Tangent Space Well Defined?

Question

Question:

Let $M$ be a $k$-manifold of class $C^r$ in $\mathbb R^n$. Let $p\in M$. Show that the tangent space to $M$ at $p$ is well-defined, independent of choice patch.

Unsure if I'm understanding what this is asking of me. What does it mean to be well defined and how do I prove it?

From Munkres Calculus on Manifolds

Thanks in advance!

Answer 1

Take two local charts around the point $p \in M $, say $(U, \varphi)$ and $(V, \psi)$ and let $\alpha, \beta : ]-\varepsilon, \varepsilon[ \to M$ two differentiable curves on $M$, such that $\alpha (0) = p = \beta(0)$. Now using the chain rule we have

$$\beta_{\psi}'(0) = (\psi \circ \beta)'(0) = D(\psi \circ \varphi^{-1})_{\varphi(p)} \cdot \beta'_{\varphi}(0)\\ \\ \alpha_{\psi}'(0) = (\psi \circ \alpha)'(0) = D(\psi \circ \varphi^{-1})_{\varphi(p)} \cdot \alpha'_{\varphi}(0)$$

Now if $\beta$ and $\alpha$ are tangent at $p$ with respecct to $\varphi$ then same happens with respect to $\psi$ because

$$\begin{align}\beta'_{\psi}(0) &= D(\psi \circ \varphi^{-1})_{\varphi(p)} \cdot \beta'_{\varphi}(0)\\&= D(\psi \circ \varphi^{-1})_{\varphi(p)} \cdot \alpha'_{\varphi}(0)\\&=\alpha_{\psi}'(0)\end{align}$$

Answer 2

What it means is this: your definition of the tangent space at $P$ involves a coordinate patch $\phi$ (whose domain includes $P$) somewhere. A certain entity is in the tangent space if is true. But you might ask "suppose a friend picked a different coordinate patch $\psi$ whose domain also included $P$. Would that friend find that the same entity is in the tangent space, or would the friend's choice rule out that entity?" If it's the latter, then you don't really have a good definition, since something can be "in" or "out" depending on a choice of coordinate patch.

Here's an example of an ill-defined concept to illustrate the idea:

"We'll now define what it means for a function $f: \mathbb R \to \mathbb R$ to be globulous. For any smooth function $g$, we examine $h = g \circ f$, and if $h$ is constant, then $f$ is globulous."

So...is $f(x) = x^2$ globulous? If you happen to pick the constant function $g(x) = 1$, then it is. But if you pick $g(x) = x$, then it isn't. Hence the "globulous-ness" of $f$ is not well-defined.