Timeline for What fields can be used for an inner product space?
Current License: CC BY-SA 4.0
24 events
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| Feb 4, 2023 at 22:54 | comment | added | Gerry Myerson | It would have been better, had OP linked to the previous discussions, instead of just mentioning their existence. | |
| Feb 4, 2023 at 14:48 | history | edited | LSpice | CC BY-SA 4.0 |
While this is on the front page, deleted salutation and signature; link to @MarkGrant's comment
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| Feb 4, 2023 at 12:51 | comment | added | Maarten Havinga | To extend the answer of wikipedia: if $\mathbb{F}$ admits a conjugate $*$ (which may be just doing nothing), you can replace $\langle x,x \rangle \gt 0$ with $\langle x,x \rangle =a^*a$ for some $a \in \mathbb{F}$. So if $*$ does nothing, it must be a square (example: finite fields). This is the $C^*$-algebra definition of positive. | |
| Feb 4, 2023 at 12:07 | answer | added | wlad | timeline score: 3 | |
| Jan 29, 2020 at 16:38 | history | edited | YCor |
edited tags
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| May 2, 2013 at 20:41 | history | edited | Stefan | CC BY-SA 3.0 |
omittted unnecessary part
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| May 2, 2013 at 20:26 | comment | added | Emil Jeřábek | @Stefan: Because the order is in general not unique. In order for the condition to make sense, you have to fix a particular order, but then you can as well come out and say that the structure you are working with is an ordered field, rather than just a formally real field. By the way, the order on a formally real field is unique if and only if every element or its negation can be written as a sum of squares, but I don’t know if such fields have a special name. | |
| May 2, 2013 at 15:48 | comment | added | Stefan | @Gerald : regarding your suggestion, if we take a formally real field, we can just impose an order on it and make it an ordered field. Why can't we keep the condition $\langle \mathbf{x}, \mathbf{x} \rangle > 0$ for nonzero $\mathbf{x}$? And do we have the "additional structure" that wikipedia says we need (please see my edited question)? | |
| May 2, 2013 at 15:36 | history | edited | Stefan | CC BY-SA 3.0 |
fixed $\LaTeX$ formatting
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| May 2, 2013 at 15:32 | comment | added | Stefan | @darij: regarding your first comment, I just reread Wikipedia's article on formally real fields and they mention an alternative definition concerning $-1$ being a sum of squares, so I understand your first comment now. | |
| May 2, 2013 at 15:24 | comment | added | Stefan | @darij : Thanks. Apparently "ordered field" refers to a field together with the order relation while a "formally real" field is a field that can be endowed with an order relation with the required properties. I don't really understand your first comment. I don't see an obvious order relation for formal real-valued rational functions in real variables $x$ and $y$ similar to the standard order relation for rational functions in just $x$. | |
| May 2, 2013 at 15:21 | answer | added | Gerald Edgar | timeline score: 8 | |
| May 2, 2013 at 14:46 | comment | added | darij grinberg | Also, $\mathbb Q\left[\sqrt 2\right]$ can be ordered in two ways (by making $\sqrt 2$ negative and by making $\sqrt 2$ positive), neither of which is "better" than the other. This already shows that it isn't a good idea to identify "formally real" with "ordered" because the latter requires a choice of ordering while the former does not. | |
| May 2, 2013 at 14:44 | comment | added | darij grinberg | Stefan: Any rational function field over $\mathbb R$ in a finitely many variables (for example, $\mathbb R\left(x,y\right)$) is formally real (for rather simple reasons: if you had some rational functions whose squares sum up to $-1$, you could substitute values into them in such a way that the denominators don't vanish, so you'd get a sum of real squares summing up to $-1$, which is absurd), while not being totally ordered in a natural way (unless it has $0$ variables). I don't know if it can be ordered in non-natural ways, but this already shows that the two definitions aren't the same. | |
| May 2, 2013 at 14:36 | comment | added | Henry Cohn | It's also worth keeping in mind that people often look at analogues of symmetric and even Hermitian inner products in cases without positive definiteness. (For example, to define unitary groups over finite fields.) This is certainly somewhat different from the real/complex/quaternionic case, but it's not like there's a sharp dividing line conceptually. Instead, you just give up more properties. | |
| May 2, 2013 at 14:30 | comment | added | Henry Cohn | In the Hermitian case "field" is overly restrictive (the skew field of quaternions is important too). Even in the symmetric case, it can be convenient to look at rings: for example, integral lattices are often thought of as free abelian groups with integer-valued, positive-definite, symmetric bilinear forms. (Of course integers are real numbers, so this can be thought of as a special case of real inner products, but the restriction to a subring changes how things feel.) | |
| May 2, 2013 at 14:08 | comment | added | Stefan | @Mark Grant : Thanks. I guess I didn't read the Wikipedia article closely enough. The article is helpful but not 100% satisfactory. I edited my question. | |
| May 2, 2013 at 14:01 | history | edited | Stefan | CC BY-SA 3.0 |
added to question
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| May 2, 2013 at 13:46 | comment | added | Stefan | @Gerald : thanks, I never heard of a "formally real field". I read the link, and it does not explain whether there is any difference between "formally real field" and "ordered field", that is, it is unclear whether there exists a field that belongs to one class and not the other. In addition, I want to keep the requirement $\langle \mathbf{x},\mathbf{x}\rangle > 0$ for all nonzero vectors $\mathbf{x}$. | |
| May 2, 2013 at 13:35 | comment | added | Mark Grant | @Stefan: I was referring to the second paragraph of the "Remark" at the end of the "Definition" section, which discusses exactly this point. | |
| May 2, 2013 at 13:29 | comment | added | Stefan | @Mark Grant : not really. I have read that article several times and I just looked at it again. They seem to use only $\mathbb{R}$ or $\mathbb{C}$ until they discuss "generalizations", of inner product spaces , and I am not interested in generalizations. | |
| May 2, 2013 at 12:55 | comment | added | Gerald Edgar | One could easily define an inner product for a vector space over a formally real field en.wikipedia.org/wiki/Formally_real_field ... Your condition would then be $\langle \mathbf{x},\mathbf{x}\rangle \ne 0$ for nonzero vectors $\mathbf x$. | |
| May 2, 2013 at 12:31 | comment | added | Mark Grant | Doesn't the wiki page en.wikipedia.org/wiki/Inner_product_space#Definition go a long way towards clearing this up? | |
| May 2, 2013 at 12:19 | history | asked | Stefan | CC BY-SA 3.0 |