text:
\begin{document}

\title[Extensions with estimates of cohomology classes]{Extensions with estimates of cohomology classes}

\author{Vincent Koziarz}
\address{IECN, Nancy-Universit\'e, CNRS, INRIA, Boulevard des Aiguillettes B. P. 70239, F-54506
Vand\oe uvre-l\`es-Nancy, France}
\email{koziarz@iecn.u-nancy.fr}

\date{\today}

\sloppy

\begin{abstract}  We prove an extension theorem of ``Ohsawa-Takegoshi type'' for Dolbeault $q$-classes of cohomology ($q\geq 1$) on smooth compact hypersurfaces in a weakly pseudoconvex K\"ahler manifold.
\end{abstract}

\maketitle

\section{Introduction}

Let $Y$ be a complex submanifold of a K\"ahler manifold $X$ and let $L'$ be a Hermitian line bundle on $X$. First consider the following

\medskip

\noindent {\bf Problem.} {\it Let $f$ be a smooth $D''$-closed section of $\Lambda^{0,q}T^\star_X\otimes L'$ over $Y$ satisfying a suitable $L^2$ condition. Can we find a smooth $D''$-closed extension $F$ of $f$ to $X$ together with a good $L^2$ estimate for $F$ on $X$?}

\medskip

The first result of this kind was obtained by T.~Ohasawa and K.~Takegoshi~\cite{OT} in the case when $Y$ is a hyperplane of a bounded pseudoconvex domain ~$X$ in $\C^n$, $L'$ is the trivial bundle and $q=0$. It was further generalized by L.~Manivel~\cite{Ma} (with a simplified proof by J.-P.~Demailly~\cite{De3}) in the following setting: $X$ is a weakly pseudoconvex manifold, $Y$ is the zero set of a holomorphic section of a rank $r$ Hermitian bundle over $X$, $L'=K_X\otimes L$ where $L$ is a Hermitian line bundle whose curvature satisfies appropriate positivity properties, $K_X$ is the canonical bundle of $X$, and $q=0$. When $q\geq 1$, the method leads to a new technical difficulty occurring in the regularity argument for $(0,q)$ forms. In \cite{De3}, Demailly suggests an approach to overcome this difficulty but, to our knowledge, the complete arguments did not appear anywhere. In this paper, we rather consider the

\medskip

\noindent {\bf Modified problem.} {\it Let $q\geq 1$ and $f$ be a smooth $D''$-closed section of $\Lambda^{0,q}T^\star_X\otimes L'$ over $Y$ satisfying a suitable $L^2$ condition. Can we find a smooth $D''$-closed extension $F$ of $f$ to $X$ as a cohomology class (i.e.$[F_{|Y}]=[f]\in H^q(Y,L')$) together with a good $L^2$ estimate for $F$ on $X$?}

\medskip

Observe that if $Y$ is a Stein submanifold (this happens e.g. when $X$ is a Stein manifold), the modified problem is not relevant when $q\geq 1$ since the Dolbeault group $H^q(Y,L')$ vanishes. In contrast, we will focus here on the case when $Y$ is a smooth compact hypersurface of a weakly pseudoconvex K\"ahler manifold $X$. This situation naturally happens, for example, when $X$ is a compact K\"ahler manifold, or when $X$ is a holomorphic family of projective algebraic manifolds fibered over the unit disc.

\begin{theo}\label{theop}
Let $(X,\om)$ be a weakly pseudonconvex $n$-dimensional K\"ahler manifold, and let $Y\subset X$ be the zero set of a holomorphic section $s\in H^0(X,E)$ of a Hermitian line bundle $(E,h_E)$; the subvariety $Y$ is assumed to be compact and nonsingular.
Let $L$ be a line bundle endowed with a smooth Hermitian metric $h_L$ such that
\begin{eqnarray}
&&\rmo\Theta(L)+\sqrt{-1}\dr\db \log|s|^2\geq 0,\label{assone}\\
&&\rmo\Theta(L)+\sqrt{-1}\dr\db\log|s|^2\geq \alpha^{-1}\rmo\Theta(E) \hbox{ for some }\alpha\geq 1,\label{asstwo}\\
&&|s|^2\leq e^{-\alpha}\label{assthree}
\end{eqnarray}
on $X$. Let $0<\k\leq 1$ and let $\Omega\subset X$ be a relatively compact open subset containing $Y$. Then, for any $q\geq0$ and every smooth $D''$-closed $(0,q)$-form $f$ with values in $K_X\otimes L$ over $Y$, there exists a smooth extension $F$ of $f$ to $\Omega$ as a cohomology class (i.e. $[F_{|Y}]=[f]\in H^q(Y,K_X\otimes L)$) such that
$$\int_{\Omega}\frac{|F|^2}{|s|^{2(1-\k)}}dV_\om\leq \frac{C}{\k}\int_Y\frac{|f|^2}{|ds|^2}dV_{Y,\om}
$$
where $C$ is a numerical constant depending only on $\Omega$, $E$, $L$ and $q$.
\end{theo}

The norm of the forms with values in bundles will always be computed with respect to the one induced by $\om$, $h_E$ and $h_L$. Also, $\Theta(E)$ (resp. $\Theta(L)$) will always denote the curvature of the Hermitian line bundle $(E,h_E)$ (resp. $(L,h_L)$). When the metrics will be twisted by some positive functions, the weights will appear explicitely in the formulae.

\bigskip

Our proof follows many of the ideas outlined in~\cite{De3}. First,
using the weight bumping technique (and the adapted Bochner-Kodaira-Nakano inequality) initiated by Ohsawa and Takegoshi, for all $\e>0$, we build extensions of $f$ of class $C^1$ whose $L^2$ norm is controlled, and which are ``approximately'' $D''$-closed (in the sense that the $L^2$ norm of their $D''$-derivative is bounded by a constant times $\e$). Philosophically, passing to the limit as $\e\rightarrow 0$ should provide the desired extension but the limiting elliptic differential system is singular along $Y$ and this forbids the direct use of elliptic regularity arguments. Then, at this point, our strategy differs from Demailly's. Instead, we construct ``approximate'' $q$-cocycles $\zeta_\e$  in \v Cech cohomology corresponding to the previous extensions via an effective Leray's isomorphism, in a similar fashion as Y.-T.~Siu in \cite{Si}. During the process, we solve local $D''$-equations by standard techniques of L.~Hörmander.
Then, we can take the limit as $\e\rightarrow 0$ and use the ellipticity of the Laplacian in bidigree $(0,0)$ to ensure the smoothness of the extending cocycle $\zeta$. Finally, reversing the process, we get a smooth extension $F$ of $f$ as a cohomology class.
Notice that the constant $C$ in Theorem~\ref{theop} is mainly related to a finite covering of $\Omega\supset Y$ by Stein open subsets (which is used to apply Leray's isomorphism) and the norm of the derivatives of a partition of unity subordinate to this finite covering. This explains in part why we need $Y$ to be compact.

\bigskip

A consequence of Theorem~\ref{theop} is a qualitative surjectivity theorem for restriction morphisms in Dolbeault cohomology:
\begin{coro}
Let $X$, $Y$, $E$ and $L$ be as in Theorem~\ref{theop} i.e. satisfying $(\ref{assone})$, $(\ref{asstwo})$ and $(\ref{assthree})$. Then the restriction morphism
$$H^q(X,K_X\otimes L)\fd H^q(Y,(K_X\otimes L)_{|Y})
$$
is surjective for any $q\geq 0$.
\end{coro}

Applying Theorem~\ref{theop} to $E=\C$ and to any semi-positive line bundle $L$ (for instance $L=\C$), we also easily get the following corollary which contains a special case of the invariance of the Hodge numbers for a family of compact K\"ahler manifolds (a result due to K.~Kodaira and D.~Spencer):
\begin{coro}Let $\pi:{\mathfrak X}\rightarrow \Delta$ be a proper holomorphic submersion over the unit disc and $L$ a semi-positive line bundle on ${\mathfrak X}$. Assume that ${\mathfrak X}$ is a K\"ahler manifold of dimension $n+1$. Then, for any $q\geq0$, $h^{n,q}(X_t,L):={\rm dim}\, H^{n,q}(X_t,L)$ is independent of $t\in\Delta$ (where $X_t=\pi^{-1}(t)$).
\end{coro}

\noindent{\bf Acknowledgments.} {I would like to thank Mihai P\u aun for many valuable discussions. I would also like to thank Jean-Pierre Demailly for explaining me details on his article~\cite{De3}, as well as Benoît Claudon and Dror Varolin for useful comments on an earlier version of this paper.}

\section{Preliminary material}
From now on, we assume that $X$, $Y$, $L$ and $E$ satisfy the hypotheses of Theorem~\ref{theop}.

Let $c\in\R$ such that $\overline{\Omega}\subset X_c:=\{x\in X\,,\,\psi(x)<c\}$, where $\psi$ is the plurisubharmonic exhaustion of $X$. Let ${\cal U}=\{U_j\}_{j\in J}$ be a finite covering of the closure of ${\Omega}$ by
coordinate charts $\phi_j:B\fd U_j$ where $B$ is the unit ball in $\C^n$, and such that $U_j\subset {X_c}$ for all $j$.
Denoting by $\nu$ the standard Hermitian norm on $\C^n$, we assume that the functions $\varphi_j:=\nu\circ\phi^{-1}_j$ satisfy
\begin{equation}\label{positiv}
\rmo\Theta(L)-\b\rmo\Theta(E)+\rmo\dr\db\varphi_j\geq \omega
\end{equation}
on $U_j$ for any $\b\in[0,1]$ (this is always possible if the $U_j$'s are chosen small enough). For any multi-index $(j_0,\dots,j_\ell)$, we shall denote by $\varphi_{j_0,\dots, j_\ell}$ the function $\sum_{i=0}^\ell\varphi_{j_i}$ which is defined on the intersection $U_{j_0,\dots,j_\ell}:=U_{j_0}\cap\dots\cap U_{j_\ell}$.

If $F$ is a smooth Hermitian vector bundle over $X$ and $U\subset X$ is an open subset then, for all integer $k$, we denote by $\E^k(U,F)$ the space of sections of $F$ over $U$ which are of class $C^k$ and by $\E^k_c(U,F)$ those with compact support. We also denote by $W^k(U,F)$ the Sobolev space of sections whose derivatives (in the sense of distribution theory) up to order $k$ are in $L^2$.

Let us recall three useful results taken from~\cite{De1} (Remark 1.6, Lemma 3.3
and Lemma 6.9):
\begin{prop}\label{usef}
\begin{enumerate}[{\rm (}a{\rm )}]
\item $X_c\backslash Y$ is complete K\"ahler.
\item Let $\om$ and $\om'$ be two Hermitian forms on $T_X$ such that $\om\leq\om'$. Let $E$ be a Hermitian vector bundle on $X$. Then, for any $q\geq 0$ and any $u\in\L^{n,q} T^\star_X\otimes E$, $|u|_{\om'}^2\, dV_{\om'}\leq |u|^2_\om\, dV_\om$.
\item Let $\Omega$ be an open subset of $\C^n$ and $Y$ a complex analytic subset of $\Omega$. Assume that $v$ is a $(p,q-1)$-form with $L^2_{\rm loc}$ coefficients and $w$ a $(p,q)$-form with $L^1_{\rm loc}$ coefficients such that $\db v=w$ on $\Omega\backslash Y$ (in the sense of distribution theory). Then $\db v=w$ on $\Omega$.
\end{enumerate}
\end{prop}

The following lemma is a consequence of a classical result (see \cite{De2}, Corollary 5.3):
\begin{lemma}\label{loc}
Let $m$ and $p$ be positive integers.
\begin{enumerate}[{\rm (}a{\rm )}]
\item Let $v\in W^m(U_{j_0,\dots,j_\ell},\Lambda^{n,p}T^\star_X\otimes L\otimes E^{-1})$ such that  $D''v=0$ and
$$\int_{U_{j_0,\dots,j_\ell}} |v|^2 e^{-\varphi_{j_0,\dots,j_\ell}}\, dV_{\omega}<+\infty.
$$
Then there exists a $(n,p-1)$ form $u\in W^{m+1}(U_{j_0,\dots,j_\ell},\Lambda^{n,p-1}T^\star_X\otimes L\otimes E^{-1})$ such that $D''u=v$ and
$$\int_{U_{j_0,\dots,j_\ell}} |u|^2 e^{-\varphi_{j_0,\dots,j_\ell}}\, dV_{\omega}\leq \frac{1}{p}\int_{U_{j_0,\dots,j_\ell}} |v|^2e^{-\varphi_{j_0,\dots,j_\ell}}\, dV_{\omega}.
$$
\item Let $0<\k\leq 1$ and $\e>0$. Let $v\in W^m(U_{j_0,\dots,j_\ell},\Lambda^{n,p}T^\star_X\otimes L)$ such that  $D''v=0$ and
$$\int_{U_{j_0,\dots,j_\ell}} \frac{|v|^2}{(|s|^2+\e^2)^{1-\k}}e^{-\varphi_{j_0,\dots,j_\ell}}\, dV_{\omega}<+\infty.
$$
Then there exists a $(n,p-1)$ form $u\in W^{m+1}(U_{j_0,\dots,j_\ell},\Lambda^{n,p-1}T^\star_X\otimes L)$ such that $D''u=v$ and
$$\int_{U_{j_0,\dots,j_\ell}} \frac{|u|^2}{(|s|^2+\e^2)^{1-\k}}e^{-\varphi_{j_0,\dots,j_\ell}}\, dV_{\omega}\leq \frac{1}{p}\int_{U_{j_0,\dots,j_\ell}} \frac{|v|^2}{(|s|^2+\e^2)^{1-\k}}e^{-\varphi_{j_0,\dots,j_\ell}}\, dV_{\omega}.
$$

\end{enumerate}
\end{lemma}

\begin{demo}
We only check the hypotheses of Corollary~5.3 in~\cite{De2}. The open subset $U_{j_0,\dots,j_\ell}\subset X$ is Stein and on $U_{j_0,\dots,j_\ell}$, the line bundle $L\otimes E^{-1}$, resp. $L$, endowed with its metric twisted by $e^{-\varphi_{j_0,\dots,j_\ell}}$, resp. $(|s|^2+\e^2)^{-(1-\k)}e^{-\varphi_{j_0,\dots,j_\ell}}$, has curvature

$$\rmo\Theta(L)-\rmo\Theta(E)+\rmo d'd''\varphi_{j_0,\dots,j_\ell},
$$
resp.
$$\rmo\Theta(L)+(1-\k)\rmo d'd''\log(|s|^2+\e^2)+\rmo d'd''\varphi_{j_0,\dots,j_\ell},
$$
which, by inequality (\ref{bel}) and assumption~(\ref{positiv}), is bounded from below by
$$\rmo\Theta(L)-\rmo\Theta(E)+\rmo d'd''\varphi_{j_0}\geq \om,
$$
resp.
$$\rmo\Theta(L)-(1-\k)\frac{\la\rmo\Theta(E)s,s\ra}{|s|^2+\e^2}+\rmo d'd'' \varphi_{j_0}\geq \om.
$$
The fact that $u$ can be chosen in the Sobolev space $W^{m+1}$ comes from the ellipticity of the Laplacian. Let us explain why in the case $(a)$, the case $(b)$ being completely similar. In fact, the $D''$-equation is solved using complete metrics $\om_\e$ on $U_{j_0,\dots,j_\ell}$, such that $\om_\e\geq\om$ and $\om_\e\rightarrow\om$ as $\e\rightarrow 0$. For any $\e>0$, the corresponding minimal solution  $u_\e$ (i.e. the one satisfying $D'' u_\e=v$ and $u_\e\in (\Ker D'')^{\perp_{\om_\e}}$) is such that
$$\int_{U_{j_0,\dots,j_\ell}} |u_\e|_{\om_\e}^2 e^{-\varphi_{j_0,\dots,j_\ell}}\, dV_{\omega_\e}\leq \frac{1}{p}\int_{U_{j_0,\dots,j_\ell}} |v|_{\om_\e}^2e^{-\varphi_{j_0,\dots,j_\ell}}\, dV_{\omega_\e}\leq \frac{1}{p}\int_{U_{j_0,\dots,j_\ell}} |v|^2e^{-\varphi_{j_0,\dots,j_\ell}}\, dV_{\omega}$$
where the latter inequality comes from Proposition \ref{usef} $(b)$. Then, there exists a sequence $(\e_\mu)$ converging to 0 such that $u_{\e_\mu}$ converges weakly to some $u$ in $L^2_{\rm loc}$ as $\mu\rightarrow+\infty$: $u$ satisfies $D''u=v$,
$$\int_{U_{j_0,\dots,j_\ell}} |u|^2 e^{-\varphi_{j_0,\dots,j_\ell}}\, dV_{\omega}\leq \frac{1}{p}\int_{U_{j_0,\dots,j_\ell}} |v|^2e^{-\varphi_{j_0,\dots,j_\ell}}\, dV_{\omega}
$$
but also $u\in(\Ker D'')^{\perp_\om}=\overline{\Im (D'')^{\star_\om}}$ (and therefore $(D'')^{\star_\om} u=0$). Indeed, $L^2(\om)\subset L^2({\om_\e})$ because $|\,.\,|^2_{\om_\e}\, dV_{\om_\e}\leq |\,.\,|^2\, dV_{\om}$ and since $\om_\e\rightarrow\om$, we get $u\in(\Ker D'')^{\perp_\om}$ by the dominated convergence theorem. Finally, $u$ satisfies $D''u=v$ and $(D'')^\star u=0$ which is an elliptic differential system and standard arguments give $u\in W^{m+1}$ if $v\in W^m$.

Notice that we can skip the extraction of a weak limit if we only need a solution with the same estimate on a slightly smaller relatively compact open subset of $U_{j_0,\dots,j_\ell}$: all we have to do is take a complete metric on $U_{j_0,\dots,j_\ell}$ which coincides with $\om$ on the smaller subset.

\end{demo}

Finally, we also select  a smooth partition of unity $\{\sigma_j\}_{j\in J}$ subordinate to ${\cal U}$ (i.e. for each $j$, $\sigma_j\in\E^\infty_c(U_j,[0,1])$ and $\sum_{j\in J}\sigma_j(x)=1$ for any $x\in {\Omega}$).

\section{Proof of the theorem}

Recall that, by assumption, $Y\subset X$ is a smooth hypersurface so that $E\simeq{\cal O}_X(Y)$.

\medskip

\subsection{Construction of smooth extensions} In this section, we prove the following

\begin{lemma}\label{sext}
For any $k\geq 0$, there exists a smooth section
$$\widetilde f_\infty\in \E^\infty(X,\L^{n,q}T^\star_ X\otimes L)
$$
such that
\begin{enumerate}[{\rm (}a{\rm )}]
\item $\widetilde f_\infty$ coincides with $f$ in restriction to $Y$,
\item $|\widetilde f_\infty|=|f|$ at every point of $Y$,
\item $D''\widetilde f_\infty=0$ at every point of $Y$,
\item $s^{-1} D''\widetilde f_\infty\in \E^k(X,\L^{n,q+1}T^\star_ X \otimes L\otimes {\cal O}_X(-Y))$.
\end{enumerate}
\end{lemma}

\begin{demo}
Let us cover $Y$ by coordinate patches $W_j\subset X$ biholomorphic to polydiscs and with the following property:  if we denote the corresponding coordinates by $(z_j,w_j)\in\Delta\times\Delta^{n-1}$, where $w_j=(w_j^1,\dots, w_j^{n-1})$, then $W_j\cap Y=\{z_j=0\}$. On each $W_j$, we fix some holomorphic $\sigma_j\in \G(W_j,K_X\otimes L)$ which trivializes $K_X\otimes L$.

As explained in~\cite{De3}, the restriction map $(\L^{0,q}T^\star_X)_{|Y}\fd\L^{0,q}T^\star_Y$ can be viewed as an orthogonal projection onto a $C^\infty$ subbundle of $(\L^{0,q}T^\star_X)_{|Y}$. One might extend this subbundle from $W_j\cap Y$ to $W_j$ and then extend $f$ on ${W_j}$ by some smooth form $\widehat f_j\in \E^\infty(W_j,\L^{n,q}T^\star_ X\otimes L)$. Using a smooth partition of unity $\theta_j\in \E^\infty_c(W_j,\R)$, $\sum_j \theta_j=1$ on a neighbourhood of $Y$, we get a global smooth extension $\widehat f=\sum_j \theta_j\widehat f_j$ of $f$ which fulfills conditions $(a)$ and $(b)$. Since
$$(D''\widehat f)_{|Y}=(D''\widehat f_{|Y})=D'' f=0,$$
we can write $D''\widehat f=d\bar z_j\wedge g_j$ on $W_j\cap Y$ for some smooth $(0,q)$-forms $g_j$ which we extend arbitrarily to $W_j$. Then
$$\widetilde f_\infty:=\widehat f-\sum_j\theta_j\bar z_j g_j$$
coincides with $\widehat f$ on $Y$ and satisfies $(c)$.

We proceed by induction to get $(d)$. Assume that on each $W_j$,
$$D''\ext=z_j f_j(z_j,w_j) +\bar z_j^k \Bigl[d\bar z_j\wedge\sum_{|I|=q}a_I(w_j)\sigma_j d\bar w_j^I+\sum_{|I'|=q+1}b_{I'}(w_j)\sigma_j d\bar w_j^{I'}\Bigr]+\bar z_j^{k+1} h_j(z_j,w_j)
$$
for some $f_j,h_j\in \E^\infty(W_j,\L^{n,q+1}T^\star_ X\otimes L)$, $a_I,b_{I'} \in \E^\infty(\Delta^{n-1},\C)$, $k\geq 1$, and where the multi-indices $I$, $I'$ are increasing. We say that $\ext$ enjoys property $(P_k)$. Remark that such an equality implies that $s^{-1} D''\widetilde f_\infty\in \E^{k-2}(X,\L^{n,q+1}T^\star_ X\otimes L\otimes {\cal O}_X(-Y))$ if $k\geq 2$. Moreover, the extension $\widetilde f_\infty$ we just constructed satisfies property $(P_1)$ because $D''\widetilde f_\infty=0$ along $Y$.

Of course, $D''(D''\ext)=0$, but also the direct computation gives
$$D''(D''\ext)=z_jD''f_j(z_j,w_j)+k\bar z_j^{k-1} d\bar z_j\wedge\Bigl[\sum_{|I'|=q+1}b_{I'}(w_j)\sigma_j d\bar w_j^{I'}\Bigr]+\bar z_j^k h'_j(z_j,w_j)
$$
for some $h'_j\in\E^\infty(W_j,\L^{n,q+2}T^\star_ X\otimes L)$, hence
the $b_{I'}$'s must vanish identically. So if we take
$$\widetilde f'_\infty=\ext-\sum_j\theta_j\frac{\bar z^{k+1}_j}{k+1}\sum_{|I|=q}a_I(w_j)\sigma_j d\bar w_j^I,
$$
we have
$$\begin{array}{rcl}
D''\widetilde f'_\infty &= & \displaystyle D''\ext-\sum_j \biggl(\frac{\bar z^{k+1}_j}{k+1}d''\theta_j+\theta_j \bar z^k_jd\bar z_j\biggr)\wedge\sum_{|I|=q}a_I(w_j)\sigma_j d\bar w_j^I\\
& &\hskip 4cm -\displaystyle\sum_j\theta_j\frac{\bar z^{k+1}_j}{k+1}D''\biggl(\sum_{|I|=q}a_I(w_j)\sigma_j d\bar w_j^I\biggr)\\
&=& \displaystyle \sum_j\theta_j\biggl(D''\ext-\bar z_j^k d\bar z_j\wedge\sum_{|I|=q}a_I(w_j)\sigma_j d\bar w_j^I\biggr)+\sum_j\bar z_j^{k+1} h''_j(z_j,w_j)\\
& = & \displaystyle\sum_j  z_j\theta_j f_j(z_j,w_j)+\sum_j \bar z_j^{k+1}\bigl(\theta_j h_j(z_j,w_j)+ h''_j(z_j,w_j)\bigr)
\end{array}
$$
for some $h''_j\in \E_c^\infty(W_j,\L^{n,q+1}T^\star_ X\otimes L)$. Then, $\widetilde f'_\infty$ enjoys property $(P_{k+1})$.

\end{demo}

\medskip

\subsection{Construction of approximate extensions with control}

Let $\theta:\R\fd [0,1]$ be a smooth function with support in $(-\infty,1)$, such that $\theta\equiv 1$ on $(-\infty,1/2]$ and $|\theta'|\leq 4$, and consider the truncated extension of $f$
$$\widetilde f_\e:=\theta(\e^{-2}|s|^2)\widetilde f_\infty$$
where $\ext$ is the extension provided by Lemma~\ref{sext}, such that $s^{-1} D''\widetilde f_\infty\in \E^k(X,\L^{n,q+1}T^\star_ X \otimes L\otimes {\cal O}_X(-Y))$ for some $k\geq 1$ which will be determined later. We wish to solve on $X$ the equation
$$D'' u_{\e}=D'' \widetilde f_{\e}$$
with estimate, and the additional constraint that  $u_{\e}$ vanishes along $Y$. We also expect some regularity on $u_\e$ in order to justify that $\widetilde f_{\e}-u_\e$ is a ($D''$-closed) extension of $f$. In general, we are not able to get this by the method we use here, and we can only produce approximate solutions.

\medskip

The fundamental tool is the following existence result (see \cite{De3}, \cite{Pa}):
\begin{theo}\label{ape}
Let $X$ be a complete K\"ahler manifold of dimension $n$ equipped with a (not necessarily complete) K\"ahler metric $\omega$, and let $L$ be a line bundle endowed with a smooth Hermitian metric. Assume that there exist two smooth bounded functions $\eta,\lambda>0$ on $X$ satisfying
\begin{equation}\label{cond}
\eta\sqrt{-1}\Theta(L)-\sqrt{-1}\dr\db\eta-\rmo\frac{\dr\eta\wedge\db\eta}{\lambda}\geq\sqrt{-1}\tau\dr\mu\wedge\db\mu
\end{equation}
for some positive function $\tau$ and some function $\mu$. Let us consider the (densely defined) modified $D''$ operators
$$Tu:=D''(\sqrt{\eta+\lambda}u)\ \ {\rm and}\ \ Su:=\sqrt\eta(D'' u)$$
acting on forms with values in $L$. Let $g=\db\mu\wedge g_0+g_2$ be a $L^2$ form of $(n,q+1)$ type ($q\geq 0$) with values in $L$ such that
\begin{enumerate}[{\rm (}a{\rm )}]
\item $D'' g=0$,
\item $g_0\in L^2 (X,\L^{n,q}T^\star_X\otimes L)$,
\item $C(g_0,\tau):=\int_X1/\tau|g_0|^2dV_\om<+\infty$,
\item $|g_2|^2\leq \gamma\, C(g_0,\tau)$ almost everywhere for some positive constant $\gamma$.
\end{enumerate}
Then, for any $u\in{\rm Dom}\,T^\star\cap{\rm Dom}\,S$, we have
$$\Bigl |\int_X \la g,u\ra dV_\om\Bigr |^2\leq C(g_0,\tau)\bigl(\|T^\star u\|^2+\|Su\|^2+\gamma\|u\|^2\bigr).$$
In particular, there exist $v\in L^2(X,\L^{n,q}T_X^\star\otimes L)$ and $w\in L^2(X,\L^{n,q+1}T_X^\star\otimes L)$ such that
$$Tv+\gamma^{1/2}w=g
$$
together with the estimate
$$\int_X |v|^2dV_\om+\int_X |w|^2dV_\omega\leq C(g_0,\tau).$$
\end{theo}

\bigskip

As before, let $c\in \R$ be such that $\overline{\Omega}\subset X_c$. For simplicity, we will assume in the sequel that $X=X_c$. We are going to apply Theorem~\ref{ape} to $D'' \widetilde f_{\e}$ on $X\backslash Y$. By Proposition~\ref{usef} $(a)$, $X\backslash Y$ can be equipped with a complete K\"ahler metric. As for the bundle $L$, we endow it with its original metric multiplied with the weight $|s|^{-2}$ in order to force the vanishing of the approximate solution along $Y$.

For any $\e>0$, we set $\sigma_\e:=-\log(\e^2+|s|^2)$. Remark that, because of the condition $|s|\leq e^{-\a}$, this function is positive for $\e$ small enough. Let $\chi:\R_+\fd\R_+$ be any strictly concave function whose derivative satisfies $1\leq\chi'\leq 2$ and such that $\chi(-\log(\e^2+e^{-2\a}))\geq 2\a$ for any $\e>0$ small enough (in \cite{De3} and \cite{Pa}, $\chi(t)=t+\log (1+t)$ but we will choose other functions picked in~\cite{MV}).  Let us define the two positive functions (again $\e$ is assumed to be small enough)
$$\eta_\e:=\chi(\sigma_\e)\ \ {\rm and}\ \ \lambda_\e:=-\frac{\chi'(\sigma_\e)^2}{\chi''(\sigma_\e)}.$$
Although this is done carefully in \cite{De3} and \cite{Pa}, we check quickly that $\eta_\e$ and $\lambda_\e$ fulfill condition $(\ref{cond})$ in Theorem~\ref{ape}. It is easy to see that
\begin{equation}\label{bel}
-\rmo\dr\db\s_\e\geq\rmo\frac{\e^2}{|s|^2}\dr\s_\e\wedge\db\s_\e-\frac{\la\rmo\Th(E)s,s\ra}{\e^2+|s|^2}
\end{equation}
and it is straightforward that
$$\dr\eta_\e=\chi'(\s_\e)\dr\s_\e\ ,\ \ \db\eta_\e=\chi'(\s_\e)\db\s_\e\ ,\ \ \dr\db\eta_\e=\chi'(\s_\e)\dr\db\s_\e+\chi''(\s_\e)\dr\s_\e\wedge\db\s_\e.$$
Thus, since $\chi'$ is positive,
$$-\rmo\dr\db\eta_\e\geq\Bigl(\frac{1}{\chi'(\s_\e)}\frac{\e^2}{|s|^2}+\frac{1}{\l_\e}\Bigr)\rmo\dr\eta_\e\wedge\db\eta_\e -\frac{\chi'(\s_\e)}{\e^2+|s|^2}\la\rmo\Th(E)s,s\ra.
$$
If $\e$ is small enough, then for any $x\in X$, $\eta_\e(x)\geq\chi(-\log(\e^2+e^{-2\a}))\geq 2\a$. Taking into account the curvature assumptions $(\ref{assone})$ and $(\ref{asstwo})$ in Theorem~\ref{theop} as well as the fact that $\chi'\leq 2$, we obtain
$$\eta_\e(\rmo\Th(L)+\rmo\dr\db\log|s|^2)\geq\frac{\eta_\e}{\a}\rmo\Th(E)\geq \frac{\chi'(\s_\e)}{\e^2+|s|^2}\la\rmo\Th(E)s,s\ra.
$$
Finally, summing up the two latter inequalities, we get
$$\eta_\e(\rmo\Th(L)+\rmo\dr\db\log|s|^2)-\rmo\dr\db\eta_\e-\frac{\rmo}{\l_\e}\dr\eta_\e\wedge\db\eta_\e\geq\frac{1}{\chi'(\s_\e)}\frac{\e^2}{|s|^2}\rmo\dr\eta_\e\wedge\db\eta_\e
$$
which proves that $(\ref{cond})$ is fulfilled with $\displaystyle\tau=\frac{1}{\chi'(\s_\e)}\frac{\e^2}{|s|^2}$ and $\mu=\eta_\e$. Now, we can write

$$D'' \widetilde f_\e=\db\eta_\e\wedge g_\e+\theta\Bigl(\frac{|s|^2}{\e^2}\Bigr)D''\ext
$$
where
$$g_\e:=\Bigl(1+\frac{|s|^2}{\e^2}\Bigr)\theta'\Bigl(\frac{|s|^2}{\e^2}\Bigr)\frac{\widetilde f_\infty}{\chi'(\s_\e)}.
$$

A quick computation shows that
\begin{equation}\label{lim}
\lim_{\e\rightarrow 0} \,\frac{1}{\e^2}\int_{X\backslash Y} \theta'\Bigl(\frac{|s|^2}{\e^2}\Bigr)^2 |\widetilde f_\infty|^2\,dV_{\omega}=c_0\int_{Y}\frac{|f|^2}{|ds|^2} dV_{Y,\om}
\end{equation}
for some ``universal'' constant $c_0$.
Therefore, since $\theta(\e^{-2}|s|^2)$ is supported in $\{|s|<\e\}$, and since $D''\ext=0$ on $Y$, for any $\gamma>0$,
$$\biggl|\theta\Bigl(\frac{|s|^2}{\e^2}\Bigr)D''\ext\biggr|\leq \gamma\int_{X\backslash Y} \frac{1}{\tau} \frac{|g_\e|^2}{|s|^2}dV_{\om}=\frac{\g}{\e^2}\int_{X\backslash Y} \Bigl(1+\frac{|s|^2}{\e^2}\Bigr)^2\theta'\Bigl(\frac{|s|^2}{\e^2}\Bigr)^2 |\widetilde f_\infty|^2\,dV_{\omega}
$$
if $\e>0$ is small enough.

Hence, we can apply Theorem~\ref{ape}: we find $u_{\e,\gamma}=\sqrt{\eta_\e+\l_\e} v_{\e,\gamma}$ and $w_{\e,\gamma}$ which satisfy the equation
$$D'' u_{\e,\gamma}+\gamma^{1/2}w_{\e,\gamma}=D''\widetilde f_\e$$
on $X\backslash Y$ and such that

\begin{eqnarray}
\displaystyle\int_{X\backslash Y} \frac{|u_{\e,\gamma}|^2}{|s|^{2}(\eta_\e+\l_\e)}dV_{\omega} + \int_{X\backslash Y} \frac{|w_{\e,\gamma}|^2}{|s|^{2}}dV_{\omega}& \leq & \displaystyle\frac{1}{\e^2}\int_{X\backslash Y} \Bigl(1+\frac{|s|^2}{\e^2}\Bigr)^2\theta'\Bigl(\frac{|s|^2}{\e^2}\Bigr)^2\frac{|\widetilde f_\infty|^2}{\chi'(\s_\e)}dV_{\omega}\nonumber\\
& \leq &\displaystyle\frac{4}{\e^2}\int_{X\backslash Y} \theta'\Bigl(\frac{|s|^2}{\e^2}\Bigr)^2 |\widetilde f_\infty|^2\,dV_{\omega}.\label{esti}
\end{eqnarray}

\subsection{Regularization of the approximate solution}\label{regul}

Recall that $Y$ is a divisor such that $E\simeq {\cal O}_X(Y)$. Here we use a trick of Demailly:  we consider $s^{-1}u_{\e,\gamma}$ (resp. $s^{-1}w_{\e,\gamma}$) as a $L^2$ $(0,q)$-form (resp. $(0,q+1)$-form) with values in the twisted line bundle $K_X\otimes L\otimes{\cal O}_X(-Y)$ equipped with a smooth Hermitian metric. By Proposition~\ref{usef} $(c)$, we can write
$$D''(s^{-1}u_{\e,\gamma})+\gamma^{1/2}s^{-1}w_{\e,\gamma}=s^{-1}D''\widetilde f_\e
$$
not only on $X\backslash Y$ but also on $X$ because $s^{-1}u_{\e,\gamma}$ is locally $L^2$,  $s^{-1}w_{\e,\gamma}$ is $L^2$ hence locally $L^1$, and $s^{-1}D''\widetilde f_\e$ is of class $C^k$ hence locally $L^1$ (recall that $\ext$, as chosen in Lemma~\ref{sext}, is such that $s^{-1}D''\ext$ is of class $C^k$, $k\geq 1$). However, we do not know much about the regularity of $u_{\e,\g}$ and $w_{\e,\g}$.

But $\E^\infty_c(X,\Lambda^{n,q}T^\star_X\otimes L\otimes{\cal O}_X(-Y))$ is dense in ${\rm Dom}\,D''$ for the graph norm, where we consider $D''$ as an operator acting on $(n,q)$ forms on $X$ with values in $L\otimes{\cal O}_X(-Y)$. More precisely, the density holds when $X$ is endowed with a complete metric. If $X=X_c$ as we assumed above, we can work instead on $X_{c'}$ for some $c'>c$, and there exists on $X_{c'}$ some complete K\"ahler metric which coincides with $\om$ on $X_c$.

Then, we can find some $t_{\e,\gamma}\in \E^\infty(X,\L^{n,q}T^\star_{X}\otimes L\otimes{\cal O}_X(-Y))$, which is $L^2$, such that
\begin{equation}\label{control}
\left|\int_{X} \frac{|t_{\e,\gamma}|^2}{\eta_\e+\l_\e}dV_{\omega}-\int_{X} \frac{|s^{-1}u_{\e,\gamma}|^2}{\eta_\e+\l_\e}dV_{\omega}\right|\leq\e
\end{equation}
(recall that $\eta_\e$ is bounded by $2\alpha$ from below), and $D''(t_{\e,\gamma}-s^{-1}u_{\e,\gamma})$ has $L^2$ norm bounded by $\gamma^{1/2}$ from above. As a consequence,
$$D'' t_{\e,\gamma}=s^{-1}D''\widetilde f_\e+r_{\e,\gamma}
$$
on $X$, with $r_{\e,\gamma}\in \E^k(X,\L^{n,q+1}T^\star_X\otimes L\otimes E^{-1})$ since $D'' t_{\e,\gamma}$ and $s^{-1}D''\widetilde f_\e$ are of class $C^k$. Moreover, $r_{\e,\gamma}$ satisfies
\begin{equation}\label{rgbound}
\int_X |r_{\e,\g}|^2\leq C_1^2\g
\end{equation}
for some positive constant $C_1$ depending on $f$, but not on $\e$ and $\g$ (see (\ref{esti}) and (\ref{lim})). Finally, $s^{-1}D''\widetilde f_\e$ is $D''$-closed on $X\backslash Y$, hence on $X$ by Proposition~\ref{usef} $(c)$, and therefore $D''r_{\e,\gamma}=0$.

\subsection{The choice of $\eta_\e$ and $\l_\e$}\label{choice}

Let us come now to the choice of $\eta_\e$ and $\l_\e$ (see \cite{MV} for more details). For any $0<\k\leq1$, we define for $t\geq 0$ the functions
$$g_\k(t)=\k^{-1}e^{\k t}\ ,\ \  h_\k(t)=\int_0^t \frac{1}{2e^{\k y}-1} dy\ \ {\rm and}\ \ \chi_\k(t)=1+t+h_\k(t).$$
One checks immediatly that $1\leq\chi'_\k\leq 2$ and $\chi''_\k<0$. Moreover,
$$\chi_\k(-\log(\e^2+e^{-2\a}))\geq 1- \log(\e^2+e^{-2\a})\geq 1+ 2\a -\log(1+\e^2 e^{2\a})\geq 2\a
$$
when $\e$ is small enough. Clearly, $\chi_\k(t)\leq 1+2t$ and it follows that
$$\frac{\chi_\k(t)}{g_\k(t)}\leq \frac{\k(1+2t)}{e^{\k t}}\leq 2
$$
as is seen from a simple computation. Moreover,

$$-\frac{\chi'_\k(t)^2}{\chi''_\k(t)}\leq 2 g_\k(t).
$$
As a consequence, if we fix $\k\leq 1$ and take $\chi=\chi_\k$, we have
\begin{equation}\label{kbound}
\eta_\e+\l_\e\leq \frac{4}{\k(|s|^2+\e^2)^\k}.
\end{equation}

\subsection{Construction of $q$-cochains via Leray's isomorphism}

Recall that we have fixed a finite open covering ${\cal U}=\{U_j\}_{j\in J}$ of $\overline{\Omega}$. We endow the group $C^\ell_2({\cal U},\E^1(\Lambda^{n,p}T^\star_X\otimes L\otimes E^{-1}))$ of (alternate) $\ell$-cochains with values in $\E^1(\Lambda^{n,p}T^\star_X\otimes L\otimes E^{-1})$ which are $L^2$ with the norm

$$\|\varsigma^\ell\|^2=\max_{j_0<\dots<j_\ell}\int_{U_{j_0,\dots,j_{\ell}}} |\varsigma^\ell_{j_0,\dots,j_{\ell}}|^2 e^{-\varphi_{j_0,\dots,j_\ell}}\,dV_\om
$$
and for all $0<\k\leq 1$ and $\e>0$, we endow $C_2^\ell({\cal U},\E^1(\Lambda^{n,p}T^\star_X\otimes L))$ with the norm
$$\|\varsigma^\ell\|^2_{\k,\e}=\max_{j_0<\dots<j_\ell}\int_{U_{j_0,\dots,j_{\ell}}} \frac{|\varsigma^\ell_{j_0,\dots,j_{\ell}}|^2}{(|s|^2+\e^2)^{1-\k}} e^{-\varphi_{j_0,\dots,j_\ell}}\,dV_\om.
$$
Remark that in the case when $\varsigma^0$ is the $0$-cocycle associated to a section $\varsigma$ of $\E^1(\Lambda^{n,p}T^\star_X\otimes L\otimes E^{-1}))$ (resp. $\E^1(\Lambda^{n,p}T^\star_X\otimes L))$),
$$\|\varsigma^0\|^2\leq \int_{X} |\varsigma|^2 dV_\om\  \Bigl(\hbox{ resp. } \|\varsigma^0\|_{\k,\e}^2\leq \int_{X} \frac{|\varsigma|^2}{(|s|^2+\e^2)^{1-\k}}\, dV_\om\Bigr)
$$
since the $\varphi_j$'s are nonnegative.

Now, we construct a $(q+1)$-cocycle in $Z^{q+1}({\cal U},{\cal O}(K_X\otimes L\otimes E^{-1}))$ corresponding to $r_{\e,\gamma}$ via Leray's isomorphism between the Dolbeault and the \v Cech cohomology groups. In fact, we are mostly interested in the intermediate cochains which appear during the process and the control we have on their norm.
The extension $\ext$ of $f$ is supposed to be sufficiently regular (i.e. $k$ is large enough in Proposition~\ref{sext}) in order that $r_{\e,\g}$, and every cochain obtained by solving local $D''$-equations below, is at least of class $C^1$ (see section~\ref{regul}, Lemma~\ref{loc} and use Sobolev lemma: $W^m(U)\subset \E^1(U)$ for any open subset $U\subset X$ if $m>1+\frac{n}{2}$).

For notational simplicity, we denote $r^{\ell}_{\e,\gamma,(j_0,\dots,j_\ell)}\in \G(U_{j_0,\dots,j_\ell},\E^1(\Lambda^{n,q-\ell}T^\star_X\otimes L\otimes E^{-1}))$ by $r^{\ell}_{j_0,\dots,j_\ell}$.

First, we solve the equation $D'' r_j^0=r_{\e,\gamma}$ on the $U_j$'s, then we solve the equations
$$D'' r^{\ell+1}_{j_0,\dots,j_{\ell+1}}=(\delta r^\ell)_{j_0,\dots,j_{\ell+1}}\ \ {\rm on}\ \ U_{j_0}\cap\dots\cap U_{j_{\ell+1}}\ \ (0\leq\ell\leq q-1)
$$
using each time Lemma~\ref{loc} $(a)$. Finally, $\delta r^q_{\e,\gamma}\in Z^{q+1}({\cal U},{\cal O}(K_X\otimes L\otimes E^{-1}))$ is a representative in \v Cech cohomology of $[r_{\e,\g}]\in H^{q+1}(X,K_X\otimes L\otimes E^{-1})$.

\begin{lemma}\label{rbound}
For any $\ell$, we have
$$\|r^\ell_{\e,\g}\|^2\leq \frac{(\ell+1)\dots 2.1}{(q+1). q \dots(q-\ell+1)}\,\|r_{\e,\g}\|^2\leq \frac{(\ell+1)\dots 2.1}{(q+1). q \dots(q-\ell+1)}\,\int_{X} |r_{\e,\gamma}|^2 dV_\om.$$
\end{lemma}

\begin{demo}
For any $\ell\geq 1$,
$$\begin{array}{rcl}
\|r^\ell_{\e,\g}\|^2 &\leq &  \displaystyle  \frac{1}{q-\ell+1} \|\dt r^{\ell-1}_{\e,\g}\|^2\\
&\leq & \displaystyle  \frac{\ell+1}{q-\ell+1} \|r^{\ell-1}_{\e,\g}\|^2\\
\end{array}
$$
and
$$
\|r^0_{\e,\g}\|^2 \leq \frac{1}{q+1} \|r_{\e,\g}\|^2\leq \frac{1}{q+1}\int_{X} |r_{\e,\gamma}|^2 dV_\om
$$
by the estimate in Lemma~\ref{loc} $(a)$.

\end{demo}
\medskip

In a similar manner, we produce a $q$-cochain $\zeta_{\e,\g}\in C^{q}({\cal U},{\cal O}(K_X\otimes L))$ corresponding to the ``approximately'' $D''$-closed extension $\widetilde f_\e-st_{\e,\gamma}\in\G(X,\E^1(\Lambda^{n,q}T^\star_X\otimes L))$ of $f$. More precisely, on the $U_j$'s, we solve the equation $D''h^0_j=\widetilde f_\e-st_{\e,\gamma}+sr^0_{\e,\g,j}$, then we solve
$$D''h^{\ell+1}_{j_0,\dots,j_{\ell+1}}=(\delta h^\ell)_{j_0,\dots,j_{\ell+1}}+(-1)^{\ell+1} sr^{\ell+1}_{j_0,\dots,j_{\ell+1}}\ \ {\rm on}\ \ U_{j_0}\cap\dots\cap U_{j_\ell+1}\ \ (0\leq\ell\leq q-2)
$$
using Lemma~\ref{loc} $(b)$. This is indeed possible since the right-hand side is $D''$-closed: if
$$D''h^{\ell}_{j_0,\dots,j_{\ell}}=(\delta h^{\ell-1})_{j_0,\dots,j_{\ell}}+(-1)^{\ell} sr^{\ell}_{j_0,\dots,j_{\ell}}
$$
then
$$\begin{array}{rcl}
\displaystyle D''(\delta h^{\ell})_{j_0,\dots,j_{\ell+1}} =  (\delta(D'' h^\ell))_{j_0,\dots,j_{\ell+1}}& =& \displaystyle (\delta^2h^{\ell-1})_{j_0,\dots,j_{\ell+1}}+(-1)^\ell s(\delta r^\ell)_{j_0,\dots,j_{\ell+1}}\\
& = & \displaystyle 0+(-1)^\ell sD''r^{\ell+1}_{j_0,\dots,j_{\ell+1}}.
\end{array}
$$
Finally, let $\zeta_{\e,\g}:=\delta h_{\e,\g}^{q-1}+(-1)^{q}sr^q_{\e,\g}\in C^{q}({\cal U},{\cal O}(K_X\otimes L))$ (in particular $D''\zeta_{\e,\g}=0$ and $\zeta_{\e,\g}$ is actually smooth by ellipticity of $D''$ in bidegree $(0,0)$).

\medskip

In the next proposition, we denote by ${\cal V}$ the finite Stein covering $\{V_j\}_{j\in J}=\{U_j\cap Y\}_{j\in J}$ of $Y$.

\begin{prop}\label{estim}
 Let $0<\k\leq 1$. The cochain $\zeta_{\e,\g}$ enjoys the following properties:
\begin{enumerate}[{\rm (}a{\rm )}]
\item
$$\|\zeta_{\e,\g}\|^2_{\k,\e}\leq \frac{16(q+1)c_0}{\k}\int_Y\frac{|f|^2}{|ds|^2} dV_{Y,\om}+\beta(\e,\g)
$$
where $\b$ is a positive function such that $\b(\e,\g)\rightarrow 0$ as $\e,\g\rightarrow 0$ (recall that for any $\g>0$, $\zeta_{\e,\g}$ only exists if $\e>0$ is small enough).
\item For any $\e,\g$, ${\zeta_{\e,\g}}_{|Y}\in Z^q({\cal V},{\cal O}((K_X\otimes L)_{|Y}))$ is a representative in \v Cech cohomology of the cohomology class $[f]\in H^q(Y,(K_X\otimes L)_{|Y})$.

\end{enumerate}
\end{prop}

\begin{demo}
Let $\chi=\chi_\k$ be as in section~\ref{choice}. We use the corresponding $\eta_\e$ and $\l_\e$.

\begin{enumerate}[{\rm (}\it a{\rm )}]
\item We have
$$\begin{array}{rcl}
\|\zeta_{\e,\g}\|_{\k,\e} & \leq & \|\dt h^{q-1}_{\e,\g}\|_{\k,\e}+\|sr^q_{\e,\g}\|_{\k,\e}\\
&\leq&\displaystyle \sqrt{q+1}\|h^{q-1}_{\e,\g}\|_{\k,\e}+e^{-\frac{\a\k}{2}}\|r_{\e,\g}\|
\end{array}
$$
by Lemma~\ref{rbound}, since $|s|^2(|s|^2+\e^2)^{-(1-\k)}\leq |s|^\k\leq e^{-\frac{\a\k}{2}}$ according to assumption $(\ref{assthree})$. In the same way, for any $\ell\geq 1$,
$$\begin{array}{rcl}
\|h^\ell_{\e,\g}\|_{\k,\e} & \leq & \displaystyle\frac{1}{\sqrt{q-\ell}}(\|\dt h^{\ell-1}_{\e,\g}\|_{\k,\e}+\|sr^\ell_{\e,\g}\|_{\k,\e})\\
&\leq&\displaystyle\frac{\sqrt{\ell+1}}{\sqrt{q-\ell}} \|h^{\ell-1}_{\e,\g}\|_{\k,\e}+ \frac{\sqrt{(\ell+1)\dots 2.1}}{\sqrt{(q+1).q\dots(q-\ell)}}e^{-\frac{\a\k}{2}}\|r_{\e,\g}\|
\end{array}
$$
where we also used the estimate in Lemma~\ref{loc} $(b)$. Finally,
$$\begin{array}{rcl}
\|h^0_{\e,\g}\|_{\k,\e} & \leq & \displaystyle\frac{1}{\sqrt{q}}(\|\widetilde f_\e-st_{\e,\gamma}\|_{\k,\e}+\|sr^0_{\e,\g}\|_{\k,\e})\\
&\leq&\displaystyle\frac{1}{\sqrt{q}}\|\widetilde f_\e-st_{\e,\gamma}\|_{\k,\e}+ \frac{1}{\sqrt{(q+1).q}}e^{-\frac{\a\k}{2}}\|r_{\e,\g}\|.
\end{array}
$$
Collecting all these inequalities, we get
$$\|\zeta_{\e,\g}\|_{\k,\e}\leq \sqrt{q+1}\|\widetilde f_\e-st_{\e,\gamma}\|_{\k,\e}+q e^{-\frac{\a\k}{2}}\|r_{\e,\g}\|\leq \sqrt{q+1}\|\widetilde f_\e-st_{\e,\gamma}\|_{\k,\e}+q\,C_1\g^{1/2}.
$$
Now, it is easy to see that
$$\int_{X} \frac{|\widetilde f_\e-st_{\e,\gamma}|^2}{(|s|^2+\e^2)^{1-\k}}\, dV_\om\leq \int_{X} \frac{|st_{\e,\gamma}|^2}{(|s|^2+\e^2)^{1-\k}}\, dV_\om+C_2\,\e^{2\k}\\
$$
for some constant $C_2$, as $\widetilde f_\e$ is uniformly bounded with support in $\{|s|<\e\}$. Finally, by $(\ref{kbound})$,
$$\int_{X} \frac{|st_{\e,\gamma}|^2}{(|s|^2+\e^2)^{1-\k}}\, dV_\om\leq\int_{X} (|s|^2+\e^2)^\k |t_{\e,\g}|^2\, dV_\om\leq \frac{4}{\k}\int_{X}\frac{|t_{\e,\g}|^2}{\eta_\e+\l_\e}\, dV_\om
$$
and the desired inequality follows from (\ref{esti}), (\ref{control}) and (\ref{lim}).

\item It is clear since on $U_{j_0,\dots,j_\ell+1}\cap Y$, the restriction of $(\delta h^\ell)_{j_0,\dots,j_\ell}$ is always $D''$-closed, hence we construct an ``exact'' representative in restriction to $Y$.
\end{enumerate}
\end{demo}

\subsection{Passing to the limit and reversing the process to get the extension}

Now, we just have to make $\g$ and $\e$ go to zero and extract a weak limit of $\zeta_{\e,\g}$. This weak limit $\zeta$ is an element of $Z^{q}({\cal U},{\cal O}(K_X\otimes L))$ since $D''\zeta_{\e,\g}=0$ for any $\e,\g$, and $\delta \zeta_{\e,\g}=(-1)^q\delta (sr^q_{\e,\gamma})$ which, by $(\ref{rgbound})$ and Proposition~\ref{rbound}, has $L^2$ norm bounded by some constant times $\g^{1/2}$, thus $\delta\zeta=0$. Moreover, $\zeta_{|Y}\in Z^q({\cal V},{\cal O}((K_X\otimes L)_{|Y}))$ is a representative in \v Cech cohomology of the cohomology class $[f]\in H^q(Y,(K_X\otimes L)_{|Y})$ since this is the case of any ${\zeta_{\e,\g}}_{|Y}$ by Proposition~\ref{estim} $(b)$. For all $j_0,\dots,j_q$, as the $\varphi_j$'s are bounded from above by 1, we obtain
$$\displaystyle\int_{U_{j_0,\dots,j_{q}}} \frac{|\zeta|^2}{|s|^{2(1-\k)}} \,dV_\om\leq \frac{16 e^{q+1}(q+1)c_0}{\k}\int_Y\frac{|f|^2}{|ds|^2} dV_{Y,\om}
$$
if we take the limit in the inequality of Proposition~\ref{estim} $(a)$.

Finally, we construct the desired extension $F$ in the following way.
For $0\leq \ell\leq q-1$, we produce $\xi^\ell\in C^{\ell}({\cal U},\E^\infty(\Lambda^{n,q-\ell-1}T^\star_X\otimes L))$ such that
\begin{enumerate}[$(i)$]
\item $(\delta\xi^{q-1})_{j_0,\dots, j_q}=\zeta_{j_0,\dots, j_q}$ on $U_{j_0}\cap\dots\cap U_{j_q}$,
\item $(\delta\xi^{\ell})_{j_0,\dots, j_{\ell+1}}=D''\xi^{\ell+1}_{j_0,\dots, j_{\ell+1}}$ on $U_{j_0}\cap\dots\cap U_{j_{\ell+1}}$ ($0\leq \ell\leq q-2$).
\end{enumerate}
These $\dt$-equations are solved by using the partition of unity $\{\sigma_j\}_{j\in J}$ subordinate to ${\cal U}$ in the following way:

$$\begin{array}{rcl}
\xi^{q-1}_{j_0,\dots, j_{q-1}} & = & \sum_i \sigma_i\, \zeta_{i,j_0,\dots, j_{q-1}}\,,\\
\xi^{\ell}_{j_0,\dots, j_{\ell}} & = & \sum_i \sigma_i\,D''\xi^{\ell+1}_{i,j_0,\dots, j_{\ell}}\ \ \ (0\leq \ell\leq q-2).
\end{array}$$
Finally, we set
$$F=D''\xi^{0}_{j}=q!\sum_{j_1<\dots<j_{q} \atop j_i\not = j} \zeta_{j_{q},\dots, j_{1},j}\, d''\sigma_{j_1}\wedge\dots\wedge d''\sigma_{j_q}$$
on $U_{j}$. Then, $F$ defines a $D''$-closed section in $\G(\Omega,\E^\infty(\Lambda^{0,q}T^\star_X\otimes K_X\otimes L))$ such that $[F_{|Y}]=[f]\in H^q(Y,(K_X\otimes L)_{|Y})$ and the estimate
$$\int_{\Omega}\frac{|F|^2}{|s|^{2(1-\k)}} \,dV_\om\leq \frac{16 e^{q+1}(q+1)c_0 C_\sigma |J|!}{\k(|J|-q-1)!}\int_Y\frac{|f|^2}{|ds|^2} dV_{Y,\om}
$$
holds for some constant $C_\sigma$ depending only on the partition of unity $\{\sigma_j\}_{j\in J}$ and $q$ (one can take $C_\sigma=\max_{j\in J} |d''\sigma_j|^{2q}$).


TITLE: Find all eigenvalues and vectors.
QUESTION [0 upvotes]: I have found the eigenvalues of the given matrix A:
$$
        \begin{bmatrix}
        1 & -1 & 2 \\
        -1 & 1 & 2 \\
        2 & 2 & 2 \\
        \end{bmatrix}
$$
to be -2,2,4. When looking for the eigenvectors I started by assesing the matrix when lambda is = -2 and got the matrix
$$
        \begin{bmatrix}
        1 & 0 & 0 \\
        0 & 1 & 0 \\
        0 & 0 & 1 \\
        \end{bmatrix}
$$
does this mean that the eigenvector is (0,0,0) or is that impossible.

REPLY [0 votes]: If you take your matrix, assuming you computed eigenvalues correctly, then $\lambda = -2,2,4$ yields the matrices determined by $A-\lambda I$.
So, for $\lambda=-2$, it should look like:
$\begin{matrix}
        3 & -1 & 2 \\
        -1 & 3 & 2 \\
        2 & 2 & 4 \\
        \end{matrix}$


TITLE: What am I doing wrong? Finding a limit as $x$ approaches $0$
QUESTION [1 upvotes]: $$\lim_{x\to 0} {a-\sqrt{a^2-x^2}\over x^2} =$$
$${a-\sqrt{a^2-x^2}\over x^2}\cdot{a+\sqrt{a^2-x^2}\over a+\sqrt{a^2-x^2}} = $$
$$a^{2} - a^{2} - x^{2}\over ax^{2} + x^{2}\sqrt{a^{2}-x^{2}} $$
$$-x^{2}\over ax^{2}+x^{2}\sqrt{a^{2}-x^{2}}$$
$$-1\over a+\sqrt{a^{2}-x^{2}}$$
Can anyone tell me where I'm wrong?

REPLY [1 votes]: Third equality $$\left (a - \sqrt{a^2-x^2}\right )\left (a+\sqrt{a^2+x^2}\right ) \color{blue}{=x^2}$$


TITLE: if the solution of a system of ode's blows up, does it mean it's unstable in the Lyapunov sense?
QUESTION [0 upvotes]: there are many techniques and theorems which let you conclude about the stability of a system without actually solving it.
let's say for example you've got a system : $X'= AX + B, \; X(0)= X_0$
$A \in \mathbb{R^{n \times n}}, \; B \in \mathbb{R^n}$
if $\|X(t)\| \to \infty$ intuitvely it seems that the $\epsilon-\delta$ defintion of stability in the sense of lyapunov doesn't hold.
so does it mean the system is indeed unstable ?

REPLY [2 votes]: Say $x^{*}$ is an equilibrium point of your system. Then this equilibrium point is stable if for every $\epsilon>0$, there exists a $\delta>0$ such that if $||x(0) - x^{*}||<\delta$, then for all $t>0$, $||x(t)-x^{*}||<\epsilon$.
If in your case $||x(t)|| \to \infty$ no matter what the initial conditions,  then you can't satisfy the above condition, so your equilibrium point is by definition unstable.
Or in other words, in your case, there exists an $\epsilon>0$, such that for all $\delta>0$, $||x(0) - x^{*}||<\delta$ and $||x(t)-x^{*}||>\epsilon$.


TITLE: What trig identities must one know to derive the others?
QUESTION [2 upvotes]: My TA told me in problem section one day that every trig identity could be derived from just 2: the Pythagorean identity and the double-angle identity (or he might have said the half-angle identity).  I'm a bit dubious that every trig identity could be derived from just these two.  What would you say the minimum number of identities one must know to say derive every identity on the wikipedia page?
If it is possible to derive all of the identities from just 2, can anyone recommend a source that takes that approach?

REPLY [2 votes]: I think that Apostol's Calculus book might contain this idea in detail:
If you are willing to believe that sine and cosine are continuous, and have proved that a continuous function on a dense subset of an interval has a unique continuous extension to the interval, then once you know
(1) $\sin(0) = 0$, $\cos(0) = 1$, $\sin(\pi/2) = 0$; $\cos(x) > 0$ for $x \in [0, \pi/2)$,
(2) $\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)$ for all $a, b$, and
(3) $\sin(-x) = -\sin(x)$ for all $x$.
you can derive $1 = \cos^2(a) + \sin^2(a)$ by setting $b = a$ in the second formula.
You then can find that cosine is even (set $a = 0$).
Assumption 3 may not be necessary, but I confess, I forget how to show that the sine is odd without it.
Then you can set $b = -a$ to get
$$cos(2a) = \cos^2(a) - \sin^2(a) = 2\cos^2(a) - 1.$$
Applying this to $a = x/2$, you get
$$
\cos(x) = 2\cos^2(x/2) - 1 \\
\cos(x/2) = \sqrt{\frac{\cos(x) + 1}{2}}
$$
from this, you can determine cosine of all numbers of the form $\frac{\pi}{2} \frac{1}{2^n}$; using the addition formula, you can determine cosine at all points of the form
$$\frac{\pi}{2} \frac{k}{2^n}$$
where $k$ is an integer. These form a dense subset of the interval $[0, \pi/2]$.
You can then also show that for $x$ small, $\sin x$ is also small, so that (using the addition/subtraction formulas) cosine is continuous on this dense subset; it therefore has a unique continuous extension to $[0, \pi/2]$. The same goes for sine, and you're on your way.
@Semiclassical suggested, in comments above, that an addition formula, together with $\sin^2 + \cos^2 = 1$, might suffice, but that a half-angle formula might not. The conjecture about half-angle formulas is correct, as the following shows:
Let
\begin{align}
Sin(x) = \begin{cases}
\sin(x) & x = \frac{p}{2^k} \pi & \text{for $p, k, \in \mathbb  Z$} \\
0 & \text{otherwise}
\end{cases}\\
Cos(x) = \begin{cases}
\cos(x) & x = \frac{p}{2^k} \pi & \text{for $p, k, \in \mathbb  Z$} \\
0 & \text{otherwise}
\end{cases}
\end{align}
Then $Sin$ and $Cos$ satisfy $Sin^2 + Cos^2 = 1$ and the half-angle formulas, but they are not the same functions as $\sin$ and $\cos$, and hence need not satisfy the other formulas. In particular, they fail to satisfy the addition formula.


TITLE: If $f: X \longrightarrow Y$ is uniformly continuous and $(x_n)$ is Cauchy in $X$, then $(f(x_n))$ is Cauchy in $Y$
QUESTION [1 upvotes]: $1)$ If $(x_n)$ is Cauchy in $(X,d)$, then $x_n \longrightarrow x_0$. Or, equivalently, $\forall \epsilon > 0, \exists N \in \mathbb{N}$ s.t: $d(x_n,x_0) < \epsilon , \forall n \geq N$.
$2)$ Defn: A function $f: X \longrightarrow Y$ is called uniformly continuous if, $\forall \epsilon > 0, \exists \delta > 0$ s.t: $\rho(f(x),f(y)) < \epsilon, \forall x,y \in X$ satisfying $d(x,y) < \delta$.
Proof: It suffices to show that $(f(x_n))$ converges to some $f(x_0)$ in $(Y,\rho)$. Indeed, since $(x_n)$ is Cauchy, it converges in $(X,d)$ and so $\forall \delta > 0, \exists N \in \mathbb{N}$ s.t: $d(x_n,x_0) < \delta$, whenever $n \geq N$. So fix $\delta > 0$. Then, in view of the uniform continuity of $f$, we have that $\forall \epsilon > 0, \exists \delta > 0$ s.t: $\rho(f(x_n),f(x_0)) < \epsilon, \forall x_n,x_0 \in X$ satisfying $d(x_n,x_0) < \delta$.
In other words, $(f(x_n)) \longrightarrow f(x_0)$, and so $(f(x_n))$ is a Cauchy sequence in $(Y,\rho)$ since it converges. QED.
How did I do?

REPLY [3 votes]: Let $\epsilon$ be a positive real number. Thus exist $\delta>0$ such that $$d(x_{n},x_{m})<\delta\Rightarrow\rho(f(x_{n})f(x_{m}))<\epsilon.$$ Since $(x_{n})$ is Cauchy, there is $n_{0}\in\mathbb{N}$ such that $$n,m>n_{0}\Rightarrow d(x_{n},d_{m})<\delta.$$ It follow that $$n,m>n_{o}\Rightarrow\rho(f(x_{n}),f(x_{m}))<\epsilon$$ that is, $(f(x_{n}))$ is Cauchy.


TITLE: Could Cauchy sequence have no accumulation points?
QUESTION [0 upvotes]: I started a course on Analysis last month, and we have a question "How many accumulation points could Cauchy sequence possibly have?". I will take this example:
$$a_n= ({1 + {1\over n}})^n$$
And examine it in it metric space $\mathbb Q$.
From the textbook I see that every Cauchy sequence in metric space X is bounded, therefore by BW theorem it must have at least one accumulation point(in this case it's $e$ ). But when I go back in my textbook I see that accumulation point has to be the element of given metric space X (in this case it's $\mathbb Q$). It's obvious that this sequence is Cauchy, but I can't see that it has accumulation point in $\mathbb Q$.
So can someone explain to me what am I doing wrong?

REPLY [2 votes]: Yes, and your example is a good exemple. Your sequence is strictly increasing and do not converge, therefore it can't have accumulation point. But if the space is complete (for example $\mathbb R$) it's impossible since in a complete space every Cauchy sequence converge (and thus has a unique accumulation point).
Notice that $$\text{BW true } \iff\text{ the space is  complete},$$
since $\mathbb Q$ is not complete, BW is not useable.


TITLE: Do we only get conics when we do the Dupin indicatrix procedure?
QUESTION [0 upvotes]: Calculation of the Dupin Indicatrix involves taking a plane close by to the tangent plane at a point on the surface and seeing the curve intersected. This curve intersected, depending on the type of conic, we say the point which we started with is elliptical, hyperbolic , parabolic etc.
Question: Can Dupin indicatrix classify any point on any surface? i.e: Will the curve produced by this procedure ALWAYS be a conic? How to prove?

REPLY [1 votes]: Assume a regular surface $S$ with the parametrization
\begin{equation}
    \sigma: \left(u,v\right) \longmapsto \left(\sigma_1,\sigma_2,\sigma_3\right).
\end{equation}
Furthermore, let $N$ be the unit normal to $S$. If $T_p\,S$ is tangent to the surface at the point $p$ and $q = \sigma(u + du, v + dv)$ is another point on the surface, then the distance from $q$ to $T_p\,S$ is:
\begin{equation}
    h := \langle\,q - p\,,\,N\,\rangle.
    \label{distance:to:tangent:plane}
\end{equation}
Note that in this way $q - p$ will be a vector in $\mathbb{R}^3$. Having in mind that each $\sigma_i:U\subset\mathbb{R}^2 \longrightarrow \mathbb{R}$ and using the multi-variable Taylor expansion we get
\begin{equation*}
    q - p = \sigma\left(u + du, v + dv\right) - \sigma\left(u, v\right) =\\[1 cm]
\underbrace{ \left(\begin{array}{>{\displaystyle}c}
    \frac{\partial \sigma_1}{\partial u}\,du + \frac{\partial \sigma_1}{\partial v}\,dv\\[0.5 cm] \frac{\partial \sigma_2}{\partial u}\,du + \frac{\partial \sigma_2}{\partial v}\,dv\\[0.5 cm]
    \frac{\partial \sigma_3}{\partial u}\,du + \frac{\partial \sigma_3}{\partial v}\,dv
    \end{array}\right)}_{\star}
    +\frac{1}{2}\,\underbrace{
    \left(\begin{array}{>{\displaystyle}c}
    \frac{\partial^2\sigma_1}{\partial u^2}\,du^2 + 2\,\frac{\partial^2\sigma_1}{\partial u\,\partial v}\,du\,dv + \frac{\partial^2\sigma_1}{\partial v^2}\,dv^2\\[0.5 cm]
    \frac{\partial^2\sigma_2}{\partial u^2}\,du^2 + 2\,\frac{\partial^2\sigma_2}{\partial u\,\partial v}\,du\,dv + \frac{\partial^2\sigma_2}{\partial v^2}\,dv^2\\[0.5 cm]
    \frac{\partial^2\sigma_3}{\partial u^2}\,du^2 + 2\,\frac{\partial^2\sigma_2}{\partial u\,\partial v}\,du\,dv + \frac{\partial^2\sigma_3}{\partial v^2}\,dv^2
    \end{array}\right)}_{\star\star}
    + \dots.
\end{equation*}
Now, in the light of the above equation, $h$ simplifies to ($\star$ disappears due to $N$ being normal to $T_p\,S$):
\begin{equation}
    h = \frac{1}{2}\left(\,\langle\,N\,,\,\sigma_{uu}\,\rangle\,du^2 + \langle\,N\,,\,\sigma_{uv}\,\rangle\,du\,dv + \langle\,N\,,\,\sigma_{vv}\,\rangle\,dv^2\right) + \ldots = \frac{1}{2}\,\mathbf{II}\,(u,v) + \ldots
\end{equation}
Neglecting the higher order terms results in
\begin{equation}
    2\,h = \mathbf{II}\,(u,v).
    \label{eq:dis:2nd:fundamental:form}
\end{equation}
Now, if on $T_p\,S$ we choose the basis ${e_1}$ and ${e_2}$ such that they are the eigenvectors of $dN_p$ then this is equal to thinking of the surface (locally) to be parameterized by curvature lines ($F = f = 0$). This results in $k_1 = e/E$ and $k_2 = g/G$. Hence, $h$ simplifies to
\begin{equation*}
    2\,h = \kappa_1{E}\,du^2 + \kappa_2{G}\,dv^2,
\end{equation*}
where $\kappa_i$ are the principal curvatures. Now, thinking of $\sqrt{E}du$ and $\sqrt{G}dv$ as $x$ and $y$ then you'll see the result is a quadric unless if one of your principal curvatures vanish (parabolic points) or both of them vanish, which gives a planar point.
Now, having in mind that the Dupin indicatrix is the set of points $(x,y)$ fitting into the equation
$$\kappa_1 x^2 + \kappa_2 y^2 = \pm 1$$
you can see the similarity between the two equations. Therefore, the Dupin indicatrix tells you how the surface "behaves" around the point under consideration. You can use this on all regular surfaces to know the local behavior of the surface around a point.


TITLE: Prove that $|f(x)|\leq (Mx^2)/2$
QUESTION [3 upvotes]: It is given that $|f''(x)|<=M$ for all $x$, and $f(0)=f'(0)=0$. Then prove that $|f(x)|\leq\frac{Mx^2}{2}$ for all $x$.
What I think about was by using the Mean Value Theorem twice, and then I will get $|f(x)|\leq Mx^2$ with no $\frac{1}{2}$. And I do not know if I use the MVT twice whethere there would be any mistake.
Is there any way other than the MVT to solve this problem? What I also considered were the Fundamental Theorem of Calculus and the Taylor's Theorem. But still, feel not too good about them.
Anyone offers a hint?

REPLY [2 votes]: It is an immediate consequence of Taylor's theorem.
You have $f(x) = f(0) + f'(0)x + \int_0^x f''(t) (x-t)dt$. Using the facts that $f(0) = f'(0) = 0$ and $|f''(t)| \leq M$, you have $|f(x)| \leq M \int_0^x (x-t) dt = \frac{1}{2} M x^2$, as required.


TITLE: I stuck on this step on the chapter about Lagrange multipliers.
QUESTION [2 upvotes]: As I understood, to find critical solutions for some system of equations, we assume if giving a function f() restricted to some level set of other functions gi() with the same domain and range, then its gradient vector will be perpendicular to the tangent vector of the level set at a critical points of f() and can be expressed as a linear combination of gradient vectors of functions gi() at that point. Also to get particular results, we should have enough information for our linear equation.
The book suggests to just put our linear combinations to restriction functions and solve for lambdas. That is the part that I'm struggling. I can't describe the meaning of this step and why we can acquire the answer without even using information from the distance function.

REPLY [2 votes]: I can't describe the meaning of this step and why we can acquire the answer without even using information from the distance function.

It is using information about $f$. we have $\nabla f = (2v, 2w, 2x, 2y, 2z)$. Then the book "absorbs the $2$ into the $\lambda$'s" and just writes $(v, w, x, y, z)$ instead.
The general formula they're using here is:
$$ \nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2 .$$


TITLE: What is global Lorentz transformation and what is local Lorentz transformation?
QUESTION [2 upvotes]: I will consider $\textbf{spacetime}$  as $(M,\eta)$ where $M$ is a four dimensional $\textbf{manifold}$ and $\eta$ the metric which in this coordinates
 $$
\begin{align*}
  x \colon M &\longrightarrow \mathbb{R}^4\\
  p &\mapsto x(p)=(x_0,x_1,x_2,x_3).
\end{align*}
$$
 is given by
$$\eta=dx^0\otimes dx^0-dx^1\otimes dx^2-dx^2\otimes dx^1-dx^3\otimes dx^3 \tag1$$
An $\textbf{observer}$ is a worldline $\gamma$ with together with a choice of basis
$
O=v_{\gamma,\gamma(\lambda)} \equiv e_0(\lambda) , e_1(\lambda), e_2(\lambda), e_3(\lambda)
$
of each $T_{\gamma(\lambda)}M$ where the observer worldline passes, if
$$
\eta(e_a(\lambda), e_b(\lambda)) = \eta_{ab} = \left[ \begin{matrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{matrix} \right]_{ab} \tag2
$$
$v_{\gamma,\gamma(\lambda)}$  is the tangent vector of the the curve $\gamma$ at the point $\gamma(\lambda)$
In text books i've found three definition of $\textbf{Lorentz transformation} \quad \Lambda$

$\Lambda \colon \mathbb{R}^4 \longrightarrow \mathbb{R}^4$ is a group of coordinate transformations that leave eq.1 in the same form ,that is $\Lambda \cdot x(p)=y(p)=(y_0,y_1,y_2,y_3)$ such that in this coordinate  $$\eta=dy^0\otimes dy^0-dy^1\otimes dy^2-dy^2\otimes dy^1-dy^3\otimes dy^3 $$
$\Lambda \colon M \longrightarrow M$ a Spacetime diffeomorphism such that $\Lambda_* \eta=\eta$ where $\Lambda_* \eta$ is the pullback of the metric $\eta$
$\Lambda \colon T_pM \longrightarrow T_pM$ such that $\Lambda O=O'=e'_0(\lambda) , e'_1(\lambda), e'_2(\lambda), e'_3(\lambda)$ satisfy the eq.2 that is
$$
\eta(e'_a(\lambda), e'_b(\lambda)) = \eta_{ab} = \left[ \begin{matrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{matrix} \right]_{ab}
$$

My question is which is of these transformation is global Lorentz transformation and which is local?

REPLY [0 votes]: The basic definition of the Lorentz transformation is, given a vector space $V$ equipped with a Minkowski metric $\eta$, it is the group that leaves the norm invariant (in other words, definition 1).
In a spacetime, every tangent space can be considered as a copy of Minkowski space, ie for every $p \in M$, $T_p M \cong V$, since by Sylvester's law, we can always put the metric tensor $g_p$ at that point in the appropriate form by a change of basis (that basis being an orthonormal basis $\{ e_\mu \}$). Then at each point, you can perform a Lorentz transform of that basis.
Diffeomorphisms on a manifold are a rather large class, but there exists a subset of diffeomorphisms such that, if $\phi \in \mathrm{Diff}(M)$, the pushforward on a vector  at $p$, $\phi^* v$, corresponds to a Lorentz transform. Just by having say
\begin{equation}
\frac{\partial x^\mu(p)}{\partial y^\nu(p)} \in \mathrm{SO}(3,1)
\end{equation}
Locally, in the Riemann/Fermi coordinates, this is roughly equivalent to a Lorentz transform, since the normal neighbourhood is diffeomorphic to $T_pM$.


TITLE: Prove that $\frac{a + b + c}{b - a} > 3$ with $ab < b^2 < 4ca$.
QUESTION [2 upvotes]: $a$, $b$, $c$ are three positives such that $ab < b^2 < 4ca$. Prove that $$\large \dfrac{a + b + c}{b - a} > 3.$$

I can't think of a way to get around this problem. Although I can see that based on the given condition, $ax^2 + bx + c = 0$ has no roots, which adds almost no information whatsoever.
If you have written an answer below, thanks for that!

REPLY [1 votes]: $\displaystyle a+b+c>a+b+\frac{b^2}{4a}=\bigg(4a+\frac{b^2}{4a}\bigg)+b-3a\geq 3(b-a)$
$$\displaystyle \frac{a+b+c}{b-a}>3$$


text:
\section{Multivariate extension}
\label{sec:multi-d-odes}

The GMRK model can be extended to the multivariate case analogously to
Runge-Kutta methods.  All equations in the RK framework also work with
vector-valued function values, and all derivations presented in the
paper and in this supplement carry over to the non-scalar case without
modification: Consider dimension $j \in \{1, \dotsc, N\}$. The
projected outputs are the same as if $j$ were an independent
one-dimensional problem, which can be modeled with a separate Gaussian
process. For a joint notation, vectorize the $N$ dimensions with a
Kronecker product: if $k(t,t')$ is an one-dimensional covariance
function, the function
\begin{equation}
\bar{k}(\vec{t},\vec{t'}) = \mat{D}_{ij} k(t_i,t'_j),
\end{equation}
where $\mat{D}$ is an $N\times N$ positive semi-definite matrix,
defines a covariance function over $N$ dimensions, if $\vec{t}$ and
$\vec{t'}$ are $N$ dimensional. Choosing $\mat{D} = \Id$ results in a
$N$-dimensional GP model where output dimensions are independent of
each other as required.

config: arxiv


TITLE: Predicting the Nth combination in sequence of A and B
QUESTION [0 upvotes]: First I need a way of generating every possible combination of As and Bs in an array of 2048 items.
Once I have that would it be possible to row N without action generating every row between 0 and N?

REPLY [0 votes]: [Edited] If I understand your question correctly, all you have to do is count from 0 to 2^2048 -1 in binary, then replace 0 and 1 in the binary string with A and B, respectively.  (I'll be waiting in the car while you do this.)  To get row N, just convert N to binary, assuming you start counting the rows with 0.