/sci/ - Science & Math

>>11436306
Because we need enough regularity in [math]F[/math] to make statements about and characterize its level sets, which constitute solutions to the DE. Generally [math]F[/math] is a functional derivative [math]\delta_u \mathcal{F}[v][/math] of an energy functional [math]S[u] = \int dx \mathcal{F}[u(x)][/math] and to optimize [math]S[/math] by enforcing the stationarity condition [math]\delta S = \int \delta \mathcal{F} = 0[/math] we need [math]\mathcal{F}[/math] to be at least Frechet differentiable.

>>11312681
Use Stone-Weierstrass. Sample some number of points of the graph and use polynomial interpolation.

>>11082568
You don't, because what you've said is nonsense.
>>11082569
Thank you again for the hard work.
>>11085261
For a discrete group [math]G[/math] we have [math]BG \cong K(G,1)[/math]. So if [math]\pi_1 G \cong G[/math] then [math]G \simeq BG[/math] weakly. In addition, as all higher homotopy groups are trivial for both, their suspension spectra are equivalent in [math]+{\bf hSpec}[/math]. As [math]G[/math] is CW, this implies by Whitehead's theorem that [math]G \simeq BG[/math], which is impossible unless [math]G[/math] is trivial.
>>11086472
Weird question because, as you've said, [math]\ln z[/math] has a branch-cut on [math]\mathbb{R}_{<0}[/math]. This puts you on different branches of [math]\ln z[/math] whenever the neighborhood intersects [math]\mathbb{R}_{<0}[/math].
If you notice that [math]\ln z[/math] is a conformal map taking the punctured plane to the cylinder, however, you can define the Cauchy principal value by [math]forcing[/math] the "regularized" [math]\ln z[/math] to cancel the [math]2\pi[/math] jump across [math]\mathbb{R}_{<0}[/math] whenever it changes the branch.
>>11087473
Try understanding this for free groups first. The unique homomorphism basically tells you that there is essentially only one way [math]F[/math] is "embedded" in [math]M[/math].
>>11088534
A group action is a map [math]G\times A\rightarrow A[/math], so for [math]A = \emptyset[/math] we have the action is a map [math]G\times \emptyset = \emptyset \rightarrow \emptyset[/math], which actually exists (at least if you accept that the category [math]{\bf Set}[/math] has an initial object).
>>11098363
Spins are irreps of [math]SU(2)[/math].

>>11065652
Sure it does, especially in the QFT context where you Legendre transform the interaction vertex to get the quantum action. This requires a symplectic (or even Poisson) structure on the infinite-dimensional jet bundle.

>>10959458
https://en.wikipedia.org/wiki/Exterior_algebra#Linear_algebra

>>10899024
>>10899067
The intensive/extensive variables form symplectic coordinates in phase space with the thermodynamic potential giving rise to a contact structure. Legendre transforms then constitute contactomorphisms that preserve the Gibbs contact form.
https://arxiv.org/abs/physics/0604164

>>10898716
https://arxiv.org/abs/1511.00063

>>10836423
It seems that you're confused. The so-called measurement "problem" has nothing to do with whether if the duality transformation is unitary or not.
Perhaps it would benefit you to read Griffiths as a first primer into quantum mechanics.

>>10783333
https://ncatlab.org/nlab/show/persistent+homology

>>10735736
>superalgebra
No. It's graded over [math]\mathbb{Z}_2[/math], not graded by [math]\mathbb{Z}_2[/math].

>>10713657
Here: https://arxiv.org/abs/0707.1889, https://arxiv.org/abs/1303.1202..
In general the choice of fusion rules in your CFT for the anyons determines the semisimples objects in the (multi-)metaplectic modular category, the properties of which is related to classification problems in TQFT via the Levin-Wen construction. This can produce more powerful/novel invariants for 2D TQFT https://arxiv.org/abs/1702.07113 and generalizations to consider defects in 3D Reshetikhin-Turaev state sum TQFTs https://arxiv.org/abs/1710.10214..
The moral of the story is that these physical applications tell the math where to go, and can lead to powerful techniques that'd be otherwise unthought of by mathematicians. As my previous supervisor liked to say: "mathematicians work on problems that have been solved by physicists 20 years ago."

>>10344338
Notice [math]d(f,g) = \int_{[0,1]}|f-g| [/math] defines a metric on [math]C^0([0,1]) = C[0,1][/math]. Think of [math]\mathcal{M}(f,\epsilon)[/math] as balls of radius [math]\epsilon[/math] centred at [math]f \in C^0([0,1])[/math].
>>10344451
The probability distribution is the Boltzmann factor [math]p(x,N) = e^{ -\beta (H(x)-\mu N)}[/math]. Let [math]X_N[/math] denote the configuration space of [math]N[/math] particles with sigma algebra [math]\Sigma_N[/math] defined by Borel subsets of [math]X_N[/math], then [math]p:\coprod_N \Sigma_N \rightarrow [0,1][/math] defines a probability distribution.
This leads to the fact that the [math]n[/math]-point correlation functions [math]\langle x_1,\dots,x_n\rangle[/math] can be written as [math]\frac{1}{Z}\operatorname{tr}(x_1\dots x_n e^{-\beta(H-\mu N)})[/math].
>>10345298
First of all QFT does not say any such thing. Gravity is the fluctuation of the metric about [math]\eta_\mu^\nu[/math]; it's basically the theory of principal [math]\operatorname{Diff}(M)[/math]-bundles on [math]M[/math]. Second of all the fact that matter are fields does not originate from gravity; matter fields have been studied ever since the inception of QFT in the 1920's as operator-valued distributions. It'd do you well to read an actual QFT book.
>>10346585
Write [math]|f|[/math] as [math]f^+ - f^-[/math] and you can show both parts are absolutely continuous by restricting to their disjoint supports. Then reconstruct [math]f[/math] as [math]f^+ + f^-[/math] and viola, you're done.

>>10197134
Draw the phase diagram and do linearization around hyperbolic fixed points using Hartman-Grobman theorem.
>>10197345
What you have done is essentially showed the Cauchy-Riemann equation for the complex function [math]F(z) = u(z) + iv(z)[/math] under the complexification [math]\mathbb{R}^2 \rightarrow \mathbb{C}, ~(x,y) \mapsto x+iz = z[/math]. This means that [math]F[/math] is holomorphic and hence Cauchy integral formula gives [eqn]\int_C dz F(z) = \frac{1}{2\pi i}\sum_{\operatorname{Res}z_i}F(z_i).[/eqn] Since [math]F[/math] has no poles the integral is 0. Alternatively you have also showed that the 1-form [math]\omega = {\bf F} \cdot d{\bf x} = d\sigma[/math] is exact. Stokes's theorem in [math]\mathbb{R}^2[/math] then says [eqn]\int_C d\omega = \int_{\partial C}\sigma = 0[/eqn] since [math]\partial C = \emptyset[/math].
>>10197659
The vector field being even means that the path integral takes the same value on the two semicircles [math]C_+[/math] so [math]\int_C d{\bf r}\cdot{\bf F} = 2\int_{C_+}d{\bf r}\cdot {\bf F} \neq 0[/math]. What you're thinking is when [math]{\bf F}[/math] is odd, which the vector field in question isn't.