@alarge They responded consistent with the law and with their training. Somebody could argue the law and training are wrong, and they are encouraged to vote for the changes they would like to see
@tpg2114 I agree that the police in the States seem badly trained and very trigger-happy. That is really the point of the article as well (comparing deaths from police shootings).
@tpg2114 Fair enough. Why do you think, though, that deaths from police shootings in the States is much higher than pretty much anywhere else? I'd attribute it largely to training (in the many police videos the officers in the States are very quick to shoot), but as you disagree, I'd like to hear what you consider feasible alternatives.
Sometimes it happens that you answer a particular question (or set of related questions) that ends up creating in the OP the impression that you have become their personal tutor, and they ask for your email address.
A few approaches one can use: ignore, be rude, be polite, give in.
I lean towar...
@alarge I think it's two things -- first, the number of violent crimes in the US is much higher than other places and so the police need to be trained to respond to that. And it's difficult/impossible to know when somebody taking a swing at you has a gun they will pull when they knock you down. So the police need to be ready to respond faster than there is time to think. Like we were discussing earlier, 70+% of violent crimes in the US involve firearms
The second issue is that we really only hear of reports of disproportionate responses and they get blown up quite a bit by the media. There are probably millions of police-citizen interactions a year that may or may not result in arrest, which may or may not require force by the police, where the response is entirely reasonable
Police all of the world have a very tough job to do. They are in constant threat of bodily harm, and this is more true in the US than other places for many reasons. And so they have to be trained to handle and respond to situations rapidly where you may not be able to assess everything
I disagree with some of the police training, in particular the para-militarization of many local/regional police departments. But that's not the type of situations that we're talking about here
And even there, it's difficult to fault them when they have to go serve warrants on drug dealers who booby trap their houses and are armed to the teeth with military weaponry.
@KyleKanos I'd never heard that!I didn't want to follow the trend and be an independent investigator (like everyone on my facebook wall was trying to be) so I really didn't look into it.
@tpg2114 Is there not a vicious cycle at play here, though, for if the criminals know the police will come guns blazing, they will also be more prone to shoot? Even with all the statistics (more firearms, say), the police in the States do seem to shoot and kill disproportionately many (but I'd be very interested to see if you were to compound all the effects you talked about in numbers of some kind; could you show that the deaths are indeed comparable to other Western nations).
@alarge I don't think so actually. There are far too many historical reasons to just take away guns in the US. It will just never happen in any meaningful way. And so the criminals will always be armed. Look at the gangsters in the 30's. They were on the forefront of weapon technology and the police couldn't even come close to keeping up
And so the mobsters ran the show. The police just had to let them fight it out
@Danu It's a reference to this SMBC and the note under the red button that says "Yes, if someone figures out how to manufacture these, I will sell them".
Point is -- it's all in the eye of the beholder. And it's also up to that community to define their own successes or failures. At least until there is a panel of wise Internet judges who can decide
Arguably it is a success from our stand point because it means far fewer repetitive meta discussions. And it's a success from their stand point because they have a site they can design and craft the rules to their liking
> he Q&A category is dried up, there is almost nobody on the site that is able to give answers and nobody anymore writes reviews. We are near to have more moderators than users.
KK yeah saw some of the desperate-tone thread titles over there eg that one (some cited by alarge a few wks ago), did not have nerve to read much further
And I have a sneaking suspicion that they won't be able to agree on it. I got the impression the only thing that united them was their anti-phys.SE feelings. And once they got their own sandbox to play in and didn't have a common enemy to fight, things got rocky
did notice the infamous ron maimon resigned as mod. (someone recently mentioned him which reminded me of my wondering about others opinions on this site.)
> Regarding "open problems", I am willing to offer money bounties on both of them, and if we did allowed people to put money bounties, it would get attention. I have proposed this before.
It's not really a reserve judgment so much as it's a "It's not my place to judge them" judgment. Just like I don't sit around and worry about how reddit runs their site -- I don't participate, it's none of my business
ok. yeah. neutral. impartial. yeah web sites are a dime-a-dozen. although sites that get the social media mix/ formula well seem quite tricky/ rare. esp for intellectual pursuits/ focus. so have some big appreciation for se generally, which seemingly generally succeeds on average.
I will say that I do enjoy SE sites in general and I like popping onto the random beta sites/SE 2.0 sites and reading the really interesting questions ask about topics I hardly knew existed
That's part of what makes physics.SE interesting for me -- there's the hot bar down on the site that lets me explore other ideas/topics. I think if it was just physics in a bubble, I wouldn't be as interested
se has a great diversity & continues to grow. was impressed that google now has a close connection wrt patent analysis site. iirc KK cited that in this chat room.
← always thought "microcurrency" might succeed somewhere on internet from long ago & hyped predictions (back in dotcom era!), seems still(!) nobody has cracked it
@Jiminion the former is relatively young & started by "protestors" of the latter. theres also an old history of a failed physicsoverflow attempt on se.
A community ad caught my eye, and I visited http://www.physicsoverflow.org. What are the links between Physics SE and Physics Overflow and how are they different from each other?
@ACuriousMind what is 'radial quantization' explicitly? I mean, how do I give an actual definition? It should involve like a Schrodinger/Heisenberg equation and commutation relations right? But none of the books explicitly state any of this
I've been looking into the 3-D Ising lol it's related to non-critical string theory shockingly
@bolbteppa Radial quantization means we consider theory on the plane instead of the cylinder, where the Hamiltonian is just the dilatation $L_0 + \bar L_0$. States are associated to circles, then, and evolve, as usual, by $\mathrm{e}^{-\mathrm{i}Ht}$ (this is all the Schrödinger/Heisenberg equations tell you, after all). There are no "canonical commutation relations" because we usually do not quantize a specific action, so there are no fields and their conjugate momenta.
Essentially, quantum CFT is already "fixed" by demanding that it carries representations of the Virasoro algebra - there is no need to tediously quantize classical systems to get CFTs
(This is for 2D, you can also do the conformal map "cylinder" -> "plane" in higher dimensions, but there you need to give a field theory on the cylinder as a starting point, and then get a field theory on the plane, and "radial quantization" is then just canoncial quantization of the theory on the plane)
Interesting, when formulating CFT it seems we basically take something 'global', like the conformal group, and examine it locally, and use local properties to talk about the global picture. It kind of looks like we are defining a quantum field theory locally using special condition X as a direct way to satisfy a bunch of global properties without ever even looking at them, but it only holds for cft's
@Sean The area51 still exists for historical purposes (i.e., if someone tries reviving it, they can be pointed to that and told Been done & failed). But I believe that all of the questions from that site were migrated to this site.
ps re theoretical physics, "failed" or "crashed" is the harsh way of putting it. "didnt work out" or "didnt achieve traction" is maybe more diplomatic. :|
(would personally like to see at least one such site succeed.) there is likely enough audience worldwide to support it but it seems still not many hard-core scientists/ physicists "hang out" in cyberspace :|
@Danu RIP :(
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5:40 PM
@vzn I cannot begin to imagine any of my professors spending time online. In fact, I'd hesitate to mention to them that I'm on this site, because it would come across as an enormous waste of time I could be using writing papers.
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Unlike mathematicians, physicists don't just sit around chatting and reflecting on their subject all day long :P
@ChrisWhite yes have noticed & it seems maybe distinctly real how the math culture/ psychology is possibly somewhat different & mathematicians have really taken to MathOverflowen masse but its not happening as much/ strongly in other fields eg Physics or Computer Science, [cstheory] etc... so far...
@vzn Yes, although I don't have a particular interpretation I've chosen to subscribe to
I was taught Copenhagen, as are almost all undergrads I think. But I don't find it... satisfying
And I think an interpretation is necessary, because I find "Shut up and Calculate" to leave too many intriguing possibilities at the door
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@vzn That much is true. In fact, I'd say astronomers rival even mathematicians and comp. scientists for their addiction to arxiv (my department formally gathers to discuss new postings each and every morning)
When I was working on a technical paper for publication, I was strongly urged by my co-author to put only ONE new concept in the paper. That is because one needed to publish as many papers as possible for one's career, so it would be wasteful to have more than one new thing per paper. Compare this with Shannon's paper on Information Theory, which was basically the only major paper he ever wrote. But a good one!
@Sean yes also see "many intriguing possibilities at the door"... hopefully with a little luck & lots of (hard) work a (currently) non-popperian-falsifiable qm interpretation will evolve enough into something falsifiable... within our lifetime. lets fairly note that large swathes of physics have been strongly critiqued even by "insiders" as non-falsifiable (and therefore similar status as the much-maligned "qm interpretations"), eg maybe foremost, string theory.
@Sean re "intriguing possibilities at the door" fyi :)
@ChrisWhite now if only se or a similar social networking platform could be made as "sticky" (engaging) to (senior) scientists as arxiv! kinda wish arxiv might advance some/ further to consider stuff like peer review, social networking etc., but its probably "done" at this pt. :|
@bolbteppa have studied solitons at a high-level pov. would like to dig further into details. conjecture a 3d model of particle physics mixed with soliton dynamics holds great promise. :)
@bolbteppa Then you have a different definition of a soliton, because mine is as special kind of waves. Of course they are described by certain PDEs, but what is suprising about that?
A soliton is just one of many ways of representing the family of solutions to a non-linear pde, from the wiki: "Many exactly solvable models have soliton solutions, including the Korteweg–de Vries equation, the nonlinear Schrödinger equation, the coupled nonlinear Schrödinger equation, and the sine-Gordon equation. The soliton solutions are typically obtained by means of the inverse scattering transform and owe their stability to the integrability of the field equations. The mathematical theory of these equations is a broad and very active field of mathematical research."
these things are definitely not being ignored, a lot of my professors are linked to the people who invented half this shit lol
@vzn I don't think its very suprising that actual wavepackets behave like the generalized wavepackets that are particles. It's interesting, but not surprising.
Also, it's no coincidence - particles are described as generalized wavepackets, after all.
Solitons are one of the most mainstream thing ever, the Bethe ansatz is the quantum version of this stuff, it's all so unbelievably famous, the best people in the world work on this stuff man, maybe to non-mathematical physicists they seem obscure but I mean even Tong has notes on them damtp.cam.ac.uk/user/tong/tasi.html
We were given solitons on our 2nd year exam lol
The shocking thing is that the theory in mathematical books is like a carbon copy of a CM/QM/QFT book if you tear your eyes out searching it out
@bolbteppa that is exactly what have been saying for years & still nobody is really paying attn to that. its going to be revolutionary at some pt, mark my words... just not this second
haha well that is like advanced soliton stuff, this arxiv.org/pdf/hep-th/0110125v1.pdf is the older CM/QM/QFT style soliton stuff, Tong has the string soliton stuff
"string solitons". wow. ah was thinking/ conjecturing many yrs ago those two fields were very likely to be unified. havent gotten into that yet. thx. also very cool to hear it was a part of your basic edu. maybe next generation has a chance after all :)
@KyleKanos Ron's writing is juicy as usual: "The story here is essentially the same policy as anywhere, with different people on top. There's no "community moderation", there's moderation by arbitrary fiat, and official warnings, and official blocks. That's not what I signed up for. I have refrained from advertising the site further, and I think it's time to call the experiment a failure. You really can't trust anyone at all. "
@ChrisWhite @Danu: Haha...lab course, and it would be nice to have something to compare our measured values to (I know they're crap, but quantifying crappiness is good).
@Danu There were so many kinds of problems in that class and he just picked four. We also covered Monte Carlo methods and there wasn't a monte carlo method problem on it!
@ACuriousMind Is an interval $I$ that is open in $\mathbb{R}$ neither open nor closed in $\mathbb{R}^2$ because there it takes the form $I\times\{0\}$?
@0celo7 In grad math classrooms maybe. But I don't think many physicists care about the topology $\{\emptyset,\mathbb{R}\}$ so saying "metric" is nice because it emphasizes what kind of topology it is.
@ACuriousMind I think I found the true meaning of radial quantization... So, thinking of the Witt algebra as a Hilbert space (local Lie algebra) representation of a conformal symmetry group of functions, and then the Virasoro algebra as a projective operator representation of the Witt algebra, we managed to express the conformal classical field/function information in a bunch of operators, yet we just interpret these operators as states, the way a ray in a Hilbert space represents a state.
Quantization in QFT consists of expressing the Hamiltonian in terms of the states expressed as operators, and since $H = L_0 + \bar{L}_0$ we see this is the true meaning of radial quantization, it is the culmination of expressing multivariable functions (conformal fields) as operators using derivatives (Lie/Witt algebra) then taking a projective representation to express the rays as operators (Virasoro) & finally expressing the Hamiltonian in terms of the fields expressed as operators. WTF...
@bolbteppa: I believe your terminology is off. The Witt algebra is the Lie algebra of the 2D (local) conformal group, and hence carries naturally the adjoint representation. The Virasoro algebra is not a projective representation of the Witt algebra, but a central extension. I have no idea what you mean when you say "Quantization in QFT consists of expressing the Hamiltonian in terms of the states expressed as operators". Quantization means promoting the fields to operators.
@ACuriousMind The only reason you do promote fields to operators is so that you can express the Hamiltonian of your system as a sum of creation and annihilation operators (Weinberg), it's called quantum field theory, not quantum operator theory after all... Also, the Virasoro algebra is the Lie algebra analogue of a projective representation (Weinberg) and is the origin of the central extension definition in the first place...
That is absofuckinglutely beautiful
@0celo7 I think you should read my post above about the true meaning of radial quantization!
@bolbteppa: I have not really an idea what you are talking about. I'm inclined to not call the thing with the Hamiltonian and the operators false (although that only works that way for free theories), but the Virasoro algebra is not a projective representation of the Witt algebra, it is the algebra of its covering group (obtained as the central extension by the fundamental group), and its linear representations are precisely the projective representations of the Witt algebra.
@bolbteppa Yes. I agree with the idea that the central charge arises from considering projective representations, it is just not correct to say that the algebra itself is the projective representation, I think.
@ACuriousMind I think we might both be right in a sense, mathphysseminar.blogspot.ie/2014/02/virasoro-algebra.html I am almost right except the short-exact-sequence and fiathfulness stuff I don't fully understand may fuck it up and justify what you're saying?
@bolbteppa Faithfulness means just that the representation map is injective, and the short exact sequence that the algebra that is being extended can be recovered by taking the quotient of the extension by the subgroup of the center that was "added". The way the $\hat{\mathfrak{g}}$ is defined shows that, indeed, every projective representation of the original algebra is equivalently a linear representation of the extended one.
Still, this does not turn the central extension into a projective representation itself - since we can embed it into its extension, the natural adjoint representation of it on itself (and its extension) is still linear.
@ACuriousMind are you just fishing for counterexamples? You are basically telling me that when I take a projective representation of a group and find it's Lie algebra I might not end up with a central extension of the Witt algebra to get the Virasoro algebra
@bolbteppa Oh, I think we might have a definition problem here. For me, a representation of a group is a space $V$ together with an algebra homomorphism $\rho: \mathfrak{g}\to \mathrm{End}(V)$. In this language, "taking the projective representation of a group and finding its Lie algebra" is not a sensical thing to say. You also can't take the Lie algebra of $\rho(\mathfrak{g})$, that's also not the central extension, you have to define the $\hat{\mathfrak{g}}$ as it is done in the link.
As the link shows, you will, for the Witt algebra, always find that $\hat{\mathfrak{g}}$ is the Virasoro algebra, but $\hat{\mathfrak{g}}$ is not "a projective representation of the Witt algebra", but its linear reps give you the projective reps.
@KyleKanos remarkable & yet prob not too different than other top schools (eg caltech) but maybe not as visible elsewhere. and even somewhat avg schools have suicides... also reminds me of foxconn (apple supplier)... "the shadow side!"
also relevant to all the young high achievers/ near perfectionists on here. aka "work life balance" :|
@0celo7 The linear reps of the central extension descend to (all) projective reps of the algebra that was extended.
(This happens here because the central extension is unique - if it were not unique, you would have to look at the linear rep of all the possible extensions to get all the projective reps)
Ugh, what's a projective rep? Wiki doesn't help and I thought it was a representation where the commutator has an extra term in it.
But apparently that's just a central extension.
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@vzn Caltech had a small string of suicides while I was there. Academic stress was never directly implicated, and the ones I'm familiar with had much deeper personal issues. Everyone always tries to connect overworking with suicide when it happens at such schools, but I'm unconvinced at the strength of that connection.
@0celo7 Linear unitary reps are given by maps into the unitary group of a vector space. The projective reps are given by maps into the projective unitary group of a vector space, essentially saying it is a linear representation "up to phase"
@ACuriousMind A symmetry group $G$ of maps $T_i$ mapping rays into rays forms a Lie group. A unitary representation $U : G \rightarrow U(G)$ such that $U(T_2)U(T_1) = e^{i\theta(T_2,T_1)}U(T_2T_1)$ forms a projective representation of the Lie group $G$. The set $U(G)/\{e^{i\theta}\}$ forms a Lie group too right? Therefore it's Lie algebra should give the central extension, and because of the modding out aspect we're gonna have to consider direct sums...
@vzn From what I remember, but do correct me if I am wrong, the incidence of suicide among FoxConn employees is (and was) still much lower than the national average. It is an enormous corporation after all. If you're looking for high suicide rates (among students), look up Japan and Korea.
@bolbteppa No, it's lie algebra is just the unextended algebra. The algebra sequence is $\mathbb{C}\to\hat{\mathfrak{g}}\to\mathfrak{g}$, and the integrated group sequence is $\mathbb{C}\to\hat{G}\to G$. The projective rep is a group homomorphism from $G$ to the projective unitary group, and its image is what you write as $U(G)/\{\mathrm{e}^{\mathrm{i}\theta}\}$, and if the rep is faithful, the image has the Lie algebra of $G$, not of $\hat{G}$.
$\hat{G}$ is linearly represented by the group homomorphism into $U$ that is uniquely defined by the map from $G$ by which you defined your projective representation of $G$.
(the fact that the projective rep defines that morphism uniquely is the reason that linear reps of the extension are in bijection to projective reps of the unextended group)
@ACuriousMind Yesterday, we were discussing the nature of the wedge product and the exterior algebra (I think lol). You were saying something like if we had a metric, we could raise or lower indices for our vectors. For instance, given a vector $v$ we could create a covector by plugging it into $g$ yielding $g(v,-)$. Correct? And that this was one way to think about how to apply the tensor product (and therefore) wedge product to vectors.
@ACuriousMind Since the original formulation of tensor products you gave was with $1$-tensors. Did I follow that correctly?
@StanShunpike Yes, although that is rather an artifact of the fact that I restricted to multilinear maps on the tangent space. With a metric, you have the raising/lowering maps, and this suffices to define tensor acting on arbitrary collections of vectors and covectors. If you don't have a metric, you have to build the tensors as multiplinear maps on the tangent as well as the cotangent space, and then you can tensor and wedge vectors without metric, because vectors are linear maps on the covectors
(since the double dual is the original space again)
@alarge that sounds like near-propaganda to me :( ... alas subj was highly politicized/ corporatized... did you hear at 1 pt how foxconn put up "suicide nets" in the buildings?
@ChrisWhite think thats highly mistaken, think there is strong correlation between (academic) stress/ suicides & elite schools, think it hasnt been studied much, think that there is incentive for schools to sweep issue under rug, much like campus rapes (which is getting more exposure lately). agreed some of the stress can be from other factors eg social isolation/ ostracism etc.
@ACuriousMind like if I have a tangent space at a point $p$, it would make sense to construct a multilinear map out of multiple tangent spaces unless they were the same vector space...right?
Yes, I know this ;) @ACuriousMind Okay, now you were saying something about how the wedge product can define both a plane and it's orientation. And I wasn't following this. You said that our vectors in question $v$ and $w$ were $n$ dimensional and spanned the plane, which they can do for obvious reasons.
Yes. And then I said that the object $v\wedge w$ is an equivalent encoding of the oriented plane, because it sends all vectors that are orthogonal to the plane to zero - it projects into it, in a sense#
@StanShunpike think of a vector space $V$ as representing vectors as linear combinations of basis elements $\hat{e}_i$ only, you're never allowed to write coordinate vectors like $(3,2)$ etc... okay? Now think of the dual space $V^*$ as representing the coordinates. This is why the dual space is a set of linear functionals $f : V \rightarrow \mathbb{R}$ mapping vectors to real numbers... So given $\vec{v} = 2\hat{e}_1 + 3\hat{e}_2$ we see the linear functional $\hat{e}^2$ is $3$ since
$\hat{e}^{2}(\vec{v}) = \hat{e}^{2}(2hat{e}_1 + 3\hat{e}_2) = 2 \hat{e}^{2}(\hat{e}_1) + 3\hat{e}^{2}(\hat{e}_2) = 3$ If your space has a metric/inner-product we could *also* have gotten $3$ by noticing $<\hat{e}^2,\vec{v}> = <\hat{e}^2,2\hat{e}_1 + 3\hat{e}_2> = 3$ So when you have a metric you can exploit the $<\hat{e}_1,\hat{e}_1> = <\hat{e}_1,\hat{e}^1> = <\hat{e}^1,\hat{e}_1> = <\hat{e}^1,\hat{e}^1>$ symmetry to go between vectors and coveectors as you please, messing around with defining things like $<\hat{e}^1,\hat{e}_2> = \hat{e}^1(\hat{e}_2} = \delta^1_2$
@bolbteppa Depends on how sloppy you want to be ;) Your internal picture of this is probably correct, but the way you are phrasing simply sounds technically incorrect to me.
@ACuriousMind bolbteppa Raises a good point about the components of the vectors. I am used to working with components and you aren't specifying any. Is all the information I need encoded in the orthogonality you mentioned some how? @bolbteppa mentioned the dual space as like representing coordinates. I don't know what that means. But I think this distinction sounds important. Why aren't we specifying coordinates?
@StanShunpike Because...we don't need them? Vectors are simply things you can add to each other and multiply by scalars, and that's all I am using.
Might also have to do with my fetish for coordinate-free notation. You could do this in coordinates, but I personally don't find that enlightening in any way
@StanShunpike if you picture a vector in space you are interpreting it as a linear combination of three vectors that specify the three directions in space, you do not have any components you just have multiples of arrows pointing in different directions that join together
You want to talk about the space of coordinates in such a way that nothing drastic will change when you change your basis, you want to talk about a space of coordinates that allows you to explicitly consstruct the new coordinates when you change the basis
@StanShunpike if you picture a vector as an arrow in space, that is a geometric quantity independent of the mathematics we use to describe it. To mathematically describe it we need to set up a coordinate system, and furthermore we need to find a set of (basis) vectors that allow us to build any vector out of them. Once we've established a basis, we can describe the arrow with reference to this basis.
These are the coordinates, a man-made construction depending on the form of our basis and the coordinates we used. What happens if we rotate the basis? The numbers would change
@StanShunpike Well, actually, everything on the manifold is "coordinate independent" in the sense that the abstract object doesn't change when we switch our coordinate systems.
Coordinate-free usually means that you don't even refer to any coordinates when doing stuff. Coordinate-independent merely means that whatever you write down works in any coordinate system, but it can contain coordinates
But the vector stays the same geometrically, just our description of it has to change We have to form new linear combinations of the new basis to get the same thing
@StanShunpike an exercise for you is to find out a) How to set up a basis in $\mathbb{R}^3$, b) How to set up it's dual basis, c) How to describe a vector as a linear combination of those basis elements, d) How to find the dual to this vector, e) How to change the basis, f) How to express the vector in terms of the new basis (like 2/3 ways to do this), g) How to change the dual basis, h) What happens when you apply the old dual vector to the vector in a new basis
You really have to do examples in linear algebra tbh, really, so this post math.stackexchange.com/a/1077144/82615 should set you right theory wize, then schaums L.A. 4'th edition or the 3,000 solved problems version is good for problems
@bolbteppa fyi think this recent interesting work by bush/ MIT is probably closely connected to solitons but nobody realizes it yet, you might find it interesting
@bolbteppa think the connection to bohmian mechanics is not (very) tight. it uses it as a semi convenient historical jumping off point/ reference.
ie something like an "interpretive/ conceptual/ high-level/ abstract/ bkg framework/ context" (because a lot of bohmian mechanics was connected to/ motivated by that)
@vzn I imagine you notice the link to all the 'waves' lol, the article I linked you to on classical integrability would be helpful since it's so similar to QM
@bolbteppa lol "its so similar" to QM is an understatement (of the century?). QM has not been proven to be distinct from it. and actually a handful are now arguing (persuasively) that it is not.
@bolbteppa motl (2009) well summarizes std/ valid historical objections to bohmian mechanics. but newer discoveries are now seen as overcoming some of the indeed sizeable (previously thought intractable) obstacles/ hurdles.
imho there is no known/ theoretical barrier to this (admittedly long term/ highly ambitious/ previously insufficient) research program continuing to evolve/ advance & eventually succeeding.
@ACuriousMind Let's suppose I have a vector space $V$ and a dual space $V^*$. Then obviously an element $w \in V^*$ can accept as its input a vector from $V$ and spit out a real number.
@ACuriousMind but what happens if I take and element $\omega \in V^{**}$ and plug in $w$ to that? Do I get a real number. OH! Sure! Because $w$ is just a vector until I plug in a $v$ into it. And thus, we wouldn't be mapping s number but a vector, right?
@StanShunpike Yes. By definition $V^{**}$ is the space of linear maps $V^*\to\mathbb{R}$, so by definition, $\omega$ takes a $w\in V^*$ as argument to produce a number.
So how does linear independence of vectors influence the tensor product? If I plug in two linearly independent vectors, why would the wedge be nonzero?
@0celo7 First, the assumption that $G$ is finitely generated is missing. Second, look at $G\cap H = (\mathbb{Z}\oplus\dots\oplus\mathbb{Z})\cap H$ and see that the intersection of $H$ with each summand has to be a subgroup of the summand, and the subgroups of $\mathbb{Z}$ are precisely the $n\mathbb{Z}$.
@StanShunpike "basic tensor" is something you sometimes hear.
Sometimes it happens that you answer a particular question (or set of related questions) that ends up creating in the OP the impression that you have become their personal tutor, and they ask for your email address.
A few approaches one can use: ignore, be rude, be polite, give in.
I lean towar...
A community ad caught my eye, and I visited http://www.physicsoverflow.org. What are the links between Physics SE and Physics Overflow and how are they different from each other?
