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When one first encounters the concept of vector field, especially in physics, it is often presented just as n-tuple of numbers $(x_1, x_2, \ldots , x_n)$ prescribed to each point. In this manner $n$ is allowed to accept arbitrary value.

However, when one proceeds with vector fields in differential geometry, the dimension of a manifold dictates the dimension of a tangent space:

$$\dim M = \dim T_x(M)$$

There are other ways to attach a vector space to each point of manifolds --- tensor products of tangent spaces, p-forms, etc.

Now I ask, given an n-dimensional manifold, what are allowable dimensions for vector spaces on it? Are all of them allowed, or there is no way to naturally construct a k-dimensional vector field (in a broad sense) for certain k?

Another (stronger) version for this question, given n-dimenstional vector space, can we naturally construct k-dimensional vector space from it for arbitrary k?

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$M\times \mathbb{R}^k$ is a $k$ dimensional vector bundle over an $\dim M$ dimensional manifold. This construction always works, for any manifold $M$ and any dimension $k$. You have to specify what "natural" means. This has a precise mathematical meaning, and might be what you are after, but I am not sure... –  Thomas Rot Jul 17 '12 at 8:04
    
Can you help me with this naturality, but without severe category theory? I would call natural what can be encountered in physics --- you have a manifold (and that's all you have), vector bundles on it with what dimensions are physically plausible? –  Yrogirg Jul 23 '12 at 7:23
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Yes, a vector field is a section of the tangent bundle. Thus it has to have the same dimension as the manifold.

More generally, one can consider sections of any vector bundle over the manifold. Their dimension is not related to the dimension of the manifold.

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What vector bundles are natural, that is all needed for their existence is the manifold itslef? –  Yrogirg Jul 23 '12 at 7:24
    
@Yrogirg: Do you mean "needed for their construction" instead of "needed for their existence"? Anyway, I think you are asking the wrong question. The construction of the spinor bundle, for instance, depends on more than just the underlying manifold. –  Rasmus Jul 23 '12 at 9:26
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