Let me first confess this a copy/paste of a question I asked on another forum; I trust it's not against the rules.(adsbygoogle = window.adsbygoogle || []).push({});

Let [tex]M[/tex] be a [tex]C^{\infty}[/tex] manifold, and, for some neighbourhood [tex]U\ni p \subsetneq M[/tex] let there be local coordinates [tex]x^i [/tex] such that [tex]p=(x^1,\,x^2,...,x^n)[/tex]

Suppose that [tex]T_pM[/tex] is a tangent vector space at [tex]p[/tex], and define a coordinate basis for [tex]T_pM[/tex] as [tex]\frac{\partial}{\partial x^i}[/tex].

By modeling on "ordinary" linear algebra, suppose that any [tex]v \in T_pM = \sum\nolimits_ i \alpha^i \frac{\partial}{\partial x^i}[/tex], where the [tex]\{\alpha^i\}[/tex] are scalar.

I want to prove that [tex]\alpha^i = v x^i[/tex].

My thoughts, based on inner product spaces......

Suppose [tex]V[/tex] is a vector space with inner products. Let the set [tex]\{e_j\} [/tex] denote the basis vectors. Then any [tex]v \in V[/tex] can be expressed as [tex]v = \sum \nolimits_j a^j e_j[/tex], where the [tex]\{a^j\}[/tex] are scalar.

Now the inner product of an arbitrary basis vector with an arbitrary vector will be denoted by [tex](v,e_i) = \sum \nolimits_j(a^j e_j, e_i)=\sum \nolimits_j a^j( e_j, e_i)[/tex] (since inner products are bilinear) hence [tex] (v, e_i) = \sum \nolimits_j a^j(e_j,e_i) = \sum \nolimits_j a^j \delta _{ij} = a^i[/tex]

This looks promising, except we don't have an IP on [tex]T_pM[/tex], and moreover, the [tex]\{x^i\}[/tex] arecoordinates, not a basis!

Where do I go from here? I tried the simple operation [tex]vx^i = \sum \nolimits_i \alpha^j \frac{\partial}{\partial x^j}x^i = \alpha^j\delta_{ij}= \alpha^i[/tex] but I am told this is no proof

Any thoughts out there? Have I effed up somewhere?

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# Tangent space question.

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