@0celo7 Sure - you just take the preimage under the covering map of the disk you're cutting out from $M$, and that gives you two disjoint disks in $\bar{M}$ you can use to do $\#T\#T$, no?
@ACuriousMind For reference, I am studying manifolds of the form $M\# T^n$ where $M$ is orientable. If it isn't, then I can do that unfolding to work with $(\tilde M\# T^n)\# T^n$ instead and it works just as well.
$M\# T^n$ is a natural compactification for AF spaces
@Semiclassical So I've managed to wrap my head around the positive mass theorem literature. I'm about to offer to give more talks next semester. Is that a mistake?
allo... just wonderin’ if the chat community could identify subfields of physics where the majority of submissions for peer review are NOT in LaTeX. I mean: are there subfields of physics where the bulk of manuscripts are submitted in MSWord for instance? Anecdotally LaTeX is completely dominant in theoretical physics, and a simple survey of arXiv submissions suggests most of these are in LaTeX... but there might be exceptions.
I asked this question Zero Radiation Nuclear Bomb Possible? and a moderator completely understood my question, provided an answer and mention he would like to see other answers, but it still got put on hold for being unclear after he posted. How is this? I believe that it was too quickly dismis...
After three years I returned to stackexchange to ask one question...Because it was not recieved well, I am now banned from asking with this account. Really ? Three years on and there is no room for one honest question? I have had a few poorly recieved questions in the past but I have also had a ...
If you're wondering if it's normal to be fascinated by abstruse maths then then the answer is clearly no, but by that measure we're all too many standard deviations from the mean
The sad reality is that people don't do it on their own. It's not that we're all bastards, but that there is always something more important to spend our money on closer to home.
It’s all about the initiation of violence. And in this case, net neutrality is violence against the ISPs, whereas the converse is...no one really knows.
The ISPs are just companies. They take money from us, spend some of on running their business and keep the remainder as profit. Net neutrality isn't doing violence against the ISPs, but you can argue it raises the charges the ISPs levy on us.
There is a kind of purity to libertarian theorizing that is very appealing—I desperately want to believe in it myself and think that more than a few thing in the way the US is organized could be improved by a applying a more libertarian approach.
And of course real working governances violate the deepest core ideas at some level. (Or like most of the real life pockets of anarchy that do or have existed in place they are embedded in a polty with such a governace and get some support services from it).
@JohnRennie The theororizers like to imagine a social structure that reinforces the good tendencies and thereby controls the ill. See Vinge's "The Ungoverned" for instance.
But at best we don't have examples of those structures that work at scales bigger than a few time the human monkey group size.
Dunbar's number is a suggested cognitive limit to the number of people with whom one can maintain stable social relationships—relationships in which an individual knows who each person is and how each person relates to every other person. This number was first proposed in the 1990s by British anthropologist Robin Dunbar, who found a correlation between primate brain size and average social group size. By using the average human brain size and extrapolating from the results of primates, he proposed that humans can comfortably maintain only 150 stable relationships. Dunbar explained it informally...
Vinge, Heinlein et al have proposed lots of social arrangements that would work brilliantly if all humans were like Vinge/Heinlein's protagonists. But we aren't.
@dmckee the sleeping part. Who is sleeping well? You or She?
Anonymous
@0celo7 For the sequence space $s$, if $|\xi_i-\eta_i|$ blows up as $i\to\infty$, does the metric $d=\sum_i\frac{1}{2^{i}}(|\xi_i-\eta_i|)/(1+|\xi_i-\eta_i|)$ still remain defined (always)? I mean the limit as $i\to\infty$, could be infinite (for $|\xi_i-\eta_i|$), and we'd then get a $\frac{\infty}{\infty}$ for the distance function's terms (ending terms)...which may either be finite or undefined. Is there something more to the definition of the metric (which is implicit) ?
I managed to spill some epoxy hardener while repairing the laptop I bought yesterday, and it's horrible sticky stuff. So to clean it up I tried (in increasing order of desperation): - alcohol - white spirit - trichloroethylene and nothing worked.
Then I do what I should have done at the outset and Google for how to clean epoxy hardener and ...
... it turns out it's water soluble. Warm water does the trick! Doh :-)
@Phase I was going to use it for experiments with upgrading the CPU. I had thouht it would be a bit battered so ideal as a testbed, but now I find it's in very nice condition. Now I'm not sure whether to use it for testing or not.
I'll probably use it as my main workstation for a couple of weeks just to convince myself it's really all perfect, then decide what to do.
@ACuriousMind what I m confused about is, in that example, where essentially we re dealing with an explicitly time dependent hamiltonian, it seems that we lose the equivalence between heisenberg and schrodinger picture, and as a result the "interaction picture" becomes a necessity to remedy the situation. Is that at all true? or is there still a way to show the equivalence of H-SCH pictures that holds for time dependent and independent cases?
@user929304 No, we don't lose equivalence. They're still unitarily equivalent by $H_H \leftrightarrow U^\dagger H_SU$ at each point in time.
I think you're confused about what equivalence of the pictures means, but I can't quite tell what you think it means so I don't know how to convince you of the correct notion ;)
@ACuriousMind well i understood equivalence in practical terms, that either theory, when it comes to predict average values of observables, they would turn out in agreement with experiment, so it would not matter which picture I used to express that average, the result will be the same
$\langle \psi_H\vert A_H\vert \psi_H\rangle = \langle \psi_S\vert A_S\vert \psi_S\rangle = \langle \psi_I \vert A_I\vert \psi_I\rangle$ for all states $\psi$ and all operators $A$.
I just gave the Fiend lady needle and thread, but now all the doors seem to be locked and I cant find any body with the number 42 which is my only lead
@ACuriousMind makes sense. was my gut feeling as well. Do you happen to know of nice texts (paper or book) where this equivalence is carefully explained? (among the main 3 pictures of QM)
@user929304 I'm not sure what there is to explain that would justify a text devoting much time to it. At the level of ordinary QM, you're just taking $\langle \psi\vert U^\dagger A U\vert \psi \rangle$ and defining $\psi_{H/S/I}$ and $A_{H/S/I}$ differently, but the actual value of the expression never changes.
Switching between the pictures is basically just a time-dependent basis change of your Hilbert space by the unitary time evolution operator(s)
The only potential problem, which is the problem that essentially leads to Haag's theorem in QFT, is that this unitary basis change may not respect the commutation relations of the algebra of observables, but for ordinary QM, the Stone-von Neumann theorem guarantees that nothing can happen at least for the usual algebra of $x^i$s and $p_i$s
@ACuriousMind in retrospect, was what Zachos saying in that post true then? I mean he seemed to suggest "we lose the equivalence" in the time-dependent case
@Slereah Yes, the SvN theorem really only works for the exponentiated/bounded version of the commutation relations, but I don't think that's a good detail to go into right now ;P
@user929304 I don't see where he would suggest that
@ACuriousMind "The crucial point is that, as you implicitly noticed, the Heisenberg Hamiltonian is not the Schroedinger Hamiltonian, by contrast to the time-independent case"
@user929304 Time dependence means that the Hamiltonian no longer commutes with the time evolution $U$, since $U$ is no longer a simple exponential of $H$. In the time-independent case, they commute, so $H_H = U^\dagger H_S U = U^\dagger U H_S = H_S$, which is not valid in the time-dependent case
@ACuriousMind ah indeed, sorry silly me :) many thanks for your patient explanations and your time, I ll carefully go through these again. My knowledge of QM seems completely shaken at the moment now that I ve learned H and U don't always commute...
Baldur's Gate's first few maps are essentially an exercise in reloading unless you start abusing the AI e.g. by kiting with one character while shooting at the enemy with the other