18.117 Problem Set #3
1. Let V be the volume of X . Show that if
max |K (x, y )| <
then I TK is invertible and its inverse is of the form I TL , L C (X X ).
2. TK is a nite rank smoothing operator if K is of the form
K (x, y ) =
fi (x)gi (y )
Well just list a bunch of denitions. X a topological Hausdor space, second countable.
Denition. A chart is a trip (, U, V ), U open in X , V open in C and : U V a homeomorphism.
If you consider two charts (i , Ui , Vi ), i = 1, 2 we get an overl
Again, V = V n and B : V V R a non-degenerate bilinear form. A few properties of we have not
/ / 1 =
Computing the -operator
We now present a couple of applications to computation
(a) B symmetric and positive denite. Let v1 ,
IF (U, z1 , . . . , zn ) is a coordinate patch, then this splitting agrees with our old splitting. Son on a complex
manifold we have the bicomplex (, , , ). Again, we have lots of interesting subcomplexes.
Ap (X ) = Ap = ker : p,0
We have a manifold CP n . Take
P (z0 , . . . , P zn ) =
a homogenous polynomial. Then
1. P (z ) = m P (z ), so if P (z ) = 0 then P (z ) = 0
2. Eulers identity holds
Lemma. The following are equivalent
1. For all z C
18.117 Assignment # 4
1. Let Rd be a simple mdimensional convex polytope, and let Rd .
Assume that , v v = 0 for every pair of adjacent vertices v and v of .
For v Vert(), dene
ind v = #cfw_vi , vi v , < 0,
where v1 , ., vm are the vertices adjacent to
First, lets prove a holomorphic version of the inverse and implicit function theorem.
For real space the inverse function theorem is as follows: Let U be open in Rn and f : U Rn a C
map. For p U and
U open Cn . Make the convention that r (U ) = r . We showed that r = p+q=r p,q , i.e. its bigraded.
And we also saw that d = + , so the coboundary operator breaks up into bigraded pieces.
: p,q p+1,q
r , s . Then
: p,q p,q+1
d( ) = d +