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I am reading Manfredo Carmo's book on differential geometry and I am slightly confused by something. It says that the definition of a parametrized differentiable curve is a map $\alpha: I \rightarrow \mathbb{R}^3$ for an interval $I$ of $\mathbb{R}.$ However, it lists a map $\alpha: \mathbb{R} \rightarrow \mathbb{R}^2$ as an example. My guess is that this is equivalent to a map $\alpha: \mathbb{R} \rightarrow \mathbb{R}^3$ except we send the $z$ coordinate to 0 at all points. Is this the correct way of looking at this?

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    $\begingroup$ A "differentiable curve" is really a map $I \to \Bbb{R}^n$ for any $n$. $\endgroup$ – Nick Jan 12 '18 at 4:29
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If you chose to embed $\iota:\mathbf{R}^2\to\mathbf{R}^3$ as $(x,y)\mapsto (x,y,0)$ then this indeed works out naturally. A parameterized curve $\alpha:I\to \mathbf{R}^2$ given by $\alpha(t)=(x(t),y(t))$ can be identified with the curve $\tilde{\alpha}=\iota\circ \alpha.$ More concretely, we have that $\tilde{\alpha}(t)=(x(t),y(t),0).$

On the other hand, we really needn't worry about the dimensionality. The definition holds for any choice of $n$. We can define a differentiable curve $\gamma:I\to \mathbf{R}^n$ for any $n\in \mathbf{N}_{\ge 1}$ by requiring that the component functions $\gamma_i(t)$ for $1\le i\le n$ be smooth as functions $\gamma_i:I\to\mathbf{R}$. Note that here I'm thinking of $\gamma$ as $$ \gamma(t)=(\gamma_1(t),\ldots, \gamma_n(t)).$$

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  • $\begingroup$ Great answer. Thank you for your help! $\endgroup$ – green frog Jan 12 '18 at 4:33

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