Annulus and its boundary
by Hidenori
Proposition
For $R > 0$ and $n$ an integer, define the singular 1-cube $c_{R, n}: [0, 1] \rightarrow \mathbb{R}^2 \setminus 0$ by $c_{R, n}(t) = (R\cos 2\pi nt, R\sin 2\pi nt)$. Show that there is a singular 2-cube $c:[0, 1]^2 \rightarrow \mathbb{R}^2 \setminus 0$ such that $c_{R_1, n} - c_{R_2, n} = \partial c$.
Solution
Define $c$ such that $c(r, t) = ((R_2 + hr)\cos(2\pi nt), (R_2 + hr)\sin(2\pi nt))$ where $h = R_1 - R_2$.
\[\begin{align*} \partial I^2 &= \sum_{i=1}^{2} \sum_{\alpha = 0, 1} (-1)^{i + \alpha} I^2_{(i, \alpha)} \\ &= -(I^2_{(1, 0)} - I^2_{(1, 1)}) + (I^2_{(2, 0)} - I^2_{(2, 1)}) \\ &= -(0, x^1) + (1, x^1) + (x^1, 0) - (x^1, 1). \end{align*}\] \[\begin{align*} \partial c(t) &= c(\partial I^2(t)) \\ &= -(R_2\cos(2\pi nt), R_2\sin(2\pi nt)) + (R_1\cos(2\pi nt), R_1\sin(2\pi nt)) + (R_2 + ht, 0) - (R_2 + ht, 0) \\ &= c_{R_1, n} - c_{R_2, n}. \end{align*}\]Subscribe via RSS