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I'm reading the h-cobordism by Milnor and he claims that we can alter a map to make it orientation preserving(second paragraph in page 58). I'll give a detailed description in the following:

Suppose $f:\mathbb{R}^n\to \mathbb{R}^n$ an orientation reversing diffeomorphism. And $\xi, \hat{\xi}$ two vector fields in $\mathbb{R}^n$($n\geq2$).

My question is: if $f_*\hat{\xi}=\xi$, is there exist an orientation preserving diffeomorphism $g$ from $\mathbb{R}^n$ to itself such that $g_*\hat{\xi}=\xi$?

I think if I can take a map that change the direction of a vector field orthogonal to the one above, this problem will be done. Nevertheless I'm not sure there a correspondent modification of map to the modification in tangent space.

Thanks in advance.

edit: In the situation about the Milnor's book, it's enough to consider only the gradient like vector field in a morse chart, more explicitly $(-x_1,…,-x_\lambda, x_{\lambda+1},…,x_n)$so reflection about first coordinate will be an answer. Maybe the answer to this question is negative in general case.

taiat
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  • You should assume $n\ge 2$ and also write about the origin of the problem and your thoughts about it. – Moishe Kohan Feb 09 '22 at 09:24
  • @MoisheKohan Thanks! I've edited it. – taiat Feb 09 '22 at 09:53
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    Is Milnor talking about arbitrary vector field or ones which satisfy some further properties (e.g. have no zeroes). Note also that it suffices to ask about a single vector field $\xi$: "Is there a diffeomorphism which preserves $\xi$ ad reverses orientation?" – Moishe Kohan Feb 09 '22 at 10:48
  • @MoisheKohan these two vector fields exactly transverse to zero section of tangent bundle, however in this case they all have only one zero point. – taiat Feb 09 '22 at 11:08

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If there are no restrictions on the vectorfields then there is a counter example in dimension 2. Let $f(x,y)=(x,-y)$, $\xi_1(x,y)=(-y,x)$ ( or if you prefer $\xi_1(x,y)=-y\partial_{x}+x\partial_{y}$ ) and $\xi_2(x,y)=(y,-x)$. Then $f_*(\xi_1)=\xi_2$, and $f$ is orientation reversing. Now suppose that we have an orientation preserving diffeomorphism $g$ such that $g_*(\xi_1)=\xi_2$. Notice, that $g$ has to map the origin to the origin. Since $g_*$ maps $\xi_1$ to $\xi_2$ we can check that if $\gamma(t)$ is an integral curve of $\xi_1$, then $g(\gamma(t))$ is an integral curve of $\xi_2$. Now the integral curves of $\xi_1$ are just circles around the origin going counterclockwise and the integral curves of $\xi_2$ are circles around the origin going clockwise. Now let $\gamma(t)$ be an integral curve of $\xi_1$, then $g(\gamma)$ is an integral curve of $\xi_2$, thus it has winding number $-1$ around the origin, but $\gamma$ has winding number $1$ around the origin, and $g$ is orientation preserving so the winding number of $g(\gamma)$ has to be non-negative, and we have a contradiction.

Laci
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