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I am somewhat confused about the connection between the dimension of tangent spaces and the definition of tangent vectors:

Let's say we have a d-dimensional differentiable manifold $M^d$. Further, consider its tangent space $T_p M^d$ at a point $p \in M^d$. According to Königsberger (p. 120) $T_p M^d$ is a vector space with dimension $d$. That is, $T_p M^d$ has the same dimension as $M^d$.

Example: Consider a surface in the $\mathbb{R}^3$, i.e. $M^2$ is an open set embedded in $\mathbb{R}^3$. Following Königsberger, every $T_p M^2$ is 2-dimensional. Hence, $\forall X \in T_p M^2: X \in \mathbb{R}^2$. This makes sense as we know the (affine) tangent space coincides with the tangent plane in this case.

Bredon (p.76f.) defines the tangent vector as a differential operator on the germ of a smooth function $f: U \subset M \rightarrow \mathbb{R}$:

$$ D_{\gamma}(f) = \sum_{i=1}^n \frac{d \gamma_i}{d t} \frac{\partial }{\partial x_i} f\, \Bigg|_{t=0},$$

where $\gamma = (\gamma_i(t), ..., \gamma_n(t))$ represents a smooth curve in $M$. He defines the tangent space as

$$T_p M = \{D_{\gamma}\, |\, \gamma(0) = p\}$$

Point of confusion: Because $f \in C^{\infty}(M, \mathbb{R})$ it follows that $D_{\gamma}(f) \in \mathbb{R}$ irrespective of the dimensionality of $M$. This contradicts what Königsberger said (and also the example).

My thought on that: Define $D_{\gamma}$ on the germ of functions $f \in C^{\infty}(M, \mathbb{R}^d)$ such that $D_{\gamma}(f) \in \mathbb{R}^d$. However, as I am fairly new to this topic, I am unsure whether something like this would be possible.

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  • $\begingroup$ $D_{\gamma}(f) \in \mathbb{R}$ is irrelevant, that $\mathbb{R}$ isn't the tangent space. Here it is not at all obvious that the dimension of the tangent space ought to work out to still be $d$ but it will; work in local coordinates to see this. $\endgroup$ Aug 11 at 11:29

3 Answers 3

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Of course $D_{\gamma}$ evaluated on a function is a number, but the members of tangent space are actually the differential operators on the function. Roughly speaking if the space is n-dimensional, then you can move in n-different direction in the space, that means n- different directions you can take the derivative.

Hence, dimension of manifold=dimension of Tp

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  • $\begingroup$ Okay I see. I try figure out how $D_{\gamma}$ is associated with e.g. a tangent vector from the tangent plane. Therefore I thought, when the tangent plane coincides with the tangent space, the elements have to be the same. Following that thought, it should be possible to interpret $D_{\gamma}$ as a vector. So is it fair to say that if we assume the embedding space to be the $\mathbb{R}^n$ the $\partial / \partial x_i$ are given by the canonical basis? $\endgroup$
    – theDude
    Aug 11 at 13:04
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    $\begingroup$ Yeah excellent question. I had pretty much the same. There is a bit of trickery going on with the association. I'll link in a moment @theDude $\endgroup$ Aug 11 at 13:18
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    $\begingroup$ This one here @theDude $\endgroup$ Aug 11 at 13:19
  • $\begingroup$ The question you've linked has a nice way of elaborating the term: $$D_{\gamma}(f) = \sum_{i=1}^n \frac{d \gamma_i}{d t} \frac{\partial }{\partial x_i} f\, \Bigg|_{t=0}$$ However, I am actually looking for something like an isomorphism as described here. Something that established a one-to-one correspondence between the differential operators $D_{\gamma} \in T_p M$ and, to stick to the example, a vector in a tangent plane. $\endgroup$
    – theDude
    Aug 11 at 16:07
  • $\begingroup$ Oh right, the philosophy is all you're going to use vectors for is taking directioncal derivative, so think directly of the vector as the directional derivative operator through the isomorphism. So for instance, i+j+k -> partial_x + partial_y + partial_z @theDude $\endgroup$ Aug 11 at 16:13
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Perhaps the following observation may help:

for any $f \in C^{\infty}(\mathbb{R}^n)$ it is true that $$\dfrac{\partial f}{\partial \vec{v} }\bigg|_p=\sum_i \dfrac{\partial f}{\partial x_i }\bigg|_pv^i$$

where $\dfrac{\partial f}{\partial \vec{v} }$ denotes the directional derivative of $f$ in the direction of $\vec{v}$.

This is worth dwelling over in detail: It means that the space of directional derivatives at a point $p$ is a finite dimensional vector space whose dimension exactly equals to the dimenion of the underlying space $\mathbb{R}^n$.

This is $\underline{not}$ always the case. If you take instead of $C^{\infty}(\mathbb{R}^n)$ the space of all functions whose directional derivatives exist in all directions. Fix $f$ and a point $p$, then the map $$\vec{v} \to \dfrac{\partial f}{\partial \vec{v} }\bigg|_p$$ may fail to be linear and hence the space of all directional derivative operators will be an infinite dimensional vector space. The deep fact underlying manifold theory is that if we take our space to be $C^{\infty}(\mathbb{R}^n)$ (although $C^{1}(\mathbb{R}^n)$ will also suffice in this case) then we observe the following crucial phenomena: "the directional derivative in the direction $\vec{v}=(v_1,...,v_n)$ is a linear combination of directional derivatives in the cartesian directions for any function $f$ in that space". this means that the set of linearly independent directional derivative operators on $\mathbb{R}^n$ at a point $p$ is exactly equal to $n$. When we work with general manifolds the above theorem is still true and hence the number of linearly independent "derivations" (which are directional derivative operators in charts) is exactly equal to the independent degrees of freedom in the underlying manifold. If the manifold is $n$-dimensional then there are exactly $n$ linearly independent differential operators at the point $p$.

You may have taken for granted the fact that you can write the gradient of a function $f$ as a row vector; well, this follows exactly from the above fact: that at a fixed point $p$ the directional derivative is linear in the direction you are trying to derive; which can actually be false in some cases.

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You might have $D_{\gamma}(f) \in \mathbb{R}$ but this a priori doesn't give you any information on the dimension of $T_p M$. Your domain has high dimension after all! Recall for example from linear algebra that for a finite-dimensional $k$-vector space $V$ and its dual space $V^* = \operatorname{Hom}(V, k)$ we have $\dim{V} = \dim{V^*}$.

By the way, your definition of a surface is not correct, surely $M \subseteq \mathbb{R}^3$ is not open: Open submanifolds of $\mathbb{R}^3$ are $3$-dimensional.

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  • $\begingroup$ I have not thought about interpreting $T_p M$ as the dual space of $M$. But it actually helps me understanding the dimension of the tangent space as it matches with what Königsberger says! (And thank you for the note about the surface definition) $\endgroup$
    – theDude
    Aug 11 at 12:41
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    $\begingroup$ @theDude It's not the dual space of $M$, rather we have $T_p M \subseteq C^{\infty}(M, \mathbb{R})^*$. Glad it helped :) $\endgroup$
    – Qi Zhu
    Aug 11 at 13:18

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