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Vertical and horizontal subspace of a vector space T_pP.

  1. Aug 24, 2010 #1
    suppose we have a principle fiber bundle P

    at a point p \in P

    we have the decomposition T_pP=V_pP + H_pP

    it is said that the vertical subspace V_pP is uniquely defined while H_pP is not

    i cannot understand this point

    the complement to a unique subspace is surely unique, i think.

    it is a basic fact in linear algebra.
  2. jcsd
  3. Aug 24, 2010 #2
    I assume you mean a vector bundle, right.? (or maybe you use a different name for it),
    and I guess you want a direct sum decomposition and uniqueness up to isomorphism.?

    Where did you read that the complement was not unique (and unique up to what)?.

    Then, you are correct: given an n-dimensional subspace S of an m-dim
    vector space V , we can always define a subspace S' (e.g., by extending the
    basis B_S of S to a basis B_V for V , so that S' is the subspace with basis
    B_V-B_S )so that


    And dimensions add up, so DimS'=m-n . so S' is unique up to vector space isomorphism.

    And all
  4. Aug 24, 2010 #3
    no, it is about principle fiber bundle

    it is in the context of horizontal lift of a curve in the base manifold M to a curve in the bundle P

    it is said that some connect is needed to determine the horizontal subspace.
  5. Aug 24, 2010 #4
    The orthogonal complement is unique, but that needs an inner product. In general, given a vector space V with a subspace U, there are many choices of subspaces W such that V is the direct sum of U and W.
  6. Aug 24, 2010 #5
    But there is only one such W up to isomorphism, by invariance of dimension.

    I have always been curious about "solving" for quotients, or direct sums, i.e.,

    If we know V=S(+)S' , and we know S, how do we find S' up to isomorphism;

    similarly, if we know that a group G is the quotient of two groups H,K, i.e.,

    G~ H/K , and we only know either H or K, but not both, can we find the other.?
  7. Aug 24, 2010 #6
    But the isomorphism problem for vector spaces is nearly trivial (two vector spaces are isomorphic iff they have the same dimension), making it uninteresting. If V = S ⊕ S', then S' is always isomorphic to V/S, but who cares?

    Back to the original problem: It wasn't saying unique up to isomorphism; just unique.

    edit: I did some reading about the objects in the original problem. A fibre bundle p: E -> M gives a canonically defined (i.e. uniquely satisfies a certain property) linear map of vector bundles Tp: TE -> TM; the vertical bundle VE is defined as the kernel of Tp. A horizontal bundle is then an arbitrary subbundle HE of TE such that TE = VE ⊕ HE. Even though any two horizontal bundles may be isomorphic, they may still be different subbundles of TE.

    Generally speaking, if you're given a fixed object, you aren't interested in the various subobjects only up to isomorphism.
    Last edited: Aug 24, 2010
  8. Aug 25, 2010 #7
    to the point

    thanks a lot!
  9. Aug 25, 2010 #8


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    Spelling nitpick: It's principal, not principle.
  10. Oct 8, 2010 #9
    Re: vertical and horizontal subspace

    where in the definition of vertical subspace we understand that the notion of canonical vertical vector: a vertical vector is a vector tangent to the fiber. ?
  11. Oct 8, 2010 #10


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    The vertical space is tangent to the fiber. But a complementary subspace of the tangent space to the fiber is not uniquely determined.
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