Timeline for Why does the field norm on the field extension $ \mathbb C/\mathbb R $ induce a vector space norm?
Current License: CC BY-SA 4.0
14 events
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| Mar 7, 2022 at 12:47 | comment | added | Denis Nardin | @UriBader Unless I'm mistaken, if you use the sup instead of the Haar measure, the only place where you use local compactness is to show that the sup is finite. so if one could show that $\chi$ is bounded on the unit ball, this would prove the result without local compactness. I am not sure how one would do that, but maybe it is doable in the special case of the norm. | |
| Aug 29, 2021 at 6:00 | comment | added | Uri Bader | In fact, we can define the $L^*/K^*$-invariant norm by taking supremum rather then integrating, which makes the proof elementary. | |
| Aug 27, 2021 at 13:02 | comment | added | Johannes Hahn | Wow, that's a nice proof! | |
| Aug 27, 2021 at 12:53 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
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| Aug 27, 2021 at 12:37 | history | edited | Uri Bader | CC BY-SA 4.0 |
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| Aug 27, 2021 at 11:54 | comment | added | Uri Bader | Thank you, Ege. In particular, thank you for making me thinking of this. I will correct this equality symbol. Also, I intend to update my other answer and to emphasize the aspect that, unlike this one, the method there holds for arbitrary complete fields. | |
| Aug 27, 2021 at 11:50 | vote | accept | Ege Erdil | ||
| Aug 27, 2021 at 11:49 | comment | added | Ege Erdil | Very nice! This is the kind of argument I was looking for when I asked the question. One small correction: I assume you meant that $ \chi $ is an absolute value if $ \chi(x+y) \leq \chi(x) + \chi(y) $, but you used the equality symbol instead. | |
| Aug 27, 2021 at 10:00 | history | edited | Uri Bader | CC BY-SA 4.0 |
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| Aug 26, 2021 at 21:27 | comment | added | Uri Bader | This is stupid, I know, but yet, I want to note that you don't really need complex numbers for the existence of Haar measures, so this proof that the real absolute value extends to the complex numbers is not circular. | |
| Aug 26, 2021 at 18:29 | comment | added | David E Speyer | Got it, thanks. | |
| Aug 26, 2021 at 18:15 | comment | added | Uri Bader | @DavidESpeyer Thanks you. Indeed, the space of norms is a cone, but in the parenthetical remark, it is $L$ that is considered as vector space over $K$, so norms are $K$-norms, not $L$-norms :). The integration is pointwise: the new norm at $v\in L$ is $\int \|xv\|dx$, so measurability is not a problem. | |
| Aug 26, 2021 at 17:52 | comment | added | David E Speyer | Interesting. One correction, and one question: (1) The space of norms is a convex cone, not a vector space and (2) Should I be worried about whether your action is a measurable function of $x$, so that I can make your averaging argument work? | |
| Aug 26, 2021 at 17:19 | history | answered | Uri Bader | CC BY-SA 4.0 |