Re: Partitions of unity proof roadblock

On Dec 4, 2:14 pm, smn <smnewber...@xxxxxxxxxxx> wrote:
On Dec 3, 3:39 pm, smn <smnewber...@xxxxxxxxxxx> wrote:



On Nov 30, 10:59 am, Jeff Rubin <JeffBRu...@xxxxxxxxx> wrote:

I recently posted about the trouble I was having understanding
a proof in Lang's Fundamentals of Differential Geometry relating
to partitions of unity. So I tried turning to another text,
Manifolds, Tensor Analysis, and Applications by Abraham,
Marsden, and Ratiu. This text had a different approach but
my understanding ran into the same roadblock. I was wondering
if someone could help me out and explain.

The proof is of proposition 5.5.17 where M is a paracompact manifold
modeled on a Banach space E. It shows the equivalence of a number
of statements, the last two of which are:
(iv) every chart domain of M admits C^k partitions of unity
subordinate
to any open covering
(v) E admits C^k partitions of unity subordinate to any open covering
of E.

I had no problem with the implications (i)=>(ii)=>(iii)=>(iv). But
the proof
of (iv)=>(v) has got me stumped:

I wanted to describe a simplified version of the proof which pointed
out the problem I'm having, but I can't figure out how to do that
without sounding very confused. So please bear with me; here
is the full proof directly from the text: (I use /\ for intersection
and \/ for union)

"Consider now any open covering {U_\alpha}_{\alpha \in A} of E and let
(U, \phi) be an arbitrary chart of M. Refine first the covering of E
by taking
the intersections of all its elements with all translates of \phi(U).
Since E
is paracompact, refine again to a locally finite open covering {V_
\beta}.
The inverse images by translations and \phi of these open sets are
subsets of U,
hence chart domains, and thus by (iv) they admit partitions of unity
subordinate
to any open covering, for example to {V_\beta /\ U_\alpha : \alpha \in
A};

This is rediculous! V_beta is already contained in some U\alpha so is
a member of the covering of itself.so you might as well take the one
element cover V_\beta =V_beta/\U_alpha_1 for some alpha_1
(g_1\beta=1 (identically (all other g_i\betas =0 Identically.
So you don't have a double sum at all .Give up on this proof entirely
and consider the problem itself.Even the modification where you then
just showed me how to extend the g's (good job) doesnt help as you
said.Anyway the proof in Marsden is wrong .

Here is what I think the problem really is.Given a Banach space E with
B an open ball -say the unit open ball about 0 such that(1) for any
open cover of B there is a C^k (you might as well take k=oo) partition
of 1 subordinate to the given open cover ; then show(3) that there is
C^k partition of 1 for any open cover of E .By using (1) and the 2
element partition ||x||<2/3 ;and 1/3<||x||<1
shows (2) there is a C^k nonnegtive function which is 1 for ||x||<1/3
and 0 for ||x ||>2/3 .By fiddling (translation and scalr multiples 1/3
<2/3 can be replaced by ant r<s and B any open ball.
It may be that (2) is enough to establish (3) but it is not clear.One
has to look at the proofs of the construction of of partitions of 1
for the finite dimensional case to see whether the local compactness
(missing here 0 was just used to get (2) )It might be that
separability (countable base for the norm topology is needed ) I
havent looked at it for a while so don't remember .Regards smn- Hide quoted text -

- Show quoted text -


Excellent. Although I think there are a few small problems with the
proof as you have it, the idea you've used works.

Hi,I think I have a correct proof .The situation is as
follows.b=beta .V_b ,b in B is a locally finite open cover of a Banach
space E ,and each V_b admits C^k partitions of unity for any open
cover .You have to shrink V_b twice.That is ,choose an open cover of
E , V_1,b ,b in B with clV_1,b in V_b where cl is the closure in
E .Now choose V_2,b an open cover of E with clV_2,b in V_1,b.

Consider the 2 element open coverA1= V_1,b and A2= V_b \ ClV_2,b of
V_b and choose a 2 element nonnegative C^k partition of 1(defined and
C^k on V_b) ,f1 supported is A1 and f2 supported on A2.

You can't quite have the support of f1 being in A1 but you can have
the "carrier" of f1 being in A1 (the carrier is the set of points
where
f1 is nonzero). Similarly the carrier of f2 is contained in A2.
But this doesn't alter the rest of the proof.

and f1+f2 =1 on V_b.
f2 is 0 on A1 (which contains V_2,b) hence f1 is 1 on V_2,b .

f2 doesn't have to be 0 on A1, but it does have to be 0 on V_2,b
which is contained in cl(V_2,b) = V_b \ A2. Therefore f1 = 1
on V_2,b.

Since f1
is 0 on A2 ,

Not quite true. What is true is that f1 is 0 on V_b \ A1
and hence also on V_b \ Cl(A1)

if we define f1=0 onA3=te E \ClV_2b

We should define f1 = 0 on A3 = E \ Cl(A1)

the extended f1 is 0
on the open set A3 and C^k on the open set V_b and since A3 union V_b
is all of E we have the extended f1 C^k on all of E

Actually we need to comment that on the intersection
(E \ Cl(A1)) /\ V_b = V_b \ Cl(A1), f1 is 0 either way,
and this is what makes f1 C^k on all of E.

and 1 on V_2,b and
supported in V_b .

This is true. f1 is 0 outside of A1 so its carrier is contained
in A1 and therefore its support is contained in Cl(A1) which
is contained in V_b.

Let f_b = this extended f1 .f =Sum(binB)(f_b) is C^k
on E by local finiteness of the V_b 's and positive everywhere since
V_2 ,b ,b in B ,is a cover of E.The f_b/f ,b in B is a C^k partition
of 1 in E subordinate to the open cover, V_b ,b in B .

Yes, it works! Wonderful!

Regards smn

Thanks for all your help.

Jeff
.

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