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I'm watching a series of lectures on differential geometry, and I've run into a bit of a problem with the definition of the tangent space. We first defined a tangent space as $\{(p,v) | v \in \mathbb{R}^n\}$, which makes sense to me: it's the set of all vectors attached at point $p$. We then defined the directional derivative as

$$ (Df)(p,v) = \lim_{t \rightarrow 0} \frac{f(p + tv) - f(p)}{t} $$

We expanded that to this:

$$ (Df)(p,v) = \left( \sum_{i = 0}^{n}v_i \left.\frac{\partial}{\partial x_i}\right|_{p} \right) f$$

This makes sense to me; we have defined the directional derivative as an operator that is applied to the function.

Here's the part where I lose the plot. I'm then told that, if I think about it, the portion inside the parentheses is really interchangeable with $(p,v)$. I'm afraid that I've thought about it, and I can't see the equivalence. $\sum_{i = 0}^{n}v_i \left.\frac{\partial}{\partial x_i}\right|_{p}$ is an operator (isn't it?) whereas $(p,v)$ is a ordered pair of elements of $\mathbb{R}^n$. Does that mean that the expression $(p,v)(f)$ makes sense? What would that mean?

I must be thinking about this the wrong way; can someone clarify?

narip
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anjruu
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  • Think about the information contained in the expression inside the parentheses. It is exactly the information conveyed by the ordered pair $\left(p,v\right)$. Then, we can define the notation $\left(p,v\right)\left(f\right)$ to "mean" $\left(Df\right)\left(p,v\right)$. It's a matter of formal notation, as far as I can make out. – udit.m Dec 03 '15 at 14:29
  • It is essentially a reconceptualization of tangent vectors, shifting from viewing them as directions on the manifold to viewing them as the directional derivative operators themselves. (Acting on scalar fields on the manifold.) – Matt Dickau Dec 03 '15 at 15:03

2 Answers2

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I think the phrase "the portion inside the parentheses is really interchangeable with $(p,v)$" is quite misleading. This is not really true -- as you correctly observed, $(p,v)$ is an ordered pair of elements of $\mathbb R^n$, while the expression in parentheses is an operator on functions.

What is true is that there is a linear map from the set $\{(p,v)|v\in\mathbb R^n\}$ (let's call that the geometric tangent space) into the set of linear differential operators on functions, which takes the pair $(p,v)$ to the "directional derivative operator" that you wrote down. The image of this map is the set of derivations at $\boldsymbol p$, which is the set of all linear maps $X\colon C^\infty(\mathbb R^n)\to \mathbb R$ that satisfy this product rule: $$ X(fg) = f(p)X(g) + g(p)X(f). $$ The geometric tangent space is thus canonically isomorphic to the set of derivations at $p$.

Why does this matter? Because on an abstract manifold, the geometric tangent space doesn't have any coordinate-independent meaning, but the space of derivations at $p$ does. So we take the space of derivations at $p$ as our definition of the tangent space to $M$ at $p$.

Once you get comfortable with the canonical isomorphism between the geometric tangent space and the space of derivations at $p$, then you might start thinking of them as "interchangeable." But when you're first trying to learn this stuff, it's more productive to think of them as canonically isomorphic.

Jack Lee
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  • Hi Prof. Lee, I am reading your textbook Introduction to Smooth Manifolds and have a question about the definition of tangent spaces. You defined the geometric tangent space to $\mathbb{R}^n$ at $p$ to be the set ${(p,v)|v\in\mathbb R^n}$ in the text. So I guess the geometric tangent space to $\mathbb{S}^2$ at $p$ would be defined as ${(p,v)|v\in\mathbb S^2}$, which is a unit sphere centered at $p$. But shouldn't a tangent space to a ball at some point be a plane? I have been stuck here for a while. Thanks for your answer in advance. – Sam Wong Nov 06 '20 at 06:10
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    @SamWong: I didn't define the geometric tangent space for a submanifold of $\mathbb R^n$; only for $\mathbb R^n$ itself. As you observed, mimicking the same definition would not make any sense for a submanifold. Tangent spaces to submanifolds are treated in Chapter 5 of my book. – Jack Lee Nov 06 '20 at 23:30
  • Oh ok, I will read the chapter 5 later. Thanks! – Sam Wong Nov 07 '20 at 04:39
  • Hi Professor Lee. I am a bit confused on the definition of the tangent space as derivations. One important property of that definition is to write a tangent vector $v \in T_p M$ as $v = \sum_{\mu =1}^{m} v(x^{\mu}) \cdot \left. \frac{\partial}{\partial x^{\mu}} \right|_p \ \ $, where $x=(x^1,...,x^m)$ is a local chart around $p$. But generaly $dom(x) \neq M$ and $v$ is supposed to act just in functions of $C^{\infty} M$. Can you clarify for me, please? – Gustavo Feb 12 '24 at 15:30
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    @Gustavo: Because the value $v(f)$ depends only on the values of $f$ in an arbitrarily small neighborhood of $p$, a vector $v\in T_pM$ can act on functions that are defined only in a neighborhood of $p$. This is discussed on page 56 of my Introduction to Smooth Manifolds (2nd ed.). – Jack Lee Feb 12 '24 at 16:08
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The usual notation for the tangent space at a point $p$ of a differentiable manifold $M$ is $T_pM$. By the definition you can see that this space is a vector space that has the same dimension $n$ as the manifold $M$. The elements of $T_pM$ are not ''all vectors attached at point $p$'' as you say, but the vectors, attached at $p$, that stay in the tangent plane to the manifold at $p$.

So, in a notation as $ T_pM=\{(p,v):v\in \mathbb{R}^n\}$ $p$ really is simply a parameter, that specifies the point where we take the tangent plane.

The vectors in $T_pM$ can be represented, as you notice, as $$ \sum_{i = 0}^{n}v_i \left.\frac{\partial}{\partial x_i}\right|_{p} $$ and we can can think at $T_pM$ as a vector space of linear operators with basis $$ \left(\frac{\partial}{\partial x_i}|_{p}\right) \qquad i=1,2\cdots n \quad p \in M $$ We have one such space for every $p$ and we can consider the set of all such tangent spaces: $$ TM=\bigcup_{p\in M}\{(p,v):v\in T_pM \} $$ In this notation the presence of $p$ has clearly the sense of parametrize all the tangent spaces, but note that the element of $T_pM$ is $v$, not the couple $(p,v)$, and $TM$ is the tangent bundle of $M$.

Emilio Novati
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