This file explains one theorem of complex geometry, slowly, from scratch.

The theorem: if a compact Kahler manifold has no nonconstant entire curves,
then its canonical bundle is ample, and the manifold is projective.

That sentence is meaningless without about forty definitions, and every one of
them is built out of the ones before. So this file is a chain: each block asks
one question and answers it completely, and each block uses only blocks above.

Read it top to bottom. Skipping ahead will not work.

Notation used throughout: D = {z in C : |z| < 1} is the open unit disc.
C is the complex plane. n is always the complex dimension of the manifold.
X is always a compact connected complex manifold. k is always a fixed Kahler
metric on X. P^n is complex projective space. c_1 is the first Chern class.
K_X is the canonical bundle. i is the square root of minus one.
d^c and dd^c are defined in the block where they first appear.

Exercise: Roadmap. What are the five big steps of the proof?
(1) A compact no-entire-curves manifold bounds the derivative of every
holomorphic disc at its centre by a constant D. (2) Build certain "extremal"
discs, discs that maximise a carefully chosen functional, and prove they are
regular enough to vary at the boundary of the disc. (3) Compute the second
derivative of the functional along those variations. (4) Read two inequalities
off that second derivative: one bounds the Ricci form along discs and gives
nefness, the other gives a lower bound for the volume form of any Kahler metric
with R_h >= -h. (5) Feed the volume bound into the Aubin-Yau equation to get
positive canonical top self-intersection, then use the holomorphic Morse
inequalities, Moishezon's theorem, and the cone theorem to reach ampleness.
Steps (1) and (2) are the new analysis. Steps (4) and (5) are the payoff.

Exercise: Start from the end. What is a complex manifold?
A topological space covered by open sets each identified with an open subset of
C^n, with overlapping identifications holomorphic in both directions. Real
dimension is 2n, because C^n is R^{2n}. A chart is such an identification.
"Holomorphic" means locally a convergent power series in z_1,...,z_n, no
conjugates. C itself is the case n = 1. The disc D is a complex manifold.
X compact means X is compact as a topological space.
X connected means it does not split into two disjoint open pieces.
X Kahler means, next block.
X projective means much later, at the Ample block.

Exercise: Kaehler. What is a Kaehler metric, without the usual fuss?
A Riemannian metric on a complex manifold is a smoothly varying inner product
on each real tangent space. It is Hermitian if it is compatible with
multiplication by i. A Hermitian metric h has an associated real 2-form omega,
given by omega(v, x) = h(i v, x). The metric is a Kaehler metric if d omega = 0,
that is, if that 2-form is closed. Complex projective space P^n with its
Fubini-Study metric is the standard example; complex tori and smooth
hypersurfaces in P^n are also Kahler. The payoff: closedness forces local
potentials and a strong Hodge theory, and it is what lets "curvature" be a
cohomology class rather than just a tensor.

Exercise: Local potential. Why is every Kaehler form locally dd^c of a function?
Fix the operator first, because signs are where this subject eats people.
Put d^c = (i/2)(dbar - d) and put dd^c = i d' dbar, so that on a smooth
function dd^c f = i sum_{j,k} (d^2 f/dz_j dzbar_k) dz_j ^ dzbar_k, a real
(1,1)-form. These are the conventions of this file and of the paper.
The dd^c-Poincare lemma says: every closed real (1,1)-form q on a polydisc
equals dd^c f for some smooth real f, the local Kaehler potential. Sketch:
Poincare gives q = d(eta) for a complex 1-form eta; split eta into (1,0) and
(0,1) parts, and the absence of (0,2) and (2,0) parts in q kills dbar of the
(0,1) part and d' of the (1,0) part, leaving exactly dd^c of an unknown
function. The potential is unique up to adding a pluriharmonic function,
meaning one whose dd^c vanishes, for instance the real part of a holomorphic function. That
non-uniqueness looks like a nuisance and is the central technical point,
because gluing local potentials means controlling pluriharmonic differences.

Exercise: Line bundle. What is a holomorphic line bundle, and a section of one?
On X, a holomorphic line bundle L is a family of one-dimensional complex
vector spaces, one fibre over each point, glued by holomorphic transition
functions: over the overlap of two charts, the two trivializations differ by
multiplication by a nowhere zero holomorphic function. A holomorphic section s
picks a vector in each fibre, holomorphically; locally it is a holomorphic
function, and on overlaps the two local descriptions differ by the transition
function. Isomorphism classes of line bundles form the Picard group, which is
H^1(X, O^*). The first Chern class c_1(L) in H^2(X, Z) is the image of that
class, and it only sees the topology, not the holomorphic structure. On P^n the
basic example is O(m): its sections are the homogeneous polynomials of degree
m, with m < 0 interpreted by duality. Line bundles are the algebra of this
subject; forms and metrics are the analysis.

Exercise: Canonical bundle. What is K_X, and why that bundle?
Take the top exterior power of the holomorphic cotangent bundle,
K_X = det(T^{1,0}X)^*. Its sections are holomorphic n-forms: a rule assigning
a complex number to each n-tuple of holomorphic tangent vectors, alternating
and complex-linear, varying holomorphically over X. Why the top power: this is
the bundle whose local sections transform by exactly the Jacobian determinant
under a change of coordinates, so an n-form is the thing you can integrate, and
the thing that measures how a volume element twists. For a Riemann surface,
K_X is the cotangent bundle itself and its sections are the holomorphic
1-forms. Chern classes: c_1(K_X) = -c_1(T^{1,0}X), because det dualizes and
duality inverts a determinant. So K_X negative means the tangent bundle is
positive, which is the direction of P^n; K_X positive is the direction of a
hyperbolic space.

Exercise: Ample. What does "K_X is ample" mean, and what does it deliver?
Let L be a line bundle with global sections. Choose a basis s_0,...,s_N of
H^0(X,L). The map x -> (s_0(x): ... : s_N(x)) goes X to P^N wherever the
sections are not all zero at x. Call L very ample if this map is defined
everywhere and is a holomorphic embedding, meaning it is injective and its
derivative is injective at every point; call L ample if some positive tensor
power L^m is very ample. A power is needed only because sections can be too
sparse locally, not because of anything deep. Sections of K_X^m are the same as
holomorphic m-tuples of n-forms, called pluricanonical forms. So "K_X ample"
means: for some m >= 1, pluricanonical forms embed X into projective space.
That gives projectivity, algebraic coordinates, and every consequence of being
a smooth projective variety, including the Kodaira embedding of X in P^N. The
conjecture the paper proves is that no-entire-curves forces exactly this.

Exercise: Entire curve. What is an entire curve, and what is Brody hyperbolicity?
An entire curve is a holomorphic map f : C -> X, defined on the whole complex
line. Holomorphic maps from the disc D into X always exist in droves, so the
disc alone cannot impose positivity; the line is the demanding source, because
a map from C can escalate, wind up, and run off to infinity inside a compact
target only by covering something. Define: X is Brody hyperbolic if EVERY
entire curve C -> X is constant. For compact complex manifolds this is a
theorem of Brody to be equivalent to Kobayashi hyperbolicity, the statement
that the intrinsic Kobayashi pseudodistance is a distance. The main theorem's
hypothesis is exactly Brody hyperbolicity on a compact connected Kahler
manifold. Note what is NOT assumed: no negativity of curvature, no projective
algebraicity, no metrics of special sign. That is what makes the statement
hard, and it is why the proof spends all its technique on discs rather than
on curvature assumptions.

Exercise: Brody. Why does no-entire-curves bound the derivative of every disc?
This is the paper's first lemma, and it is pure rescaling. Suppose not: there
are discs f_j : D -> X with |f_j'(0)|_k -> infinity. For each j pick a point z_j
in the closed disc of radius 1/2 where z -> (1/2 - |z|) |f_j'(z)|_k is largest;
then |z_j| < 1/2 for large j, and the value there is positive. Put
lambda_j = |f_j'(z_j)|_k, R_j = (1/2 - |z_j|) lambda_j, and
g_j(w) = f_j(z_j + w/lambda_j). Then g_j is defined on |w| < R_j, and
R_j >= (1/2) |f_j'(0)|_k -> infinity, while |g_j'(0)|_k = 1 by construction.
The maximality of z_j gives |g_j'(w)|_k <= 2 on |w| < R_j/2. Compactness of X
plus Arzela-Ascoli on shrinking discs plus a diagonal subsequence gives a
local uniform limit g : C -> X, holomorphic by the uniform limit theorem with
Cauchy's formula for derivatives, and |g'(0)| = 1, so g is a nonconstant
entire curve. Contradiction. Hence D = sup |f'(0)|_k < infinity. The point to
remember: rescaling is the ONLY place hyperbolicity enters the argument.
Everything after this block runs on a compact Kahler manifold with no further
hypothesis than the number D.

Exercise: Finite area disc. Why "finite area" and not "smooth up to the boundary"?
A holomorphic disc is just a holomorphic map f : D -> X; it has no reason to
extend to the closed disc, and the extremal discs we want do not. What a disc
does always give is its area, computed with any smooth Hermitian metric:
A_h(f) = (1/pi) integral over D of h(f'(z), f'(z)) dS, where f'(z) = df(d/dz)
and dS is ordinary Euclidean area. Because X is compact, two smooth Hermitian
metrics are uniformly comparable, so the property A_h(f) < infinity is
independent of h; those are the finite-area discs. Regularity at the boundary
is a result to be earned later, never an assumption. The reason the paper
insists on this: the extremal discs are constructed by compactness on shrinking
subdiscs, and compactness on shrinking subdiscs gives boundary behaviour only
after a separate argument.

Exercise: The three integrals. What are A_q, P_q and d?
Fix a smooth closed real (1,1)-form q on X. Evaluate it on the derivative of f,
getting a real-valued function z -> q(f'(z), f'(z)) on the disc, and integrate:
A_q(f) = (1/pi) integral over D of q(f', f') dS, the area functional;
P_q(f) = (2/pi) integral over D of log(1/|z|) q(f', f') dS, the Poisson
functional, weighted by the Green kernel; and d(f) = k(f'(0), f'(0)), the
squared length of the central derivative with respect to the fixed Kahler
metric k. All three are finite for a finite-area disc: the weight log(1/|z|)
blows up only at the centre, where f' is bounded and the weight is
integrable in two dimensions. A_q is signed in general, because q need not be
positive; only the case q = h, a Kahler form, is nonnegative. Write
A_{q,r}(f) for the same area integral restricted to |z| < r, with 0 < r < 1.

Exercise: Sign conventions. Which sign does the power of the weight carry?
Every sign in this subject comes from one formula, so pin it once. With
dd^c = i d' dbar and the quadratic form convention that the tensor of a real
function u is sum u_{j kbar} v^j vbar^k, one has q(f', f') = d_z dbar_z (u o f)
whenever q = dd^c u locally. So a positive q reports that u o f is
plurisubharmonic, and the Green formula below produces a plus sign. The paper's
R_h is R_h = -dd^c log det h, chosen so that R_h represents
2 pi c_1(T^{1,0}X). Consistently, c_1(K_X) = -c_1(T^{1,0}X), so a Kahler
metric with R_h >= -h is one whose canonical direction is nonnegative after a
twist by h, which is the hypothesis of the volume estimate.

Exercise: Green. Why is log(1/|z|) exactly the right weight?
Let u be smooth on the disc and let <u>_r be the mean of u over the circle of
radius r. Two identities hold, and both are used later verbatim.
Poisson version: <u>_r - u(0) = (2/pi) integral over |z| < r of
log(r/|z|) d_z dbar_z u dS. Area version: (1/pi) integral over |z| < r of
d_z dbar_z u dS = (r/2) <d_r u>_r, where d_r is the outward radial derivative.
Why the logarithm: log(1/|z|) is the Green function of the Laplacian in the
plane, the function whose Laplacian is a point mass. Consequence used
constantly: if q = dd^c u locally, then the mean of u o f over the unit circle
equals u(f(0)) plus P_q(f), assuming enough regularity to pass to r = 1. That
converts a boundary average into the value at the centre plus a Green-weighted
interior integral. The limit r -> 1 is legal when u o f is continuous on the
closed disc and d_z dbar_z (u o f) lies in L^a for some a > 1, by Holder and
dominated convergence against the integrable weight.

Exercise: The functional we maximise. Why maximise this and not something else?
Fix a point x in X, a Kahler metric h, a smooth closed real (1,1)-form q, and a
number delta > 0. Consider the functional of a finite-area holomorphic disc f
with f(0) = x:
J(f) = d(f) + P_q(f) - delta A_h(f).
The three terms have three different jobs. The term d(f) makes the problem
anchored at the centre, so the disc cannot shrink away. The term P_q(f) is the
one that sees a global potential: it is a boundary average minus a central
value, so it can be evaluated on a potential and controlled near the boundary.
The term -delta A_h(f) is negative and controls total area, which is what buys
compactness. Why a maximum: a maximiser is a critical point, and at a critical
point the second derivative in any allowed direction is <= 0. That inequality,
computed once, is the engine of the whole paper; both positivity results are
faces of it. This is the single idea to hold on to.

Exercise: Which q do we plug in, and why those two?
Two choices carry the proof. First q = R_k and h = k the fixed metric, and
then the second derivative bounds a Poisson integral of the Ricci form along
every disc; feeding that through a disc envelope gives locally bounded
potentials for K_X, hence nefness. Second q = -h with delta = 1, so the
functional is d(f) - P_h(f) - A_h(f); comparing with the constant disc and
using the pointwise inequality R_h >= -h turns the same second-derivative bound
into a lower bound for h^n in terms of k^n, efficient in n and in the constant
D. The constants differ in nature: the first application sends delta to zero
after fixing the test disc, the second keeps delta = 1 and produces a uniform
constant independent of h. Neither application needs the maximiser itself to be
bounded in any norm uniformly.

Exercise: Existence. Why does a maximiser of J exist at all?
Because the negative area term makes the problem coercive, and that has to be
checked carefully since the functional involves an unbounded weight. Fix
C_q >= 0 with |q(v,v)| <= C_q h(v,v) for all tangent v, which exists because q
is a smooth form and h a smooth metric on a compact manifold. Choose
r_0 in [1/2, 1) so close to 1 that 2 C_q log(1/r_0) <= delta/2. Write
a_f(z) = h(f'(z), f'(z)) and b_f(z) = q(f'(z), f'(z)). On the outer annulus
r_0 < |z| < 1 put T_f(z) = delta a_f(z) - 2 log(1/|z|) b_f(z). Comparing
|b_f| against C_q a_f and using the choice of r_0 gives
T_f(z) >= (delta/2) a_f(z) >= 0. That single inequality does both jobs: it
bounds J above, because the annulus contribution to P_q minus delta times its
contribution to A_h is minus the integral of a nonnegative quantity, and it
bounds the OUTER area of every approximately maximising disc. On the inner
disc there is no such control, but there the weight log(1/|z|) is integrable
and f' is bounded by the Brody constant on compacts, so the inner contribution
is bounded by a constant K independent of f. Hence
J(f) <= D + K - (delta/(2 pi)) times the outer annulus area integral. The
supremum is therefore finite, and nonnegative because the constant disc centred
at x gives J = 0 exactly: its derivative vanishes, so d = 0, and q, h vanish on
it. Values of J are thus in [0, D+K].

Exercise: Attainment. How does a maximising sequence converge to a disc?
Take a maximising sequence f_j with J(f_j) >= -1. The estimate of the previous
block bounds their outer-annulus areas uniformly; the interior derivative bound
bounds their inner areas; so sup_j A_h(f_j) < infinity. Brody's lemma gives
local uniform convergence on compacts of a subsequence to a holomorphic disc f
centred at x, and derivative convergence on compacts gives pointwise convergence
of a_{f_j} and b_{f_j}. Fatou then gives A_h(f) <= liminf A_h(f_j) < infinity,
so the limit is a legitimate finite-area disc. Three terms, three convergence
arguments. The central term d converges because derivatives converge. The inner
integral converges by dominated convergence, using the uniform inner bound and
the integrable logarithmic weight. On the outer annulus the functions T_{f_j}
are nonnegative and converge pointwise to T_f, so Fatou gives
limsup of minus the annulus integral of T_{f_j} <= minus the annulus integral of
T_f. Adding the three gives J(f) >= limsup J(f_j), which is upper
semicontinuity of J, and that forces attainment. Note what was NOT used: no
area bound uniform in delta, no boundary regularity of the f_j. The discs are
regular only on the interior at this stage.

Exercise: Dilation. Why compare a maximiser with its own scalings?
Because a maximiser is a maximum over ALL finite-area centred discs, so it
beats its own dilates, and that comparison is a differential inequality in
disguise. Put f_r(z) = f(rz) for 0 < r < 1. Change of variables gives
d(f_r) = r^2 d(f), A_h(f_r) = A_{h,r}(f), and A_{q,s}(f_r) = A_{q,rs}(f).
Next, the identity log(1/|z|) = integral from |z| to 1 of ds/s, plus Fubini,
gives the key rewrite P_q(f) = 2 times integral from 0 to 1 of
A_{q,s}(f) ds/s, and therefore
P_q(f) - P_q(f_r) = 2 times integral from r to 1 of A_{q,s}(f) ds/s.
Fubini is legal even though the integrand is signed: near the centre the
derivatives are bounded, and away from the centre the weight is bounded while
q(f',f') is controlled by the finite h-area. Since f_r is centred at x and of
finite area, maximality gives J(f) >= J(f_r); substituting the three
computations and using h >= 0 gives the tail inequality
0 <= delta (A_h(f) - A_{h,r}(f)) <= (1 - r^2) d(f) + 2 times integral from r
to 1 of A_{q,s}(f) ds/s.

Exercise: Two corollaries of the tail inequality. What do we actually get?
First, divide the right-hand side by 1 - r and let r tend to 1. The quantity
(1-r^2)/(1-r) tends to 2, and the integral term divided by 1 - r tends to
2 A_q(f) by the fundamental theorem, as long as the integrand has a limit,
which it does because the outer area converges. The left-hand side is
nonnegative for every r, so the limit is nonnegative: A_q(f) >= -d(f) >= -D.
This is the pointwise lower bound for the q-area of a maximiser, and it is
what converts an inequality about a boundary integral into an inequality about
the Ricci form later. Second, for r >= 1/2 the right-hand side of the tail
inequality is at most 2 d(f)(1-r) + 4 C_q A_h(f)(1-r), so the h-area tail is
bounded by a constant C_f times (1-r), with C_f depending on the disc.
Constant in r, not uniform in f, not uniform in h: that is legitimate, because
the regularity hypothesis for the frame construction is a statement about one
disc at a time. The Brody constant D, by contrast, is uniform in everything,
and it is the only uniform constant in the paper.

Exercise: Boundary regularity. Why does maximality control the derivative at the edge?
This is the passage from a statement about AREA to a statement about POINTWISE
size, and it is entirely local. The area tail says the part of the disc near
the boundary carries area proportional to the distance to the boundary. The
derivative at a point is controlled by the average of its square over a small
ball around that point. Combining the two gives a pointwise bound that blows up
like the square root of the distance to the boundary, which is exactly the
integrability needed later. Concretely, let t = 1 - |z|. Choose a small
number c in (0, 1/4), independent of z and of f, and consider the ball
B(z, ct) of radius ct about z in the disc. The straight segment from z to any
point of that ball stays at distance at least (1-c)t from the boundary, so
Brody's gradient bound bounds the length of its image by a constant times
sqrt(D) times c/(1-c). Choosing c small enough makes that less than the radius
of the charts we are using. That is the reason for introducing nested charts
V in W in U at the start: on W the fixed metric k and the Euclidean coordinate
metric are comparable, with one constant for the whole finite family. Inside
such a chart, the squared Euclidean norm of the derivative vector is
subharmonic, because it is a sum of squared moduli of holomorphic functions.
The mean-value inequality for subharmonic functions then gives
|f'(z)|_k^2 <= (C_1/t^2) times the integral of |f'|_k^2 over B(z, ct).
Every point of B(z, ct) has modulus bigger than 1 - (1+c)t, so for t small
enough the area tail bounds that integral by C_2 t. Divide, and the conclusion
is |f'(z)|_k <= C_f times (1-|z|)^{-1/2} near the boundary, which is the
paper's boundary regularity statement. Notice the constant is C_f, depending on
the disc, and the exponent 1/2 is forced by two-dimensional area: area tail
linear in t, mean value dividing by t^2, leaves 1/t.

Exercise: Boundary continuity. How does a pointwise derivative bound produce a limit?
By integration along radii, and by completeness of X. Let zeta be a point of
the unit circle and integrate the derivative estimate along the radial segment
from r zeta to s zeta, with 1/2 <= r < s < 1. The length of the image is at most
C_3 times the integral from r to s of (1-u)^{-1/2} du, which is at most
2 C_3 sqrt(1-r). The bound is uniform in zeta, which is the whole point: the
right-hand side depends on r only. Since X is compact and therefore complete as
a metric space, each radial path has a limit as r tends to 1, so f extends to
each boundary point. Moreover the continuous maps zeta -> f(r zeta) on the
circle converge uniformly as r tends to 1, because the same estimate bounds
their oscillation; a uniformly Cauchy sequence of continuous functions on a
complete space has a continuous limit. That limit is the boundary value, and it
extends f continuously to the closed disc. The estimate also shows the boundary
map is not wild: it is the uniform limit of real analytic circle maps at
controlled speed. This block is the only place where compactness of X enters
the regularity argument, and it enters only as completeness.

Exercise: Sobolev exponent. Why do we need W^{1,p} for some p > 2 and not just continuity?
Continuity says the disc does not escape at the boundary. The frame
construction needs more: it needs the derivative to be integrable to a power
strictly above two, because that is the regime where Sobolev functions on a
two-dimensional domain form an algebra and where point evaluation is bounded.
The improvement is a counting argument on annuli. For large integers j let
E_j = {z : 2^{-j-1} < 1 - |z| < 2^{-j}}, a thin annulus of thickness about
2^{-j}. The area tail bounds the L^2 norm of the derivative on E_j by
C_4 2^{-j}, and the pointwise bound bounds its sup norm there by C_5 2^{j/2}.
Interpolating between the two, for 2 < p < 4,
the L^p norm on E_j is at most (sup)^{p-2} times the L^2 norm squared, which is
at most C_p 2^{-j(4-p)/2}. That is summable in j precisely because p < 4, and
the interior compact disc contributes a bounded amount, so the derivative lies
in L^p of the disc for every p < 4. Hence the maximiser is in W^{1,p} for every
2 < p < 4, in the sense of local target coordinates, and this is
the same boundary regularity statement again. The exponent 4 is not a wall
that the argument barely
misses; it is where the two-dimensional scaling stops helping. Read the three
blocks just ended as one sentence: maximising forces the area tail, the area
tail forces the square-root derivative bound, the square-root derivative bound
forces continuity and W^{1,p} for p in (2,4).

Exercise: Sobolev conventions. What exactly is meant by "f is in W^{1,p} up to the boundary"?
Since f maps into a manifold, the phrase must be read in charts. Continuity on
the closed disc lets us cover the image of the closed disc by finitely many
target coordinate charts, each relatively compact; shrink the source disc a
little around the preimages so that the images land in those charts. In each
chart, the components of f are ordinary functions on a piece of the disc,
bounded, smooth in the interior, with first derivatives in L^p by the estimate
just proved. Those classical derivatives are the weak derivatives, because
testing against compactly supported smooth functions and integrating by parts
uses only interior smoothness. So each component lies in W^{1,p} up to the
boundary, with the continuous boundary values already constructed serving as
its trace. Two facts about p > 2 are used repeatedly and both are standard:
W^{1,p} on a bounded planar domain embeds continuously into the continuous
functions on the closure, and it is a Banach algebra under pointwise
multiplication. The second follows from the first plus the product rule, and it
is what lets us multiply frames, take inverses of matrices with entries in
W^{1,p}, and compose with smooth functions, all without leaving W^{1,p}.
Everything in the frame and variation sections is an application of that one
algebra property.

Exercise: Frame. What is a frame of the pulled-back tangent bundle, and why is one needed?
The disc maps into X, so over each point z of the disc sits the vector space
T^{1,0}_{f(z)}X of dimension n, and these spaces vary with z. Their disjoint
union is a holomorphic vector bundle E = f^*T^{1,0}X on the disc. A frame is a
choice of n holomorphic sections e_1,...,e_n of E that are independent at every
point; equivalently, an invertible n by n matrix of functions whose columns are
those sections in some local trivialization. Why bother: with a frame, a family
of discs through f corresponds to a family of vector-valued functions on the
disc, and anything geometric becomes matrix algebra. Varying the disc in
directions b_1,...,b_n means writing the variation as a coefficient matrix
times the frame. The second derivative of the functional then becomes a
computation with matrices, and the curvature appears as a matrix-valued
function on the disc. Without a frame, all of that is coordinate soup.

Exercise: First frame. Why does a holomorphic frame exist despite f being merely W^{1,p}?
Because the base is a disc, and holomorphic vector bundles on a disc are
trivial. Topologically it is easy: the closed disc is contractible, so the
continuous pullback bundle is trivial, and by approximating the coefficients
uniformly in finitely many local frames and gluing with a partition of unity one
gets a continuous frame v that is of class W^{1,p} and stays pointwise
independent. In that frame the Dolbeault operator on the open disc becomes
dbar + a, with a a matrix of (0,1)-forms whose coefficients lie in L^p, because
f is in W^{1,p} and the transition functions of X are smooth. A frame is
holomorphic exactly when dbar e = 0, which in the frame v reads dbar U = -a U
for the change-of-frame matrix U. To solve that, extend a by zero to a slightly
larger disc and, on a small disc U of radius r, use the solid Cauchy transform
T_U g(z) = (1/pi) integral over U of g(w)/(z-w) dS(w), which satisfies
dbar T_U g = g. The estimates are ||T_U g||_sup <= C_p r^{1-2/p} ||g||_p and
||T_U g||_{W^{1,p}} <= C_{p,r} ||g||_p, the first from Holder applied to the
kernel and the second from the L^p bound for the planar Cauchy singular
integral. So U_0 = I - T_U(a U_0) is solved by contraction in the sup norm once
r is small, and the solution stays within distance 1/2 of the identity, hence is
invertible, and its W^{1,p} norm is controlled. For two such local solutions the
ratio U_i^{-1} U_j has zero dbar on the overlap, hence is holomorphic there, so
the local solutions glue into a holomorphic vector bundle on the larger disc.
A holomorphic vector bundle on a disc is holomorphically trivial, by the
Oka-Grauert principle, so it has a global holomorphic frame. Writing that frame
against the original trivialization gives U continuous, invertible and W^{1,p}
on the closed disc with dbar U = -a U, and e = v U is the holomorphic frame we
wanted, continuous and nonsingular up to the boundary. That is the paper's
lemma producing the first frame. The
moral: holomorphicity is cheap on a disc, and the price is paid in Sobolev
regularity of the change of frame, which is exactly what we can afford.

Exercise: Boundary-unitary frame. Why impose B^* h B = I on the circle?
A frame B is called boundary-unitary for a Hermitian metric h on X if
B(z)^* h(f(z)) B(z) = I for every z on the unit circle. In words: on the
boundary the columns of B are an orthonormal basis for the metric h. Why that
condition: the second-derivative computation produces a matrix N = B^* h B, and
its determinant is exactly the Poisson integral we want to bound; if N equals
the identity on the boundary, the determinant is a scalar function vanishing on
the boundary and the Green formula gives the answer directly. Without boundary
normalization the determinant would carry a boundary term nobody can control.
So the whole point of this section is: take the holomorphic frame from the
previous block, and change it by a matrix S, holomorphic and invertible, so that
N = I on the circle. That is a factorization problem for a positive definite
matrix-valued function on the circle, and it has a classical answer from
prediction theory. Two words of warning before we do it. The matrix N is only
W^{1,p} and continuous, not smooth, so the classical smooth theory has to be
redone with regularity bookkeeping. And the result must stay holomorphic in the
interior and nonsingular at the boundary, which is not automatic from a
factorization that only controls almost-everywhere boundary values.

Exercise: Hardy space. What is H^2, and why is it where this factorization lives?
The Hardy space H^2 is the space of holomorphic functions on the disc whose
Taylor coefficients are square summable; equivalently, via boundary values, the
closed subspace of L^2 of the circle consisting of functions whose Fourier
series has only nonnegative frequencies. Take it vector valued for C^n, so
each element is an n-tuple of holomorphic functions. Its inner product is
<u,v> = integral over the circle of u^* v dsigma with dsigma = dtheta/(2 pi).
The shift operator is multiplication by z, and it is an isometry on H^2 because
|z| = 1 on the circle; its image zH^2 is closed, and by the Beurling-type
argument its orthogonal complement W has dimension exactly n. Concretely: a
vector is in the orthogonal complement of zH^2 exactly when it is orthogonal to
z times everything, which pins down its Fourier coefficients to just the
constant term, and constants form an n-dimensional space. Now the trick. Given
a positive definite matrix function M on the circle, bounded between m I and
M_0 I for positive constants, keep the same vector space H^2 but change the
inner product to <u,v>_M = integral of u^* M v dsigma. This is an equivalent
Hilbert norm. Multiplication by z is still an isometry for it, because M does
not see z on the circle; and zH^2 is still closed with codimension n. Let
W = H^2 minus zH^2 with respect to that weighted inner product, and let
s_1,...,s_n be an M-orthonormal basis of W. Form the matrix S with those
columns. Then the vectors z^j s_a with j >= 0 and 1 <= a <= n are
M-orthonormal, because iterating H^2 = W orthogonally direct sum zH^2 shows
that their span has orthogonal complement inside every z^j H^2, and the
intersection of those is zero. Pairing z^j s_a with z^k s_b and using the
isometry: for j = k the inner product reduces to
integral of s_a^* M s_b dsigma, and orthonormality says that equals the
identity. That is the whole factorization: S^* M S = I almost everywhere on the
circle. No smoothness was used, only boundedness and positive definiteness.
Note what S is: n holomorphic H^2 functions, i.e. n holomorphic functions on
the disc, whose boundary values orthonormalize M.

Exercise: The factor is invertible. Why is S nonsingular in the interior?
Because point evaluation is a continuous surjection of H^2 onto C^n, and the
shift-invariant span of the columns fills the whole space. Precisely: for a
point zeta in the disc, evaluating an H^2 function at zeta is a bounded linear
functional, so the vector of evaluations of the columns is well defined, and
the range of S(zeta) contains the evaluations of all vectors z^j s_a. Those
vectors are dense in H^2, and their evaluations therefore span C^n, because
evaluation is onto. Hence S(zeta) is surjective, hence invertible, for every
zeta in the open disc. Two consequences worth stating separately. First, S is a
holomorphic matrix on the disc with holomorphic inverse. Second, the identity
S^* M S = I a priori holds only almost everywhere, since it came from an L^1
computation. At this stage we cannot yet say S is invertible at boundary
points, because we only know its boundary values exist almost everywhere and we
do not know they are continuous. That gap is what the next block closes, and
the paper is explicit that nonsingularity at the boundary is a conclusion, not
an assumption.

Exercise: Regularity of the factor. How do we get continuity and W^{1,p_0} for S at the boundary?
By measuring how fast the boundary values of M can move, and reading that rate
off against the Toeplitz operator. Step one: since M is in W^{1,p} on the disc
with p > 2, its boundary values have a modulus of continuity in L^p. Writing
M_theta for M composed with rotation by theta, for |theta| <= 1/4,
||M_theta - M||_p <= C |theta|^{1-1/p}. This is elementary, not a trace
theorem: pick a radius rho between 1-2t and 1-t at which the angular L^p norm of
the radial derivative is at most C t^{-1/p}, integrate radially to compare the
boundary values with the circle of radius rho, use Holder at that radius to
compare with a rotation, and add. Step two: consequently M restricted to the
circle lies in the fractional Sobolev space H^s for every s with
1/2 < s < 1 - 1/p, because squaring the rotation difference and integrating
against |theta|^{-1-2s} is finite, which by Parseval is the same as
sum over integers l of (1 + |l|^{2s}) |M_l|^2 being finite. Step three: the
Toeplitz operator T_M v = Pi(Mv), where Pi is the orthogonal Hardy projection,
is bounded, self-adjoint, and bounded below by m, so it is invertible; its
range is closed by the lower bound and dense because the orthogonal complement
of the range is the kernel of the adjoint. Step four: a column v of S is
characterized by the property that T_M v is a constant vector, which is just a
rewriting of membership in W. Rotating that identity and subtracting gives
T_M(v_theta - v) = Pi((M - M_theta) v_theta). Boundedness of v and
invertibility of T_M turn that into ||v_theta - v||_2 <= C ||M_theta - M||_2, so
v is in H^s as well. Step five: since s > 1/2, the Taylor series of v
converges absolutely and uniformly on the closed disc, so v is continuous
there, and the boundary values are honest continuous functions. Step six: the
derivative estimates. For 1/2 < r < 1,
||v'(r .)||_2^2 <= C (1-r)^{2s-2} times the weighted sum of coefficients, and
||v'(r .)||_infinity <= C (1-r)^{s-3/2} times the square root of the same sum.
Interpolating between these two bounds gives
||v'(r .)||_{p_0}^{p_0} <= C (1-r)^{p_0(s-3/2)+1}, which is integrable in r
exactly when 2 < p_0 < 2/(3/2 - s). That interval is nonempty because s > 1/2,
and we also take p_0 <= p. Conclusion: S is in W^{1,p_0} on the disc and
continuous on the closed disc. Now the identity S^* M S = I holds at every
boundary point by continuity, not just almost everywhere, so S is invertible at
every boundary point as well; compactness then gives uniform nonsingularity on
the closed disc. Finally B = e S is continuous, nonsingular, W^{1,p_0}, and
holomorphic on the disc, with B^* h B = I on the circle. That is
the paper's boundary-unitary frame theorem. Read the sequence of quantifiers:
continuity first, identity
at every boundary point second, invertibility third. Reverse that order and the
argument is circular, which is why the paper spells it out.

Exercise: Variations. Why can we not simply differentiate in the direction z b_j?
Because z b_j is a section of the pulled-back bundle, and sections are not
derivatives of discs. To apply the second derivative test we need actual
holomorphic discs F_t with F_0 = f, with the centre fixed, F(t, 0) = x for all
small t, and with d/dt of F_t at t = 0 equal to z b_j(z). That last condition is
the interesting one: the factor z matters because the disc must stay pinned at
the centre, and a variation vanishing at the origin in the parameter cannot move
the point z = 0. So the question is whether the entire infinite-dimensional
family of sections z b_j can be integrated into a finite-dimensional family of
genuine discs. For smooth discs this is a classical construction, known as a
spray. Here f is only W^{1,p_0} up to the boundary, so the construction has to
be redone with Sobolev bookkeeping throughout. The payoff is that the space of
variations is only n-dimensional: n complex parameters t_j, each moving the
disc in one direction. The second derivative sum in those n directions is what
the paper calls the trace of the complex Hessian.

Exercise: Fiber coordinates. How do we flatten the geometry near the graph of f?
We build a family of coordinates transverse to f, one for each point of the
disc, with Sobolev dependence on the point. Cover the closed disc by finitely
many relatively open sets U_i and choose target charts phi_i : V_i -> C^n with
f of the closure of U_i compactly inside V_i. Let rho_i be a smooth partition of
unity subordinate to the U_i. Two matrices appear, and their roles must be kept
apart. The frame e from the earlier lemma trivializes the tangent bundle; the
chart derivative D phi_i(f(z)) converts tangent vectors into coordinate
vectors. Their product A_i(z) = D phi_i(f(z)) e(z) is holomorphic in the
interior, continuous and invertible up to the boundary, and W^{1,p_0}, being a
product of such things. Define the chart-relative coordinate of a point y near
f(z) by C_i(z, y) = A_i(z)^{-1} ( phi_i(y) - phi_i(f(z)) ). This is holomorphic
in y, and for fixed admissible y it is holomorphic in z in the interior, because
both phi_i(y) and phi_i(f(z)) are, and the matrix inverse of a holomorphic
invertible matrix is holomorphic. Now average with the partition of unity:
Phi(z, y) = sum over i of rho_i(z) C_i(z, y), each term used only where its
chart is defined, extension by zero being harmless because the support of rho_i
sits inside its own source patch. Two identities come out, and everything
downstream rests on them: Phi(z, f(z)) = 0, because every C_i(z, f(z)) = 0; and
D_y Phi(z, f(z)) e(z) = I, because the average of A_i^{-1} D phi_i(f(z)) e(z)
is the average of A_i^{-1} A_i, which is the average of the identity matrix with
weights summing to one. So Phi is a coordinate system transverse to the graph,
normalized to have unit derivative in the frame e. What is spoiled: Phi is not
holomorphic in z, because the rho_i are not; that defect is repaired several
blocks below, and it is the only reason the nonlinear dbar equation appears at
all.

Exercise: Inverting fiber coordinates. How do we get E with Phi(z, E(z, xi)) = xi?
Because Phi has the identity as its fiber derivative on the graph, the inverse
function theorem applies fiberwise, with a uniform radius by compactness; the
work is making the dependence on z Sobolev rather than merely continuous, since
every later substitution has to land back in W^{1,p_0}. Work on one small source
patch and fix one target chart psi, and put c(z) = psi(f(z)) and
e_psi(z) = D psi(f(z)) e(z). Normalize the forward map by
K_z(v) = Phi(z, psi^{-1}(c(z) + e_psi(z) v)). Then K_z(0) = 0 and
D_v K_z(0) = I, both inherited from the two normalization identities. Continuity
of all fiber derivatives, and compactness of the closed disc, give a single
radius R > 0 such that these maps are defined for |v| <= R and satisfy
||D_v K_z(v) - I|| <= 1/4 there. For |xi| < R/2 the equation
v = xi - (K_z(v) - v) maps the closed ball of radius R into itself, because
|K_z(v) - v| <= |v|/4, and it is a contraction with constant 1/4. Its fixed
point V(z, xi) is continuous in both variables and holomorphic in xi, as a
uniform limit of contraction iterates, each of which is holomorphic in xi.
Define E by psi(E(z, xi)) = c(z) + e_psi(z) V(z, xi). Local inverses agree near
xi = 0 by uniqueness and then on their whole common fiber domain by holomorphy;
taking the minimum of the finitely many radii gives one fiber neighbourhood with
Phi(z, E(z, xi)) = xi, and with E(z, 0) = f(z) and E_xi(z, 0) = e(z).
Differentiating the identity Phi(z, E(z, xi)) = xi in the fiber variable gives
E_xi = (D_y Phi)^{-1}. Now for the Sobolev dependence, which is the subtle part.
The chart formula for K_z depends smoothly on the finite coefficient list
a(z) = (rho_i(z), A_i(z)^{-1}, phi_i(f(z)), c(z), e_psi(z)) over i. Every entry
of that list is W^{1,p_0} with values in a compact set, since f is W^{1,p_0} and
the charts are smooth. Fix a smaller closed fiber polydisc inside |xi| < R/2.
The strict inequalities in the contraction argument survive a small perturbation
of the coefficient list, so the finite-dimensional implicit function theorem,
applied with those coefficients regarded as real parameters, makes V a smooth
function of the list and of xi, holomorphic in xi, with uniformly bounded first
derivatives in the parameters on the fixed polydisc. The Sobolev chain rule then
shows that z -> E_psi(z, xi) is in W^{1,p_0} with the estimate
|E_psi(z, xi)| <= C and |grad_z E_psi(z, xi)| <= C b(z) for almost every z,
uniformly in xi, for some single b in L^{p_0}; one may take b to be a constant
plus the sum of absolute values of the first weak derivatives of the coefficient
list. Derivatives in xi obey the same kind of bound after one more shrink, by
Cauchy's formula on the polydisc. This is the estimate that makes everything
else possible, and it is worth restating in words: varying the fiber variable
costs nothing in Sobolev regularity on the disc.

Exercise: Substitution is holomorphic. Why does u -> E(z, u(z)) preserve W^{1,p_0}?
By expanding in the fiber variable and using the geometric bounds above. Write
E_psi(z, xi) = sum over multi-indices alpha of E_alpha(z) xi^alpha. Cauchy's
formula on the fixed fiber polydisc, differentiating under its contour integral
in the weak z variable, gives ||E_alpha||_infinity + ||grad E_alpha||_{p_0}
<= C r_1^{-|alpha|} for some positive r_1. Now measure functions with the
algebra norm ||a|| = ||a||_infinity + ||grad a||_{p_0}, which is
submultiplicative, since W^{1,p_0} is a Banach algebra for p_0 > 2. Then the
series sum over alpha of E_alpha u^alpha converges normally in the local
W^{1,p_0} space whenever ||u|| is small, because the geometric decay
r_1^{-|alpha|} beats the polynomial growth in the number of multi-indices of a
given degree. Each term is a continuous homogeneous polynomial in u, so the sum
is holomorphic as a map from a neighbourhood of zero in W^{1,p_0} into the local
W^{1,p_0} space. Immediately: varying the fiber variable costs nothing, and
varying the parameter costs nothing, so all the parametrized discs below are
continuous up to the boundary automatically. One caution about the meaning:
this is a statement in fixed local target coordinates, and no linear structure
on X is being presumed; the charts are the linear structure, and there are
finitely many of them.

Exercise: The dbar defect. Why is E(z, u(z)) not holomorphic even when u is?
Because dbar_z of the fiber coordinate does not vanish: the partition of unity
made Phi non-holomorphic in z, and that defect survives in E. The chain rule in
local target coordinates gives
dbar( E(z, u(z)) ) = E_xi(z, u(z)) dbar u + dbar_z E(z, u(z)),
and there is no term involving dbar of u-bar, because E is holomorphic in its
fiber variable. So E(z, u(z)) is weakly holomorphic precisely when
dbar u + Q(z, u) = 0, where Q(z, xi) = E_xi(z, xi)^{-1} dbar_z E(z, xi). The
definition is independent of the chosen target chart, because a holomorphic
change of chart multiplies both terms by the same invertible derivative. Now
compute Q honestly instead of by definition-chasing. Differentiate the identity
Phi(z, E(z, xi)) = xi in zbar at fixed xi:
0 = dbar Phi(z, E) + D_y Phi(z, E) dbar_z E. Since
E_xi = (D_y Phi)^{-1}, this reads dbar_z E = -E_xi dbar Phi, so
Q = -dbar_z Phi(z, E(z, xi)) = -sum over i of (dbar rho_i)(z) C_i(z, E(z, xi)),
using that C_i is holomorphic in z at fixed y. Three properties follow at once.
Q is holomorphic in xi. Q is jointly continuous and bounded on the closed disc
times a smaller closed fiber polydisc, because it is a finite sum of continuous
functions times the (dbar rho_i). And its Taylor coefficients decay
geometrically, ||Q_alpha||_infinity <= C r_2^{-|alpha|} by Cauchy's formula, so
the substitution map u -> Q(., u(.)) is holomorphic from a neighbourhood of zero
in W^{1,p_0} into L^{p_0}, using the embedding of W^{1,p_0} into the sup norm.
Finally the constant and linear terms vanish, and the reason is the
normalization. Since C_i(z, f(z)) = 0, Q(z, 0) = 0. Since
D_y C_i(z, f(z)) e(z) = A_i(z)^{-1} D phi_i(f(z)) e(z) = I, and
E_xi(z, 0) = e(z),
Q_xi(z, 0) = -sum (dbar rho_i)(z) D_y C_i(z, f(z)) e(z) = -sum (dbar rho_i)(z) I,
and that is zero because the derivatives of a partition of unity sum to zero.
So Q(z, 0) = 0 and Q_xi(z, 0) = 0: the equation is not merely solvable, it is
solvable with a vanishing linear part, which is exactly the condition that puts
the implicit function theorem in its easy regime.

Exercise: Implicit function theorem. How is dbar u + Q(u) = 0 solved?
By the Banach space implicit function theorem, applied to a correctly centered
operator. Let T be a right inverse of dbar on the disc, given by the solid
Cauchy transform of the whole disc; it is bounded from L^{p_0} to W^{1,p_0}.
Point evaluation is bounded on W^{1,p_0}, so H g = T g - (T g)(0) is a bounded
operator with dbar H g = g and (H g)(0) = 0; the centering is what keeps the
disc pinned at x. Seek a solution of the form
u(t, z) = z S(z) t + H g(t)(z), with g(t) in L^{p_0}. Since S is holomorphic and
in W^{1,p_0}, the equation becomes G(g, t) = g + Q(., z S t + H g) = 0, a
holomorphic equation with values in L^{p_0}. The vanishing of the constant and
linear parts of Q gives G(0,0) = 0 and D_g G(0,0) = I, the identity operator.
So the implicit function theorem applies in its easiest form and produces a
holomorphic solution g(t) for small t. Differentiating the equation at zero
gives g(0) = 0 and D_t g(0) = 0, so the linear part of u comes entirely from
the explicit term. Concretely, u(t, 0) = 0, u(0, z) = 0, and
du/dt_j at t = 0 equals z S(z) e_j. Set F(t, z) = E(z, u(t, z)). Then the three
spray identities hold: F(0, z) = f(z), F(t, 0) = x, and
dF/dt_j at t = 0 equals E_xi(z, 0) z S(z) e_j = e(z) z S(z) e_j = z (e S)(z) e_j,
which is z b_j(z) as required, because B = e S. Each F(t, .) is weakly
holomorphic, since u(t, .) solves the dbar equation, hence holomorphic; each is
continuous on the closed disc and W^{1,p_0} by the substitution estimate; and
each is holomorphic in t by boundedness of point evaluation on W^{1,p_0}
together with holomorphy of E in the fiber variable. Local separate holomorphy
then gives joint holomorphy on the product. Finally each F(t, .) has compact
image and is W^{1,p_0} with p_0 > 2, so its area is finite for every metric, and
small parameters give admissible centered discs. That is the paper's spray proposition. Note
what was never needed: no condition on the boundary map of F(t, .), and no
extension across the circle.

Exercise: Derivatives at the centre. Is it legitimate to differentiate the central derivative in t?
Yes, but not by the naive argument, and the paper flags the point. Evaluation of
the derivative at a point is not a bounded functional on W^{1,p_0}: a function
can be small in the Sobolev norm and still oscillate quickly near the centre.
However, on the HOLOMORPHIC subspace, and on any strictly smaller disc, the
Cauchy formula makes derivative evaluation bounded. Since t -> F(t, .) is
holomorphic into the holomorphic subspace, the map t -> df/dt of F(t, .) at z = 0
is holomorphic in t, and can be differentiated as often as we like. That is the
whole content of the paper's remark after that proposition, and it is what allows the
second derivative of the central term d(F_t) to be computed by ordinary calculus
on a fixed finite-dimensional space, namely the tangent space T^{1,0}_x X.

Exercise: Determinant identity. Why is log det(B^*hB)(0) = P_{R_h}(f)?
This is the identity that converts a matrix at a point into a boundary integral,
and the boundary-unitary normalization is what makes it clean. Let
N(z) = B(z)^* h(f(z)) B(z). N is smooth in the open disc, continuous and
uniformly positive definite on the closed disc, in W^{1,p_0}, and equals the
identity on the unit circle. In a holomorphic target chart the matrix of B is
holomorphic and nonsingular, so log det N and log det h(f(z)) differ by
log |det B|^2, which is harmonic. Hence
d_z dbar_z log det N = d_z dbar_z log det h(f(z)) = -R_h(f', f'),
where the second equality is the definition R_h = -dd^c log det h rewritten on
the disc, using the convention that q(f',f') = d_z dbar_z(u o f) whenever
q = dd^c u. The right-hand side is in L^{p_0/2}, which is integrable to a power
above one because p_0 > 2, so the Green identity may be passed to the limit
r = 1. The boundary value of log det N is zero, so Green gives
0 - log det N(0) = P_{-R_h}(f), that is, log det N(0) = P_{R_h}(f). Written out:
the value of a harmonic-type quantity at the centre equals a Green-weighted
average of minus the Ricci form along the disc. The chart independence of the
formula is part of the statement: a holomorphic change of chart adds a harmonic
term to log det N. Keep this identity in mind as the "determinant half" of the
Hessian computation; the next blocks supply the "trace half".

Exercise: The local Hessian identity. Why is a second derivative equal to a d_z dbar_z?
This is a chain rule computation, and its shape is the reason the whole method
works. Write L for the sum over j of the second derivative d^2/dt_j dbar_t_j
evaluated at t = 0, the trace of the complex Hessian in the n parameter
directions. Suppose q = dd^c phi locally, with phi smooth and real. Joint
holomorphy of F in (t, z) gives
q(F_z, F_z) = d_z dbar_z ( phi o F ), while summing the second derivatives in
the parameters gives
L (phi o F) = sum over j of q(F_{t_j}, F_{t_j}) at t = 0.
Commuting the parameter derivatives with the z derivatives, and using
F_{t_j}(0, z) = z b_j(z), produces the identity
L q(F_z, F_z) = d_z dbar_z ( |z|^2 tr( B^* q(f) B ) ).
Everything on the right is evaluated at the original disc f; the factor |z|^2
appears because the variation was chosen to vanish at the centre, and the trace
appears because we summed over the n parameter directions. Three facts about the
right-hand side matter. It is intrinsic, hence chart independent. The mixed
source derivative belongs to L^{p_0/2}, by the differentiability in t of
t -> q(F_z(t,.), F_z(t,.)) into L^{p_0/2} and the product rule. And integration
against 1 or against log(1/|z|) is a continuous functional on that space,
because p_0/2 > 1. That last point is what justifies differentiating the
functionals under their integrals, and it is the only place where the Sobolev
exponent above two gets used a second time.

Exercise: The central and Poisson traces. What do we get when q is arbitrary?
Reading two different functionals off the single identity. The central term:
since F(t, 0) = x, the vector d_z F(t, 0) is holomorphic in t and lies in the
fixed finite-dimensional space T^{1,0}_x X, and its t_j derivative at zero is
b_j(0); differentiating its squared k(x) norm gives
L d(F_t) = tr( B(0)^* k(x) B(0) ). Note this is a trace over the same n
directions, and note that everything is evaluated at the single point x.
The Poisson term: integrate the local identity against the Green weight
log(1/|z|), which is the same as applying the functional P to both sides, and
use Green on the scalar function |z|^2 tr(B^* q(f) B). That function is
continuous on the closed disc and vanishes at z = 0, so its Poisson integral is
its boundary average. On the boundary the columns of B are h-orthonormal, so
tr(B^* q B) is the h-trace of q, written tr_h q. The conclusion is
L P_q(F_t) = (1/(2 pi)) times the integral over the circle of
(tr_h q)(f(e^{i theta})) d theta. Read that as: the second derivative of the
Poisson functional is the average of the h-trace of q over the boundary circle
of the disc, evaluated on the boundary values of f. No regularity of f beyond
continuity is used here, because the trace of q is a smooth function on X
composed with a continuous map.

Exercise: The area trace. Why is L A_h(F_t) = n - A_{R_h}(f) the delicate one?
Because h is a Kahler form and thus positive, and the identity at the identity
matrix must be combined with a radial derivative that we do not yet know has a
limit. Here is the computation. Put T = tr N, note that T = n on the boundary.
Integrating the local identity with q = h over the disc of radius r gives
(1/pi) integral over |z| < r of L h(F_z, F_z) dS = r^2 <T>_r + (r^3/2) <T_r>_r,
the second term because the radial derivative of r^2 <T>_r produces both a
derivative of the weight and a derivative of T. The left side tends to
L A_h(F_t) by integrability, and the first right-hand term tends to n. The
second right-hand term is the problem: we would like <T_r>_r to tend to
<(log det N)_r>_r, because the curvature calculation identifies
(r/2) <(log det N)_r>_r with -A_{R_h,r}(f), and then the limit is immediate. At
the identity matrix the trace and the log determinant have the same
differential, so the error is T_r - (log det N)_r = tr( (I - N^{-1}) N_r ),
bounded by C ||N(r,.) - I||_2 ||N_r(r,.)||_2. The radial Sobolev estimate plus
the boundary value N = I gives
||N(r,.) - I||_2 <= (1-r)^{1/2} ( integral from r to 1 of ||N_s||_2^2 ds )^{1/2},
first on almost every radial line by the fundamental theorem for Sobolev
functions, then by Cauchy-Schwarz in the angle. Since the integral of
||N_r||_2^2 over (1/2, 1) is finite, there are radii r_nu tending to 1 along
which (1 - r_nu) ||N_r(r_nu, .)||_2^2 stays bounded; otherwise that integrability
would fail. Along those radii the error is o(1). Substituting and using
(r/2) <(log det N)_r>_r = -A_{R_h,r}(f), whose right side tends to -A_{R_h}(f),
the left side has a limit independent of the sequence, and hence so does the
right. The result is L A_h(F_t) = n - A_{R_h}(f). Two remarks. First, the
equality is genuinely about a weak boundary limit; the disc is not assumed to
extend holomorphically across the circle, only continuously. Second, the term n
is just the dimension, arising as the limit of r^2 <T>_r, and it is what makes
the final inequality have the shape "something <= delta(n - nonnegative)".

Exercise: Maximizing inequality. What does the second derivative test actually say?
It says the sum of the two traces is at most delta times the third. Since
t = 0 is a local maximum of the real-valued twice differentiable function
t -> J(F_t) on a neighbourhood of 0 in C^n, the trace of its complex Hessian at
0 is <= 0, that is, L J(F_t) <= 0. Substituting the three trace identities of
the previous blocks gives the master inequality: for a maximiser of
d + P_q - delta A_h centred at x, with B the boundary-unitary frame along it,
tr( B(0)^* k(x) B(0) )
  + (1/(2 pi)) times the integral over the circle of (tr_h q)(f(e^{i theta}))
  <= delta ( n - A_{R_h}(f) ).
Three features are worth naming because every later step uses one of them.
The left side has no delta and no maximiser-dependent constant: it is a sum of a
positive definite trace at the centre and a boundary average of a scalar
function. The right side is delta times (dimension minus a quantity that the
dilation inequality bounds below by -D). And the whole inequality is available
for the two specific choices of (h, q). Alongside it we retain the determinant
identity log det N(0) = P_{R_h}(f) and the dilation corollary A_q(f) >= -d(f).
Those three facts, valid for every maximiser, are the entire quantitative output
of the analysis so far. Everything in the rest of the paper is the standard
machinery of complex geometry applied to them.

Exercise: Nefness, step one. How do we bound the Poisson Ricci integral of an ARBITRARY disc?
By comparing it with a maximiser chosen for its own centre. Fix the background
metric k and take the functional with h = k and q = R_k, for a number delta with
0 < delta <= 1; the functional is d + P_{R_k} - delta A_k. Let f be a maximiser
for this delta and this centre, which exists by the existence theorem. Put
M = sup over X of |tr_k R_k|, a finite number because X is compact and R_k is a
smooth form. The dilation corollary gives A_{R_k}(f) >= -d(f) >= -D. The master
inequality, applied with h = q-free terms as above, gives
tr N(0) <= delta ( n - A_{R_k}(f) ) + M <= delta(n + D) + M.
Since delta <= 1 that is at most n + D + M. Now the determinant identity and the
arithmetic-geometric mean inequality for the eigenvalues of the positive matrix
N(0) turn that trace bound into a determinant bound:
P_{R_k}(f) = log det N(0) <= n log( (n + D + M)/n ) =: C_1.
Consequently the maximal value of the functional is at most D + C_1: the middle
term is at most C_1, the first is at most D, and the last is <= 0. Now fix an
ARBITRARY finite-area disc g, and run the maximisation at the centre g(0). Since
the maximiser beats g,
d(g) + P_{R_k}(g) - delta A_k(g) <= D + C_1,
and A_k(g) is finite, so letting delta tend to zero gives
P_{R_k}(g) <= D + C_1 =: C. That is the paper's uniform Ricci-disc bound.
Notice the order of
quantifiers, which is where this argument is easy to misread: for each delta we
get a different maximiser, with no uniform bound on its area or its regularity,
and we never need one; we only need its VALUE, which is bounded uniformly in
delta and in the centre. The limiting step happens after the arbitrary disc g is
fixed, not before.

Exercise: Disc envelope. How does a bound on P_{R_k} become a plurisubharmonic potential?
Work on the total space of the determinant bundle, where the Ricci form is an
exact curvature, and let an infimum over discs do the work. Put L = det T^{1,0}X,
so K_X = L^*, and let pi : Y -> X be L with its zero section removed. On Y
define ell(s) = log |s|_k^2, the log squared norm with respect to the metric k
induced on L. This is smooth and finite valued on the connected complex manifold
Y. In a local trivialization s = lambda s_i(x) it reads
ell(s) = log|lambda|^2 + a_i(x) with a_i = log|s_i|_k^2, and since lambda is
never zero the first summand is pluriharmonic; hence d d^c ell = -pi^* R_k as
forms on Y. Now define v to be the Poletsky-Rosay envelope of ell: the infimum
of the boundary mean of ell o G over discs G in Y centred at s that are
holomorphic past the closed unit disc. Two facts. For such a G the projection
f = pi o G is a finite-area disc in X, and Green applied to ell o G, whose
d d^c equals -R_k(f',f'), gives mean of ell o G = ell(s) - P_{R_k}(f). Since
the nefness bound says P_{R_k}(f) <= C, every admissible mean is at least
ell(s) - C, while the constant disc gives mean exactly ell(s). So
ell - C <= v <= ell, which both makes v real valued and rules out the
identically minus infinity alternative. The Poletsky-Rosay theorem then makes v
plurisubharmonic, and it applies here without any boundedness of ell because the
test discs are required to be holomorphic past the closed disc. Homogeneity is
the last step: for lambda in C^*, multiplication by lambda is a biholomorphism
of Y and bijects test discs centred at s with those centred at lambda s, while
ell(lambda s) = ell(s) + log|lambda|^2. Therefore
v(lambda s) = v(s) + log|lambda|^2, so v - ell is invariant under the
C^*-action and descends to a function u on X with v = ell + u o pi, where
-C <= u <= 0. Finally, pulling v back along the holomorphic local section s_i,
which avoids the zero section, gives a_i + u plurisubharmonic and locally
bounded, because a_i is smooth and u is bounded. Those are the local weights of
a singular Hermitian metric on K_X with nonnegative curvature.

Exercise: Line bundle sign. Why does the resulting current live on K_X and not on L?
Because the weights are taken for the dual. If s_i = g_ij s_j is a change of
frame of L, then a_i - a_j = log|g_ij|^2, and the dual frame of K_X = L^*
transforms by s_i^* = g_ij^{-1} s_j^*. Prescribe |s_i^*|^2 = exp(-(a_i + u)).
That prescription is compatible with the transformation law precisely because of
the sign flip, and its curvature is d d^c (a_i + u) = -R_k + d d^c u, which is
nonnegative and represents -[R_k] = 2 pi c_1(K_X). Had we kept the plus sign we
would have built a metric on L with positive curvature, which would say that
c_1(L) is pseudoeffective, the opposite statement. This is the whole content of
the paper's bounded-potential proposition, and the point of doing it on the
total space rather than on X is
that a global potential for R_k need not exist on X: what exists is a bounded
potential after the change of sign, and it is produced by an infimum over discs,
not by solving an equation.

Exercise: Smoothing. Why is a bounded plurisubharmonic potential enough to conclude nefness?
Because bounded potentials have zero Lelong numbers, so their Monge-Ampere
measure has no atoms, and standard regularization loses arbitrarily little
positivity. The paper includes a direct argument, and its one real idea is worth
isolating: bounded plurisubharmonic functions are not continuous, so uniform
convergence of smoothings is unavailable, and the replacement is a comparison
theorem at fixed multiples of the same small radius, together with a
regularized maximum that can safely discard a branch. Set up: choose finitely
many coordinate patches U_i, smooth real a_i with alpha = d d^c a_i on
neighbourhoods of the closure of U_i, and open sets V_i compactly inside W_i
compactly inside U_i covering M; the functions psi_i = a_i + u are
plurisubharmonic and bounded above and below with bounds uniform over the finite
family, and psi_i - psi_j = a_i - a_j on overlaps. Define the sup-convolution
S_{i,b}(x) = sup of psi_i over the closed ball of radius b about x, and let
phi_{i,b} be S_{i,b} convolved with a kernel rescaled to radius b/4. Then phi
is smooth and plurisubharmonic, and comparing balls of radius b/2, b and 2b
gives S_{i,b/2} <= phi_{i,b} <= S_{i,2b}. The comparison theorem is this: for a
plurisubharmonic psi with m <= psi <= M on balls of radius up to R, and any
A > 1, one has 0 <= S_x(Ar) - S_x(r) <= (M-m) log A / log(R/r). Proof: the
function t -> S_x(e^t) is nondecreasing and convex, because on each complex line
the maximum principle compares psi on an annulus with the function affine in
log|z| taking the two boundary values, and convexity at three logarithmic radii
gives the statement. Two consequences. First, for fixed small b the map
x -> S_{i,b}(x) is plurisubharmonic: it is the supremum of the
plurisubharmonic translates psi_i(x + zeta) over |zeta| <= b, so upper
semicontinuity plus local boundedness does it. Second, the overlap discrepancy
sup over the intersection of W_i and W_j of |(phi_{i,b} - a_i) - (phi_{j,b} - a_j)|
tends to zero as b does, at rate O(1/|log b|) + O(b): coordinate balls of
radius b in one chart contain and are contained in balls of comparable fixed
radii in the other, a_i - a_j varies by O(b) across such a ball, and the
comparison theorem does the rest, uniformly because there are finitely many
pairs.

Exercise: Gluing. How do the local smoothings become one global function?
By a regularized maximum with a cutoff that makes the unwanted branches
provably inactive before their patches end. Choose smooth chi_i with
-3 <= chi_i <= 0, chi_i = 0 on V_i, chi_i = -3 near the boundary of W_i, and a
fixed constant C_0 with d d^c chi_i >= -C_0 k for every i. Given sigma > 0,
pick b so small that every discrepancy from the previous block is less than
sigma, and set f_i = phi_{i,b} - a_i + sigma chi_i on W_i. For each nonempty
finite index set I define the regularized maximum of (t_i) by averaging
max(t_i + s_i) against a smooth even probability density on R^I supported in a
ball of radius sigma/4. It is smooth, convex, nondecreasing in each variable,
it commutes with adding a constant to all arguments, and, crucially, a branch
that sits more than sigma/2 below the maximum of the others never attains the
shifted maximum, so deleting it leaves exactly the regularized maximum on the
smaller index set. Now put u_sigma(x) equal to that regularized maximum of the
f_i(x) over the set of patches containing x. This is globally smooth: at a
boundary point of W_i pick j with that point in V_j; nearby chi_i = -3 and
chi_j = 0, so the overlap comparison gives f_i < f_j - 2 sigma, and the exact
deletion property removes branch i before its patch runs out. For the curvature,
near a point in a patch with local potential a_j, each active branch has
a_j + f_i = phi_{i,b} + (a_j - a_i) + sigma chi_i, and a_j - a_i is
pluriharmonic since both are potentials for alpha, so
d d^c (a_j + f_i) >= -C_0 sigma k. A regularized maximum of branches is a
convex combination of their complex Hessians plus a nonnegative form, since the
first derivatives are nonnegative and sum to one and the Hessian is positive
semidefinite. Hence alpha + d d^c u_sigma >= -C_0 sigma k, and choosing sigma
with C_0 sigma <= epsilon finishes. Applied with alpha = -R_k and the bounded u
of the envelope, this gives smooth u_epsilon with
-R_k + d d^c u_epsilon >= -epsilon k, and therefore 2 pi c_1(K_X) is nef. Note
the shape of the loss: not a small change of the potential but a small NEGATIVE
curvature allowance proportional to the smoothing scale, which is exactly the
definition of nef.

Exercise: Volume bound. How does the second choice of q produce a lower bound for h^n/k^n?
By paying with the assumption R_h >= -h. Fix a point x and maximise
J(f) = d(f) - P_h(f) - A_h(f), the case q = -h and delta = 1. The constant disc
centred at x has J = 0, so the maximiser satisfies P_h(f) + A_h(f) <= d(f) <= D.
Both left terms are nonnegative, so the assumed Ricci inequality gives
A_{R_h}(f) >= -A_h(f) >= -D and P_{R_h}(f) >= -P_h(f) >= -D. Now the master
inequality with tr_h(-h) = -n reads
tr( B(0)^* k(x) B(0) ) - n <= n - A_{R_h}(f), that is,
tr( B(0)^* k(x) B(0) ) <= 2n - A_{R_h}(f) <= 2n + D.
The determinant identity gives
det( B(0)^* h(x) B(0) ) = exp(P_{R_h}(f)) >= e^{-D}, and the
arithmetic-geometric mean inequality applied to the positive matrix
B(0)^* k(x) B(0), whose trace is at most 2n + D, gives
det( B(0)^* k(x) B(0) ) <= ((2n + D)/n)^n. Take the quotient: the factor
|det B(0)|^2 appears in both determinants and cancels, leaving
h^n/k^n (x) >= e^{-D} (n/(2n + D))^n = c_{n,D}. This is
the paper's uniform volume bound. Two things to notice. The constant depends
only on n and D, not
on h, x, or the maximiser; the maximiser's own regularity and its dependence on h
are irrelevant because the identities hold for each h separately. And the proof
is a two-sided squeeze: the h-determinant is bounded BELOW by the Ricci
assumption, the k-determinant is bounded ABOVE by the trace bound, and the trace
bound came from the boundary normalization. Without the boundary-unitary frame
neither half would be available.

Exercise: Aubin-Yau. How does a volume bound become positive top self-intersection?
By producing, in each nearby cohomology class, a Kahler metric to which the
volume bound applies, then letting the class degenerate. Nefness says that for
each t > 0 there is a smooth real a_t with -R_k + d d^c a_t >= -(t/2) k, so
gamma_t = -R_k + t k + d d^c a_t is Kahler, being at least (t/2)k. Apply the
negative-sign complex Monge-Ampere theorem, which for a compact Kahler manifold,
a Kahler form gamma and a smooth positive volume form Omega produces a smooth w
with gamma + d d^c w Kahler and (gamma + d d^c w)^n = e^w Omega, with no
normalization of Omega imposed on the class. Take Omega = e^{a_t} k^n and
h_t = gamma_t + d d^c w_t; then h_t^n = e^{a_t + w_t} k^n. Taking -d d^c of the
logarithm of the ratio of the two volume forms gives
R_{h_t} = R_k - d d^c (a_t + w_t) = -h_t + t k, because
h_t = -R_k + t k + d d^c (a_t + w_t). In particular R_{h_t} >= -h_t, so the
volume bound applies with the SAME constant for every t:
the integral of h_t^n is at least c_{n,D} times the integral of k^n. On the
other hand [h_t] = alpha + t[k] with alpha = [-R_k] = 2 pi c_1(K_X), so by
closedness and Stokes the integral of h_t^n equals the integral of
(alpha + t[k])^n, which is a polynomial in t with coefficients the intersection
numbers of alpha and k. The left side is bounded below uniformly in t, so the
polynomial, being continuous at 0, has nonnegative value there; being a limit of
positive quantities it has positive value there. Hence the integral of alpha^n is
at least c_{n,D} times the integral of k^n, which is positive. Three remarks.
The argument uses cohomology only: no convergence of the metrics h_t, and no
bounds on their potentials, are needed. The sign is the negative one throughout,
which is why c_1(K_X) and not c_1(T^{1,0}X) appears. And the exponent n matters:
this produces the top self-intersection and nothing smaller.

Exercise: Morse. How do nefness and positive top self-intersection give bigness?
By perturbing the curvature form into a positive definite form and accounting for
the negative part. Let L be nef with beta = 2 pi c_1(L) and V = integral of
beta^n > 0. For 0 < epsilon <= 1 choose a smooth metric on L with curvature
Theta_eps >= -epsilon k; then P_eps = Theta_eps + 2 epsilon k is positive
definite. At a point where Theta_eps has a negative eigenvalue, write its
eigenvalues relative to k as lambda_1,...,lambda_n and pick j with lambda_j < 0.
Then |lambda_j| <= epsilon, while for i different from j, lambda_i + 2epsilon
dominates |lambda_i|. Multiplying the n - 1 other factors gives the pointwise
bound |Theta_eps^n| <= n epsilon P_eps^{n-1} wedge k on the whole locus where an
eigenvalue is negative, and the integral of the right side is
n epsilon times the integral of (beta + 2epsilon[k])^{n-1} wedge [k], which is
O(epsilon) with a constant independent of the chosen metric because what remains
is a fixed cohomological polynomial in epsilon, bounded for 0 <= epsilon <= 1.
The degenerate locus contributes zero, so removing the loci of index at least
two changes the total integral by at most O(epsilon); since that total is V, the
integral of Theta_eps^n over the locus where Theta_eps is nondegenerate with at
most one negative eigenvalue is at least V - O(epsilon) > 0 for a small fixed
epsilon. Demailly's strong holomorphic Morse inequalities then say that a
positive integral of Theta^n over that locus forces L to be big, with the
manifold Moishezon. Two remarks. The number of negative eigenvalues is
independent of the background metric, so the locus is well defined. And in
dimension one the locus of index at least two is empty and the same computation
with P_eps^0 = 1 applies unchanged, which is why the lemma needs no separate
treatment of curves.

Exercise: Projectivity. Where does X become a projective variety?
From Moishezon plus Kahler, by a theorem that needs no input from us. Applying
the previous block to L = K_X, using nefness and positive top self-intersection,
gives that K_X is big and X is Moishezon. Moishezon's projectivity theorem says
that a compact Moishezon manifold which admits a Kahler metric is projective; a
stronger form covering Moishezon spaces with 1-rational singularities is due to
Namikawa, and a smooth manifold satisfies that hypothesis trivially. So at this
point X is a smooth projective variety. That matters because the last step uses
the projective cone theorem, which is unavailable on a merely Kahler manifold.
This is the exact place where the Kahler hypothesis is spent: it was not needed
for the disc analysis, and it is needed here to convert bigness into algebraicity.

Exercise: Cone theorem. Why does no rational curve turn bigness into ampleness?
By subtracting an effective divisor and then testing a perturbed canonical class
against the cone of curves. Kodaira's lemma gives an integer m > 0, an ample
divisor H and an effective divisor E with m K_X linearly equivalent to H + E.
For the needed form of the lemma take H to be a smooth very ample hyperplane
section; bigness gives h^0 of m K_X growing like m^n along divisible powers,
whereas the weak Morse inequality gives h^0 of the restriction of m K_X to H
growing only like m^{n-1}, so the restriction sequence produces a nonzero section
of m K_X - H for some m, which is E. In dimension one H is a finite set and the
same count is O(1). Now choose a small positive rational eta so that the pair
(X, eta E) is Kawamata log terminal; this is possible even when E is singular or
nonreduced, because on a log resolution the discrepancies have the form
a_i - eta b_i with a_i >= 0 from smoothness of X and b_i >= 0 from effectivity of
E, and finitely many inequalities of the type "greater than -1" and "less than
one" are satisfied for all small enough eta. The log cone theorem says that a
projective Kawamata log terminal pair with effective rational boundary has a
rational curve of negative (K_X + Delta)-degree whenever K_X + Delta is not nef.
Since X contains no rational curves, K_X + eta E is therefore nef in the
numerical sense. Combine with the Kodaira decomposition:
(1 + eta m) K_X is numerically equivalent to (K_X + eta E) + eta H, a sum of a
nef and an ample rational divisor, which is ample by Kleiman's criterion, since
it is positive on every nonzero element of the closed cone of curves. Clear the
positive rational coefficient: a positive integral multiple of K_X is ample,
hence K_X itself is ample. The main theorem follows: a rational curve would
normalize to a nonconstant map from P^1, whose restriction to C inside P^1 is
nonconstant by the identity theorem, contradicting the hypothesis. So X has no
rational curve, K_X is big, and the previous block gives ampleness.

Exercise: Freeness. Why do pluricanonical bundles become globally generated at m = n+2?
Once K_X is ample, Fujita's freeness statement applies in its all-powers form:
for an ample line bundle L on a manifold of dimension n, the adjoint bundle
K_X tensor L^{m-1} is globally generated for every m - 1 >= n + 1. Take
L = K_X, so the adjoint bundle is K_X^{m}, and m >= n + 2 is exactly the
condition. "Globally generated" means that at every point the sections span the
fibre, which is weaker than very ample but is what the theorem gives with this
uniform exponent. Note the shape of the statement: the exponent n + 2 is uniform
in X, not merely finite for each X. That is a feature of the multivariate Fujita
conjecture, of which this is a case, and it is the reason the corollary is stated
with an explicit bound rather than as "for some m".

Exercise: Universal covers. What does a semialgebraic universal cover buy?
A finite cover that splits as a domain quotient times a compact factor. Suppose
the ordinary universal cover of X is biholomorphic to a semialgebraic open subset
of a complex projective variety, with openness in the ordinary complex topology.
The main theorem first gives projectivity of X. The semialgebraic-cover
classification then gives the universal cover in the form D_0 x C^m x F with F
simply connected, normal and projective. Smoothness of the product forces F to be
smooth, since a product manifold has smooth factors. Brody hyperbolicity passes
to covers and to factors, so m = 0 and F is Brody hyperbolic. If F has positive
dimension, the main theorem gives K_F ample, and Kobayashi's finite-automorphism
theorem for manifolds with ample canonical bundle says Aut(F) is finite; the
statement is also true when F is a point. Now let a deck transformation act. As a
biholomorphism of D x F, its first coordinate is constant on each compact
connected fibre F, and applying the same observation to the inverse shows that
the base map is an automorphism of D and the fibre maps are automorphisms of F,
independent of the base point because D is connected and Aut(F) is finite. So the
deck group acts by products. The kernel of its homomorphism to Aut(F) has finite
index, acts freely on D and properly discontinuously there, the latter by
applying the deck action to K x F for compact K in D; in particular it is
discrete. Quotienting the universal cover by that kernel gives a connected finite
unramified holomorphic cover isomorphic to (D/Gamma_0) x F, which is a finite
etale cover of the projective manifold X, and compactness of the cover forces
D/Gamma_0 to be compact. Either factor may be a point.

Exercise: Inventory. What in this argument is new, and what is borrowed?
Borrowed, in order of use: Brody's rescaling lemma; the solid Cauchy transform
and its L^p bounds; the Oka-Grauert principle for holomorphic bundles on a disc;
Wiener-Masani and Helson-Lowdenslager spectral factorization for the weighted
Hardy space construction; the complex Banach implicit function theorem in the
form due to Gloeckner; the Poletsky-Rosay disc envelope theorem on manifolds; the
regularization theory of closed positive currents, with a self-contained bounded
potential case in the appendix; the negative-sign complex Monge-Ampere theorem of
Aubin and Yau; Demailly's strong holomorphic Morse inequalities; Moishezon's
projectivity theorem in Namikawa's form; the log cone theorem and Kleiman's
criterion; and Fujita freeness for the pluricanonical corollary. New, and the
only new thing: the disc analysis. Existence and, crucially, boundary regularity
of the maximisers; the boundary-unitary holomorphic frame at exactly the Sobolev
regularity the maximisers supply; the integration of the sections z b_j into
genuine center-fixed holomorphic discs; and the resulting two trace identities,
one giving a uniform bound on the Poisson Ricci integral of an arbitrary disc,
the other giving a uniform lower bound for h^n in terms of k^n. Everything else
is scaffolding around those two identities. It is also worth being clear about
what the proof does not do: no curvature sign is assumed at any point; the
constants are not uniform in the manifold, only in n and D; the Kahler
hypothesis is spent exactly once, at the passage from Moishezon to projective;
and the projectivity conclusion does not come from any explicit embedding.