# Loop concatenation

**Stub grade: A***

## Contents

## Definition

Loop concatenation is a special case of path concatenation. We use [ilmath]I:=[0,1]\subset\mathbb{R}[/ilmath] to denote the closed unit interval in [ilmath]\mathbb{R} [/ilmath].

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]\ell_1:I\rightarrow X[/ilmath] and [ilmath]\ell_2:I\rightarrow X[/ilmath] be loops in [ilmath](X,\mathcal{ J })[/ilmath] based at [ilmath]b\in X[/ilmath]^{[Note 1]}; then we can *concatenate* the loops:

- [ilmath]\ell_1*\ell_2:I\rightarrow X[/ilmath] by [ilmath](\ell_1*\ell_2):t\mapsto\left\{\begin{array}{lr}\ell_1(2t) & \text{for }t\in[0,\frac{1}{2}] \\ \ell_2(2t-1)&\text{for }t\in[\frac{1}{2},1]\end{array}\right.[/ilmath]
^{[Note 2]}- we claim this is also a loop based at [ilmath]b[/ilmath] (see**Claim 1**)- In words: the loop [ilmath]\ell_1*\ell_2[/ilmath] first does [ilmath]\ell_1[/ilmath] but at double the speed, thus completing [ilmath]\ell_1[/ilmath] by [ilmath]t=\frac{1}{2}[/ilmath]. Then, as [ilmath]\ell_1[/ilmath] ends at [ilmath]b[/ilmath] we're in a position to start [ilmath]\ell_2[/ilmath]. We do this at double speed, thus completing [ilmath]\ell_2[/ilmath] by time [ilmath]t=\frac{1}{2}[/ilmath].

Loops also lend themselves to other concatenations, all permutations of concatenations of [ilmath]\ell_1[/ilmath], [ilmath]\ell_1^{-1} [/ilmath], [ilmath]\ell_2[/ilmath] and [ilmath]\ell_2^{-1} [/ilmath] exist.

## Caveats

Loop concatenation is not associative, that is:

- [ilmath](\ell_1*\ell_2)*\ell_3\ne\ell_1*(\ell_2*\ell_3)[/ilmath]

Notice the loop [ilmath](\ell_1*\ell_2)*\ell_3[/ilmath] does [ilmath]\ell_1[/ilmath] at 4x the normal speed, completing it by [ilmath]t=\frac{1}{4}[/ilmath], then embarks on [ilmath]\ell_2[/ilmath] at 4x the speed also, completing that by [ilmath]t=\frac{2}{4}=\frac{1}{2}[/ilmath], then embarks on [ilmath]\ell_3[/ilmath] at double speed, completing it by [ilmath]t=1[/ilmath].

Whereas, [ilmath]\ell_1*(\ell_2*\ell_3)[/ilmath] does [ilmath]\ell_1[/ilmath] at double speed, completing it by [ilmath]t=\frac{1}{2}[/ilmath], then embarks on [ilmath]\ell_2[/ilmath] at 4x speed, completing it by [ilmath]t=\frac{3}{4}[/ilmath], then embarks on [ilmath]\ell_3[/ilmath] at 4x speed, completing it by [ilmath]t=1[/ilmath].

Although the image of both loops is the same (that is: [ilmath]\big((\ell_1*\ell_2)*\ell_3\big)(I)=\big(\ell_1*(\ell_2*\ell_3)\big)(I)[/ilmath], they are clearly different. However [ilmath](\ell_1*\ell_2)*\ell_3[/ilmath] and [ilmath]\ell_1*(\ell_2*\ell_3)[/ilmath] are path homotopic, or homotopic [ilmath]\text{rel }\{0,1\} [/ilmath]

See: The fundamental group for more information.

## See also

## Notes

- ↑ That is: TODO: Put definition of loop based at [ilmath]b\in X[/ilmath] here
- ↑ We include [ilmath]t=\frac{1}{2}[/ilmath] in both parts as a nod to the pasting lemma.