Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Why is it defined like that? Because it makes sense ... –  Hagen von Eitzen May 8 '13 at 11:21

3 Answers 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.

share|improve this answer

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.

share|improve this answer
    
Hi, thanks for your reply. I may be wrong but your working shows that the tangent space of $SO(n,R)$ at $I$ is the group on anti-symmetric matrices. The bit I am stuck on is why the $\varphi$ function in the question is a map from the tangent space onto itself. I think this comes from a general result on smooth maps and manifolds but an example-specific proof would work as well. Thanks again –  Madeline W May 8 '13 at 11:03
    
you have $\phi$. you know that the diffrential at a point of a function goes from the tanget in the point to the tangent to the image of the point ($T_{\phi(x)N}$, not $T_xN$). Now the application you defined sends $I \to I$ therefore you just need to prove that the tangent to I is the group of antisymm matrices to to prove that $D_I\varphi$ goes from $so(n)$ to $so(n)$. Is it clearer? –  user01123581321345589144... May 8 '13 at 11:31
    
"you know that the diffrential at a point of a function goes from the tanget in the point to the tangent to the image of the point". This is the point I do not understand. I can see why it is the case but I cannot formulate a proof from the definitions I have seen of the tangent space. Books on the topic treat the fact as part of the definition of a derivative on a tangent space. Here, however, my derivative is defined conventionally and I want to show that it maps the tangent spaces. –  Madeline W May 8 '13 at 14:29
    
try with Lee's Introduction to smooth manifolds, pag 132. I hope this clarifies all your doubts... –  user01123581321345589144... May 8 '13 at 15:03

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.