Vector Fields and Vector Bundles


by AiRAVATA
Tags: bundles, fields, vector
AiRAVATA
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#1
Oct29-06, 10:49 AM
P: 173
I need help solving the following problem:

Let [itex]M,N[/itex] be differentiable manifolds, and [itex]f\in C^\infty(M,N)[/itex]. We say that the fields [itex]X\in \mathfrak{X}(M)[/itex] and [itex]Y \in \mathfrak{X}(N)[/itex] are f-related if and only if [itex]f_{*p}(X(p))=Y_{f(p)}[/itex] for all [itex]p\in M[/itex].

Prove that:

(a) [itex]X[/itex] and [itex]Y[/itex] are f-related if and only if [itex]X(g \circ f)=Y(g) \circ f[/itex], for all [itex]g\in C^\infty(M)[/itex].

(b) If [itex]X_i[/itex] is f-related with [itex]Y_i[/itex], [itex]i=1,2[/itex], then [itex][X_1,X_2][/itex] is f-related with [itex][Y_1,Y_2][/itex].


I know this is silly, but my main problem is that i dont know how the identity [itex]f_{*p}(X(p))=Y_{f(p)}[/itex] looks like. What i mean is the following:

If [itex]X=\sum_{i=1}^m X_i \frac{\partial }{\partial x_i}[/itex], and [itex]Y=\sum_{j=1}^m Y_j \frac{\partial }{\partial y_j}[/itex], then

[tex]f_{*p}(X(p))=f_{*p}(\sum_{i=1}^m X_i(p) \frac{\partial }{\partial x_i})=\sum_{i=1}^m f_{*p}(X_i(p) \frac{\partial }{\partial x_i})[/tex]

so

[tex]f_{*p}(X(p))=\sum_{i=1}^m f_{*p}(X_i(p))\frac{\partial }{\partial x_i}+X_i(p) f_{*p}(\frac{\partial }{\partial x_i})=\sum_{j=1}^m Y_j (f(p))\frac{\partial }{\partial y_j}.[/tex]

Is this correct?

Another question. Does anyone know some good online notes regarding vector bundles?
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mathwonk
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#2
Oct29-06, 12:10 PM
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these are just fancy versions of the chain rule.


th standard intro to bundles was milnors princeton notes, later written up and published by stasheff.

and the early parts of atiyahs k theory book.

oh and spivaks dif geom vol 1.

these are not free anymore i guess. have you tried searching google under "notes on vector bundles"?
mathwonk
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Oct29-06, 12:34 PM
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a vector bundle is a continuous family of vector spaces. a basic example is the family of tangent spaces to the sphere, one plane for each point, thought oif as disjoint. thus a point of the tangent bundle is a pair, (p,v) where p is a point of the sphere and v is a vector tangent to the sphere at p. it makes sense to say a pair (p,v) is near the pair (q,w) if p is near q, and v is near w, i.e. if both head and foot of v are near respectively both head and foot of w.

given a vector bundle on a manifold like the sphere, then the basic question is whether it is possible to find a continous family of vectors, one at each point of the sphere, i.e. a continuoius map from the sphere to the tangent bundle that takes p to some pair (p,v).

such a family is called a continuous vector field.

given a manifold, a basic question is what are all the vector bundles on that manifold. that is called k theory. for some reason.

the most fundamental, or tautological, vector field is the one on a grassmannian, like the projective plane. by definition the projective plane is the family of all lines through the origin of euclidean 3 spoace.

hence every "poibnt" is anturally associated to a line, anmely itwelf. so projective space comes equipped with a natural or tautological lien bundle, as follows:


let E be euclidean 3 space, and let P be the projective plane, i.e. P = (E-0)/k* where k* = the non zero scalars acting on E-0 by scalar multiplication. i.e. E-0 is the punctured euclidean space, and P is the set of equivalence classes of points, where points on the same punctured line through the origin are equivalent.

then each point p of E-0 determines a line, namely the one it spans togetehr with 0. thus consider all pairs (p,L) in ExP, where p lies on the line L.

this is aline bundle over P, with the inverse image of each "point" L of P being the line of points on L.

similarly, one can define the grassmannian Gr(k,n) of k planes through the origin of n space. this space of equivalence classes of points on the same k plane, carries a natural tautological k plane bundle.

niow the basic theory of vector bundles and their classification is this: given any manifold M and map M--->Gr(k,n) there is anatrual k plane bundle on M pulled back from the tautological one on Gr.

And conversely, given any k plane bundle on M, there is a map M-->Gr, such that the pullback bundle is equivalent to the original one.

moreover homotopic maps from m to Gr pullback equivalent bundles.

hence the classification of bundles on M is the same as the analysis of homotopy clases of maps from M to Gr.

moreover, any one bundle has a natural sequnec of obstruction classes in tis cohomology that emasure the possibility of finding independent r tuples of vector fields, called characteristic classes.

sinbce the theory of vector bundle clasification says all bundles are pullbacks of the tautological ones, it follows that for all bundles the natural cahracteristic classes can be regarded as pullbacks of the concrete classes on Gr.

thus the cohomology of grasmannians is basic to th stduy of vectorbundles.


this much is about what everyone nows abut vectro bun\dles, and rougly what irecall from second year in gradschool. there are also comutational techniques like whitneys product formula, and the trick of comouting classes of a hyopersurface from an embedding, using the fact that the sum of the the normal bundle and tangent bundle equals the ambient trivial bundle of euclidean space.

this stuff is nicely described in [an exercise?] in bott- tu.

AiRAVATA
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#4
Oct29-06, 01:11 PM
P: 173

Vector Fields and Vector Bundles


Thx a lot for the reply mathwonk.

I have with me Spivaks book. Sadly, the notes of Milnor wherent abailable in my school library (I guess someone esle in my course took them).

I'll keep reading Spivaks and anything on the web i can find.
mathwonk
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#5
Oct29-06, 03:36 PM
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try thesew free notes from allen hatcher:

http://www.math.cornell.edu/~hatcher/VBKT/VBpage.html

what i have described seems to be the first parts of chap 1 and of chap 3 of hatcher.

hatchers book is apparently just his synthesis of the other books i have mentioned where the originators of the theory described their own work.

hence the other books are preferable as original sources, but his books are well regarded and free.
mathwonk
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#6
Oct29-06, 07:50 PM
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spivak is milnors student, so his version will have benefited from milnors.
AiRAVATA
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Oct29-06, 10:08 PM
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Thx a lot :)
mathwonk
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Oct29-06, 10:35 PM
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my pleasure.


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