Why is the differential of a map between manifolds a map between the tangent spaces?

Question

In the books that I have seen, given a smooth map $\phi: M \rightarrow N$ where $N$ and $M$ are manifolds, the differential at a point $x$ is defined as $d \phi_x: T_x M \rightarrow T_x N$. Why is it the case that the differential is defined as a map of the tangent spaces?

Is it possible to show that this is true taking, for example, the definition of the tangent space of $M$ at $x$ to be $(dF_x)^{-1}(0)$ if $M=F^{-1}(0)$?

For example, the problem I am working on treats $SO(n, \mathbb{R})$ as a manifold of $M(n, \mathbb{R})$. Given $w_1, \ldots w_n \in SO(n, \mathbb{R})$ I define the function

\begin{align} \varphi: SO(n, \mathbb{R}) &\rightarrow SO(n, \mathbb{R})\\ g & \mapsto gw_1g^{-1}w_1^{-1} \ldots g w_n g^{-1}w_n^{-1} \end{align}

I need to show that $d \varphi_I$ (where $I$ is the identity matrix) is a map from $SO(n, \mathbb{R})$ to itself where $SO(n, \mathbb{R})$ is the set of anti-symmetric matrices and is the tangent space of $SO(n, \mathbb{R})$ at $I$.

Thanks in advance.

Answer 1

Let $F(A) =\ ^tAA - I$. $F \colon \mathbb{R}^{n^2} \to \text{Sim}(n)$, indeed $\ ^t(\ ^tAA-I) = \ ^tAA - I$. (you have probably encountered $F$ to see that $SO(n)$ is a smooth manifold...)

Let's calculate $D_AF(H)$:

$$\lim_{\|H\| \to 0} \frac{\|F(A + H) - F(A) - D_AF(H)\|}{\|H\|} = \lim_{\|H\| \to 0} \frac{\|\ ^t(A+H)(A+H) - I -\ ^tAA + I -D_AF(H) \|}{\|H\|} = \lim_{\|H\| \to 0} \frac{\|\ ^tAA +\ ^tAH +\ ^tHA +\ ^tHH - I -\ ^tAA + I - D_AF(H)\|}{\|H\|} = \lim_{\|H\| \to 0} \frac{\|\ ^tAH +\ ^tHA +\ ^tHH - D_AF(H)\|}{\|H\|}$$

Let $D_AF(H) = \ ^tAH +\ ^tHA$, then $$\lim_{\|H\| \to 0} \frac{\|\ ^tHH\|}{\|H\|} \le \lim_{\|H\| \to 0} \frac{\|\ ^tHH\|}{\|H\|} \le \lim_{\|H\| \to 0} \frac{\|H\|^2}{\|H\|} \to 0.$$

Now let $A = I$ and the calculation over this constant path yields $0 = D_IF(H) = \ ^tH + H$.

I hope this helps!

EDIT: why this applies to your situation? You have $\varphi$ defined as in the question. $D_x\varphi \colon T_xSO(n) \to T_{\varphi(x)}SO(n)$. If $x = I$, then $\varphi(x) = \varphi(I) = I$ and $T_ISO(n) = T_{\varphi(I)}SO(n) = so(n)$, as proved above.

Answer 2

Let's think of the problem locally. Think of a linear continuous map $F:R^{n}\rightarrow R^{n}$. $R^{n}$ is considered equipped with two structures, a topological one and a linear one. In this case $F$ applies to both structures. In the general case, though, the "linear part" is represented by the differential as a first step of approximation to the "topological part". And the tangent space "approximates" the underlying topological one.

Answer 3

This is easy if you use the definition of the tangent space $T_xM$ as the set of equivalence classes of smooth curves in $M$ passing through $x$.

Suppose you have a smooth curve $\gamma : I \to M$, where $I$ is an open interval in $\mathbb{R}$ containing $0$, and suppose that $\gamma(0) = x \in M$. Then by definition the equivalence class $[\gamma]$ is an element of the tangent space $T_xM$.

Now suppose we have a smooth map $\phi : M \to N$. Then we can consider the composite map $\phi \circ \gamma : I \to N$. This composite map is a smooth curve in $N$, and $\phi \circ \gamma(0) = \phi(x)$, so the equivalence class $[\phi\circ\gamma]$ is an element of the tangent space $T_{\phi(x)}N$.

Now, I think that this is not really the best definition of the tangent space for theoretical purposes (in particular it is not clear how to add tangent vectors) but it is equivalent to all of the other definitions, and it captures the intuition very well. Chapter 3 of John Lee's book Introduction to Smooth Manifolds discusses the different definitions of the tangent space in a very clear manner, so I would recommend that you read that.