diff f6bf760..6fbd3ac symplectic_ball_packings_v2.txt
diff --git a/symplectic_ball_packings_v2.txt b/symplectic_ball_packings_v2.txt
index fde85de..ca13e16 100644
--- a/symplectic_ball_packings_v2.txt
+++ b/symplectic_ball_packings_v2.txt
@@ -23,7 +23,7 @@ THEOREM (Siegel-Yao, Conjecture A, now proved).
There are no other conditions, for any n >= 3, any k >= 1,
and any positive capacities.
-The two inequalities are obstructions, so they are necessary
+The two ineualities are obstructions, so they are necessary
(Section 3). The whole rest of the file proves they are also
sufficient (Sections 4 through 12). The strict "<" reflects a
convention: the closed balls must land inside the OPEN interior
@@ -96,7 +96,7 @@ is just ordinary area and is boring.
3. WHY THE TWO INEQUALITIES ARE NECESSARY
----------------------------------------------------------------
-We prove that if a packing exists, both inequalities must hold.
+We prove that if a packing exists, both ineualities must hold.
This direction is short.
3.1 Volume.
@@ -110,7 +110,7 @@ This direction is short.
so a little of the target's volume is never used.)
3.2 The two-ball obstruction (Gromov 1985).
- This is the surprising inequality R_i + R_j < R. It comes
+ This is the surprising ineuality R_i + R_j < R. It comes
from pseudo-holomorphic ("J-holomorphic") curves, a genuine
theorem of symplectic topology. We state exactly the input
we use; this is the one deep tool imported into the proof.
@@ -134,7 +134,7 @@ This direction is short.
Since the two ball images are disjoint, this sphere cannot
enter the second ball. Tracking the largest such sphere, and
the smallest one the second ball forces, gives a + b <= R.
- The point of the inequality is that it is NOT visible to
+ The point of the ineuality is that it is NOT visible to
volume: two balls of capacities 0.6 and 0.6 have total
volume 2*0.6^n/n!, which for large n is far below 1/n!, yet
they cannot fit because 0.6+0.6 > 1.
@@ -151,7 +151,7 @@ This direction is short.
remaining images are still disjoint from these two). Apply
3.2 to that pair:
R_i + R_j < R for every i != j.
- That gives all the pairwise inequalities at once.
+ That gives all the pairwise ineualities at once.
Together, 3.1 and 3.3 prove the necessity half of the theorem:
volume and all pairs are forced. The hard direction -- that
@@ -347,7 +347,7 @@ so the model keeps its shape.
<= (M_lambda + d) - d
= M_lambda ,
where we used lambda M_lambda <= M_lambda + d, an immediate
- rearrangement of M_lambda >= -d/(1-lambda). The inequality is
+ rearrangement of M_lambda >= -d/(1-lambda). The ineuality is
strict because P > d there.
INNER CORE (p = tau v in tau Delta, v in Delta). Convexity of
K_lambda gives
@@ -395,7 +395,7 @@ so the model keeps its shape.
black box with error eps.
Choose q_j for j=1,...,m with errors eps_1,...,eps_m whose
total is < d/2. Let q = q_1 o q_2 o ... o q_m (applied in
- this order). Chaining the inequalities,
+ this order). Chaining the ineualities,
(K_lambda - P)(q u) <= R(K_lambda - P)(u) + sum eps_j
< K_lambda(u) - d + d/2
< K_lambda(u).
@@ -456,7 +456,7 @@ so the model keeps its shape.
H(y, z_j) = l(y) * B_t( y, z_j / sqrt(l(y)) ).
Because H does not involve the angles of the OTHER
coordinates, those moments are unchanged; only the j-th angle
- moves. On slices of length <= l(y) the inequality
+ moves. On slices of length <= l(y) the ineuality
F(q u) <= R_j F + eps holds by Lemma 6.10; on the outer strip
l(y) < p_j <= L(y) we can take the identity map (there R_j F
already dominates F to within the modulus of continuity of F,
@@ -1063,7 +1063,7 @@ captures essentially all the volume.
with sum rho_i^n < 1. Increase them slightly to RATIONAL
r_i > rho_i with r_* := max r_i < 1/2 and sum r_i^n < 1 (open
conditions, so rationals are dense enough). By 9.6 and the
- strict volume inequality choose RATIONAL parameters
+ strict volume ineuality choose RATIONAL parameters
r_* < h < d < 1/2, with J_m(h) > (1/n!) sum_i r_i^n . (9.7)
(First push h close to 1/2 to make J_m(h) exceed the sum, then
insert d strictly between h and 1/2.)
@@ -1133,7 +1133,7 @@ HYPERPLANE inside T = Bl_o PP^n, then a QUADRICS CURVE inside that
hyperplane.
10.1 Choosing the parameters.
- The inequalities are strict and finite in number, so we can
+ The ineualities are strict and finite in number, so we can
perturb up to RATIONAL numbers
A < a < 1, 1/2 < a, rho_i < r_i < s := 1 - a,
sum_i r_i^n < 1 - a^n. (10.1)
@@ -1749,7 +1749,7 @@ We now combine everything.
THEOREM 12.2. Theorem 1 holds: for every n >= 3, k >= 1, and
positive R, R_1,...,R_k, the packing exists iff both the volume
- inequality and all pairwise inequalities hold.
+ ineuality and all pairwise ineualities hold.
Proof. Necessity is Section 3 (volume plus Gromov's two-ball
rigidity, then all pairs). Sufficiency is Claim 12.1. Together
they prove the "if and only if" in every n >= 3. QED.
@@ -1765,7 +1765,7 @@ n = 1 (real dimension 2).
"symplectic" = "area preserving". Packing is pure area:
the k disks fit iff R_1 + ... + R_k < R. No pairwise condition
is needed (2-dim packing of round disks is flexible). The
- second inequality is vacuous. Nothing deep happens.
+ second ineuality is vacuous. Nothing deep happens.
n = 2 (real dimension 4).
This is the rich case. Gromov (1985) proved the two-ball
@@ -1780,7 +1780,7 @@ n = 2 (real dimension 4).
balls fill almost the whole volume); Buse-Hind (2011, 2013)
extended equal-ball stability to higher dimensions and to all
closed rational symplectic manifolds; Buse-Hind-Opshtein (2016)
- proved strong stability for unequal balls, provided every
+ proved strong stability for uneual balls, provided every
capacity is sufficiently small. But the EXACT criterion (which
configurations fit) is still open in dimension four.
Our proof uses, in Cases 2 and 3, a base surface of positive