TITLE: Chemical potential, Boltzmann
QUESTION [5 upvotes]: The electrochemical potential of an ion i in an electrolytic solvent is given by:
\begin{align}\mu_i(\vec r) &= \mu_i^0 + RT \ln(a_i(\vec r))\\& = \mu_i^0 + RT \ln(\gamma_i(\vec r) c_i(\vec r)/c_0) \\&=  \mu_i^0 + RT \ln(c_i(\vec r)/c_0) + RT \ln(\gamma_i(\vec r)) \\&= \mu_i^0 + RT \ln(c_i(\vec r)/c_0)+z_i e \phi(\vec r)
\end{align}
where $\mu_i^0$ is the chemical potential of the ion in an ideal solution with ionic concentration $c_0$, $\gamma$ is the activity coefficient, $\phi$ the electrostatic potential of the ions in solution (given by the Poisson equation).
Is this expression correct? The authors in this paper (I. Rubinstein, Electro-Diffusion of Ions (1990) (SIAM link)) claim the following:

... the entire system, including the boundary layers. This implies in turn the establishment of a certain degree of smoothness on the microscopic scale $r_d$ ($\sqrt{\varepsilon}$ dimensionless terms) of some function of concentration and the electric potential, termed the electrochemical potential and defined as
$$\tilde{\mu}_i = RT \ln{\tilde{C}_i} + z_iF\tilde{\varphi} \tag{1.19a}$$
or in dimensionless terms
$$\mu_i = \ln{C_i} + z_i\varphi . \label{eq:1.19b} \tag{1.19b}$$
Indeed, by $\eqref{eq:1.19b}$
$$j_i = -\alpha_iC_i\nabla\mu_i, \label{eq:1.19c} \tag{1.19c}$$
so that $j_i = O(1)$ implies that the variation $(\delta \mu)|_{\varepsilon^{1/2}}$ on the length $\sqrt{\varepsilon}$ is of the order
$$(\delta \mu)|_{\varepsilon^{1/2}} = O\left(\varepsilon^{1/2}\right).$$
[...]
To illustrate some of the notions introduced so far, let us consider the $C_i$ and $\varphi$ fields in an electrodialysis cell at equilibrium. For simplicity, let us limit our consideration to a $1,1$ valent electrolyte at bulk (feed) concentration $C_0$. Assume a constant fixed charge density $\tilde{N}(-\tilde{N})$ for the an- (cat-) ion membrane.
Recall that equilibrium is a steady state without macroscopic fluxes. By $\eqref{eq:1.19c}$ this implies constancy of ionic electrochemical potentials in the system. Thus,
$$\mu_i = \ln{C_i} + z_i\varphi = 0 \tag{1.23a}$$
or
$$C_i = e^{-z_i\varphi} \label{eq:1.23b} \tag{1.23b}$$
(For the univalent electrolyte under consideration $M = 2$, $z_1 = 1$, $z_2 = -1$ and the dimensionless bulk concentration at $\varphi = 0$ is normalized to unity.)
Equation $\eqref{eq:1.23b}$ is the equilibrium Boltzmann distribution in a potential field. Substitution of $\eqref{eq:1.23b}$ into (1.9c) yields the Poisson-Boltzmann ...

When you assume equilibrium, the ionic flux $j_i \propto \nabla \mu_i$ should be zero. But what people do in this paper, is setting
$$
RT \ln(c_i(\vec r)/c_0)+z_i e \phi(\vec r)=0
$$
and therefore we get:
$$c_i(\vec r) = c_0 \exp(-z_i e/(RT) \phi(\vec r)),$$
which is the Boltzmann distribution of the ions according o the electrostatic potential. Does anyone understand this logic? Why can we set the additional term to zero in equilibrium? And why do we neglect $\mu_i^0$?

REPLY [2 votes]: You state

Then you can assume equilibrium, thus the ionic flux $j_i\propto\nabla\mu$ should be zero

So it seems to me that the authors you read are not defining
$$ RT \ln(c_i(\mathbf r)/c_0)+z_i e \phi(\mathbf r)\equiv0 $$
but instead defining
$$
\mu_i(\mathbf r)=\mu_i^0.
$$
That is, the chemical potential doesn't change.


text:
\begin{document}

\pagenumbering{arabic}

\begin{center}
\Large{{\bf SPECIAL STANDARD STATIC SPACE-TIMES
\footnote{{\bf Key Words and Phrases:} Space-time, Warped
Product, Einstein Manifold, Ricci-flat, Scalar Curvature.\\
{\bf Mathematics Subject Classification (2000):} 53C25, 53C50,
53C80.}}  }
\end{center}

\vspace{0.5 cm}

\begin{center}
\large{Fernando Dobarro \footnote{Author partially supported by
funds of the National Group 'Analisi Reale' of the Italian
Ministry of University and Scientific Research at the University
of Trieste.} \qquad B\"ulent \"Unal }
\end{center}

\vspace{0.5 cm}

\begin{abstract}
Essentially, some conditions for the Riemannian factor and the
warping function of a standard static space-time are obtained in
order to guarantee that no nontrivial warping function on the
Riemannian factor can make the standard static space-time
Einstein.
\end{abstract}

\vspace{0.5 cm}

\renewcommand{\thepage}{\arabic{page}}
\setcounter{page}{1}

\section{Introduction}

In order to obtain general solutions to Einstein's field
equations, Lorentzian warped product manifolds were introduced
in general relativity (see \cite{BEE,ON}). Generalized
Robertson-Walker space-times and standard static space-times are
two well known important examples. The former are clearly a
generalization of Robertson-Walker space-times and the latter a
generalization of the Einstein static universe. In this paper,
we basically focus on properties of the Ricci tensor and scalar
curvature of a standard static space-time.

We recall the definition of a warped product of two
pseudo-Riemannian manifolds $(B,g_B)$ and $(F,g_F)$ with a
smooth function $b \colon B \to (0,\infty)$ (see \cite{BEE,ON}).
Suppose that $(B,g_B)$ and $(F,g_F)$ are pseudo-Riemannian
manifolds and also suppose that $b \colon B \to (0,\infty)$ is a
smooth function. Then the (singly) warped product, $B \times
{}_b F$ is the product manifold $B \times F$ equipped with the
metric tensor $g=g_B \oplus b^{2}g_F$ defined by $$
g=\pi^{\ast}(g_B) \oplus (b \circ \pi)^2 \sigma^{\ast}(g_F)$$
where $\pi \colon B \times F \to B$ and $\sigma \colon B \times
F \to F$ are the usual projection maps and ${}^\ast$~denotes the
pull-back operator on tensors. Here, $(B,g_F)$ is called as the
base manifold and $(F,g_F)$ is called as the fiber manifold and
also $b$ is called as the warping function. There are also
different generalizations of warped products such as warped
products with more than one fiber manifold, called multiply
warped products (see \cite{MWP}) or warped products with two
warping functions acting symmetrically on the fiber and base
manifolds, called doubly warped products (see \cite{DWP}).
Finally, a warped product is said to be a twisted product if the
warping function defined on the product of the base and fiber
manifolds (see \cite{TWP}).

Basically, a standard static space-time can be considered as a
Lorentzian warped product where the warping function is defined
on a Riemannian manifold and acting on the negative definite
metric on an open interval of real numbers. More explicitly, a
standard static space-time, ${}_f(a,b) \times F$ is a Lorentzian
warped product furnished with the metric $g=-f^2{\rm d}t^2
\oplus g_F,$ where $(F,g_F)$ is a Riemannian manifold, $f \colon
F \to (0,\infty)$ is smooth, and $-\infty \leq a < b \leq
\infty.$ This class of space-times has been previously
considered by many authors. Now, we give a summary of some major
work about standard static space-times. In \cite{ON}, it was
shown that any static space-time is locally isometric to a
standard static space-time. Kobayashi and Obata \cite{KO} stated
the geodesic equation for this class of space-times and the
causal structure and geodesic completeness was considered in
\cite{AD}, where sufficient conditions on the warping function
for nonspacelike geodesic completeness of the standard static
space-time was obtained (see also \cite{RASM} and \cite{DWP}).
In \cite{AD1}, conditions are found which guarantee that
standard static space-times either satisfy or else fail to
satisfy certain curvature conditions from general relativity.
The existence of geodesics in standard static space-times have
been studied by several authors. S\'{a}nchez \cite{mS2} gives a
good overview of geodesic connectedness in semi-Riemannian
manifolds, including a discussion for standard static
space-times. In \cite{GES}, geodesic structure of standard
static space-times is studied and conditions are found on the
warping function to imply non-returning and pseudo-convex
geodesic systems on a standard static space-time.

The Minkowski space-time and the Einstein static universe are
two most famous examples of standard static space-times (see
\cite{BEE,HE}) which is $\mathbb R \times \mathbb S^3$ equipped
with the metric
$$g=-{\rm d}t^2+({\rm d}r^2+\sin^2 r {\rm d}\theta^2+
\sin^2 r \sin^2 \theta {\rm d} \phi^2)$$ where $\mathbb S^3$ is
the usual 3-dimensional Euclidean sphere and the warping
function $f \equiv 1.$ Another well-known example is the
universal covering space of anti-de Sitter space-time, a
standard static space-time of the form ${}_f \mathbb R \times
\mathbb H^3$ where $\mathbb H^3$ is the 3-dimensional hyperbolic
space with constant negative sectional curvature and the warping
function $f \colon \mathbb H^3 \to (0,\infty)$ defined as
$f(r,\theta,\phi)=\cosh r$  (see \cite{BEE,HE}). As a final
example, we can also mention the Exterior Schwarzschild
space-time (see \cite{BEE,HE}), a standard static space-time of
the form ${}_f \mathbb R \times (2m,\infty) \times \mathbb S^2,$
where $\mathbb S^2$ is the 2-dimensional Euclidean sphere, the
warping function $f \colon (2m,\infty) \times \mathbb S^2 \to
(0,\infty)$ is given by $f(r,\theta,\phi)=\sqrt{1-2m/r},$ $r>2m$
and the line element on $(2m,\infty) \times \mathbb S^2$ is
$${\rm d}s^2=\bigl(1-\frac{2m}{r} \bigl)^{-1} {\rm
d}r^2+r^2({\rm d} \theta^2+\sin^2 \theta {\rm d} \phi^2).$$

In literature, it is not known in full generality that whether
there exists a nontrivial warping function $f$ for which the
warped product pseudo-Riemannian (or Riemannian) manifold $_fB
\times F$ is Einstein for given pseudo-Riemannian (or
Riemannian) manifolds $(F,g_F)$ and $(B,g_B)$.  In fact, the
answer of this question depends on whether there exists a
nontrivial common solution $f$ to some differential equations on
$(F,g_F)$. (See page 267 of \cite{B}). This problem was
considered especially for Einstein Riemannian warped products
with compact base and some partial answers were also provided
(see \cite{GA,KD,KK, KC}). In \cite{KK}, it is proven that an
Einstein Riemannian warped product with a non-positive scalar
curvature and compact base is just a trivial Riemannian product.
Constant scalar curvature of warped products was studied in
\cite{CDM,DD,EJK,EJKS} when the base is compact and of
generalized Robertson-Walker space-times in \cite{EJK}.
Furthermore, partial results for warped products with
non-compact base were obtained in \cite{BD} and \cite{CB}. The
physical motivation of existence of a positive scalar curvature
comes from the positive mass problem. More explicitly, in
general relativity the positive mass problem is closely related
to the existence of a positive scalar curvature (see \cite{ZX}).
As a more general related reference, one can consider \cite{KJ}
to see a survey on scalar curvature of Riemannian manifolds.

The problem of existence of a warping function which makes the
warped product Einstein was already studied for special cases
such as generalized Robertson-Walker space-times and a table
given summarizing different cases of Einstein Ricci tensor of a
generalized Robertson-Walker when the Ricci tensor of the fiber
is Einstein in \cite{ARS} (see also references therein). In this
paper, we consider this problem for standard static space-times.
In fact, we essentially investigate the conditions for $(F,g_F)$
so that there exists no nontrivial function $f$ on $F$
guaranteing that the standard static space-time $_f(a,b) \times
F$ is Einstein. Although the results in this paper remain valid
for the Riemannian setting, that is, when $(B,g_B)$ is the
Euclidean interval $((a,b),{\rm d}t^2)$, we prefer to state
these results in Lorentzian setting since there are certain
standard static space-times in Lorentzian geometry, such as,
Einstein static universe, Schwarzschild exterior space-time and
(universal) anti-de Sitter space-time, which are of interest in
Relativity Theory and appear as examples to the results of this
paper.  See {\it Remark \ref{rem3}} for the summary of the
results of this paper from the view of this Introduction.

Einstein Ricci tensor and constant scalar curvature of standard
static space-times with perfect fluid were already considered in
(see \cite{KO,MM}). Note that a matter is called a perfect fluid
if the energy-momentum tensor ${\rm T}$ has the form
$${\rm T}=(\mu+p)W \otimes W+pg,$$ where $W$ is a 1-form with
$g(W,W)=-1$ and $\mu$ and $p$ are called the energy density and
the pressure, respectively. In \cite{KO}, it is shown that a
standard static space-time ${}_f(a,b) \times F$ has a perfect
fluid if $${\rm Ric}_F-\frac{\tau_F}{s}g_F=\frac{1}{f}
\Bigl({\rm H}_F^f -\frac{\Delta_F(f)}{s}g_F \Bigl),$$ where
${\rm Ric}$ is the Ricci tensor, ${\rm H}$ is the Hessian form
and $\tau_F$ is the scalar curvature on $(F,g_F)$ and also ${\rm
dim}(F)=s$ (see also \cite{HE,ON,MM}). Moreover, in \cite{KO2},
the conformal tensor on standard static space-times with perfect
fluid is studied and it is shown that a standard static
space-time with perfect fluid is conformally flat if and only if
its fiber is Einstein and hence of constant curvature.

In this paper, we will consider arbitrary standard static
space-times, i.e., we will not assume the existence of a perfect
fluid, our results can be considered as extensions of the
results in \cite{KO,KO2,MM} where standard static space-times
with perfect fluid were considered. Duggal studied the scalar
curvature of 4-dimensional triple Lorentzian products of the
form $L \times B \times {}_fF$ and obtained explicit solutions
for the warping function $f$ to have a constant scalar curvature
for this class of products (see \cite{DK}). We also discuss
conditions on the warping function or on the fiber of a standard
static space-time to have a constant scalar curvature on the
space-time. Especially, we show that the Einstein static
universe cannot be generalized to a standard static space-time
$_f\mathbb{R}\times \mathbb{S}^s$ modelling a ``static universe"
with a nonconstant warping function $f$ on $(S^s,d\sigma^2)$,
that is, to a standard static space-time with constant scalar
curvature in {\it Theorem \ref{thm1}}.

\section{Preliminaries}

In this section, we give the formal definition of a standard
static space-time and state some curvature formulas (see
\cite{BEE,ON}).

\begin{dfn} \label{dfna} Let $(F,g_F)$ be an $s$-dimensional
Riemannian manifold and $f \colon F \to (0,\infty)$ be a smooth
function. Then $n(=s+1)$-dimensional product manifold
$(a,b)\times F$ furnished with the metric tensor $g=-f^2{\rm
d}t^2 \oplus g_F$ is called a standard static space-time and is
denoted by $_f (a,b) \times F$, where ${\rm d}t^2$ is the
Euclidean metric tensor on $(a,b)$ and $-\infty \leq a < b \leq
\infty.$
\end{dfn}

Throughout the paper the fiber $(F,g_F)$ of a standard
static space-time of the form $_f(a,b) \times F$ is always
assumed to be connected. Now we state some curvature
formulas for standard static space-times to be used later
in the proofs. Note that, since $_f (a,b) \times F$ and
$F\times_f(a,b)$ are isometric, the curvature formulas
below can easily be obtained from the well known curvature
formulas for warped product metric tensors by making
suitable substitutions (see for example,
\cite{BEE,B,ON,DWP}).

Here, we use sign convention for the Laplacian in \cite{ON},
i.e., defined by or $\Delta={\rm tr}({\rm H}),$ (see page 86 of
\cite{ON}) where ${\rm H}$ is the Hessian form (see page 86 of
\cite{ON}) and ${\rm tr}$ denotes for the trace, or
equivalently, $\Delta={\rm div}({\rm grad}),$ where ${\rm div}$
is the divergence and ${\rm grad}$ is the gradient (see page 85
of \cite{ON}).

\begin{prp} \label{prpa} Let $_f(a,b)\times F$ be a standard
static space-time. If $\tau$ and $\tau_F$ denote the scalar
curvatures of the space-time and fiber, respectively, then
$$\tau=\tau_F-2\frac{\Delta_F(f)}{f},$$
where $\Delta_F$ denotes the Laplace operator on $F.$
\end{prp}

\begin{rem} \label{rema} Note that, in the above Proposition,
since $\tau (t,p)$ is independent of $t\in (a,b)$, $\tau$ is a
lift of a unique function $\tilde\tau$ on $F$ to $(a,b)\times F$.
For brevity in expressions, sometimes we abuse the notation and
write $\tau$ instead of $\tilde\tau$ on $F$ to avoid taking the
lifts of functions on $F$ to $(a,b)\times F$.
\end{rem}

\begin{prp} \label{prpb} Let $_f(a,b)\times F$ be a standard
static space-time and also let $V$ and $W$ be vector fields on
$F.$ If ${\rm Ric}$ and ${\rm Ric}_F$ denote the Ricci tensors
of the space-time and the fiber, respectively, then
$${\rm \Ric}(\frac{\partial}{\partial t}+V, \frac{\partial}{\partial
t}+W)={\rm Ric}_F(V,W) + f \Delta_F(f)- \frac{1}{f}{\rm
H}^f_F(V,W),$$ where ${\rm H}^f_F$ is the Hessian form of $f$ on
$(F,g_F).$
\end{prp}

\section{Standard Static Space-times of Constant Scalar
Curvature}

In this Section, we essentially investigate geometric and
topological conditions on a Riemannian manifold $(F,g_F)$
which yield the constancy of the warping function $f$ on
$F$ of a constant scalar curvature standard static
space-time $_f(a,b)\times F$.

\begin{thm} \label{thm1} Let $_f(a,b)\times F$ be a standard
static space-time where $s\geq 2$. Assume that $(F,g_F)$ is
compact and the scalar curvature $\tau$ of the space-time is
constant. Then, $(F,g_F)$ is of constant scalar curvature
$\tau_F$ if and only if $f$ is constant on $F$. In either case,
$\tau=\tau_F.$
\end{thm}

\begin{proof} First note that, $\int_F(\tau_F-\tau)f=0$ by
{\it Proposition \ref{prpa}} and page 104 of \cite{NPR}. Thus,
since $f>0$ on $M$, $\tau_F(p_0)=\tau$ at some $p_0\in F$. Now,
if $\tau_F$ is constant on $F$ then $\tau_F(p)=\tau$ for all
$p\in F$, and it follows from {\it Proposition \ref{prpa}} that
$\Delta_F(f)=0$ on $(F,g_F)$. Thus, $f$ is constant on $F$.
Conversely, if $f$ is constant on $F$ then, by {\it Proposition
\ref{prpa}}, $\tau_F(p)=\tau$ for all $p\in F$, and hence
$\tau_F$ is constant on $F$.
\end{proof}

Note that, one of the ingredients of the physical concept
of a ``static universe" (see \cite{F}) is the constancy of
the scalar curvature of the space-time modelling a ``static
universe", which in fact, corresponds to the constancy of
the trace of the stress-energy tensor of the space-time via
Einstein equation.  Note that, the Einstein static universe
$\mathbb{R}\times\mathbb{S}^s$ is of constant scalar
curvature, where $\mathbb{S}^s=(S^s,d\sigma^2)$ is the unit
Euclidean sphere. (See page 189 of \cite{BEE}).  Thus, by
{\it Theorem \ref{thm1}}, the Einstein static universe
cannot be generalized to a standard static space-time
$_f\mathbb{R}\times \mathbb{S}^s$ modelling a ``static
universe" with a nonconstant warping function $f$ on
$(S^s,d\sigma^2)$, that is, to a standard static space-time
with constant scalar curvature.

It is noticed that one cannot obtain a non-trivial standard
static space-time of a constant scalar curvature when it has a
compact fiber of a constant scalar curvature. Thus we should
focus on standard static space-times with compact fibers of
nonconstant scalar curvatures. In \cite{EJKS}, a similar problem
was considered on a wider class of warped products (see also
\cite{EJK}). In order to make use of \cite{EJKS}, we introduce a
linear operator ${\rm L} \colon {\rm H}^{1,2}(F) \to {\rm
H}^{1,2}(F)$ on a compact Riemannian manifold $(F,g_F)$ defined
by $${\rm L}(v)=-\Delta_F(v)+\frac{\tau_F(q)}{2}v,$$ where $v$
in the Sobolev space ${\rm H}^{1,2}(F).$ Then we are ready to
state the following result.

\begin{thm} \label{thm1a} Let $(F,g_F)$ be a compact Riemannian
manifold with variable scalar curvature $\tau_F \colon F \to
\mathbb R$ where $s\geq 2.$ Then there exists a smooth function
$f \colon F \to (0,\infty)$ such that the corresponding standard
static space-time $_f(a,b) \times F$ is of constant scalar
curvature $\tau$.
\end{thm}

\begin{proof} From {\it Proposition \ref{prpa}}, like in \cite{DD}
and \cite{EJKS}, we look for $\tau \in \mathbb R$ and $f \in
\mathcal C^\infty(F)$ such that ${\rm L}f=\tau f.$ It is well
known that this type of eigenvalue problem has only one
eigenvalue, $\lambda_1(\tau_F),$ (which is simple) such that the
corresponding eigenfunction is strictly positive (see
\cite{NPR}). Note the centrality of the compactness of $F.$
\end{proof}

\begin{rem} The constant scalar curvature of the standard
static space-time $\tau=2 \lambda_1$ where $\lambda_1$ is
the first eigenvalue of the operator ${\rm L}$ on ${\rm
H}^{1,2}(F).$
\end{rem}

Thus it is possible to produce a non-trivial standard static
space-time of constant scalar curvature when the fiber is
compact and has a nonconstant scalar curvature. Now we turn our
attention to the complete case (not necessarily compact),
roughly speaking, under suitable hypothesis for the curvature of
the fiber, we will give a necessary condition for constant
scalar curvature in a standard static space-time.

\begin{thm} \label{nccase-s} Let $(F,g_F)$ be a complete
manifold without boundary where $s \geq 2$ . Suppose the Ricci
curvature of $F$ is non-negative, and suppose $\Delta_F(\tau_F)
\leq 0$ and also $\|\nabla_F(\tau_F) \|=o(r(x)),$ where $r(x)$
is the distance from $x$ to some fixed point $p \in F.$ If
${}_f(a,b) \times F$ is a standard static space-time with
constant scalar curvature $\tau,$ then $\tau \leq \inf_F
(\tau_F).$
\end{thm}

\begin{proof} By contradiction, suppose that $\tau>\inf_F(\tau_F).$
{\it Proposition \ref{prpa}} implies that $$\Delta_F(f)-
\bigl(\frac{\tau_F}{2}-\frac{\tau}{2} \bigl)f=0,$$ with $\tau$
constant and $f$ is positive. Let $q=(\tau_F-\tau)/2,$ since
$\tau>\inf_F(\tau_F)$ there results $\inf_F(q)<0.$ thus by
Corollary 1.1 of \cite{LY}, we obtain a contradiction.
\end{proof}

\begin{rem} In the previous theorem, we may also require
$(F,g_F)$ be compact and in this case, by the variational
structure of the principal eigenvalue of the operator
$-\Delta_F+\tau_F/2,$ namely $$\frac{\tau}{2}=\inf_{{\rm
H}^{1,2}(F)}\frac{\int_F \|\nabla_F(u)\|^2+
\frac{\tau_F}{2}u^2}{\int_F u^2},$$ there results
$$\frac{\tau}{2} \geq \inf_{{\rm H}^{1,2}(F)} \frac{\int_F
\|\nabla_F(u)\|^2+ \frac{\inf_F(\tau_F)}{2}u^2}{\int_F u^2}=
\frac{\inf_F(\tau_F)}{2}.$$ Here, notice that $\inf_{{\rm
H}^{1,2}(F)}\frac{\int_F \|\nabla_F(u)\|^2+
\frac{\inf_F(\tau_F)}{2}u^2}{\int_F u^2}$ is the principal
eigenvalue of $-\Delta_F+\inf_F(\tau_F)/2$ on the compact
manifold $F.$ So, $\tau \geq \inf_F(\tau_F).$ But by the above
theorem $\tau \leq \inf_F(\tau_F),$ thus $\tau=\inf_F (\tau_F)
\geq 0,$ since the latter inequality holds because of the
non-negative Ricci curvature of $F.$
\end{rem}

We now state a simple result for 2-dimensional standard
static space-times with constant scalar curvatures. Note
that if $M=_f(a,b) \times (c,d)$ is a 2-dimensional
standard static space-time with the metric tensor
$g=-f^2{\rm d}t^2+{\rm d}x^2,$ then its scalar curvature
$\tau$ is given by $\tau=-2f^{\prime \prime}/f,$ where
$-\infty \leq a < b \leq \infty$ and $-\infty \leq c < d
\leq \infty$ also $f \colon (c,d) \to (0,\infty)$ is
smooth.

\begin{prp} Let $_f(a,b) \times (c,d)$ be a 2-dimensional
standard static space-time. Then the scalar curvature
$\tau$ is constant if $f$ satisfies one of the followings
\begin{enumerate}
\item $\displaystyle{f(x)=c_1x+c_2,}$ for some
$c_1,c_2 \in \mathbb R$ when $\tau=0,$
\item $\displaystyle{f(x)=c_1\exp(\sqrt{-2\tau}x)+
c_2\exp(-\sqrt{-2\tau}x),}$ for some $c_1,c_2 \in \mathbb
R$ when $\tau<0,$
\item $\displaystyle{f(x)=c_1\cos(\sqrt{2\tau}x)+
c_2\sin(\sqrt{2\tau}x),}$ for some $c_1,c_2 \in \mathbb R$
when $\tau>0.$
\end{enumerate}
\end{prp}

One can compare the previous result with the characterization of
constant Gauss curvature revolution of surfaces embedded in
$\mathbb R^3$ (see the examples in page 66-67 of \cite{KW},
chapter 3 of \cite{PP} and page 169 of \cite{doC}).

\section{Einstein Standard Static Space-times}

We now concentrate on the Ricci tensor of standard static
space-times. More precisely, we will try to determine
conditions on the warping function of a standard static
space-time so that the space-time becomes Einstein or
Ricci-flat when the fiber is Einstein or Ricci-flat.

Recall that an arbitrary $n$-dimensional pseudo-Riemannian
manifold $(M,g)$ is said to be Einstein with $\lambda$ if there
exists a smooth map $\lambda \colon M \to \mathbb R$ such that
$\Ric=\lambda g.$ Furthermore, if $(M,g)$ is Einstein with
$\lambda$ and $\dim(M)=n \geq 3,$ then $\lambda$ is constant and
$\lambda=\tau/n,$ where $\tau$ is the (constant) scalar
curvature of $(M,g).$ Also note that for a 2-dimensional
Einstein manifold $(M,g)$ with $\lambda,$ one cannot necessarily
conclude the constancy of $\lambda$ (see \cite{ON}).

\begin{prp} \label{prp1}  Let $(F,g_F)$ be an $s$-dimensional
Riemannian manifold with scalar curvature $\tau_F$ where $s\geq
2$ and let the standard static space-time $_f(a,b)\times F$ be
Einstein with constant scalar curvature $\tau$. Then,
\begin{enumerate}
\item[a)] $\Delta_F(f)=-\frac{\tau}{n}f$ and
${\rm Ric}_F=\frac{1}{f}{\rm H}_F^f+\frac{\tau}{n}g_F$ on
$(F,g_F)$, where $n=s+1$.
\item[b)] $(s+1)\tau_F=(s-1)\tau$.
\end{enumerate}
\end{prp}

\begin{proof} By assumption, since
${\rm Ric}=\frac{\tau}{n}(-f^2dt^2\oplus g_F)$, we obtain from
{\it Proposition \ref{prpb}} that,
$${\rm Ric}_F(V,W)+f\Delta_F(f)-\frac{1}{f}{\rm H}_F^f(V,W)=-
\frac{\tau}{n}f^2+\frac{\tau}{n}g_F(V,W)$$ for all $V,W\in \Gamma
TF$.
\begin{enumerate}
\item[{\it a)}] If we set $V=0=W$ in the above expression, we obtain
$\Delta_F(f)=-\frac{\tau}{n}f$ and hence, it also follows
that ${\rm Ric}_F(V,W)=\frac{1}{f}{\rm
H}_F^f(V,W)+\frac{\tau}{n}g_F(V,W)$ for all $V,W\in\Gamma
TF$.
\item[{\it b)}] Note that, if we take the trace of the equation
${\rm Ric}_F=\frac{1}{f}{\rm H}_F^f+\frac{\tau}{n}g_F$ with
respect to $g_F$ on $F$, we obtain
$\tau_F=\frac{1}{f}\Delta_F(f)+\frac{s}{n}\tau$, and hence,
it follows from {\it (a)} that $(s+1)\tau_F=(s-1)\tau$.
\end{enumerate}
\end{proof}

\begin{rem} Let $(F,g_F)$ be a Riemannian manifold of
scalar curvature $\tau_F$ where $s\geq 2$ and let the
standard static space-time $_f(a,b)\times F$ be Einstein
with constant scalar curvature $\tau$.  Then, by {\it
Theorem \ref{thm1}} and {\it Proposition \ref{prp1}}, we
conclude the following:
\begin{enumerate}
\item[a)] $(F,g_F)$ is of constant scalar curvature $\tau_F$.
\item[b)] If $(F,g_F)$ is compact then $\tau=0,$ \,
$\tau_F=0$ and $f$ is constant on $F$.
\end{enumerate}
\end{rem}

\begin{rem} \label{rem1} Note that, in {\it Proposition
\ref{prp1}}, if we further assume that $(F,g_F)$ is
Einstein with scalar curvature $\tau_F$ then, since ${\rm
Ric}_F=\frac{\tau_F}{s}g_F$ on $(F,g_F)$, we obtain by
using {\it Proposition \ref{prp1}} that ${\rm
H}_F^f=(\frac{\tau_F}{s}-\frac{\tau}{n})fg_F=-\frac{\tau_F}{s(s-1)}g_F
=-\frac{\tau}{s(s+1)}g_F$ on $(F,g_F)$.  (Note that, by
{\it Proposition \ref{prp1}-b}, $\tau_F$ is also constant
when $dim\,F=s=2$).
\end{rem}

\begin{thm} \label{thm2} Let $(F,g_F)$ be a complete
Riemannian manifold with nonnegative Ricci curvature where
$s\geq 2$. If the standard static space-time $_f(a,b)\times
F$ is Ricci flat then $f$ is constant on $F$.
\end{thm}

\begin{proof} Since $_f(a,b)\times F$ is Einstein with scalar
curvature $\tau = 0$, it follows from {\it Proposition
\ref{prp1}} that $\Delta_F(f)=0$ on $(F,g_F)$.  Since $f$
is positive on $F$, $f$ is constant by {\it Corollary 1} of
\cite{Y}.
\end{proof}

Note that, by {\it Theorem \ref{thm2}}, if a standard
static space-time $_f(a,b)\times F$ is Ricci flat with a
nonconstant warping function $f$ on $F$ then $(F,g_F)$ is
either incomplete or not of nonnegative Ricci curvature (or
both).  Hence, for the Schwarzschild exterior space-time
$_f\mathbb{R}\times F$ (see page 367 of \cite{ON}), we
conclude that $(F,g_F)$ is either incomplete or not of
nonnegative Ricci curvature, where $N=(2m,\infty)\times
S^2$ and $g_F=\bigl(1-\frac{2m}{r} \bigl)^{-1}
dr^2+d\sigma^2$. Indeed, for the Schwarzschild exterior
space-time, $(F,g_F)$ is both incomplete and not of
nonnegative Ricci curvature.

\begin{rem} Note that the converse of {\it Theorem
\ref{thm2}} is not true in general.  For example, the
$n(\geq 3)$-dimensional Einstein static universe is an
counterexample to this case.
\end{rem}

\begin{thm} \label{thm3} Let $(F,g_F)$ be a complete
Einstein $s(\geq 2)$-dimensional Riemannian
manifold with scalar curvature $\tau_F$. If the standard
static space-time $_f(a,b)\times F$ is Einstein with scalar
curvature $\tau$ then $\tau\leq 0$. Furthermore,
\begin{enumerate}
\item[a)] if $\tau =0$ then $f$ is constant on $F$,
\item[b)] if $\tau <0$ then $f$ is nonconstant on $F$ and
$(F,g_F)$ is a warped product of the Euclidean line and a complete
Riemannian manifold with warping function $\psi$ on the real line
satisfying the equation
$\frac{d^2\psi}{dt^2}+\frac{\tau}{s(s+1)}\psi=0$, $\psi >0$.
\end{enumerate}
\end{thm}

\begin{proof} Suppose that $\tau >0$.  Then by {\it Proposition
\ref{prp1}}, $\tau_F=\frac{s-1}{s+1}\tau >0$ and it follows
from Myers theorem that $(F,g_F)$ is compact.  Hence, by
{\it Theorem \ref{thm1}}, $f$ is constant on $F$.  But this
conflicts with $\Delta_F(f)=-\frac{\tau}{n}f$ since $f>0$
on $F$ (see {\it Proposition \ref{prp1}}).  Thus $\tau \leq
0$.  Furthermore,
\begin{enumerate}
\item[{\it a)}] if $\tau =0$ then, by {\it Proposition \ref{prp1}},
$\Delta_F(f)=0$ on $(F,g_F)$ and ${\rm
Ric}_F=\frac{\tau_F}{s}g_F=0$ since $\tau_F=0$.  Thus,
since $f$ is positive on $F$, $f$ is constant by {\it
Corollary 1} of \cite{Y}.
\item[{\it b)}] if $\tau <0$ then, by {\it Proposition
\ref{prp1}} and {\it Remark \ref{rem1}}, $f$ is nonconstant
and ${\rm H}_F^f=-\frac{\tau_F}{s(s-1)}g_F$ on $(F,g_F)$,
where $\tau_F=\frac{s-1}{s+1}\tau <0$.  Hence, it follows
from {\it Corollary E} of \cite{KA} that $(F,g_F)$ is a
warped product of the Euclidean line and a complete
Riemannian manifold with warping function $\psi$ on the
real line satisfying the equation
$\frac{d^2\psi}{dt^2}+\frac{\tau}{s(s+1)}\psi=0$, $\psi
>0$.
\end{enumerate}
\end{proof}

Note that, the (universal) anti-de Sitter space-time
$_f\mathbb{R}\times\mathbb{H}^s$ of constant sectional
curvature $-1$ is an example to {\it Theorem \ref{thm3}-b}
for $s\geq 2$. (See page 183 of \cite{BEE}).  Indeed, note
that the Riemannian hyperbolic space $\mathbb{H}^s$ can be
written as a warped product of the Euclidean line and the
Euclidean space with warping function $\psi =e^{\pm t}$ on
the real line.

As we did in {\it Section 3} we will consider the same type of
problems on a standard static space-time by using \cite{LY} when
the fiber is complete without boundary.

\begin{thm} \label{nccase-e1} Let $(F,g_F)$ be a complete
manifold without boundary where $s \geq 2.$ Suppose the Ricci
curvature of $F$ is non-negative, and suppose $\Delta_F(\tau_F)
\leq 0$ and also $\|\nabla_F(\tau_F) \|=o(r(x)),$ where $r(x)$
is the distance from $x$ to some fixed point $p \in F.$ If
${}_f(a,b) \times F$ is an Einstein standard static space-time
then $\tau \leq \frac{s+1}{s}\inf_F(\tau_F).$
\end{thm}

\begin{proof} First note that the space-time is Einstein this
means that ${\rm \Ric}=\lambda g,$ where $(s+1)\lambda=\tau$ and
$\tau$ is constant. Considering the trace in $(F,g_F)$ in {\it
Proposition \ref{prp1}(a)}, there results a positive solution
$f$ for the Schr\"odinger equation $-\Delta_F(f)+q_E(f),$ where
$q_E=\tau_F-\frac{s}{s+1} \tau.$ On the other hand, $q_E$
verifies $\Delta_F(q_E) \leq 0$ and $\|\nabla_F(q_E)\|=o(r(x))$
with $r(x)$ as in the hypothesis. So by Corollary 1.1 of
\cite{LY}, $0 \geq \inf_F(q_E)=\inf_F-\frac{s}{s+1}\tau,$ or
equivalently $\tau \leq \frac{s+1}{s} \inf_F(\tau_F).$
\end{proof}

\begin{thm} \label{nccase-e2} Let $(F,g_F)$ be a complete
manifold without boundary where $s \geq 2.$ Suppose the Ricci
curvature of $F$ is non-negative. If ${}_f(a,b) \times F$ is an
Einstein standard static space-time then $\tau=\tau_F=0$ and
hence the space-time is Ricci-flat.
\end{thm}

\begin{proof} As again in the proof of the previous theorem,
note that the space-time is Einstein this means that ${\rm
\Ric}=\lambda g,$ where $(s+1)\lambda=\tau$ and $\tau$ is
constant. By  {\it Proposition \ref{prp1}(a)},
$\tau_F=\frac{s-1}{s+1} \tau$ is constant. Thus the hypothesis
of {\it Theorem \ref{nccase-e1}} are satisfied and this leads
$\tau \leq \frac{s+1}{s} \inf_F(\tau_F)=\frac{s-1}{s} \tau.$ On
the other hand, as Ricci curvature of $F$ is non-negative,
$\tau=\frac{s+1}{s-1}\tau_F \geq 0.$ So, $\tau=0.$
\end{proof}

\begin{cor} \label{nccase-e3} Let $(F,g_F)$ be a complete
manifold without boundary where $s \geq 2.$ Suppose the Ricci
curvature of $F$ is non-negative. If ${}_f(a,b) \times F$ is an
Einstein standard static space-time then $f$ is constant and
$(F,g_F)$ is Ricci-flat.
\end{cor}

\begin{proof} It just follows {\it Theorem \ref{thm2}}
and {\it Theorem \ref{nccase-e2}} and also {\it Proposition
\ref{prp1}(a)}.
\end{proof}

As a consequence, we obtain the result that follows:

\begin{thm} \label{nccase-e4} Let $(F,g_F)$ be a complete
manifold without boundary where $s \geq 2.$ If ${}_f(a,b) \times
F$ is an Einstein standard static space-time then either
$(F,g_F)$ is Ricci-flat or the Ricci curvature of $F$ cannot be
non-negative.
\end{thm}

In the remark below, we collect the results in this section from
viewpoint of the nonexistence of a nonconstant warping function
$f$ on a connected, complete Riemannian manifold $(F,g_F)$ for
which the standard static space-time $_f(a,b)\times F$ is
Einstein.

\begin{rem} \label{rem3} Let $(F,g_F)$ be a Riemannian
manifold and $_f(a,b)\times F$ be a standard static space-time
where $s\geq 2.$
\begin{enumerate}
\item[a)] If $(F,g_F)$ is compact and of constant scalar curvature
(or Einstein) then there exists no nonconstant function $f$ on
$F$ for which $_f(a,b)\times F$ is of constant scalar curvature
(or Einstein). (See {\it Theorem \ref{thm1}}).
\item[b)] If $(F,g_F)$ is complete and of nonnegative Ricci
curvature then there exists no nonconstant function $f$ on $F$
for which $_f(a,b)\times F$ is Ricci flat. (See {\it Theorem
\ref{thm2}}).
\item[c)] If $(F,g_F)$ is complete and Einstein then there exists
no nonconstant function $f$ on $F$ for which $_f(a,b)\times F$
is Ricci flat. (See {\it Theorem \ref{thm3}}).
\item[d)] If $(F,g_F)$ is complete and of non-negative Ricci
curvature, then there exists no nonconstant function $f$ on $F$
for which $_f(a,b)\times F$ is Einstein. (See {\it Corollary
\ref{nccase-e3}}).
\end{enumerate}
\end{rem}

Here note that, the above Remark remains valid (as well as other
results in this section) in the Riemannian setting, that is,
when we take $(a,b)$ as an (Euclidean) interval with metric
tensor ${\rm d}t^2$.

We now consider 2-dimensional Einstein standard static
space-times. Let $M=_f(a,b) \times (c,d)$ be a 2-dimensional
standard static space-time with the metric tensor $g=-f^2{\rm
d}t^2+{\rm d}x^2.$ If $\frac{\partial}{\partial t} \in \mathfrak
X(a,b)$ and $\frac{\partial}{\partial x} \in \mathfrak X(c,d),$
then
$${\rm Ric}(\frac{\partial}{\partial t}+
\frac{\partial}{\partial x})=ff^{\prime \prime}-\frac{f^{
\prime \prime}}{f},$$ where $-\infty \leq a < b \leq
\infty$ and $-\infty \leq c < d \leq \infty$ also $f \colon
(c,d) \to (0,\infty)$ is smooth. Thus we can easily state
the following result.

\begin{prp} Let $_f(a,b) \times (c,d)$ be a 2-dimensional
standard static space-time. Then
\begin{enumerate}
\item the space-time is Einstein with $\lambda$ if and only
if $\displaystyle{f^{\prime \prime}=-\lambda f},$
\item the space-time is Ricci-flat if and only
if $\displaystyle{f(x)=c_1x+c_2}$ on $(c,d),$ for some
$c_1,c_2 \in \mathbb R.$
\end{enumerate}
\end{prp}

In the previous result, since we cannot conclude the constancy
of $\lambda,$ it is impossible for us to obtain explicit
solutions. However, we can only say that $\lambda \colon (a,b)
\times (c,d) \to \mathbb R$ depends only on the second variable,
i.e., for each $x \in (c,d),$ we have $\lambda(t_1,x)=
\lambda(t_2,x)$ for any $t_1, t_2 \in (a,b).$

config: arxiv


TITLE: The Finishing Step to Showing that $1/(1-4x)^{1/2}$ generates the sequence $\binom{2r}{r}$
QUESTION [2 upvotes]: The Full Question:
Show that $(1-4x)^{-\frac{1}{2}}$ generates the sequence $\binom{2n}{n}$, $n\in
\mathbb N$
My Research
How to show that $1 \over \sqrt{1 - 4x} $ generate $\sum_{n=0}^\infty \binom{2n}{n}x^n $
Show $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$
Both these questions seem similar, but I found they used a very different approach (more calculus based) to find their answers whereas my question requires me to use more algebra and definitions to get the answer.
My Work
1) Let $y = -4x$
2) We know the $r^{th}$ term of$(1-y)^{-1/2}$ will have the coefficient of $\binom{-1/2}{r}(-4^r)$
3) $\binom{-1/2}{r}(-4^r) = \frac{(-1/2)(-1/2-1)(-1/2-2)\cdots(-1/2-r+1)}{r!}(-4)^r$
4) If we factor out $(-1/2)$ from the whole thing, we have: $\frac{(1)(1+2)(1+4)\cdots(1+2r-2)}{r!}(2^r)=\frac{(2r-1)(2r-3)\cdots(5)(3)(1)}{r!}2^r$
5) If we multiply top and bottom by $r!$ we get:$\frac{(2r-1)(2r-3)\cdots(5)(3)(1)2^rr!}{r!r!}$
6) All that remains is to show the numerator to equal $(2r)!$ and I'm golden, however I don't really know how to do this.
My Question
Can anyone help me with step 6? I feel I'm very close but haven't quite been able to see it.

REPLY [1 votes]: Note: Avoiding a small calculation error in OPs work in 4.) and considering a small trick in 6.) will close the gap. Here's a calculation:
\begin{align*}
\binom{-\frac{1}{2}}{r}
&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)\cdot\ldots\cdot
\left(-\frac{1}{2}-(r-1)\right)}{r!}\\
&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdot\ldots\cdot
\left(-\frac{2r-1}{2}\right)}{r!}\\
&=\left(-\frac{1}{2}\right)^r\cdot\frac{(2r-1)(2r-3)\cdot\ldots\cdot5\cdot3\cdot1}{r!}\tag{1}\\
&=\left(-\frac{1}{2}\right)^r\cdot\frac{(2r)(2r-1)(2r-2)(2r-3)\cdot\ldots\cdot5\cdot4\cdot3\cdot2\cdot1}
{r!(2r)(2r-2)(2r-4)\cdot\ldots\cdot4\cdot2}\tag{2}\\
&=\left(-\frac{1}{2}\right)^r\cdot\frac{(2r)!}
{r!2^r\cdot r(r-1)\cdot\ldots\cdot2\cdot1}\\
&=\left(-\frac{1}{4}\right)^r\frac{(2r)!}{r!r!}\\
&=\left(-\frac{1}{4}\right)^r\binom{2r}{r}\\
\end{align*}

Comment:

In (1) we factor out $\left(-\frac{1}{2}\right)^r$. Please note that $2^r$ is part of the denominator (check corr. part (4) in your calculation)
In (2) we fill in even factors to complete the numerator to $(2r)!$ and we also write these factors in the denominator as compensation.

Clever notation: In (1) it's convenient to use the compact notation of double factorials.We observe
  $$(2r)!=(2r)!!(2r-1)!!=2^rr!(2r-1)!!$$
  and using this notation in (1):
\begin{align*}
\binom{-\frac{1}{2}}{r}
&=\ldots\\
&=\left(-\frac{1}{2}\right)^r\frac{(2r-1)!!}{r!}\\
&=\left(-\frac{1}{2}\right)^r\frac{(2r)!}{r!(2r)!!}\\
&=\left(-\frac{1}{2}\right)^r\frac{(2r)!}{r!2^rr!}\\
&=\left(-\frac{1}{4}\right)^r\binom{2r}{r}\\
\end{align*}


text:
\begin{document}

\title{higher auslander's formula}

\author{Ramin Ebrahimi}
\address{Department of Pure Mathematics\\
Faculty of Mathematics and Statistics\\
University of Isfahan\\
P.O. Box: 81746-73441, Isfahan, Iran}
\email{ramin69@sci.ui.ac.ir}

\author{Alireza Nasr-Isfahani}
\address{Department of Pure Mathematics\\
Faculty of Mathematics and Statistics\\
University of Isfahan\\
P.O. Box: 81746-73441, Isfahan, Iran\\ and School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box: 19395-5746, Tehran, Iran}
\email{nasr$_{-}$a@sci.ui.ac.ir / nasr@ipm.ir}

\subjclass[2010]{{18E10}, {18E20}, {18E99}}

\keywords{abelian category, $n$-abelian category, $n$-cluster tilting,  Auslander's formula}

\begin{abstract}
Let $\mathcal{M}$ be a small $n$-abelian category. We show that the category of finitely presented functors $\modd$-$\mathcal{M}$ modulo the subcategory of effaceable functors $\modd_0$-$\mathcal{M}$ has an $n$-cluster tilting subcategory which is equivalent to $\mathcal{M}$. This  gives a higher-dimensional version of Auslander's formula.
\end{abstract}

\maketitle

\section{Introduction}
For a positive integer $n$, $n$-cluster tilting subcategories of abelian categories were introduced by Iyama \cite{I1, I2} (see also \cite{I3} and \cite{I4}) to construct a higher Auslander correspondence and develop the higher-dimensional analogs of Auslander-Reiten theory. Recently, Jasso \cite{J} introduced $n$-abelian categories as a higher-dimensional analogs of abelian categories from the viewpoint of higher homological
algebra. $n$-abelian categories are an axiomatization of $n$-cluster tilting subcategories. In this subcategories kernels and cokernels don't necessarily exist and are replaced with $n$-kernels and $n$-cokernels. Jasso in \cite{J} proved that any $n$-cluster tilting subcategory of an abelian category is $n$-abelian.

Let $\mathcal{M}$ be a projectively generated small $n$-abelian category, $\mathcal{P}$ be the
subcategory of projective objects in $\mathcal{M}$ and $F:M\rightarrow \modd$-$\mathcal{P}$ be the functor defined
by $F(X)=\mathcal{M}(-, X)|_{\mathcal{P}}$. Jasso in \cite{J} proved that if there exists an exact duality $D:
\modd$-$\mathcal{P}\rightarrow \modd$-$\mathcal{P}^{op}$, then $F$ is fully faithful and the essential image of $F$ is an $n$-cluster tilting subcategory of $\modd$-$\mathcal{P}$. Kvamme in \cite{Kv} proved that the existence of exact
duality is unnecessary. In this paper we show that the projectively generated assumption is also unnecessary.

Let $\mathcal{M}$ be a small $n$-abelian category. In \cite{EN} we showed that there exist an abelian category $\mathcal{A}$ and a fully faithful functor $\mathcal{M}\hookrightarrow \mathcal{A}$ which preserve and reflects $n$-exact sequences (see Definition \ref{d1}). Using these result we prove the following theorem, which is one of the main result of this paper.

\begin{theorem a*}$($Theorem \ref{M1}$)$
Let $\mathcal{M}$ be a small $n$-abelian category. Then $\mathcal{M}$ is equivalent to an $n$-cluster tilting subcategory of an abelian category $\mathcal{B}$.
\end{theorem a*}

Let $\mathcal{C}$ be a small abelian category and $\mathcal{G}$ be the category of abelian groups. All contravariant additive functors $F:\mathcal{C}\rightarrow \mathcal{G}$ and natural transformations between them, form an abelian category which is denoted by $\Mod$-$\mathcal{C}$ or $(\mathcal{C}^{op},\mathcal{G})$. A contravariant additive functor $F$ is called finitely presented if there exist $C_1, C_2\in \mathcal{C}$ and an exact sequence $(-,C_1)\rightarrow (-,C_2)\rightarrow F\rightarrow 0$ in $\Mod$-$\mathcal{C}$. The full subcategory of $\Mod$-$\mathcal{C}$ consisting of all finitely presented functors denoted by $\modd$-$\mathcal{C}$. A finitely presented functor $F$ is called effaceable if there exists a presentation $(-,C_1)\rightarrow (-,C_2)\rightarrow F\rightarrow 0$ such that $C_1\rightarrow C_2$ is an epimorphism. The full subcategory of $\modd$-$\mathcal{C}$ consisting of all effaceable functors, denoted by $\modd_0$-$\mathcal{C}$, is a Serre subcategory of $\modd$-$\mathcal{C}$. A very famous and important formula due to Auslander, which is known as Auslander's formula, shows that we have the following equivalence of abelian categories

\begin{equation}
\frac{\modd\text{-}\mathcal{C}}{\modd_0\text{-}\mathcal{C}}\simeq \mathcal{C}. \notag
\end{equation}

This equivalence shows that to study the abelian category $\mathcal{C}$ it is enough to study the category of finitely presented functors that has better homological behaviour. This leads to functorial approach in representation theory, and as Lenzing said in \cite{L}, "a substantial part of
Maurice Auslander's work on the representation theory of Artin algebras can be linked to the formula". Krause in \cite{K} proved a derived version of Auslander's formula.

In this paper we prove a higher-dimensional version of Auslander's formula for small $n$-abelian categories. More precisely we prove the following theorem.

\begin{theorem b*}$($Theorem \ref{M2}$)$
Let $\mathcal{M}$ be a small $n$-abelian category. The category of finitely presented functors $\modd$-$\mathcal{M}$ modulo the subcategory of effaceable functors $\modd_0$-$\mathcal{M}$ has an $n$-cluster tilting subcategory which is equivalent to $\mathcal{M}$.
\end{theorem b*}

The paper is organized as follows. In section 2 we recall the definitions of $n$-abelian categories, $n$-cluster tilting subcategories and recall some results that we need in the paper. In \cite[Theorems 4.2 and 4.3]{EN} we showed that for a small $n$-abelian category $\mathcal{M}$ there exist an abelian category $\mathcal{A}$ and the embedding
\begin{equation}
\widetilde{H}:\mathcal{M}\rightarrow \mathcal{A}. \notag
\end{equation}

 In Section 3 we prove that the essential image of $\widetilde{H}$ is $n$-rigid. In Section 4 we first show that there exists an abelian subcategory $\mathcal{B}$ of $\mathcal{A}$ such that $\mathcal{M}$ is equivalent to an $n$-cluster tilting subcategory of $\mathcal{B}$. Finally, following Auslander, we describe the category of finitely presented functors on $\mathcal{M}$ modulo the subcategory of effaceable functors and show that $\mathcal{M}$ is equivalent to an $n$-cluster tilting subcategory of this quotient category.

\subsection{Notation}
Throughout this paper, unless otherwise stated, $n$ always denotes a fixed positive integer, $\mathcal{M}$ is a fixed small $n$-abelian category and $\mathcal{G}$ is the abelian category of all abelian groups.

\section{preliminaries}
In this section we recall the definition of $n$-abelian category, $n$-cluster tilting subcategory and recall some results that we need in the rest of paper. For further information and motivation of definitions the readers are referred to \cite{I1, I2, I3, J}.

\subsection{$n$-abelian categories}
Let $\mathcal{M}$ be an additive category and $f:A\rightarrow B$ a morphism in $\mathcal{M}$. A weak cokernel of $f$ is a morphism $g:B\rightarrow C$ such that for all $C^{\prime} \in \mathcal{M}$  the sequence of abelian groups
\begin{equation}
\mathcal{M}(C,C')\overset{(g,C')}{\longrightarrow} \mathcal{M}(B,C')\overset{(f,C')}{\longrightarrow} \mathcal{M}(A,C') \notag
\end{equation}
is exact. The concept of weak kernel is defined dually.

Let $d^0:X^0 \rightarrow X^1$ be a morphism in $\mathcal{M}$. An $n$-cokernel of $d^0$ is a sequence
\begin{equation}
(d^1, \ldots, d^n): X^1 \overset{d^1}{\rightarrow} X^2 \overset{d^2}{\rightarrow}\cdots \overset{d^{n-1}}{\rightarrow} X^n \overset{d^n}{\rightarrow} X^{n+1} \notag
\end{equation}
of objects and morphisms in $\mathcal{M}$ such that for each $Y\in \mathcal{M}$
the induced sequence of abelian groups
\begin{align}
0 \rightarrow \Hom(X^{n+1},Y) \rightarrow \Hom(X^n,Y) \rightarrow\cdots\rightarrow \Hom(X^1,Y) \rightarrow \Hom(X^0,Y) \notag
\end{align}
is exact. Equivalently, the sequence $(d^1, \ldots, d^n)$ is an $n$-cokernel of $d^0$ if for all $1\leq k\leq n-1$
the morphism $d^k$ is a weak cokernel of $d^{k-1}$, and $d^n$ is moreover a cokernel of $d^{n-1}$ \cite[Definition 2.2]{J}. The concept of $n$-kernel of a morphism is defined dually.
\begin{definition}(\cite[Definition 2.4]{L})\label{d1}
Let $\mathcal{M}$ be an additive category. A right $n$-exact sequence in $\mathcal{M}$ is a complex
\begin{equation}
X^0 \overset{d^0}{\rightarrow} X^1 \overset{d^1}{\rightarrow} \cdots \overset{d^{n-1}}{\rightarrow} X^n \overset{d^n}{\rightarrow} X^{n+1} \notag
\end{equation}
such that $(d^1, \ldots, d^{n})$ is an $n$-cokernel of $d^0$. The concept of left $n$-exact sequence is defined dually. An $n$-exact sequence is a sequence which is both a right $n$-exact sequence and a left $n$-exact sequence.
\end{definition}

Let $\mathcal{M}$ be a category and $A$ be an object of $\mathcal{M}$. A morphism $e\in \mathcal{M}(A, A)$ is idempotent if $e^2 = e$. $\mathcal{M}$ is called idempotent complete
if for every idempotent $e\in \mathcal{M}(A, A)$ there exist an object $B$ and morphisms $f\in\mathcal{M}(A, B)$ and
$g\in\mathcal{M}(B, A)$ such that $gf = e$ and $fg = 1_B$.

\begin{definition}$($\cite[Definition 3.1]{J}$)$
An $n$-Abelian category is an additive category $\mathcal{M}$ which satisfies the following axioms:
\begin{itemize}
\item[(A0)]
The category $\mathcal{M}$ is idempotent complete.
\item[(A1)]
Every morphism in $\mathcal{M}$ has an $n$-kernel and an $n$-cokernel.
\item[(A2)]
For every monomorphism $d^0:X^0 \rightarrow X^1$ in $\mathcal{M}$ and for every $n$-cokernel $(d^1, \ldots, d^n)$ of $d^0$, the  following sequence is $n$-exact:
\begin{equation}
X^0 \overset{d^0}{\rightarrow} X^1 \overset{d^1}{\rightarrow} \cdots \overset{d^{n-1}}{\rightarrow} X^n \overset{d^n}{\rightarrow} X^{n+1}. \notag
\end{equation}
\item[(A3)]
For every epimorphism $d^n:X^n \rightarrow X^{n+1}$ in $\mathcal{M}$ and for every $n$-kernel $(d^0, \ldots, d^{n-1})$ of $d^n$, the following sequence is $n$-exact:
\begin{equation}
X^0 \overset{d^0}{\rightarrow} X^1 \overset{d^1}{\rightarrow} \cdots \overset{d^{n-1}}{\rightarrow} X^n \overset{d^n}{\rightarrow} X^{n+1}. \notag
\end{equation}
\end{itemize}
\end{definition}

\begin{remark}
Let $\mathcal{M}$ be an additive category, and consider the following morphism of complexes $f:X\rightarrow Y$
\begin{center}
\begin{tikzpicture}
\node (X1) at (-4,1) {$X$};
\node (X2) at (-2,1) {$X^0$};
\node (X3) at (0,1) {$X^1$};
\node (X4) at (2,1) {$\ldots$};
\node (X5) at (4,1) {$X^{n-1}$};
\node (X6) at (6,1) {$X^n$};
\node (X7) at (-4,-1) {$Y$};
\node (X8) at (-2,-1) {$Y^0$};
\node (X9) at (0,-1) {$Y^1$};
\node (X10) at (2,-1) {$\ldots$};
\node (X11) at (4,-1) {$Y^{n-1}$};
\node (X12) at (6,-1) {$Y^n$};
\draw [->,thick] (X1) -- (X7) node [midway,left] {$f$};
\draw [->,thick] (X2) -- (X8) node [midway,left] {$f^0$};
\draw [->,thick] (X3) -- (X9) node [midway,left] {$f^1$};
\draw [->,thick] (X5) -- (X11) node [midway,left] {$f^{n-1}$};
\draw [->,thick] (X6) -- (X12) node [midway,left] {$f^n$};
\draw [->,thick] (X2) -- (X3) node [midway,above] {$d_0^X$};
\draw [->,thick] (X3) -- (X4) node [midway,above] {$d_1^X$};
\draw [->,thick] (X4) -- (X5) node [midway,above] {$d_{n-2}^X$};
\draw [->,thick] (X5) -- (X6) node [midway,above] {$d_{n-1}^X$};
\draw [->,thick] (X8) -- (X9) node [midway,above] {$d_0^Y$};
\draw [->,thick] (X9) -- (X10) node [midway,above] {$d_1^Y$};
\draw [->,thick] (X10) -- (X11) node [midway,above] {$d_{n-2}^Y$};
\draw [->,thick] (X11) -- (X12) node [midway,above] {$d_{n-1}^Y$};
\end{tikzpicture}
\end{center}
The mapping cone $C(f)$ is the following complex
\begin{equation}
X^0\overset{d_C^0}{\longrightarrow} X^1\oplus Y^0 \overset{d_C^1}{\longrightarrow} \ldots \overset{d_C^{n-2}}{\longrightarrow} X^n\oplus Y^{n-1}\overset{d_C^n}{\longrightarrow} Y^n, \notag
\end{equation}
where $$d_C^k=\left(\begin{array}{cc}
-d_k^X &0 \\
f^k &d_{k-1}^Y\\
\end{array}\right):X^k\oplus Y^{k-1}\longrightarrow X^{k+1}\oplus Y^k,$$ for each $0\leq k\leq n$. In particular, $$d_C^0=\binom{-d_0^X}{f^0} \text{ }\text{and}\text{ } d_C^n=(f^n \text{ }\text{ } d_{n-1}^Y).$$
\end{remark}

\begin{definition}$($\cite[Definition 2.11]{J}$)$
Let $\mathcal{M}$ be an additive category, $X: X^0 \overset{d^0}{\rightarrow} X^1 \overset{d^1}{\rightarrow}\cdots \overset{d^{n-2}}{\rightarrow} X^{n-1}\overset{d^{n-1}}{\rightarrow} X^n$ a complex in $\mathcal{M}$ and $f^0:X^0\rightarrow Y^0$ a morphism in $\mathcal{M}$. An $n$-pushout diagram of $X$ along $f^0$ is a morphism of complexes
\begin{center}
\begin{tikzpicture}
\node (X1) at (-4,1) {$X$};
\node (X2) at (-2,1) {$X^0$};
\node (X3) at (0,1) {$X^1$};
\node (X4) at (2,1) {$\ldots$};
\node (X5) at (4,1) {$X^{n-1}$};
\node (X6) at (6,1) {$X^n$};
\node (X7) at (-4,-1) {$Y$};
\node (X8) at (-2,-1) {$Y^0$};
\node (X9) at (0,-1) {$Y^1$};
\node (X10) at (2,-1) {$\ldots$};
\node (X11) at (4,-1) {$Y^{n-1}$};
\node (X12) at (6,-1) {$Y^n$};
\draw [->,thick] (X1) -- (X7) node [midway,left] {$f$};
\draw [->,thick] (X2) -- (X8) node [midway,left] {$f^0$};
\draw [->,thick] (X3) -- (X9) node [midway,left] {};
\draw [->,thick] (X5) -- (X11) node [midway,left] {};
\draw [->,thick] (X6) -- (X12) node [midway,left] {};
\draw [->,thick] (X2) -- (X3) node [midway,above] {};
\draw [->,thick] (X3) -- (X4) node [midway,above] {};
\draw [->,thick] (X4) -- (X5) node [midway,above] {};
\draw [->,thick] (X5) -- (X6) node [midway,above] {};
\draw [->,thick] (X8) -- (X9) node [midway,above] {};
\draw [->,thick] (X9) -- (X10) node [midway,above] {};
\draw [->,thick] (X10) -- (X11) node [midway,above] {};
\draw [->,thick] (X11) -- (X12) node [midway,above] {};
\end{tikzpicture}
\end{center}
such that the mapping cone $C=C(f)$ of $f$ is a right $n$-exact sequence. The concept of $n$-pullback diagram is defined dually.
\end{definition}

\begin{theorem}$($\cite[Theorem 3.8]{J}$)$ \label{theorem1}
Let $\mathcal{M}$ be an additive category which satisfies axioms $(A0)$ and $(A1)$ of $n$-abelian category,
\begin{equation}
X: X^0 \overset{d^0}{\rightarrow} X^1 \overset{d^1}{\rightarrow}\cdots \overset{d^{n-1}}{\rightarrow} X^n  \notag
\end{equation}
a complex in $\mathcal{M}$ and $f^0:X^0\rightarrow Y^0$ a morphism in $\mathcal{M}$. Then the following statements hold:
\begin{itemize}
\item[(i)]
There exists an $n$-pushout diagram
\begin{center}
\begin{tikzpicture}
\node (X2) at (-2,1) {$X^0$};
\node (X3) at (0,1) {$X^1$};
\node (X4) at (2,1) {$\ldots$};
\node (X5) at (4,1) {$X^{n-1}$};
\node (X6) at (6,1) {$X^n$};
\node (X8) at (-2,-1) {$Y^0$};
\node (X9) at (0,-1) {$Y^1$};
\node (X10) at (2,-1) {$\ldots$};
\node (X11) at (4,-1) {$Y^{n-1}$};
\node (X12) at (6,-1) {$Y^n$};
\draw [->,thick] (X2) -- (X8) node [midway,left] {$f^0$};
\draw [->,thick] (X3) -- (X9) node [midway,left] {};
\draw [->,thick] (X5) -- (X11) node [midway,left] {};
\draw [->,thick] (X6) -- (X12) node [midway,left] {};
\draw [->,thick] (X2) -- (X3) node [midway,above] {$d_X^0$};
\draw [->,thick] (X3) -- (X4) node [midway,above] {};
\draw [->,thick] (X4) -- (X5) node [midway,above] {};
\draw [->,thick] (X5) -- (X6) node [midway,above] {};
\draw [->,thick] (X8) -- (X9) node [midway,above] {$d_Y^0$};
\draw [->,thick] (X9) -- (X10) node [midway,above] {};
\draw [->,thick] (X10) -- (X11) node [midway,above] {};
\draw [->,thick] (X11) -- (X12) node [midway,above] {};
\end{tikzpicture}
\end{center}
\item[(ii)]
Suppose, moreover, that $\mathcal{M}$ is an n-abelian category. If $d_X^0$ is a monomorphism, then $d_Y^0$ is also a monomorphism.
\end{itemize}
\end{theorem}

A subcategory $\mathcal{B}$ of an abelian category $\mathcal{A}$ is called cogenerating if for every object
$X\in \mathcal{A}$ there exists an object $Y\in\mathcal{B}$ and a monomorphism $X\rightarrow Y$. The concept of
generating subcategory is defined dually.

Let $\mathcal{A}$ be an additive category and $\mathcal{B}$ be a full subcategory of $\mathcal{A}$. $\mathcal{B}$ is called
covariantly finite in $\mathcal{A}$ if for every $A\in \mathcal{A}$ there exists an object $B\in\mathcal{B}$ and a morphism
$f : A\rightarrow B$ such that, for all $B'\in\mathcal{B}$, the sequence of abelian groups $\mathcal{A}(B, B')\rightarrow \mathcal{A}(A, B')\rightarrow 0$ is exact. Such a morphism $f$ is called a left $\mathcal{B}$-approximation of $A$. The notions of contravariantly
finite subcategory of $\mathcal{A}$ and right $\mathcal{B}$-approximation are defined dually. A functorially
finite subcategory of $\mathcal{A}$ is a subcategory which is both covariantly and contravariantly finite
in $\mathcal{A}$.

\begin{definition}$($\cite[Definition 3.14]{J}$)$
Let $\mathcal{A}$ be an abelian category and $\mathcal{M}$ be a generating-cogenerating full subcategory of $\mathcal{A}$. $\mathcal{M}$ is called an $n$-cluster tilting subcategory of $\mathcal{A}$ if $\mathcal{M}$ is functorially finite in $\mathcal{A}$ and
\begin{align}
\mathcal{M}& = \{ X\in \mathcal{A} \mid \forall i\in \{1, \ldots, n-1 \}, \Ext_{\mathcal{A}}^i(X,\mathcal{M})=0 \}\notag \\
& =\{ X\in \mathcal{A} \mid \forall i\in \{1, \ldots, n-1 \}, \Ext_{\mathcal{A}}^i(\mathcal{M},X)=0 \}.\notag
\end{align}

Note that $\mathcal{A}$ itself is the unique $1$-cluster tilting subcategory of $\mathcal{A}$.

\end{definition}
\begin{remark} {\rm Let $\mathcal{A}$ be an abelian category and $\mathcal{M}$ be an $n$-cluster tilting subcategory of $\mathcal{A}$. Since $\mathcal{M}$ is a generating-cogenerating subcategory of $\mathcal{A}$, for each $A\in \mathcal{A}$, every left $\mathcal{M}$-approximation of $A$ is a monomorphism and every right $\mathcal{M}$-approximation of $A$ is an epimorphism.}
\end{remark}

A full subcategory $\mathcal{M}$ of an abelian category $\mathcal{A}$ is called $n$-rigid, if for every two objects $M,N\in \mathcal{M}$ and for every $k\in \{1,\ldots,n-1\}$, we have $\Ext_{\mathcal{A}}^k(M,N)=0$. Any $n$-cluster tilting subcategory $\mathcal{M}$ of an abelian category $\mathcal{A}$ is $n$-rigid.

The following result gives a rich source of $n$-abelian categories. Note that our main result in this paper proves another direction of this theorem.
\begin{theorem}$($\cite[Theorem 3.16]{J}$)$
Let $\mathcal{A}$ be an abelian category and $\mathcal{M}$ be an $n$-cluster tilting subcategory of $\mathcal{A}$. Then $\mathcal{M}$ is an $n$-abelian category.
\end{theorem}

\section{the higher-dimensional version of the Freyd-Mitchell embedding
theorem}

Let $\mathcal{M}$ be a small $n$-abelian category. In \cite[Theorems 4.2 and 4.3]{EN} we constructed an embedding $\widetilde{H}:\mathcal{M}\rightarrow \mathcal{L}_2(\mathcal{M},\mathcal{G})$, where $\mathcal{L}_2(\mathcal{M},\mathcal{G})$ is the category of absolutely pure covariant group valued functors over $\mathcal{M}$. We also showed that $\widetilde{H}$ preserve and reflect $n$-exact sequences. In this section first we dualize $\mathcal{M}$ and show that we have an embedding $\widetilde{H}:\mathcal{M}\rightarrow \mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$, then we prove that the essential image of $\widetilde{H}$ is $n$-rigid in $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$.

We denote by $(\mathcal{M}^{op},\mathcal{G})$ the category of all contravariant additive functors from $\mathcal{M}$ to the category of abelian groups. First we recall some basic properties of the category $(\mathcal{M}^{op},\mathcal{G})$. The reader can find proofs in \cite{Fr}.

A category is called complete if all small limits exist in it. Dually, a category is called
cocomplete if all small colimits exist in it. A cocomplete abelian category with a generator in which direct limit of exact sequences is exact, is called a Grothendieck category. The category $(\mathcal{M}^{op},\mathcal{G})$ is a complete abelian category that all limits compute pointwise. It is easy to see that $(\mathcal{M}^{op},\mathcal{G})$ is a Grothendieck category. If for every $X \in \mathcal{M}$ we denote the functor $\Hom(-,X) \in (\mathcal{M}^{op},\mathcal{G})$ by $H_X$, then the Yoneda's lemma states that $H_X$ is a projective object in $(\mathcal{M}^{op},\mathcal{G})$ and $\Sigma_{X\in \mathcal{M}} H_X$ is a generator for $(\mathcal{M}^{op},\mathcal{G})$,  where $\Sigma_{X\in \mathcal{M}} H_X$ is the direct sum of all representable functors $H_X$ in $(\mathcal{M}^{op},\mathcal{G})$.

\subsection{The subcategory of mono functors}
A functor $F\in (\mathcal{M}^{op},\mathcal{G})$ is called a mono functor if it preserves monomorphisms (i.e. send epimorphisms in $\mathcal{M}$ (monomorphisms in $\mathcal{M}^{op}$) to monomorphisms). We denote by $\mathbb{M}(\mathcal{M}^{op},\mathcal{G})$, the full subcategory of $(\mathcal{M}^{op},\mathcal{G})$ consisting of all mono functors.

\begin{proposition}\label{propositionf1}$($\cite[Proposition 3.7]{EN}$)$
The inclusion functor $I:\mathbb{M}(\mathcal{M},\mathcal{G}) \hookrightarrow (\mathcal{M},\mathcal{G})$ has a left adjoint $\mathbb{M}:(\mathcal{M},\mathcal{G}) \rightarrow \mathbb{M}(\mathcal{M},\mathcal{G})$ such that for each $F\in (\mathcal{M},\mathcal{G})$ the morphism $F\rightarrow \mathbb{M}(F)$ is an epimorphism and kernel of this morphism is the maximal torsion subobject of $F$ (i.e. the maximal subobject that has no nonzero morphism to $\mathbb{M}(\mathcal{M},\mathcal{G})$).
\end{proposition}

$\mathbb{M}(\mathcal{M}^{op},\mathcal{G})$ is not in general an abelian category. There may be a monomorphism in $\mathbb{M}(\mathcal{M}^{op},\mathcal{G})$ which is not a kernel of a morphism in $\mathbb{M}(\mathcal{M}^{op},\mathcal{G})$. To fix this, following \cite{Fr} we introduce the subcategory of absolutely pure functors.

\begin{definition}$($\cite[Page 144]{Fr}$)$
A subfunctor $F^{\prime}\subseteq F$ in $\mathbb{M}(\mathcal{M}^{op},\mathcal{G})$ is said to be pure subfunctor if the quotient functor $\dfrac{F}{F^{\prime}}\in \mathbb{M}(\mathcal{M}^{op},\mathcal{G})$. A mono functor is called absolutely pure if and only if whenever it appears as a subfunctor of a mono functor it is a pure subfunctor. We denote by $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})\subseteq \mathbb{M}(\mathcal{M}^{op},\mathcal{G})$ the full subcategory of absolutely pure functors.
\end{definition}

In the following proposition and theorem we summarize the basic properties of the category $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$.

\begin{proposition}\label{propositionf2}$($\cite[Proposition 3.9]{EN}$)$
The inclusion functor $J:\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})\hookrightarrow \mathbb{M}(\mathcal{M}^{op},\mathcal{G})$ has a left adjoint $R:\mathbb{M}(\mathcal{M}^{op},\mathcal{G})\rightarrow \mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$ such that for each $M\in \mathbb{M}(\mathcal{M}^{op},\mathcal{G})$ the morphism $M\rightarrow R(M)$ is a monomorphism.
\end{proposition}

\begin{theorem}\label{theoremab}$($\cite[Theorem 3.10]{EN}$)$
$\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$ is an abelian category and every object of $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$ has an injective envelope.
\end{theorem}

A functor $F\in \mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$ is called torsion functor if for every mono functor $F'$, $\Hom(F,F')=0$. We say that $F$ is effaceable in $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$ if for every $X\in \mathcal{M}$ and each $x\in F(X)$ there exists an epimorphism $f:Y\rightarrow X$ such that $F(f)(x)=0$. These two concepts are equivalent, see \cite[Exercise A of chapter 7]{Fr}.

\begin{remark}\label{Re}$($\cite[Remark 3.11]{EN}$)$
In this remark we recall that how kernels and cokernels in $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$ are constructed. Let $L_1\rightarrow L_2$ be any morphism in $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$. If $K\rightarrow L_1$ is a kernel of $L_1\rightarrow L_2$ in $(\mathcal{M}^{op},\mathcal{G})$ it is also kernel in $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$ and if $L_2\rightarrow F$ is a cokernel of $L_1\rightarrow L_2$ in $(\mathcal{M}^{op},\mathcal{G})$ then the composition $L_2\rightarrow F\rightarrow \mathbb{M}(F)\rightarrow R(\mathbb{M}(F))$ is a cokernel in $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$. Thus a morphism in $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$ is epimorphism if and only if its cokernel in $(\mathcal{M}^{op},\mathcal{G})$ is torsion.
\end{remark}

Let $\mathcal{A}$ be an abelian category, $k$ a positive integer and $A, C\in \mathcal{A}$. An exact sequence
\begin{equation}
0\rightarrow A\rightarrow X_{k-1}\rightarrow \cdots \rightarrow X_0\rightarrow C\rightarrow 0 \notag
\end{equation}
in $\mathcal{A}$ is called a $k$-fold extensions of $C$ by $A$. Two $k$-fold extensions of $C$ by $A$
\begin{equation}
0\rightarrow A\rightarrow B_{k-1}\rightarrow \cdots \rightarrow B_0\rightarrow C\rightarrow 0 \notag
\end{equation}

\begin{equation}
0\rightarrow A\rightarrow B'_{k-1}\rightarrow \cdots \rightarrow B'_0\rightarrow C\rightarrow 0 \notag
\end{equation}

are said to be Yoneda equivalent if there is the following commutative diagram
\begin{center}
\begin{tikzpicture}
\node (X1) at (-6,1) {$0$};
\node (X2) at (-4,1) {$A$};
\node (X3) at (-2,1) {$B_{k-1}$};
\node (X4) at (0,1) {$\ldots$};
\node (X5) at (2,1) {$B_0$};
\node (X6) at (4,1) {$C$};
\node (X7) at (6,1) {$0$};
\node (X8) at (-6,-1) {$0$};
\node (X9) at (-4,-1) {$A$};
\node (X10) at (-2,-1) {$B''_{k-1}$};
\node (X11) at (0,-1) {$\ldots$};
\node (X12) at (2,-1) {$B''_0$};
\node (X13) at (4,-1) {$C$};
\node (X14) at (6,-1) {$0$};
\node (X15) at (-6,-3) {$0$};
\node (X16) at (-4,-3) {$A$};
\node (X17) at (-2,-3) {$B'_{k-1}$};
\node (X18) at (0,-3) {$\ldots$};
\node (X19) at (2,-3) {$B'_0$};
\node (X20) at (4,-3) {$C$};
\node (X21) at (6,-3) {$0$};
\draw [->,thick] (X1) -- (X2) node [midway,left] {};
\draw [->,thick] (X2) -- (X3) node [midway,left] {};
\draw [->,thick] (X3) -- (X4) node [midway,left] {};
\draw [->,thick] (X4) -- (X5) node [midway,left] {};
\draw [->,thick] (X5) -- (X6) node [midway,left] {};
\draw [->,thick] (X6) -- (X7) node [midway,above] {};
\draw [->,thick] (X8) -- (X9) node [midway,above] {};
\draw [->,thick] (X9) -- (X10) node [midway,above] {};
\draw [->,thick] (X10) -- (X11) node [midway,above] {};
\draw [->,thick] (X11) -- (X12) node [midway,above] {};
\draw [->,thick] (X12) -- (X13) node [midway,above] {};
\draw [->,thick] (X13) -- (X14) node [midway,above] {};
\draw [->,thick] (X15) -- (X16) node [midway,above] {};
\draw [->,thick] (X16) -- (X17) node [midway,above] {};
\draw [->,thick] (X17) -- (X18) node [midway,above] {};
\draw [->,thick] (X18) -- (X19) node [midway,above] {};
\draw [->,thick] (X19) -- (X20) node [midway,above] {};
\draw [->,thick] (X20) -- (X21) node [midway,above] {};
\draw [double,-,thick] (X2) -- (X9) node [midway,above] {};
\draw [->,thick] (X10) -- (X3) node [midway,above] {};
\draw [->,thick] (X12) -- (X5) node [midway,above] {};
\draw [double,-,thick] (X6) -- (X13) node [midway,above] {};
\draw [double,-,thick] (X9) -- (X16) node [midway,above] {};
\draw [double,-,thick] (X13) -- (X20) node [midway,above] {};
\draw [->,thick] (X10) -- (X17) node [midway,above] {};
\draw [->,thick] (X12) -- (X19) node [midway,above] {};
\end{tikzpicture}
\end{center}
, where the middle row is a $k$-fold extension of $C$ by $A$ as well. $\Ext_{\mathcal{A}}^k(C,A)$ is defined as the set of Yoneda equivalence classes of $k$-fold extensions of $C$ by $A$ \cite{Mac, Mi}.

\begin{lemma}\label{L1}
Let $\mathcal{A}$ be an abelian category and assume that we have the following two commutative diagrams with exact rows in $\mathcal{A}$.
\begin{center}
\begin{tikzpicture}
\node (X1) at (-6,1) {$X:$};
\node (X2) at (-4.5,1) {$\cdots$};
\node (X3) at (-3,1) {$X_{k+1}$};
\node (X4) at (-1.5,1) {$X_k$};
\node (X5) at (0,1) {$X_{k-1}$};
\node (X6) at (1.5,1) {$\cdots$};
\node (X7) at (3,1) {$X_1$};
\node (X8) at (4.5,1) {$C$};
\node (X9) at (6,1) {$0$};
\node (X10) at (-6,-1) {$Y:$};
\node (X11) at (-3,-1) {$0$};
\node (X12) at (-1.5,-1) {$A$};
\node (X13) at (0,-1) {$Y_{k-1}$};
\node (X14) at (1.5,-1) {$\cdots$};
\node (X15) at (3,-1) {$Y_1$};
\node (X16) at (4.5,-1) {$C$};
\node (X17) at (6,-1) {$0$};
\draw [->,thick] (X2) -- (X3) node [midway,left] {};
\draw [->,thick] (X3) -- (X4) node [midway,left] {};
\draw [->,thick] (X4) -- (X5) node [midway,left] {};
\draw [->,thick] (X5) -- (X6) node [midway,left] {};
\draw [->,thick] (X6) -- (X7) node [midway,above] {};
\draw [->,thick] (X7) -- (X8) node [midway,left] {};
\draw [->,thick] (X8) -- (X9) node [midway,above] {};
\draw [->,thick] (X11) -- (X12) node [midway,above] {};
\draw [->,thick] (X12) -- (X13) node [midway,above] {};
\draw [->,thick] (X13) -- (X14) node [midway,above] {};
\draw [->,thick] (X14) -- (X15) node [midway,above] {};
\draw [->,thick] (X15) -- (X16) node [midway,above] {};
\draw [->,thick] (X16) -- (X17) node [midway,above] { };
\draw [->,thick] (X4) -- (X12) node [midway,left]{$f$};
\draw [->,thick] (X5) -- (X13) node [midway,above] {};
\draw [->,thick] (X7) -- (X15) node [midway,above] {};
\draw [double,-,thick] (X8) -- (X16) node [midway,above] {};
\end{tikzpicture}
\end{center}
and
\begin{center}
\begin{tikzpicture}
\node (X1) at (-6,1) {$X:$};
\node (X2) at (-4.5,1) {$\cdots$};
\node (X3) at (-3,1) {$X_{k+1}$};
\node (X4) at (-1.5,1) {$X_k$};
\node (X5) at (0,1) {$X_{k-1}$};
\node (X6) at (1.5,1) {$\cdots$};
\node (X7) at (3,1) {$X_1$};
\node (X8) at (4.5,1) {$C$};
\node (X9) at (6,1) {$0$};
\node (X10) at (-6,-1) {$Z:$};
\node (X11) at (-3,-1) {$0$};
\node (X12) at (-1.5,-1) {$A$};
\node (X13) at (0,-1) {$Z_{k-1}$};
\node (X14) at (1.5,-1) {$\cdots$};
\node (X15) at (3,-1) {$Z_1$};
\node (X16) at (4.5,-1) {$C$};
\node (X17) at (6,-1) {$0$};
\draw [->,thick] (X2) -- (X3) node [midway,left] {};
\draw [->,thick] (X3) -- (X4) node [midway,left] {};
\draw [->,thick] (X4) -- (X5) node [midway,left] {};
\draw [->,thick] (X5) -- (X6) node [midway,left] {};
\draw [->,thick] (X6) -- (X7) node [midway,above] {};
\draw [->,thick] (X7) -- (X8) node [midway,left] {};
\draw [->,thick] (X8) -- (X9) node [midway,above] {};
\draw [->,thick] (X11) -- (X12) node [midway,above] {};
\draw [->,thick] (X12) -- (X13) node [midway,above] {};
\draw [->,thick] (X13) -- (X14) node [midway,above] {};
\draw [->,thick] (X14) -- (X15) node [midway,above] {};
\draw [->,thick] (X15) -- (X16) node [midway,above] {};
\draw [->,thick] (X16) -- (X17) node [midway,above] {};
\draw [->,thick] (X4) -- (X12) node [midway,left] {$f$};
\draw [->,thick] (X5) -- (X13) node [midway,above] {};
\draw [->,thick] (X7) -- (X15) node [midway,above] {};
\draw [double,-,thick] (X8) -- (X16) node [midway,above] {};
\end{tikzpicture}
\end{center}
Then the $k$-fold extensions $Y$ and $Z$ of $C$ by $A$ are Yoneda equivalent.
\begin{proof}
See \cite[Proposition 5.1 of chapter III]{Mac}.
\end{proof}
\end{lemma}

\begin{lemma}\label{L2}
Let $\mathcal{M}$ be a small $n$-abelian category, $X^0 \overset{d^0}{\rightarrow} X^1 \overset{d^1}{\rightarrow} \cdots \overset{d^{n-1}}{\rightarrow} X^n \overset{d^n}{\rightarrow} X^{n+1}$ be an $n$-exact sequence in $\mathcal{M}$, $1\leq k\leq n-1$ and $f:X^k\rightarrow A$ be a morphism such that $fd_X^{k-1}=0$. Then we can construct the following $n$-push out diagram, such that the bottom row is contractible.
\begin{center}
\begin{tikzpicture}
\node (X1) at (-6,1) {$X^0$};
\node (X2) at (-4.5,1) {$X^1$};
\node (X3) at (-3,1) {$\cdots$};
\node (X4) at (-1.5,1) {$X^{k-1}$};
\node (X5) at (0,1) {$X^k$};
\node (X6) at (1.5,1) {$\cdots$};
\node (X7) at (3,1) {$X^n$};
\node (X8) at (4.5,1) {$X^{n+1}$};
\node (X9) at (-6,-1) {$0$};
\node (X10) at (-4.5,-1) {$0$};
\node (X11) at (-3,-1) {$\cdots$};
\node (X12) at (-1.5,-1) {$0$};
\node (X13) at (0,-1) {$A$};
\node (X14) at (1.5,-1) {$\cdots$};
\node (X15) at (3,-1) {$Y^n$};
\draw [->,thick] (X1) -- (X2) node [midway,above] {$d^0$};
\draw [->,thick] (X2) -- (X3) node [midway,above] {$d^1$};
\draw [->,thick] (X3) -- (X4) node [midway,above] {$d^{k-2}$};
\draw [->,thick] (X4) -- (X5) node [midway,above] {$d^{k-1}$};
\draw [->,thick] (X5) -- (X6) node [midway,above] {$d^{k}$};
\draw [->,thick] (X6) -- (X7) node [midway,above] {$d^{n-1}$};
\draw [->,thick] (X7) -- (X8) node [midway,above] {$d^{n}$};
\draw [->,thick] (X9) -- (X10) node [midway,above] {};
\draw [->,thick] (X10) -- (X11) node [midway,left] {};
\draw [->,thick] (X11) -- (X12) node [midway,above] {};
\draw [->,thick] (X12) -- (X13) node [midway,above] {};
\draw [->,thick] (X13) -- (X14) node [midway,above] {};
\draw [->,thick] (X14) -- (X15) node [midway,above] {};
\draw [->,thick] (X1) -- (X9) node [midway,above] {};
\draw [->,thick] (X2) -- (X10) node [midway,above] {};
\draw [->,thick] (X4) -- (X12) node [midway,above] {};
\draw [->,thick] (X5) -- (X13) node [midway,right] {f};
\draw [->,thick] (X7) -- (X15) node [midway,above] {};
\end{tikzpicture}
\end{center}
\begin{proof}
The proof is easy and is left to the reader.
\end{proof}
\end{lemma}

Let $0\rightarrow K\rightarrow B_0\rightarrow C\rightarrow 0$ and $0\rightarrow A\rightarrow B_1'\rightarrow K'\rightarrow 0$ be two short exact sequences in an abelian category $\mathcal{A}$ and $\sigma:K\rightarrow K'$ be a morphism. Then by taking pull back and push out along $\sigma$, we have the following commutative diagram with exact rows.
\begin{center}
\begin{tikzpicture}
\node (X1) at (-4.,1) {$0$};
\node (X2) at (-2.5,1) {$A$};
\node (X3) at (-1,1) {$W_1$};
\node (X4) at (0,0) {$K$};
\node (X5) at (1,1) {$B_0$};
\node (X6) at (2.5,1) {$C$};
\node (X7) at (4,1) {$0$};
\node (X8) at (-4,-1) {$0$};
\node (X9) at (-2.5,-1) {$A$};
\node (X10) at (-1,-1) {$B'_1$};
\node (X11) at (0,-2) {$K'$};
\node (X12) at (1,-1) {$W_0$};
\node (X13) at (2.5,-1) {$C$};
\node (X14) at (4,-1) {$0$};
\draw [->,thick] (X1) -- (X2) node [midway,left] {};
\draw [->,thick] (X2) -- (X3) node [midway,left] {};
\draw [->,thick] (X3) -- (X4) node [midway,left] {};
\draw [->,thick] (X3) -- (X5) node [midway,left] {};
\draw [->,thick] (X4) -- (X5) node [midway,left] {};
\draw [->,thick] (X5) -- (X6) node [midway,left] {};
\draw [->,thick] (X6) -- (X7) node [midway,above] {};
\draw [->,thick] (X8) -- (X9) node [midway,above] {};
\draw [->,thick] (X9) -- (X10) node [midway,above] {};
\draw [->,thick] (X10) -- (X11) node [midway,above] {};
\draw [->,thick] (X10) -- (X12) node [midway,left] {};
\draw [->,thick] (X11) -- (X12) node [midway,above] {};
\draw [->,thick] (X12) -- (X13) node [midway,above] {};
\draw [->,thick] (X13) -- (X14) node [midway,above] {};
\draw [double,-,thick] (X2) -- (X9) node [midway,above] {};
\draw [->,thick] (X3) -- (X10) node [midway,above] {};
\draw [->,thick] (X4) -- (X11) node [midway,left] {};
\draw [->,thick] (X5) -- (X12) node [midway,above] {};
\draw [double,-,thick] (X6) -- (X13) node [midway,above] {};
\end{tikzpicture}
\end{center}
These two rows are Yoneda equivalent. Two $k$-fold extensions $S$ and $S'$ in $\mathcal{A}$ with the same start and the same end are Yoneda equivalent if and only if $S'$ can be obtained from $S$ by a finite sequence of replacements of this form and replacements of a short exact sequence by another short exact sequence equivalent to it (see \cite[Section 5 of chapter III]{Mac}).

\begin{lemma}\label{L3}
Consider the following commutative diagram in an abelian category $\mathcal{A}$ with exact rows.
\begin{center}
\begin{tikzpicture}
\node (X1) at (-6,1) {$X:$};
\node (X2) at (-4.5,1) {$\cdots$};
\node (X3) at (-3,1) {$X_{k+1}$};
\node (X4) at (-1.5,1) {$X_k$};
\node (X5) at (0,1) {$X_{k-1}$};
\node (X6) at (1.5,1) {$\cdots$};
\node (X7) at (3,1) {$X_1$};
\node (X8) at (4.5,1) {$C$};
\node (X9) at (6,1) {$0$};
\node (X10) at (-6,-1) {$Y:$};
\node (X11) at (-3,-1) {$0$};
\node (X12) at (-1.5,-1) {$A$};
\node (X13) at (0,-1) {$Y_{k-1}$};
\node (X14) at (1.5,-1) {$\cdots$};
\node (X15) at (3,-1) {$Y_1$};
\node (X16) at (4.5,-1) {$C$};
\node (X17) at (6,-1) {$0$};
\draw [->,thick] (X2) -- (X3) node [midway,left] {};
\draw [->,thick] (X3) -- (X4) node [midway,left] {};
\draw [->,thick] (X4) -- (X5) node [midway,left] {};
\draw [->,thick] (X5) -- (X6) node [midway,left] {};
\draw [->,thick] (X6) -- (X7) node [midway,above] {};
\draw [->,thick] (X7) -- (X8) node [midway,left] {};
\draw [->,thick] (X8) -- (X9) node [midway,above] {};
\draw [->,thick] (X11) -- (X12) node [midway,above] {};
\draw [->,thick] (X12) -- (X13) node [midway,above] {};
\draw [->,thick] (X13) -- (X14) node [midway,above] {};
\draw [->,thick] (X14) -- (X15) node [midway,above] {};
\draw [->,thick] (X15) -- (X16) node [midway,above] {};
\draw [->,thick] (X16) -- (X17) node [midway,above] {};
\draw [->,thick] (X7) -- (X15) node [midway,above] {};
\draw [double,-,thick] (X8) -- (X16) node [midway,above] {};
\end{tikzpicture}
\end{center}
Then we can obtain the following commutative diagram with exact rows such that $Y$ and $Z$ are Yoneda equivalent.
\begin{center}
\begin{tikzpicture}
\node (X1) at (-6,1) {$X:$};
\node (X2) at (-4.5,1) {$\cdots$};
\node (X3) at (-3,1) {$X_{k+1}$};
\node (X4) at (-1.5,1) {$X_k$};
\node (X5) at (0,1) {$X_{k-1}$};
\node (X6) at (1.5,1) {$X_{k-2}$};
\node (X7) at (3,1) {$\cdots$};
\node (X8) at (4.5,1) {$X_1$};
\node (X9) at (6,1) {$C$};
\node (X10) at (7.5,1) {$0$};
\node (X11) at (-6,-1) {$Z:$};
\node (X12) at (-3,-1) {$0$};
\node (X13) at (-1.5,-1) {$A$};
\node (X14) at (0,-1) {$Z_{k-1}$};
\node (X15) at (1.5,-1) {$X_{k-2}$};
\node (X16) at (3,-1) {$\cdots$};
\node (X17) at (4.5,-1) {$X_1$};
\node (X18) at (6,-1) {$C$};
\node (X19) at (7.5,-1) {$0$};
\draw [->,thick] (X2) -- (X3) node [midway,left] {};
\draw [->,thick] (X3) -- (X4) node [midway,left] {};
\draw [->,thick] (X4) -- (X5) node [midway,left] {};
\draw [->,thick] (X5) -- (X6) node [midway,left] {};
\draw [->,thick] (X6) -- (X7) node [midway,above] {};
\draw [->,thick] (X7) -- (X8) node [midway,left] {};
\draw [->,thick] (X8) -- (X9) node [midway,above] {};
\draw [->,thick] (X9) -- (X10) node [midway,above] {};
\draw [->,thick] (X12) -- (X13) node [midway,above] {};
\draw [->,thick] (X13) -- (X14) node [midway,above] {};
\draw [->,thick] (X14) -- (X15) node [midway,above] {};
\draw [->,thick] (X15) -- (X16) node [midway,above] {};
\draw [->,thick] (X16) -- (X17) node [midway,above] {};
\draw [->,thick] (X17) -- (X18) node [midway,above] {};
\draw [->,thick] (X18) -- (X19) node [midway,above] {};
\draw [double,-,thick] (X6) -- (X15) node [midway,above] {};
\draw [double,-,thick] (X8) -- (X17) node [midway,above] {};
\draw [double,-,thick] (X9) -- (X18) node [midway,above] {};
\end{tikzpicture}
\end{center}
\begin{proof}
By the above argument the proof is straightforward and is left to the reader.
\end{proof}
\end{lemma}

Now we can prove the main theorem of this section.
\begin{theorem}\label{TR}
The essential image of $\mathcal{M}\hookrightarrow \mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$ is $n$-rigid.
\begin{proof}
For every two objects $Y$ and $Z$ in the essential image of $\mathcal{M}\hookrightarrow \mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$ and every $k\in \{1,\ldots,n-1\}$ we show $Ext^k(Y,Z)=0$. Let $A_k, X^{n+1}\in \mathcal{M}$ such that $H_{A_k}=Z$ and $H_{X^{n+1}}=Y$. Let $\xi: 0\rightarrow H_{A_k}\rightarrow F_{k-1}\rightarrow \cdots \rightarrow F_0\rightarrow H_{X^{n+1}}\rightarrow 0$ be a $k$-fold extension of $Y$ by $Z$ in $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$. Then the cokernel of $F_0\rightarrow H_{X^{n+1}}$, denote by $C$, is effaceable in $(\mathcal{M}^{op},\mathcal{G})$. In particular, there exist $X^n\in \mathcal{M}$ and an epimorphism $X^n\rightarrow X^{n+1}$ in $\mathcal{M}$,  such that $C(X^{n+1})\rightarrow C(X^n)$ carries the image of $1_{X^{n+1}}$ to $0$. This means that
there is a commutative diagram of the following form for an $n$-exact sequence $X:X^0\rightarrow X^1\rightarrow \cdots \rightarrow X^n\rightarrow X^{n+1}$ in $\mathcal{M}$.
\begin{center}
\begin{tikzpicture}
\node (X1) at (-8,1) {$0$};
\node (X2) at (-6.5,1) {$H_{X^0}$};
\node (X3) at (-5,1) {$\cdots$};
\node (X4) at (-3,1) {$H_{X^{n-k}}$};
\node (X5) at (-.5,1) {$H_{X^{n-k+1}}$};
\node (X6) at (1.5,1) {$\cdots$};
\node (X7) at (3.5,1) {$H_{X^n}$};
\node (X8) at (5.5,1) {$H_{X^{n+1}}$};
\node (X9) at (7,1) {$0$};
\node (X10) at (-5,-1) {$0$};
\node (X11) at (-3,-1) {$H_{A_k}$};
\node (X12) at (-.5,-1) {$F_{k-1}$};
\node (X13) at (1.5,-1) {$\cdots$};
\node (X14) at (3.5,-1) {$F_0$};
\node (X15) at (5.5,-1) {$H_{X^{n+1}}$};
\node (X16) at (7,-1) {$0$};
\draw [->,thick] (X1) -- (X2) node [midway,left] {};
\draw [->,thick] (X2) -- (X3) node [midway,left] {};
\draw [->,thick] (X3) -- (X4) node [midway,left] {};
\draw [->,thick] (X4) -- (X5) node [midway,left] {};
\draw [->,thick] (X5) -- (X6) node [midway,left] {};
\draw [->,thick] (X6) -- (X7) node [midway,above] {};
\draw [->,thick] (X7) -- (X8) node [midway,left] {};
\draw [->,thick] (X8) -- (X9) node [midway,above] {};
\draw [->,thick] (X10) -- (X11) node [midway,left] {};
\draw [->,thick] (X11) -- (X12) node [midway,above] {};
\draw [->,thick] (X12) -- (X13) node [midway,above] {};
\draw [->,thick] (X13) -- (X14) node [midway,above] {};
\draw [->,thick] (X14) -- (X15) node [midway,above] {};
\draw [->,thick] (X15) -- (X16) node [midway,above] {};
\draw [->,thick] (X7) -- (X14) node [midway,above] {};
\draw [double,-,thick] (X8) -- (X15) node [midway,above] {};
\end{tikzpicture}
\end{center}
First assume that $k=1$. Since kernels in $\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$ and $(\mathcal{M}^{op},\mathcal{G})$ are coincide by Remark \ref{Re}, we obtain the following commutative diagram
\begin{center}
\begin{tikzpicture}
\node (X3) at (-4,1) {$0$};
\node (X4) at (-2.5,1) {$H_{X^0}$};
\node (X5) at (-.5,1) {$\cdots$};
\node (X6) at (1.5,1) {$H_{X^{n-1}}$};
\node (X7) at (3.5,1) {$H_{X^n}$};
\node (X8) at (5.5,1) {$H_{X^{n+1}}$};
\node (X9) at (7,1) {$0$};
\node (X10) at (-0.5,-1) {$0$};
\node (X11) at (1.5,-1) {$H_{A_k}$};
\node (X12) at (3.5,-1) {$F_0$};
\node (X13) at (5.5,-1) {$H_{X^{n+1}}$};
\node (X14) at (7,-1) {$0$};
\draw [->,thick] (X3) -- (X4) node [midway,left] {};
\draw [->,thick] (X4) -- (X5) node [midway,left] {};
\draw [->,thick] (X5) -- (X6) node [midway,left] {};
\draw [->,thick] (X6) -- (X7) node [midway,above] {};
\draw [->,thick] (X7) -- (X8) node [midway,left] {};
\draw [->,thick] (X8) -- (X9) node [midway,above] {};
\draw [->,thick] (X10) -- (X11) node [midway,left] {};
\draw [->,thick] (X11) -- (X12) node [midway,above] {};
\draw [->,thick] (X12) -- (X13) node [midway,above] {};
\draw [->,thick] (X13) -- (X14) node [midway,above] {};
\draw [->,thick] (X6) -- (X11) node [midway,above] {};
\draw [->,thick] (X7) -- (X12) node [midway,above] {};
\draw [double,-,thick] (X8) -- (X13) node [midway,above] {};
\end{tikzpicture}
\end{center}
By Lemmas \ref{L1} and \ref{L2} the bottom row is a split short exact sequence. This means that $Ext^1(H_X,H_Y)=0$ for every $X,Y\in \mathcal{M}$. Now assume that $k>1$. By Lemma \ref{L3}, we have the following commutative diagram
\begin{center}
\begin{tikzpicture}
\node (X1) at (-8,1) {$0$};
\node (X2) at (-6.5,1) {$H_{X^0}$};
\node (X3) at (-5,1) {$\cdots$};
\node (X4) at (-3,1) {$H_{X^{n-k}}$};
\node (X5) at (-.5,1) {$H_{X^{n-k+1}}$};
\node (X6) at (1.5,1) {$H_{X^{n-k+2}}$};
\node (X7) at (3.5,1) {$\cdots$};
\node (X8) at (5.5,1) {$H_{X^{n+1}}$};
\node (X9) at (7,1) {$0$};
\node (X10) at (-5,-1) {$0$};
\node (X11) at (-3,-1) {$H_{A_k}$};
\node (X12) at (-.5,-1) {$\bar{F}_{k-1}$};
\node (X13) at (1.5,-1) {$H_{X^{n-k+2}}$};
\node (X14) at (3.5,-1) {$\cdots$};
\node (X15) at (5.5,-1) {$H_{X^{n+1}}$};
\node (X16) at (7,-1) {$0$};
\draw [->,thick] (X1) -- (X2) node [midway,left] {};
\draw [->,thick] (X2) -- (X3) node [midway,left] {};
\draw [->,thick] (X3) -- (X4) node [midway,left] {};
\draw [->,thick] (X4) -- (X5) node [midway,left] {};
\draw [->,thick] (X5) -- (X6) node [midway,left] {};
\draw [->,thick] (X6) -- (X7) node [midway,above] {};
\draw [->,thick] (X7) -- (X8) node [midway,left] {};
\draw [->,thick] (X8) -- (X9) node [midway,above] {};
\draw [->,thick] (X10) -- (X11) node [midway,left] {};
\draw [->,thick] (X11) -- (X12) node [midway,above] {};
\draw [->,thick] (X12) -- (X13) node [midway,above] {};
\draw [->,thick] (X13) -- (X14) node [midway,above] {};
\draw [->,thick] (X14) -- (X15) node [midway,above] {};
\draw [->,thick] (X15) -- (X16) node [midway,above] {};
\draw [double,-,thick] (X6) -- (X13) node [midway,above] {};
\draw [double,-,thick] (X8) -- (X15) node [midway,above] {};
\end{tikzpicture}
\end{center}
where the bottom row is Yoneda equivalent to $\xi$. We want to show that the Yoneda equivalence class of the bottom row is zero. It is enough to show that the induced short exact sequence $0\rightarrow H_{A_k}\rightarrow \bar{F}_{k-1}\rightarrow C_{n-k+1}\rightarrow 0$ splits. Applying the functor $Hom(-,H_{A_k})$ to the following short exact sequence, obtained from the top row,
\begin{equation}
0\rightarrow C_{n-k}\rightarrow H_{X^{n-k+1}}\rightarrow C_{n-k+1}\rightarrow 0, \notag
\end{equation}
we have the exact sequence
\begin{align*}
0\rightarrow &Hom(C_{n-k+1},H_{A_k})\rightarrow Hom(H_{X^{n-k+1}},H_{A_k})\rightarrow Hom(C_{n-k},H_{A_k})\rightarrow \\
&Ext^1(C_{n-k+1},H_{A_k})\rightarrow Ext^1(H_{X^{n-k+1}},H_{A_k})=0.
\end{align*}
So it remains to show that $Hom(H_{X^{n-k+1}},H_{A_k})\rightarrow Hom(C_{n-k},H_{A_k})$ is an epimorphism. Since $X$ is an $n$-exact sequence, using the Yoneda lemma the statement follows from the following diagram.
\begin{center}
\begin{tikzpicture}
\node (X1) at (-3,1) {$H_{X^{n-k-1}}$};
\node (X2) at (0,1) {$H_{X^{n-k}}$};
\node (X3) at (3,1) {$H_{X^{n-k+1}}$};
\node (X4) at (1.5,0) {$C_{n-k}$};
\node (X5) at (1.5,-1.5) {$H_{A^k}$};
\draw [->,thick] (X1) -- (X2) node [midway,left] {};
\draw [->,thick] (X2) -- (X3) node [midway,left] {};
\draw [->>,thick] (X2) -- (X4) node [midway,left] {};
\draw [>->,thick] (X4) -- (X3) node [midway,left] {};
\draw [->,thick] (X4) -- (X5) node [midway,left] {};
\draw [->,dotted] (X3) -- (X5) node [midway,left] {};
\end{tikzpicture}
\end{center}
\end{proof}
\end{theorem}

\section{The main Theorem}

Let $\mathcal{M}$ be a small $n$-abelian category. We have a full and faithful functor $\mathcal{M}\hookrightarrow \mathcal{A}$ for the abelian category $\mathcal{A}=\mathcal{L}_2(\mathcal{M}^{op},\mathcal{G})$, that preserve and reflect $n$-exact sequences and its essential image is $n$-rigid in $\mathcal{A}$. We identify $\mathcal{M}$ with the essential image of the functor. Recall that a functor $F\in (\mathcal{M}^{op},\mathcal{G})$ is called finitely presented if there exist $X, Y\in \mathcal{M}$ and an exact sequence $(-,Y)\rightarrow (-,X)\rightarrow F\rightarrow 0$ in $(\mathcal{M}^{op},\mathcal{G})$. The full subcategory of $(\mathcal{M}^{op},\mathcal{G})$ consisting of all finitely presented functors denote by $\modd$-$\mathcal{M}$. Since $\mathcal{M}$ has weak kernels, $\modd$-$\mathcal{M}$ is an abelian category \cite{Au}. The following proposition is a special case of \cite[Proposition 2.1]{Au}, which used by Auslander for the proof of the Auslander's formula. We follow Auslander's approach to prove a higher-dimensional version of the Auslander's formula.

\begin{proposition}$($\cite[Proposition 3.1]{AHK}$)$ \label{prop}
Consider the Yoneda embedding $Y:\mathcal{M}\hookrightarrow \modd$-$\mathcal{M}$. Then given any abelian
category $\mathcal{D}$, the induced functor $Y^{\mathcal{D}}:(mod$-$\mathcal{M},\mathcal{D})\rightarrow (\mathcal{M},\mathcal{D})$ has a left adjoint $Y_{\lambda}^{\mathcal{D}}$ such that for every $F\in (\mathcal{M},\mathcal{D})$, $Y_{\lambda}^{\mathcal{D}}(F)$ is right exact and $Y_{\lambda}^{\mathcal{D}}(F)Y=F$.
\end{proposition}

By the same setting as in the above proposition, following Auslander \cite{Au}, set $\mathcal{D}=\mathcal{A}$ and let $l:\mathcal{M}\rightarrow \mathcal{A}$ be the inclusion. Then $l$ can be extended to a right exact
functor $Y_{\lambda}^{\mathcal{D}}(l):mod$-$\mathcal{M}\rightarrow \mathcal{A}$. For simplicity we denote $Y_{\lambda}^{\mathcal{D}}(l)$ by $\mathcal{V}$.

\begin{lemma}$($See \cite[Theorem 7.27]{Fr}$)$ \label{theoremc}
$\mathcal{V}:\modd$-$\mathcal{M}\rightarrow \mathcal{A}$ is an exact functor.
\begin{proof}
Let $F\in \modd$-$\mathcal{M}$. Since $\mathcal{M}$ is an $n$-abelian category, we have the exact sequence $(-,X_3)\rightarrow (-,X_2)\rightarrow (-,X_1)\rightarrow F\rightarrow 0$ in $\modd$-$\mathcal{M}$ such that $X_3\rightarrow X_2\rightarrow X_1$ is exact in $\mathcal{A}$. So by applying $\mathcal{V}$, we have the following exact sequence
\begin{equation}
\mathcal{V}((-,X_3))\rightarrow \mathcal{V}((-,X_2))\rightarrow \mathcal{V}((-,X_1)) \notag
\end{equation}
which is isomorphic to $X_3\rightarrow X_2\rightarrow X_1$ by Proposition \ref{prop}. This means that the first left derived functor of $\mathcal{V}$ vanishes. Therefore $\mathcal{V}$ is exact.
\end{proof}
\end{lemma}

By the above lemma it is easy to determine the functor $\mathcal{V}$. Consider $F\in \modd$-$\mathcal{M}$ and a projective presentation $(-,X_2)\rightarrow (-,X_1)\rightarrow F\rightarrow 0$ of $F$. Because $\mathcal{V}$ is exact, after applying $\mathcal{V}$ we have the exact sequence $X_2\rightarrow X_1\rightarrow \mathcal{V}(F)\rightarrow 0$ in $\mathcal{A}$.

Because $\mathcal{V}:\modd$-$\mathcal{M}\rightarrow \mathcal{A}$ is exact, the essential image of $\mathcal{V}$, denote by $\mathcal{B}$, is an abelian subcategory of $\mathcal{A}$. Then the induced functor, denote by $\mathcal{V}:\modd$-$\mathcal{M}\rightarrow \mathcal{B}$, is essentially surjective.

Let's summarize briefly what we have proved up to now. For a small $n$-abelian category $\mathcal{M}$, we found an abelian category $\mathcal{B}$ such that $\mathcal{M}$ is a full subcategory of $\mathcal{B}$ satisfying the following conditions:
\begin{itemize}
\item[(i)] The inclusion functor preserve and reflect $n$-exactness.
\item[(ii)] $\mathcal{M}$ is $n$-rigid in $\mathcal{B}$.
\item[(iii)] Every object $B\in \mathcal{B}$ is a cokernel of a morphism in $\mathcal{M}$.
\end{itemize}

Now by these information we can prove our main result in this paper.

\begin{theorem}\label{M1}
$\mathcal{M}$ is an $n$-cluster tilting subcategory of $\mathcal{B}$.
\begin{proof}
Let $B\in \mathcal{B}$. There exists a morphism $f:X^0\rightarrow X^1$ in $\mathcal{M}$ such that $B$ is a cokernel of $f$. Let $X^1\rightarrow X^2\rightarrow\cdots\rightarrow X^n\rightarrow X^{n+1}$ be an $n$-cokernel of $f$ in $\mathcal{M}$. Then we have the following commutative diagram with an exact row in $\mathcal{B}$.
\begin{center}
\begin{tikzpicture}
\node (X0) at (-3,1) {$X^0$};
\node (X1) at (-1,1) {$X^1$};
\node (X2) at (1,1) {$X^2$};
\node (X3) at (3,1) {$\cdots$};
\node (X4) at (5,1) {$X^{n+1}$};
\node (X5) at (7,1) {$0$};
\node (X6) at (0,0) {$B$};
\draw [->,thick] (X0) -- (X1) node [midway,left] {};
\draw [->,thick] (X1) -- (X2) node [midway,left] {};
\draw [->,thick] (X2) -- (X3) node [midway,left] {};
\draw [->,thick] (X3) -- (X4) node [midway,left] {};
\draw [->,thick] (X4) -- (X5) node [midway,left] {};
\draw [->,thick] (X1) -- (X6) node [midway,left] {};
\draw [->,thick] (X6) -- (X2) node [midway,above] {};
\end{tikzpicture}
\end{center}

The diagram tells that $\mathcal{M}$ is a generating-cogenerating in $\mathcal{B}$. Also it is easy to see that $X^1\rightarrow B$ is a right $\mathcal{M}$-approximation of $B$. Since $B\rightarrow X^2$ is a kernel of $X^2\rightarrow X^3$, by duality process, $B$ has a left $\mathcal{M}$-approximation. Then $\mathcal{M}$ is a functorially finite subcategory of $\mathcal{B}$. Since $\mathcal{M}$ is $n$-rigid by Theorem \ref{TR}, we only need to show that it is maximal $n$-rigid. Let $B$ be an object of $\mathcal{B}$ such that for every $M\in \mathcal{M}$ and every $k\in \{1,2,\ldots,n-1\}$ we have $\Ext_{\mathcal{B}}^k(M,B)=0$. We need to show that
 $B\in \mathcal{M}$. By the previous paragraph we have an exact sequence
 \begin{equation}
 0\rightarrow B\rightarrow X^2\rightarrow X^3\rightarrow \cdots \rightarrow X^n\rightarrow X^{n+1}\rightarrow 0,  \notag
 \end{equation}
 where $X^2, \cdots, X^{n+1}\in \mathcal{M}$. Set $C^1=B$ and $C^r=\Imm(X^r\rightarrow X^{r+1})$ for $r\in \{2,\ldots n-1\}$. Inductively we show that for every $r\in \{1,2,\ldots n-1\}$, $k\in \{1,2,\ldots n-r\}$ and every $M\in \mathcal{M}$, $\Ext_{\mathcal{B}}^k(M,C^r)=0$. If $r=1$, then the claim is obvious. Suppose that $\Ext_{\mathcal{B}}^k(M,C^r)=0$ for each $k\in \{1,2,\ldots n-r\}$. Applying the functor $\Hom_{\mathcal{B}}(M,-)$ to the exact sequence $0\rightarrow C^r\rightarrow X^{r+1}\rightarrow C^{r+1}\rightarrow 0$ gives an exact sequence
\begin{equation}
\Ext_{\mathcal{B}}^k(M,X^{r+1})\rightarrow \Ext_{\mathcal{B}}^k(M,C^{r+1})\rightarrow \Ext_{\mathcal{B}}^{k+1}(M,C^r) \notag
\end{equation}
which proves the claim. Then the sequence $0\rightarrow C^{n-1}\rightarrow X^n \rightarrow X^{n+1}\rightarrow 0$ is an split short exact sequence and since $\mathcal{M}$ is idempotent complete, $C^{n-1}\in \mathcal{M}$. Again because $\Ext_{\mathcal{B}}^1(C^{n-1},C^{n-2})=0$, we have $C^{n-2}\in \mathcal{M}$. Repeating this process we obtain that $C^2\in \mathcal{M}$. Since $\Ext_{\mathcal{B}}^1(C^2,B)=0$, $B$ is a direct summand of $X^2$ and so $B\in\mathcal{M}$. By a similar argument we can see that for every $B\in \mathcal{B}$ such that $\Ext_{\mathcal{B}}^k(B,M)=0$ for each $M\in \mathcal{M}$ and all $k\in \{1,2,\ldots,n-1\}$, $B \in\mathcal{M}$. Therefore $\mathcal{M}$ is an $n$-cluster tilting subcategory of $\mathcal{B}$.
\end{proof}
\end{theorem}

A functor $F$ in $\modd$-$\mathcal{M}$ is called effaceable if in the presentation $(-,X_2)\rightarrow (-,X_1)\rightarrow F\rightarrow 0$, $X_2\rightarrow X_1$ is an epimorphism. The full subcategory of $\modd$-$\mathcal{M}$ consisting of all effaceable functors is denoted by $\modd_0$-$\mathcal{M}$.

Now we are ready to prove a higher-dimensional version of the Auslander's formula.

\begin{theorem}\label{M2}
Let $\mathcal{M}$ be a small $n$-abelian category. Then the quotient category $\frac{\modd\text{-}\mathcal{M}}{\modd_0\text{-}\mathcal{M}}$ has an $n$-cluster tilting subcategory which is equivalent to $\mathcal{M}$.
\begin{proof}
By Theorem \ref{M1} it is enough to show that abelian categories $\frac{\modd\text{-}\mathcal{M}}{\modd_0\text{-}\mathcal{M}}$ and $\mathcal{B}$ are equivalent. By Lemma \ref{theoremc}, $\mathcal{V}:\modd$-$\mathcal{M}\rightarrow \mathcal{B}$ is an exact functor and $\Ker(\mathcal{V})=\{F |\mathcal{V}(F)=0\}=\modd_0\text{-}\mathcal{M}$. Therefore by \cite{G} it is enough to show that there is a full and faithful functor $\mathcal{U}:\mathcal{B}\rightarrow \modd$-$\mathcal{M}$ which is a right adjoint of $\mathcal{V}$. Define a functor $\mathcal{U}:\mathcal{B}\rightarrow \modd$-$\mathcal{M}$ given by $\mathcal{U}(b)=(-, b)|_{\mathcal{M}}$ for each $b\in \mathcal{B}$. Let $b\in \mathcal{B}$. Since $\mathcal{M}$ is a contravariantly finite and generating subcategory of $\mathcal{B}$, we have the following commutative diagram with an exact row
\begin{center}
\begin{tikzpicture}
\node (X0) at (3,1) {$m_2$};
\node (X1) at (5,1) {$m_1$};
\node (X2) at (7,1) {$b$};
\node (X4) at (9,1) {$0$};
\node (X3) at (4,0) {$k$};
\draw [->,thick] (X0) -- (X1) node [midway,left] {};
\draw [->,thick] (X2) -- (X4) node [midway,left] {};
\draw [->,thick] (X1) -- (X2) node [midway,above] {f};
\draw [->,thick] (X0) -- (X3) node [midway,left] {g};
\draw [->,thick] (X3) -- (X1) node [midway,right] {h};
\end{tikzpicture}
\end{center}
, where $f$ and $g$ are right $\mathcal{M}$-approximations and $h$ is a kernel of $f$. Then the sequence $(-, m_2)\longrightarrow(-, m_1)\longrightarrow(-, b)|_{\mathcal{M}}\longrightarrow 0$ is exact and hence $(-, b)|_{\mathcal{M}}\in \modd$-$\mathcal{M}$. Now by this exact sequence and Yoneda's lemma for every $b, b'\in \mathcal{B}$ we have $\Hom_{\mathcal{B}}(b, b')\cong \Hom_{\modd\text{-}\mathcal{M}}((-, b)|_{\mathcal{M}}, (-, b')|_{\mathcal{M}})$. Therefore $\mathcal{U}$ is full and faithful. Now we show that $\mathcal{U}$ is a right adjoint of $\mathcal{V}$. Let $F\in \modd$-$\mathcal{M}$ and $(-, m_2)\longrightarrow(-, m_1)\longrightarrow F\longrightarrow 0$ be the presentation $F$, where $m_1, m_2\in \mathcal{M}$. Since $\mathcal{V}$ is exact, after applying $\mathcal{V}$ we have an exact sequence $m_2\longrightarrow m_1\longrightarrow \mathcal{V}(F)\longrightarrow 0$. Now applying $\Hom_{\modd\text{-}\mathcal{M}}(-, (-, b)|_{\mathcal{M}})$ to the presentation of $F$, by Yoneda's lemma, we have an exact sequence $0\longrightarrow\Hom_{\modd\text{-}\mathcal{M}}(F, (-, b)|_{\mathcal{M}})\longrightarrow(m_1, b)\longrightarrow (m_2, b)$. Therefore we have $\Hom_{\modd\text{-}\mathcal{M}}(F, (-, b)|_{\mathcal{M}})\cong \Hom_{\mathcal{B}}(\mathcal{V}(F), b)$ and the result follows.
\end{proof}
\end{theorem}

\section*{acknowledgements}
During the preparation of this paper Sondre Kvamme informed us that he has a different proof that any $n$-abelian category is equivalent to an $n$-cluster tilting subcategory of an abelian category, but he hasn't written up the result yet. The research of the second
author was in part supported by a grant from IPM (No. 99170412).


TITLE: Determining whether given functions are uniformly convergent - Solution Feedback
QUESTION [1 upvotes]: Working through Spivak's Calculus - Ch 24 - 2) (i) and (ii), I've been asked to determine whether the following functions converge pointwise and to determine if they are uniformly convergent. The functions are:
i) $f_{n}(x) = x^{n} - x^{2n}$ on $[0,1]$
ii) $f_{n}(x) = \frac{nx}{1 + n + x}$ on $[0, \infty)$
Hints were given to solve the expressions and they are the following:

So solving (i). The first thing to observe after a little playing around is that:
$$f(x) = \lim_{n \to \infty} f_{n}(x) = 0$$
Now using the hint one works out that the maximum of $|f-f_{n}|$ on $[0,1]$ occurs when:
$$x = \frac{1}{\sqrt[n]{2}}$$
If you plug this value into the expression of $|f(x) - f_{n}(x)|$ it will always work out to be:
$$|\frac{1}{2} - \frac{1}{4}| = \frac{1}{4}$$ for all values of $n$. From this I concluded that the sequence of functions does not converge uniformly to $f(x)$ because there will always be a distance of $\frac{1}{4}$ between the sequence and $f(x) = 0$.
So now working on (ii) with some fiddling around again one can conclude that:
$$f(x) = \lim_{n \to \infty}f_{n}(x) = x$$
Now using the hint given one works out:
$$\bigg|\frac{nx}{1+n+x} - x \bigg| \Rightarrow \bigg|\frac{-1(x^{2} + x)}{1 + n + x}\bigg| =  \frac{(x^{2} + x)}{1 + n + x}$$
Now here is where my question lies. I was going to conclude that this illustrates that the sequence of functions does converge uniformly. I was basing this off of the fact that the distance between $f_{n}$ and $f(x)$ will always be dependent on $n$ and as $n \to \infty$ my expression will go to $0$ (i.e. the distance between $f(x)$ and $f_{n}(x)$). This was the reasoning I used behind why (i) does not uniformly converge because the distance is not dependent on $n$ and is instead always constant.
Looking at the solutions my conclusion for (ii) is wrong. I wanted to know why that is the case and I also wanted to know if my reasoning for why (i) is not uniformly convergent is the right approach as well.

REPLY [1 votes]: The reason is that your distance depends on $x$. It is not true that you can make it uniformly small (by choosing $n$ large) for all $x\ge 0$. If you assume that you can make the distance say less than $1$ for $n\ge N>2$, you get a contradiction for $x=N$, where the distance is greater than $N/2>1$.


TITLE: Moment generating function and Expected value
QUESTION [0 upvotes]: I need to find the moment generating function and $E(x)$ for the function where the pdf of x for ranges
$x\geq 1$ is $ap^x$ but $P(X=0)=1-\sum_1^\infty ap^x$
so when you add them together you get $P(X=x)=1-\sum_2^\infty ap^x$
In order to find the mgf I need to $\sum_0^\infty e^{tx}(1-\sum_2^\infty ap^x)$
I cant seem to get past this point for the mgf which would lead me to the expected value.

REPLY [1 votes]: If I am interpreting the description correctly, $X$ is a random variable that takes on values $0,1,2,3,\dots$.  For $n\ge 1$ we have $\Pr(X=n)=ap^n$, while $\Pr(X=0)=1-\sum_1^\infty ap^n$.  So $\Pr(X=0)=1-\frac{ap}{1-p}$ (sum of geometric series).
Note that the information we are given does not identify $a$ even when we know $p$. But let us continue.
We want $E(e^{tX})$. When $n=0$, we have $e^{tX}=1$. This gives a contribution of $$1-\frac{ap}{1-p}\tag{1}$$ to the mgf.
For the rest, we want $\sum_{n=1}^\infty e^{tn}ap^n$. This sum is an infinite geometric series with first term $ae^tp$ and common ratio $e^tp$. The sum is $$\frac{ae^tp}{1-e^tp}.\tag{2}$$
Finally, add Terms (1) and (2).
For finding $E(X)$ we do not even need Term 1, since it is a constant. Differentiate Term 2, and set $t=0$.