Bundle morphisms and automorphism

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The discussion centers on the definition of isomorphisms in vector bundles, particularly emphasizing that a map f: B -> M must be a diffeomorphism for such an isomorphism to exist. It addresses the linear isomorphism between fibers of a vector bundle E and its pullback, clarifying that this arises from the definition of induced bundles. The conversation also distinguishes between two types of bundle automorphisms: those that preserve the base and those that induce a diffeomorphism on the base. The importance of maintaining a vector space structure in the fibers is highlighted, ensuring that the pullback remains a vector bundle. Understanding these concepts is crucial for navigating the complexities of differential geometry and vector bundles.
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In an article of differential Geometry righted by ALEXI KOVALEV , he said that to define an isomorphism of vector bundle covering a map f: B-> M ( B and M are two manifolds ) we need that f must be a diffeomorphism.

then an other question he consider an exemple of morphism vector bundle F between a vector bundle E and his pull back . why we are certain that we have a linear isomorphism between any pairs of fibres (E) and a fiber of the pull back of E.

Finally , when we take trivial bundle E=BXV ( B manifold and V typical fibre of E) any automorphism of E is defined by a smooth map B->G ( when G= group of invertible matrix)

thnx a lot to explain me this point .
 
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math6 said:
then an other question he consider an exemple of morphism vector bundle F between a vector bundle E and his pull back . why we are certain that we have a linear isomorphism between any pairs of fibres (E) and a fiber of the pull back of E.

Finally , when we take trivial bundle E=BXV ( B manifold and V typical fibre of E) any automorphism of E is defined by a smooth map B->G ( when G= group of invertible matrix)

thnx a lot to explain me this point .

The first question can be answered directly from the definition of the induced bundle.

The second is obvious.
 
math6 said:
Finally , when we take trivial bundle E=BXV ( B manifold and V typical fibre of E) any automorphism of E is defined by a smooth map B->G ( when G= group of invertible matrix).

This is not very precise. First, there are two kinds of bundle automorphisms: those preserving the base and those that induce a diffeomorphism on the base. Evidently in your question we are dealing with the first case. Then the only change is in the fibers, and this, for each point x of the base, should be an endomorphism (invertible linear transformation) of the fiber. It is only when we endow V with a linear basis that such an endomorphism is described by an invertible matrix. Change the basis and the matrix representation will change
 
for the first question how we see clearly the answer from the definition ? if we look to the definition of induced bundle we must so have a linear map g between a fibre E\f(p) ( a fibre of pullback bundle) and a fibre also E\f(p) to have a morphisme between the vector bundle E nad his pull back ?

why it is clearly g is a linear map ?
 
If you would give an exact place in Kovalev's notes where you are having a problem - it would be easier to help you.
 
thnx arkajad for your first answer you help me really to understand the meaning . just i would ask you about something .
you say " First, there are two kinds of bundle automorphisms: those preserving the base and those that induce a diffeomorphism on the base "
can you explain more these ?
 
Kovalev in his "Pulling back" section does not say exactly how the fibers of the pullback are defined. He refers to a "commutative diagram". But when, after (2.5) he says "isomorphism onto a fibre" he means a linear isomorphism.
 
why we must have linear isomorphisme onto a fibre ?
 
  • #10
math6 said:
why we must have linear isomorphisme onto a fibre ?

Well, we want to stay in vector bundles category. So, we want a pullback of a vector bundle to be again a vector bundle. Or, alternatively: we can take any pullback and then define a vector space structure in the fibers over M by defining

p+q=F^{-1}(F(p)+F(q))

ap= F^{-1}(aF(p))

where a is a number and p,q are two points in the same fibre over M. Then we get the isomorphism by the very construction.
 
  • #11
ohhhh i get it it is very simple i just didn't concentrate :shy:
it is just because we have the same fibre so we have the map id( fibre ) that'is clearly linear isomorphism .
thnx arkajad. but you didn't answer me for

"First, there are two kinds of bundle automorphisms: those preserving the base and those that induce a diffeomorphism on the base."

if you look to definition in the article you always preserve the same base ?
 
  • #12
math6 said:
if you look to definition in the article you always preserve the same base ?

Different authors may have different terminology. For instance Husemoller in his "Fibre Bundles" will use the name B-automorphism for an automorphism that keeps the points at the base fixed. And it nice to be able to say, for instance, that every diffeomorphism of B lifts to an automorphism of its tangent bundle TM.

But many authors define a bundle automorphism demanding that it induces the identity map on the base. So, as long as you know the definition in a given book - you are ok. But when you move to a different book or a paper, it is better to be prepared for a possible change.
 
  • #13
thnx arkajad .
 

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