Concatenation of two oriented curves

Question

I've encountered a question with a notation I've never seen before, and I was wondering if I could get some clarification as to what it could mean. Here's an excerpt from the question:

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Given two oriented curves $\mathbf{C}_1 : R_1(t) = (\sin t, \cos t, 0), 0 \leq t \leq \pi$ and $\mathbf{C}_2 : R_2(t) = (0, -\cos t, \sin t), 0 \leq t \leq \pi$, let $\mathbf{C}$ be the concatenation $\mathbf{C}_1 \cdot \mathbf{C}_2$.

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Is the dot notation here ($\cdot$) effectively being used as a union ($\cup$) of the two curves? I noticed that the end points of the two curves match to form a loop, so I'm wondering if that's relevant here too.

Thank you!

Answer 1

What you noticed is exactly correct.

Here's the general definition of concatenation. Suppose that you have two parameterized curves of the form $$c_1 : [0,a_1] \to \mathbb R^n, \qquad c_2 : [0,a_2] \to \mathbb R^n $$ Suppose also that $c_1(a_1)=c_2(0)$, so the terminal endpoint of curve $c_1$ is the same as the initial endpoint of curve $c_2$. The concatenation of $c_1$ with $c_2$, which I will denote $c_1 * c_2$ instead of re-using that over-used "dot" notation, is the curve $$c_1 * c_2 : [0,a_1 + a_2] \to \mathbb R^n $$ defined by $$c_1 * c_2(t) = \begin{cases} c_1(t) &\quad\text{if $0 \le t \le t_1$} \\ c_2(t-a_1) &\quad\text{if $t_1 \le t \le t_1 + t_2$} \end{cases} $$ One thing that's nice about this notation is that you can then use it in path integral settings, and you get equations like this: $$\int_{c_1 * c_2} \text{[something]} \, dt = \int_{c_1} \text{[something]}\, dt + \int_{c_2} \text{[something]}\, dt $$ I'll mention also that this concatenation notation becomes very important in more advanced courses, particularly topology.

Answer 2

Yes, it denotes concatenation, but keep in mind that it is inaccurate to think of it as their union. It is true that the image of the concatenation of $C_1$ and $C_2$ is the union of the images of $C_1$ and $C_2$, and for this reason sometimes concatenation is inaccurately written using the union symbol. To avoid confusion it is better to write this $C_1+C_2$ or $C_1 * C_2$ or $C_1 \cdot C_2$ as in your question.

The formal definition of the concatenation of $\alpha(t)$ and $\beta(t)$, both defined on $[0,1]$ and with $\alpha(1)=\beta(0)$, is to travel first on $\alpha(t)$ (at twice the speed) and then on $\beta(t)$ (again at twice the speed), that is, $$\alpha(t)+\beta(t) = \begin{cases}\alpha(2t)&t\in [0,\frac{1}{2}]\\\beta(2t-1)&t\in[\frac{1}{2},1]\end{cases}$$ Note the distinction between the curve itself and its image; the image loses all information of direction, and for instance, the image of a curve traveling around the unit circle once is the same image as that of a curve traveling around it twice.