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I am reading Milnor's ''Topology from the differentiable viewpoint'' and in the chapter about vector fields there is a Lemma that states that given $f:\mathbb{R}^n \to \mathbb{R}^n$ an orientation preserving diffeomorphism, then $f$ is isotopic to the identity.

He starts the proof (somewhat) as follows:

Without loss of generality, you can suppose $f(0)=0$, Then you can define the derivative at $0$ by $d_0f(x)=lim_{t\to 0} \frac{f(tx)}{t}$.

Then define $F:\mathbb{R}^n \times [0,1] \to \mathbb{R}^n$ as $F(x,t)=\frac{f(tx)}{t}$ for noncero $t$. And $F(x,0)=d_0f(x)$.

Now from here on out, my doubts start, so I am going to try to be loyal to the text on the proof:

To prove that $F$ is smooth, even as $t$ tends to $0$, you can write $f$ of the form: $f(x)= \sum_{i=1}^{n} x_ig_i(x)$

(So I am guessing this is looking at $f$ as in $f=(g_1,...,g_n)$. If this is not it please correct me)

Then it justs says: ''Not that $F(x,t)=\sum_{i=1}^{n} x_ig_i(tx)$ for all values of t.

Thus $f$ is isotopic to the linear mapping $d_0f$, which is clearly isotopic to the identity.''

So, why does this proves $F$ is smooth? And when did you use the hypothesis of $f$ preserving orientation?

Any help would be appreciated, thanks in advanced.

Bajo Fondo
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  • What is $g_i$ defined to be? I don't think "$\sum_i x_ig_i(x)$" means "$(x_1g_1(x), x_2g_2(x),\ldots, x_ng_n(x))$". – Neal Nov 19 '18 at 21:10
  • I was guessing $g_i: \mathbb{R}^n \to \mathbb{R}$ smooth functions. i.e the $g_i$'s as the coordinate functions $f(x)=(g_1,...,g_n)$ and $x_1=(1,0,..,0) , x_2=(0,1,0....0)$ etc. – Bajo Fondo Nov 19 '18 at 21:13

1 Answers1

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To get $f(x)=\sum x_i g_i(x)$ you can use $f(x)=\int_0^1\frac{d}{ds}f(sx)ds=\sum_i\int_0^1x_i\frac{\partial}{\partial x_i}f(sx)ds$, so you just set $g_i(x)=\int_0^1\frac{\partial}{\partial x_i}f(sx)ds$. The function $F(x,t)=\sum_{i=1}^{n} x_ig_i(tx)$ is evidently smooth as $g_i$'s are smooth. $F$ is an isotopy between $f$ (for $t=1$) and the linear map $d_0f$. So you still need to verify that $d_0f$ is isotopic to the identity, and that's because we suppose that $\det (d_0f)>0$.

user8268
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  • Ok.. Thanks. One thing though, how come $F$ is smooth when $t=0$? – Bajo Fondo Nov 20 '18 at 09:52
  • @BajoFondo it's a composition of smooth functions (it is a smooth function of $x$'s and of $t$) – user8268 Nov 20 '18 at 11:42
  • Maybe I am not getting something, here you have that in $\mathbb{R^n} \times (0,1]$, $F$ can be seen as that sum. But in $t=0$, you have that $F(x,0)=d_0f(x)$. How come $F$ is smooth here? Sorry if I am missing something, I cannot quite get this. – Bajo Fondo Nov 20 '18 at 12:54
  • @BajoFondo use $F(x,t)=\sum_{i=1}^{n} x_ig_i(tx)$ to see that $F$ is smooth – user8268 Nov 20 '18 at 17:40
  • The definition of $F(x,0)$ is equivalent to $\lim_{t\to 0}\frac{f(xt)}{t}$, substituting by the combination of the $g_i$'s we get $F(x,0)=$ $\sum_{n=0}^{i} \lim_{t\to 0}x_i g(tx)$$ = \sum_{n=0}^{i}x_i g(0)$. Thus $F(x,t)$ is truly equal to that sum everywhere. – lou Jun 28 '22 at 15:54