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Starting with these definitions

  • A curve on a manifold $\mathcal M$ is a smooth (i.e. $C^{\infty}$) map $\sigma $ from some open interval $(-\epsilon,\epsilon)$ of the real line into $\mathcal M$

  • Two curves $\sigma_1$ and $\sigma_2$ are tangent at a point $p$ in $\mathcal M$ if (a) $\sigma_1(0) = \sigma_2(0) = p$ and (b) In some local coordinate system $(x^1,x^2,\ldots,x^m)$ around $p$, two curves are tangent in the usual sense as curves in $\mathbb R^m$, $$ (x^i \circ \sigma_1)'(0) = (x^i \circ \sigma_2)'(0) $$ here, $i=1,\ldots,m$

  • The tangent vector is defined as the equivalence class of curves in $\mathcal M$ where the equivalence relation between two curves is that they are tangent at point $p$.

  • The tangent space is $T_p\mathcal M$ to $\mathcal M$ at point $p$ is the set of all tangent vectors at point $p$

I am trying to prove the tangent space at the point $p$ in a manifold $\mathcal M$ is a vector space.

I'm starting with $v_1 \in T_p\mathcal M$, and $v_2 \in T_p\mathcal M$, and I have the following definitions $$ v_1 + v_2 := [\phi^{-1}\circ \ (\phi\ \circ \sigma_1 + \phi\ \circ \sigma_2 )] \\ r \ v_1 := [\phi^{-1}\circ \ (r \phi\ \circ \sigma_1)]\ \forall r \in \mathbb R $$

I want to show that $v_1 + v_2 \in T_p \mathcal M$ and $r \ v_1 \in T_p \mathcal M$

As $v_1 ,v_2 \in T_p\mathcal M$, then $$ \sigma_1(0) = \sigma_2(0) = p $$

Now, for $v_1 + v_2$ be a vector at $p$ , $\phi^{-1}\circ \ (\phi\ \circ \sigma_1 + \phi\ \circ \sigma_2 )(0) = p$ $$ \phi^{-1}\circ \ (\phi\ \circ \sigma_1 + \phi\ \circ \sigma_2 )(0) = \phi^{-1} \ (\phi\ ( \sigma_1(0)) + \phi\ (\sigma_2(0)) ) \\ = \phi^{-1}((\phi\ ( p) + \phi\ (p) )) \\ = \phi^{-1}( \ 2\phi\ ( p) ) \neq p $$

I can't prove the closure relations starting from the definitions, what I'm doing wrong?

Edit:

The book I am following "Isham, Chris J. Modern differential geometry for physicists. Vol. 61. World Scientific, 1999.", takes a special chart $(U,\phi)$ such that $\phi(p) = \mathbf 0 \in \mathcal M$, using this choice

$$ \phi^{-1}\circ \ (\phi\ \circ \sigma_1 + \phi\ \circ \sigma_2 )(0) = \phi^{-1}( \ 2\phi\ ( p) ) = \phi^{-1}(0) = p $$ So, the closure is proven under addition. But this chart is a special choice. But the definitions hold for any charts around $p$, so another choice of charts should give the same result.

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  • 1
    $\begingroup$ The definition of a manifold means it's always possible to find a chart that maps any point in the manifold to a defined point in the underlying vector space, so without loss of generality we can always choose a chart that maps $p$ to $0$. See Definition 2.1 in the book and figure 2.2 $\endgroup$
    – PeteBabe
    Aug 21, 2020 at 12:52
  • $\begingroup$ @PeteBabe Now, I get it, the functions $\phi : U \to \mathbb R^m$ is defined as per my need. So I can choose $\phi(p) = \mathbf 0$ $\endgroup$
    – Galilean
    Aug 21, 2020 at 14:56

1 Answer 1

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Tangent vectors a $p \in M$ are equivalence classes of smooth curves $\sigma : (-\epsilon,\epsilon) \to M$ such that $\sigma(0) = p$ ("smooth curves in $M$ through $p$"). Here $\epsilon = \epsilon (\sigma)$ is a parameter which may vary from curve to curve. The equivalence relation is given by $\sigma_1 \sim \sigma_2$ if $(\phi \sigma_1)'(0) = (\phi \sigma_2)'(0)$ for some chart $\phi$ around $p$. It is easy to verify that $\sigma_1 \sim \sigma_2$ iff $(\phi \sigma_1)'(0) = (\phi \sigma_2)'(0)$ for all charts $\phi$ around $p$.

Given a smooth curve $\sigma : (-\epsilon,\epsilon) \to M$ through $p$, you can of course define $r \cdot \sigma : (-\epsilon/\lvert r \rvert,\epsilon/\lvert r \rvert) \to M, (r \cdot \sigma)(t) = \sigma (rt)$. Unfortunately there is no similar definition of $\sigma_1 + \sigma_2$ for curves $\sigma_i$ in $M$ trough $p$. You try to add them via the definition $$\sigma_1 + \sigma_2 = \phi^{-1}(\phi\sigma_1 + \phi \sigma_2).$$ This exploits the fact that the chart $\phi : U \to V \subset \mathbb R^n$ take values in $\mathbb R^n$, but in general it does not work because you cannot be sure that $\phi\sigma_1(t) + \phi \sigma_2(t) \in V$ for $\lvert t \rvert$ sufficiently small. Not even $\phi\sigma_1(0) + \phi \sigma_2(0) = \phi(p) + \phi(p) = 2\phi(p)$ is in general contained in $V$.

The solution is to consider only charts such that $\phi(p) = 0$. This can always be achieved if we replace an arbitrary chart $\phi$ by $T\phi$ where $T$ is the translation by $-\phi(p)$. The same holds for your definition of $r \cdot \sigma$.

Doing so, you will see that you get in fact the structure of a vector space on $T_p M$. Formally I suggest to proceed as follows:

  1. Show that $\phi_* : T_pM \to T_0V, \phi_*([\sigma]) = [\phi\sigma]$, is a bijection.

  2. Show that $T_0V$ becomes a vector space via $[\tau_1] + [\tau_2] = [\tau_1 + \tau_2]$ and $[r \cdot \tau] = [r \cdot \tau]$, where $(\tau_1 + \tau_2(t) = \tau_1(t)+ \tau_2(t)$ and $(r \cdot \tau)(t) = r \cdot \tau(t)$. Note that there always exist a maximal interval on which $\tau_1(t)+ \tau_2(t) \in V$ and $r \cdot \tau(t) \in V$; we take these intervals as the domains of $\tau_1 + \tau_2$ and $r \cdot \tau$. It is then easy to see that the map $\mathbb R^n \to T_0V, v \mapsto \tau_v$ with $\tau_v(t) = tv$, gives an isomorphism of vector spaces whichs shows that $\dim T_0V = n$.

  3. Observe that $\phi_*$ induces a unique structure of a vector space on $T_pM$ such that $\phi_*$ becomes an isomorphism of vector spaces.

  4. At first glance it seems that the vector space structure on $T_pM$ depends on the choice of $\phi$. The final step will therefore be to prove that any two charts $\phi_1, \phi_2$ around $p$ with $\phi_i(p) = 0$ produce the same vector space structure on $T_pM$.

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  • $\begingroup$ wouldn't it be possible to, instead, define the sum of equivalence classes by sending $[\tau_1]+[\tau_2]=[t\mapsto \phi^{-1}(\phi(\tau_1(t))+\phi(\tau_2(t))-\phi(p))]$? In other words, using an affine transformation to sum the $\mathbb R^n$ vectors rather than just their sum. This would at least give a path mapping $0\mapsto p\in M$, as required $\endgroup$
    – glS
    Dec 23, 2021 at 17:16
  • $\begingroup$ @gls Yes, you can do that. It results in the same what I have done: You replace the chart $\phi$ by the translated chart $\psi = \phi - \phi(p)$ which satisfies $\psi(p) = 0$. Then you take the sum $\psi(\tau_1(t)) + \psi(\tau_2(t))$ and compute $\psi^{-1}(\psi(\tau_1(t)) + \psi(\tau_2(t)))$. $\endgroup$
    – Paul Frost
    Dec 24, 2021 at 8:04

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