/sci/ - Science & Math

>>11407916
Regular algebraic functions on [math]\mathbb{R}[/math] admit an analytic continuation to [math]\mathbb{C}[/math]. However, to do this we must leverage the holomorphicity (Laurent regularity) in a neighborhood, which requires us to consider them as [math]\mathbb{C}[/math]-valued. In other words, if you want to include complex parameters in your domain by analytic continuation, then you have to make your image [math]\mathbb{C}[/math] as well.
>>11404064
Consider "fruitfulness" as a map [math]f: X\rightarrow B\text{Fruits}[/math] then endow on each n-cobordism [math]M\in {\bf Bord}_n[/math] the tangent structure [math]f^*\xi[/math], where [math]\xi:E\text{Fruits}\rightarrow B\text{Fruits}[/math] is the classifying bundle. In this way we form [math]{\bf Bord}_n^{f}[/math] and we can gauge the theory [math]Z_f:{\bf Bord}_n^f \rightarrow {\bf Vect}_\mathbb{C}[/math] by averaging [math]Z(M) = \sum_{f\in [X,B\text{Fruits}]}\frac{1}{|\text{Fruits}|}Z_f(M)[/math].
>>11416800
Generators form the basis of the Lie algebra. Whenever you take [math]\exp[/math] you're moving to the (connected component) of the Lie group. By interpreting these Lie groups as "physical symmetries", the Lie algebra generators are the integrals of motion. The Hamiltonian [math]H[/math] is the total energy and [math]{\bf J}[/math] are the angular momenta.

>>11407152
If that's even a concern are you really trans in the first place?

>>11406058
In relativistic QFT, observables forms the even subalgebra of the local operator net of fields satisfying C/ACR. Because of this non-commtativity, there is no notion of "observing particle at a point" due to the usual uncertainty principle. What you [math]can[/math] measure, however, is scattering processes and the propagation of particles; this is given by [math]\langle G(x,y)\rangle[/math] where [math]G = \square^{-1}[/math] is the (renormalized) Green function/inverse Minkowski Laplacian.
Now in relativistic QFT, [math]G[/math] and its expectation are Lorentz invariant, since [math]\square[/math] is. What this means is that you observe the same scattering amplitude [math]\langle G(x,y)\rangle = \langle G(\Lambda x,\Lambda y)\rangle[/math] no matter what frame you're in, for every [math]\Lambda\in \mathbb{P}(1,3)[/math] (including boosts and rotations). The particle doesn't "know" when it's being measured, it just "knows" when and how it propagates from [math]x[/math] to [math]y[/math], and this process, which is what is actually observed, is by definition local and Lorentz invariant.
This is the whole basis behind Bohm's philosophical idea that processes are more fundamental than the states themselves, btw. Read more in his book "Wholeness and the Implicate Order".

>>11378793
>I was told it's entirely fine to not have anything very specific;
Yeah I've heard that too, but I put my interest (TQFT) and my desired direction of research in my application anyways and that got me into a math PhD at PI onmy first pass despite not having all A's on my transcript and only applied to 5 schools.
>undergrad is not expected to know enough to have a field of specialization in mind
That might be true for the average applicant, but they're not looking for average applicants. They're looking for ones that are exceptional.
>I mentioned how I found functional analysis, spectral theory, cohomology, homological algebra and category theory to be very interesting topics
See, to me this just seems like you have no direction in terms of research, unless you have interest in something that unifies them such as generalized Atiyah-Singer or some novel application of Lusternik-Schnirelmann in analysis (but that's extremely ambitious), in which case you should mention that in your app anyway. My undergrad background is in theoretical physics and the profs all emphasized to me that having a clear direction in my application is tantamount; I doubt that plays a minor role in math.
>insert specialty of school I'm applying to
You should target the profs, not the institution. And believe me, they can tell that you copypaste your letters of intent.
Personally I did thorough research of the profs at the institutions I was looking at, specifically their pubs and thought about how I can add to their work, both for my UG [math]\rightarrow[/math] MSc and MSc [math]\xrightarrow[\text{physics}\rightarrow \text{math}]{}[/math] PhD transitions and completely rewrote my apps accordingly. This can definitely go a long way.
Try applied math too, they have more capacity and money to go around.

>>11327783
>1
In general given a measure-preserving ergodic map [math]T\in\operatorname{End}(\Omega,\mu)[/math], Koopman's lemma says that the unitary [math]U_T = T^* \in \mathcal{B}(L^2(\Omega,\mu))[/math] has [math]1\in\sigma(U_T)[/math] with multiplicity 1; namely [math]\operatorname{dim}\operatorname{ker}(U_T - 1)=1[/math]. Now ergodicity, by deifinition, also implies that [math]\{T^n x\mid x\in\Omega\}\subset\Omega[/math] is dense for some [math]n\in\mathbb{N}[/math], whence at some point in the application of [math]U_T^n[/math] you will hit the one-dimensional eigenspace and stay there; convergence in finite time occurs because the singular dimensionality means that there is essentially "no where for the image to go". See also Poincare recursion.
>2
I presume you're referring to the time evolution operator [math]\alpha_t[/math] generated by the complete SA Hamiltonian vector field [math]X_H[/math]. By the criterion above, [math]\alpha_t[/math] is ergodic when the induced map [math]U_\alpha(t)\in \mathcal{B}(L^2(\Omega,\mu))[/math] has a simple eigenvalue 1. Hence we examine [math](U_\alpha(t)-1)\psi = 0[/math] whence by Stone's theorem implies [math]\psi\in\operatorname{ker}X_H[/math]. If [math]X_H[/math] is elliptic, then the existence of [math]\psi[/math] can be proven with the spectral theorem. By Fredholm alternative, either this [math]\psi[/math] is unique or [math]X_H^{-1}\in\mathcal{B}(L^2(\Omega,\mu))[/math] exists, but typically the Green function [math]G = X_H^{-1}[/math] isn't bounded.
>>11328853
That's not how it works hun.

>>11319215
First of all, photons are bosons with spin-0 and classically obey the Klein-Gordon equation [math]\square A = 0[/math], while quarks are fermions with spin-1/2 obeying Dirac equation [math](\not\partial -im)\psi = 0[/math]. There is no concept of "flavour" for photons, or any other gauge fields for that matter.
Second of all, massless bosons exhibit full internal gauge symmetry while massless fermions only exhibit a [math]U(1)[/math] axial symmetry. Mass terms gap out the spectrum and leads to cross terms [math]m^2\psi_f^\dagger \gamma^5 \psi_{f'}[/math] that contribute to scattering between quarks of different flavours. No such thing occurs for photons; pair creation does not require a spectrum-gapping mass term to occur,. It's like the first vertex [math]\psi^\dagger \not A \psi[/math] you see in QED.