Partitions of Unity (proof)
Finally we can prove what we’ve asserted: given any open cover 
of a smooth manifold 
we can find a countable smooth partition of unity 
subordinate to it.
So, as we’ve seen we can find a countable atlas 
; we use 
for the coordinate maps since we’ll want the 
free. We’ve also seen that we have a smooth bump function 
between the two cubes 
and 
in 
. So let’s define
Now, it’s easily verified that the furthest points from the origin in are the corners, each of which is 
away. Thus we can tell that 
, and the support of is contained entirely within 
. This means that the support of 
is entirely contained within 
— by the time we get to the edge it’s already smoothly tailed off to zero, and so even though we define it piecewise, is a smooth function defined on all of .
Now set up the sum
Since is locally finite and is supported within , this sum is guaranteed to be finite at each point, which makes 
a smooth function on all of . The ball 
is contained within the cube , so takes the constant value 
on this ball. Since the preimages 
form an open cover of , there is always at least one 
for which 
. In particular, for which it’s not zero, and thus the whole sum is nonzero at 
.
Since is never zero, we can divide by it. We define 
. These are smooth functions on all of , and their sum is everywhere exactly . Thus the form a partition of unity subordinate to 
. And since refines , the partition is subordinate to this cover as well.
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March 14, 2011 -
Posted by John Armstrong |
Differential Topology, Topology
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