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#1 2009-04-03 06:25:10

ben
Full Member

Offline

Push-me, pull you

This is so much fun! Look

We will suppose that

are vector spaces, and that the linear operators (aka transformations)
. Then we know that the composition
as shown here.

http://i246.photobucket.com/albums/gg98/quarkH/pushfp4.png

Notice the rudimentary (but critical) fact, that this only makes sense because the codomain of
is the domain of


Now, it is a classical result from operator theory that the set of all operators
is a vector space (you can take my word for it, or try to argue it for yourself).

Let's call the vector space of all such operators
etc. Then I will have that
are vectors in these spaces.

The question naturally arises: what are the linear operators that act on these spaces? Specifically, what is the operator that maps
onto
?

By noticing that here the
is a fixed domain, and that
, we may suggest the notation
. But, for reasons which I hope to make clear, I will use a perfectly standard alternative notation
.

Now, looking up at my diagram, I can think of this as "pushing" the tip of the f-arrow along the g-arrow to become the composite arrow. Accordingly, I will call this the push-forward of
on
, or, by a horrid abuse of English as we normally understand it, the push-forward of


So, no real shocks here, right? Ah, just wait, the fun is yet to begin, but this post is already over-long, so I'll leave you to digest this for a while.........

 

#2 2009-04-04 08:22:08

ben
Full Member

Offline

Re: Push-me, pull you

Recall that, given

as a linear operator on vector spaces, we found
as the linear operator that maps
onto
, and called it the push-forward of
. In fact let's make that a definition:
defines the push-forward.

This construction arose because we were treating the space
as a fixed domain. We are, of course, free to treat
as as fixed codomain, like this.
http://i246.photobucket.com/albums/gg98/quarkH/pullcz3.png
This seems to make sense, certainly domains and codomains come into register correctly, and we easily see that
.

Using our earlier result, we might try to write the operator
, but something looks wrong;
is going "backwards"!

Nothing daunted, let's adopt the convention
. (We will see this choice is no accident)

Looking up at my diagram, I can picture this a pulling the "tail" of the h-arrow back along the g-arrow onto the composite arrow, and accordingly (using the same linguistic laxity as before), call
the pull-back of
, and make the definition:
defines the pullback

(Compare with the pushforward)

This looks weird, right? But it all makes beautiful sense when we consider the following special case of the above.
http://i246.photobucket.com/albums/gg98/quarkH/pull2kc7.png
where I have assumed that
is the base field for the vector spaces
. As before, the composition makes sense, and I now have
, and the pullback
. But, hey, lookee here....

is the vector space of all linear maps - functionals -
, so we quite simply have that
, the dual vector spaces.

Putting this all together I find that, for
I will have
as my pullback.

I say this is just about as nice as it possibly could be. What say you?

 

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