Quick questions and answers
I know I'm not a mod here, but this might alleviate some of the thread strain around these parts. In this thread you can ask quick questions that don't really warrant a thread. Something like "what do you guys think of *****". That would be a good question to ask in here. Also "do you guys use pedals on your bikes?" would also be appropriate. If this thread relieves the front page from two threads a day then it will be well worth keeping around.
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yes, what do you guys think of***** ?
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I really don't get why so many people are so anal about "unnecessary" or "redundant" threads.
Why can't people just ignore the threads that don't interest them and allow those who want to start their own thread to do so - regardless of the topic. I don't understand the need to "relieve the front page from two threads a day." So, what do you all think of that? |
Some people need to type less and ride more.
I tried to ride today, but the 4+ inches of snow put a damper on that. I need some skis now...or a Puglsey! |
So what bag goes with a pugsley?
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paper or plastic?
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yeah, just ignore the threads you don't like...
i'm not sure how to put this philosophically...people waste their time writing replies because they CAN... just like i'm doing now. |
Threads that you have no knowledge to contribute to or have anothing to do with... DO NOT REPLY.
Easy! ps. yes... kinda like what I just did... don't do that! |
Check out my new "I Already Knew That" thread. It seeks to alieviate some of the stress of the "Quick Question" thread by inviting people to shed light on the fact that they already know the answer to whatever question you intend to ask (as oppoesed to actually answering 'em). Perhaps the mods will make it a "sticky."
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I've always had some quick questions in mind, now i have chance to ask them:
• If a stripper gets breast implants can she write it off on her taxes as a business expense? • How do "do not walk on grass" signs get there? • If there was a crumb on the table and you cut it in half, would you have two crumbs or two halves of a crumb? • Do illiterate people get the full effect of Alphabet soup? • In that song, she'll be coming around the mountain, who is she? and lastly • Why do all superheroes wear spandex? |
Originally Posted by ostro
I've always had some quick questions in mind, now i have chance to ask them:
• If a stripper gets breast implants can she write it off on her taxes as a business expense? • How do "do not walk on grass" signs get there? • If there was a crumb on the table and you cut it in half, would you have two crumbs or two halves of a crumb? • Do illiterate people get the full effect of Alphabet soup? • In that song, she'll be coming around the mountain, who is she? and lastly • Why do all superheroes wear spandex?
next. |
ok...
• If you die and you have a broken leg do they take the cast off? • Since there is a rule that states "i" before "e" except after "c", isnt "science" spelled wrong? • Why isn't the word 'gullible' in the dictionary? •*Why did Superman wear his briefs on the outside of his tights? • How many licks does it take to get to the center of a tootsie pop? • Why do British people never sound British when they sing? |
Originally Posted by ostro
ok...
• If you die and you have a broken leg do they take the cast off? • Since there is a rule that states "i" before "e" except after "c", isnt "science" spelled wrong? • Why isn't the word 'gullible' in the dictionary? •*Why did Superman wear his briefs on the outside of his tights? • How many licks does it take to get to the center of a tootsie pop? • Why do British people never sound British when they sing?
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How many cyclists can legaly ride abrest? ;)
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Originally Posted by JoshFrank
How many cyclists can legaly ride abrest? ;)
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oops, to quick with my bad pun...
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damn, the quick question thread works.
What is the geometrical meaning of the central extension of the algebra of diffeomorphisms of a circle? |
of the whole group, or for a given circle?
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Originally Posted by dolface
of the whole group, or for a given circle?
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ok, i haven't looked at the problem in a long time, so all i've got is a general explanation (you can plug in the numbers for whatever circle you have).
(if you don't care about the math [which sorta makes it pointless] this is a very basic description: there is a lie algebra called the virasoro algebra; its almost just the lie algebra of the group of diffeomorphisms of the circle, but it's actually just one dimension bigger, being a "central extension" thereof; projective representations of the lie algebra of the group of diffeomorphisms of the circle correspond to honest representations of the virasoro algebra.) i'm not gonna re-type my notes, but this post (from google groups) covers it pretty well: "In 2D the conformal group is infinite-dimensional - it consists of transformations which are analytic in z = x+iy and z* = x-iy. If we concentrate on the former, we see that an infinitesimal conformal transformation is generated by L_m = z^{m+1} d/dz, and they obey the algebra [L_m, L_n] = (n-m) L_{m+n}. However, it is easy to see that this is the algebra of vector fields (= infinitesimal diffeomorphisms) on the circle, vect(1). If the circle coordinate is x, the generators are L_m = -i exp(imx) d/dx. vect(1) has a central extension, known as the Virasoro algebra: [L_m, L_n] = (n-m) L_{m+n} - c/12 (m^3 - m) delta_{m+n,0}, where c is a c-number known as the central charge or conformal anomaly. This means that the Virasoro algebra is still a Lie algebra - anti-symmetry and the Jacobi identities still hold. The term linear in m is unimportant, because it can be removed by a redefinition of L_0. The m^3 term is a genuine quantum effect, which simply is there when you quantize a string. When string theorists criticize Thiemann's LQG string, they are basically complaining that he does not get this term, which simply must be there. There is some confusion about anomaly freedom here, because at the end people want to eliminate the conformal symmetry. The nice way to do this is to introduce a ghost pair b_m, c_n, satisfying fermionic brackets { b_m, c_n } = delta_{m+n,0}. One can now write down the BRST operator, which looks something like (double dots denote normal ordering) Q = sum_m :L_{-m} c_m: + 1/2 sum_{m,n} (m-n) :b_{-m-n} c_m c_n: If the BRST operator is nilpotent, Q^2 = 0, we can identify physical states with BRST cohomology. A state is physical if it is BRST closed, Q|phys> = 0, and two states are equivalent if they differ by a BRST exact term, |phys> ~ |phys'> if Q( |phys> - |phys'> ) = 0. It turns out that the ghost has central charge c = -26, so the BRST operator is nilpotent if the L_m's have c = 26; this is where the 26 dimensions of they bosonic string comes from. However, the important thing from my viewpoint is not that the end result is anomaly free, but that an anomaly exists, even if only in intermediate calculations. Thiemann does not have an anomaly even intermediately, and from this string theorists (and myself) conclude that LQG is wrong. Let us now return to the math. A lowest-weight representation (LWR) of the Virasoro algebra is characterized by a lowest-weight state |h,c> satisfying L_0 |h,c> = h |h,c>, L_{-m} |h,c> = 0, for all -m < 0. It is known that the only unitary LWR of vect(1) is the trivial one. However, the Virasoro algebra has many unitary LWRs: the discrete series with 0 <= c < 1, where c = 1 - 6/m(m+1), m positive integer and h = h_{p,q}(c) = (pm^2 - q(m+1)^2) / 4m(m+1) (or something similar, I'm quoting from memory), and also all c > 1, h > 0. Anyway, the important thing is that the only acceptable value of (h,c) with c = 0 is h = 0 - this is the trivial representation. That Thiemann obtains a non-trivial unitary representation of the 1D diffeomorphism group with c = 0 is thus very strange. It is hard to see how that could be compatible with quantum theory. The Virasoro algebra can be generalized to several dimensions - an extension of the diffeomorphism algebra on the N-dimensional torus, say. The generators are L_k(m) = i exp(im.x) d/dx^k, where m = (m_i) and m.x = m_i x^i and I use the summation convention. The algebra depends on two parameters (abelian charges) c_1 and c_2, [L_i(m), L_j(n)] = n_i L_j(m+n) - m_j L_i(m+n) + (c_1 m_j n_i + c_2 m_i n_j) m_k S^k(m+n), [L_i(m), S^j(n)] = n_i S^j(m+n) + delta^j_i m_k S^k(m+n), [S^i(m), S^j(n)] = 0, m_k S^k(m) = 0. This is the Virasoro algebra in 1D because the last condition then becomes m S(m) = 0, which only has the solution S(m) ~ delta(m). The Virasoro extension is not central (does not commute with everything) except in 1D. It is straightforward but somewhat tedious to check that these relations indeed do define a Lie algebra. Just as the Virasoro algebra is anomalous when c != 0, its higher- dimensional sibling is anomalous unless both c_1 = c_2 = 0. And just as for the Virasoro algebra, there is no non-trivial, unitary LWRs unless the algebra is anomalous. The correct definition of lowest-weight is more subtle in several dimensions. Let me just say that the right definition does not introduce any anisotropy." :D |
I <heart> dolface.
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*point - dolface*
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dol, that made me nauseous.
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Originally Posted by steaktaco
dol, that made me nauseous.
i need another beer |
dolface, if you didnt copy/paste that, you need to get out on your bike ;)
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