# Why defining tangent bundle?

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?

• The tangent bundle has a natural structure of smooth manifold. We need this structure to define smooth vector fields – Ottavio Bartenor Jan 27 at 12:29
• 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. – Shalop Jan 27 at 12:43

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 $$x$$.

• I think you need to elaborate a bit more on "why" is all of this necessary. – user8469759 Jan 27 at 12:03
• See edits please – 808GroundState Jan 27 at 12:08
• Why can't we talk about "points" and "curves" on a base manifold? what are we lacking of? – user8469759 Jan 27 at 12:11
• 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 – 808GroundState Jan 27 at 12:15

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.

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.

• 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). – user8469759 May 23 at 15:13