# A Tangent Bundle questions about commutative diagram

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1. Aug 31, 2016

### orion

I don't know how to create a commutative diagram here so I'd like to refer to Diagram (1) in this Wikipedia article. I need to discuss the application of this diagram to the tangent bundle of a smooth manifold because there are some basic points that are either glossed over or conflict in the literature.

So first I'd like to examine the left hand side of that diagram and in particular $\pi$. Some books say the following: $\pi(p,v) = p$. Other books say $\pi(v) = p$. Which is correct? And if it is $\pi(p,v) = p$ then how does $\pi$ differ from $\text{proj}_1$ in that diagram? And doesn't that conflict with $\pi^{-1}(p) = T_pM$ and not $\{ p \} \times T_pM$?

I'm sorry for all the technical questions, but this is really bothering me that I can't understand that diagram as applied to the tangent bundle.

2. Aug 31, 2016

### Lucas SV

It depends on what you mean by $v$. The case $\pi(p,v)=p$ (this notation should be made more precise as I did below) only makes sense if you are talking about a local trivialization, and $v$ here means an element of the fibre. The case $\pi(v)=p$ is more general and here $v$ is an element of the fibre bundle.

In this case, we are talking about a local trivialization. More precisely, we should write $\pi(\phi^{-1}(p,v))$, and the statement that this is equal to $\mathbf{proj_1}$ is precisely the statement of commutativity.

I make an analogy with smooth manifold theory. Instead of local trivializations we use charts. We should make chart independent definitions, even though often we do calculation in charts. In calculations, it is common to relax the notation. I believe this is the reason why some authors may choose to use the more relaxed notation for a point in the tangent bundle within a local trivialization.

3. Aug 31, 2016

### Staff: Mentor

I admit it's hard not to think of $\pi$ as $\text{proj}_1$ and of $E$ as $B \times F$. And in addition homeomorphisms all over the place. So the details are in the generality of the definition or in the difference between a trivial and a non-trivial bundle.
$E$ is more flexible than just $B \times F$. It behaves locally as the direct product, but not globally. That is the main point.
Of course in many cases (and this is how I try to get hold of it), it is just a manifold where at each point the tangent space of it is attached.
I think it's a good idea to keep the Möbius strip in mind.

4. Aug 31, 2016

### orion

My next question concerns $\pi^{-1}(p)$. This should be $T_pM$ as I am told. Suppose $v \in T_pM$ for some $p \in M$. How exactly is $p$ associated with $T_pM$ in the fiber? I mean other than notationally as $T_pM$. In other words, how does $v$ "know" that it is associated with $p$? Again, I'm thinking without appealing to the bundle chart.

5. Aug 31, 2016

### Lucas SV

$\pi(v)$ is exactly the answer to the question: to which $p$ is $v$ associated to?

You may be thinking: 'I need to define $v$ in terms of $p$ ...' but you may think of $v$ as an abstract point in a tangent bundle which you randomly pick.

The example as orion stated is the Mobius strip. It is a fibre bundle. But it is also a manifold, which you can imagine, is embedded in $\mathbb{R}^3$. So there is certainly nothing wrong with picking a point $v$ in $\mathbb{R}^3$ which lies in the mobius strip submanifold, and asking the question: To which $p$ in the circle (which is also embedded in $\mathbb{R}^3$), should I assign to $v$? The answer is again $\pi(v)$.

Last edited: Aug 31, 2016
6. Aug 31, 2016

### orion

I understand, but where does the information reside? In $\pi$ or in $v$ itself? To complicate matters, in the book by JEFFREY Lee, he defines a vector at a point $p$ by $v_p:=(p,v)$.

7. Aug 31, 2016

### Staff: Mentor

If you don't like Wiki's hair brush, you can also think of $E$ as a corn field. I prefer wheat, but anyway. $p$ is where the grains $v$ meat the ground.
Personally I find it better to denote $\pi^{-1}(p)$ as $\{p\} \times TM$, i.e. the entire vector field where $T_pM$ is the specification at $p$. We do the same with the gradient all the time without even thinking about it. How often did you use $\nabla$ without evaluating it at a special point? Do you really always write $\nabla_p$? How does the gradient know, where it has to be evaluated?

$\pi$ is the projection along the grains to the ground point. Locally it is perfectly alright to imagine a direct product, $U \times F$, resp. $\{p\} \times F$. You simply cannot chose $U$ to be the entire base as long as your bundle might be non-trivial.

8. Aug 31, 2016

### Staff: Mentor

Because it's a vector field. The points are important at such an abstract approach. (See my $\nabla$ example above.)

9. Aug 31, 2016

### Lucas SV

The answer is probably either. If you fix a $\pi$ you can perturb $v$ to give different values of $p$. If you fix a $v$, you can perturb $\pi$ (by having chosen a slightly different definition of the fibre bundle) to give different values of $p$.

10. Aug 31, 2016

### orion

Ok. thanks guys. You have been a big help.

11. Aug 31, 2016

### Staff: Mentor

P.S.: The picture with a field of wheat is the better one when it comes to sheaves and sections.
(Edit: corrected a technical term.)

Last edited: Sep 1, 2016
12. Sep 1, 2016

### lavinia

- Whenever one has a function between two sets $f:S →T$ then the pre-image $f^{-1}(p)$ of a point in $T$ is all points that $f$ maps onto $p$. Each point in the pre-image is associated with $p$ by the function $f$ itself. In your example $T_{p}M$ is the pre-image of the point $p$ in $M$ under the bundle projection map $π$. A point $v$ in the bundle $TM$ "knows"" that it is associated with $p$ because $π(v) = p$. $T_{p}M$ is called the "fiber" of $TM$ over $p$.

- Whenever one has a Cartesian product of sets $U×V$( not necessarily topological spaces but just plain old sets) then one can represent any point as a pair $(u,v)$. The maps $ρroj_{1}(u,v) = u$ and $ρroj_{2}(u,v) = v$ are called projection maps.

The map $π$ is also called a projection map but it does not mean the same thing since the bundle $TM$ is generally not a Cartesian product. The local trivialization described in the Wikipedia page says that a bundle generalizes the idea of a Cartesian product. Locally around any point it is a product but globally it may not be.

- Whenever one has a function $f:M →N$ between two sets the "graph" of $f$ is the function $F:M →M×N$ defined by $F(p) = (p,f(p))$. Note that $proj_{1}(F(p)) = p$ . Conversely any function $F$ from $M$ into $M×N$ such that $proj_{1}(F(p))=p$ is the graph of a function from $M$ into $N$. For a bundle, a map $f:M →TM$ such that $π(f(p)) = p$ generalizes the idea of a graph. Locally around any point, $f$ is a graph. For a bundle, $f$ is usually called a "section" of the bundle. An important example of a section of a bundle is a vector field on smooth manifold.

- It is true that $T_{p}M$ is homeomorphic to $p×T_{p}M$. But this just says that a space is homeomorphic to its Cartesian product with a point. The point could be any point anywhere and does not have to be the point $p$.

BTW: Usually the notation $TM$ means the tangent bundle of a smooth manifold.

Last edited: Sep 2, 2016
13. Sep 1, 2016

### lavinia

Here is an example that you might find helpful. Suppose $M$ is a closed smooth manifold without boundary in $R^{n}$. Define $TM$ to be the subset of $M×R^{n}$ of pairs $(p,v)$ where $v$ is tangent to $M$ at $p$. $TM$ is a subset of a Cartesian product but in general will not be a Cartesian product itself. But each of its points $(p,v)$ has a natural projection onto the first factor $(p,v) →p$. This is the bundle projection map π. So here it is obvious how $v$ is connected to $p$.

$TM$ is isomorphic to the tangent bundle of $M$.

One could also take the vectors $v$ to be orthogonal to $M$. Then one gets another bundle called the normal bundle of the embedding of $M$ in $R^{n}$.

Last edited: Sep 2, 2016
14. Sep 2, 2016

### orion

This is very appealing to me except I have one question. Take, for example, the simple real-valued function $f(x)=x^2$. Now this function associates the number 3 with the number 9. How do I know? Because the function says take the number given and square it. I know exactly the prescription.

But $\pi$ is something else. We are told that it is a bundle projection. A projection of what onto what? My question is how does $\pi$ take an element of $T_pM$ and output $p$? How does it determine the point $p$ associated with $v$?

I realize that I could just accept it as it is written in the books and move on, but these sorts of details drive me crazy because I feel like there is something lacking in my understanding.

lavinia, I thank you so much for your help and thorough replies. I feel like I'm making progress, like I'm 90% where I need to be.

15. Sep 2, 2016

### lavinia

Think about the example of the tangent bundle in my second post. There I tried to make it crystal clear.

One can always take a local trivialization and project onto the first factor. This projection is independent of which local trivialization you choose.

For instance you can think of $R^2$ as a trivial bundle of the real line with fiber $R$. Any pair $(x,y)$ has a natural bundle projection π defined by π(x,y) = x. Here (x,y) is a point in the fiber above x.

Last edited: Sep 2, 2016
16. Sep 2, 2016

### Lucas SV

Not every function is like the square function. In particular the $\pi$ function does not take real numbers as arguments. Even for real valued functions, it is not always the case you can write down a concrete number for all values of the function.

In the definition of fibre bundle, we do not specify the $\pi$ map. But if we want to check a certain structure $(E, \pi, B)$, we need to check $\pi$ is a continuous surjection from $E$ to $B$ with a local triviality condition. So, when we talk about a specific example of a fibre bundle, we do need to specify $\pi$.

17. Sep 2, 2016

### lavinia

Orion

As Lucas SV explained the definition of a bundle just says that there must be a projection function. There is no need to specify one since it is a general definition that applies to all possible bundles. When one defines a specific bundle then you need to define the projection. Otherwise it is not specific.

I have given you some examples of specific projections.

A function can be talked about in general terms just like anything else. For instance I can speak of the integers. Do I need to list all of them to make sense? I can speak of differentiable functions. Do I need to list say which specific ones I am talking about.? The same goes for bundle projections or for that matter pretty much anything else.

18. Sep 2, 2016

### Lucas SV

A.K.A. the axiom of choice.

19. Sep 2, 2016

### orion

Thanks to both of you.

I need to digest all this before I can say whether I have any other questions, but I do have a rather simple, silly question that came up in lavinia's first reply because I've seen it stated in textbooks as well. Given $T_pM$ why can we say that $T_pM$ is homeomorphic to $\{ p \} \times T_pM$?

20. Sep 2, 2016

### Lucas SV

Consider the map $f:T_p M \rightarrow \{p\} \times T_p M$ defined by $f(V)=\{p,V\}$ for a vector $V\in T_p M$. You can easily show that $f$ is a homeomorphism.