# Intuition behind tangent space to a point on a manifold

Let $$M$$ be a smooth manifold and let $$p \in M$$. We have a notion of a "tangent space" of $$p$$, i.e. a vector space structure around $$p$$ to give us the idea, roughly, or "directions we can travel in" from $$p$$, which an abstract manifold need not have inherently. I want to get a sense of the intuition behind what exactly a tangent vector is and how it's defined, and I break this up into four questions.

1) What is the purpose behind defining a tangent space?

As I will write below, tangent vectors are defined in terms of directional derivative operators evaluated at $$p$$. Is the only use behind tangent vectors to be able to take directional derivative? We might define a "direction" in our tangent space to be an operator that produces the directional derivative of a $$C^{\infty}$$ function in that "direction". Intuitively, this notion of direction doesn't look useful for doing anything other than taking directional derivatives; is that indeed the case?

2) Geometric Interpretation

How would one visualize a tangent space? Say, for simplicity of picturing, that our manifold is actually a $$k$$-submanifold in Euclidean space. In this case, isn't the tangent space every single vector in $$\mathbb{R}^{k}$$? How does this compare with visualizing the tangent space as a parallelepiped?

3) Definition 1: Smooth Curves

We might define the tangent space as the equivalence class of all smooth curves $$\gamma: \mathbb{R} \to M$$ with $$\gamma(0) = p$$, where two smooth curves $$\gamma_{1}, \gamma_{2}$$ are equivalent if $$(\varphi \circ \gamma_{1})'(0) = (\varphi \circ \gamma_{2})'(0)$$. In this sense, each equivalence class defines a "direction" about $$p$$, which helps us take directional derivatives. If $$f: M \to \mathbb{R}$$ is a smooth function, then $$(f \circ \gamma)'(0)$$ (differentiated in the ordinary sense, which makes sense here) is the directional derivative of $$f$$ in direction $$\gamma$$. I again come back to my question of what use direction $$\gamma$$ is serving other than giving us directional derivatives. Now, I give the other definition, and want to know why these two definitions are exactly the same:

4) Definition 2: Directional Derivative Operator

Note - This is often given in terms of "derivations" (linear maps that satisfy a generalized product rule, or Liebniz's rule): But a (non-trivial) result tells us that derivations are nothing but directional derivatives, so I stick to talking about directional derivatives here.

Let $$\mathcal{C}$$ denote $$C^{\infty}(M, \mathbb{R}$$), i.e. smooth functions $$M \to \mathbb{R}$$. Let $$D_{\gamma}: \mathcal{C} \to \mathbb{R}$$ be the operator s.t. $$D_{\gamma}(f) = (f \circ \gamma)'(0)$$, where $$\gamma: \mathbb{R} \to M$$ is a smooth curve with $$\gamma(0) = p$$, as above .We can define an equivalence relation (similar to what we did above) and define our tangent space to be all these "directional derivative operators" (that take a function and spit out its derivative in the direction of a smooth curve). In this sense, each "direction" in our tangent space is basically one of these operators. How is our notion of direction here same as the notion of direction we obtained in 3)? In one case, a curve (under equivalence relation) is our direction, while in this case, an operator (defined using a curve, but nevertheless different) is our direction. Further, this again brings me back to my question on whether direction and directional derivative can be used synonymously in this context.

Thank you!

• The book of Spivak on differential geometry, first volume, has a treatment of this. Essentially, it says and proves that "no matter how you define the tangent space, you'll always obtain the same thing", up to isomorphism, of course. Apr 6 '20 at 12:50

1) Taking differential derivatives allows you to do differential calculus on manifolds. One explicit example could be defining tangent fields, i.e. maps $$X:M\to TM:=\sqcup_{p\in M}T_pM$$ such that $$\pi\circ X=\mathrm{id}_M$$ where $$\pi:TM\to M$$ is the canonical projection, and integrating them in order to get flow maps, i.e. maps $$\varphi:\mathbb{R}\times M\to M$$ such that $$\varphi(0,\cdot)=\mathrm{id}_M$$ and $$\left.\frac{\partial\varphi(\cdot,x)}{\partial t}\right|_t=X_{\varphi(t,x)}$$. Thus, from linear data ($$X$$), you recover a family of diffeomorphisms of $$M$$ with a certain behaviour.

2) If your manifold $$S$$ is a submanifold of an ambient one $$M$$, the inclusion $$i:S\to M$$ induces a map $$di_p:T_pS\to T_pM$$ which allows you to consider the tangent space of $$S$$ at $$p$$ as a linear subspace of the tangent space of $$M$$ at $$p$$. There is an other identification for tangent vectors of affine manifolds (that is $$M=\mathbb{R}^n$$ with the maximal atlas induced by $$\mathcal{A}=\{(\mathrm{id}_{\mathbb{R}^n},\mathbb{R}^n)\}$$) in order to identify them with actual vectors of $$\mathbb{R}^n$$: this identification is given by $$\mathbb{R}^n\ni v\mapsto\partial_v\in T_p\mathbb{R}^n$$, where $$\partial_v$$ acts on functions $$f\in C^\infty_p(\mathbb{R}^n)$$ by

$$\partial_vf=\lim\limits_{t\to 0}\frac{f(p+tv)-f(p)}{t}.$$

In other words, you identify the vector $$v$$ with the directional derivative in the direction $$v$$. So when you have a submanifold $$S$$ of an affine one, you can:

1. Identify a tangent vector of $$S$$ as a tangent vector of $$\mathbb{R}^n$$

2. Identify the tangent vector of $$\mathbb{R}^n$$ with an actual vector of $$\mathbb{R}^n$$.

3) Again, taking directional derivatives on a manifold is authorizing himself to do differential calculus on manifolds, allowing the use of useful theorems as implicit function theorem or inverse function theorem. For the identification of the two definitions, I will answer it in 4).

4) You answer your question by pointing the identification $$[\gamma]\mapsto D_\gamma$$, but you have to be carful that this does not depend of the choice of the representant $$\gamma$$. But since

$$(f\circ\gamma)'(0)=(f\circ\varphi^{-1}\circ\varphi\circ\gamma)'(0)=d(f\circ\varphi^{-1})_{\varphi\circ\gamma(0)}\left((\varphi\circ\gamma)'(0)\right)$$

by the chain rule, it is clear by the definition of the equivalence relation that is will be the case.