Timeline for In the quaternions, "any imaginary unit may be called i"
Current License: CC BY-SA 2.5
10 events
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| Apr 14, 2021 at 9:50 | comment | added | Incnis Mrsi | Just a quibble: $u\bar v + v\bar u$ is an inner product on ℍ, namely the standard inner product by the factor of 2. | |
| Feb 26, 2011 at 5:51 | history | edited | Emerton | CC BY-SA 2.5 |
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| Feb 1, 2011 at 18:34 | comment | added | Qfwfq | Dear Emerton, you're totally right! I've been too sloppy in it. | |
| Feb 1, 2011 at 3:48 | history | edited | Emerton | CC BY-SA 2.5 |
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| Feb 1, 2011 at 3:09 | vote | accept | Daniel Briggs | ||
| Jan 31, 2011 at 17:52 | history | edited | Emerton | CC BY-SA 2.5 |
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| Jan 31, 2011 at 14:21 | comment | added | Emerton | Dear unkonwngoogle, I'm probably being dense, but I don't follow your argument. For example, conjugation is not an automorphism, but an anti-automorphism (i.e. it switches the order of multiplication). And even given this, is it a priori clear that this anti-automorphism commutes with all automorphisms? This is true in the end, but I don't see how to prove it without making some argument (although it's likely that what I wrote is not the most elegant argument --- it's just what came to mind first!). Best wishes, Matthew | |
| Jan 31, 2011 at 13:03 | comment | added | Qfwfq | Concerning your third paragraph: i think you can just say that 1) the automorphisms of $\mathbb{H}$ (as an $\mathbb{R}$-algebra) preserve $1$, hence they preserve $\mathbb{R}$, 2) the conjugation on $\mathbb{H}$ is an automorphism, 3) the scalar product on $\mathbb{H}$ is defined in terms of conjugation hence the orthogonality condition is invariant under automorphisms, 4) imaginary quaternions are just $\mathbb{R}^{\perp}$. Conclusion: $\mathrm{Aut}(\mathbb{H})$ preserves imaginary quaternions. | |
| Jan 31, 2011 at 3:46 | history | edited | Emerton | CC BY-SA 2.5 |
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| Jan 31, 2011 at 3:40 | history | answered | Emerton | CC BY-SA 2.5 |