Let $v$ be a constant vector field $v=\sum_{i=1}^n c_i\dfrac{\partial}{\partial x_i}$, and let $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be a bijective linear map. What is the vector field $f_*v$?

So, the vector field $v$ maps each point $(x_1,\ldots,x_n)$ to the vector $(x_1,\ldots,x_n,c_1,\ldots,c_n)\in T_p\mathbb{R}^n$. Hence, the push-forward vector field $f_*v$ must map $f(x_1,\ldots,x_n)$ to $$(f(x_1,\ldots,x_n),Df(x_1,\ldots,x_n)\cdot (c_1,\ldots,c_n))\in T_p\mathbb{R}^n$$

What else can we say about it?


The total differential $Df$ of a linear map $f$ is just the map itself. This follows directly from the definition of the derivative as a linear map given e.g. in this article.

  • $\begingroup$ I don't think so. Say $f:\mathbb{R}\rightarrow\mathbb{R}$ is given by $f(x)=x$. Then $Df=1$. $\endgroup$ – Mika H. Dec 8 '13 at 7:09
  • $\begingroup$ This is a bit confusing. In ordinary calculus, we have $f'(x)$ as usual: for each $x$, $f'(x)$ is a number which gives the slope of the function $f$ at the point $x$. Another way to think of this number $f'(x)$ is as a linear map from the tangent space at $x$ to the tangent space at $f(x)$ using the standard basis $\frac{\partial}{\partial x}$ since linear maps in one dimension are just multiplication by a single number. Then if we identify the tangent space to $x$ with $\mathbb{R}$ via $a\frac{\partial}{\partial x} \mapsto a$, our linear map is $a \mapsto f'(x)a$. $\endgroup$ – Brian Klatt Dec 8 '13 at 18:16
  • $\begingroup$ For $f(x)=x$ as above, the derivative is $1$ everywhere, but this translates to $a \mapsto f'(x)a = a$ on the tangent space to $x$, and $a \mapsto a$ is "the same as" the map $f$. $\endgroup$ – Brian Klatt Dec 8 '13 at 18:20
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    $\begingroup$ Perhaps I should summarize my answer here, in case it is unclear: The total derivative $Df$ is an assignment of a linear map to each $x \in \mathbb{R}^n$, which takes the tangent space to $x$ to the tangent space to $f(x)$. In the case where $f$ is a linear map, $Df(x)$ is a linear map which is the same as the map $f$ by canonically identifying the tangent space at $x$ with $\mathbb{R}^n$. $\endgroup$ – Brian Klatt Dec 8 '13 at 18:28

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