uncool

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CalabiYau manifolds are Kahler manifolds.
Kahler manifolds are symplectic manifolds.
Symplectic manifolds are always evendimensional.
So no.
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9 hours ago, studiot said:
And I did ask for an example of this 'anything' , whilst at the same time pointing to a well respected treatment of QM, involving plenty of imaginary and/or complex numbers.
Sure.
Instead of saying "the product of 1+2i and 2+5i is 8 + 9i", you could instead say "the product of (1, 2) and (2, 5) under the operation (a, b)*(c, d) = (ac  bd, ad  bc) is (8, 9)".
Yes, it's a trivial renaming of the complex numbers; that's part of my point.
9 hours ago, studiot said:But this is taking the discussion away from the key statement
But QM commutation is about the commutation of two different operators.
OK, here are two real operators (as I understand it; I never received an answer as to what you meant) that fail to commute:
0 1 0 0
0 0 1 0
Possibly more relevant to QM, here are some real operators that act like the spin operators, if I haven't made a stupid mistake:
0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0
0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0
1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
where multiplication by i is replaced by multiplication by the matrix
0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0
All of the components of each operator are now real.
9 hours ago, studiot said:12 hours ago, studiot said:How is (1+√5)/2 an element of an integer ring, when division is not closed with the integers ?
Yes I am aware of extension algebras and arithmetic and I know tha there are some pretty exotic creations floating around there, but I am not very adept at them.
I was always given to understand that for every gain in one direction you tend to loose something in another when you add to algebraic complexity.You seem to have quoted yourself; I'm not sure I see how this responds to my post.
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2 hours ago, studiot said:
So can you detail your real operators in QM that do not commute ?
The dreaded 'i' seems to appear lots in this treatment.I'd need you to first define "real operators".
As I said earlier, anything that can be defined in terms of complex numbers can instead be defined in terms of pairs of real numbers, etc.
2 hours ago, studiot said:Maybe the factorisation was stretching things a bit far as composition is involved, but I can't see what you mean by this
I meant that I have no idea what connection you are trying to draw between unique factorization and quantum physics.
2 hours ago, studiot said:How is (1+√5)/2 an element of an integer ring, when division is not closed with the integers ?
Why not? Why should √5 be considered "integer", while (1+√5)/2 can't be? How are you defining "integer" outside of the actual integers?
It turns out that (1+√5)/2 actually is quite integerlike, in that it can solve monic polynomials; such numbers are actually called "algebraic integers", and act somewhat like integers do for rational numbers.
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1 hour ago, studiot said:
The twist in the tail of using complex numbers in QM lies in non cummutativity
Err, what?
Complex numbers are entirely commutative, and any way in which noncommutativity appears in QM can appear in a "real" explanation just as much as for a "complex" one.
1 hour ago, studiot said:Such rings of complex numbers fail to satisfy unique factorisation  essentially a quantum behaviour.
Um.
Two things.
First, there are plenty of subrings of the real numbers that fail to satisfy unique factorization. Consider Z[√5], and factor 4. That can be fixed in a somewhat natural way by adding (1+√5)/2; however, the "natural" fix doesn't work for Z[√7] (and this failure  among others  is a reason for the modern study of algebraic number theory, especially ideal class groups). Failure of unique factorization is not inherently related to complexity.
Second, I see no reason why a failure of unique factorization is "essentially a quantum behavior".
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Strictly speaking, no, you do not "need" the complex numbers for anything.
You can write everything you do in term of pairs of complex numbers, which just so happen to have the properties of the complex numbers.
You could go even further; real numbers are equivalence classes of sequences of rational numbers (under an equivalence relation of, approximately, "limits to the same number").
And further still; rational numbers are equivalence classes of pairs of integers (under an equivalence relation of "same ratio").
And yet further; integers are equivalence classes of pairs of rational numbers (under an equivalence relation of "same difference").
But it's far more convenient to not constantly be going through this entire hierarchy to make even simple statements. It's much, much easier to simply define a set of numbers once, and then deal with all of the ramifications by repeatedly using that definition. And the complex numbers turn out to be useful enough to do so  much more than you might expect, in fact.
1 
Funny enough, for Catholics it apparently doesn't necessarily have to be pure for the purposes of baptism.
https://christianity.stackexchange.com/questions/65635/howmuchcanholywaterbediluted
If the baptismal water has so diminished that it is foreseen it will not suffice, unblessed water may be added even repeatedly, but in lesser quantity than the blessed each time this is done.
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On 10/3/2022 at 7:34 PM, NTuft said:
I don't think we actually know that. It may be that the complex numbers have a lot more transcendentals than the reals, don't you suppose???
No. They are, in fact, in bijection, and both are in bijection with the set of all real numbers, and both are in bijection with the set of all complex numbers.
On 10/3/2022 at 7:34 PM, NTuft said:Ok, but to reinforce: the reals have cardinality in part defined by having all transcendentals you claimed, and the naturals have none.
Again, what is your point?
Again: there is nothing inherently uncountable about individual transcendental numbers. You keep bringing up that some of these numbers are transcendental as if that, by itself, implies uncountability. It does not.
On 10/3/2022 at 7:34 PM, NTuft said:First: what is your argument for uncountability?
Second: that set is very clearly countable. It is very clearly the image of a countable set, namely the set of pairs (x, y) where x and y are both in i', under the map of exponentiation.
Once again: emphasizing that you are getting results which are transcendental numbers does not tell you anything about whether the output you get is uncountable. At the absolute best, it removes a single, overlysimplistic proof of countability; that does not constitute a proof of uncountability.
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It was almost certainly the tensor product. The statement "these two particles are entangled" is, more formally, "the combined state of these two particles is not a pure tensor element in the tensor product of the state spaces of each element separately".
While I don't know the history itself, entanglement itself is a rather natural aspect of the fact that quantum physics is fundamentally linear, and that the tensor product is the "natural" way of dealing with bilinearity.
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5 hours ago, studiot said:
They are just ordinary cartesian x,y,z... 'spaces' that have an infinite count of dimensions.
Bit more than that; don't forget the inner product. I'd say that that's the central aspect of Hilbert spaces.
7 hours ago, geordief said:Does that mean that every aspect (,or attribute?) of a quantum system (I think one has to use the term "system" rather than "object") exists in something like it's own dimension?
Depends on what you mean by "aspect" or "attribute"; I'd also be careful about what you mean when you say "something like its own dimension".
If you mean that each particle (and simplifying to quantum mechanics, where there is no particle creation or annihilation), then kindof, but not really. To explain further, and in some formality: the state space for a combination of particles comes from the (completion of the) tensor product of the state spaces of the individual particles, not a direct sum. Sorry, but I don't really have a less technical explanation for what that means.
3 hours ago, exchemist said:Regarding @geordief's question about QM entities and dimensions, I suppose eigenstates being orthogonal means each state a QM entity can be in is in a different dimension, doesn't it?
I'd say that each eigenstate a QM "entity" can be in can be thought of somewhat separately, yes (until you get some time evolution that mixes between eigenstates). To my recollection (which is admittedly rusty), the use of "state" can be somewhat ambiguous; it can either be used to mean a single eigenstate (of an unspecified operator), or can mean an arbitrary mixture. In the latter case, I'd say that the answer to your question is "no"; for example, the the states a>, (3/5) a> + (4/5) b>, and b> cannot really be thought of as in "different dimensions".
1 
On what level do you want an answer?
Because on one level, it was a bunch of wars and abolition movements. But that doesn't answer how all those wars and abolition movements came together.
On another, it was a bunch of philosophical movements that led to ideas of liberty. But that doesn't answer why those movements came to the forefront.
On another, it was a result of economic forces giving rise to systems that outcompeted slavery. But that doesn't answer why they hadn't happened earlier.
And so on and so forth.
2 
20 minutes ago, joigus said:
But Bob would learn about it no sooner than d/c, where d is the distance between him and Alice in their inertial system. I didn't say Bob would know it instantly. This makes sense, doesn't it?
Yes, that makes sense; I had read your statement as "So he could determine in principle that Alice has performed a measurement [instantly], but he wouldn't be able to tell which outcome Alice got until he made the measurement", because the second half of the statement was qualified by timing while the first half was unqualified. Sorry about that.
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On 9/9/2022 at 6:15 PM, joigus said:
So he could determine in principle that Alice has performed a measurement, but he wouldn't be able to tell which outcome Alice got until he made the measurement.
...if Bob could determine whether Alice has made a measurement, then Alice could send a message faster than light purely by choosing whether to perform measurements, if I'm understanding what you're saying correctly.
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1 hour ago, iNow said:
The history here seems relevant. Certain classes of people are considered “protected,” not privileged, and there’s good reason this had to be.
For decades / centuries they were discriminated against merely for how they were born.
..."protected class" is a designation based on various US laws, most prominently the Civil Rights Act of 1964. And in that law, the classes are not restricted to the "oppressed" classes. Both "men" and "women" are classes that are protected, all religions (including Christianity) are protected, and so on. Explicitly:
"All persons shall be entitled to the full and equal enjoyment of the goods, services, facilities, privileges, advantages, and accommodations of any place of public accommodation, as defined in this section, without discrimination or segregation on the ground of race, color, religion, or national origin.[...]
All persons shall be entitled to be free, at any establishment or place, from discriniination or segregation of any kind on the ground of race, color, religion, or national origin, if such discrimination or segregation is or purports to be required by any law, statute, ordinance, regulation, rule, or order of a State or any agency or political subdivision thereof."
0 
...or until it failed under internal stress, or until it was shattered by random space debris, or ...
it would be an absurd undertaking on logistical terms alone, even before considering actual use. the largest artificial satellite that we have ever launched is the international space station, which is 73 meters by 109 meters. you're talking about a mile wide object. that's 1609 meters.
then you have to consider the feasibility of using a milewide mirror. that is, you need a properly curved, smooth, reflective surface that achieves a specified position in orbit above earth.
and I'm pretty sure swansont meant 6000 degrees as an absurd upper limit, not a reasonable estimate, although i'm not sure that all of the necessary assumptions for that theorem are met (i haven't checked thoroughly, though, so i'd default to trusting him!)
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https://www.law.cornell.edu/uscode/text/18/1512#c_2
QuoteWhoever corruptly [...] obstructs, influences, or impedes any official proceeding, or attempts to do so, shall be fined under this title or imprisoned not more than 20 years, or both.
"Corruptly" isn't statutorily defined, as far as I can tell; there is some case law on the definition.
https://www.ce9.uscourts.gov/juryinstructions/node/732
Quotethe term "corruptly" must reflect some consciousness of wrongdoing.
Arthur Andersen LLP v. United States
If he urged people to the Capitol in an attempt to impede the counting of the electoral votes (which multiple courts have ruled is an official proceeding), and if he did so with some consciousness of wrongdoing (which is something that might be inferred), then yes, that is enough, if I've read it right.
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8 minutes ago, iNow said:
I believe that distinction is both relevant and necessary since the original question was whether congress could bring charges and enforce jail time against January 6 insurrectionists... unless you're suggesting those who stormed the capitol are being held in contempt for their treason?
I read the question as a general question of whether Congress has any power to bring charges and punish under its own authority. But even with the interpretation of charges being specifically about the Jan 6 insurrection, I think a case could be made for Congress having the power to punish the insurrectionists itself. Not that that is what is happening, or even that it is likely to happen, but that that is within its power.
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On 6/18/2022 at 6:05 AM, swansont said:
Finding someone in contempt is not the same thing as bringing criminal charges for the acts being brought to light, which was the context of the discussion.
While they are not the same, part of my point with the link is that Congress's power is not limited to finding someone in contempt, but may include the ability to prosecute those charges itself. The link I provided gives examples of both houses of Congress themself trying cases where someone was accused of attempting to bribe a member. From the link: "Ultimately, Mr. Anderson was found in contempt of Congress and was ordered to be reprimanded by the Speaker for the “outrage he committed” and discharged into the custody of the SergeantatArms. Mr. Anderson subsequently filed suit against Mr. Thomas Dunn, the SergeantatArms of the House, alleging assault, battery, and false imprisonment. [...] The Anderson decision [by the Supreme Court] indicates that Congress’s contempt power is centered on those actions committed in its presence that obstruct its deliberative proceedings. "
Unless you are making a distinction between "bringing criminal charges" and imprisoning someone for contempt, I don't see how the context changes my point  not that this is likely to happen, but that Congress may have the power to do the latter.
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8 hours ago, swansont said:
My understanding is no; the criminal charges have to come from the department of justice. The J6 hearings are bringing information to light, some of which (apparently) was previously unknown to the DoJ.
As an explanation of what is likely to happen, I think this is correct; as an explanation of what can happen, I think this is incorrect.
Congres has the inherent power to hold people in contempt, and can do so without involving the executive branch whatsoever, and this can include fining or imprisoning the accused contemnor for the length of that session of Congress. And this can be for either a coercive purpose  trying to force the accused contemnor to do something  or a punitive one  punishing them for failing to do so.
https://sgp.fas.org/crs/misc/RL34097.pdf
With that said, it apparently hasn't happened since 1935, and was rare before then as well.
0 
I don't think that anyone really answered the question, because the argument the question is about uses assumptions that aren't that obvious.
The statement "Force is proportional to mass" is somewhat imprecise; the fully precise statement is "Force is proportional to mass when acceleration is held constant." That is, if you have two masses which are accelerating in the same way, then the force on one divided by its mass will equal the force on the other divided by its mass. Alternatively, for any situation with forces and masses, F/m is a function of the acceleration.
Similarly, the statement "Force is proportional to acceleration" is imprecise, with a more precise version being "Force is proportional to acceleration when mass is held constant." If you have two objects with the same mass, the force on one divided by its acceleration will equal the force on the other divided by its acceleration. F/a is a function of the mass.
So we have that F/m = f(a) for some function f, and F/a = g(m) for some function g. Then F/ma = f(a)/a = g(m)/m. f(a)/a is also a function of acceleration independent of mass, and g(m)/m is also a function of mass independent of acceleration. And the only way that can happen is if f(a)/a = g(m)/m is constant  is independent of both mass and acceleration. Call that constant C. Then F = C*ma  in other words, force is proportional to the product of mass and acceleration.
1 
…no, NTuft, that is not why e^(i pi) = 1. Your blind substitution is not correct.
You are claiming that e^(x*i*pi) = x cos(pi) + i x sin(pi) = x, if I’ve remade your equations correctly. This simply isn’t true. It has nothing to do with the actual justification of the original equation (which has to do with McLaurin series), and is selfcontradictory with a bit of thought.
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…none of this is accurate, and this can be checked directly by e.g. wolfram alpha.
e^(i’ pi) = e^ (i sqrt(2) pi) = cos(sqrt(2) pi) + i sin(sqrt(2) pi) ~ 0.266  0.963 i
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4 hours ago, Airbrush said:
When a woman decides that she would rather abort her fetus than give birth, it could indicate that perhaps she would not be the ideal parent for that child. Unwanted children often grow up to be criminals. Tens of millions of abortions have happened since Row v Wade. How many of them would have been unwanted, abused, neglected, and sent to the street to be educated as criminals.
…I have to admit, I’m…uneasy with this reasoning. It suggests that there is policy interest in who gets to reproduce based on speculation on future criminality, which has some connections with eugenics. Those connections strengthen when you add in the conservative framing of abortion being connected to the right of the fetus to life  under that framing, this argument suggests that some people don’t have a right to live because of speculative criminal activity.
I’m not suggesting that this is what you meant to suggest. Just that it may have unintended implications, especially among people with other framings.
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Worth noting it was used on both “sides”: by slavery supporters to say that their “rights” to own slaves shouldn’t end when they visited free states (or when slaves successfully ran away to free states) and by abolitionists to say that African Americans should have the same right to be free in all states
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Additionally, the Senate could refuse to confirm any extra Supreme Court justices if it changed its mind after allowing more.
It would be very unlikely, but the Senate has both direct power over the number of justices (through legislation) and indirect control (through confirmation).
3 minutes ago, MigL said:That being said, most of those rules should apply to the greater society of the country, and not be a mix'n'match at the state level.
An American should not have one set of rights in one state, and another set in the next. Nor should travel between states be a problem.
That is the absurd situation.What are you defining as “rights” here? Does e.g. the right to buy alcohol count as a “right”? The right to buy guns? The existence of a patchwork of rights sounds like the same thing as different states having different laws.
What you are looking for is probably covered by the “Full Faith and Credit” clause of the Constitution.
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Mathematics is Inconsistent!
in Analysis and Calculus
Posted
It doesn't need to be a solid; the "solid sphere" version of the theorem is proven based on the "hollow sphere" version of it, as shown in the Wiki page: