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Notations are from John M.Lee's book Introduction to smooth manifolds. $\mathbb{R}_a^n=\{a\}\times \mathbb{R}^n=\{(a,v):v \in\mathbb{R}^n\}$ $D_v|a:C^\infty(\mathbb{R}^n\to\mathbb{R})$,$f\mapsto \frac{d}{dt}|_{t=0}f(a+tv)$.

Since the directional derivative is the same with respect to some direction $v$ and $2v$, i.e., the map from geometric tangent space to tangent space of $\mathbb{R}^n$ is not injective, it cannot be an isomorphism. However, a prove can be found on any book that it is isomorphism. Where I am wrong? thanks

Brooks
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You claim the directional derivative is the same with respect to $v$ and $2v$. But doesn't that contradict when you replace $v$ for $2v$ in : $f\mapsto \frac{d}{dt}|_{t=0}f(a+tv)$ ?
You would get : $\frac{d}{dt}|_{t=0}f(a+2tv)= 2\frac{d}{dt}|_{t=0}f(a+tv)$.

Rutger Moody
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  • so this directional derivative is not the same with the directional derivative in calculus? – Brooks Apr 19 '17 at 12:08
  • @Brooks No, it is defined in Lee's book as the above derivative. So the size of $v$ does matter in this case. – Rutger Moody Apr 19 '17 at 12:10
  • in calculus $\frac{d}{dt}|{t=0}f(a+2tv)= \frac{d}{dt}|{t=0}f(a+tv)$. – Brooks Apr 19 '17 at 12:10
  • @Brooks I can imagine a definition of directional derivative that calculates derivative along a unit vector in the direction of $v$. But what you wrote above : $\frac{d}{dt}|{t=0}f(a+2tv)= \frac{d}{dt}|{t=0}f(a+tv)$ is not correct. That is just not the derivative of some function of $t$. – Rutger Moody Apr 19 '17 at 12:14
  • @Brooks see for example here : http://mathworld.wolfram.com/DirectionalDerivative.html . I think this is what you know as the definition of directional derivative. It specifically states that the vector ($v$ in our case ) should be a unit vector. Lee's definition is simply different. Maybe that's why on page 43 he puts "directional derivative" between quotes. It's just not the same, length is absolutely a factor here and $v$ is not necessarily a unit vector. – Rutger Moody Apr 19 '17 at 12:36
  • thanks. there is little difference of these definition in manifolds and the one you said in terms of unit vector. It is really trick for me and you have been help me out of it. – Brooks Apr 19 '17 at 12:37