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Put \(J^{2}=\tau_{1}+\tau_{2}\), where
\tau_{1}=\left[\mathbf{0}, \mathbf{e}_{1}, \mathbf{e}_{1}+\mathbf{e}_{2}\right], \quad \tau_{2}=-\left[\mathbf{0}, \mathbf{e}_{2}, \mathbf{e}_{2}+\mathbf{e}_{1}\right] .
Explain why it is reasonable to call \(J^{2}\) the positively oriented unit square in \(R^{2}\). Show that \(\partial J^{2}\) is the sum of 4 oriented affine simplexes. Find these. What is \(\partial\left(\tau_{1}-\tau_{2}\right) ?\)
in Mathematics by Diamond (55.6k points) | 126 views

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Although \(J^{2}\) is really a collection of two affine mappings, the ranges of these mappings cover the unit square, the diagonal from \((0,0)\) to \((1,1)\) being covered twice with opposite orientations in the two mappings. In both cases, the sense of orientation is such that the cross product of the last two vertices of the simplex is \(e_{3}\), which is a reasonable definition of the positive orientation on the unit square.
By routine computation,
\partial J^{2}=&\left(\left[\mathbf{e}_{1}, \mathbf{e}_{1}+\mathbf{e}_{2}\right]\right.\\
&\left.\quad-\left[\mathbf{0}, \mathbf{e}_{1}+\mathbf{e}_{2}\right]+\left[\mathbf{0}, \mathbf{e}_{1}\right]\right)-\left(\left[\mathbf{e}_{2}, \mathbf{e}_{1}+\mathbf{e}_{2}\right]-\left[\mathbf{0}, \mathbf{e}_{1}+\mathbf{e}_{2}\right]+\left[\mathbf{0}, \mathbf{e}_{2}\right]\right) \\
=& {\left[\mathbf{e}_{1}, \mathbf{e}_{1}+\mathbf{e}_{2}\right]+\left[\mathbf{e}_{1}+\mathbf{e}_{2}, \mathbf{e}_{2}\right]+\left[\mathbf{e}_{2}, \mathbf{0}\right]+\left[\mathbf{0}, \mathbf{e}_{1}\right] }
Again, by routine computation,
\partial\left(\tau_{1}-\tau_{2}\right)=&\left(\left[\mathbf{e}_{1}, \mathbf{e}_{1}+\mathbf{e}_{2}\right]-\left[\mathbf{0}, \mathbf{e}_{1}+\mathbf{e}_{2}\right]+\left[\mathbf{0}, \mathbf{e}_{1}\right]\right) \\
&+\left(\left[\mathbf{e}_{2}, \mathbf{e}_{1}+\mathbf{e}_{2}\right]-\left[\mathbf{0}, \mathbf{e}_{1}+\mathbf{e}_{2}\right]+\left[\mathbf{0}, \mathbf{e}_{2}\right]\right) \\
=& {\left[\mathbf{e}_{1}, \mathbf{e}_{1}+\mathbf{e}_{2}\right]-\left[\mathbf{e}_{1}+\mathbf{e}_{2}, \mathbf{e}_{2}\right]-\left[\mathbf{e}_{2}, \mathbf{0}\right]+\left[\mathbf{0}, \mathbf{e}_{1}\right]-2\left[\mathbf{0}, \mathbf{e}_{1}+\mathbf{e}_{2}\right] }
by Diamond (55.6k points)

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