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Let $M$ be a manifold and $p \in M$, then we have the tangent space $T_pM$ that has a basis $\left\{ \left(\frac{\partial}{\partial x^i}\right)_p \right \}$. The cotangent space $T^\ast_p M$ has then a basis $\{ (dx^i)_p \}$.

I'm trying to figure out why the cotangent space has the proposed base. If $V$ is a finite dimensional vector space with base $\{e_1, \dots e_n \}$, then $V^\ast$ has a basis $\{f^1, \dots f^n \}$, where $f^i : V \to \Bbb R$ and $f^i(e_j) = \delta_i^j$.

In the cotangent space case $\{ (dx^i)_p \}$ would be a basis if $$(dx^i)_p\left(\frac{\partial}{\partial x^j}\right)_p = \delta_i^j.$$

Now $(dx^i)_p : T^\ast_p M \to \Bbb R$ is just any linear map so how can I know anything about $(dx^i)_p\left(\frac{\partial}{\partial x^j}\right)_p$ without defining what $(dx^i)_p$ does first?

I'm told that $dx^i$ is "just a notation", but it should apparently be defined to be something in order for this to work out?

Walker
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  • What's your definition of $dx^i$? – Didier Jan 10 '23 at 11:21
  • Which definition of $T_pM$ do you use? Via derivations? Which textbook do you use? – Paul Frost Jan 10 '23 at 11:24
  • I'm assuming you've already defined $\frac{\partial}{\partial x^i}$ beforehand? In that case you can define $dx^i$ via this orthogonality relation you've given - it uniquely identifies the basis in the finite dimensional case. Note that this isn't something that any basis of the cotangent space has to statisfy: it's only true for the dual basis to the basis of the tangent space you've given. – SV-97 Jan 10 '23 at 11:28
  • @Didier I think that's exactly the question here. The definition for it at the moment for me is that it's only a linear map. – Walker Jan 10 '23 at 11:31
  • @PaulFrost Via derivations or tangent curves should both work here right? – Walker Jan 10 '23 at 11:32
  • @Walker Yes, both work, but if you want to define the $(dx^i)_p$ as in point 1. of Didiers's answer, you need to know what the elements of $T_pM$ look like. – Paul Frost Jan 10 '23 at 11:38

1 Answers1

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Given local coordinates $(x^1,\ldots,x^n)$ on an open subset $U \subset M$, there are essentially two ways to define the coframe $(dx^1,\ldots,dx^n)$ on $U$:

  1. For $i\in \{1,\ldots,n\}$, we have a smooth function $x^i \colon U \to \Bbb R$. Its differential $dx^i$ is a $1$-form on $U$. With this definition, it is not a priori obvious that $(dx^1,\ldots,dx^n)$ is the dual coframe of $(\partial/\partial x^1,\ldots,\partial/\partial x^n)$. But here is the reason: the integral curve of $\partial/\partial x^j$ through $p\in U$ is the curve given in coordinates by $t\mapsto (x^1(p), \ldots, x^j(p)+t,\ldots,x^n(p))$. From this, you can easily derive that $dx^i(\partial/\partial x^j) = \delta^i_j$.
  2. The second way is the following: it is defined as the dual coframe of $(\partial/\partial x^1,\ldots,\partial/\partial x^n)$ pointwisely by the relation $dx^i_p(\partial/\partial x^j|_p) = \delta^i_j$ for $p\in U$. Here, the $d$ in $dx^i$ is just a notation, and it is not a priori obvious that $dx^i$ is the differential of the smooth function $x^i$. But the proof is not that complicated: it looks like that mentionned in the first point.
Didier
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  • What do you mean by an integral curve of $\partial/\partial x^j$ through $p \in U$? Wikipedia seems to relate this to differential equations. @didier – Walker Jan 10 '23 at 11:53
  • @Walker An integral curve of a vector field $X$ is a curve $\gamma$ such that $\gamma'=X\circ \gamma$. In other words, the tangent vector to the curve $\gamma$ at the point $\gamma(t)$ is precisely the vector $X(\gamma(t))$. – Didier Jan 10 '23 at 13:10
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    @Walker You can replace my argument as follows: The curve $\gamma(t) = (x^1(p),\ldots,x^j(p)+t,\ldots,x^n(p))$ (in coordinates) satisfies $\gamma(0) = p$ and $\gamma'(0) = \partial/\partial x^j|_p$. By the chain rule, you have $(x^i\circ \gamma)'(0) = dx^i_p(\partial/\partial x^j|_p)$. Since $(x^i\circ \gamma)(t) = x^i(p) + t\delta^i_j$, the result follows. – Didier Jan 10 '23 at 13:13
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    I prefer definition 1 because it is consistent with the standard notation for the differential of a function and derive the other property as a consequence. All very nice and clean. – Deane Jan 10 '23 at 16:22
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    @Deane I completely agree with you – Didier Jan 11 '23 at 13:52