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Let $M$ be a manifold, $T_xM$ its tangent space with basis {$\frac{\partial}{\partial x_i}$}, with $x_i$ being the $i$-th coordinate function of a chart. A cotangent space $T_x^\star X$ is defined as its dual, so as a vector space of linear functionals i.e. mappings from $T_xM$ to $\mathbb{R}$. As I search through literature, the basis of $T_x^\star X$ is shown to be {$dx_i$}, where $dx_i$ is a derivative of $x_i$ and $dx_i: T_x X \rightarrow T_{x_i(x)}\mathbb{R} \simeq \mathbb{R}$. Which makes sense, when $dx_i v$ is defined as

$$dx_i v = v(x_i)$$

for $v=\sum_i v_i \frac{\partial}{\partial x_i}$, $v_i \in \mathbb{R}$. Then $dx_i$ is a linear functioal and $v(x_i)$ a real number. Here we consider $\frac{\partial}{\partial x_i}x_j=\delta_{i,j}$

But another notion that i have come across is such: $$(dx_i v)f = v(f \circ x_i)$$

I don't really see how does that make sense, since $dx_i$ is supposed to be a functional, so $dx_i v$ must be a real number, but when defined as before, $dx_i v$ acts on a function, thus it cannot be a real number.

Filip
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  • Can you point to the source where you've come across the second notion? I'm not too sure what you mean as typed. – Ken Jan 15 '23 at 14:53
  • There is an example https://math.stackexchange.com/questions/193258/basis-of-cotangent-space – Filip Jan 15 '23 at 14:56
  • Oh that's right, I'm still drinking my morning coffee! I'll type up an answer shortly. – Ken Jan 15 '23 at 14:57
  • You have $v=\sum_i v_i \frac{\partial}{\partial x_i} \in \mathbb{R}$ but this is not true, $v$ is a vector field. To be a real number you need to fix it at one point, say $q \in M$ so that $v_q \in T_qM$ is given by $v_q = \sum_iv_i(q)\frac{\partial}{\partial x_i}\mid_q$. Now, this tangent vector acts on smooth functions as $v_q(f) = \sum_iv_i(q)\frac{\partial f}{\partial x_i}\mid_q$ thus yielding a real number. – user57 Jan 15 '23 at 15:44

1 Answers1

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Let $M$ be a smooth $n$-manifold and $(U,\phi)$ be a coordinate chart where $\phi$ has component functions $(x^i)$. Recall that $\phi$ is a smooth function $\phi:U\to\mathbb{R}^n$ whose component functions are smooth maps $x^i:U\to\mathbb{R}$. In particular, for any point $p\in M$, we may look at the differential $dx^i_p:T_pM\to T_{x^i(p)}\mathbb{R}\cong\mathbb{R}$. So for any tangent vector $v\in T_pM$, its image under $dx^i_p$ is indeed a real number.

But also recall that $dx^i_p(v)\in T_{x^i(p)}\mathbb{R}^n$ was defined to be the tangent vector such that, for any smooth $f:\mathbb{R}\to\mathbb{R}$, we have the equality $$ dx^i_p(v)f=v(f\circ x^i). $$ Thus, these equivalences of $dx^i$ essentially boil down to the fact that the tangent spaces of the smooth manifold $\mathbb{R}$ are isomorphic to $\mathbb{R}$.

Ken
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  • But how can $dx^i_p$ form a basis of the dual space, when defined as that? – Filip Jan 15 '23 at 15:20
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    @Filip The dual space $T_p^M$ is defined to be the set of linear functionals on $T_pM$, namely, $\omega\in T_p^M$ if $\omega:T_pM\to\mathbb{R}$ linear. So as defined, we see that $dx^i_p:T_pM\to T_{x^i(p)}\mathbb{R}$, but the codomain is isomorphic to $\mathbb{R}$. So $dx^i_p$ is definitely an element of the dual space. Now, you just have to check that they are linearly independent, span, and are the dual basis to $\partial_{x^i}$. – Ken Jan 15 '23 at 15:26
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    @Filip: you do not need to regard elements of the tangent space at $p$ as derviations. Please see my answer for a perhaps more intutive approach. I am going to delete it after you have had a chance to read it. – Matematleta Jan 15 '23 at 16:23