2

In Tu's book, the tangent vector at $p \in U \subseteq \mathbb{R}^n$ is defined as an arrow emanating from $p$. However, when passing to manifolds, such an approach cannot be used, since we should fix a chart $(U,\phi)$ about $p$ and, next, to decree that a tangent vector is an arrow emanating at $\phi(p)$. In such a case, changing the chart around $p$ (take for instance $(V,\psi)$, a priori the arrow emanating from $\psi(p)$ is different from that emanating from $\phi(p)$, so the definition of tangent vector so given is ambiguous.

My question is: $\mathbb{R}^n$ is itself a manifold. On it we can take another atlas, instead of the obvious one. Why does the previous definition of tangent space work? Also in this case one should have the same problem.

TheWanderer
  • 5,126

1 Answers1

0

One of the main takeaways of linear algebra is that it doesn't make sense to speak of a vector by itself, vectors only make sense in the broader context of a vector space. So the definition that Tu makes is that you define the tangent space in the chart, and a tangent vector is a vector in that space. The definition indeed a priori depends on the chart, but you then show that it doesn't essentially depend on chart. For this, you take the inverse of the first chart and compose it with the second chart (everything on their intersection, of course). This gives you a diffeomorphism from an open set in $\mathbb R^n$ to an open set in $\mathbb R^n$. Now the tangent spaces are identified by a linear isomorphism, which is the differential at the point $\phi(p)$ of this diffeomorphism.

This definition has the advantage of giving an intuition for what the tangent space is, but the disadvantage that it makes the tangent space practically useless. For the purpose of usefulness, it is better to see the tangent space as the vector space of derivations of locally defined smooth functions around $p$ (of germs of smooth functions, to be a bit technical). You'll probably want to pass through the tangent space as equivalence classes of curves passing through $p$ which are smooth in any chart, equivalence being given by the curves having the same tangent vector in any chart, to get at this definition. Tu probably covers these definitions as well, but in either case, you can also look at Lee's Introduction to Smooth Manifolds.

Regarding your question in the end. The definition indeed depends on the chart you take for $\mathbb R^n$. In this case, this chart is the identity. But you may take other atlases for $\mathbb R^n$ (which may not give you the same smooth structure; see here) and then you have to use the general definition you have given first.

rosecabbage
  • 1,645
  • $T_p(\mathbb{R}^n)$ canonically identifies with the space of derivations of $\mathcal{C}^\infty_p(\mathbb{R}^n)$. Equip $\mathbb{R}$ with a non-standard atlas (denote it by $\widehat{\mathbb{R}^n}$). Let $(U,\phi)$ be a chart about $p$. Define $T_p(\widehat{\mathbb{R}^n})$ as the space of derivations as before. Then it canonically identifies with $T_p(\mathbb{R}^n) \cong \mathbb{R^n}$ (again canonically). By the last considerations, may I see geometrically a vector in $T_p(\widehat{\mathbb{R}^n})$ as an arrow emanating from $p$? What can I do with $\phi(p)$? Which is its role? – TheWanderer Mar 19 '24 at 10:25
  • Here, the problem is the following: is it true that taking $\widehat{\mathbb{R}^n}$, its tangent space at $p$ canonically identifies with $T_p(\mathbb{R}^n)$? If not, can you provide an example, please? – TheWanderer Mar 19 '24 at 10:27
  • As a vector space, the tangent space is $T_p(\mathbb R^n)$ and it also is $\mathbb R^n$. AN identification of the tangent space with these, however, is equivalent with choosing a basis, and this is not canonical. Each chart gives you a different basis, so a different identification; explicityl, if $(x_1, \ldots, x_n)$ are the coordinates in your chart, then $(\frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial x_n})$ is the basis corresponding with the chart. – rosecabbage Mar 19 '24 at 10:52
  • Seeing a vector in $T_p(\mathbb R^n)$ as "an arrow emanating from $p$" is not very mathematically rigorous. In fact, to make it rigorous, you need to suppose that your manifold is embedded into some ambient space, and while this is always possible by the Whitney (or Nash if you prefer) embedding theorem(s), the dimension of the ambient space could be very big, so it's not something really visualisable. That's why it's better (in my mind) to see vectors intrinsically, as derivations of functions, which does away with the need to see your space as embedded. – rosecabbage Mar 19 '24 at 10:55