1
$\begingroup$

Assume $U\subset\mathbb{R}^n$ open and $$ \gamma\colon (-\varepsilon,\varepsilon)\to U,~~\gamma(0)=p\in U. $$ continously differentiable.

Then, it is said that $\gamma'(0)$ is tangent vector of $p$.

One naive question:

Isn't $\gamma'(0)$ a linear map and not a vector, i.e. $$ \gamma'(0)=d\gamma(0)\colon (-\varepsilon,\varepsilon)\to U $$ and by evaluating this map at some $t\in (-\varepsilon,\varepsilon)$ we then get a vector in $U$?

(I am thinking of multidimensional Analysis, where the total derivative or differential is a linear map.)

1 Answers 1

1

You are right in saying that $\gamma'(0) \colon \mathbf R \to \mathbf R^n$ (note that a linear map is defined on the whole of $\mathbf R$ and can have values outside $U$) is a linear map. But a linear map on $\mathbf R$ is uniquely determined by its value at $1$, an hence usually identified with this value. Hence, when we talk of the vector $\gamma'(0)$, we mean $\gamma'(0)1$.

  • 0
    Thanks! I guess with $1$ you mean the canoncical base vector of the reals and that a linear map is determined by the Images of ist base vectors, here $\gamma'(0)1$. – 2017-01-06