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Span of a vector space question

  1. Aug 29, 2011 #1

    WannabeNewton

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    Hey guys, this question is more or less related to the way Frobenius' theorem is presented in my text. Consider an n - manifold M, an m - dimensional submanifold S of M, and a set of k linearly independent vector fields [itex]V^{\mu }_{(a)}[/itex] such that [itex]k \geq m[/itex]. In order for S to be an integral submanifold of the [itex]V^{\mu }_{(a)}[/itex]'s, each vector field in [itex]V^{\mu }_{(a)}[/itex] must be tangent to S everywhere. The text states that, if this condition is met, then at each [itex]p\in S[/itex], each [itex]V^{\mu }_{(a)}(p)[/itex] will be an element of the tangent space [itex]T_{p}(S)[/itex] and since they are linearly independent, they will span the tangent space. The conclusion then follows that since each vector field is tangent to S everywhere, the vector fields themselves will span each tangent space so as to span the tangent bundle to S. I can agree with this if k = m but if k > m I don't see how the vector fields can be linearly independent so as to span the space. Isn't the linearly independent set maximal when it spans a space? If the dimension of each tangent space is m, how can the vector fields span each tangent space when there are k > m linearly independent vector fields? The argument only makes sense to me when k = m. I am not disputing the claims any way by the way, in fact the condition in Frobenius' theorem that in order for a submanifold to be an integral submanifold of a set of vector fields the vector fields must form a lie algebra is easily seen by the set of three killing vector fields, on [itex]S^{2}[/itex], which [itex]S^{2}[/itex] is an integral submanifold of. I am just asking for a clarification because I am obviously overlooking something. Thanks all and sorry in advance if this is the wrong section.
     
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  3. Aug 31, 2011 #2

    Fredrik

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    Suppose that X is an n-dimensional vector space, and that Y is an m-dimensional subspace of X with m<n. Then any linearly independent set in Y with m members spans Y, and there's no linearly independent set in Y with m+1 members. (That's what dim Y=m means). But a linearly independent set in X can have up to n members.

    It seems to me that you have understood this correctly, so I think the only thing you have overlooked is that you're supposed to conclude that k≥m and k≤m together imply that k=m. (If all the k≥m members of a linearly independent set in X are in the m-dimensional subspace Y, then k≤m).
     
  4. Aug 31, 2011 #3

    WannabeNewton

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    Hey Fredrik, thanks for the reply. I'm a little more sure now. Going back to the example on [itex]S^{2}[/itex], we have the three killing vector fields [itex]R = \partial _{\phi }, S = cos\phi \partial _{\theta} - cot\theta sin\phi \partial _{\phi }, T = -sin\phi \partial _{\theta} - cot\theta cos\phi \partial _{\phi }[/itex] and these three vector fields satisfy [itex]L _{V(a)}V_{(b)} = [V_{(a)}, V_{(b)}] = \alpha ^{c}V_{(c)}[/itex] where c runs through all k. Here we have k = 3, and a submanifold of dimension 2 but [itex]S^{2}[/itex] is still an integral submanifold of the set. I used a different text as reference this time and it made it clearer: we choose a set of k vector fields from the original manifold M such that the k vector fields span a subbundle of TM where the subbundle is of dimension m <= k. We don't assume they are all linearly independent here so it is ok if the number of vector fields in the set is greater than the dimension of the subbundle. According to the text, at each point on the submanifold we pick k = m vector fields from the set that are linearly independent so that they form a basis for each tangent space at each point and all the members of the set can be expressed as a linear combination of them so that the members of the set form a lie algebra. I guess the other text forgot to mention that. Thanks Fredrik.
     
    Last edited: Aug 31, 2011
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