I don't come from a Differential Geometry background but I have been trying to read a bit about Lie algebras. I am using Humphrey's as my main source but just to get a glimpse of the correspondance between Lie groups and Lie algebras, I am using a particular set of course notes. However, there is something that I cannot really understand about tangent spaces.

Firstly, we will make a major simplification by saying that our manifold $M$ lives inside some $\mathbb{R}^n$. (i.e. $M\subset \mathbb{R}^n$)

For a given point $m\in M$, the tangent space that I am given is:

$T_mM=\{v\in R^n: \exists \textrm{ curve }\gamma:(-\epsilon,\epsilon)\to M \textrm{ with } \gamma(0)=m\textrm{ and } \frac{d}{dt}\gamma(t)|_{t=0}=v\}$

I kind of convinced myself that this definition could be generalised to one involving charts by replacing the last part with $\frac{d}{dt}\phi\gamma(t)|_{t=0}=v$ where $\phi$ is a chart on the manifold containing $m$.

It was left as an exercise to show that $T_mM$ is a vector space. Naively, I thought given $v=\frac{d}{dt}{\gamma_1}|_{t=0}$ and $w=\frac{d}{dt}{\gamma_2}|_{t=0}$, then we have $\gamma=\frac{1}{2}[\gamma_1(2t)+\gamma_2(2t)]$ as a curve that has derivative $v+w$ and $\gamma(0)=m$. But clearly, there is a huge problem that this $\gamma$ may not be a curve in $M$ as $M$ need not be a vector space and in general space manifolds are not vector subspaces of $\mathbb{R}^n$.

I did a bit of searching around and found the definitions to be quite different and defined linear maps called tangent vectors. Now, clearly, linear maps will sidestep this problem about not being a vector space.

How can I modify this definition to circumvent this problem?

(The notes I was using is http://pi.math.cornell.edu/~dmehrle/notes/partiii/liealg_partiii_notes.pdf Page 10)

  • $\begingroup$ Before all else, be careful that Definition 2.3 in the notes (smooth functions between smooth manifolds) is not quite standard. You could not even prove that every smooth function is continuous the way it is defined in the notes. One standard definition of smooth function is to add continuity condition to the function in addition to the property stated in the notes. I did not read through the notes however, and I don't know if there will be some more "mistakes". $\endgroup$ – edm Nov 11 '18 at 10:47
  • $\begingroup$ Could you be more specific about what is wrong with Def. 2.3? Do you mean that on top of the what is mentioned, the function $f$ needs to be continuous in the sense of a mapping between topological spaces? And I take it that you agree that something is wrong with this definition of a tangent space $\endgroup$ – daruma Nov 11 '18 at 11:07
  • $\begingroup$ To be simple, the correct definition of smooth function is a function $f:M\to N$ that is continuous in topological sense and for all charts $\phi:U\to\Bbb R^m$ on $M$, $\psi:V\to\Bbb R^n$ on $N$, the function $\psi\circ f\circ\phi^{-1}$ is smooth at where it is defined. The definition of tangent space should be using curves $\gamma:(-\varepsilon,\varepsilon)\to M$ considered as smooth functions from $(-\varepsilon,\varepsilon)$ to $M$$. $\endgroup$ – edm Nov 11 '18 at 11:32
  • $\begingroup$ @edm I am confused about your latest comment. So you are saying that the definition of the tangent space is correct or incorrect? $\endgroup$ – daruma Nov 11 '18 at 12:05
  • $\begingroup$ The problem is, I don't see "curves" being defined before Definition 2.11. So you need a definition of "curves" to make sense of the definition of tangent space, and I provided you the definition of "curves". $\endgroup$ – edm Nov 11 '18 at 12:19

An equivalent definition of tangent space at a point is to use a chart and its derivative. First we recap on the notion of charts.

Let $p$ be a point on the manifold $M$, and let $\phi:U\to V$ be a chart from $U\subseteq M$ to $V\subseteq\Bbb R^m$, with $U$ open in $M$, $V$ open in $\Bbb R^m$. By definition of a chart, $\phi$ is a homeomorphism from $U$ to $V$, and its inverse $\phi^{-1}$ is smooth as a map $\phi^{-1}:V\to\Bbb R^n$ (remember, $M\subseteq\Bbb R^n$ so we can see the codomain of $\phi^{-1}$ as $\Bbb R^n$ just fine), and the derivative of $\phi^{-1}$ at any point is an injective linear map. Write the derivative of $\phi^{-1}$ at $\phi(p)$ as $(d\phi^{-1})_{\phi(p)}:\Bbb R^m\to\Bbb R^n$.

The following proposition gives you an easy way to show that the tangent space is a vector space.

Proposition: The tangent space $T_pM$ is exactly equal to the range of the linear map $(d\phi^{-1})_{\phi(p)}$. As a corollary, $T_pM$ is a vector space.

You may try to prove this yourself. If you cannot, I can provide a reference or a proof sketch.


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