does the linear isomorphism between the tangent space at a point and $ \mathbb{R}^n $ depend on charts?

Question

I am taking a class in analysis and differential topology. We were learning about the tangent bundle $$TM := \sqcup_{p \in M}T_pM $$ of a $C^1$ manifold M. In class our definition of the tangent space $T_pM$ at a point $p \in M$ was the definition involving a space of equivallence classes of curves. We were given the task of writing up how to topologize $TM$ and turn it into a $C^1$ manifold. I think I understand most of this but there is a part that I am confused about.

It is my understanding that the tangent space at a point doesn't depend on charts in the sense that I don't choose a chart and get a tangent space and then choose a different chart and get a different tangent space. This means that given a point $p \in M$ we should be able to let $\rho_p: T_pM \rightarrow \mathbb{R}^n$ be a linear isomorphism that is independent of any charts.

I topologized $T_pM$ by taking the domain of a chart $(\phi_U, U)$ in the atlas $ \mathcal{A} $ for $M$ and constructing a bijection $$ \Phi_U: \pi^{-1}(U) \rightarrow \phi_U(U) \times \mathbb{R}^n $$ I then pulled back the topology on $ \phi_U(U) \times \mathbb{R}^n $ and did this for all charts $(\phi_U,U) \in \mathcal{A} $ and showed that they were compatible. To give $TM$ a $C^1$ structure one can let the collection of all $(\Phi_U, \pi^{-1}(U)) $ be the charts for $TM$. I though that the following excerpt from m homework assignment would be enough to show that the transition maps for these charts were $C^1$:

"Lastly we must show that these charts define a $C^1$ structure. To do this we must show that the transition maps between any two charts $(\Phi_U, \pi^{-1}(U))$,$(\Phi_{U'},\pi^{-1}(U'))$ is $C^1$. Let $(x,v) $ be a vector in $\mathbb{R}^{2n}$ with $x,v \in \mathbb{R}^n$. Then $$ \Phi_U \circ \Phi_{U'}^{-1}|_{\Phi_{U'}(U)}(x,v) = \Phi_U(\phi_{U'}^{-1}(x),\rho_{\phi_{U'}^{-1}(x)}(v)) = ((\Phi_U \circ \Phi_{U'}^{-1})(x),v) $$ This allows us to see that this map must be $C^1$ because it is the identity in the last $n$ coordinates and a $C^1$ map in the second $n$ coordinates since $(\Phi_U \circ \Phi_{U'}^{-1})$ must be $C^1$ since they are both in the $C^1$ atlas $ \mathcal{A} $."

The grader for my class told me that I made a mistake and that actually this transition map isn't the identity in the last $n$ coordinates. He said that it should be "the differential of the transition map on M, since the linear isomorphism actually depends on your chart". I don't understand this because like I said earlier the tangent space doesn't depend on any specific chart so I don't see how the linear isomorphism between it and $\mathbb{R}^n$ could depend on a chart.

I am looking for some clarification on this topic and any errors in my thinking. Normally I would just ask the grader but this homework wasn't graded until the class ended and I'm sitting here during winter break being really bothered by the fact that I don't feel like I understand tangent spaces and tangent bundles very well. For anyone that answers thanks for your help!

Answer 1

As I mentioned in the comments, if $M$ is a $C^{k +1}$ manifold ($k \geq 0$ being an integer) then $TM$ is only guaranteed to be a $C^k$ manifold (of twice the dimension).

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My aim is to emphasize to you that it is the construction of tangent spaces and tangent bundle which is chart independent, but as the answer by @Sam Freedman points out: the isomorphism between $T_pM$ and $\Bbb{R}^n$ is NOT chart independent.

First of all, let's say we have a smooth $C^{k+1}$ manifold $M$; now fix a point $p \in M$. First thing we do is define $C_pM$ to be the set of all $C^{k+1}$ curves $\gamma: I \to M$ ($I$ being an open interval in $\Bbb{R}$ containing $0$) such that $\gamma(0) = p$. Then, we define an equivalence relation on $C_pM$ (for instance) as follows: fix a chart $(U, \phi)$ around the point $p$; then we call two curves $\gamma_1, \gamma_2$ equivalent with respect to $(U, \phi)$ if and only if \begin{align} (\phi \circ \gamma_1)'(0) = (\phi \circ \gamma_2)'(0) \end{align} and let's write this as $\gamma_1 \sim_{\phi} \gamma_2$. Then, using the chain rule you can show that for any pair of charts $(U, \phi), (V, \psi)$ (with $p \in U \cap V$), $\gamma_1 \sim_{\phi} \gamma_2$ if and only if $\gamma_1 \sim_{\psi} \gamma_2$.

1.) So, the equivalence relation is chart independent.

Now, $T_pM$ can be defined as $C_pM/ \sim$ (i.e the set of equivalence classes under the above equivalence relation). So far, $T_pM$ is only a set. We can construct a bijection between $T_pM$ and $\Bbb{R}^n$ as follows: using a chart $(U, \phi)$ about the point $p$, define $\rho_{\phi, p}: T_pM \to \Bbb{R}^n$ by \begin{align} \rho_{\phi, p}\left([\gamma]\right) &:= (\phi \circ \gamma)'(0) \end{align} (this map is well defined because of the equivalence relation). It is also injective due to the equivalence relation, and is easily shown to be surjective (the inverse is also easy to calculate directly). Hence, using the bijection $\rho_{\phi, p}$, we can "pull-back" the vector space structure of $\Bbb{R}^n$ to $T_pM$. However, this is all with respect to the chart $(U, \phi)$.

It is a straightforward exercise (again, there will be chain rule somewhere) to check that the vector space operations of addition and scalar multiplication doesn't actually depend on the choice of chart.

2. The vector space structure on $T_pM$ obtained by pulling back via $\rho_{\phi, p}$ is actually chart independent.

Hence, it is the construction of the vector space $T_pM$ which is chart-independent. Also, by construction, it follows that for each chart $(U, \phi)$ about $p$, the map $\rho_{\phi, p}: T_pM \to \Bbb{R}^n$ defined above is a linear isomorphism (but again, this isomorphism clearly depends on the chart).

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Now I'll briefly sketch the relevant results for proving $TM$ is a $C^k$ manifold (with some modified notation, which I think will be useful). Given a chart $(U, \phi)$ of $M$, we shall construct a chart $(TU, T \phi)$ on $TM$, where $TU := \pi^{-1}[U]$, and $T \phi: TU \to \phi[U] \times \Bbb{R}^n$ is defined by \begin{align} T \phi\left( [\gamma]\right) &:= \left( (\phi \circ \pi)([\gamma]), \rho_{\phi, \pi([\gamma])}([\gamma])\right)\\ &:= \left( (\phi \circ \gamma)(0), (\phi \circ \gamma)'(0)\right) \end{align} i.e it's the coordinate representation of the base point, and the velocity vector

You should be able to prove this is a bijection easily. If $(V, \psi)$ is an overlapping chart, then (on the respective domain), you should find that \begin{align} (T \psi) \circ (T \phi)^{-1}(x,v) &= \left( (\psi \circ \phi^{-1})(x), D(\psi \circ \phi^{-1})_{x}(v)\right) \end{align}

By the way, it is the presence of the derivative in the last $n$ coordinates which reduces the smoothness degree by $1$.

Answer 2

I think a source of confusion is coming from your second paragraph, where you say that there is some linear isomorphism $\rho_p : T_p M \to \mathbb{R}^n$ that is independent of choice of chart.

In analogy with linear algebra, while $n$-dimensional $\mathbb{R}$-vector spaces are all isomorphic to $\mathbb{R}^n$, this isomorphism is non-canonical in the sense that one must choose a basis. In our situation, we are choosing a chart $(\phi_U , U)$ around $p$, which lets us express an (equivalence class of a) curve $\gamma_p : I \to M$ in some real coordinates given by $\phi_U(U) \cong \mathbb{R}^n$, so that its "velocity vector" or derivative $\gamma'_p$ is really some vector in $\mathbb{R}^n$. But changing the coordinate chart $U$ changes the local equations for $\gamma_p$, consequently changing the specific vector in $\mathbb{R}^n$ corresponding to its velocity vector.

However, all is not lost: if $p$ is contained in two charts $U$ and $U'$, then the transition mapping $\phi_{U'} \circ \phi_U^{-1}$ induces an isomorphism between the two $\mathbb{R}^n$s of velocity vectors. Intuitively, the transition map changes the local expression of $\gamma_p$ from $U$-coordinates to $U'$-coordinates, letting us write down a matrix expressing how the representation of the velocity vector changes.

Does this clarify what your grader said?

Answer 3

I think it's just a matter of following more closely the action of the maps. $\Phi_U$ sends a pair $(p,v)$ to the tuple $(\phi(p), a_i)$ where the $a_i$ are the coefficients of the expression for $v\ \textit{in the coordinates}\ x^i$. That is, $v=\sum^n_{i=1}a_i\frac{\partial }{\partial x^i}.$ Now take another chart $(U',\phi')$ and you get another expression for $v$ on $U\cap U'$. That is, $v= \sum^n_{i=1}b_i\frac{\partial }{\partial y^i}$.

So, $\Phi_U \circ \Phi_{U'}^{-1}|_{\Phi_{U'}(U)}(\phi'(p),v)$ is a composition that first sends the tuple in $\mathbb R^{2n},\ (\phi'(p),b_i)$ back to $(p,v)\in TM$ and then $\Phi_U$ sends $(p,v)$ to the tuple $(\phi(p),a_i)\in \mathbb R^{2n}.$ Overall then, the composition just maps $(\phi'(p),b_i)$ to $(\phi(p),a_i).$

The transition map is $f_{UU'}:\mathbb R^{n}\to \mathbb R^{n}:(b_i)\mapsto (a_i)$. And since $v=\sum^n_{i=1}a_i\frac{\partial }{\partial x^i}=\sum^n_{i=1}b_i\frac{\partial }{\partial y^i}$, we have $b_j=\sum^n_{i=1}a_i\frac{\partial y^j }{\partial x^i}$ so $f_{UU'}$ is a linear transformation whose matrix representation is just the Jacobian of $\phi\circ\phi'^{-1}$ on the Euclidean space $\mathbb R^n.$