@Mr.Feynman i think stuff is like in usual QFT when you map to the cylinder via the transformation $w = e^z$. While you work on the two punctured sphere it has some differences
oh but maybe you are already working on the cylinder since you speak of fourier modes
otherwise you would have coefficients of the Laurent series
@DIRAC1930 i guess its because those are the stuff you are most interested in. It is the same for me. also welcome back!
The analogy with the KG scalar product was not working because of the normalization choice in the sum. That $i$ factor ruined it
Basically, because of the $i$, I had to replace $$f\overset{\leftrightarrow}{\partial_0}g=f\partial_0g-g\partial_0f\mapsto f\partial_0g+g\partial_0f$$
Other than that the analogy works perfectly fine, just noting that the role of negative/positive energy modes of QFT is played by left/right movers here, so we take the scalar product with $\mathrm{e}^{im\sigma^\pm}$, depending on whether we want $\tilde{\alpha}_m$ or $\alpha_m$
I mean yeah technically the X are just massless scalars in 2d, it's more than an analogy it's actually a Klein Gordon field. Just you have way more symmetry due to the fact that you're in two dimensions and the field is massless
also ACM: I think I understand what you were helping with yesterday better now. I guess the two short exact sequences can be summarized as being consequences of (1) the identity component $G_0 \subseteq G$ is normal and (2) if $G$ is compact, then $G_0$ is also compact. Since it is a connected component, $G_0$ is also connected. And, there is a well known classification of compact, connected Lie groups.
i wonder why the notes put it that way :P i guess i will see hopefully
@SillyGoose for people used to the notation, writing "$1\to G_0 \to G \to \pi_0(G)\to 1$" is shorter and cleaner than spelling it all out, e.g. saying "We have an inclusion $G_0\to G$, a projection $G\to\pi_0(G)$ and $\pi_0(G) \cong G / G_0$"
you shouldn't suspect some deep reason why sequences are "better" here, it's just notation
@ACuriousMind in this claim by F&H is it implied that the $G_i$ are connected? I am trying to match the claim stated in F&H to the theorem seemingly applied in the chern-simons notes
@SillyGoose if the group you're decomposing is connected, then so are the $G_i$ - the discrete quotient cannot turn a disconnected Lie group into a connected one
you will almost never see a form and take its "differential" in the ordinary sense, to the degree that whoever wrote this didn't even consider this a potential confusion :P
also note that the differential of a 0-form $M\to \mathbb{R}$ is the same as its exterior derivative, so that's why one might also call the exterior derivative "differential"
@lucabtz as forms are sections $\omega : M\to T^\ast M$, you could take their differential in the ordinary sense of smooth maps, viz. $\mathrm{D}\omega : TM \to T(T^\ast M)$. I assumed that was the notion of differential the silly goose meant
I recommend to let go of this pedantry with respect to notation/terminology :P
when someone says that a form is $\mathfrak{g}$-valued, what they mean is what you understood as "$\theta_g$ being $\mathfrak{g}$-valued"
there is no potential for confusion here because there is no meaningful interpretation of the phrase "$\theta$ is $\mathfrak{g}$-valued" other than this
@SillyGoose My what is the argument at which we can make newton's law of gravitation (NLG) and newton's second law N2L are equal
Like we know N2L F = ma where F is the summation of all external forces. These external forces could be friction, drag etc.. how law of gravitation be external force
is it a nontrivial result that if an equality holds in local coordinates that it holds in general (i.e. when re-expressed in coordinate free langauge)?
I mean, it's not trivial, but it's a foundational part of differential geometry
it's part of why physicists - always working in coordinates - emphasize that all the things with indices have to "transform properly", since equations where one (or both) sides transform "improperly", i.e. not as tensors, will not hold in all coordinate systems just because they hold in one
hm i think im still not understanding $\theta$ conceptually. at a high level it is defined to take in a single element $g \in G$ and spit out a $\mathfrak{g}$-valued 1-form. so then i am trying to parse your written local coordinate form of (what seems to be the reasonable interpretation) $d\theta_g$. okay the $dx_i$ form a basis of the cotangent space (?) of $G$, so $d\theta$ eats a vector in the tangent space of $T_gG$ and spits out a vector in $\mathfrak{g}$?
Since the pilot wave would have to move through whatever slit the particle-component of the photon doesn't move through, then return to interact with the photon. All while the photon moves in a more-or-less straight line at c.
@Mr.Feynman that's one reason i am excited to work through these notes :D it seems to be a pointer towards relevant lie and bundle theory and differential geometry relevant for the physics i am interested in hehe
@SillyGoose Yes, that's how forms work - the $\mathrm{d}x^i$ eats a tangent vector and returns a number, and since the $\theta_i^a T^a$ in front of it is an element in $\mathfrak{g}$, the full result is something in $\mathfrak{g}$
Also, so if i wanted to prove that $\theta$ satisfies the maurer-cartan equation, would i prove that in local coordinates the equation holds for $\theta_g$ for all $g \in G$?
because sooner or later you need to have a firm grasp on how to convert beautiful abstraction into coordinate messes in order to connect any of the formalism to what you will see in examples and applications
while I don't like doing stuff in coordinates, it is very important in a practical sense
is $[\theta \wedge \theta]$ is literally just the wedge product and [] is solely for decoration? from your answer yesterday this seems to make sense, but i am wondering why there is a distinct wiki page dedicated to these $\mathfrak{g}$-valued forms: en.wikipedia.org/wiki/… and a section for their wedge product
well i guess there is no the wedge product it has to be specified
@Obliv log is intimately related to the trig functions, via the complex numbers. In fact, log is an inverse tan. For historical reasons, we normally start learning trig before we learn complex numbers, so we don't learn about these important connections until later.
The key equation is De Moivre's identity. $e^{i\theta}=\cos\theta+i\sin\theta$ The exponential function is well-behaved over the entire complex plane, so we can use that identity to extend the definitions of the trig functions to the complex plane. When we do that, we find that sin & cos are also well-behaved over the entire plane.
Let $(t+1)(x+1)=2$, so $t=(1-x)/(1 + x)$. Then it can be shown that $\ln(x) = 2i \tan^{-1}(it)$
So using complex arithmetic you can calculate logs using an arctan algorithm. Or vice versa.
In practice, if you're working with real values, you can do the computations using real numbers, you just have to tweak the algorithm slightly. Eg, take a look at the Taylor series for arctan(x) and for log((1-x)/(1+x))
Almost all of the discussion I've seen has focused on massive particles. But the double-slit works with both light and massive particles, so the same underpinnings should apply.
Also, weak measurements show Bohmian-like trajectories for light: doi:10.1126/science.1202218
@WaveInPlace standard Bohmian mechanics is not for relativistic particles, and even in standard relativistic QM the question of what constitutes a "wavefunction" in the positional sense for massless particles is a subtle and thorny issue
The double slit can be explained with the classical wave theory of light much easier than by talking about the wavefunctions of individual photons
@ACuriousMind is the second equality obtained by splitting the coefficients into $\frac{1}{2}$ of itself plus $\frac{1}{2}$ of itself and then doing some relabeling and using $dx^i \wedge dx^j = -dx^j \wedge dx^i$?
@ACuriousMind, it's more a question of validity, I guess. I understand that they are usually understood through different frameworks, but if there's no real difference in outcomes between relativistic and non-relativistic particles it seems like the theories that work for one should work for the other (with tweaks, of course).
@ACuriousMind okay last thing before i sleep :P how do i make sense of $(\partial_j \theta^a_i)$? it seems we would like to show that this expression is equal to $-\frac{1}{2}[\theta_i, \theta_j]$ and probably using one of the defining properties of $\theta$
$\partial_j\theta_i$ is contracted with an object antisymmetric in $i$ and $j$. What those lines imply is that $1/2[\theta_i,\theta_j]$ is the antisymmetric part
Hi everyone, does anyone has some good references for starting to study quantum spin liquids? I have found and red mainly review articles which are great for general knowledge but nothing that actually explains results and stuff.
to clarify for @SillyGoose: the spirit of @Mr.Feynman is saying is right, I was just confused with the factor of 1/2. The reason in more mathematical terms is that $dx^i \wedge dx^j$ are not all linearly independent but they are independent only for say $i > j$ so you can't conclude $(\partial_i \theta_j +\frac{1}{2}[\theta_i, \theta_j]) dx^i \wedge dx^j = 0 \implies (\partial_i \theta_j +\frac{1}{2}[\theta_i, \theta_j]) = 0$
you can though express the form in the basis $dx^i \otimes dx^j$ to get $(\partial_i \theta_j - \partial_j \theta_i + [\theta_i, \theta_j])dx^i \otimes dx^j = 0$ and thus $\partial_i \theta_j - \partial_j \theta_i + [\theta_i, \theta_j] = 0$
which is the correct formula for the vanishing of the curvature
however i think there should be a factor of 1/2 difference wrt what Feynman said