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Tangent bundle

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Tangent space

From calc 1 we know that can we use the derivative to find the tangent line for the graph of the function. In higher dimensions we talk instead of finding the tangent vectors to the graph.

Now, suppose the domain of our function is a manifold (we'll keep the target ${\bf R}$ for the sake of simplicity). Then we may use the exterior derivative of the function, which is a differential form of degree $1$, for the same purpose. This form is defined on the tangent vectors to the domain manifold. We now investigate the nature of this set.

Recall the setup in dimension $1$:

UniverseNegativeCurvature.png

Suppose the $x$-axis is a curve $C$ and $f \colon C \rightarrow {\bf R}$ is a function. Let's parametrize $C$ as $p \colon [a,b] \rightarrow C$, where $p$ is differentiable.

Suppose $p(s_0)=a$ and $p'(s_0) \neq 0$. Since $y = f(x)$, we have $y = f(p(t))$, so $y$ is a parametric curve in the $y$-axis. Then by the Chain Rule, $$y'(t)=f'(p(t))p'(t).$$ In particular, $$y'(s_0)=f'(p(s_0))p'(s_0) = f'(a)p'(s_0).$$ Now let's define:

  • $dx=p'(s_0)$ is a tangent vector to $C$ at $a$, and
  • $dy=y'(s_0)$ is a tangent vector to the $y$-axis at $f(a)$.

Then we have $$dy=f'(a)dx.$$ Therefore, the derivative $f'(a)$ is a linear operator, as before.

But on what space?

The construction works for all parametric curves like that. So, let take into account all of them. Then we have all possible tangent vectors $dx$ at $a$. Together, they form the tangent space, $T_aC$ of $C$ at $a$.

In dimension $1$, we only need one vector, since the space is $1$-dimensional: $${\rm span} \{p'(s_0)\} = T_aC$$ at $a$.

TangentSpace.png

What if we are dealing with a $k$-manifold? How do we find all tangent vectors?

TangentSpaceKManifold.png

We again take all parametric curves through $a$ in $M$ and again the set of their tangent vectors at $a$ is a vector space, $T_aM$:

TangentSpaceTaM.png

Exercise. Prove that.

What happens to $f$? Suppose $M$ is a surface and $f:M \rightarrow N$ is a differentiable map, where $N$ is another surface.

Then

  • $T_{f(a)}N $ is the image of $f'(a)$,

provided

TangentSpaceKManifold2.png

Here the curved piece is the image of $f$. And $\{f'(a)(e_1), f'(a)(e_2)\}$ is a basis of the tangent plane for $N$.

As a linear operator, $f'(a)$ works like this:

LinearOperatorDiagram.png

Then $$\dim T_{f(a)}N = \dim T_a M = \dim M = 2 .$$

Tangent bundle

Now, where is the whole image of the derivative, $f'(a)$ over all $a \in M$?

We need to combine all the tangent spaces together:

TangentBundle.png

The tangent bundle of a manifold $M$ is the disjoint union of all tangent spaces: $$TM = \displaystyle\bigsqcup_{a \in M} T_aM,$$ (How they are glued together is discussed elsewhere.)

Note: These pictures should be understood as if there are no intersections.

TangentBundle2.png

Another way to understand it is to think of it as the set of pairs: $$\{ (a,v), a \in M, v \in T_aM \}.$$

Exercise. Prove that the tangent bundle of a simple curve is homeomorphic to a strip:

Tangent bundle of a simple curve.png

Example: $T{\bf S}^1 = ?$

TangentBundle3.png

Try to turn the tangents, vertically, and stand them up.

TangentBundle4.png

We end up with ${\bf S}^1 \times {\bf R}$, the cylinder.

It's OK to illustrate this way, sometimes.

Example: It's not OK for the sphere ${\bf S}^2$! $$T{\bf S}^2 \ne {\bf S}^$ \times {\bf R}.$$ There is no way to turn the planes.

As a corollary: there are no flat maps of the Earth, ${\bf S}^2$, without distortions. This is non-Euclidean geometry!

TangentToSphere.png

In the definition, $$TM^n = \displaystyle\bigsqcup_{a \in M^n} T_aM^n,$$ each term is a copy ${\bf R}^n$. This does suggest that the whole thing is "like" the product $M \times {\bf R}^n$. However it might have a "twist" -- think the Mobius band.

Mobius band as fiber bundle.png

Tangent bundles are a particular case of fiber bundles and the Mobius band above is shown as a fiber bundle over the circle.

In 3d, compare these hairbrushes -- with and without a twist:

Hair brush w twist.png Hair brush wo twist.png

Exercise: What is the tangent bundle of the cylinder?

The most important operation on the tangent bundle is its projection $$p_M:TM \to M,$$ which operates very similarly to the the projection of products: $$p_M(a,v)=a.$$

Just as $n$-manifolds are locally ${\bf R}^n$, tangent bundles are locally $U × {\bf R}^n$, where $U$ is an open subset of ${\bf R}^n$.