Timeline for Which real Pin groups agree?
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| Jun 13, 2020 at 23:18 | comment | added | LSpice | If your argument were correct, it would show en passant that $\tau\bigl(\phi\bigl(\prod_{i = 1}^n v_i\bigr)\bigr) = \tau\bigl(\prod_{i = 1}^n \phi(v_i)\bigr) = \phi\bigl(\prod_{i = 1}^n v_{n - i}\bigr) = \phi\bigl(\tau\bigl(\prod_{i = 1}^n v_i\bigr)\bigr)$, which is why @TheoJohnson-Freyd mentioned $\phi$ not commuting with $\tau$. The problem is that your equality $\tau\bigl(\prod_{i = 1}^n \phi(v_i)\bigr) = \prod_{i = 1}^n \phi(v_i)$ assumes that $\phi(v_i)$ are still anisotropic elements of $V$. | |
| Oct 24, 2014 at 7:31 | history | edited | Name | CC BY-SA 3.0 |
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| Oct 24, 2014 at 7:17 | history | edited | Name | CC BY-SA 3.0 |
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| Oct 24, 2014 at 0:53 | comment | added | Theo Johnson-Freyd | Why does $\phi$ commute with $\tau$? Indeed, let $\phi: \mathrm{Cliff}(4,0) \to \mathrm{Cliff}(0,4)$ be the isomorphism. It identifies a generator $x$ with a cubic $yzw$ (for some arbitrary ortho(normal up to sign) bases). Then $\phi(\tau(x)) = \phi(x) = yzw$, whereas $\tau(\phi(x)) = \tau(yzw) = wzy = -yzw$. | |
| Oct 23, 2014 at 18:41 | history | answered | Name | CC BY-SA 3.0 |