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The construction of tangent bundle on a $C^\infty$ manifold, as for example in the book of Warner, uses the existence of double derivatives. Of course the tangent space for a point is first constructed in case of an open set in a Euclidean space and then the whole setup is glued up. But then in the neighborhood of a point the second derivatives might be different for a different $C^1$-chart around the point. So I suppose the tangent space/tangent bundle construction is for $C^2$-manifolds.

Question is:

Since $C^2$ is apparently the minimum condition for existence of tangent bundle, is the tangent bundle just $C^1$, or it is again $C^2$?

flag	asked May 17 2010 at 12:56 Akela 1,773●6●19

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I haven't looked at Warner in a long time, but I'm pretty sure he is working only with infinitely differentiable manifolds, so he does not try to minimize the number of derivatives used in any of his constructions. So you shouldn't rely on his constructions to calculate the minimal required differentiability of an object. – Deane Yang May 17 2010 at 13:34

André Henriques · Accepted Answer · 2010-05-17 13:19:47Z UTC

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The tangent bundle of a $C^1$ manifold exists: it's a $C^0$ manifold.
Similarly, for every $n$, the tangent bundle of a $C^n$ manifold is a $C^{n-1}$ manifold.

answered May 17 2010 at 13:19

André Henriques
14.9k●2●30●83

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To elaobrate: a manifold is $C^k$, if the co-ordinate transition functions are $C^k$. A vector bundle is $C^k$ if the fiber transition functions are $C^k$. The fiber transition functions of a tangent bundle are just the derivatives of the co-ordinate transition functions. So a $C^k$ manifold has a $C^{k-1}$ tangent bundle. – Deane Yang May 17 2010 at 13:32

What is the order of differentiability of the tangent bundle of a C^2- manifold?

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