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As a follow-up to this question I would like to clarify whether the tangent bundle on a sphere in $\mathbb R^3$ spans $\mathbb R^3$ to make sure I get the concept.

The tangent bundle is the set of the tangent planes at every single point on the surface of the 2-sphere $S$ and would be defined as

$$TS := \{(p, v):\ p\in S, v\in T_pS\}$$

If I get the idea correctly, there would be tangent planes through each point on the surface like the following ones in the drawing representing 3 single points ($P, S, Q)$:

enter image description here

Each plane would be translated to go through the origin to construct a vector space of tangent planes:

enter image description here

If the above is correct, the intuition is clear: the tangent bundle on the sphere would enable us to find a plane in any possible orientation, and hence, the disjoint union of these tangent planes would span $\mathbb R^3.$ Or is the disjoint piece a game changer?

QUESTIONS:

  1. Does this "fan" of translated tangent planes span $\mathbb R^3$?
  2. And how are the addition and scalar multiplication of a vector space defined on this tangent bundle?
Antoni Parellada
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  • Do you mean span as in span of a subset of $\Bbb R^3$ or just that the union of the tangent planes gives $\Bbb R^3$? The latter is false; just take any plane passing through the sphere. The former is true; consider the 3 tangent planes each containing one of the canonical (standard) basis vectors of $\Bbb R^3$. – Prasun Biswas Apr 01 '18 at 02:55
  • @PrasunBiswas I meant it as in the second idea: would the (disjoint) union of these infinity planes shooting out from zero in all directions "reproduce" $\mathbb R^3$? – Antoni Parellada Apr 01 '18 at 02:58
  • I don't think so. None of the tangent planes contain points that are in the interior of the sphere. – Prasun Biswas Apr 01 '18 at 03:05
  • @Prasun Biswas What about the fact that the planes are translated through the origin? – Antoni Parellada Apr 01 '18 at 03:18
  • Does it matter? Any plane of $\Bbb R^3$ can be seen as a translation of some plane passing through origin. – Prasun Biswas Apr 01 '18 at 03:26

1 Answers1

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The tangent bundle $TS$ of the two dimensional sphere $S$ is the set you mentioned in the question. I'm not sure in what sense you ask if it spans $\mathbb R^3$ though ($TS$ is not even a subset of $\mathbb R^3$). The tangent bundle does not, in general, have the structure of a vector space. Each tangent space $T_pS$ can be given one, but their disjoint union can't.

The fact that $TS$ is the disjoint union of the $T_pS$ and not simply their union is key. You want to keep track of the point the vectors in the tangent space are tangent to in order to keep vectors coming from different tangent spaces separate. So when you form the tangent bundle you prefix every tangent vector with the point it's tangent to, from which the definition $TS=\{(p, v):\ p\in S, v\in T_pS\}$. Since the manifold you are starting from, $S$, is a $2$-dimensional object, and you are attaching $2$-dimensional objects $T_pS$ at every point $p\in S$, it's reasonable to think that $TS$ should be a $4$-dimensional object. And in fact it is, it's a $4$ dimensional manifold, but not, in general, a $4$-dimensional vector space.

That is generally speaking because, in some cases, like $T\mathbb R^n$, there is the structure of a $2n$ dimensional vector space since $T\mathbb R^n$ is essentially $\mathbb R^n\times \mathbb R^n$, but in general there is no meaningful vector space structure on a tangent bundle.

Note on the disjoint union: the disjoint union is a general construction. If for every element $a$ in a set $A$ you have another set $B_a$ (that may depend on $a$), the disjoint union is $$ \biguplus_{a\in A} B_a:=\bigcup_{a\in A} \{a\}\times B_a = \{(a, b):\ a\in A, b\in B_a\} $$

Fra
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  • Regarding the idea of 4 dimensional manifold, how is it then that if I had an uncountably infinite amount of time I could have drawn all the tangent planes - one for every point, not just 3 exanples - within the confines of a 3D representation of the diagrams in the OP? – Antoni Parellada Apr 01 '18 at 03:58
  • Also, what is the point of translating these planes to the origin if we are not going to get a vector space (identity element)? – Antoni Parellada Apr 01 '18 at 04:00
  • @justquestions well, translating the planes to the origin is actually pointless. – edm Apr 01 '18 at 04:01
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    You could have, but that doesn't tell you what the "dimension" of the tangent bundle should be. In your drawing tangent spaces of different points on the sphere may intersect, in the tangent bundle they are kept separate. If you want another way of looking at it, to pick a point on the tangent bundle you need to pick a point $p$ on the sphere (2 degrees of freedom) and then you need to pick a vector on the tangent space to $p$ (another 2 degrees of freedom). Total: 4 degrees of freedom to determine a point in $TS$ – Fra Apr 01 '18 at 04:02
  • The identification of $T_pS$ with a plane through the origin of $\mathbb R^3$ just gives $T_pS$ a vector space structure. You could imagine another copy of $\mathbb R^3$ with the origin $0$ at $p$ instead of translating $T_pS$ – Fra Apr 01 '18 at 04:04
  • So the drawing is wrong? - there is no way for these planes not to intersect... – Antoni Parellada Apr 01 '18 at 04:05
  • So the tangent plane is a vector space, but not the tangent bundle? – Antoni Parellada Apr 01 '18 at 04:06
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    @justquestions the dimension of $\Bbb R^3$ is too low to contain the information about the tangent spaces properly. And yes, the tangent planes are vector spaces, and no, the tangent bundle is not a vector space. – edm Apr 01 '18 at 04:08
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    Draw arrows coming out of $p$ in $T_pS$, those arrows are your vectors. The tangent bundle is the union of all the arrows with their originating point from all the tangent planes, so no no vector space structure on the tangent bundle in general, unless you can tell me how you sum points of a sphere and multiply them by a scalar. – Fra Apr 01 '18 at 04:09
  • @Fra You closed the loop in your last sentence, because that was precisely my second question in the OP. I couldn't see that, but now it is clearer - the vector space formed by all the vectors tangent to the point is simply a subspace of $\mathbb R^3$ with the usual sum and scalar multiplication. Hence the TS is a vector space, but not the bundle. – Antoni Parellada Apr 01 '18 at 04:13
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    You can also see it this way: you can sum the arrows if they come out of the same point $p$ of the sphere, but an arrow attached to a point $p$ and one attached to another point $q$ have no meaningful sum, since they literally live on different vector spaces. – Fra Apr 01 '18 at 04:19