatrix01.png}
\caption{Geometrical interpretation of a special square root of $I_2$}\label{Fig5}
\end{center}
\end{figure}

\subsection{Geometrical interpretation: Householder transformation}

The matrix given by Eq.\eqref{Householder} is a Householder matrix.

\begin{figure}[!t]
\begin{center}
\includegraphics{SquareRootOfIdentityMatrix10.png}
\caption{A special square root of $I_2$ as a Householder transformation}\label{SquareRootOfIdentityMatrix10}
\end{center}
\end{figure}

Let us recall the definition of a Householder matrix. Let $\mathbf v$ be a column unit vector in $\mathbb C^n$ or $\mathbb R^n$ and $\mathbf v^H$ the Hermitian transpose of $\mathbf v$. Let $I$ be the identity matrix of order $n$. Then a matrix of the form
$$
P = I - 2 \mathbf v \mathbf v^H
$$
is called a Householder matrix. It is interesting to note that the identity matrix is not a Householder matrix. The map $\mathbf x \mapsto P\mathbf x$ is called a Householder transformation (see \cite{Householder} and \S 13.3 \cite{Datta}). Householder matrices are extensively used in numerical linear algebra.

The matrix $P$ arises in the following way: Let $\mathbf x$ represent a point in space. Then the vector $P\mathbf x$ is the reflection of the point $\mathbf x$ in the hyperplane orthogonal to the unit vector $\mathbf v$.

Now to see that $X$ given by Eq.\eqref{Householder} is a Householder matrix,
let
$$
\mathbf v = \begin{bmatrix} -\sin \frac{\phi}{2} \\ \phantom{-}\cos\frac{\phi}{2}\end{bmatrix}
$$
Then
$$
\mathbf v^H = \begin{bmatrix} -\sin\frac{\phi}{2} & \cos\frac{\phi}{2}\end{bmatrix}
$$
Now, we have

\begin{align*}
P & = I - 2 \mathbf v\mathbf v^H \\
& = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - 2 \begin{bmatrix}\sin^2\frac{\phi}{2} & -\sin\frac{\phi}{2}\cos\frac{\phi}{2}  \\
\sin\frac{\phi}{2}\cos\frac{\phi}{2} & \cos^2\frac{\phi}{2}
\end{bmatrix}  \\
& = \begin{bmatrix}\cos \phi & \phantom{-}\sin \phi \\ \sin \phi & - \cos \phi \end{bmatrix} \\
& = X \qquad \text{(see Eq.\eqref{Householder})}.
\end{align*}

\section{Square roots of $-I_2$}

Since $I_2$ and $-I_2$ are closely related, in this section we briefly discuss the square roots of $-I_2$.

As in Section 2, let $R=\begin{bmatrix} a & b \\ c& d \end{bmatrix}$ be a square root of $-I_2$ so that $R^2=-I_2$. So we must have
\begin{align}
a^2+bc & = -1 \label{N1}\\
(a+d)b & =\phantom{-1}0 \label{N2}\\
(a+d)c & = \phantom{-1} 0 \label{N3}\\
d^2 + bc & = -1 \label{N4}
\end{align}

Proceeding as in Section 2, it can be seen that the real square roots of $-I_2$ are of the form
$$
R=\begin{bmatrix} a & \phantom{-}b \\ -\left(\frac{1+a^2}{b}\right) & -a
\end{bmatrix}
$$

where $a$ and $b\ne 0$ are real numbers. The matrix $R$ as given above is a square root of $-I_2$ even in the case when $a$ and $b$ complex numbers. If we allow $a$ to be a complex number, then we must have $a^2+1\ne 0$.

The following are some special complex square roots of $-I_2$. All of them can be obtained as in Section 2.

$$
iI_2, \quad -iI_2,\quad
\begin{bmatrix} \phantom{-}i & \phantom{-}b \\ \phantom{-}0 & -i \end{bmatrix},\quad
\begin{bmatrix} -i & \phantom{-}b \\ \phantom{-}0 & \phantom{-}i \end{bmatrix},\quad
\begin{bmatrix} \phantom{-}i & \phantom{-}0 \\ \phantom{-}c & -i \end{bmatrix},\quad
\begin{bmatrix} -i & \phantom{-}0 \\ \phantom{-}c & \phantom{-}i \end{bmatrix}.
$$

\section{$S(\alpha, \beta)$: A generalisation of the set of square roots of $I_2$}\label{SectionA}

The set of all square roots of $I_2$ is  the set of all matrices $X$ satisfying the condition $X^2=I_2$. Similar such sets have been considered in the literature.

For example, the geometry of the set $E_1$  of idempotent matrices of order $2$ has been  studied briefly in \cite{Krish01} and extensively in \cite{Krish02}. (Recall that a matrix $X$ is an idempotent if $X^2=X$.)  The set $E_1$ has been shown to be a hyperboloid of one sheet embedded in the four-dimensional space $\mathbb R^4$. The set of nilpotent matrices, that is, the set of matrices $X$ for which $X^2=0$,  is known to be a right circular cone. Some of these results on the set of idempotent matrices of order $2$ have been extended to sets of arbitrary order idempotent matrices (see \cite{Krish03}).

The conditions defining the various sets of matrices described above are all
special cases of  the following general condition:

\begin{equation}\label{Eq5}
X^2 - \alpha X + \beta I_2=0.
\end{equation}

We shall now consider the geometry of the set of matrices $X$ satisfying Eq.\eqref{Eq5}. There are two special matrices $X$ satisfying Eq.\eqref{Eq5}.
Let $t$ be a solution of the equation

\begin{equation}\label{Eq6}
t^2 - \alpha t +\beta =0.
\end{equation}

 then it can be seen that the matrix  $X=tI_2$ satisfies Eq.\eqref{Eq5}. The two solutions of Eq.\eqref{Eq6} yield two such matrices. In the sequel, when we consider solutions of Eq.\eqref{Eq5} we exclude these special solutions.

To simplify the discussions, we introduce the following notation:

\begin{equation}\label{Eq7}
S(\alpha, \beta)= \left\{ X : X^2 -\alpha X + \beta I_2 =0 \text{ and } X\ne tI_2 \text{ for some } t\in \mathbb R\right\}.
\end{equation}

\subsection{Special cases}

We have the following interesting special cases:
\begin{itemize}
\item
$S(0,-1)$: This is the set of square roots of $I_2$ excluding the two roots $\pm I_2$.
\item
$S(1,0)$: This is the set $E_1$ of idempotent matrices of order 2 excluding $I_2$ and the zero matrix, studied in \cite{Krish02}.
\item
$S(0,0)$: This is the set of nilpotent matrices of order 2 excluding the zero matrix.
\end{itemize}

\section{A simple characterization of the set $S(\alpha, \beta)$}

We now let
$$
X=\begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}
$$

and identify $X$ with the point $(x_1, x_2, x_3, x_4)$ in the four-dimensional space $\mathbb R^4$.

\begin{lemma}\label{Lemma1}
Let $X$ and $S(\alpha, \beta)$ be as above. Then $X\in S(\alpha, \beta)$ if and only if the following two conditions are satisfied:
\begin{align}
x_1+x_4 & = \alpha \label{Eq6a}\\
x_1x_4 - x_2x_3 & =\beta\label{Eq6b}
\end{align}

\end{lemma}

\paragraph{Proof.}
Let $X$ be in $S(\alpha, \beta)$ and so $X$ satisfies Eq.\eqref{Eq5}.

If we carry out the indicated operations in Eq.\eqref{Eq5} and equate the corresponding elements, we get the following system of equations:

\begin{align}
x_1^2+x_2x_3 - \alpha x_1 + \beta & =0 \label{Eq7}\\
(x_1+x_4)x_2 -\alpha x_2 & = 0 \label{Eq8}\\
(x_1+x_4)x_3 - \alpha x_3 & = 0 \label{Eq9}\\
x_4^2 +x_2x_3 -\alpha x_4 +\beta & =0 \label{Eq10}
\end{align}

Assume that $x_1+x_4\ne 2\alpha$.
Subtracting Eq.\eqref{Eq10} from Eq.\eqref{Eq7}, we have
$$
(x_1-x_4)(x_1+x_4-\alpha)=0
$$
which implies that $x_1-x_4=0$. Also, from the remaining two equations we get $x_2=0$ and $x_3=0$. All these imply that $X=tI_2$ where $t=x_1=x_4$ which is a contradiction. Thus we must have $x_1+x_4=\alpha$.
Further, replacing $x_1-\alpha$ by $- x_4$ in Eq.\eqref{Eq7} and simplifying we get
$$
x_1x_4-x_2x_3=\beta.
$$

Thus if $X\in S(\alpha, \beta)$ then Eq.\eqref{Eq6a} and Eq.\eqref{Eq6b} are satisfied.

Conversely, it is straight forward to verify that if for a matrix $X$ the equations Eq.\eqref{Eq6a} and Eq.\eqref{Eq6b} are satisfied, then the Eqs.\eqref{Eq7}--\eqref{Eq10} are also satisfied so that $X$ is in $S(\alpha, \beta)$. $\blacksquare$

Since, $x_1+x_4=\tr (X)$, the trace of $X$, and $x_1x_4-x_2x_3=\det (X)$, the determinant of $X$, Lemma \ref{Lemma1} implies that $X\in S(\alpha,\beta)$ if and only if
$$
\tr (X)=\alpha, \quad \det(X)=\beta.
$$
Thus we have
$$
S(\alpha, \beta) = \{ X : \tr(X)=\alpha\} \cap \{ X : \det(X)=\beta \}
$$

To simplify the discussions we introduce the following notation:
$$
P(\alpha) = \{ X : \tr(X)=\alpha\}.
$$

\section{The hyperplane $P(\alpha)$ in $\mathbb R^4$}

As we have already indicated in the beginning of Section \ref{SectionA}, the square matrix $X=\begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}$ of order $2$ can be identified with the point $(x_1, x_2, x_3, x_4)$ in the four-dimensional Euclidean space $\mathbb R^4$. This correspondence is a bijection. Now, the set of points $X$ in $\mathbb R^4$ which satisfies the equation $\tr(X)=\alpha$  defines a hyperplane in this four-dimensional space. A hyperplane in $\mathbb R^4$ is a three-dimensional Euclidean space. Thus the set of matrices $S(\alpha, \beta)$ is actually a set of points in a three-dimensional Euclidean space $P(\alpha)$. We now investigate the geometry of this set of points.

\begin{figure}[!h]
\begin{center}
\includegraphics{SquareRootOfIdentityMatrix09.png}
\caption{An orthogonal Cartesian coordinate system in the hyperplane $x_1+x_4=\alpha$}\label{Fig9}
\end{center}
\end{figure}

\subsection{A coordinate system in the hyperplane $P(\alpha)$}

To describe the geometry of the set $S(\alpha, \beta)$ we introduce a rectangular cartesian coordinate system in the hyperplane $P(\alpha)$.
Consider the following points in the hyperplane (see Figure \ref{Fig9}):

\begin{align*}
&O^\prime\left( \alpha/2, 0, 0, \alpha/2\right)\\
& A\left(\alpha/2+1/\sqrt{2}, 0, 0, \alpha/2 -1/\sqrt{2}\right)\\
& B\left( \alpha/2, 1/\sqrt{2}, 1/\sqrt{2}, \alpha/2\right)\\
& C\left( \alpha/2, -1/\sqrt{2}, 1/\sqrt{2},\alpha/2\right)
\end{align*}

These points define the following ordinary vectors in the three-dimensional space $P(\alpha)$:
\begin{align*}
\overrightarrow{O^\prime A} & = \overrightarrow{OA} - \overrightarrow{OO^\prime} \\
& = \left(\alpha/2+1/\sqrt{2}, 0, 0, \alpha/2 -1/\sqrt{2}\right) - \left( \alpha/2, 0, 0, \alpha/2\right)\\
& = \left(1/\sqrt{2}, 0, 0, -1/\sqrt{2}\right) \\
\overrightarrow{O^\prime B} & = \left( 0, 1/\sqrt{2}, 1/\sqrt{2}, 0 \right)\\
\overrightarrow{O^\prime C} & = \left( 0, - 1/\sqrt{2}, 1/\sqrt{2}, 0 \right)
\end{align*}

Using the usual inner product in $\mathbb R^4$ it can be seen that these vectors form a set of mutually perpendicular unit vectors in the three-dimensional space $P(\alpha)$. We choose $O^\prime$ as the origin and the lines $O^\prime A, O^\prime B, O^\prime C$ as the coordinate axes, the positive directions of the the coordinate axes being along the directions of the vectors $\overrightarrow{O^\prime A}, \overrightarrow{O^\prime B}, \overrightarrow{O^\prime C}$. As a tribute to R.J.T. Bell, author of a well known textbook on classical analytical three-dimensional geometry, we call this coordinate system the Bell coordinate system in $P(\alpha)$. Let $X(x_1,x_2,x_3,x_4)$ be any point in this hyperplane and let its coordinates relative to the Bell coordinate system be $(x,y,z)$. Then we have:

\begin{align*}
\overrightarrow{O^\prime X}
& = x \overrightarrow{O^\prime A} + y\overrightarrow{O^\prime B} + z \overrightarrow{O^\prime C} \\
& = x \left(1/\sqrt{2}, 0, 0, -1/\sqrt{2}\right) +
y \left( 0, 1/\sqrt{2}, 1/\sqrt{2}, 0 \right) +
z \left( 0, - 1/\sqrt{2}, 1/\sqrt{2}, 0 \right) \\
& = \left( x/\sqrt{2} , (y-z)/\sqrt{2},  (y+z)/\sqrt{2}, -x/\sqrt{2}\right)
\end{align*}

We also have:

\begin{align*}
\overrightarrow{O^\prime X} & = \overrightarrow{OX} - \overrightarrow{OO^\prime} \\
& = (x_1,x_2,x_3,x_4) - \left( \alpha/2, 0, 0, \alpha/2\right)\\
& = \left(x_1 - \alpha/2, x_2, x_3, x_4-\alpha/2\right)
\end{align*}

Thus we have

\begin{align*}
x_1 - \alpha/2 & = x/\sqrt{2} \\
x_2 & = (y-z)/\sqrt{2}\\
x_3 & = (y+z)/\sqrt{2}\\
x_4 - \alpha/2 & = -x/\sqrt{2}
\end{align*}

and therefore, we have

\begin{align}
x_1 & = \alpha/2 + x/\sqrt{2} \label{Eqx1}\\
x_2 & = (y-z)/\sqrt{2}\\
x_3 & = (y+z)/\sqrt{2}\\
x_4 & = \alpha/2 -x/\sqrt{2}\label{Eqx4}
\end{align}

\section{Geometry of the set $S(\alpha, \beta)$}

The following result characterises the geometry of the set $S(\alpha, \beta)$.
\begin{lemma}\qquad
\begin{enumerate}
\item
If $\alpha^2 - 4 \beta > 0$, then $S(\alpha, \beta)$ is a hyperboloid of one sheet.
\item
If $\alpha^2 - 4 \beta = 0$, then $S(\alpha, \beta)$ is a right circular cone.
\item
If $\alpha^2 - 4 \beta < 0$, then $S(\alpha, \beta)$ is a hyperboloid of two sheets.
\end{enumerate}
\end{lemma}

\paragraph{Proof.}

By Eq.\eqref{Eq6a} and Eq.\eqref{Eq6b}, $X=\begin{bmatrix}x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} =(x_1,x_2,x_3,x_4)$ is in $S(\alpha, \beta)$ if and only if $X$ is in $P(\alpha)$ and Eq.\eqref{Eq6b} is satisfied. Now $X$ is in $P(\alpha)$ if and only if $x_1, x_2, x_3$ and $x_4$ can be expressed in the form given by Eqs.\eqref{Eqx1}-\eqref{Eqx4}, where $(x,y,z)$ are the coordinates of $X$ relative to the Bell coordinate system in $P(\alpha)$.

Thus $X$ is in $S(\alpha, \beta)$ if and only if its coordinates relative to the Bell coordinate sysytem satisfies the following equation:
\begin{equation*}
\left(\alpha/2 + x/\sqrt{2}\right) \times \left(\alpha/2 -x/\sqrt{2}\right) - \left((y-z)/\sqrt{2}\right)\times \left((y+z)/\sqrt{2}\right) = \beta
\end{equation*}

This can be simplified to

\begin{equation}\label{Eqhy}
x^2+y^2-z^2=\frac{\alpha^2}{2}- 2 \beta.
\end{equation}

This equation represents a hyperboloid of one sheet when $\alpha^2/2 -2\beta \ne 0$, that is,  when $\alpha^2 -4\beta \ne 0$ (see \S 64 \cite{Bell}). Also Eq.\eqref{Eqhy} represents a hyperboloid of two sheets when $\alpha^2 - 4\beta<0$ (see \S 64 \cite{Bell}).  When $\alpha^2 -4\beta=0$, the equation Eq.\eqref{Eqhy} represents a right circular cone with semi-vertical angle $\frac{\pi}{4}$ (see \S 59 \cite{Bell}). $\blacksquare$

\begin{figure}
\begin{center}
\begin{tabular}{m{4cm}m{3.5cm}m{4cm}}
\centering \includegraphics[width=4.25cm, height=5cm]{HyperboloidOfOneSheet} &
\centering \includegraphics[width=3cm, height=5cm]{Cone} &
{\centering \includegraphics[width=3.5cm, height=5cm]{HyperboloidOfTwoSheets}} \\
\centering (a) & \centering (b) &  \begin{center}(c)\end{center} \\
\centering Hyperboloid of one sheet &  \centering Right circular cone &  {\centering Hyperboloid of two sheets} \\
\end{tabular}
\caption{Images of the set $S(\alpha, \beta)$}\label{Sab}
\end{center}
\end{figure}

Figure \ref{Sab}(a) shows a sketch of a hyperboloid of one sheet, Figure \ref{Sab}(b) shows a sketch of a right circular cone and Figure \ref{Sab}(c) shows a sketch of a hyperboloid of two sheets.

\subsection{Special cases}
\begin{itemize}
\item
$S(1, 0)$ is a hyperboloid of one sheet. This has already been noted in \cite{Kris01}. This hyperboloid lies in the hyperplane $x_1+x_4=1$.
\item
$S(0,0)$ is a right circular cone with semi-vertical angle $\frac{\pi}{4}$. This fact has also been noted in \cite{Kris02}. This cone lies the hyperplane $x_1+x_4=0$.
\item
$S(0,-1)$ is a hyperboloid of one sheet. Thus the set of square roots of the identity matrix $I_2$ defines a hyperboloid of one sheet in a four-dimensional space. This hyperboloid lies in the hyperplane $x_1+x_4=0$.
\end{itemize}

\begin{cor}
The set of all square roots of the identity matrix $I_2$ minus the two special square roots $\pm I_2$ forms a hyperboloid of one sheet in $\mathbb R^4$. The hyperboloid lies in the hyperplane $x_1+x_4=0$.
\end{cor}

\section{The hyperboloid of square roots of $I_2$}

In this section we closely examine the hyperboloid $S(0,-1)$ formed by the
set of all square roots of $I_2$. The surface $S(0,-1)$ lies in the hyperplane $P(0)$, that is, the hyperplane represented by the equation
$$
x_1+x_4=0.
$$
To define the Bell coordinate system in this hyperplane, we choose $O^\prime(0,0,0,0)$, that is, the zero-matrix $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ as the origin. If $X(x_1,x_2,x_3,x_4)$ is any point in the plane $P(0)$ and if the Bell coordinates of $X$ are $x,y,z)$ we have

\begin{align}
x_1 & = \phantom{-1} x/\sqrt{2} \label{Eqx1b}\\
x_2 & = (y-z)/\sqrt{2}\\
x_3 & = (y+z)/\sqrt{2}\\
x_4 & = -x/\sqrt{2}\label{Eqx4a}
\end{align}
and, we also have

\begin{align}
x & = \sqrt{2}x_1 \\
y & = \sqrt{2} (x_2+x_3) \\
z & = \sqrt{2} (x_3 - x_2)\label{Eqy4}
\end{align}

The equation of $S(0,-1)$ now simplifies to

\begin{equation}
x^2+y^2-z^2 = 2.
\end{equation}

We make the following observations about this hyperboloid.

\begin{enumerate}
\item
{\bf Centre of $S(0,-1)$}

The centre of the hyperboloid $S(0,-1)$ is the origin of the Bell coordinate system in $P(0)$ which is the zero matrix in $M_2$.
\item
{\bf Two special points  on $S(0,-1)$}

The matrix $\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$ is a square root of $I_2$. The point which corresponds to this is $(\sqrt{2},0,0)$. Similarly,
the point which corresponds
to $\begin{bmatrix}-1 & 0 \\ 0 & 1\end{bmatrix}$  is  $(-\sqrt{2},0,0)$.
\item
{\bf The principal section}

The principal section of the hyperboloid $S(0,-1)$ is the intersection of the hyperboloid with the plane $z=0$. This contains all $X$ in $S(0,-1)$ which satisfies the additional condition
$$
x_3-x_2=0\quad \text{(see Eq.\eqref{Eqy4}).}
$$
Thus the principal section of $S(0,-1)$ consists of all symmetric matrices which are square roots of $I_2$.

Recalling the notations of Section 2, let $R=\begin{bmatrix} a & b \\ c& d \end{bmatrix}$ be an element of the principal section of $S(0,-1)$. Then we have
\begin{align*}
a^2 + bc & =0 \\
(a+d)b & = 0 \\
(a+d)c & = 0\\
d^2 + bc & = 0 \\
b -c & =0
\end{align*}

This system reduces to

\begin{align*}
a^2 + b^2 & = 1 \\
(a+d)b & =0 \\
d^2 - a^2 & = 0
\end{align*}

Let $ a = \cos \phi$ and $b = \sin \phi$. If $a+d\ne 0$ then $b=0$ and $a=\pm 1$ and then $R=\pm I_2$.
If $a+d=0$ then we get
$$
R=\begin{bmatrix} \cos\phi & \sin\phi \\ \sin\phi & -\cos\phi \end{bmatrix}
$$
which is a Householder transformation. Thus, the principal section of $S(0,-1)$ represents the set of square roots of $I_2$ which are Householder transformations.

\item
{\bf The principal axis}

The principal axis is the $z$-axis specified by the equations $x=y=0$.
These equations give the matrix $\begin{bmatrix}0 & x_2 \\ -x_2 & 0 \end{bmatrix}$ which is a skew-symmetric matrix. It follows that thw principal axis of $S(0,-1)$ is the set of skew-symmetric matrices in $M_2$.

\item
{\bf The asymptotic cone from centre}

The asymptotic cone from the centre has the equation
$$
x^2 + y^2 -z^2 = 0.
$$
This yields the equation
$$
x_1^2+4x_2x_3=0.
$$
Hence the asymptotic cone from the centre consists of all matrices  in the following set:
$$
\left\{ \begin{bmatrix}x_1 & x_2 \\ x_3 & -x_1\end{bmatrix} : x_1^2+4x_2x_3=0\right\}.
$$

\item
{\bf The generators}

One distinguishing feature of the hyperboloid of one sheet is the existence of two systems of generators on the surface. The generators are straight line lying completely on the surface. Through every point on the hyperboloid there passes one generator of each system. Let us consider the generators of the hyperboloid $S(0,-1)$. The following lemma can be used to determine the generators of this surface.

\begin{lemma}
Let $A\in S(0,-1)$ and let $U, V\in M_2$ be such that
\begin{gather}
AU=U,\quad UA=-U,\quad U^2 =0\\
AV=-V, \quad VA=V,\quad V^2 =0.
\end{gather}

Then the two generators through $A$ are the lines specified by the sets

\begin{align*}
L_1& = \{ A+tU:t\text{ a scalar } \}\\
L_2 & = \{ A+tV: t\text{ a scalar} \}
\end{align*}
\end{lemma}
\paragraph{Proof.} We have

\begin{align*}
(A+tU)^2
& = A^2 + tAU+tUA+U^2\\
& = I_2 + tU -tU +0\\
& + I_2
\end{align*}

Hence $A+tU\in S(0,-1)$ and therefore $L_1\subseteq S(0,-1)$. Since $L_1$ is parameterized by a single parameter in the first degree, it represents a
line. By a similar argument it can be shown that $L_2$ is also a line lying completely in $S(0,-1)$.

Now, if possible let
$$
A+t_1U=A+t_2V.
$$
Then we have
$$
t_1U=t_2V.
$$
Pre-multiplying by $A$ we get
$$
t_1U=-t_2V.
$$
These together imply that $t_1=t_2=0$. Thus we must have $L_1\cap L_2=\emptyset$.

It follows that $L_1$ and $L_2$ are the two generators through $A$. $\blacksquare$

It is rather easy to find $U$ and $V$ satisfying the conditions of the lemma. Let $X$ be any element of $M_2$ then the following matrices have the required properties:

\begin{align*}
U& = (A+I_2)X(A-I_2)\\
V& = (A-I_2)X(A+I_2)
\end{align*}

In particular, let us find the generators through points on the principal section. A general element on the principal section is given by
$$
P=\begin{bmatrix}\cos \phi & \sin\phi \\ \sin\phi & -\cos\phi\end{bmatrix}
$$
Let us take
$$
X=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}
$$
Then we get
\begin{align*}
U & =
\begin{bmatrix} -\sin^2\phi & \sin\phi(\cos\phi -1) \\
\sin\phi(\cos\phi+1) & \sin^2\phi \end{bmatrix} \\
V & =\begin{bmatrix} -\sin^2\phi & \sin\phi(\cos\phi +1) \\
\sin\phi(\cos\phi -1) & \sin^2\phi \end{bmatrix}
\end{align*}

Removing the common factor $\sin \phi$ and let it absorb with the scalar $t$, we may take
\begin{align*}
U & = \begin{bmatrix} -\sin\phi & \cos\phi-1 \\ \cos\phi+1 & \sin\phi\end{bmatrix}\\
V & = \begin{bmatrix} -\sin\phi & \cos\phi + 1 \\ \cos\phi -1 & \sin\phi\end{bmatrix}
\end{align*}

Therefore the two generators through $P$ are the lines represented by

\begin{gather}
P+t\begin{bmatrix} -\sin\phi & \cos\phi-1 \\ \cos\phi+1 & \sin\phi\end{bmatrix}\\
P+t\begin{bmatrix} -\sin\phi & \cos\phi + 1 \\ \cos\phi -1 & \sin\phi\end{bmatrix}
\end{gather}

Incidentally, we can write $U$ as
$$
U=\begin{bmatrix}-\sin\phi & \cos\phi \\ \cos\phi & \sin\phi\end{bmatrix} + \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.
$$
The first term is a Householder transformation and the second term is a square root of $-I_2$. There is similar representation for $V$ also.
\end{enumerate}

\begin{figure}
\begin{center}
\includegraphics[height=5cm]{h1s}
\caption{The hyperboloid $S(0,-1)$ showing the principal section and the generators through a point on the principal section}
\end{center}
\end{figure}

\section{Matrix function approach}

\subsection{Introduction}

In this approach we begin with the concept of a function of a matrix, and then consider a definition of matrix function. We then apply this definition to the square root function of a matrix. This square root function is then specialized to compute the square roots of $I_2$.

The term ``function of a matrix'' has several meanings. Following Nigham, we look upon a function of matrix as a generalization of the concept of complex function $f(z)$ of a complex variable $z$. Accordingly, in our definition of a  matrix function, we start with some $f(z)$ and assign a meaning to $f(A)$ for any arbitrary matrix $A$. In this approach the determinant, the trace, the transpose and the adjoint are all not functions of matrices.

\subsection{Definition of function of matrix}

Let $f(z)$ be a complex function of a complex variable $z$ and let $A$ be an arbitrary square matrix of order $n$.
\begin{itemize}
\item
Let $Z$ be a nonsingular matrix and $J$ a matrix in the Jordan canonical form such that
$$
A=ZJZ^{-1}.
$$
(Given $A$, $Z$ and $J$ exist.)
\item
Let
$$
J=\text{diag\,}(J_1, J_2, \ldots, J_p)
$$
where
$$
J_k = J(\lambda_k) = \begin{bmatrix}
\lambda_k & 1 &  & \\
          & \lambda_k & \ddots & \\
          &           & \ddots & 1 \\
          &            &      & \lambda_k
\end{bmatrix}\in \mathbb C^{m_k\times m_k}.
$$
\item
We have
$$
m_1+m_2+\cdots+m_k=n.
$$
\item
Let $\lambda_1, \lambda_2,\ldots, \lambda_s$ be the distinct eigenvalues of $A$. Let $n_i$ be the order of the
largest Jordan block in which $\lambda_i$ appears.
\item
Assume that the values
$$
f^{(j)}(\lambda_i)\text{ for }j=0,\ldots, n_i-1,\quad i=1,\ldots, s
$$
exist.
\end{itemize}

Now $f(A)$ is defined as
$$
f(A) = Z\,\,\text{diag}\, (f(J_k))Z^{-1}
$$
where we define
$$
f(J_k)=
\begin{bmatrix}
f(\lambda_k) & f^\prime(\lambda_k) & \cdots & \dfrac{f^{(m_k-1)}(\lambda_k)}{(m_k-1)!} \\
             & f(\lambda_k) & \ddots & \vdots \\
             &              & \ddots & f^\prime(\lambda_k) \\
             &              &         & f(\lambda_k)
\end{bmatrix}
$$

\subsection{Computation of square roots of $I_2$}

To compute the square roots of $I_2$ we have to consider the function
$$
f(z)=\sqrt{z}
$$
and then we have to compute the value of $f(I_2)$. For clarity of discussion, let us temporarily write $A=I_2$. The Jordan canonical form of $A$ is
$$
J =
\begin{bmatrix}
J_1 & 0 \\ 0 & J_2
\end{bmatrix}
\text{ where } J_1 = [ 1 ], \quad J_2 = [1].
$$
For any nonsingular matrix $Z$ we have
$$
A=ZJZ^{-1}.
$$
Now, by definition, we have
\begin{align*}
f(A)& = Z\, \text{diag}\,(f(J_1), f(J_2)) Z^{-1}\\
& = Z \begin{bmatrix} [\sqrt{1}] & 0 \\ 0 & [\sqrt{1}]  \end{bmatrix} Z^{-1}\\
& = Z \begin{bmatrix} \sqrt{1} & 0 \\ 0 & \sqrt{1} \end{bmatrix} Z^{-1}.
\end{align*}

The function $f(z)$ has two branches and the value $f(1)=\sqrt{1}$ can be in any one of these branches. Since $f(1)$ appears twice in $f(A)$, the two values may be chosen as lying in the same branch or in two different branches. If w choose them as lying in the same branch we get the {\em the primary values} of $f(A)$. These are given by
$$
Z\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}Z^{-1} = I_2,\quad Z\begin{bmatrix}-1 & \phantom{-}0 \\ \phantom{-} 0 & -1 \end{bmatrix} Z^{-1}= -I_2.
$$

If we choose them as lying two different branches, we get {\em non-primary values} of $f(A)$. These are  given by

$$
Z\begin{bmatrix} \phantom{-}1 & \phantom{-}0 \\ \phantom{-}0 & - 1 \end{bmatrix}Z^{-1} ,\quad Z\begin{bmatrix}-1 & \phantom{-}0 \\ \phantom{-}0 & \phantom{-}1 \end{bmatrix} Z^{-1}.
$$

\section{Square roots of $\pm I_2$ via split-quaternions}

\subsection{Split-quaternions}

The split-quaternions, also called coquaternions, are elements of a 4-dimensional associative algebra introduced by James Cockle in 1849. They form a four dimensional real vector space equipped with a multiplication  operation. Unlike the quaternions, the algebra of split-quaternions contains zero divisors, nilpotent elements, and nontrivial idempotents. We denote the algebra of split-quaternions by $\mathsf P$.

The set $\{1, i, j, k\}$ forms a basis for the algebra $\mathsf P$. The products of these elements are given below:

\begin{alignat*}{2}
ij & = \phantom{-}k &  & = -ji \\
jk & = -i & & = -kj\\
ki & = \phantom{-} j & & = -ik \\
i^2 & = -1 &&\\
j^2 & = +1 &&\\
k^2 & = +1 &&
\end{alignat*}
These are equivalent to the following set of equations:
$$
i^2 =-1, \quad j^2=k^2=ijk=1.
$$
The conjugate of the split-quaternion
$$
q=w+xi+yj+zk
$$
is
$$
q^*=w-xi-yj-zk
$$
and its modulus is
$$
qq^*=w^2+x^2-y^2-z^2.
$$
If $qq^*\ne 0$ then $q$ has an inverse, namely, $\dfrac{q^*}{qq^*}$.
A split-quaternion is spacelike,  lightlike or timelike according as $qq^*<0$, $qq^*=0$ or $qq^*>0$.

\subsection{Isomorphism of $\mathsf P$ with the ring of $2\times 2$ real matrices}

Let $M_2$ be the set of real $2\times 2$ matrices. It can be easily verified that the map
$$
w+xi+yj+zk \mapsto
\begin{bmatrix}
w+z & x+y \\ y-x & w-z
\end{bmatrix}
$$
defines a bijection from $\mathsf P$ to $M_2$ which is also a ring isomorphism.
The inverse of the map is
$$
\begin{bmatrix} a & b \\ c & d \end{bmatrix}
\mapsto
\frac{1}{2}(a+d) + \frac{1}{2}(b-c)i + \frac{1}{2}(b+c)j + \frac{1}{2}(a-d)k.
$$

If
$$
X=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \mapsto q
$$
then
$$
qq^* = ad-bc = \det(X).
$$
Also, the identity matrix $I_2$ maps to the identity $1$ in $\mathsf P$.

\subsection{Square roots of $1$ in $\mathsf P$}

Let
$$
q=w+xi+yj+zk\in\mathbf P
$$
be such that
$$
q^2 =1.
$$
Post-multiplying by $q^*$ we have
$$
qqq^* = q^*.
$$
Clearly $qq^*\ne 0$, because otherwise we must have $q^*=0$ and so $q=0$ which is obviously a contradiction. So we must have

$$
q=\frac{q^*}{qq^*}.
$$

This yields the following equations:

\begin{align}
w\left(1 - \frac{1}{qq^*}\right)& = 0 \label{SQ1}\\
x\left(1 + \frac{1}{qq^*}\right)& = 0 \label{SQ2}\\
y\left(1 + \frac{1}{qq^*}\right)& = 0 \label{SQ3}\\
z\left(1 + \frac{1}{qq^*}\right)& = 0 \label{SQ4}
\end{align}

\subsubsection*{Case 1}

If $w\ne 0$ then $1-\frac{1}{qq^*} =0$ and so $1 + \frac{1}{qq^*}\ne 0$. Tis would then imply that $x=y=z=0$. Hence $q=w$ and $w^2=1 $ and hence $q=\pm 1 \in \mathbf P$.

\subsubsection*{Case 2}

Let $w=0$. If $1-\frac{1}{qq^*} =0$ then as above we have $x=y=z=0$ which leads to the contradiction that $q=0$.

Now we must have
$$
1+\frac{1}{qq^*} =0;
$$

that is,
\begin{equation}\label{ABC}
x^2 - y^2 - z^2 = -1.
\end{equation}
Thus, the square roots of $1\in \mathsf P$, other than $\pm 1$, are split-quaternions of the form
$$
q=xi+yj+zk
$$
where $x,y,z$ satisfy Eq.\eqref{ABC}. If we identify $x,y,z$ as the coordinates of a point $(x,y,z)$ in ordinary three-dimensional euclidean space, then the set of points defined by Eq.\eqref{ABC} form a hyperboloid of one sheet. It follows that the set of square roots of $1\in \mathsf P$ can be identified with a hyperboloid of one sheet.

We can find real numbers $t$ and $\phi$ such that
\begin{align*}
x & = \sinh t \\
y & = \cosh t \sin \phi\\
z & = \cosh t \cos \phi
\end{align*}
Thus we have the parametrized form for the square roots of $1$:
$$
q=i \sinh t +(j \sin\phi + k\cos \phi)\cosh t
$$

\subsection{Square roots of $I_2$}

The square roots of $1\in \mathbf P$ are
$$
1, \quad -1, \quad i \sinh t +(j \sin\phi  + k\cos \phi                )\cosh t
$$
By the isomorphism between $\mathbf P$ and $M_2$, we see that the roots of $I_2$ are the following matrices:
$$
\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}, \quad
\begin{bmatrix}- 1 & \phantom{-}0 \\ \phantom{-}0 & -1\end{bmatrix}, \quad
\begin{bmatrix}
 \cosh t \cos\phi  &  \cosh t \sin\phi  +\sinh t \\
\cosh t \sin \phi  - \sinh t & -  \cosh t \cos \phi  \end{bmatrix}
$$
The third form for the square root can be expressed in the following form:
$$
 \cosh t
\begin{bmatrix} \cos\phi  & \phantom{-}\sin\phi  \\ \sin\phi  & -\cos\phi  \end{bmatrix} +\sinh t \begin{bmatrix} \phantom{-}0 & 1 \\ -1 & 0 \end{bmatrix}
$$

The first term is a scalar multiple of a Holder transformation which is a square root of $I_2$ and the second term is a scalar multiple of a square root of $-I_2$.

\subsection{Square roots of $-1\in \mathsf P$ and $-I_2$}

Using an argument similar to one used to find the square roots of $1 \in \mathsf P$, we can see  that the square roots of $-1\in \mathsf P$ are split-quaternions of the form

$$
q=xi+yj+zk
$$

where $x,y,z$ satisfy the condition
\begin{equation}\label{2S}
x^2 - y^2 -z^2 =1.
\end{equation}
Eq.\eqref{2S} represents a hyperboloid of one sheet in $\mathbb R^3$. Thus the set of square roots of $-1$ forms a hyperboloid of one sheet.

We can parameterize $x,y,z$ as follows:
\begin{align*}
x & = \sec t \\
y & = \tan t \sin \phi \\
z & = \tan t \cos \phi
\end{align*}
Then, we have
$$
q=i \sec t + j \tan t\sin\phi + k \tan t \cos\phi.
$$
Using the isomorphism between $M_2$ and $\mathsf P$ we see that  the square roots of $-I_2$ are of the form
$$
\begin{bmatrix}
\tan t \cos \phi & \tan t \sin\phi + \sec t \\
\tan t \sin\phi - \sec t & - \tan t \cos \phi
\end{bmatrix}.
$$
This can be expressed as

$$
\tan t
\begin{bmatrix}
\cos \phi & \phantom{-}\sin\phi \\ \sin\phi & -\cos \phi
\end{bmatrix}
+
\sec t
\begin{bmatrix}
\phantom{-}0 & 1 \\ -1 & 0
\end{bmatrix}
$$

The first term is a scalar multiple of a Holder transformation and the second term is a multiple of a special square root of $-I_2$.

\section{$2\times 2$ matrices with infinite number of square roots}

We have been stressing the fact that $\pm I_2$ have infinite number of square roots. Now we investigate which other $2\times 2$ matrices have infinite number of square roots. In the sequel, we shall use the following notation: For any $2\times 2$ matrix $A$, we write
$$
S(A)=\{ X \in M_2: X^2 = A\}
$$
We have to determine for which $A$'s are the sets $S(A)$ infinite.

The following results are fairly obvious.

\begin{lemma}
For any $A\in M_2$ and any positive real number $\alpha$,
\begin{enumerate}
\item
$S(\alpha A ) = \sqrt{\alpha} S(A)$.
\item
For any nonsingular matrix $X\in M_2$, we have $S(Z^{-1}AZ)=Z^{-1}S(A)Z$.
\end{enumerate}
\end{lemma}

\paragraph{Proof.} We shall give a proof of the latter result. Let
$X\in S(Z^{-1}AZ)$ Then we have
$$
X^2 = Z^{-1}AZ.
$$
Let $Y=ZXZ^{-1}$.
\begin{align*}
Y^2 & = (ZXZ^{-1})(ZXZ^{-1})\\
& = ZX^2 Z^{-1} \\
& = Z(Z^{-1}AZ)Z^{-1}\\
& = A
\end{align*}
Therefore, $X=Z^{-1}YZ$ with $Y\in S(A)$. This implies that
$$
S(Z^{-1}AZ)\subseteq Z^{-1}S(A)Z.
$$
Similarly we also have
$$
Z^{-1}S(A)Z\subseteq S(Z^{-1}AZ).
$$
It follows that $S(Z^{-1}AZ)=Z^{-1}S(A)Z$. $\blacksquare$

Let $A\in M_2$ and let its eigenvalues $\lambda_1, \lambda_2$ be real.  Let $J$ be the Jordan canonical form of $A$ and let $Z$ be a non-singular
matrix such that $A=Z^{-1}JZ$. By Lemma ,
$$
S(A)=Z^{-1}S(J)Z.
$$
Thus to find $A$'s having infinite number of square roots, it is enough to find $J$'s having infinite number of square roots. We consider the various forms of $J$ one by one.
\begin{enumerate}
\item
Let $0<\lambda_1<\lambda_2$ so that

$$
J=
\begin{bmatrix}
\lambda_1 & 0 \\ 0 & \lambda_2
\end{bmatrix}
$$

Clearly $S(J)$ contains only the following four elements.
$$
\begin{bmatrix}
\pm\sqrt{\lambda_1} & 0 \\
0 & \pm\sqrt{\lambda_2}
\end{bmatrix}
$$
Thus $S(A)$ is finite and contains precisely four elements.

\item
Let $0<\lambda_1=\lambda_2=\lambda$, say.
We have to consider two sub-cases:
\begin{enumerate}
\item
Let
$$
J =
\begin{bmatrix}
\lambda & 0 \\ 0 & \lambda
\end{bmatrix}
=\lambda I_2
$$
Therefore by Lemma we have
$$
S(J) =\sqrt{\lambda} S(I_2).
$$
Since $S(I_2)$ is an infinite set, $S(J)$ is also an infinite set.
\item
Let
$$
J=
\begin{bmatrix}
\lambda & 1 \\ 0 & \lambda
\end{bmatrix}
$$
Now we can find the square roots of $J$ as in Section 2. It can be shown that the following are the only two square roots of $J$ in this case:
$$
\begin{bmatrix}
\sqrt{\lambda} & 1/(2\sqrt{\lambda}) \\ 0 & \sqrt{\lambda}
\end{bmatrix},\quad
\begin{bmatrix}
- \sqrt{\lambda} & - 1/(2\sqrt{\lambda}) \\ 0 & - \sqrt{\lambda} \end{bmatrix}
$$
\end{enumerate}

\item
Let $0=\lambda_1 <\lambda_2 = \lambda$, say. Then we have
$$
J =
\begin{bmatrix}
\lambda & 0 \\ 0 & 0
\end{bmatrix}
$$
In this case the only two square roots of $J$ are
$$
\begin{bmatrix}
\pm\sqrt{\lambda} & 0 \\ 0 & 0
\end{bmatrix}.
$$

\item
Let $0=\lambda_1=\lambda_2=0$. Again we have to consider two cases:
\begin{enumerate}
\item
Let

$$
J =
\begin{bmatrix}
0 & 1 \\ 0 & 0
\end{bmatrix}.
$$

This matrix has no square roots.
\item
Let
$$
J=
\begin{bmatrix}
0 & 0 \\ 0 & 0
\end{bmatrix}
$$

Any nilpotent matrix is a square root of $J$. Since there are infinite number of nilpotent matrices, $S(J)$ is an infinite set.
\end{enumerate}

\end{enumerate}

\begin{lemma}
An element $A\in M_2$ with real eigenvalues has an infinite number of square roots if and only if $A=\lambda I_2$ for some $\lambda\ge 0$.
\end{lemma}

config: arxiv


TITLE: Contribution of quadrupole term using Quadrupole moment
QUESTION [2 upvotes]: I read this derivation in  Griffiths that if we obtain dipole moment vector $\vec{p}$ , then dipole term in the potential (due to general distribution far away at $\vec{r}$ ) can be written as $$\frac{\vec{p}.\hat{r}}{4\pi\epsilon_or^2}$$
I was wondering if quadrupole contribution can be written like this in terms of the quadrupole moment matrix. I mean of course, it wont be a dot product but maybe some other way.  (Couldn't find it on net!)

REPLY [1 votes]: Hey the potential term you seek is the following (for gravitational potential. The functional form is exactly the same in the electric potential case.):
$$\phi_{quad}(\vec r)=-G\frac{Q_{ij}\hat r_i\hat r_j}{r^3},$$
$$Q_{ij}=\int\frac{3x_ix_j-\delta_{ij}r^2}{2}_{}\rho(\vec r)d^3\vec r.$$
The quantity $\rho(\vec r)d^3\vec r$ is just the element $dm(\vec r)$ of the mass distribution generating the gravitational potential. This should correspond to the electric charge density generating the electric potential, in your case.


TITLE: Is Collatz' conjecture the only stable solution of its type?
QUESTION [17 upvotes]: The Collatz Conjecture is well known with the sequence
$$f(n) = \begin{cases} n/2 &;\text{if } n \equiv 0 \pmod{2}\\ k\,n+1 &; \text{if } n\equiv 1 \pmod{2} \end{cases}$$
and $k=3$; the sequence converging $1$ (so called oneness).
Is there any conjecture/theorem on whether the sequence would converge for any other value of $k$; or could it be shown that the sequence diverges for values of $k$ other than $3$?
By convergence here I mean that the sequence after finite steps ends with a stable fixed number such as in case of Collatz it is the case with the number $1$.
Append: In the mean time I wrote out a conjecture on this over here >>>, for those who might be interested.

REPLY [3 votes]: I find the following formalism helpful for my own understanding of the problem, so I think I'll add this here too.
To have this transformation generalized let's write:
$$ T_M(x) = (M \cdot x+1) / 2^A \qquad \small \text  { where A equals the exponent at 2 in the numerator }$$
and we look only at the odd numbers (because all even numbers can be seen as "trivially" dealt with.
For easier notation I like also to write it in this form
$$ a_{k+1}= {M \cdot a_k +1 \over 2^{A_k}}$$
or, for linear writing, using consecutive letters instead of numerical indexes
$$ b = {M\cdot a +1 \over 2^A}, c={M\cdot b +1 \over 2^B }, \text{ and so on} $$
Then to have a one-step-cycle we must have the result equalling the input
$$ b = {M\cdot a +1 \over 2^A} =a$$
and obviously we must have
$$ 2^A =M + \frac 1a $$
To have such a one-step-cycle on the number $a$ we must have that $M$ is a mersenne number $M=2^A-1$ and because of its simpliness it usually denotes the "trivial cycle".
$M=3,7,15,31,...$ are such mersenne numbers and they allow "trivial cycles".
If we allow negative integers, then for $a =-1$, we have a similar pattern, only that the allowed $M$ ar $M=3,5,9,17,...$. For $ |a| \ne 1$ the expression $ \frac 1a$ would introduce fractional values and thus other one-step-cycles are not possible.
To look at two-step-cycles it suffices to look at the generalization
$$ b\cdot c = {M\cdot a +1 \over 2^A}\cdot {M\cdot b +1 \over 2^B} = b \cdot a$$
It is too complicated for the place here to discuss this in generality. But to show one helpful pattern I'll show this for $M=3$ and $M=5$
For $M=3$ we get
$$ bc =  {3a+1 \over 2^A}\cdot {3b+1 \over 2^B} = ba $$
Rearranging gives
$$ 2^{A+B} = (3+ {1 \over a})\cdot (3+{1 \over b})  $$
We see immediately, that the rhs can only fall in the interval $9 .. 16$ (where $16$ can only occur if $a=1$ and thus for the trivial case). But there is no other exact power of 2 in this interval, so the lhs-part cannot be satisfied.
If we allow negative integers we get instead
$$ 2^{A+B} = (3- {1 \over a})\cdot (3-{1 \over b})  $$
which covers now the interval $4..9$. It gives for $4$ the trivial cycle in the negatives, and for $a=5,b=7$ another solution compatible with the lhs which means thus a two-step cycle (and we must not forget that the sign must reverted to the negatives).
For $M=5$ this looks like
$$ 2^{A+B} = (5+ {1 \over a})\cdot (5+{1 \over b} ) $$
The range for the rhs is now $25..36$ and there is only one number (=32) compatible with the lhs.
The smallest numbers for a $M=5$ cycle are $1$ and $3$ . We try these numbers
$$ 2^{A+B} = (5+ {1 \over 1})\cdot (5+{1 \over 3} )=6\cdot {16 \over 3} =32 $$
Bingo!
So we just found the smallest cycle-length for the $M=5$-problem.
Now again in the negative integers integ
$$ 2^{A+B} = (5- {1 \over a})\cdot (5-{1 \over b} ) $$
The range for the rhs is 16..25 where 16 indicates the trivial cycle in the negatives. There is not other perfect power of 2 in that interval, so no 2-step-cycle in the negatives for the $M=5$ problem.

This expositions for the one- and for the two-step cycles can generalized for multistep-cycles in the obvious way. However, with more steps the view at it through the modular glasses does not lead very far. But there is another tool (which leads to another numbertheoretic tool with which finally R.Steiner proved the non-existence of the "1-cycle" (arbitrary many steps with exponents $A=1$, and only one exponent $B \gt 1$)): we arrange the $M$'s out of the rhs getting
$$ {2^S\over M^N} = (1+{1 \over M\cdot a})\cdot (1+{1 \over M\cdot b}) \cdots $$
where I denote the sum-of-exponents (=number of divisions by 2) as $S$ and the number of steps (=number of multiplications by 3 = number of parentheses on the rhs) as $N$
Now with high $a,b,c,...$ on the rhs and high $M$ this is very near to 1, in many cases much nearer than the quotient on the lhs, and one can look at the approximation of $2^S / M^N$ for many $N$ and discover a trend which suggest a required minimal cycle length for a set of integers all larger than its minimal element $a$


TITLE: Determinant of an endomorphism (Proof)
QUESTION [0 upvotes]: I Need to prove That for every a there is an endomorphism phi such that det(phi)=a.
I habe no idea how to do this. Can anybody help me?

REPLY [0 votes]: If we are in a $n$-dimensional vector space $V$ over a field $\mathbf F$, consider the endomorphism $\phi : V \to V$ such that, in some basis, its matrix representation is
$$\begin{pmatrix}
a & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{pmatrix}.$$