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A group of 8 volunteers are invited to spend 10 days in south London. But it won't feel as if you are in a city: volunteers will stay at a Scout Centre in the midst of woodland with no houses in sight! Dulwich is a pleasant, quiet corner of London.


text:
California cuts electric-car rebates, drops luxury models
California's rebate program to entice more drivers to purchase electric vehicles has gotten less generous, especially for those looking to buy luxury models …
California's rebate program to entice more drivers to purchase electric vehicles has gotten less generous, especially for those looking to buy luxury models
California's rebate program to entice more drivers to purchase electric vehicles has gotten less generous, especially for residents looking to buy luxury models.
The San Francisco Chronicle reports that effective Tuesday, regulators have stopped offering rebates for buyers of electric cars or plug-in hybrid vehicles that cost more than $60,000.
The state Air Resources Board, which regulates the program, also reduced the standard rebate by $500 per vehicle, from $2,500 to $2,000 for all-electric cars. In addition it eliminated rebates for plug-in hybrid cars with an electric-battery range of less than 35 miles.
Agency officials said the changes will ensure that more people will receive rebates, particularly those with low incomes.
Some advocates for clean cars have criticized the move, saying reducing rebates could deter buyers at a crucial time, as electric car sales spike following years of sluggish growth. Electric cars are typically more expensive than comparable gas-powered models. The state rebates, combined with a federal tax break, can cut thousands of dollars off the sticker price.
Melanie Turner, a spokeswoman for the Air Resources Board, told the newspaper that the new rules put an emphasis on rebates for disadvantaged communities, including areas with higher pollution and lower incomes.
Under the rules, low-income buyers will still be eligible for higher rebates, which are unchanged: $4,500 for all-electric cars and $3,500 for hybrids that run on a combination of gas and plug-in battery.
"California simply cannot meet its clean air or climate goals without transforming and cleaning up all our cars and trucks with an emphasis on putting ultra-clean vehicles in communities that need them the most, which are those most impacted by pollution," Turner said in an email.
California leads the nation in electric-car sales, and the state has issued rebates to 354,064 car buyers since its program began in 2011, according to the Chronicle. About 5% of rebates have been earmarked for low- and moderate-income people, currently defined as those earning less than $50,730 for a household of two.
Assemblyman Phil Ting, a San Francisco Democrat and vocal advocate for electric cars in the state Legislature, criticized the changes. He said the new rules create confusion for buyers.
Ting sponsored a bill this year that could have tripled rebates, but it died in committee. He told the newspaper that the state should ramp up incentives and decrease them over time with a specific end date, so people have incentive to buy now.
The Air Resources Board plans to spend $238 million on rebates this year.
Joel Levin, executive director of Plug In America, an advocacy group for electric-car drivers, said some of the changes are discouraging, including smaller rebates and a more complicated maze of regulations.
But he said the changes are positive because low-income people get priority.
"It's a myth that they're only expensive cars for rich people," Levin said of electric vehicles. "The truth is they're fundamentally better cars — we just need to get pricing to the right point for people."
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Interpave will be celebrating the work of sustainable drainage systems (SuDS) designers Robert Bray Associates on the British Precast stand G44 at Futurebuild, London Excel, from 5th – 7th March.
The 2018 winner of the 'British Precast Creativity in Concrete Award', Robert Bray Associates (RBA) has been at the forefront of SuDS for over 20 years. RBA has realised the potential of concrete block permeable paving for SuDS and continues to explore new developments with this key, multi-functional SuDS technique.
The Award recognised RBA's inspiring and innovative work combining SuDS with hard landscape. An exemplar of this approach is RBA's regeneration of Australia Road, London – winner of several awards including the 'Winner of Winners', Landscape Institute President's Award for 2017 – which will be featured at Futurebuild.
Case studies covering various RBA projects – including Australia Road, St George's Primary School Kidderminster, Parkside Civic Centre Bromsgrove, Bewdley School Worcestershire and Hazeley Academy Milton Keynes – can be viewed here. The principles behind good SuDS design with precast concrete paving are explored in the 'Innovations with SuDS & Hard Landscape' guide, download here.


Homes for sale in Lake Stevens the number of buyers searching for their dream homes greatly outnumbers the number of homes for sale. This has led to a competitive marketplace where buyers often need to stand out. One way to show you are serious about buying your dream home is to get pre-qualified or pre-approved for a mortgage before starting your search.
Mike Kellogg and Sheli Kellogg are real estate brokers in Lake Stevens. Give us a call if you have any questions.


Symrise has expanded its fine fragrance genealogy to the digital world. The fine fragrance Genealogy app provides users with 40 years of the world's most popular fragrances.
The fragrances are categorized in an alphabetical index where users can search directly for a particular fragrance. Fragrances are grouped into the nine fragrance families and feminine and masculine categories.
The app includes detailed descriptions of fragrance families and single fragrances. Users will also be able to "favorite" scents as well as share via e-mail.


Vital statistics includes the issuing of marriage licenses, recording of birth and death certificates, and issuing certified copies of the same. To obtain a copy of birth and death certificates, an application must be completed. The County Clerk is closely supported by the Texas Department of State Health Services / Vital Statistics Unit in Austin, Texas. View out-of-state vital statistic information.
A birth record becomes public information and is available for public viewing on and after the 75th anniversary of the date of birth as shown on the record filed with the Texas Department of Vital Statistics Unit or local registration official.
To obtain an informal marriage, a man and woman must agree to be married, live together in this state as husband and wife, and represent to others that they are married.


text:
\section{Introduction}

In the theory of surfaces, the \emph{Gauss-Bonnet Theorem} is one of the most profound and fundamental results:
\begin{align}\label{eqn:Gauss-Bonnet}
\int_M K_g dv_g = 2\pi \chi (M),
\end{align}
 for any closed surfaces $(M^2 ,g)$ with Euler characteristic $\chi (M)$. \\

On the other hand, the \emph{Uniformization Theorem} assures that any metric on $M^2$ is locally conformally flat. Thus an interesting question is that whether we can find a scalar type curvature quantity such that it generalizes the classic \emph{Gauss-Bonnet Theorem} in higher dimensions? \\

For any closed $4$-dimensional Riemannian manifold $(M^4, g)$, we can define the $Q$-curvature as follows
\begin{align}\label{Q_4}
Q_g = - \frac{1}{6} \Delta_g R_g - \frac{1}{2} |Ric_g|_{g}^2 + \frac{1}{6} R_g^2,
\end{align}
which satisfies the \emph{Gauss-Bonnet-Chern Formula}
\begin{align}\label{Gauss_Bonnet_Chern}
\int_{M^4} \left( Q_g + \frac{1}{4} |W_g|^2_g \right) dv_g = 8\pi^2 \chi(M).
\end{align}
Here $R_g$, $Ric_g$ and $W_g$ are scalar curvature, Ricci curvature and Weyl tensor for $(M^4, g)$ respectively.\\

In particular, if $W_g = 0$, \emph{i.e.} $(M^4, g)$ is locally conformally flat, we have
\begin{equation}\label{total_Q}
\int_{M^4} Q_g dv_g = 8\pi^2 \chi(M),
\end{equation}
which can be viewed as a generalization of (\ref{eqn:Gauss-Bonnet}).\\

Inspired by Paneitz's work (\cite{Paneitz}), Branson (\cite{Branson}) extended (\ref{Q_4}) and defined the $Q$-curvature for arbitrary dimension $n \geq 3$ to be
\begin{align}
Q_{g} = A_n \Delta_{g} R_{g} + B_n |Ric_{g}|_{g}^2 + C_nR_{g}^2,
\end{align}
where $A_n = - \frac{1}{2(n-1)}$ , $B_n = - \frac{2}{(n-2)^2}$ and
$C_n = \frac{n^2(n-4) + 16 (n-1)}{8(n-1)^2(n-2)^2}$.\\

In the study of conformal geometry, there is an $4^{th}$-order differential operator closely related to $Q$-curvature, called Paneitz operator, which can be viewed as an analogue of conformal Laplacian operator:
\begin{align}
P_g = \Delta_g^2 - div_g \left[(a_n R_g g + b_n Ric_g) d\right] + \frac{n-4}{2}Q_g,
\end{align}
where $a_n = \frac{(n-2)^2 + 4}{2(n-1)(n-2)}$ and $b_n = - \frac{4}{n-2}$.\\

We leave the discussion of conformal covariance of $Q$-curvature and Paneitz operator in the appendix at the end of the article for readers who are interested in it.\\

The most fundamental motivation in this article is to seek the connection between $Q$-curvature and scalar curvature both as scalar-type curvature quantities. Intuitively, they should share some properties in common since both of them are generalizations of Gaussian curvature on surfaces. Of course, as objects in conformal geometry, a lot of successful researches have revealed their profound connections in the past decades. (See the appendix for a brief discussion.) However, when beyond conformal classes there are only very few researches on $Q$-curvature from the viewpoint of Riemannian geometry.
Motivated by the early work of Fischer and Marsden (\cite{F-M}) on the deformation of scalar curvature, we started to consider generic deformation problems of $Q$-curvature. \\

In order to study deformations of scalar curvature, the central idea of \cite{F-M} is to investigate the kernel of $L^2$-formal adjoint for the linearization of scalar curvature. To be precise, regard the scalar curvature $R(g)$ as a second-order nonlinear map on the space of all metrics on $M$,
\begin{align*}
\notag
R: &\mathcal{M} \rightarrow C^{\infty}(M); \
g\mapsto R_{g}.
\end{align*}
Let $\gamma_g : S_2(M) \rightarrow C^\infty(M)$ be its linearization at $g$ and $\gamma_g^*: C^\infty(M) \rightarrow S_2(M)$ be the $L^2$-formal adjoint of $\gamma_g$, where $S_2(M)$ is the space of symmetric $2$-tensors on $M$. \\

A crucial related concept is the so-called \emph{vacuum static space}, which is the spatial slice of a type of special solutions to \emph{vacuum Einstein equations} (see \cite{Q-Y}). A vacuum static space can also be defined as a complete Riemannian manifold with $\ker \gamma_g^* \neq \{0\}$ (see the last section for an explicit definition). Typical examples of vacuum static spaces are space forms. In this sense, we can regard the notion of vacuum static spaces as a generalization of space forms. Of course, there are many other interesting examples of vacuum static spaces besides space forms, say $S^1\times S^2$ \emph{etc}.
 The classification problem is a fundamental question in the study of vacuum static spaces, even in the field of mathematical general relativity. We refer the article \cite{Q-Y} for readers who are interested in it.\\

In fact, being vacuum static or not is the criterion to determine whether the scalar curvature is being linearized stable at this given metric. It was shown in \cite{F-M} that a closed non-vacuum static manifold is linearized stable and hence any smooth function sufficiently close to the scalar curvature of the background metric can be realized as the scalar curvature of a nearby metric. This result was generalized to non-vacuum static domains by Corvino (\cite{Corvino}). On the other hand, for a vacuum static space, rigidity results are expected. In \cite{F-M}, authors also showed that there is a local rigidity phenomenon on any torus. Later, Schoen-Yau and Gromov-Lawson (\cite{S-Y_1, S-Y_2, G-L_1, G-L_2}) showed that global rigidity also holds on torus. Inspired by the \emph{Positive Mass Theorem}, many people made a lot of progress in understanding the rigidity phenomena of vacuum static spaces (c.f. \cite{Min-Oo, A-D, A-C-G, Miao, S-T, B-M}). It was demonstrated that local rigidity is a universal phenomena in all vacuum static spaces (see \cite{Q-Y_1}).  \\

Along the same line, we can also consider $Q$-curvature as a $4^{th}$-order nonlinear map on the space of all metrics on $M$,
\begin{align}
\notag
Q: &\mathcal{M} \rightarrow C^{\infty}(M); g\mapsto Q_g.
\end{align}
Due to the complexity of $Q$-curvature, this map is extremely difficult to study. However, by linearizing the map we may still expect some interesting results to hold. We denote $$\Gamma_g : S_2(M) \rightarrow C^\infty(M)$$ to be the linearization of $Q$-curvature at metric $g$. \\

Now we give the following notion of $Q$-singular space introduced by Chang-Gursky-Yang in \cite{C-G-Y}, which plays a crucial role in this article.

\begin{definition}[Chang-Gursky-Yang \cite{C-G-Y}]
We say a complete Riemannian manifold $(M, g)$ is $Q$-singular, if $$\ker \Gamma_g^* \neq \{ 0 \},$$ where
$\Gamma_g^* : C^\infty(M) \rightarrow S_2(M)$ is the $L^2$-formal adjoint of $\Gamma_g$. We refer the triple $(M, g, f)$ as a $Q$-singular space as well, if $f (\not\equiv 0)$ is in the kernel of $\Gamma_g^*$.
\end{definition}

By direct calculations, we can obtain the precise expression of $\Gamma_g^*$ (see Proposition \ref{prop:Gamma^*}). Then follow their argument, we observed that the results in \cite{C-G-Y} for dimension $4$ can be extended to any other dimensions:

\begin{theorem}[Chang-Gursky-Yang \cite{C-G-Y}]\label{thm:Q_const}
A $Q$-singular space $(M^n, g)$ has constant $Q$-curvature and
\begin{align}
\frac{n+4}{2}Q_g \in Spec(P_g).
\end{align}

\end{theorem}

As in the study of vacuum static spaces, the collection of all $Q$-singular spaces is also expected to be a ''small'' set in some sense, which lead us to the classification problems naturally. In fact, when restricted in the class of closed $Q$-singular Einstein manifolds with nonnegative scalar curvature, Ricci flat and spherical metrics are the only possible ones:

 \begin{theorem}\label{Classificaition_Q_singular_Einstein}
Suppose $(M^n,g,f)$ is a closed $Q$-singular Einstein manifold. If the scalar curvature $R_g \geq 0$, then
\begin{itemize}
\item $f$ is a non-vanishing constant if and only if $(M,g)$ is Ricci flat;
\item $f$ is not a constant if and only if $(M,g)$ is isometric to a round sphere with radius $r = \left( \frac{n(n-1)}{R_g}\right)^{\frac{1}{2}}$.
\end{itemize}
\end{theorem}

For non-$Q$-singular spaces, we can show that they are actually linearized stable, which answers locally prescribing $Q$-curvature problem for this category of manifolds:
\begin{theorem}\label{Q_stability}
Let $(M,\bar{g})$ be a closed Riemannian manifold. Assume
$(M,\bar{g})$ is not $Q$-singular, then the Q-curvature is linearized stable at $\bar g$ in the sense that $Q : \mathcal{M} \rightarrow C^{\infty}(M) $ is a submersion at $\bar{g}$.
Thus, there is a neighborhood $U \subset C^{\infty}(M)$ of $Q_{\bar{g}}$ such that for any $\psi \in U$, there exists a metric $g$ on $M$ closed to $\bar{g}$ with $Q_g = \psi$.
\end{theorem}

In particular, with the aid of Theorem \ref{Classificaition_Q_singular_Einstein}, we know a generic Einstein metric with positive scalar curvature has to be linearized stable:

\begin{corollary}\label{cor:stab_pos_Einstein}
Let $(M,\bar{g})$ be a closed positive Einstein manifold. Assume $(M,\bar{g})$ is not spherical, then the Q-curvature is linearized stable at $\bar g$.
\end{corollary}

Theorem \ref{Q_stability} actually provides an answer to global prescribing $Q$-curvature problem:
\begin{corollary}\label{cor:prescribing_zero_Q}
Suppose $(M, \bar g)$ is a closed non-$Q$-singular manifold with vanishing $Q$-curvature. Then any smooth function $\varphi$ can be realized as a $Q$-curvature for some metric $ g$ on $M$.
\end{corollary}

As a direct corollary of Theorem \ref{Q_stability}, we can obtain the existence of a negative $Q$-curvature metric as the following:
\begin{corollary}\label{cor:pos_Y_negative_Q}
Let $M$ be a closed manifold with positive Yamabe invariant $Y(M) > 0$. There is a metric $g$ with $Q$-curvature $Q_g < 0$ on $M$.
\end{corollary}

We also investigate the stability of Ricci flat metrics, which are in fact $Q$-singular. Due to the speciality of such metrics, we can give a sufficient condition for prescribing $Q$-curvature problem.
\begin{theorem}\label{Ricci_flat_stability}
Let $(M,\bar{g})$ be a closed Ricci flat Manifold. Denote $$\Phi:=\{\psi \in C^{\infty}(M): \int_M \psi dv_{\bar{g}} = 0\}$$ to be the set of smooth functions with zero average. Then for any $\psi \in \Phi$, there exists a metric $g$ on $M$ such that $$Q_g = \psi.$$

\end{theorem}

On the other hand, we are interested in the question of how much the generic stability fails for flat metrics. Applying perturbation analysis, we discovered the positivity of $Q$-curvature is the obstruction for flat metrics exactly as what scalar curvature means for them. That is, flat metrics are locally rigid for nonnegative $Q$-curvature. As a special case, we proved a local rigidity result for $Q$-curvature on torus $T^n$ ($n \geq 3$), which is  similar to the non-existence result of positive scalar curvature metrics on torus due to Schoen-Yau and Gromov-Lawson (\cite{S-Y_1, S-Y_2, G-L_1, G-L_2}).

\begin{theorem}\label{flat_local_rigidity}
For $n \geq 3$, let $(M^n,\bar{g})$ be a closed flat Riemannian manifold and $g$ be a metric on $M$ with $$Q_g \geq 0.$$ Suppose $||g - \bar{g}||_{C^2(M, \bar{g})}$ is sufficiently small, then $g$ is also flat.
\end{theorem}

As for global rigidity, we have the following interesting result:

\begin{theorem}\label{thm:rigidity_tori}
Suppose $g$ is a locally conformally flat metric on $T^n$ with $$Q_g \geq 0,$$ then $g$ is flat.

In particular, any metric $g$ on $T^4$ with nonnegative $Q$-curvature has to be flat.
\end{theorem}

It would be interesting to compare notions of $Q$-singular and vacuum static spaces. In fact, we have the following observation:

\begin{theorem}\label{Q-static_R-static}
Let $\mathcal{M}_R$ be the space of all closed vacuum static spaces and $\mathcal{M}_Q$ be the space of all closed $Q$-singular spaces.
Suppose $(M, g, f) \in \mathcal{M}_R \cap \mathcal{M}_Q$, then $M$ is Einstein necessarily. In particular, it has to be either Ricci flat or isometric to a round sphere.
\end{theorem}

This article is organized as follows: We explain some general notations and conventions in Section 2. We then give a characterization of $Q$-singular spaces and proofs of Theorem \ref{thm:Q_const}, \ref{Classificaition_Q_singular_Einstein} in Section 3. In Section 4, we investigate several stability results, including Theorem \ref{Q_stability}, Corollary \ref{cor:stab_pos_Einstein}, \ref{cor:prescribing_zero_Q}, \ref{cor:pos_Y_negative_Q} and Theorem \ref{Ricci_flat_stability}. In Section 5, local rigidity of flat metrics, say Theorem \ref{flat_local_rigidity} and \ref{thm:rigidity_tori} will be shown. Some relevant results will also be discussed there. In Section 6, we discuss the relation between $Q$-singular and vacuum static spaces and prove Theorem \ref{Q-static_R-static} in the end of Section 6. In the end of this article, we provide a brief discussion about conformal properties of $Q$-curvature and Paneitz operator from the viewpoint of conformal geometry.\\

\paragraph{\textbf{Acknowledgement}}
The authors would like to express their appreciations to professor Sun-Yung Alice Chang, Professor Justin Corvino,  Professor Matthew Gursky, Professor Fengbo Hang, Professor Jie Qing and Professor Paul Yang, and Dr.Yi Fang for their interests in this work and inspiring discussions. Especially, we would like to thank Professor Matthew Gursky for pointing out the work in \cite{C-G-Y}. The authors are also grateful for MSRI/IAS/PCMI Geometric Analysis Summer School 2013, which created the opportunity for initiating this work. Part of the work was done when the second author visited \emph{Institut Henri Poincar\'e} and we would like to give our appreciations to the hospitality of IHP. \\

\section{Notations and conventions}

Throughout this article, we will always assume $(M^n, g)$ to be an $n$-dimensional closed Riemannian manifold ($n \geq 3$) unless otherwise stated. Here by \emph{closed}, we mean compact without boundary. We also take $\{\partial_i\}_{i=1}^n$ to be a local coordinates around certain point. Hence we also use its components to stand for a tensor or vector in the article.\\

For convenience, we use following notions:

$\mathcal{M}$ - the set of all smooth metric on $M$;

$\mathscr{D}(M)$ - the set of all smooth diffeomorphisms $ \varphi : M \rightarrow M$;

$\mathscr{X}(M)$ - the set of all smooth vector fields on $M$;

$S_2(M)$ - the set of all smooth symmetric 2-tensors on $M$.\\

We adopt the following convention for Ricci curvature tensor,
\begin{align*}
R_{jk} = R^i_{ijk} = g^{il} R_{ijkl}.
\end{align*}

As for Laplacian operator, we simply use the common convention as follow,
\begin{align*}
\Delta_g := g^{ij} \nabla_i \nabla_j.
\end{align*}

Let $h, k \in S_2(M)$, for convenience, we define following operations:
\begin{align*}
(h \times k )_{ij} := g^{kl}h_{ik}k_{jl} = h_i^lk_{lj}
\end{align*}
and
\begin{align*}
h \cdot k  := tr (h \times k) = g^{ij}g^{kl}h_{ik}k_{jl} = h^{jk}k_{jk}.
\end{align*}

Let $X \in \mathscr{X}(M)$ and $h \in S_2(M)$, we use the following notations for operators
\begin{align*}
(\overset{\circ}{Rm} \cdot h )_{jk}:= R_{ijkl} h^{il},
\end{align*}
and
\begin{align*}
(\delta_g h)_i := - (div_g h)_i = -\nabla^j h_{ij},
\end{align*}
which is the $L^2$-formal adjoint of Lie derivative (up to scalar multiple) $$\frac{1}{2}(L_g X)_{ij} = \frac{1}{2} ( \nabla_i X_j + \nabla_j X_i).$$\\

\section{Characterizations of Q-singular spaces

Let $(M,g)$ be a Riemannian manifold and $\{g(t)\}_{t \in (-\varepsilon, \varepsilon)}$ be a 1-parameter family of metrics on $M$ with $g(0) =g$ and $g'(0) = h$. Easy to see, ${g'}^{ij} = - h^{ij}$.\\

We have the following well-known formulae for linearizations of geometric quantities. (c.f. \cite{C-L-N, F-M, Yuan})
\begin{proposition}
The linearization of Christoffel symbol is
\begin{align}
{\Gamma'}_{ij}^k = \frac{1}{2} g^{kl} \left( \nabla_i h_{jl} +\nabla_j h_{il} - \nabla_l h_{ij} \right).
\end{align}
\label{1st_variation_Ricci_scalar}
The linearization of Ricci tensor is
\begin{align}
Ric' = \left.\frac{d}{dt}\right|_{t=0}  Ric(g(t))_{jk} = - \frac{1}{2}\left( \Delta_L h_{jk} +
\nabla_j \nabla_k (tr h) + \nabla_j (\delta h)_k + \nabla_k (\delta
h)_j  \right),
\end{align}
where the Lichnerowicz Laplacian acting on $h$ is defined to be
\begin{align*}
\Delta_L h_{jk} = \Delta h_{jk} + 2 (\overset{\circ}{Rm}\cdot
h)_{jk} - R_{ji} h^i_k - R_{ki}h^i_j,
\end{align*}
and the linearization of scalar curvature is
\begin{align}\label{scalar_1st_variation}
R' = \left.\frac{d}{dt}\right|_{t=0}  R(g(t)) = - \Delta (tr h) +  \delta^2 h - Ric \cdot h.
\end{align}
\end{proposition}

For simplicity, we use $'$ to denote the differentiation with
respect to the parameter $t$ and evaluated at $t=0$.\\

\begin{lemma}
The linearization of the Laplacian acting on the scalar curvature is
\begin{align}\label{laplacian_scalar_variation}
(\Delta R)' = \left.\frac{d}{dt}\right|_{t=0}  \Delta_{g(t)} R(g(t)) = - \nabla^2 R\cdot h + \Delta
R' + \frac{1}{2} dR \cdot (d( tr h ) + 2\delta h).
\end{align}
\end{lemma}

\begin{proof}
Calculating in normal coordinates at an arbitrary point,
\begin{align*}
&\left.\frac{d}{dt}\right|_{t=0}  \Delta_{g(t)} R(g(t)) \\
=& \left.\frac{d}{dt} \right|_{t=0}\left({g^{ij}(\partial_i\partial_j R-\Gamma^k_{ij}\partial_kR)}\right)\\
=& -h^{ij}\partial_i\partial_jR+g^{ij}(\partial_i\partial_j R'-(\Gamma^k_{ij})'\partial_k R-\Gamma^k_{ij}\partial_k R')\\
=& -h^{ij}\partial_i\partial_j R - g^{ij}(\Gamma^k_{ij})'\partial_k R+\Delta R'\\
=& -h^{ij}\partial_i\partial_j R - \frac{1}{2}g^{ij}g^{kl}(\nabla_ih_{jl}+\nabla_jh_{il}-\nabla_lh_{ij})\partial_kR+\Delta R'\\
=& -h^{ij}\nabla_i\nabla_jR-\nabla^ih^k_i\nabla_kR+\frac{1}{2}\nabla^k(trh)\nabla_kR+\Delta R'\\
=& - \nabla^2 R\cdot h + \Delta R' + \frac{1}{2} dR \cdot (d( tr h ) + 2\delta h).
\end{align*}

\end{proof}

Now we can calculate the linearization of $Q$-curvature (1st variation).\\

\begin{proposition}\label{Q_1st_variation}
The linearization of $Q$-curvature is
\begin{align}\label{Q_linearization}
\Gamma_g h :=& DQ_g \cdot h
= A_n \left( - \Delta^2 (tr h) +  \Delta \delta^2 h  -
\Delta ( Ric \cdot h ) + \frac{1}{2} dR \cdot (d( tr h ) + 2\delta h) - \nabla^2 R\cdot h\right)
\\ \notag & - B_n \left(  Ric \cdot \Delta_L h + Ric \cdot
\nabla^2(tr h) + 2 Ric \cdot\nabla (\delta h)   +2 (Ric\times Ric) \cdot h
\right) \\
\notag &+ 2 C_n R \left( - \Delta (tr h) +  \delta^2 h - Ric \cdot h
\right).
\end{align}
\end{proposition}

\begin{proof}
Note that
\begin{align*}
\left.\frac{d}{dt}\right|_{t=0} |Ric(g(t))|_{g(t)}^2
=&\left.\frac{d}{dt}\right|_{t=0} \left(g^{ik}g^{jl}R_{ij}R_{kl}\right)
=2g^{ik}g^{jl}R'_{ij}R_{kl}+2g'^{ik}g^{jl}R_{ij}R_{kl}\\
=&2Ric' \cdot Ric - 2(Ric\times Ric) \cdot h,
\end{align*}
then we finish the proof by combining it with (\ref{1st_variation_Ricci_scalar}),  (\ref{scalar_1st_variation}) and (\ref{laplacian_scalar_variation}).
\end{proof}

Now we can derive the expression of $\Gamma_g^*$ and hence the $Q$-singular equation $$\Gamma_g^* f = 0.$$
\begin{proposition}\label{prop:Gamma^*}
The $L^2$-formal adjoint of $\Gamma_g$ is
\begin{align*}
\Gamma_g^* f :=& A_n \left(  - g \Delta^2 f + \nabla^2
\Delta f - Ric \Delta f + \frac{1}{2} g \delta (f dR) + \nabla ( f
dR) - f \nabla^2 R  \right)\\ \notag
& - B_n \left( \Delta (f Ric) + 2 f
\overset{\circ}{Rm}\cdot Ric + g \delta^2 (f Ric) + 2 \nabla \delta (f
Ric) \right)\\ \notag
&- 2 C_n \left( g\Delta (f R) - \nabla^2 (f R) + f R
Ric \right).
\end{align*}
\end{proposition}

\begin{proof}
For any compactly supported symmetric 2-tensor $h \in S_2(M)$, we have
\begin{align*}
&\int_M f \left.\frac{d}{dt}\right|_{t=0}  {\Delta_{g(t)} R(g(t))}\ dv_g\\
=&\int_M f\left(- \nabla^2 R\cdot h + \frac{1}{2} dR \cdot (d( tr h ) + 2\delta h)+\Delta R'\right)dv_g\\
=&\int_M \left(-h \cdot f\nabla^2R+h \cdot \nabla(fdR) + \frac{1}{2}h \cdot \delta(fdR)g + \Delta f(-\Delta trh + \delta^2h-Ric\cdot h)\right)dv_g\\
=&\int_M  \left\langle-f\nabla^2R+\nabla(fdR) + \frac{1}{2}g\delta(fdR) -g\Delta^2f+\nabla^2\Delta f-Ric\Delta f,\ h\right\rangle dv_g.
\end{align*}\\

Similarly,
\begin{align*}
&\int_M f \left.\frac{d}{dt}\right|_{t=0} |Ric(g(t))|^2_{g(t)}\ dv_g\\
=&\int_M f\left(-2h^{il}g^{jk}R_{ij}R_{kl}+2R^{ij}{R'}_{ij}\right)dv_g\\
=&\int_M -2f(Ric\times Ric)\cdot h-fR^{ij}\left(\Delta_Lh_{ij} +\nabla_i\nabla_j(tr h)
+\nabla_i(\delta h)_j + \nabla_j(\delta h)_i\right) dv_g\\
=&\int_M \left\langle -\Delta_L (fRic)-2f(Ric\times Ric) - 2\nabla\delta(fRic) - g\delta^2(fRic),\ h \right\rangle dv_g\\
=&\int_M \left\langle-\Delta (fRic)-2f\overset\circ{Rm}\cdot Ric - 2\nabla\delta(fRic)-g\delta^2(fRic),\ h \right\rangle dv_g.
\end{align*}\\

And also,
\begin{align*}
&\int_M f \left.\frac{d}{dt}\right|_{t=0} (R(g(t)))^2 \ dv_g\\
=&\int_M 2 f R \left(-\Delta trh+\delta^2h-Ric\cdot h \right)dv_g\\
=&\int_M 2\left(-\Delta (fR)trh+\nabla^2(fR) \cdot h-(fR)Ric\cdot h\right)dv_g\\
=&\int_M 2 \left\langle -g\Delta (fR)+\nabla^2(fR)-(fR)Ric,\ h \right\rangle dv_g
\end{align*}

Combining all the equalities above, we have
\begin{align*}
\int_M f DQ_g \cdot hdv_g=\int_M \langle f,\Gamma_g h\rangle dv_g=\int_M \langle\Gamma_g^*f,h\rangle dv_g.
\end{align*}

\end{proof}

\begin{corollary}
\begin{align*}
\mathscr{L}_g f := tr_g \Gamma_g^* f = \frac{1}{2} \left( P_g - \frac{n + 4}{2} Q_g \right) f.
\end{align*}
\end{corollary}

\begin{proof}
Taking trace of $\Gamma_g^* f$, we have
\begin{align*}
tr \Gamma_g^* f =& -2 Q_g f - (n-1) A_n \Delta^2 f - \left( \frac{n-4}{2} A_n + \frac{n}{2}B_n + 2(n-1)C_n \right)f \Delta R \\
&- (A_n + B_n + 2(n-1)C_n) R \Delta f - \left( \frac{n-2}{2}A_n + n B_n + 4(n-1) C_n\right) df \cdot dR \\
&- (n-2) B_n Ric \cdot \nabla^2 f\\
=&  -2 Q_g f + \frac{1}{2} \Delta^2 f - \frac{(n-2)^2 + 4}{4(n-1)(n-2)} R \Delta f - \frac{n-6}{4(n-1)} df \cdot dR + \frac{2}{n-2} Ric \cdot \nabla^2 f\\
=&  -2 Q_g f + \frac{1}{2} \left( \Delta^2 f - a_n R \Delta f - \left(a_n + \frac{1}{2}b_n\right) df \cdot dR - b_n Ric \cdot \nabla^2 f \right)\\
=& \frac{1}{2} \left( \Delta_g^2 f- div_g \left[(a_n R_g g + b_n Ric_g) df \right] \right) - 2 Q_g \\
= &\frac{1}{2} \left( P_g - \frac{n+4}{2} Q_g \right) f.
\end{align*}

Here we used the fact that $\frac{n-4}{2} A_n + \frac{n}{2}B_n + 2(n-1)C_n =0$.
\end{proof}

Combining above calculations, we can justify that $Q$-curvature is a constant for $Q$-singular spaces using exactly the same argument in \cite{C-G-Y}. For the convenience of readers, we include a sketch of the proof as follows. For more details, please refer to \cite{C-G-Y}.

\begin{theorem}[Chang-Gursky-Yang \cite{C-G-Y}
A $Q$-singular space $(M^n, g)$ has constant $Q$-curvature and
\begin{align}
\frac{n+4}{2}Q_g \in Spec(P_g).
\end{align}

\end{theorem}
\begin{proof
We only need to show $Q_g$ is a constant.

For any smooth vector field $X \in \mathscr{X}$ on $M$, we have
\begin{align*}
\int_M \langle X,   \delta_g \Gamma_g^* f \rangle dv_g = \frac{1}{2}\int_M \langle L_X g, \Gamma_g^* f \rangle dv_g = \frac{1}{2}\int_M f\ \Gamma_g (L_X g)\ dv_g  = \frac{1}{2}\int_M \langle f dQ_g, X \rangle \ dv_g.
\end{align*}
Thus $$fdQ_g = 2 \delta_g \Gamma_g^* f = 0$$ on $M$.

Suppose there is an $x_0 \in M$ with $f(x_0) = 0$ and $dQ_g (x_0) \neq 0$. By taking derivatives, we can see $f$ vanishes to infinite order at $x_0$. \emph{i.e.} $$\nabla^m f(x_0) = 0$$ for any $m \geq 1$. Since $f$ also satisfies that $$\mathscr{L}_g f = \frac{1}{2} \left(P_g f -  \frac{n+4}{2}Q_g f\right) = 0,$$ by applying the Carleman estimates of Aronszajn (\cite{A}), we can conclude that $f$ vanishes identically on $M$. But this contradicts to the fact that $g$ is $Q$-singular.

Therefore, $dQ_g$ vanishes on $M$ and thus $Q_g$ is a constant. Here we obtain the proof of Theorem \ref{thm:Q_const}.
\end{proof}

\textbf{Examples of $Q$-singular spaces.}

\begin{itemize}
  \item
  Ricci flat spaces\\
  Take $f$ to be a nonzero constant, then it satisfies the $Q$-singular equation.\\

  \item
  Spheres\\
  Take $f$ to be the $(n+1)^{th}$-coordinate function $x_{n+1}$ restricted on $$S^n = \{x\in \mathbb{R}^{n+1}: |x|^2 = 1\}.$$ Then $f$ satisfies the Hessian type equation $$\nabla^2 f + g f = 0.$$ With the aid of the following Lemma \ref{Q-static_Einstein}, we can easily check that $f$ satisfies the $Q$-singular equation.\\

  \item
  Hyperbolic spaces\\
  Similarly, if we still take $f$ to be the $(n+1)^{th}$-coordinate function $x_{n+1}$ but restricted on $$H^n = \{(x',x_{n+1})\in \mathbb{R}^{n+1}: |x'|^2 - |x_{n+1}|^2 = -1, x' \in \mathbb{R}^n, x_{n+1} > 0\}.$$ Then $f$ satisfies the similar Hessian type equation $$\nabla^2 f - g f = 0,$$ and hence solves the $Q$-singular equation.
\end{itemize}

Based on the complexity of the $Q$-singular equation, one can easily see that it is very difficult to study the geometry of generic $Q$-singular spaces. However, when restricted to some special classes of Riemannian manifolds we can still get some interesting results of $Q$-singular spaces.

\begin{lemma}\label{Q-static_Einstein}
Let $(M,g)$ be an Einstein manifold and $f \in \ker \Gamma_g^*$, we have
\begin{align}
\Gamma_g^* f = A_n \left[ \nabla^2 \left(\Delta f + \Lambda_n R f
\right) + \frac{R}{n(n-1)}g \left(\Delta f + \Lambda_n R f\right)
\right] = 0,
\end{align}
where $\Lambda_n : = \frac{2}{A_n} (\frac{B_n}{n} + C_n) = - \frac{(n+2)(n-2)}{2n(n-1)} < 0$, for any $n \geq 3$.
\end{lemma}

\begin{proof}

Since $g$ is Einstein, \emph{i.e.} $Ric=\frac{R}{n}g$, we get
\begin{align*}
\Gamma_g^* f=&A_n \left[ -g\Delta^2f+\nabla^2\Delta f-\frac{R}{n}g\Delta f \right] -B_n\left[\frac{R}{n}\Delta (fg)+2f \overset\circ{Rm}\cdot \frac{R}{n}g + 2\frac{R}{n}\nabla\delta(fg) +\frac{R}{n}g\delta^2(fg) \right]\\
& \ \ -2C_n\left[Rg\Delta f-R\nabla^2f+\frac{R^2}{n}fg\right]\\
=&A_n \left[-g\Delta^2f+\nabla^2\Delta f-\frac{R}{n}g\Delta f\right] - 2B_n \left[ \frac{R}{n}g\Delta f+\frac{R^2}{n^2}gf - \frac{R}{n}\nabla^2f \right] \\
& \ \ -2C_n \left[ Rg\Delta f-R\nabla^2f+\frac{R^2}{n}fg \right]\\
=& A_n \left[-g\Delta^2f+\nabla^2\Delta f - \left(\frac{1}{n} + \Lambda_n\right) R g\Delta f + \Lambda_n R \nabla^2 f - \frac{\Lambda_n}{n}R^2 g f \right]\\
=& A_n \left[ \nabla^2 \left(\Delta f + \Lambda_n R f\right)  - g \left( \Delta \left(\Delta f + \Lambda_n R f\right) +\frac{R}{n}\left(\Delta f + \Lambda_n R f\right) \right) \right].
\end{align*}

By assuming $f\in \ker \Gamma_g^*$, \emph{i.e.} $\Gamma_g^*(f)=0$, when taking trace, we have
\begin{align*}
A_n \left[  - (n-1) \Delta \left(\Delta f + \Lambda_n R f\right) - R\left(\Delta f + \Lambda_n R f\right) \right] = 0.
\end{align*}

Thus,
\begin{align*}
\Delta \left(\Delta f + \Lambda_n R f\right) = - \frac{R}{n-1}\left(\Delta f + \Lambda_n R f\right).
\end{align*}

Now substitute it in the expression of $\Gamma_g^* f$, we have
$$
\Gamma_g^* f =A_n \left[ \nabla^2 \left(\Delta f + \Lambda_n R f\right) + \frac{R}{n(n-1)}g \left(\Delta f + \Lambda_n R f\right)\right] = 0,
$$
where $\Lambda_n= \frac{2}{A_n}\left(\frac{B_n}{n}+C_n\right) = - \frac{(n+2)(n-2)}{2n(n-1)} < 0$, for any $n \geq 3$.\\
\end{proof}

Next we show that a closed $Q$-singular Einstein manifold with positive scalar curvature has to be spherical:
\begin{proposition} \label{Q-static_sphere}
Let $(M^n,g)$ be a complete $Q$-singular Einstein manifold with positive scalar curvature. Then $(M^n,g)$ is isometric to the round sphere $(S^n(r), g_{_{S^n(r)}})$, with radius $r = \left(\frac{n(n-1)}{R_g}\right)^{\frac{1}{2}}$. Moreover, $\ker \Gamma_g^*$ is consisted of eigenfunctions of $(-\Delta_g)$ associated to $\lambda_1 = \frac{R_g}{n-1}> 0$ on $S^n(r)$ and hence $\dim \ker \Gamma_{g}^* = n+1$.
\end{proposition}

\begin{proof}

Note that $\Lambda_n < 0$ and $Spec (-\Delta_g)$ consisted of nonnegative real numbers, then $$\Lambda_n R_g \not\in Spec ( - \Delta_g).$$

Let $f \in \ker \Gamma_g^*$, $f\not\equiv 0$ and $\varphi := \Delta_g f + \Lambda_n R_g f$, then
\begin{align*}
\varphi \not\equiv 0.
\end{align*}

Therefore, by Lemma \ref{Q-static_Einstein}, we have the following equation
\begin{align}\label{Obata_type_eqn}
\nabla^2\varphi=-\frac{R_g}{n(n-1)}\varphi g
\end{align}
with a nontrivial solution $\varphi$.\\

Taking trace, we get
\begin{align}
\Delta_g \varphi = - \frac{R_g}{n-1}\varphi.
\end{align}
\emph{i.e.} $\frac{R_g}{n-1}$ is an eigenvalue of $-\Delta_g$.\\

On the other hand, by \emph{Lichnerowicz-Obata's Theorem}, the first nonzero eigenvalue of $(- \Delta_g)$ satisfies
$$\lambda_1 \geq \frac{R_g}{n-1}$$
with equality if and only if $(M^n,g)$ is isometric to the round sphere $(S^n(r), g_{S^n(r)})$ with radius $r = \left(\frac{n(n-1)}{R} \right)^{\frac{1}{2}}$. Hence the first part of the theorem follows.\\

Since $\Lambda_n R_g \not\in Spec ( - \Delta_g)$, it implies that the operator $\Delta_g + \Lambda_n R_g$ is invertible. On the other hand,
\begin{align*}
\left( \Delta_g + \Lambda_n R_g \right) \left(\Delta_g + \frac{R_g}{n-1}\right) f = \left(\Delta_g + \frac{R_g}{n-1}\right) \left( \Delta_g + \Lambda_n R_g \right) f = \left(\Delta_g + \frac{R_g}{n-1}\right) \varphi=0.
\end{align*}
Thus $$\left(\Delta_g + \frac{R_g}{n-1}\right) f = 0.$$ \emph{i.e.} $f$ is an eigenfunction associated to $\lambda_1 = \frac{R_g}{n-1}$.\\

Therefore, $\ker \Gamma_g^*$ can be identified by the eigenspace associated to the first nonzero eigenvalue $\lambda_1> 0$ of $(-\Delta_g)$ on $S^n(r)$ and hence $\dim\Gamma_g^* = n+1$.

\end{proof}

As for Ricci flat ones, we have:

\begin{theorem}\label{Static_Ricci_flat}
Suppose $(M^n,g)$ is a $Q$-singular Riemannian manifold. If $(M,g)$ admits an nonzero constant potential $f \in \ker \Gamma_g^*$, then $(M,g)$ is $Q$-flat, \emph{i.e.} the $Q$-curvature vanishes identically. Furthermore, suppose $(M,g)$ is a closed $Q$-singular Einstein manifold, then $(M,g)$ is Ricci flat if and only if $\ker \Gamma_g^*$ is consisted of constant functions.
\end{theorem}

\begin{proof}

Without lose of generality, we can assume that $1 \in \ker \Gamma_g^*$. We have
\begin{align*}
tr\Gamma^{*}_{g} 1 = \mathscr{L}_g 1  = \frac{1}{2} \left( P_g 1 - \frac{n+4}{2}Q_g\right) =  - 2 Q_g = 0,
\end{align*}
which implies that $Q_g=0$.\\

Now assume $(M,g)$ is a closed $Q$-singular Einstein manifold.\\

We have
\begin{align*}
\int_M Q_g dv_g &= \int_M \left(A_n \Delta_{g} R + B_n |Ric|^2 + C_n R^2\right) dv_g \\
&= \left(\frac{B_n}{n} + C_n \right) R^2\ Vol_g(M) \\
&= \frac{(n+2)(n-2)}{8n(n-1)^2}R^2\ Vol_g(M).
\end{align*}

Suppose $1 \in \ker \Gamma_g^*$, then $Q_g = 0$ and hence
\begin{align*}
R = 0 ,
\end{align*}
That means $(M, g)$ is Ricci flat.\\

On the other hand, if $(M,g, f)$ is Ricci flat,
\begin{align*}
\mathscr{L}_g f = \frac{1}{2}\Delta^2 f = 0.
\end{align*}

Since $M$ is compact without boundary, $f$ is a nonzero constant on $M$.

\end{proof}

\begin{remark}
It was shown in \cite{C-G-Y} that for a closed $4$-dimensional $Q$-singular space $(M, g)$, $1 \in \ker \Gamma_g^*$ if and only if $g$ is Bach flat with vanishing $Q$-curvature. But this theorem doesn't generalized to other dimensions automatically. In fact, for generic dimensions, $1 \in \ker \Gamma_g^*$ doesn't imply $g$ is Bach flat by direct calculations, although it is still $Q$-flat as shown above.
\end{remark}

Theorem \ref{Classificaition_Q_singular_Einstein} follows from combining Propositions \ref{Q-static_sphere} and \ref{Static_Ricci_flat}.\\

For Einstein manifolds with negative scalar curvature, generically they are not $Q$-singular:
\begin{proposition}\label{Q-hyperbolic}
Let $(M, g)$ be a closed Einstein manifold with scalar curvature $R < 0$. Suppose $\Lambda_n R \not\in Spec(-\Delta_g)$, then $$\ker \Gamma_g^* = \{0\}.$$ \emph{i.e.} $(M, g)$ is not $Q$-singular.
\end{proposition}

\begin{proof}
For any smooth function $f\in  \ker \Gamma_g^*$, let $\varphi := \Delta f + \Lambda_n R f$. By Lemma \ref{Q-static_Einstein}, we have $$\nabla^2 \varphi + \frac{R}{n(n-1)}g\varphi = 0.$$

Taking trace,
$$\Delta \varphi + \frac{R}{n-1}\varphi = 0.$$
Thus $$\varphi = \Delta f + \Lambda_n R f = 0$$ identically on $M$, since $0 > \frac{R}{n-1} \not\in spec(-\Delta_g)$. Thus $$\Delta f = - \Lambda_n R f .$$

By assuming $\Lambda_n R \not\in Spec(-\Delta_g)$, we conclude that $f \equiv 0$. That means, $\ker \Gamma_g^*$ is trivial.

\end{proof}

\section{Stability of Q-curvature}

In this section, we will discuss the linearized stability of $Q$-curvature on closed manifolds.
As main tools, we need following key results. Proofs can be found in referred articles.

\begin{theorem}[Splitting Theorem (\cite{B-E, F-M})]\label{Splitting_Theorem}
Let $(M,g)$ be a closed Riemannian manifold, $E$ and $F$ be vector bundles on $M$. Let $D : C^\infty (E) \rightarrow C^\infty(F)$ be a $k^{th}$-order differential operator and $D^* : C^\infty(F) \rightarrow C^\infty(E)$ be its $L^2$-formal adjoint operator.

For $k \leq s\leq \infty$ and $1< p < \infty$, let $D_s : W^{s,p}(E) \rightarrow W^{s-k,p}(F)$ and $D_s^* : W^{s,p}(F) \rightarrow W^{s-k,p}(E)$ be the bounded linear operators by extending $D$ and $D^*$ respectively.

Assume that $D$ or $D^*$ has injective principal symbols, then
\begin{align}
W^{s,p}(F) =  \Ima D_{s+k} \oplus \ker D_s^*.
\end{align}
Moreover,
\begin{align}
C^\infty(F) = \Ima D \oplus \ker D^*.
\end{align}
\end{theorem}

In particular, take the vector bundle $F$ to be all symmetric 2-tensors $S_2(M)$, we have

\begin{corollary}[Canonical decomposition of $S_2$ (\cite{B-E, F-M})]\label{Canonical_decomposition_S_2}
Let $(M,g)$ be a closed Riemannian manifold, then the space of symmetric 2-tensors can be decomposed into
\begin{align}
S_2(M) = \{ L_X g : X \in \mathscr{X}(M)\} \oplus \ker \delta_g.
\end{align}
\end{corollary}

Another result we need is

\begin{theorem}[Generalized Inverse Function Theorem (\cite{Gel'man})]\label{Generalized_Inverse_Function_Theorem}
Let $X$, $Y$ be Banach spaces and $f : U_{x_0} \rightarrow Y$ be a continuously differentiable map with $f(x_0) = y_0$, where $U_{x_0} \subset X$ is a neighborhood of $x_0$. Suppose the derivative $D f (x_0) : X \rightarrow Y$ is a surjective bounded linear map. Then there exists $V_{y_0} \subset Y$, a neighborhood of $y_0$, and a continuous map $\varphi : V_{y_0} \rightarrow U_{x_0}$ such that
\begin{align*}
f(\varphi(y)) = y, \ \ \ \forall y\in V_{y_0};
\end{align*}
and
\begin{align*}
\varphi(y_0) = x_0.
\end{align*}

\end{theorem}

Combining Corollary \ref{Canonical_decomposition_S_2} and Theorem \ref{Generalized_Inverse_Function_Theorem}, we obtain \emph{Ebin's Slice Theorem}(\cite{Ebin}), which we will use later in this article.

\begin{theorem}[Slice Theorem (\cite{Ebin, F-M})]\label{Slice_Theorem}
Let $(M,\bar{g})$ be a Riemannian manifold. For $p> n$, suppose that $g$ is also a Riemannian metric on $M$ and $||g - \bar{g}||_{W^{2,p}(M,\bar{g})}$ is sufficiently small, then there exists a diffeomorphism $\varphi \in \mathscr{D} (M)$ such that $h := \varphi^* g - \bar{g}$ satisfies that $\delta_{\bar{g}} h = 0$ and moreover,
$$||h||_{W^{2,p}(M,\bar{g})} \leq N ||g - \bar{g}||_{W^{2,p}(M,\bar{g})},$$
where $N$ is a positive constant only depends on $(M, \bar g)$.
\end{theorem}

\begin{remark}
Brendle and Marques (c.f. \cite{B-M}) proved an analogous decomposition and slice theorem for a compact domain with boundary.
\end{remark}

Now we can give the proof of the main theorem (Theorem \ref{Q_stability}) in this section.
\begin{theorem
Let $(M,\bar{g})$ be a closed Riemannian manifold. Assume
$(M,\bar{g})$ is not $Q$-singular, then the Q-curvature is linearized stable at $\bar g$ in the sense that $Q : \mathcal{M} \rightarrow C^{\infty}(M) $ is a submersion at $\bar{g}$.
Thus, there is a neighborhood $U \subset C^{\infty}(M)$ of $Q_{\bar{g}}$ such that for any $\psi \in U$, there exists a metric $g$ on $M$ closed to $\bar{g}$ with $Q_g = \psi$.
\end{theorem}
\begin{proof

The principal symbol of $\Gamma_{\bar{g}}^*$ is
\begin{align*}
\sigma_{\xi}(\Gamma_{\bar{g}}^*) = - A_n \left( g |\xi|^2 - \xi \otimes\xi \right)|\xi|^2.
\end{align*}

Taking trace, we get
\begin{align*}
tr\ \sigma_{\xi}(\Gamma_{\bar{g}}^*) = - A_n \left( n - 1 \right)|\xi|^4.
\end{align*}

Thus, $\sigma_{\xi}(\Gamma_{\bar{g}}^*) = 0$ will imply that $\xi = 0$. \emph{i.e.} $\Gamma^*_{\bar{g}}$ has an injective principal symbol.

By the \emph{Splitting Theorem} \ref{Splitting_Theorem},
$$C^\infty(M) = \Ima \Gamma_{\bar{g}} \oplus \ker \Gamma^*_{\bar{g}},$$
which implies that $\Gamma_{\bar{g}}$ is surjective, since we assume that $(M,\bar{g})$ is not $Q$-singular, \emph{i.e.} $\ker \Gamma^*_{\bar{g}} = \{0\}$.

Therefore, applying the \emph{Generalized Implicit Function Theorem} (Theorem \ref{Generalized_Inverse_Function_Theorem}), $Q$ maps a neighborhood of $\bar g$ to a neighborhood of $Q_{\bar{g}}$ in $C^\infty(M)$.

\end{proof}

As a consequence, we can derive the stability of generic positive Einstein manifolds (Corollary \ref{cor:stab_pos_Einstein}).

\begin{corollary
Let $(M,\bar{g})$ be a closed positive Einstein manifold. Assume $(M,\bar{g})$ is not spherical, then the Q-curvature is linearized stable at $\bar g$.
\end{corollary}

\begin{proof

Since $M$ is assumed to be not spherical, by Theorem \ref{Classificaition_Q_singular_Einstein}, $(M,g)$ is not $Q$-singular. Now the stability follows from Theorem \ref{Q_stability}. \\
\end{proof}

For a generic $Q$-flat manifold, we can prescribe any smooth function such that it is the $Q$-curvature for some metric on the manifold (Corollary \ref{cor:prescribing_zero_Q}).

\begin{corollary
Suppose $(M, \bar g)$ is a closed non-$Q$-singular manifold with vanishing $Q$-curvature. Then any smooth function $\varphi$ can be realized as a $Q$-curvature for some metric $ g$ on $M$.
\end{corollary}
\begin{proof
Since $(M, \bar g)$ is non-$Q$-singular, applying Theorem \ref{Q_stability}, as a nonlinear map, $Q$ maps a neighborhood of the metric $\bar g$ to a neighborhood of $Q_{\bar g} = 0$ in $C^\infty(M)$. Thus there exists an $\varepsilon_0 > 0$ such that for any smooth function $\psi$ with $||\psi||_{C^\infty(M)}< \varepsilon_0$, we can find a smooth metric $g_\psi$ closed to $\bar g$ with $Q_{g_\psi} = \psi$.

Now for any nontrivial $\varphi \in C^\infty(M)$, let $\tilde \varphi:= u_{\varepsilon_0 , \varphi}\varphi$, where $u_{\varepsilon_0 , \varphi} = \frac{\varepsilon_0 }{2 ||\varphi||_{_{C^\infty(M)}}} > 0$. Clearly, $||\tilde \varphi||_{C^\infty(M)} < \varepsilon_0$ and hence there is a metric $g_{\tilde \varphi}$ closed to $\bar g$ such that $Q_{g_{\tilde \varphi}} = \tilde \varphi$. Let
$$g = u_{\varepsilon_0 , \varphi}^{\frac{1}{2}} g_{\tilde \varphi},$$ then we have $$Q_g = u_{\varepsilon_0 , \varphi}^{-1} \cdot Q_{g_{\tilde \varphi}} =  \varphi.$$
\end{proof}

Now we can prove the Corollary \ref{cor:pos_Y_negative_Q}.
\begin{corollary}
Let $M$ be a closed manifold with positive Yamabe invariant $Y(M) > 0$. There is a metric $g$ with $Q$-curvature $Q_g < 0$ on $M$.
\end{corollary}
\begin{proof
By Matsuo's theorem (see Corollary 2 in \cite{Mat14}), on a closed manifold $M$ with dimension $n\geq 3$ and positive Yamabe invariant, there exists a metric $g$ with scalar curvature $R_g = 0$ but $Ric_g \not\equiv 0$ on $M$. Thus the $Q$-curvature satisfies $$Q_g = - \frac{2}{(n-2)^2} |Ric_g|^2 \leq 0.$$
If $|Ric_g|>0$ pointwisely, then $Q_g < 0$ on $M$. Otherwise, there is a point $p \in M$ such that $|Ric_g(p)|^2 = 0$ and hence $Q_g$ is not a constant on $M$. This implies that the metric $g$ is not $Q$-singular. Therefore, by Theorem \ref{Q_stability}, we can perturb the metric $g$ to obtain a metric with strictly negative $Q$-curvature. This gives the conclusion.
\end{proof}

For the Ricci flat case, we have a better result since we can identify $\ker \Gamma_{\bar{g}}^*$ with constants.

\begin{theorem
Let $(M,\bar{g})$ be a closed Ricci flat Manifold. Denote $$\Phi:=\{\psi \in C^{\infty}(M): \int_M \psi dv_{\bar{g}} = 0\}$$ to be the set of smooth functions with zero average. Then for any $\psi \in \Phi$, there exists a metric $g$ on $M$ such that $$Q_g = \psi.$$
\end{theorem}
\begin{proof
In the proof of Theorem \ref{Q_stability}, we have showed that the principal symbol of $\Gamma_{\bar{g}}^*$ is injective and the decomposition
$$C^\infty(M) = \Ima \Gamma_{\bar{g}} \oplus \ker \Gamma^*_{\bar{g}}$$
holds.

On the other hand, by Theorem \ref{Static_Ricci_flat}, $\ker \Gamma^*_{\bar{g}}$ is consisted by constant functions and hence $ \Ima \Gamma_{\bar{g}} = \Phi$. By identifying $\Phi$ with its tangent space, we can see the map $Q$ is a submersion at $g$ with respect to $\Phi$. Therefore, by \emph{Generalized Inverse Function Theorem} (Theorem \ref{Generalized_Inverse_Function_Theorem}), we have the local surjectivity. That is, there exists a neighborhood of $\bar{g}$, say $U_{\bar{g}} \subset \mathcal{M}$ and a neighborhood of $0$, say $V_0 \subset C^\infty(M)$ such that $ V_0|_\Phi \subset Q_{g}(U_{\bar{g}})$, where $g\mapsto Q_{g}$ is a map.\\

Now for any $\psi \in \Phi$, let $r > 0$ be a sufficiently large constant such that $\frac{1}{r^4}\psi \in V_0$. There exists a metric $g_r \in U_{\bar{g}}$, such that $Q_{g_r} = \frac{1}{r^4} \psi$. Let $g := \frac{1}{r^2}g_r$, we have $$Q_g = r^4 \cdot Q_{g_r} = r^4 \cdot \frac{1}{r^4}\psi = \psi.$$
Thus we prove Theorem \ref{Ricci_flat_stability}.
\end{proof}

Similarly, we can also give an answer to the prescribing $Q$-curvature problem near the standard spherical metric by noticing the fact that
$$\ker \Gamma_{\bar{g}}^* = E_{\lambda_1}$$ from Theorem \ref{Q-static_sphere}:

\begin{theorem}\label{thm:stability_sphere}
Let $(S^n, \bar{g})$ be the standard unit sphere and $E_{\lambda_1}$ be the eigenspace of $(-\Delta_g)$ associated to the first nonzero eigenvalue $\lambda_1 = n$. Then for any $\psi \in E_{\lambda_1}^\perp$ with $||\psi - Q_{\bar{g}}||_{C^\infty(S^n,\bar{g})}$ sufficiently small, there exists a metric $g$ near $\bar{g}$ such that $$Q_g = \psi.$$
\end{theorem}

\section{Rigidity Phenomena of Flat manifolds}

Let $(M,g)$ be a closed Riemannian manifold. Consider a deformation of $(M,g)$, $g(t)=g+th$, $t \in (-\varepsilon, \varepsilon)$.\\

We have calculated the first variation of $Q$-curvature (see equation (\ref{Q_linearization})). In order to study the local rigidity of $Q$-curvature, we are going to calculate the second variation of $Q$-curvature.\\

First, we recall the following well known $2^{nd}$-variation formulae, which can be found in \cite{F-M}. For detailed calculations, we refer to the appendices of \cite{Yuan}.

\begin{proposition}
We have the following $2^{nd}$-variation formulae for metrics,
\begin{align}
g''_{ij} = \left.\frac{d^2}{dt^2}\right|_{t=0} (g(t))_{ij} = 0,
\end{align}
and
\begin{align}
g''^{ij} = \left.\frac{d^2}{dt^2}\right|_{t=0} (g(t))^{ij} = 2 h_k^j h^{ik}.
\end{align}

Also for Christoffel Symbols,
\begin{align}
{\Gamma''}_{ij}^k = \left.\frac{d^2}{dt^2}\right|_{t=0} \Gamma(g(t))^k_{ij} = - h^{kl} \left( \nabla_i  h_{jl} + \nabla_j h_{il}
- \nabla_l h_{ij} \right).
\end{align}
\end{proposition}

\begin{lemma}
The second variation of $Q$-curvature is
\begin{align}
\left.\frac{d^2}{dt^2}\right|_{t=0} Q(g(t)) =& A_n [ \Delta_g R'' + 2 \nabla^2 R \cdot ( h \times h ) - 2 h \cdot \nabla^2 R' + ( 2 \delta h + d (tr h) ) \cdot dR' \\ \notag &\ \ \ \ \ \ +  h^{ij} ( 2 \nabla_i h_j^k - \nabla^k h_{ij} ) \nabla_k R - h \cdot ((2 \delta h + d(tr h))  \otimes d R )  ]\\ \notag
& + B_n [ 4 (Ric \times Ric) \cdot ( h \times h ) + 2 | Ric \times h |^2 - 8 Ric' \cdot ( Ric \times h )\\ \notag &\ \ \ \ \ \ + 2 Ric''\cdot Ric + 2 | Ric' |^2  ]\\ \notag
& + C_n [ 2 R R'' + 2(R')^2], \notag
\end{align}
where $Ric'$, $Ric''$; $R'$, $R''$ are the first and second variations of Ricci tensor and scalar curvature at $g$ respectively.
\end{lemma}

\begin{proof}
Choose normal coordinates at any point $p \in M$, \emph{i.e.} $\Gamma_{jk}^i = 0$ at $p$, then
\begin{align*}
&\left.\frac{d^2}{dt^2}\right|_{t=0}(\Delta_{g(t)}R(g(t)))\\
=& (g^{ij}(\partial_i\partial_j R - \Gamma_{ij}^k \partial_k R))''\\
=& g''^{ij} \partial_i\partial_j R  + 2 g'^{ij}(\partial_i\partial_j R' - {\Gamma'}_{ij}^k \partial_k R) + g^{ij}(\partial_i\partial_j R'' - {\Gamma''}_{ij}^k \partial_k R - 2{\Gamma'}_{ij}^k \partial_k R')\\
=& 2 h_k^j h^{ik} \nabla_i \nabla_j R - 2 h^{ij} (\nabla_i\nabla_j R' - {\Gamma'}_{ij}^k \nabla_k R) + \Delta R'' - g^{ij} {\Gamma''}_{ij}^k \nabla_k R - 2 g^{ij} {\Gamma'}_{ij}^k \nabla_k R'\\
=& \Delta R'' + 2 \nabla^2 R \cdot (h \times h) - 2 h\cdot \nabla^2 R' + ( 2 \delta h + d trh) \cdot dR' +  h^{ij} ( 2 \nabla_i h_j^k - \nabla^k h_{ij}) \nabla_k R\\
 &- h \cdot ((2 \delta h + d tr h) \otimes d R),
\end{align*}
by substituting the expression of ${\Gamma'}_{ij}^k$ and ${\Gamma''}_{ij}^k$.\\

And
\begin{align*}
&\left.\frac{d^2}{dt^2}\right|_{t=0} |Ric(g(t))|^2_{g(t)}\\
=& \left( g^{ik} g^{jl} R_{ij} R_{kl} \right)''\\
=& 2 g''^{ik}R_{ij} R_k^j + 2 g'^{ik} g'^{jl} R_{ij} R_{kl} + 8 g'^{ik} R'_{ij} R_k^j + 2 R''_{ij} R^{ij} + 2 g^{ik} g^{jl} R'_{ij} R'_{kl}\\
=& 4 (Ric \times Ric) \cdot ( h \times h) + 2 |Ric \times h|^2 - 8 Ric' \cdot (Ric \times h) + 2 Ric'' \cdot Ric + 2|Ric'|^2. \\
\end{align*}

Also,
\begin{align*}
\left.\frac{d^2}{dt^2}\right|_{t=0} R(g(t))^2 = (R^2)'' = 2 (R')^2 + 2 R R''.
\end{align*}

We prove the lemma by combining all three parts together.
\end{proof}

Simply by taking $Ric = 0$ and $R = 0$, we get the second variation for $Q$-curvature at a Ricci flat metric.
\begin{corollary}\label{2nd_variation_Ricci_flat}
Suppose the metric $g$ is Ricci flat, then
\begin{align}
D^2 Q_g \cdot ( h, h )
=A_n(\Delta_{g}R'' - 2 h \cdot \nabla^2 R' + ( 2 \delta h + d (tr h)) \cdot dR') + 2 B_n |Ric'|^2 + 2C_n(R')^2.
\end{align}
\end{corollary}

Now we assume $(M,\bar{g}, f)$ is a $Q$-singular metric.\\

Consider the functional
\begin{align*}
\mathscr{F}(g)=\int_M Q_g \cdot f dv_{\bar{g}}.
\end{align*}
Note that here we fix the volume form to be the one associated to the $Q$-singular metric $\bar{g}$.\\

\begin{remark}
The analogous functional
\begin{align*}
\mathscr{G}(g)=\int_M R_g \cdot f dv_{\bar{g}},
\end{align*}
plays a fundamental role in studying rigidity phenomena of vacuum static spaces (c.f. \cite{F-M, B-M, Q-Y_1}, \emph{etc.}).
\end{remark}

\begin{lemma}
The metric $\bar{g}$ is a critical point of the functional $\mathscr{F}(g)$.
\end{lemma}

\begin{proof}
For any symmetric 2-tensor $h \in S_2$, let $g(t) = \bar{g} + t h$, $t \in (-\varepsilon, \varepsilon)$ be a family of metrics on $M$. clearly, $g(0) = \bar{g}$ and $g'(0) = h$. Then
\begin{align*}
\left. \frac{d}{dt}\right|_{t=0} \mathscr{F}(g(t)) = \int_M DQ_{\bar{g}} \cdot h f dv_{\bar{g}} = \int_M \Gamma_{\bar{g}} h \cdot f dv_{\bar{g}} = \int_M h \cdot \Gamma_{\bar{g}}^* f \ dv_{\bar{g}} = 0,
\end{align*}
\emph{i.e.} $\bar{g}$ is a critical point for the functional $\mathscr{F}(g)$.
\end{proof}

Furthermore, if we assume $\bar{g}$ is a flat metric, by Theorem \ref{Static_Ricci_flat}, we can take $f$ to be a nonzero constant. In particular, we can take $f \equiv 1$, since $Q$-singular equation is a linear equations system with respect to $f$.\\

\begin{lemma}\label{2nd_variation_flat}
Let $\bar{g}$ be a flat metric and $f \equiv 1$. Suppose $\delta h = 0$, then the second variation of $\mathscr{F}$ at $\bar{g}$ is given by
\begin{align}
D^2 \mathscr{F}_{\bar{g}} \cdot (h,h) = -2 \alpha_n \int_M|\Delta(tr h)|^2 dv_{\bar{g}} + \frac{1}{2}B_n\int_M(|\Delta \overset{\circ}{h}|^2)dv_{\bar{g}},
\end{align}
where $\overset{\circ}{h}$ is the traceless part of $h$ and $\alpha_n := - \frac{1}{2}\left(A_n + \frac{n+1}{2n}B_n + 2C_n \right)= \frac{(n^2 - 2) (n^2 - 2n - 2)}{8n(n-1)^2(n-2)^2} > 0$, $B_n=-\frac{2}{(n-2)^2}< 0$, for any $n \geq 3$.
\end{lemma}

\begin{proof}

With the aid of Corollary \ref{2nd_variation_Ricci_flat}, we have
\begin{align*}
&\left. \frac{d^2}{dt^2}\right|_{t=0} \mathscr{F}(g(t))\\
 =& \int_M Q''_{\bar{g}} dv_{\bar{g}}\\
 =& \int_M \left[A_n(\Delta R'' - 2 h \cdot \nabla^2 R' + ( 2 \delta h + d (tr h)) \cdot dR') + 2 B_n |Ric'|^2 + 2C_n(R')^2 \right] dv_{\bar{g}}\\
 =& \int_M \left[A_n(- 2 \delta h \cdot d R' + ( 2 \delta h + d (tr h)) \cdot dR') + 2 B_n |Ric'|^2 + 2C_n(R')^2 \right] dv_{\bar{g}}\\
 =& \int_M \left[A_n(   - \Delta (tr h)) \cdot R' + 2 B_n |Ric'|^2 + 2C_n(R')^2 \right] dv_{\bar{g}}\\
 =& \int_M \left[A_n(   - \Delta (tr h)) \cdot (- \Delta (tr h) + \delta^2 h - Ric \cdot h) + 2 B_n |Ric'|^2 + 2C_n(R')^2 \right] dv_{\bar{g}}\\
 =& \int_M \left[A_n( \Delta (tr h))^2 + 2 B_n |Ric'|^2 + 2C_n(R')^2 \right] dv_{\bar{g}},
\end{align*}
where the last step is due to the assumptions of $\bar{g}$ being a flat metric and divergence-free property of $h$.\\

For exact the same reasons, by Proposition \ref{1st_variation_Ricci_scalar}, we have
\begin{align*}
|Ric'|^2 = \frac{1}{4} | \Delta h  + \nabla^2 (tr h)|^2 = \frac{1}{4} \left( |\Delta h|^2 + 2 \Delta h \cdot \nabla^2 (trh) + |\nabla^2 (tr h)|^2 \right)
\end{align*}
and
\begin{align*}
(R')^2 = ( \Delta (tr h))^2.
\end{align*}

Thus
\begin{align*}
  \int_M |Ric'|^2 dv_{\bar{g}}=& \frac{1}{4} \int_M\left[\left( |\Delta h|^2 + 2 \Delta h \cdot \nabla^2 (trh) + |\nabla^2 (tr h)|^2 \right)\right] dv_{\bar{g}}\\
 =& \frac{1}{4}\int_M \left[ \left( |\Delta h|^2 + 2 \delta\Delta h \cdot d(trh) + \delta\nabla^2 (tr h) \cdot d(tr h) \right)\right] dv_{\bar{g}}\\
 =& \frac{1}{4}\int_M \left[\left( |\Delta h|^2 + 2 \Delta \delta h \cdot d(trh) + \delta d (tr h) \cdot \delta d(tr h) \right)\right] dv_{\bar{g}}\\
 =& \frac{1}{4}\int_M \left[\left( |\Delta h|^2  + |\Delta (tr h)|^2 \right)\right] dv_{\bar{g}}\\
 =& \frac{1}{4}\int_M \left[ |\Delta \overset{\circ}{h}|^2 + \frac{n+1}{n}( \Delta (tr h))^2  \right] dv_{\bar{g}}.
\end{align*}

Now
\begin{align*}
\left. \frac{d^2}{dt^2}\right|_{t=0} \mathscr{F}(g(t)) =& \int_M \left[(A_n + \frac{n+1}{2n}B_n + 2 C_n)( \Delta (tr h))^2 +\frac{1}{2}B_n |\Delta \overset{\circ}{h}|^2 \right] dv_{\bar{g}}\\
 =& -2 \alpha_n \int_M|\Delta(tr h)|^2 dv_{\bar{g}} + \frac{1}{2}B_n\int_M(|\Delta \overset{\circ}{h}|^2)dv_{\bar{g}}.
\end{align*}

This gives the equation we claimed.
\end{proof}

Now we are ready to prove Theorem \ref{flat_local_rigidity}.

\begin{theorem
For $n \geq 3$, let $(M^n,\bar{g})$ be a closed flat Riemannian manifold and $g$ be a metric on $M$ with $$Q_g \geq 0.$$ Suppose $||g - \bar{g}||_{C^2(M, \bar{g})}$ is sufficiently small, then $g$ is also flat.
\end{theorem}

\begin{proof

Since $g$ is $C^2$-closed to $\bar{g}$, by the \emph{Slice Theorem} (Theorem \ref{Slice_Theorem}), there exists a diffeomorphism $\varphi \in \mathscr{D}( M ) $ such that $$h := \varphi^* g - \bar{g}$$ is divergence free with respect to $\bar{g}$ and $$||h||_{C^2(M,\bar{g})} \leq N ||g-\bar{g}||_{C^2(M,\bar{g})},$$ where $N > 0$ is a constant only depends on $(M,\bar{g})$.\\

Apply Lemma \ref{2nd_variation_flat}, we can expand $\mathscr{F}(\varphi^*g)$ at $\bar{g}$ as follows,
\begin{align}\label{eqn:F_expansion}
\mathscr{F}(\varphi^*g) &= \mathscr{F}(\bar g) + D\mathscr{F}_{\bar g} \cdot h + \frac{1}{2}D^2 \mathscr{F}_{\bar g} \cdot (h, h) + E_3\\
\notag   &=   - \alpha_n \int_M|\Delta(tr h)|^2 dv_{\bar{g}} + \frac{1}{4}B_n\int_M |\Delta \overset{\circ}{h}|^2 dv_{\bar{g}} + E_3,
\end{align}
where $$|E_3| \leq C_0 \int_M |h|\ |\nabla^2 h|^2 \ dv_{\bar{g}}$$ for some constant $C_0=C_0 (n, M, \bar g)> 0$.\\

We know $$\mathscr{F}(\varphi^*g) = \int_M Q_g\circ\varphi\ dv_{\bar{g}} \geq 0$$ since $Q_{g}\geq0$ and $\varphi$ is a diffeomorphism near the identity. Since
$$\alpha_n =-\frac{1}{2}\left(A_n + \frac{n+1}{2n}B_n + 2C_n \right) > 0,\ \ B_n = - \frac{2}{(n-2)^2} < 0$$ for $n\geq 3$, we can take
$\mu_n > 0$, a sufficiently small constant only depends on the dimension $n$ such that
$$\min\{ \alpha_n - \frac{\mu_n}{n} , - \frac{1}{4}B_n- \mu_n\} > 0.$$

Therefore, with the aid of equation (\ref{eqn:F_expansion}), we have
\begin{align*}
&\mu_n \int_M |\Delta h|^2 dv_{\bar{g}}\\
\leq& \left( \alpha_n - \frac{\mu_n}{n}\right)\int_M|\Delta(tr h)|^2 dv_{\bar{g}} + \left( - \frac{1}{4}B_n- \mu_n \right)\int_M |\Delta \overset{\circ}{h}|^2 dv_{\bar{g}} + \mu_n \int_M |\Delta h|^2 dv_{\bar{g}}\\
=&  \alpha_n \int_M|\Delta(tr h)|^2 dv_{\bar{g}} - \frac{1}{4}B_n\int_M |\Delta \overset{\circ}{h}|^2 dv_{\bar{g}}\\
=& - \mathscr{F}(\varphi^*g) + E_3\\
\leq& |E_3|\\
\leq& C_0 \int_M |h|\ |\nabla^2{h}|^2 \ dv_{\bar{g}}.
\end{align*}

Suppose $g$ is sufficiently $C^2$-closed to $\bar g$, say $||g-\bar{g}||_{C^2(M,\bar{g})} < \frac{\mu_n}{2NC_0}$, then $$||h||_{C^0(M,\bar{g})} \leq ||h||_{C^2(M,\bar{g})} \leq N ||g-\bar{g}||_{C^2(M,\bar{g})} < \frac{\mu_n}{2C_0}$$ and therefore,
\begin{align*}
\mu_n\int_M |\nabla^2 h|^2dv_{\bar{g}} = \mu_n\int_M |\Delta h|^2dv_{\bar{g}} \leq C_0 \int_M |h|\ |\nabla^2{h}|^2 \ dv_{\bar{g}} \leq \frac{\mu_n}{2}\int_M |\nabla^2 h|^2dv_{\bar{g}},
\end{align*}
which implies $\nabla^2 h = 0$ on $M$. \\

Now we have
\begin{align*}
\int_M |\nabla h |^2 dv_{\bar{g}} = - \int_M h \Delta h\  dv_{\bar{g}} = 0.
\end{align*}
That is $\nabla h = 0$.\\

Since $\bar{g}$ is flat, then on a neighborhood $U_p$ for any $ p \in M$, we can find a local coordinates, such that $\bar{g}_{ij} = \delta_{ij}$ and $\partial_k \bar{g}_{ij} = 0$, $i,j,k = 1, \cdots, n$ on $U_p$.\\

Under the same coordinates, Christoffel symbols of $\varphi^*g$ are
\begin{align*}
\Gamma^k_{ij}(\varphi^*g) &= \frac{1}{2} (\varphi^*g)^{kl} \left(\partial_i (\bar{g}_{jl} + h_{jl}) + \partial_j (\bar{g}_{il} + h_{il}) - \partial_l (\bar{g}_{ij} + h_{ij}) \right) \\
&= \frac{1}{2} (\varphi^*g)^{kl} (\nabla_i  h_{jl} + \nabla_j h_{il} - \nabla_l h_{ij}) = 0
\end{align*}
on $U_p$, since $h$ is parallel with respect to $\bar{g}$. \\

Thus the Riemann curvature tensor of $\varphi^* g$ vanishes identically on $U_p$ for any $p \in M$, which implies that the metric $\varphi^*g$ is flat and so is $g$.
\end{proof}

\begin{remark}
Fischer and Marsden proved an analogous result for the scalar curvature. (see \cite{F-M})
\end{remark}

As an application, we can get the local rigidity of $Q$-curvature on compact domain of $\mathbb{R}^n$, which can be thought as an analogue of rigidity  part of \emph{Positive Mass Theorem}.

\begin{corollary}\label{flat_domain_rigidity}
Suppose $\Omega \subset \mathbb{R}^n$ is a compactly contained domain. Let $\delta$ be the flat metric and $g$ be a metric on $\mathbb{R}^n$ satisfying
\begin{itemize}
\item $Q_g \geq 0$,
\item $supp(g - \delta) \subset \Omega$,
\item $||g - \delta||_{C^2(\mathbb{R}^n, \delta)}$ is sufficiently small;
\end{itemize}
then $g$ is also flat.
\end{corollary}

\begin{proof}

Since $\Omega$ is compactly contained in $\mathbb{R}^n$, we can choose a rectangle domain $\Omega'$ which contains $\Omega$ strictly. Thus, the metric $g$ is identically the same as the Euclidean metric on $\Omega' - \Omega$. Now we can derive a metric with nonnegative $Q$-curvature on the torus $T^n$ by identifying the boundary of $\Omega'$ correspondingly. Clearly, this new metric on $T^n$ is $C^2$-closed to the flat metric, hence has to be flat by Theorem \ref{flat_local_rigidity}. Now the claim follows.

\end{proof}

It would be interesting to ask whether there is a global rigidity result for $Q$-curvature. No result is known so far to the best of authors' knowledge, but we observed that the global rigidity holds in some special cases.

\begin{proposition}\label{prop:conf_Ricci_flat}
Let $(M, g)$ be a closed Riemannian manifold with $$Q_g \geq 0$$ pointwisely. Suppose that $g$ is conformal to a Ricci flat metric, then $(M, g)$ is Ricci flat.
\end{proposition}

\begin{proof}
For $n = 4$, by the assumptions, there exists a smooth function $u$ such that $g = e^{2u} \bar g$, where $\bar g$ is a Ricci flat metric on $M$. By (\ref{eqn:conf_Q_4}) in the appendix, we have
$$Q_g = e^{-4u} ( P_{\bar g} u + Q_{\bar g} )= e^{-4u} \Delta_{\bar g}^2 u \geq 0.$$ Hence $ \Delta_{\bar g}u$ is subharmonic on $M$ which implies that $u$ is a constant. Therefore, $g$ is also Ricci flat. \\

Similarly, for $n \neq 4$, there exists a positive function $u > 0$ such that $g = u^{\frac{4}{n-4}}\bar{g}$, where $\bar g$ is a Ricci flat metric on $M$. Then by (\ref{eqn:conf_Q_n}), $$Q_g = \frac{2}{n-4}u^{-\frac{n+4}{n-4}} P_{\bar{g}} u =   \frac{2}{n-4} u^{-\frac{n+4}{n-4}} \Delta_{\bar{g}}^2 u. $$

For $n=3$, we have $$\Delta_{\bar{g}}^2 u \leq 0,$$ and for $n>4$,
$$\Delta_{\bar{g}}^2 u \geq 0,$$ which imply that $u$ is a constant in both cases and thus $g$ is Ricci flat.

\end{proof}

In particular, we can consider tori $T^n$. First, we need a lemma which characterizes the conformally flat structure on $T^n$.

\begin{lemma}\label{lem:conf_structure_tori}
On the torus $T^n$, any locally conformally flat metric has to be conformal to a flat metric.
\end{lemma}

\begin{proof}
Let $g$ be a locally conformally flat Riemannian metric on $T^n$. According to the solution of Yamabe problem, $g$ is conformal to a metric $\bar g$, whose scalar curvature is a constant.

Suppose $R_{\bar g} < 0$, by Proposition 1.2 in \cite{S-Y_3}, the fundamental group of $T^n$ is non-amenable. But this contradicts to the fact that $\pi_1(T^n)$ is abelian, which is amenable. Thus, $R_{\bar g} \geq 0$, which implies that $\bar g$ is flat by famous results of Schoen-Yau and Gromov-Lawson (\cite{S-Y_1, S-Y_2, G-L_1, G-L_2}).
\end{proof}

Now we can derive the rigidity of tori with respect to nonnegative $Q$-curvature (Theorem \ref{thm:rigidity_tori}):

\begin{theorem
Suppose $g$ is a locally conformally flat metric on $T^n$ with $$Q_g \geq 0,$$ then $g$ is flat.

In particular, any metric $g$ on $T^4$ with nonnegative $Q$-curvature has to be flat.
\end{theorem}
\begin{proof
By Lemma \ref{lem:conf_structure_tori}, we can see that $g$ is conformal to a flat metric. Applying Proposition \ref{prop:conf_Ricci_flat}, we conclude that $g$ has to be flat.

In particular for $T^4$, we have the \emph{Gauss-Bonnet-Chern formula},
$$\int_{T^4} \left( Q_g + \frac{1}{4}|W_g|^2 \right) dv_{g} = 8\pi^2 \chi(T^4) = 0$$
on $T^4$. Thus the non-negativity of $Q$-curvature automatically implies that Weyl tensor $W_g$ vanishes identically on $T^4$, which means $g$ is locally conformally flat. Therefore, $g$ is flat by the previous argument.
\end{proof}

As for dimension 3, we have the following result.

\begin{proposition}
The $3$-dimensional torus $T^3$ does not admit a metric with constant scalar curvature and nonnegative $Q$-curvature, unless it is flat.
\end{proposition}

\begin{proof}

Suppose such a metric $g$ exists, its scalar curvature is non-positive and it is flat only if $R_g = 0$ (c.f. \cite{ S-Y_1, S-Y_2, G-L_1, G-L_2}).

If it is non-flat, without loss of generality, we can assume $$R_g = -1.$$ Then we have $$Q_g = - \frac{1}{4}\Delta_g R_g - 2 |Ric_g|^2 + \frac{23}{32} R_g^2 = - 2 |Ric_g|^2 + \frac{23}{32} R_g^2 \geq 0.$$

That is $$|Ric_g|^2 \leq \frac{23}{64}R_g^2 = \frac{23}{64}.$$

At any point $p \in M$, choose an orthonormal basis $\{e_1, e_2, e_3 \}$ for $T_p M$ such that the Ricci tensor at $p$ is diagonal. Let $\lambda_i$, $i = 1,2,3$ be the eigenvalues of $Ric_g (p)$. Then we have $$\lambda_1 + \lambda_2 + \lambda_3 = -1$$ and $$\lambda_1^2 + \lambda_2^2 + \lambda_3^2 \leq \frac{23}{64}.$$

Hence for $i \neq j$, we have
\begin{align*}
0 \geq& \lambda_i^2 + \lambda_j^2 + ( 1+ (\lambda_i + \lambda_j) )^2 - \frac{23}{64} \\
=& (\lambda_i + \lambda_j)^2 + 2 (\lambda_i + \lambda_j) + \lambda_i^2 + \lambda_j^2 + \frac{41}{64} \\
\geq &\frac{3}{2}(\lambda_i + \lambda_j)^2 + 2(\lambda_i + \lambda_j) + \frac{41}{64},
\end{align*}
where the last step was achieved by the mean value inequality $$\lambda_i^2 + \lambda_j^2 \geq \frac{(\lambda_i + \lambda_j)^2}{2} .$$
That is,
$$(\lambda_i + \lambda_j)^2 + \frac{4}{3}(\lambda_i + \lambda_j) + \frac{41}{96} \leq 0,$$
which implies that $\lambda_i + \lambda_j < - \frac{2}{3} + \frac{\sqrt{10}}{24}  < -\frac{1}{2}$.

Then the sectional curvature of the plane spanned by $e_i$ and $e_j$ satisfies
\begin{align*}
K_{ij} = R_{ijji} = R_{ii} g_{jj} + R_{jj} g_{ii} - \frac{1}{2} R g_{ii}g_{jj} = \frac{1}{2} + (\lambda_i + \lambda_j) < 0.
\end{align*}

Thus $g$ has negative sectional curvature. But the torus does not admit a metric with nonpositive sectional curvature by Corollary 2 in \cite{B-B-E}, which is a contradiction.
\end{proof}

\begin{remark}
From the last inequality in the above argument, we can easily see that the conclusion actually allows a perturbation on the metric. That is, any metric on $T^3$ with scalar curvature sufficiently $C^4$-closed to a negative constant can not have a nonnegative $Q$-curvature, unless it is flat.
\end{remark}

\begin{remark}
According to the solution of Yamabe problem, there is a metric with constant scalar curvature in each conformal class on any closed Riemannian manifold. Thus we can conclude that for any metric $g$ on $T^n$, if $g$ is not conformally flat, then it must be conformal to a metric with constant negative scalar curvature. However, $Q$-curvature cannot stay nonnegative when performing conformal changes on metrics in general. We hope to develop some techniques to resolve this issue in the future.
\end{remark}

According to the \emph{Bieberbach Theorem} (See Theorem 10.33 in \cite{C-L-N}), any $n$-dimensional flat Riemannian manifold is a finite quotient of the torus $T^n$. So based on Theorem \ref{flat_local_rigidity}, \ref{thm:rigidity_tori} and above observations, we give the following conjecture:

\begin{conjecture}\label{Positive_Q_on_torus}
For $n \geq 3$, let $(M^n,\bar{g})$ be a connected closed flat Riemannian manifold, then there is no metric with pointwisely positive $Q$-curvature. Moreover, if $g$ is a metric on $M$ with $$Q_g\geq 0,$$ then $g$ is flat.
\end{conjecture}

\begin{remark}
This is a higher order analogue of the famous result due to Schoen-Yau and Gromov-Lawson (\cite{S-Y_1, S-Y_2, G-L_1, G-L_2}), which says that there is no nonnegative scalar curvature metric on tori unless they are flat. \end{remark}

\section{Relation between vacuum static spaces and $Q$-singular spaces}

In this section, we will discuss the relation between $Q$-singular spaces and vacuum static spaces.

\begin{definition
We say a complete Riemannian manifold $(M,g)$ is vacuum static, if there is a smooth function $f (\not\equiv 0)$ on $M$ solving the following \emph{vacuum static equation}
\begin{align}
\gamma_g^* f := \nabla^2 f - \left(Ric_g - \frac{R_g}{n-1} g \right) f = 0.
\end{align}
We also refer $(M,g,f)$ as a \emph{vacuum static space}, if $f (\not\equiv 0) \in \ker \gamma_g^*$.
\end{definition}

Now we can prove the main result (Theorem \ref{Q-static_R-static}) in this section.

\begin{theorem
Let $\mathcal{M}_R$ be the space of all closed vacuum static spaces and $\mathcal{M}_Q$ be the space of all closed $Q$-singular spaces.
Suppose $(M, g, f) \in \mathcal{M}_R \cap \mathcal{M}_Q$, then $M$ is Einstein necessarily. In particular, it has to be either Ricci flat or isometric to a round sphere.
\end{theorem}
\begin{proof
If $(M,g,f)$ is vacuum static, $M$ has constant scalar curvature necessarily (c.f. \cite{F-M}). Then the $Q$-singular equation can be reduced to
\begin{align}\label{eqQ_and_Rstatic}
\Gamma_g^* f =& A_n \left( - g \Delta^2 f + \nabla^2
\Delta f - Ric \Delta f \right)- 2 C_n R \left( g\Delta f - \nabla^2 f + f
Ric \right)\\
&\notag - B_n \left( \Delta (f Ric) + 2 f
\overset{\circ}{Rm}\cdot Ric + g \delta^2 (f Ric) + 2 \nabla \delta (f
Ric) \right)\\ =&
\notag 0.
\end{align}

By \emph{Contracted $2^{nd}$ Bianchi Identity}
\begin{align*}
\delta Ric = - \frac{1}{2} dR = 0,
\end{align*}
we can simplify (\ref{eqQ_and_Rstatic}) furthermore,
\begin{align*}
&-\frac{1}{B_n}\Gamma_g^* f \\
=& -\frac{A_n}{B_n} \left( \nabla^2 \Delta f - g \Delta^2 f - Ric \Delta f \right) - \frac{2C_n}{B_n}R \left( \nabla^2 f - g \Delta f - Ric f \right) \\ &+ f (\Delta Ric + 2 \overset\circ{Rm} \cdot Ric) + g (Ric \cdot \nabla^2 f) + Ric \Delta f - 2 \nabla^2 f \times Ric + 2 C \cdot \nabla f\\
=& -\frac{A_n}{B_n}\ \gamma_g^* (\Delta f) - \frac{2C_n}{B_n}R\ \gamma_g^* f + f \Delta_L Ric + g (Ric \cdot \nabla^2 f)  + Ric \Delta f  +\frac{2 R }{n-1}f Ric + 2 C \cdot \nabla f\\
=& 0,
\end{align*}
where $\Delta_L$ is the Lichnerowicz Laplacian and $C$ is Cotton tensor,
\begin{align*}
C_{ijk} = \left( \nabla_i R_{jk} - \frac{1}{2(n-1)}g_{jk}\nabla_i R\right) - \left( \nabla_j R_{ik} - \frac{1}{2(n-1)}g_{ik} \nabla_j R \right) = \nabla_i R_{jk} - \nabla_j R_{ik}
\end{align*}
and
\begin{align*}
(C \cdot \nabla f)_{jk} := C_{ijk} \nabla^i f.
\end{align*}

Now suppose $g$ is also vacuum static, then $\Delta f = -\frac{R}{n-1}f$ and hence
\begin{align*}
\gamma_g^* (\Delta f) = -\frac{R}{n-1} \gamma_g^* f = 0.
\end{align*}

Thus we have
\begin{align*}
\Gamma_g^* f
= f \Delta_L Ric + g (Ric \cdot \nabla^2 f)  + \frac{R}{n-1} f Ric + 2 C \cdot \nabla f = 0.
\end{align*}

Taking trace and applying the vacuum static equation,
\begin{align*}
tr\ \Gamma_g^* f
=& f \Delta R + n Ric \cdot \nabla^2 f + \frac{R^2}{n-1} f\\
=& n Ric \cdot \left( Ric - \frac{R}{n-1}g\right) f + \frac{R^2}{n-1} f\\
=& n \left( |Ric|^2 - \frac{R^2}{n} \right) f \\
=& n \left|Ric - \frac{R}{n} g\right|^2 f\\
=& 0.
\end{align*}

Since in an vacuum static space, $df \neq 0$ on $f^{-1}(0)$ (c.f. \cite{F-M}). Then $f^{-1}(0)$ is a regular hypersurface in $M$ and hence $f \neq 0$ on a dense subset of $M$, which implies that $$Ric = \frac{R}{n} g.$$ \emph{i.e.} $(M,g)$ is Einstein.

For a closed vacuum static space, its scalar curvature is nonnegative necessarily. Hence the assertion follows easily from Theorem \ref{Classificaition_Q_singular_Einstein}.

\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
home IT/Enterprise Cloud
by Joel Hruska — Wednesday, December 17, 2014, 03:04 PM EDT
Microsoft 'Arcadia' Developer Team Working On Game Streaming Service
Ever since Microsoft announced that its Xbox One would have some sort of unspecified cloud streaming capability that would allow it to offload game processing into the cloud, speculation has run rampant over precisely what that capability would look like, or what form it would take. The announcement of yet another cloud streaming project at the company has partially answered the question: Microsoft still isn't sure.
According to a new job posting, Microsoft is looking for a "self-motivated, driven, collaborative, and flexible" candidate to seek "the simplest and most elegant" solutions to difficult problems while working on what the company acknowledges is a "v1" product. In Microsoft parlance, this implies no one should consider buying in until Version 3.0.
Mary Jo Foley has some additional commentary on the situation over at ZDNet, where she describes this new project, codenamed Arcadia, as a cloud streaming platform that's supposed to drive both gaming solutions (a la Sony's PlayStation Now) as well as possible opportunities to run non-Windows products on Windows Phone platforms. Wisely, the team appears to have put this particular notion on hold, though it still lists iOS and Android experience as positive factors for a potential candidate.
Running non-Windows Phone Software On Windows Phone? Terrible Idea
Here's the problem with the idea of running non-Windows software on a Windows device: this type of software or hardware compatibility never, ever achieves the goal of pulling users back to the initial device manufacturer. Ask IBM, with its "A Better Windows Than Windows" campaign around OS/2. Look to the various CPUs that attempted to emulate x86 instruction capability in either hardware or software.
Streaming another company's software to your platform never really helps the company doing the streaming. It may provide a sliver of compatibility for developers or die-hard users, but it introduces UI oddities and disparities in design philosophy. Much of the backlash over Windows 8 centered around the fact that the two halves of the operating system have entirely different UI layouts -- a problem that would continue to exist in any streamed software from iOS or Android.
On the other hand, Microsoft does have a potential cross-platform business that could benefit tremendously from game streaming -- the Xbox / Windows split. Exactly how this would work would depend a great deal on how Microsoft wants to leverage each set of assets, but imagine if it was possible to stream certain Office capabilities or software suites to an Xbox provided you owned the appropriate Windows license. Or, alternately, stream Xbox One games to a PC provided you had an appropriate Xbox Live account and a compatible wireless controller?
This could be a killer app for Windows just waiting to be discovered. Obviously for now, Microsoft would likely tie streaming to the Xbox One hardware itself, but the ability to stream games while on a laptop or traveling across the country would be a huge boost for the platform. This could be the killer feature that blows Sony's PS Vita Remote Play option out of the water.
The expected timeline for such developments, however, is still years in the future (Foley suspects a post-Windows 10 development cycle) and Microsoft could still ruin the pie by attempting to tie it too onerous subscription services or similar measures.
Still, the company has a massive consumer gaming business in consoles and a presence on PC hardware that still dwarfs anything Sony or Nintendo could claim outside of core gaming peripherals. It would be nice to see that expertise used in a way that leveraged the strength of both assets.
Tags: Gaming, Microsoft, Streaming, Cloud, xbox one, DeLorean, arcadia, nasdaqmsft-microsoft
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The idea with this one was a ridge-level circuit of Glen Ogilvie—up on to the east side at Broom Hill, over the tops to Craigowl at the head of the glen, and then back along the west ridge to Berry Hillock.
So I parked off-road on a little patch of turf next to the turn-off from the A928 into the Denoon Glen. I figured I'd be coming off Berry Hillock somewhere on the Denoon Glen road, and it would be nice to have the car waiting conveniently nearby.
It was a strange and beautiful day. There had been unseasonable snow a few days previously and now, on a sunny April 30, the Sidlaws were still white.
I walked back down the road and turned into Glen Ogilvie. There's an obvious route through the farmland, up on to the east side of the glen—a vehicle track that starts at the end of the public road to Dryburn, and climbs gently up below Gallow Hill, before crossing the Sidlaws ridge between Gallow Hill and Craigowl. It didn't quite come out where I wanted to be, however. Instead, an open gate into a field let me climb up next to a new plantation and then out on to the open hillside through an obvious gap in the field wall at NO 380423. From there, I had a direct route to the start of my ridge at Broom Hill.
After that, it's just a matter of following the fence southeast along the ridge to Gallow Hill. Or it would be, under normal circumstances. It was at this point I began to realize just how much snow had fallen in the last couple of days, and how much it had drifted. Although the heather was showing through in wide areas, the snowy patches between were knee-deep, and softening in the spring sunshine. I wove my way from one island of heather to the next, but it was slow going.
From there, I post-holed my way over to Ironside Hill, above Lumley Den, and then post-holed back along another fence that turns into a wall behind the television mast on the south slopes of Gallow Hill. The fence that runs over Gallow Hill terminates here, against the wall behind the television mast, but there's a handy little gate in it, hard against the wall, at NO 395408. I imagine it opens, but its lower edge was so deeply embedded in the snow, I just climbed over.
Gallow Hill from Craigowl – you'd think that snow wasn't very deep, wouldn't you?
From that point, it's only about a mile to the top of Craigowl, but things began to get really, really irritating. The drifting here was pretty extreme—I'd cross two or three snow patches an inch deep, and then on the fourth I'd suddenly drop hip-deep and have to crawl out. Or a single snow patch would alternate between inch-deep and knee-deep three times in twenty steps, destroying any possible walking rhythm. The snow was getting wetter and wetter in the sunshine, and I was getting wetter and wetter as I flailed around in it.
Meanwhile, nature was going about its business all around me—little frogs, stunned by the chill of the snow, could barely manage a couple of belated flops to get out from under my approaching boots; a male hen harrier flapped sullenly out of my line of march, maintaining an altitude of about two feet; deer dithered on the skyline and disappeared again; stonechats clattered furiously at me from the higher clumps of heather; and a quite spectacular fox's earth turned up, improbably high on the side of Craigowl and slap-dab in the middle of the largest snow-free patch of heather I'd encountered. That fox obviously knows a thing or two about snow drifts.
At the top of Craigowl, I took off my gaiters and shook the compacted snow out of them. Then I took off my boots and tipped a demi-tasse of meltwater out of each. Then I wrung out my socks. For all that it was a glorious scenic day, packed with interesting things to look at, it was becoming unreasonably trying. And I had no reason to believe that the going would be any easier on the west side of the glen.
I struck straight off into the glen, aiming to reconnect with the blessed tarmac of the Glen Ogilvie road at either Nether Handwick or Dryburn. Post-hole, post-hole, post-hole. Then hip deep, again, for the fifth or sixth time. There were wire fences around the empty pasture-land in the glen, so I followed a line of snow-choked grouse butts down the hill, thinking that they would be served by a path under the snow, and that would mean there was a gate close by—but no such luck. My alternative line from this point wasn't at all clear, so I crawled under the fence where it crossed a little burn. Then I crawled under another fence. And at that moment, as I lay prostrate in the snow, a light aircraft flew low overhead. So I tried to make animated movements, for fear the pilot would make a radio call to report me as a corpse, newly thawed out of the snowdrifts.
But then … Farm tracks! Open gates! No snow! Joy! Until I remembered that I'd left my car parked uselessly far up the road. Sigh.
Snow's not so bad down here!


A pool in Newport Beach: What happens if California runs short on water?
California is having its driest winter season in a century and a half. We should not be surprised if sooner or later, somebody tries to revive the schemes of 30 or 40 years ago to tap the rivers of Oregon to help the nation's biggest state survive.
The most recent notion, if memory serves, was proposed by a Los Angeles County supervisor in the 1970s.The idea was to construct a pipeline and pumping system to capture a small fraction of the Columbia River's copious flow at its mouth, just before it churns across the bar, and pipe it south. The idea was scorned in Oregon and never got past the conceptual stage. But this may change if conditions get worse.
According to a report in the San Jose Mercury News on Jan. 30, some scientists think California may be in for a megadrought of the kind it has seen with some regularity over thousands of years. The scientists have studied tree rings and other clues in the natural record. They found many severely dry periods lasting 20 years or so. But they also found one that started in 850 and lasted 240 years, followed by another that lasted 180 years.
California now has 38 million people, nearly 10 times the population of Oregon. Sources quoted in the San Jose story speculated that the population of cities could probably cope by adopting severe conservation measures and paying much more for the water still available, but California agriculture would disappear because the price of irrigation would be more than the crops would bring.
There are 2 types of shower valves.
One type is 100% on and only a temp. control.
One slightly more expensive adjusts volume and temp.
Every CA hotel I've ever stayed at has no volume control adjustment.Wonder how much water they could save? just sayin'.
We are in a drought in Oregon, so I don't think we will be contemplating sending massive amounts of water to California that we don't have. If we say, "Ok, Hasso, yes there have been droughts and climate change in history," will that soothe your stubbornness enough to let you admit that burning large amounts of fossil fuels is not a good thing?
I was wondering the same thing which is how I came across your article. I just recommend an edit at the very beginning of the article. Both Alaska and Texas are larger than California.
I meant "most populous," of course. Thanks for the note.
The problem is too many people & immigration………………….


Midway into the North Sea, it hit me. 'There's nothing to see out here but the sea.' Miles and miles of it. So, I took out my trusty Canon 5D Mark III and captured the expanse of ocean around us. We ended up using it for this CD cover and the CD face was inspired by the only food I could consume on the waves at the time.


Indian Air Force Group C Recruitment 2017, 232 Group C LDC Jobs @indianairforce.nic (भारतीय वायु सेना के ग्रुप सी भर्ती 2017, 232 ग्रुप सी एलडीसी नौकरियां)Indian Air Force has released the employment notification inviting applicants to fill 232 Group C Vacancies. The young and eligible candidates who want to excel their career Indian Air Force can apply for Indian Air Force Group C Recruitment 2017 before the last date. Candidates must apply through online application process. In terms of more details such as age limit, pay scale, eligibility etc go through the details mentioned in the article below.
Every year Indian Air Force releases various job notifications for the recruitment of numerous job posts. Recently it has announced the recruitment notification for various post under the Group C category. Therefore through Indian Air Force 232 Group C Notification, numerous vacancies are released for the post of Safaiwala, Multi Tasking Staff (MTS), Mess staff, Lower division clerk, cook etc. All the eligible contenders intrested in IAF Group C vacancy can go through the complete details and apply for the desired job posts. Candidates must meet certain eligibility to enroll for IAF Group C jobs, which details you can get below.
Applicant must be 10th pass and must poses ITI certificate in its relevant trade. He/ She must be a graduate in the required stream from a recognized University or board with minimum pass percentage.
Candidates who gets selected in Indian Air Force LDC Group C jobs will be provided the below mentioned salary.
Store Keeper Rs. 5200/- to Rs. 20,200/- + GP Rs. 1900/-.
All the job seekers who are applying for Indian Air Force Group C Recruitment process must follow the below mentioned steps for online application form.
Currently Indian Air Force (IAF) has announced Group C recruitment notifications. You can get the upcoming notifications in our portal www.jobs.chekrs.com. Therefore the skilled candidates intrested for Defence Jobs can apply online before last date 19/3/2017. In regards to any queries or reviews in terms of IAF Group C LDC Recruitments 2017, we welcome your valuable comments in the commenting section below. Our team will solve your query in a short while.


text:
Game of Thrones/HBO
Where to find 'Game of Thrones' maps online
Westeros and Essos are brought to life through detailed maps.
Michelle Jaworski
Published Dec 23, 2018 Updated Jan 27, 2021, 1:56 am CST
No matter how vast the story of Game of Thrones and A Song of Ice and Fire may get over several seasons and books, you can always follow along with the fictional Game of Thrones maps made famous by the show's fascinating world.
We catch glimpses of the Game of Thrones map at the start of every single Game of Thrones episode, which takes you to just about every town and settlement you'll see during that hour of television along with a few staples. As the now-iconic theme song gets stuck in your head, you're getting a Westerosi geography lesson. The map of the known world—which encompasses Westeros and beyond the Wall, Essos (from the nine Free Cities and the ruins of Valyria to Meereen and Asshai and beyond), nearby islands, and the largely unexplored regions of Sothoryos and Ulthos—is practically ubiquitous with Game of Thrones.
What is the Game of Thrones map?
Sure, it's important to know (in a general sense) where things are. But a Game of Thrones houses map can provide a scope of just how big the world Martin created is. You can't just arrive halfway across the continent in almost no time at all—until you can, anyway. But they're not just fun nods for long-time book fans—GOT maps are also strategically vital.
Aegon the Conqueror planned out his conquest of Westeros on a painted table in Dragonstone, and more than three hundred years later, both Stannis Baratheon and Daenerys Targaryen would have a similar use for it. Countless lords used paper maps with painted pieces to relay where enemy houses resided ahead of a major battle. In season 7, Cersei Lannister, now queen, commissioned a detailed map of Westeros on the ground that she could observe and walk across. Even the season 8 teaser trailer, which contained no new footage, featured a map of Westeros with wooden pieces representing the Starks, Targaryens, and Lannisters being consumed by ice and fire.
For those looking to dive in further, you won't be able to by watching the show alone. The glimpses of the map we see in the opening credits aren't very conducive for theorizing or plotting. They probably won't help sort out the show's issues with time in more recent seasons—save the length of the Wall, George R.R. Martin deliberately did not reveal distances or travel times in Westeros—but it gives us all a better idea of just where things are happening. Fortunately for us, there are plenty of ways to discover Game of Thrones maps for ourselves.
Where to find the Game of Thrones map online
1) HBO's Game of Thrones viewer's guide
HBO recognized that there is a lot of information that Game of Thrones casually reveals on a regular basis, and for some fans, it can be a bit much to keep track of. Fortunately, HBO has an official interactive guide that you explore and customize no matter how far along into the show. You can explore character biographies and family trees, see a synopsis, and view a map of the known world that's available in English, Spanish, and Hodor. (Yes, really.) Depending on what episode you're viewing the map from—the most recent one is from the season 7 finale—you'll also be able to read entries specific to that particular episode.
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2) A Song of Ice and Fire (via Random House Books)
If you can't get enough of Game of Thrones, I'd highly recommend checking out A Song of Ice and Fire, Martin's high-fantasy series on which Game of Thrones is based. Five books are available to date with a sixth on its way (eventually). They're rich in detail, they include several subplots and characters that never made it to the show like far more Stark direwolf and warging action, Victarion Greyjoy, Arianne Martell, and a secret marriage pact aligning House Martell with House Targaryen.
At the front of these books are detailed maps that reveal many of Westeros' most important towns and regions. A Game of Thrones and A Cast of Kings only contained a look at Westeros, but as the series expanded in A Storm of Swords, A Feast for Crows, and A Dance With Dragons, so did the number of maps published in Martin's books.
If you can't get your hands on any of the books in ASOIAF, you can view the maps on Random House's website.
Random House Books
3) ASOIAF-adjacent books like The World of Ice and Fire and The Lands of Ice and Fire
Similarly, The World of Ice and Fire is a book that Martin co-wrote with Westeros.org's Elio García and Linda Antonsson. It's part history book and part encyclopedia of the known world, all about the smaller scale that the known world contains. The second half of the book, which looks at each of the Seven Kingdoms of Westeros and their long histories, allows you to view each town and castle with a much smaller scope. This is especially helpful when it's too hard to remember dozens of houses and family names.
But The Lands of Ice and Fire gives you the kind of detail that's hard to come by in even more detailed maps. Here, you'll find city maps of King's Landing and Braavos, not to mention a map to frame and hang on your wall.
While neither are online, either option (or both of them) offer a physical and more tangible version of something we largely only get from screens.
Penguin House Books
4) Etsy
If you want something out of the ordinary (or a physical copy of your own), Etsy will be your best bet. A Google Maps-designed version that went viral a few years ago was once on Etsy. A simple search on Etsy brought back dozens of options of all shapes and sizes that you can click on to view or buy.
You can easily find wooden maps, canvas maps, antique-inspired maps, watercolor maps, a map of Westeros that looks like New York's MTA or the London Underground. There are more cartoon-friendly maps. The only limits how much detail you want to see in your map and, if you want to purchase one, your wallet.
5) George R.R. Martin's A World of Ice and Fire app
Want a little more interactive book-focused experience? The official World of Ice and Fire app has you covered. It operates similarly to HBO's guide (albeit in mobile form) in that maps are just one of the areas of the series you can explore, but instead of episodes, you filter out information depending on how far along in ASOIAF you are. (So if you haven't gotten to something like the Red Wedding yet, you can configure the settings to reflect that.)
A World of Ice and Fire
The app, which is available for iOS and Android, is free to download, but you'll need to pay to access InfoPacks. The InfoPack for A Game of Thrones is $1.99 and each subsequent book's data is $0.99 apiece, but you can get all five InfoPacks available for $4.99. (Buying them together is cheaper than the first book in the series in ebook format.) You can not only explore towns in Westeros with sword pins to mark various locations, but there's also a detailed map of King's Landing with several areas that fans will recognize.
6) Quarter Maester
Created using Google API, Quarter Maester allows you to really dissect into character's journeys. Once you finagle your spoiler scope—you can toggle between the show and the book to determine how you want to view it—you can view just how far a character has traveled from their first appearance to their latest one.
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Who is Azor Ahai, the prophesied savior of 'Game of Thrones'?
Everything revealed so far about George R.R. Martin's 'The Winds of Winter'
Here's everything we know about the 'Game of Thrones' spin-off prequels
7) The Game Revealed
Throughout Game of Thrones' seventh season, HBO had an online series called The Game Revealed that went behind-the-scenes of each episode and featured interviews with the cast and crew. There were plenty of fascinating tidbits from the series, including a closer look at how the crew created Cersei's gigantic map of Westeros in the season 7 premiere. If you noticed similarities to the show's opening credits in the design, you're not imagining things. It was an intentional move. And although it's one of the harder maps to view, you can't ignore the level of detail in the execution.
Editor's note: This article is regularly updated for relevance.
How linguist David J. Peterson created the Dothraki and Valyrian languages for Game of Thrones
The Daily Dot (Licensed)
Introducing 2 GIRLS 1 PODCAST, a weekly comedy show where Alli Goldberg and Jen Jamula (two actors who perform bizarre internet content on stage) have hilarious and humanizing conversations with Bronies, top Reddit mods, professional ticklers, video game archaeologists, dating app engineers, adult babies, cuddling specialists, vampires, Jedi, living dolls, and more.
Subscribe to 2 GIRLS 1 PODCAST in your favorite podcast app.
*First Published: Dec 23, 2018, 6:00 am CST
Michelle Jaworski is a staff writer and TV/film critic at the Daily Dot. She covers entertainment, geek culture, and pop culture and has covered everything from the Sundance Film Festival, NYFF, and Tribeca to New York Comic Con and Con of Thrones. She is based in Brooklyn.
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