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I'm learning a bit about smooth manifolds, and currently I'm learning about tangent bundles (just definitions mainly) and vector field.

This is my reference : Tu's Introduction to Manifolds. I was also watching a video about tangent bundles (because I was struggling with the concept).

Once I understood the definition however I realized I have an issue understanding why we need the notion of tangent bundle. I can't remember where I read this but am I right when I say that tangent bundles are necessary if we want to generalize the notion of function on a manifold?

Consider a smooth manifold $M$, if we wanted to define what a vector field is to me the definition should reflect the fact that for each $p \in M$ we have a $v \in T_p M$, therefore it should be a map.

This is probably the key why such association isn't good as definition because a map needs both domain (in this case $M$) and an image space, however my naive definition involves for each $p$ a different space $T_p M$ and this is why we need the notion of tangent bundle.

Is this observation correct?

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    $\begingroup$ The tangent bundle has a natural structure of smooth manifold. We need this structure to define smooth vector fields $\endgroup$ – Ottavio Bartenor Jan 27 at 12:29
  • $\begingroup$ Here’s one reason. For a smooth map between smooth manifolds, you want to be able to talk about its derivative as a smooth map between smooth manifolds as well. Tangent bundles provide a nice way to do that. $\endgroup$ – Shalop Jan 27 at 12:43
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Remember that the tangent bundle is the disjoint union of the tangent spaces: $$TM = \coprod_{P \in M} T_P M.$$ It has the topology of a smooth manifold in the following manner. Let $(U_\alpha, \phi_\alpha)$ be an atlas for $M$, and let $\pi: TM \longrightarrow M$ be the natural projection, i.e. if $(P, v) \in T_P M \subset TM$, then $\pi(P, v) = P$.

Edit: On the Why? bit

The point of this is just to establish a mathematical framework in which we can talk about points on a base manifold, and all the possible curves through any point on the manifold. We can talk about the base points $x$, together with possible directions at $x$.

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  • $\begingroup$ I think you need to elaborate a bit more on "why" is all of this necessary. $\endgroup$ – user8469759 Jan 27 at 12:03
  • $\begingroup$ See edits please $\endgroup$ – 808GroundState Jan 27 at 12:08
  • $\begingroup$ Why can't we talk about "points" and "curves" on a base manifold? what are we lacking of? $\endgroup$ – user8469759 Jan 27 at 12:11
  • $\begingroup$ You can, although with a vertical-horizontal decomposition of each point on $M$ into two components. To be physically useful, we want the vertical-horizontal decomposition to be independent of coordinates on the tangent space. Under a change of coordinates, vertical features must stay vertical and horizontal features must stay horizontal. This leads to an abstract notion called an Ehresmann connection, which specifies how vertical/horizontal are to be delineated. The more familiar "affine connection" in Riemannian geometry is a special simplified case $\endgroup$ – 808GroundState Jan 27 at 12:15
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You are concerned that for each $p$ you need a different space $T_p(M)$ when building a vector field. By putting all the tangent spaces together (disjoint union, with a suitable topology and even a smooth manifold structure) to form the tangent bundle $TM$ you get one target space for all vector fields on $M$. For vector field on $M$ you want the tangent vector attached to $p$ to be living in $T_p(M)$ and not $T_q(M)$ for some $q \not= p$. A precise way of specifying this condition is to say a vector field on $M$ is a map $X \colon M \rightarrow TM$ where $X(p) \in T_p(M)$ for all $p \in M$. Or, in terms of the natural surjective map $\pi \colon TM \rightarrow M$ that sends a point $(p,v)$ in $T(M)$ to the point $p$ at which it is based, a vector field on $M$ is a map $X \colon M \rightarrow TM$ such that $\pi \circ X \colon M \rightarrow M$ is the identity. We call $X$ a "section" of $\pi$ (or a section of the tangent bundle). Quite generally, when $f \colon A \rightarrow B$ is a surjective map, a section of $f$ is mapping in the other direction $g \colon B \rightarrow A$ where every $g(b)$ is in the fiber $f^{-1}(b)$, which is another way of saying $f(g(b)) = b$ for all $b \in B$, or equivalently $f \circ g \colon B \rightarrow B$ is the identity.

The map $\pi\colon TM \rightarrow M$ is smooth, and we call a vector field $X \colon M \rightarrow T(M)$ continuous or smooth when $X$ is continuous or smooth as a mapping.

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Actually we know how to do calculus (differentiation, integration...) for functions on $R^n$. And if, $f:M\rightarrow N$ is smooth map then its differential $df:TM\rightarrow TN$ Will be map between tangent spaces so overall we will need knowledge of tangent spaces if we want to study smooth maps between manifolds.

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  • $\begingroup$ Stupid question. If I have a smooth function $f : M \to N$ what exactly can I NOT do without the concept of tangent bundle? What knowledge are we missing of the the tangent space (which is defined point-wise). $\endgroup$ – user8469759 May 23 at 15:13

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