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On plane we can easy to define continuous vector field: $$\mathbb{R}^2\ni x\longmapsto v(x)\in\mathbb{R}^2,$$ where real-valued function $v_1(x)$ and $v_2(x)$ are continuous.

But when we try to define field on arbitrary 2-dimensional surface we have a trouble: $$v(x_1)\in T_{x_1}S,~~~v(x_2)\in T_{x_2}S,$$ and for different points $x_1$ and $x_2$ tangent spaces $T_{x_1}S$ and $T_{x_2}S$ are different. So function $v_1$ and $v_2$ no sense.

Question: how to define continuous vector field on surface?


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@Qiaochu: Is it necessarily to consider tangent bundle? – Aspirin Sep 14 '11 at 1:30
Nope. One can also define vector fields on a smooth manifold $M$ as derivations on $C^{\infty}(M)$: – Qiaochu Yuan Sep 14 '11 at 1:36
Perhaps we should also mention the notion of a connection (see ) which is essentially a way to identify tangent spaces at different points on the base manifold. – Grumpy Parsnip Sep 14 '11 at 2:04
up vote 2 down vote accepted

Of course, it'd be nice if all tangent vectors to $M$ were elements of some big space; then we could talk about the function assigning a point to the vector at that point being "continuous", or "smooth", etc. If we were working in a submanifold of $\mathbb{R}^n$, perhaps we could try to compare the tangent vectors within $\mathbb{R}^n$; but when we're working in an abstract manifold, we have to do everything ourselves.

The answer is simply to "put all of the $T_pM$'s together", and voila, we have a big space that contains all of our tangent vectors!

The resulting object is known as the tangent bundle, and as far as I know, this is the main reason why it was invented. As a set, it's precisely $$TM=\coprod_{p\in M}T_pM$$ and it also gets a topology (though clearly it's not going to be the disjoint union topology), and it also has a natural smooth manifold structure. It comes with a natural projection map $\pi:TM\to M$ defined by taking the elements of $T_pM$ to $p$.

A continuous vector field is then just a continuous section of $\pi$.

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May I roughly say for 2-dimensional surface $S$, that vector field on $S$ is continuous if under action of (local) homeomorphism $\Psi : S\to \mathbb{R}^{2}$ field maps to continuous field no plane? – Aspirin Sep 14 '11 at 1:38
That's a correct statement; since continuity is a local condition and the charts are homeomorphisms, the vector field is continuous iff its composition with each chart is continuous. – Zev Chonoles Sep 14 '11 at 1:42
...and define tangent vectors as $\dot\gamma(t_0)$ (where $\gamma(t):\mathbb{R}\to S$ and $\gamma(t_0)=x$) – Aspirin Sep 14 '11 at 1:44
Well, that's the tangent vector to the curve $\gamma$ at $\gamma(t_0)=x$, but two different curves could have the same tangent vector at a point, so you really have to consider equivalence classes of curves (see here). – Zev Chonoles Sep 14 '11 at 1:48

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