# Tangential Space of a differentiable manifold is always $\mathbb R^n$?

Let $\mathcal M$ be a differential manifold with a point $p$. Let U be an open set, $p\in U$, on $\mathcal M$ and let $\phi,\psi:U\to \mathbb R^n$ be a charts on $\mathcal M$. I'm having diffculties arranging all the concepts of a differential manifold. The "manifold" of required spaces gives me a headache. At the moment im utterly confused with the construction of the tangent space at $p$. I'll start describing my thoughts from the start: $\require{AMScd}$

\begin{CD} U @>\phi>> \mathbb R_\phi^n \\ @V \psi VV @.\\ \mathbb R_\psi^n \end{CD}

That's the open set U with charts $\psi,\phi$. I denoted $\mathbb R^n$ with the chart pointing to it, so we can make sure to differentiate between the different $\mathbb R^n$ -spaces we will encounter.

The tangent space of $\mathcal M$ or $U$ in $p$ is a $n$-dimensional $\mathbb R$-vector space, which has isomorpisms pointing to $T_{\phi(p)} \mathbb R^n$ and $T_{\psi(p)} \mathbb R^n$. Where $T_{\phi(p)} \mathbb R^n$ and $T_{\psi(p)} \mathbb R^n$ are vector spaces we're inducing by the choice of our bases $B_\psi , B_\phi$ depending on our charts as follows after the diagram:

\begin{CD} T_p \mathcal M @> \cong >> T_{\phi(p)} \mathbb R^n \\ @V \cong VV @.\\ T_ {\psi(p)} \mathbb R^n \end{CD}

$T_{\phi(p)} \mathbb R^n$ and $T_{\phi(p)} \mathbb R^n$ are induced by the vector space bases $B_\psi , B_\phi$ using isomorphisms from the images of our charts ($\phi(U)\subset \mathbb R_{\phi}^n, \psi(U)\subset\mathbb R_{\psi}^n)$ to the tangent space $T_{\phi(p)} \mathbb R^n$. The first base vector of the base $B_\phi$ is given as $\frac{\partial}{\partial x_i}|_p := \phi^-1 (\phi(p)+ t e_i)$, the other base vectors of $B_\phi$ follow with $i \in {2,....,n}$.

Let's call the function that takes $p$ and gives us $\frac{\partial}{\partial x_i}|_p$ and $\frac{\partial}{\partial y_i}|_p$ by names $\bar{\phi}_p$ and $\bar{\psi}_p$. In a diagram we now got two choices for the base of our ismorphic tangent spaces depending on $\phi,\psi$:

\begin{CD} T_{\phi(p)} \mathbb R^n @. T_{\psi(p)} \mathbb R^n \\ @A \bar{\phi}_p AA @A \bar{\psi}_p AA\\ \mathbb R_\phi^n @. \mathbb R_\psi^n \end{CD}

First question: Why is $T_p \mathcal M$ an $\mathbb R$-vector space, i don't know why the underlying field has to be $\mathbb R$. Or is it, that we just identify $T_p M$ with $T_{\phi(p)}\mathbb R^n$, if so, why can we do that? I have the notion that $\mathcal M$ is of a abstract nature, so how comes it's tangential space is so $\mathbb R^n$ related?

Second question: Assuming i understood why $T_p \mathcal M$ is a $\mathbb R$-vector space, i can now ask myself how i can change bases. So we get to linear algebra and the pushforward as linear function between vector spaces. The image of the coordinate vector $(1,0,...,0)^t$,of the first base vector $\frac{\partial}{\partial x_i}|_p$, is the first row of the matrix $D((\psi)\circ\psi^-1)$ giving us the coordinate vector expressed in the base $B_\psi$.

\begin{CD} T_{\phi(p)} \mathbb R^n @>D((\psi)\circ\phi^-1)>> T_{\psi(p)} \mathbb R^n \end{CD}

Third question: The "last" notion that gives me a headache right now is that of the function:

$$\phi_*: T_p \mathcal M \to T_{\phi(p)} \mathbb R^n\text{ defined as } \phi_*(\gamma) := \frac{d}{dt}(\phi \circ \gamma )|_{t=0}$$

where $\gamma$ is a tangent vector, for example a curve(geometric defintion of a tangent vector).

Is that definition base independend? If so taking the first base vector of $B_\phi$ namely $\frac{\partial}{\partial x_i}|_p := \phi^-1 (\phi(p)+ t e_i)$ and using $\phi_*$ on it, gives us another tangent vector in the tangent space $T_{\phi_*} \mathbb R^n$, or is it that same as $\mathbb R_\phi^n$ or $T_{\phi(p)}R^n$?

• i was going to end the line, like usually return/enter would do! :-) – Marcel May 29 '13 at 11:11
• Sorry, I don't understand. I was merely trying to fix what was broken in a previous edit. – Martin May 29 '13 at 11:16
• oh ok, no problem ill figure this out myself. thanks martin! – Marcel May 29 '13 at 11:17
• @Martin In case your still reading this, how would i get a newline in the comment box. Or is this feature not included. I couldn't figure out the syntax to do it or the fact that it doesn't work. Thank you! – Marcel May 29 '13 at 11:30
• Newlines are disabled. You can emulate a linebreak using an empty displayed math environment  like so: , but the result is slightly awkward. – Martin May 29 '13 at 11:32

First question:

Congratulations, you have discovered algebraic varieties and their Zariski tangent space. For classical differential geometry, the reason that the tangent space is an $\mathbb{R}$ vector space is by definition: the local model of a differentiable manifold is $\mathbb{R}^n$. The whole motivation of classical differential geometry is to study stuff that, at sufficiently small neighborhoods, look like $\mathbb{R}^n$. It would be strange if its tangent space does not look like that of $\mathbb{R}^n$, but be over some other field, no? If you model the local geometry by an affine variety associated to a different base field, you will get appropriately different vector spaces as the tangent space.

Second question:

Looks like you more or less got it, except you probably meant $\psi\circ \phi^{-1}$ rather than $\psi \circ \psi^{-1}$. Again, locally you are just working with functions defined over $\mathbb{R}^n$ (which may take $\mathbb{R}^k$ values).

Third question:

$\phi_*$ pushes forward to $T_{\phi(p)}\mathbb{R}^n$, not $T_p\mathbb{R}^n$. Since you didn't explain what $\gamma$ is, I have no idea what your second equation means. But the derivative with respect to $t$ makes me wonder if you intended $\gamma:\mathbb{R}\to M$ and you are actually thinking about the expression

$$\phi_*(\frac{\mathrm{d}}{\mathrm{d}t}{\gamma}(t)) = \frac{\mathrm{d}}{\mathrm{d}t}( \phi\circ \gamma)(t)~?$$

In that case, this definition is base independent. You can explicitly verify it by composing with the smooth maps associated to changes of charts.

Your expression $\partial/\partial x_i |_p = \phi^{-1}(\phi(p) + t e_i)$ is wrong. The left hand side is an object living in $T_p M$ and the right hand side is an object living in $M$. You appear to have a type error.

• I had no time to think about your answers yet, but i'll begin by editing my type errors. So preventing other readers from missunderstanding me. I'll think about the arguments after that. Thanks already! – Marcel May 29 '13 at 10:55