\magnification=1200
\input broc.mac

\def\than{\tha\noindent}
\def\rem#1|{\remarque\noindent{\petcap#1}|}
\def\proof{\noindent{\sl{Proof.}\ }}
\def\finpr{\hfill\hbox{
\vrule height 1.453ex  width 0.093ex  depth 0ex
\vrule height 1.5ex  width 1.3ex  depth -1.407ex\kern-0.1ex
\vrule height 1.453ex  width 0.093ex  depth 0ex\kern-1.35ex
\vrule height 0.093ex  width 1.3ex  depth 0ex}}

\let\ov=\overline
\let\la=\longrightarrow
\def\ld{{,\,\ldots\,,}}
\def\op{{\overline\partial}}
\let\wt=\widetilde
\def\rank{{\rm rank}\,}
\def\ricci{{\rm Ricci}}
\def\id{{\rm Id}}
\def\tr{{\rm Tr}}
\def\oc{{\cal O}}
\def\bb{{\rm I}\!{\rm B}}

\centerline{\bftwelve Kobayashi-L\"ubke inequalities for Chern classes}
\centerline{\bftwelve of Hermite-Einstein vector bundles and}
\centerline{\bftwelve Guggenheimer-Yau-Bogomolov-Miyaoka inequalities}
\centerline{\bftwelve for Chern classes of K\"ahler-Einstein manifolds}
\vskip20pt

Let $(X,\omega)$ be a compact K\"ahler manifold, $n=\dim X$, and let $E$
be a holomorphic vector bundle over~$X$, $r=\rank E$. We suppose that $E$
is equipped with a hermitian metric, $h$ and denote by $D_{E,h}$ the Chern
connection on $(E,h)$. The Chern curvature form is
$$\Theta_h(E)=D_{E,h}^2.$$
In a (local) orthonormal frame $(e_\alpha)_{1\le\alpha\le r}$ of $E$,
we write
$$\Theta_h(E)=(\Theta_{\alpha\beta})_{1\le\alpha,\beta\le r}$$
where the $\Theta_{\alpha\beta}$ are complex valued $(1,1)$-forms
satisfying the hermitian condition
$\ov\Theta_{\alpha\beta}=\Theta_{\beta\alpha}$. We denote
$$\Theta_{\alpha\beta}=i\sum_{1\le\alpha,\beta\le r,~1\le j,k\le n}
\Theta_{\alpha\beta jk}dz_j\wedge dz_k.$$
The hermitian symmetry condition can then be read
$\ov\Theta_{\alpha\beta jk}=\Theta_{\beta\alpha kj}$.
If at some point $x_0\in X$ the coordinates $(z_j)$ are chosen so that
$(dz_j(x_0))$ is an orthonormal basis of $T_{X,x_0}^\star$, we define
$$\tr_\omega\Theta_h(E)=\Big(\sum_j\Theta_{\alpha\beta jj}\Big)\in
C^\infty(X,\hom(E,E)).$$

\than Definition|The hermitian vector bundle $(E,h)$ is said to be
Hermite-Einstein with respect to the K\"ahler metric $\omega$ if there
is a constant $\lambda>0$ such that $\tr_\omega\Theta_h(E)=\lambda\id_E$.
\fintha

Recall that the Chern forms $c_k(E)_h$ are defined by the formula
$$\det\big(I+t\Theta_h(E)\big)=\det(\delta_{\alpha\beta}+
t\Theta_{\alpha\beta})=1+t\,c_1(E)_h+\ldots+t^rc_r(E)_h.$$
This gives in particular the identities
$$\eqalign{
c_1(E)_h&=\sum_\alpha\Theta_{\alpha\alpha},\cr
c_2(E)_h&=\sum_{\alpha<\beta}\Theta_{\alpha\alpha}\wedge\Theta_{\beta\beta}
-\Theta_{\alpha\beta}\wedge\Theta_{\beta\alpha}
={1\over 2}\sum_{\alpha,\beta}\Theta_{\alpha\alpha}\wedge\Theta_{\beta\beta}
-\Theta_{\alpha\beta}\wedge\Theta_{\beta\alpha}.\cr}$$
The trace $\tr_\omega\Theta_h(E)$ can be computed by the formula
$$\Theta_h(E)\wedge{\omega^{n-1}\over (n-1)!}=\tr_\omega\Theta_h(E)\,
{\omega^n\over n!}.$$
By taking the trace with respect to the indices $\alpha$ in~$E$ and taking
the Hermite-Einstein equation into account, we find
$$c_1(E)_h\wedge{\omega^{n-1}\over(n-1)!}=\lambda r\,{\omega^n\over n!}.$$
This implies that the number $\lambda$ in the definition of
Hermite-Einstein metrics is a purely numerical invariant, namely
$$\lambda={n\over r}\int_Xc_1(E)\wedge\omega^{n-1}\Big/\int_X\omega^n.$$

\than Kobayashi-L\"ubke inequality|If $E$ admits a Hermite-Einstein
metric $h$ with respect to $\omega$, then
$$\big[(r-1)c_1(E)_h^2-2r\,c_2(E)_h\big]\wedge\omega^{n-2}\le 0$$
at every point of~$X$. Moreover, the equality holds if and only if
$$\Theta_h(E)={1\over r}c_1(E)_h\otimes\id_E.$$
\fintha

Observe that the equality holds pointwise already if we have the numerical
equality
$$\int_X\big[(r-1)c_1(E)^2-2r\,c_2(E)\big]\wedge\omega^{n-2}=0.$$
If we introduce the (formal) vector bundle $\wt E=E\otimes(\det E)^{-1/r}$
($\wt E$ is the ``normalized'' vector bundle such that $\det\wt E=\oc$),
then $c_1(\wt E)_h=0$ and
$$\Theta_h(\wt E)=\big(\Theta_h(E)-{1\over r}c_1(E)_h\otimes\id_E\Big)\otimes
\id_{(\det E)^{-1/r}}.$$
By the formula for the chern classes of $E\otimes L$,
the Kobayashi-L\"ubke inequality can be rewritten as
$$c_2(\wt E)_h\wedge\omega^{n-2}\le0,$$
with equality if and only if $\wt E$ is unitary flat. In that case, we say
that $E$ is {\sl projectively flat}.

\proof By the above,
$$(r-1)c_1(E)_h^2-2r\,c_2(E)_h=\sum_{\alpha,\beta}-\Theta_{\alpha\alpha}
\wedge\Theta_{\beta\beta}+r\,\Theta_{\alpha\beta}\wedge\Theta_{\beta\alpha}.$$
Taking the wedge product with $\omega^{n-2}/(n-2)!$ means taking the trace,
i.e. the sum of coefficients of the terms $i\,dz_j\wedge d\ov z_j\wedge
i\,dz_k\wedge d\ov z_k$ for all $j<k$. For this, we have to look at products
of the type $(i\,dz_j\wedge d\ov z_j)\wedge(i\,dz_k\wedge d\ov z_k)$ or
$(i\,dz_j\wedge d\ov z_k)\wedge(i\,dz_k\wedge d\ov z_j)$. This yields
$$\eqalign{
2\big[(r&{}-1)c_1(E)_h^2-2r\,c_2(E)_h\big]\wedge{\omega^{n-2}\over (n-2)!}\cr
&=\sum_{\alpha,\beta,j,k}-(\Theta_{\alpha\alpha jj}\Theta_{\beta\beta kk}
-\Theta_{\alpha\alpha jk}\Theta_{\beta\beta kj})
+r(\Theta_{\alpha\beta jj}\Theta_{\beta\alpha kk}
-\Theta_{\alpha\beta jk}\Theta_{\beta\alpha kj}).\cr}$$
The initial factor $2$ comes from the fact that the final sum is taken over
all unordered indices $j,k$ (terms with $j=k$ cancel). The Hermite-Einstein
condition yields $\sum_j\Theta_{\alpha\beta jj}
=\lambda\,\delta_{\alpha\beta}$, so we get
$$\eqalign{
\sum_{\alpha,\beta,j,k}-\Theta_{\alpha\alpha jj}\Theta_{\beta\beta kk}
+r\,\Theta_{\alpha\beta jj}\Theta_{\beta\alpha kk}
&=\sum_{\alpha,\beta}-\lambda^2\delta_{\alpha\alpha}\delta_{\beta\beta}+
r\lambda^2\delta_{\alpha\beta}\delta_{\beta\alpha}\cr
&=-r^2\lambda^2+r^2\lambda^2=0.\cr}$$
Hence, using the hermitian symmetry of $\Theta_{\alpha\beta jk}$, we find
$$\eqalign{
2\big[(r-1)c_1(E)_h^2&{}-2r\,c_2(E)_h\big]\wedge{\omega^{n-2}\over (n-2)!}\cr
&=\sum_{\alpha,\beta,j,k}\Theta_{\alpha\alpha jk}\ov\Theta_{\beta\beta jk}
-r\,|\Theta_{\alpha\beta jk}|^2\cr
&=-r\sum_{\alpha\ne\beta,j,k}|\Theta_{\alpha\beta jk}|^2+
\sum_{j,k}\Big(\sum_{\alpha,\beta}\Theta_{\alpha\alpha jk}
\ov\Theta_{\beta\beta jk}-r\sum_\alpha|\Theta_{\alpha\alpha jk}|^2\Big)\cr
&=-r\sum_{\alpha\ne\beta,j,k}|\Theta_{\alpha\beta jk}|^2-
{1\over 2}\sum_{\alpha,\beta,j,k}\big|\Theta_{\alpha\alpha jk}-
\Theta_{\beta\beta jk}\big|^2\le 0.\cr}$$
This proves the expected inequality. Moreover, the equality holds if and
only if we have
$$\Theta_{\alpha\beta jk}=0~~~\hbox{\rm for}~~\alpha\ne\beta,~~~~
\Theta_{\alpha\alpha jk}=\gamma_{jk}~~~\hbox{\rm for all}~~\alpha.$$
where $\gamma=i\sum_{j,k}\gamma_{jk}dz_j\wedge d\ov z_k$ is a $(1,1)$-form
(take e.g., $\gamma_{jk}=\Theta_{11jk}$). Hence $\Theta_h(E)=\gamma\otimes
\id_E$. By taking the trace with respect to $E$ in this last equality, we
get $c_1(E)_h=r\,\gamma$. Therefore the equality occurs if and only if
$$\Theta_h(E)={1\over r}c_1(E)_h\otimes\id_E.\eqno\finpr$$

\than Corollary 1|Let $(E,h)$ be a Hermite-Einstein vector bundle with
$c_1(E)=0$ and $c_2(E)=0$. Then $E$ is unitary flat for some
hermitian metric $h'=h\,e^{-\varphi}$.\fintha

\proof By the assumption $c_1(E)=0$, we can write $c_1(E)_h={i\over 2\pi}
\partial\op\psi$ for some global function $\psi$ on~$X$. The
equality case of the Kobayashi-L\"ubke inequality yields
$$\Theta_{h\,e^{\psi/r}}(E)=\Theta_h(E)-{1\over r}{i\over 2\pi}
\partial\op\psi\otimes\id_E=0.\eqno\finpr$$

\than Corollary 2|Let $X$ be a compact K\"ahler manifold with
$c_1(X)=c_2(X)=0$. Then $X$ is a finite unramified quotient
of a torus.\fintha

\proof By the Aubin-Calabi-Yau theorem, $X$ admits a Ricci-flat
K\"ahler metric~$\omega$. Since $\ricci(\omega)=\tr_\omega
\Theta_\omega(T_X)$, we see that $(T_X,\omega)$ is a Hermite-Einstein
vector bundle, and $c_1(T_X)_\omega=\ricci(\omega)=0$. By
the Kobayashi-L\"ubke inequality, we conclude that $(T_X,\omega)$ is
unitary flat, given by a unitary representation \hbox{$\pi_1(X)\to U(n)$}.
Let $\wt X$ be the universal covering of $X$ and $\wt\omega$ the induced
metric. Then $(T_{\wt X},\wt\omega)$ is a trivial vector bundle
equipped with a flat metric. Let $(\xi_1\ld\xi_n)$ be an orthonormal
parallel frame of $T_X$. Since $\nabla\xi_j=0$, we conclude
that $d\xi_j^\star=0$, and it is easy to infer from this that
$[\xi_j,\xi_k]=0$. The flow of each vector field $\sum\lambda_j\xi_j$
is defined for all times (this follows from the fact the length
of a trajectory is proportional to the time, and $\wt\omega$ is
complete). Hence we get an action of $\cb^n$ on $\wt X$, and
it follows easily that $(\wt X,\wt\omega)\simeq(\cb^n,{\rm can})$.
Now, $\pi_1(X)$ acts by isometries on this $\cb^n$. The classification
of subgroups of affine transformations acting freely (and with compact
quotient) shows that $\pi_1(X)$ must be a semi-direct product of a finite
group of isometries by a group of translations associated to a lattice
$\Lambda\subset\cb^n$. Hence there is an exact sequence
$$0\la\Lambda\la\pi_1(X)\la G\la 0$$
where $G$ is a finite group of isometries. It follows that there is a
finite unramified covering map $\cb^n/\Lambda\to \wt X/\pi_1(X)\simeq X$
of $X$ by a torus.\hfill\finpr
\vskip8pt
We now discuss the special case of the tangent bundle $T_X$ in case
$(X,\omega)$ is a compact K\"ahler-Einstein manifold. The
K\"ahler-Einstein condition means that $\ricci(\omega)=\lambda\omega$
for some real constant $\lambda$, i.e., $\tr_\omega\Theta_\omega(T_X)=
\lambda\id_{T_X}$. In particular, $(T_X,\omega)$ is a Hermite-Einstein
vector bundle. Here, however, the coefficients $(\Theta_{\alpha\beta jk}
)_{1\le\alpha,\beta,j,k\le n}$ of the curvature tensor $\Theta_\omega(T_X)$
satisfy the additional symmetry relations
$$\Theta_{\alpha\beta jk}=\Theta_{j\beta\alpha k}=
\Theta_{\alpha kj\beta}=\Theta_{jk\alpha\beta}.\leqno(\star)$$
These relations follow easily from the identity $\Theta_{\alpha\beta jk}=
-\partial^2\omega_{\alpha\beta}/\partial z_j\partial\ov z_k$ in normal
coordinates, when we apply the K\"ahler condition
$\partial\omega_{\alpha\beta}/\partial z_j=
\partial\omega_{j\beta}/\partial z_\alpha$.
It follows that the Chern forms satisfy a slightly stronger inequality
than the general inequality valid for Hermite-Einstein bundles. In fact
$(T_X,\omega)$ satisfies a similar inequality where the rank $r=n$ is
replaced by~$n+1$.

\than Guggenheimer-Yau inequality|Let $(T_X,\omega)$
be a compact $n$-dimensional K\"ahler-Einstein manifold, with constant
$\lambda\in\rb$. If $\lambda=0$, then $c_2(T_X)_\omega\wedge\omega^{n-2}\ge 0$.
If $\lambda\ne 0$, we have the inequality
$$\big[n c_1(X)^2-(2n+2)c_2(X)\big]\cdot(\lambda c_1(X))^{n-2}\le 0,$$
and the equality also holds pointwise if we replace the Chern classes
by the Chern forms~$c_k(T_X)_\omega$. The equality occurs in the
following cases:
\smallskip
\item{\rm(i)} If $\lambda=0$, then $(X,\omega)$ is a finite unramified
quotient of a torus.
\smallskip
\item{\rm(ii)} If $\lambda>0$, then $(X,\omega)\simeq
(\pb^n,\hbox{\rm Fubini Study})$.
\smallskip
\item{\rm (iii)} If $\lambda<0$, then $(X,\omega)\simeq(\bb_n/\Gamma,
\hbox{\rm Poincar\'e metric})$, i.e. $X$ is a compact unramified quotient
of the ball in~$\cb^n$.
\smallskip\fintha

\than Corollary {\rm (Bogomolov-Miyaoka-Yau)}|Let $X$ be a surface of
general type with $K_X$ ample. Then there is an inequality
$c_1(X)^2\le 3\,c_2(X)$, and the equality occurs if and only if $X$
is a quotient of the ball $\bb_2$.\fintha

Miyaoka has shown that the inequality holds in fact as soon as
$X$ is a surface with general type.

\proof As in the proof of the Kobayashi-L\"ubke inequality, we find
$$n\,c_1(T_X)_\omega^2-(2n+2)c_2(T_X)_\omega=
\sum_{\alpha,\beta}-\Theta_{\alpha\alpha}\wedge\Theta_{\beta\beta}+
(n+1)\Theta_{\alpha\beta}\wedge\Theta_{\beta\alpha}.$$
Taking the wedge product with $\omega^{n-2}/(n-2)!$, we get
$$\eqalign{
2&\big[n\,c_1(T_X)_\omega^2-(2n+2)\,c_2(T_X)_\omega\big]\wedge
{\omega^{n-2}\over (n-2)!}\cr
&=\sum_{\alpha,\beta,j,k}-(\Theta_{\alpha\alpha jj}\Theta_{\beta\beta kk}
-\Theta_{\alpha\alpha jk}\Theta_{\beta\beta kj})
+(n+1)(\Theta_{\alpha\beta jj}\Theta_{\beta\alpha kk}
-\Theta_{\alpha\beta jk}\Theta_{\beta\alpha kj}).\cr}$$
If we had the factor $r=n$ instead of $(n+1)$ in the right hand side, the
terms in $jj$ and $kk$ would cancel (as they did before). Hence we find 
$$\eqalign{
2\big[n\,c_1(T_X)_\omega^2&{}-(2n+2)\,c_2(T_X)_\omega\big]\wedge
{\omega^{n-2}\over (n-2)!}\cr
&=\sum_{\alpha,\beta,j,k}\Theta_{\alpha\alpha jk}\Theta_{\beta\beta kj}
+\Theta_{\alpha\beta jj}\Theta_{\beta\alpha kk}
-(n+1)\Theta_{\alpha\beta jk}\Theta_{\beta\alpha kj}\cr
&=\sum_{\alpha,\beta,j,k}2\,\Theta_{\alpha\alpha jk}\Theta_{\beta\beta kj}
-(n+1)\Theta_{\alpha\beta jk}\Theta_{\beta\alpha kj}.\cr}$$
Here, the symmetry relation $(\star)$ was used in order to obtain the
equality of the summation of the first two terms. Using also the hermitian
symmetry relation, our sum $\Sigma$ can be rewritten as
$$\eqalign{\Sigma
=&{}-\sum_{\alpha,\beta,j,k}|\Theta_{\alpha\alpha jk}-
\Theta_{\beta\beta jk}|^2-(n+1)|\Theta_{\alpha\beta jk}|^2
+2n\sum_{\alpha,j,k}|\Theta_{\alpha\alpha jk}|^2\cr
=&{}-\sum_{\alpha,\beta,j,k}|\Theta_{\alpha\alpha jk}-
\Theta_{\beta\beta jk}|^2-(n+1)\sum_{\alpha,\beta,j,k,\,
\hbox{\sevenrm pairwise}\,\ne}|\Theta_{\alpha\beta jk}|^2\cr
&{}-(n+1)\Big[8\sum_{\alpha\ne j<k\ne\alpha}\kern-10pt
|\Theta_{\alpha\alpha jk}|^2
+4\sum_{\alpha\ne j}|\Theta_{\alpha\alpha\alpha j}|^2
+4\sum_{\alpha<j}|\Theta_{\alpha\alpha jj}|^2
+\sum_\alpha|\Theta_{\alpha\alpha\alpha\alpha}|^2\Big]\kern-2.91pt\cr
&\phantom{{}-(n+1)}\llap{${}+2n$}\Big[2\sum_{\alpha\ne j<k\ne\alpha}
\kern-10pt|\Theta_{\alpha\alpha jk}|^2
+2\sum_{\alpha\ne j}|\Theta_{\alpha\alpha\alpha j}|^2
+2\sum_{\alpha<j}|\Theta_{\alpha\alpha jj}|^2
+\sum_\alpha|\Theta_{\alpha\alpha\alpha\alpha}|^2\Big]\kern-2.91pt\cr
=&{}-\sum_{\alpha\ne\beta,j,k}|\Theta_{\alpha\alpha jk}-
\Theta_{\beta\beta jk}|^2-(n+1)\sum_{\alpha,\beta,j,k,\,
\hbox{\sevenrm pairwise}\,\ne}|\Theta_{\alpha\beta jk}|^2\cr
&{}-(4n+8)\sum_{\alpha\ne j<k\ne\alpha}|\Theta_{\alpha\alpha jk}|^2
-4\sum_{\alpha\ne j}|\Theta_{\alpha\alpha\alpha j}|^2
-4\sum_{\alpha<j}|\Theta_{\alpha\alpha jj}|^2\cr
&{}+(n-1)\sum_\alpha|\Theta_{\alpha\alpha\alpha\alpha}|^2.\cr}$$
All terms are negative except the last one. We try to absorb this term
in the summations involving the coefficients $\Theta_{\alpha\alpha jj}$.
This gives
$$\eqalign{\Sigma
=&{}-\sum_{\alpha\ne\beta,j\ne k}|\Theta_{\alpha\alpha jk}-\Theta_{\beta
\beta jk}|^2-(n+1)\sum_{\alpha,\beta,j,k,\,\hbox{\sevenrm pairwise}\,\ne}
|\Theta_{\alpha\beta jk}|^2\cr
&{}-(4n+8)\sum_{\alpha\ne j<k\ne\alpha}|\Theta_{\alpha\alpha jk}|^2
-4\sum_{\alpha\ne j}|\Theta_{\alpha\alpha\alpha j}|^2\cr
&{}-\sum_{\alpha\ne\beta,j}|\Theta_{\alpha\alpha jj}-
\Theta_{\beta\beta jj}|^2-4\sum_{\alpha<j}|\Theta_{\alpha\alpha jj}|^2
+(n-1)\sum_\alpha|\Theta_{\alpha\alpha\alpha\alpha}|^2.\cr}$$
The last line is equal to
$$\eqalign{
&{}-\sum_{\alpha\ne\beta\ne j\ne\alpha}|\Theta_{\alpha\alpha jj}-
\Theta_{\beta\beta jj}|^2-2\sum_{\alpha\ne\beta}|\Theta_{\alpha\alpha
\beta\beta}-\Theta_{\beta\beta\beta\beta}|^2\cr
&\phantom{{}-\sum_{\alpha\ne\beta\ne j\ne\alpha}|\Theta_{\alpha\alpha jj}-
\Theta_{\beta\beta jj}|^2}
-4\sum_{\alpha<\beta}|\Theta_{\alpha\alpha\beta\beta}|^2
+(n-1)\sum_\alpha|\Theta_{\alpha\alpha\alpha\alpha}|^2\cr
=&{}-\sum_{\alpha\ne\beta\ne j\ne\alpha}|\Theta_{\alpha\alpha jj}-
\Theta_{\beta\beta jj}|^2\cr
&{}-(n-1)\sum_\alpha|\Theta_{\alpha\alpha\alpha\alpha}|^2
-8\sum_{\alpha<\beta}|\Theta_{\alpha\alpha
\beta\beta}|^2+2\sum_{\alpha\ne\beta}
\Theta_{\alpha\alpha\beta\beta}\ov\Theta_{\beta\beta\beta\beta}+
\ov\Theta_{\alpha\alpha\beta\beta}\Theta_{\beta\beta\beta\beta}\cr
=&{}-\sum_{\alpha\ne\beta\ne j\ne\alpha}|\Theta_{\alpha\alpha jj}-
\Theta_{\beta\beta jj}|^2-\sum_{\alpha\ne\beta}
|\Theta_{\alpha\alpha\alpha\alpha}-2\Theta_{\alpha\alpha\beta\beta}|^2.
\cr}$$
Therefore we find
$$\eqalign{\Sigma
=&{}-\sum_{\alpha\ne\beta,j\ne k}|\Theta_{\alpha\alpha jk}-\Theta_{\beta
\beta jk}|^2-(n+1)\sum_{\alpha,\beta,j,k,\,\hbox{\sevenrm pairwise}\,\ne}
|\Theta_{\alpha\beta jk}|^2\cr
&{}-(4n+8)\sum_{\alpha\ne j<k\ne\alpha}|\Theta_{\alpha\alpha jk}|^2
-4\sum_{\alpha\ne j}|\Theta_{\alpha\alpha\alpha j}|^2\cr
&{}-\sum_{\alpha\ne\beta\ne j\ne\alpha}|\Theta_{\alpha\alpha jj}-
\Theta_{\beta\beta jj}|^2-\sum_{\alpha\ne\beta}
|\Theta_{\alpha\alpha\alpha\alpha}-2\Theta_{\alpha\alpha\beta\beta}|^2.
\cr}$$
This proves the expected inequality $\Sigma\le 0$. Moreover, we have
$\Sigma=0$ if and only if there is a scalar $\mu$ such that
$$\Theta_{\alpha\alpha\beta\beta}=\Theta_{\alpha\beta\beta\alpha}=
\Theta_{\alpha\beta\alpha\beta}=\mu~~\hbox{\rm for}~~\alpha\ne\beta,~~~~
\Theta_{\alpha\alpha\alpha\alpha}=2\mu,$$
and all other coefficients $\Theta_{\alpha\beta jk}$ are zero.
By taking the trace $\sum_j\Theta_{\alpha\alpha jj}$, we get
$\lambda=(n+1)\mu$. We thus obtain that the hermitian form associated to
the curvature tensor is
$$\leqalignno{
\langle\Theta_\omega(T_X)(\xi\otimes\eta),\xi\otimes\eta\rangle_\omega
&={\lambda\over n+1}\sum_{\alpha,\beta}\xi_\alpha\eta_\beta
\ov\xi_\alpha\ov\eta_\beta+\xi_\alpha\eta_\beta
\ov\xi_\beta\ov\eta_\alpha\cr
&={\lambda\over n+1}\big(|\xi|^2|\eta|^2+|\langle\xi,\eta\rangle|^2\big)
&(\star\star)\cr}$$
for all $\xi,\eta\in T_X$. When $\lambda=0$, the curvature tensor
vanishes identically and we have already seen that $X$ is a finite
unramified quotient of a torus. Assume from now on that $\lambda\ne 0$.
The formula $(\star\star)$ shows that the curvature tensor
is constant and coincides with the curvature tensor of $\pb^n$ (case
$\lambda>0$), or of the ball $\bb_n$ (case $\lambda<0$),
relatively to the canonical metrics on these spaces. By a well-known result
from the theory of hermitian symmetric spaces\footnote{${}^\star$}
{See for example, F.~Tricerri et L.~Vanhecke, {\sl Vari\'et\'es
riemanniennes dont le tenseur de courbure est celui d'un espace
sym\'etrique riemannien irr\'eductible}, C.~R.~Acad.~Sc.~Paris,
{\bf 302} (1986), 233-235.}, it follows that $(X,\omega)$
is locally isometric to $\pb^n$ (resp. $\bb_n$). Since the universal
covering $\wt X$ is a complete and locally symmetric hermitian
manifold, we conclude that $\wt X\simeq\pb^n$, resp.
$\wt X\simeq\bb^n$. In the case $\lambda>0$, $X$ is a Fano manifold,
thus $X$ is simply connected and $X=\wt X$. The proof is
complete.\hfill\finpr
\vskip8pt

To compute the curvature of $\pb^n$ and $\bb^n$, we use the fact that the
canonical metric is
$$\omega={i\over 2\pi}\partial\op\log(1+|z|^2)~~\hbox{\rm on}~\pb^n,~~~~
\hbox{\rm resp.}~~\omega=
-{i\over 2\pi}\partial\op\log(1-|z|^2)~~\hbox{\rm on}~\bb^n,$$
with respect to the non homogeneous coordinates on $\pb^n$. We thus get
$$\omega={i\over 2\pi}\Big({dz\otimes d\ov z\over 1+|z|^2}-
{|\langle dz,z\rangle|^2\over(1+|z|^2)^2}\Big),~~~~\hbox{\rm resp.}~~~
\omega={i\over 2\pi}\Big({dz\otimes d\ov z\over 1-|z|^2}+
{|\langle dz,z\rangle|^2\over(1-|z|^2)^2}\Big).$$
By computing the derivatives $\Theta_{\alpha\beta jk}=
-\partial^2\omega_{\alpha\beta}/\partial z_j\partial\ov z_k$
at $z=0$, we easily see that the curvature is given by $(\star\star)$
with $\lambda=\pm(n+1)$. The equality also holds at any other point by
the homogeneity of $\pb^n$ and~$\bb^n$.

Observe that the riemannian exponential map $\exp:T_{X,0}\to X$ at the
origin of $\cb^n\subset\pb^n$ or $\bb^n$ is unitary invariant. It
follows that the holomorphic part $h$ of the Taylor expansion of $\exp$
at $0$ is unitary invariant. This invariance forces $h$ to coincide
with the identity map in the standard coordinates of $\pb^n$ and
$\bb^n$. From this observation, it is not difficult to justify
intuitively the local isometry statement used above. In fact let $X$ be
a hermitian manifold whose curvature tensor is given by $(\star\star)$,
$\lambda\ne 0$. Then $\omega$ is proportional to $c_1(T_X)_\omega$ and
so $\omega$ is a K\"ahler-Einstein metric. Since the K\"ahler-Einstein
equation is (nonlinear) elliptic with real analytic coefficients in
terms of any real analytic K\"ahler form, it follows that $\omega$ is
real analytic, and so is the exponential map. Fix a point $x_0\in X$
and let $h:\cb^n\simeq T_{X,x_0}\to X$ be the holomorphic part of the
Taylor expansion of $\exp$ at the origin. Then $h$ must provide the
holomorphic coordinates we are looking for, i.e. $h^\star\omega$ must
coincide with the metric of $\pb^n$ (resp. $\bb^n$) in a neighborhood
of the origin. 

\end

