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% In principle, this file can be redistributed and/or modified under
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\title[\ \kern-190pt\rlap{\blank{J.-P. Demailly, Sem.\ di Algebra e Geometria, October 7, 2020}}\kern183pt\rlap{\blank{Griffiths conjecture on the positivity of vector bundles}}\kern178pt\llap{\blank{\framenumbering~}}\kern-10pt]
% (optional, use only with long paper titles)
{Hermitian-Yang-Mills approach\\
to the conjecture of Griffiths on the\\
positivity of ample vector bundles}

%% \subtitle{Presentation Subtitle} % (optional)

\author[] % (optional, use only with lots of authors)
{\vskip-3mm Jean-Pierre Demailly\vskip-3mm}

\institute[]{Institut Fourier, Université Grenoble Alpes\ \ \&\ \ Académie des Sciences de Paris}
% - Use the \inst command only if there are several affiliations.
% - Keep it simple, no one is interested in your street address.

\date[]% (optional)
{Seminario di Algebra e Geometria\\
Dipartimento di Matematica, Sapienza Università di Roma\\
October 7, 2020, 16:30 pm}

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\begin{document}
%%\def\pause{}

% Delete this, if you do not want the table of contents to pop up at
% the beginning of each subsection:
%%\AtBeginSubsection[]
%%{
%% \begin{frame}<beamer>
%%    \frametitle{Outline}
%%    \tableofcontents[currentsection,currentsubsection]
%%  \end{frame}
%%}


% If you wish to uncover everything in a step-wise fashion, uncomment
% the following command: 

%\beamerdefaultoverlayspecification{<+->}

\begin{frame}
  \pgfdeclareimage[height=2cm]{uga-logo}{logo_UGA}
  \pgfuseimage{uga-logo}
  \pgfdeclareimage[height=2cm]{acad-logo}{academie_logo2}
  \pgfuseimage{acad-logo}
  \vskip-12pt
  \titlepage
\end{frame}

%%\begin{frame}
%%  \frametitle{Outline}
%%  \tableofcontents
%% You might wish to add the option [pausesections]
%%\end{frame}


% Since this a solution template for a generic talk, very little can
% be said about how it should be structured. However, the talk length
% of between 15min and 45min and the theme suggest that you stick to
% the following rules:  

% - Exactly two or three sections (other than the summary).
% - At *most* three subsections per section.
% - Talk about 30s to 2min per frame. So there should be between about
%   15 and 30 frames, all told.

\begin{frame}
\frametitle{Positive and ample vector bundles}
\vskip-5pt  
Let $X$ be a projective $n$-dimensional manifold and $E\to X$
a holomorphic vector bundle of rank $r\ge 1$.\pause
\begin{block}{Ample vector bundles}
$E\to X$ is said to be \alert{ample in the sense of Hartshorne} if the
associated line bundle $\cO_{\bP(E)}(1)$ on $\bP(E)$ is ample.\pause\vskip3pt
By Kodaira,
this is equivalent to the existence  of a\\
\alert{smooth hermitian metric
on $\cO_{\bP(E)}(1)$ with positive curvature} (equivalently, a
negatively curved Finsler metric on $E^*$).
\end{block}
\pause\vskip-4pt
\begin{block}{Chern curvature tensor}
This is $\Theta_{E,h}=i\nabla_{E,h}^2\in C^\infty
(\Lambda^{1,1}T^*_X\otimes\Hom(E,E))$, which can be written\vskip1pt
\alert{\centerline{$\displaystyle
\Theta_{E,h}=i\sum_{1\le j,k\le n,\,1\le\lambda,\mu\le r}c_{jk\lambda\mu}dz_j\wedge
d\ol z_k\otimes e_\lambda^*\otimes e_\mu
$}}\vskip2pt
in terms of an orthonormal frame $(e_\lambda)_{1\le\lambda\le r}$ of $E$.
\end{block}

\end{frame}

\begin{frame}
\frametitle{Positivity concepts for vector bundles}
\vskip-7pt  
\begin{block}{Griffiths and Nakano positivity}
One looks at the associated quadratic form on $S=T_X\otimes E$\vskip4pt
\alert{\centerline{$\displaystyle  
\widetilde\Theta_{E,h}(\xi\otimes v):=
\langle\Theta_{E,h}(\xi,\ol\xi)\cdot v,v\rangle_h=
\sum_{1\le j,k\le n,\,1\le\lambda,\mu\le r}
c_{jk\lambda\mu}\xi_j\ol\xi_k v_\lambda\ol v_\mu.
$}}
\pause\vskip0pt
Then $E$ is said to be
\begin{itemize}
\item Griffiths positive (Griffiths 1969) if at any point $z\in X$
\vskip1pt
\alert{\centerline{$\displaystyle  
\widetilde\Theta_{E,h}(\xi\otimes v)>0,\quad\forall 0\ne\xi\in T_{X,z},~
\forall 0\ne v\in E_z$}}\pause
\item Nakano positive (Nakano 1955) if at any point $z\in X$
\vskip1pt
\alert{\centerline{$\displaystyle\widetilde\Theta_{E,h}(\tau)=
\sum_{1\le j,k\le n,\,1\le\lambda,\mu\le r}
c_{jk\lambda\mu}\tau_{j,\lambda}\overline\tau_{k,\mu}
>0,\quad\forall 0\ne\tau\in T_{X,z}\otimes E_z.$}}
\end{itemize}
\end{block}
\vskip-6pt\pause
\begin{block}{Easy and well known facts}
\alert{\centerline{$E$ Nakano positive $\Rightarrow$
$E$ Griffiths positive $\Rightarrow$ $E$ ample.}}
\end{block}
\vskip-2pt
\pause In fact $E$ Griffiths positive
$\Rightarrow$ $\cO_{\bP(E)}(1)$ positive.
\end{frame}

\begin{frame}
\frametitle{Dual Nakano positivity -- a conjecture}
\vskip-7pt  
\begin{block}{Curvature tensor of the dual bundle $E^*$}
\centerline{$\displaystyle
\Theta_{E^*,h}=-{}^T\Theta_{E,h}=-
\sum_{1\le j,k\le n,\,1\le\lambda,\mu\le r}c_{jk\mu\lambda}dz_j\wedge
d\ol z_k\otimes (e^*_\lambda)^*\otimes e^*_\mu.
$}
\end{block}
\pause\vskip-3pt
\begin{block}{Dual Nakano positivity}
One requires\vskip2pt
\alert{\centerline{$\displaystyle
-\widetilde\Theta_{E^*,h}(\tau)=
\sum_{1\le j,k\le n,\,1\le\lambda,\mu\le r}
c_{jk\mu\lambda}\tau_{j\lambda}\ol\tau_{k\mu}>0
,\quad\forall 0\ne\tau\in T_{X,z}\otimes E^*_z.$}}  
\end{block}\pause\vskip-2pt
Dual Nakano positivity is \alert{clearly stronger} than
Griffiths positivity.\\
Also, it is better behaved than Nakano positivity,
\hbox{e.g.\kern-15pt}\\
$E$ dual Nakano positive\\
~~~~$\Rightarrow$ any quotient $Q=E/S$ is also dual
Nakano positive.\pause
\begin{block}{(Very speculative) conjecture}
Is it true that \alert{$E$ ample $\Rightarrow$ $E$ dual Nakano positive}~?
\end{block}
\end{frame}


\begin{frame}
\frametitle{Brief discussion around this positivity conjecture}
\vskip-7pt  
\begin{block}{If true, Griffiths conjecture would follow:}
\centerline{\alert{$E$ ample $\Leftrightarrow$ $E$ dual Nakano positive
$\Leftrightarrow$ $E$ Griffiths positive}.}
\end{block}
\pause\vskip-5pt
\begin{block}{Remark}
\centerline{\alert{$E$ ample $\not\Rightarrow$ $E$ Nakano positive}, in fact}
\vskip3pt
\centerline{\alert{$E$ Griffiths positive $\not\Rightarrow$ $E$ Nakano
positive}.}
\end{block}
\pause\vskip-3pt
For instance, $T_{\bP^n}$ is easy shown to be ample and Griffiths
positive for the Fubini-Study metric, but it is
\alert{not Nakano positive}.\pause\
Otherwise the Nakano vanishing theorem would then yield\vskip5pt
\centerline{$
H^{n-1,n-1}(\bP^n,\bC)=
H^{n-1}(\bP^n,\Omega^{n-1}_{\bP^n})=
H^{n-1}(\bP^n,K_{\bP^n}\otimes T_{\bP^n})=0 ~~!!!
$}\pause\vskip5pt
Let us mention here that there are already known subtle
relations between ampleness, Griffiths and Nakano positivity
are known to hold -- for instance, B.~Berndtsson has proved that
the ampleness of $E$ implies the Nakano positivity of
\alert{$S^mE\otimes\det E$} for every~$m\in\bN$.
\end{frame}

\begin{frame}
\frametitle{``Total'' determinant of the curvature tensor}  
If the Chern curvature tensor $\Theta_{E,h}$ is \alert{dual
Nakano positive}, then one can introduce
the $(n\times r)$-dimensional determinant of the corres\-ponding
Hermitian quadratic form on $T_X\otimes E^*$
\alert{$$
\det\nolimits_{T_X\otimes E^*}({\,}^T\Theta_{E,h})^{1/r}
:=\det(c_{jk\mu\lambda})_{(j,\lambda),(k,\mu)}^{1/r}\,
i dz_1\wedge d\ol z_1\wedge\,...\,\wedge i dz_n\wedge d\ol z_n.
$$}\pause
This $(n,n)$-form does not depend on the choice of coordinates $(z_j)$
on~$X$, nor on the choice of the orthonormal frame $(e_\lambda)$ on $E$.\pause
\begin{block}{Basic idea}
Assigning a ``matrix Monge-Amp\`ere equation''
\alert{$$
\det\nolimits_{T_X\otimes E^*}({\,}^T\Theta_{E,h})^{1/r}=f>0
$$}%
where $f$ is a positive $(n,n)$-form, may enforce the dual Nakano
positivity of $\Theta_{E,h}$ if that assignment is combined with a
continuity technique from an initial starting point where posi\-tivity
is known.
\end{block}
\end{frame}

\begin{frame}
\frametitle{Continuity method (case of rank 1)}
\vskip-3pt
For $r=1$ and $h=h_0e^{-\varphi}$, we have\vskip5pt
\alert{\centerline{$
{}^T\Theta_{E,h}=\Theta_{E,h}=-i\partial\ol\partial\log h
=\omega_0+i\partial\ol\partial \varphi,$}}\pause\vskip5pt
and the equation reduces to a standard
Monge-Amp\`ere equation\vskip-18pt
\alert{$$
(\Theta_{E,h})^n=(\omega_0+i\partial\ol\partial\varphi)^n=f.
\leqno(*)
$$}\pause
If $f$ is given and independent of $h$, Yau's theorem guaran\-tees
the existence of a unique solution $\theta=\Theta_{E,h}>0$, provided
$E$ is an ample line bundle and $\int_Xf=c_1(E)^n$.\pause\vskip4pt
When the right hand side $f=f_t$ of $(*)$ varies smoothly with respect to some parameter~$t\in[0,1]$, one then gets a smoothly varying solution\vskip4pt
\alert{\centerline{$
\Theta_{E,h_t}=\omega_0+i\partial\ol\partial\varphi_t>0,$}}\vskip6pt
and the positivity of $\Theta_{E,h_0}$ forces the
positivity of $\Theta_{E,h_t}$ for all $t$.
\end{frame}

\begin{frame}
\frametitle{Undeterminacy of the equation}
\vskip-5pt
Assuming $E$ to be ample of rank $r>1$, the equation
\alert{$$
\det\nolimits_{T_X\otimes E^*}({\,}^T\Theta_{E,h})^{1/r}=f>0\leqno(**)
$$}%
becomes underdetermined, as the real rank of the space of hermitian matrices
$h=(h_{\lambda\mu})$ on $E$ is equal to $r^2$, while $(**)$ provides only
1 scalar equation.\pause\vskip4pt
(Solutions might still exist, but lack uniqueness and
a priori \hbox{bounds.)\kern-15pt}\pause\vskip-2pt
\begin{block}{Conclusion}
In order to recover a well determined system
of equations, one needs an additional
\alert{``matrix equation'' of rank $(r^2-1)$}.
\end{block}\pause\vskip-4pt
\begin{block}{Observation 1 (from the Donaldson-Uhlenbeck-Yau theorem)}
Take a Hermitian metric $\eta_0$ on
$\det E$ so that $\omega_0:=\Theta_{\det E,\eta_0}>0$.
If~$E$ is $\omega_0$-polystable, $\exists h$ Hermitian metric
$h$ on $E$ such that\vskip6pt
\centerline{\alert{
$\omega_0^{n-1}\wedge\Theta_{E,h}={1\over r}\,\omega_0^n\otimes \Id_E
$}~~(Hermite-Einstein equation, slope ${1\over r}$).}
\end{block}
\end{frame}

\begin{frame}
\frametitle{Resulting trace free condition}
\vskip-7pt  
\begin{block}{Observation 2}
The trace part of the above Hermite-Einstein equation is
``automatic'', hence the equation is equivalent to the trace free condition
\vskip-6pt
\centerline{\alert{
$\omega_0^{n-1}\wedge\Theta_{E,h}^\circ=0,$}}\vskip4pt
when decomposing any endomorphism $u\in\Herm(E,E)$ as
\vskip6pt
\centerline{\alert{
$u=u^\circ +{1\over r}\Tr(u)\Id_E\in\Herm^\circ(E,E)\oplus\bR\Id_E,~~
\tr(u^\circ)= 0.    
$}}\vskip4pt
\end{block}\pause\vskip-5pt
\begin{block}{Observation 3}
The trace free condition is a matrix equation of \alert{rank $(r^2-1)$}~!!!
\end{block}\pause\vskip-5pt
\begin{block}{Remark}  
In case \alert{$\dim X=n=1$}, the trace free condition means that $E$ is
\alert{projectively
flat}, and the Umemura proof of the Griffiths conjecture proceeds
exactly in that way, using the
fact that the graded pieces of the Harder-Narasimhan filtration are
projectively flat.
\end{block}
\end{frame}

\begin{frame}
\frametitle{Towards a ``cushioned'' Hermite-Einstein equation}
\vskip-5pt  
In general, one cannot expect $E$ to be $\omega_0$-polystable,
but Uhlenbeck-Yau have shown that there always exists a
smooth solution $q_\varepsilon$ to a certain
\alert{``cushioned'' Hermite-Einstein equation}.\pause\vskip4pt
To make things more precise, let $\Herm(E)$ be the space
of Hermitian (non necessarily positive) forms on~$E$.
Given a reference Hermitian metric $H_0>0$, let $\Herm_{H_0}(E,E)$
be the space of $H_0$-Hermitian endomorphisms $u\in \Hom(E,E)$; denote by
\vskip6pt
\centerline{$\displaystyle
\Herm(E)\mathop{\to}\limits^{\simeq~} \Herm_{H_0}(E,E),\quad
q\mapsto\widetilde q~~\hbox{s.t.}~~
q(v,w)=\langle\kern1pt\widetilde q\kern1pt(v),w\rangle_{H_0}
$}\vskip5pt
the natural isomorphism.\pause\ Let also\vskip6pt
\alert{\centerline{$
\Herm^\circ_{H_0}(E,E)=\big\{q\in\Herm_{H_0}(E,E)\,;\;\tr(q)=0\big\}
$}}\vskip6pt
be the subspace of ``trace free'' Hermitian endomorphisms.\pause\vskip4pt
In the sequel, we fix $H_0$ on $E$ such that\vskip4pt
\centerline{\alert{$\Theta_{\det E,\det H_0}=\omega_0>0$.}}
\end{frame}

\begin{frame}
\frametitle{A basic result from Uhlenbeck and Yau}
\vskip-7pt
\begin{block}{Uhlenbeck-Yau 1986, Theorem~3.1}
For every $\varepsilon>0$, there \alert{always exists} a (unique)
smooth Hermitian metric $q_\varepsilon$
on $E$ such that\vskip3pt
\alert{\centerline{$\displaystyle
\omega_0^{n-1}\wedge\Theta_{E,q_\varepsilon}=\omega_0^n\otimes
\bigg({1\over r}\Id_E-\varepsilon\,\log\widetilde q_\varepsilon\bigg),
$}}\vskip6pt
where $\widetilde q_\varepsilon$ is computed with respect to~$H_0$,
and $\log g$ denotes the logarithm of a positive Hermitian endomorphism~$g$.
\end{block}\pause
The reason is that the term $-\varepsilon\,\log\widetilde q_\varepsilon$
is a ``friction term'' that prevents the explosion of the a priori
estimates, similarly what happens for Monge-Amp\`ere equations
$(\omega_0+i\partial\ol\partial\varphi)^n=e^{\varepsilon\varphi+f}\omega_0^n$.
\pause\vskip4pt
The above matrix equation is equivalent to prescribing
$\smash{\det q_\varepsilon}=\det H_0$ and the trace~free equation of rank
$(r^2-1)$\vskip6pt
\centerline{\alert{$
\omega_0^{n-1}\wedge\Theta^\circ_{E,q_\varepsilon}=
-\varepsilon\,\omega_0^n\otimes\log\widetilde q_\varepsilon.
$}}
\end{frame}

\begin{frame}
\frametitle{Search for an appropriate evolution equation}
\vskip-6pt  
\begin{block}{General setup}  
In this context, given $\alpha>0$ large enough,
it is natural to search for a time dependent family of
metrics $h_t(z)$ on the fibers $E_z$ of $E$, $t\in [0,1]$, satisfying a 
generalized Monge-Amp\`ere equation
\alert{$$
~~\det\nolimits_{T_X\otimes E^*}\big({\,}^T\Theta_{E,h_t}+
(1-t)\alpha\,\omega_0\otimes\Id_{E^*}\big)^{1/r}=
f_t\,\omega_0^n,~~~f_t>0,\leqno(D)
$$}\pause%
and trace free, rank $r^2-1$, Hermite-Einstein conditions\vskip5pt
\alert{$\displaystyle(T)~~~
\omega_t^{n-1}\wedge\Theta^\circ_{E,h_t}=g_t$}\vskip7pt
with smoothly varying families of functions $f_t\in C^\infty(X,\bR)$,
Hermitian metrics $\omega_t>0$ on~$X$ and sections\vskip4pt
\alert{\centerline{$
g_t\in C^\infty(X,\Lambda^{n,n}_\bR T_X^*\otimes\Herm_{h_t}^\circ(E,E))$,~~
$t\in[0,1]$.}}
\end{block}\pause
Observe that this is a determined (not overdetermined!) system.
\end{frame}

\begin{frame}
\frametitle{Choice of the initial state ($t=0$)}
We start with the Uhlenbeck-Yau
solution $h_0=q_\varepsilon$ of of the ``cushioned'' trace free
Hermite-Einstein equation, so that $\det h_0=\det H_0$,
and take $\alpha>0$ so large that\vskip7pt
\alert{\centerline{
${\,}^T\Theta_{E,h_0}+\alpha\,\omega_0\otimes\Id_{E^*}>0$ in the sense
of Nakano.}}\pause\vskip7pt
If conditions $(D)$ and $(T)$ can be met for all $t\in[0,1]$,
thus without any explosion of the solutions $h_t$, we infer from $(D)$ that
\vskip7pt
\alert{\centerline{
${\,}^T\Theta_{E,h_t}+
(1-t)\alpha\,\omega_0\otimes\Id_{E^*}>0
\quad\hbox{in the sense of Nakano}$}}
\vskip7pt
for all $t\in[0,1]$.\pause\vskip5pt
\begin{block}{Observation}
At time $t=1$, we would then get a Hermitian
metric $h_1$ on $E$ such that \alert{$\Theta_{E,h_1}$ is dual Nakano
positive}~!!
\end{block}
\end{frame}

\begin{frame}
\frametitle{Possible choices of the right hand side}
\vskip-6pt
One still has the freedom of adjusting $f_t$, $\omega_t$
and $g_t$ in the general setup. There are in fact many
possibilities:\pause\vskip-1pt
\begin{block}{Proposition} Let $(E,H_0)$ be a smooth Hermitian
holomorphic vector bundle such that $E$ is ample and
$\omega_0=\Theta_{\det E,\det H_0}>0$.
Then the system of determinantal and trace free equations\vskip3pt
\hbox{\alert{$\displaystyle(D)~\,
\det\nolimits_{T_X\otimes E^*}\!\big({}^T\Theta_{E,h_t}+
(1-t)\alpha\,\omega_0\otimes\Id_{E^*}\big)^{1/r}\!=\,
F(t,z,h_t,D_zh_t)
$\kern-15pt}}\vskip5pt
\alert{$\displaystyle(T)~~
\omega_0^{n-1}\wedge\Theta^\circ_{E,h_t}=
G(t,z,h_t,D_zh_t,D^2_zh_t)\in\Herm^\circ(E,E)$}\vskip7pt
(where $F>0$), is a well determined system of PDEs.
\pause\vskip4pt
It is \alert{elliptic} whenever the symbol $\eta_{h}$ of the linearized operator
$u\mapsto DG_{D^2h}(t,z,h,Dh,D^2h)\cdot D^2u$ has an
Hilbert-Schmidt norm\vskip4pt
\centerline{\alert{$\displaystyle
\sup_{\xi\in T^*_X,|\xi|_{\omega_0}=1}\Vert\eta_{h_t}(\xi)\Vert_{h_t}\le
(r^2+1)^{-1/2}\,n^{-1}$}}\vskip-1pt
for any metric $h_t$ involved, e.g.\ if $G$ does not depend on $D^2h$.
\end{block}
\end{frame}

\begin{frame}
\frametitle{Proof of the ellipticity}
\vskip-5pt  
The (long, computational) proof consists of analyzing the linearized
system of equations, starting from the curvature tensor formula\vskip6pt
\alert{\centerline{
$\Theta_{E,h}=i\ol\partial(h^{-1}\partial h)=
i\ol\partial(\widetilde h^{-1}\partial_{H_0}\widetilde h)$,}}\vskip6pt
where $\partial_{H_0}s=H_0^{-1}\partial(H_0s)$ is the $(1,0)$-component of
the Chern connection on $\Hom(E,E)$ associated with~$H_0$ on~$E$.
\pause\vskip4pt
Let us recall that the ellipticity of an operator\vskip6pt
\alert{\centerline{$
P:C^\infty(V)\to C^\infty(W),~~~
f\mapsto P(f)=\sum_{|\alpha|\le m}a_\alpha(x)D^\alpha f(x)$}}\vskip6pt
means the invertibility of the principal symbol\vskip6pt
\alert{\centerline{
$\sigma_P(x,\xi)=\sum_{|\alpha|\le m}a_\alpha(x)\,\xi^\alpha\in \Hom(V,W)$}}
\vskip6pt
whenever $0\ne\xi\in T^*_{X,x}$.\pause\vskip4pt
For instance, on the torus $\bR^n/\bZ^n$, 
$f\mapsto P_\lambda(f)=-\Delta f+\lambda f$
has an~invertible symbol $\sigma_{P_\lambda}(x,\xi)=-|\xi|^2$, but
$P_\lambda$ is invertible only for~$\lambda>0$.
\end{frame}

\begin{frame}
\frametitle{A more specific choice of the right hand side}
\vskip-10pt
\begin{block}{Theorem} 
The elliptic differential system defined by\vskip-1pt
\hbox{\alert{$\displaystyle
\det\nolimits_{T_X\otimes E^*}\big({\,}^T\Theta_{E,h}+
(1-t)\alpha\,\omega_0\otimes\Id_{E^*}\big)^{1/r}=
\bigg({\det H_0(z)\over\det h_t(z)}\bigg)^\lambda\,a_0(z)$,}}\vskip0pt
\hbox{\alert{$\displaystyle
\omega_0^{n-1}\wedge\Theta_{E^\circ,h}=
-\varepsilon\,\bigg({\det H_0(z)\over\det h_t(z)}\bigg)^\mu\,
(\log\widetilde h^\circ)\,\omega_0^n$}}\vskip6pt
possesses an \alert{invertible
elliptic linearization} for $\varepsilon\ge\varepsilon_0(h_t)$ and
$\lambda\ge\lambda_0(h_t)(1+\mu^2)$,~ with $\varepsilon_0(h_t)$ and
$\lambda_0(h_t)$ large enough.
\end{block}\vskip-8pt\pause
\begin{block}{Corollary}
Under the above conditions, starting from the Uhlenbeck-Yau solution
$h_0$ such that $\det h_0=\det H_0$ at $t=0$, the PDE system \alert{still has
  a solution for $t\in [0,t_0]$} and $t_0>0$ small.\pause\
\alert{(What for $t_0=1$~?)}
\end{block}\pause\vskip-5pt
Here, the proof consists of analyzing the \alert{total symbol} of the
linearized operator, and the rest is just linear algebra.
\end{frame}

\begin{frame}
\frametitle{Monge-Amp\`ere volume for vector bundles}
\vskip-6pt
If $E\to X$ is an ample vector bundle of rank $r$ that is dual Nakano positive,
one can introduce its \alert{Monge-Amp\`ere volume} to be\vskip4pt
\alert{\centerline{$\displaystyle
\MAVol(E)=\sup_h\int_X\det\nolimits_{T_X\otimes E^*}\big(
(2\pi)^{-1}{\,}^T\Theta_{E,h}\big)^{1/r},
$}}\vskip6pt
where the supremum is taken over all smooth metrics $h$ on $E$ such that
${}^T\Theta_{E,h}$ is Nakano positive.\pause\vskip4pt
This supremum is always finite, and in fact\vskip-2pt
\begin{block}{Proposition} For any dual Nakano positive vector bundle $E$, 
one has\vskip6pt
\centerline{\alert{$\MAVol(E)\leq r^{-n}c_1(E)^n.$}}
\end{block}\pause\vskip-4pt
Taking $\omega_0=\Theta_{\det E}$, the proof is a consequence of
the inequality
$(\prod \lambda_j)^{1/nr}\leq {1\over nr}\sum\lambda_j$ between geometric and
arithmetic means, for the eigenvalues $\lambda_j$ of
$(2\pi)^{-1}{\,}^T\Theta_{E,h}$, after raising to power $n$.
\end{frame}

\begin{frame}
\frametitle{Concluding remarks}
\vskip-6pt
\begin{itemize}
\item Siarhei Finski (PostDoc at Institut Fourier right now) has observed
that the equality holds iff \alert{$E$ is projectively flat}.\pause
\item In the split case \alert{$E=\bigoplus_{1\le j\le r}E_j$ and
$h=\bigoplus_{1\le j\le r}h_j$}, the inequality reads\vskip4pt
\centerline{\alert{$\displaystyle
\Bigg(\prod_{1\le j \le r}c_1(E_j)^n\Bigg)^{1/r}\le r^{-n}c_1(E)^n,
$}}\vskip4pt\pause
with equality iff $c_1(E_1)=\cdots=c_1(E_r)$.\pause
\item In the split case, it seems natural to conjecture that\vskip4pt
\centerline{\alert{$\displaystyle
\MAVol(E)=\Bigg(\prod_{1\le j \le r}c_1(E_j)^n\Bigg)^{1/r},
$}}\vskip4pt
i.e.\ that the supremum is reached for split metrics $h=\bigoplus h_j$.\pause
\item The Euler-Lagrange equation for the maximizer is 4th order.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{The end}
\strut\vskip3mm
\centerline{\huge\bf Thank you for your attention}
\pgfdeclareimage[height=6.5cm]{CalabiYau}{CalabiYau}
\strut\kern2cm\pgfuseimage{CalabiYau}

\end{frame}

\end{document}

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