# Tangent space question.

1. Oct 23, 2008

Let me first confess this a copy/paste of a question I asked on another forum; I trust it's not against the rules.

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

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

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

I want to prove that $$\alpha^i = v x^i$$.

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

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

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

This looks promising, except we don't have an IP on $$T_pM$$, and moreover, the $$\{x^i\}$$ are coordinates, not a basis!

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

Any thoughts out there? Have I effed up somewhere?

2. Oct 23, 2008

### CompuChip

This is not really my cup of tea, but can't you simply define the inner product? After all the tangent space is just a linear vector space like Rn, just define a map $$(e_i, e_j) \mapsto \delta_{ij}$$ and extend it linearly. Would that solve your problem?

Also I don't really get the last remark. Isn't $\partial_i = \frac{\partial}{\partial x^i}$ by definition the basis dual to $x_i$, such that $\partial_i x^j \equiv \delta_{ij}$?

You might want to ask someone more knowledgable, but that's my thought on it.

3. Oct 23, 2008

### quasar987

Well,

$$vx^i = \sum \nolimits_i \alpha^j \frac{\partial}{\partial x^j}x^i = \alpha^j\delta_{ij}= \alpha^i$$

it right, and is the classical way to do it that you will find in every book existing on differentiable manifold theory, provided you define properly what you mean by that x^i being differentiated...

4. Oct 24, 2008