# Tangent space question.

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?

Related Differential Geometry News on Phys.org

#### CompuChip

Homework Helper
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.

#### quasar987

Homework Helper
Gold Member
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...

I thank you both for your responses.
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
I was told that this is sufficient to prove uniqueness, but not existence. Maybe I was told wrong; again I thank you both

#### atyy

I have no idea about rigour, but I think is it has nothing to do with inner products. It starts by defining a manifold as something on which you can put coordinates. Coordinates are functions from the manifold to the real numbers. Since we already put coordinate functions, it's no problem if we put other functions F. Then we introduce curves which are maps from the real numbers into the manifold (say like time t along a trajectory). Then whatever functions on the manifold will change along the curve parameter F(t), for which we can just do normal differentiation. The partial derivative of x wrt x is the special case of differentiating a coordinate function along a coordinate curve.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving