/sci/ - Science & Math

>>11431775
Dunno

>>11403748
>can you give an elementary proof that the only topological group structure on [math]S^1[/math] is angle sum?
Sure. [math]S^1[/math] is a smooth manifold so any continuous group structure is homotopically a smooth group structure, which makes [math]S^1[/math] into an Lie group [math]U(1)[/math]. It is 1D, compact, simple and Abelian, so its Lie algebra [math]\operatorname{Lie}U(1) \cong \mathbb{R}[/math].
>is there any intuition for a Poisson structure that doesn't appeal to symplectic geometry or physics?
Poisson algebras are in general infinite-dimensional, which is affine Lie only when you also have a compatible holomorphic structure. Because of its non-vanishing [math]H^2[/math], aside from the case on the torus any attempt at finding a full quantizations/*-reps into an actual infinite-dimensional Lie had been fruitless due to Groenewald-Van Hove.

>>11347448
If [math]V[/math] is an inner product space, that's true only when Riesz's lemma [math]V\cong V^*[/math] holds and [math]T[/math] is Ferdholm. An inner product lets us decompose the space into orthogonal complements such that [math]0\rightarrow \operatorname{im}T \rightarrow V \rightarrow\operatorname{ker}T\rightarrow 0[/math] is exact in [math]({\bf Vect},\oplus)[/math]. Now as rank-nullity [math]\operatorname{ker}T^*= \operatorname{im}T = \operatorname{coker}T[/math] holds for Fredholm [math]T[/math], we can identify [math]\operatorname{coker}T \cong \operatorname{ker}T[/math] by Riesz [math]T\psi \mapsto (T\psi)^*[/math].
>>11348066
In the non-interacting scenario, a smooth Bloch Hamiltonian [math]H(k)[/math] defines as its image a finite dimensional Hilbert space [math]\mathcal{H}_k = \operatorname{im}H(k)[/math]. The external part can be taken to be the space [math]l^2(\widehat{\Lambda})[/math] of square summable sequences on the BZ [math]\widehat{\Lambda}[/math] as the Pontrjagyn dual of the lattice [math]\Lambda[/math], whence [math]\mathcal{H} = \left[\bigoplus \mathcal{H}_{k\in\widehat{\Lambda}}\right] \otimes l^2(\widehat{\Lambda))[/math]. Fourier transform is certainly an isometry here, and you can write down whatever diagram you want (though that wouldn't be very useful).
But the point is that, because of the underlying [math]U(1)[/math] (charge) or [math]\mathbb{Z}_2[/math] (particle-hole) symmetry of TIs/TSCs, the Chern number [math]c_1 = 0[/math] vanishes if you take as fibre your entire space [math]\mathcal{H}[/math] for your Bloch bundle; the topological property is only exhibited by the [math]filled[/math] Bloch states below the Fermi level. Projection [math]P_\mu[/math] onto these states may not commute with the Fourier transform.

>>11332031
>what if the cener of mass is NOT where the thing is spinnig around
This cannot happen unless the Casimir [math]L^2[/math] is not an integral. Given the configuration space [math]Q[/math] as a homogeneous space under the Euclidean group [math]\mathbb{E}(3)[/math] of rigid body motions, the phase space [math]T^*Q[/math] can be identified with the adjoint Lie algebra [math]\mathfrak{e}^*(3)[/math] with a natural adjoint action of [math]\mathfrak{e}(3)[/math]. The generators of moments [math]p[/math] and angular moments [math]L[/math] span [math]\mathfrak{e}(3)[/math] with integrals [math]p^2,L\cdot p[/math] spanning [math]TS^2 \cong \mathbb{R}^2\times S^2\subset T^*Q[/math], on which the action of [math]\mathbb{E}(3) \cong \mathbb{R}^3 \times_{p^2,L\cdot p \text{ const.}} SO(3)[/math] decomposes. By modding out the translations we have a projection [math]TS^2 \xrightarrow{\pi} S^2[/math] with a distinguished point [math]p_0 \in S^2[/math] fixed by the [math]SO(3)[/math] action. The canonical projection [math]T^*Q\rightarrow Q[/math] then maps [math]p_0[/math] to the CoM.
>>11332401
Pair production is described by the QED scattering amplitude [math]\mathcal{M} = \int_M \operatorname{tr}(\psi^\dagger\not A^{-1}\psi)^2[/math]. Now QED is a [math]U(1)[/math] gauge theory on a spin 4-manifold [math]M[/math], where the spin [math]\operatorname{Spin}(1,3)[/math] and charge [math]U(1)[/math] symmetries constitutes the internal symmetries; since fermion bilinears of the form [math]\psi^\dagger\psi[/math] is invariant under [math]\operatorname{Spin}(1,3)[/math], and [math]\gamma A[/math] is invariant under [math]U(1)[/math], the pair production amplitude is not forbidden by symmetry. [math]\operatorname{tr}\psi^\dagger \not A^{-1}\psi[/math] (or anything involving a trace over an odd number of [math]\gamma[/math]'s) is forbidden by Furry's theorem, however, but its square isn't, so there is nothing in the theory that forces [math]\mathcal{M}=0[/math].

>>11186559
Brezis.
>>11186696
Isometries are unitaries, not just special unitaries.
>>11186876
The coordinate function in the second term is symmetric under [math]x\leftrightarrow y[/math].
>>11187041
Sure. Just linearize around hyperbolic fixed points.

>>11101626
Nope, you get notifs about LGBTQ+ events/conferences. That's about it really.

>>11067452
Adding the condition of local convexity, you can leverage the spectral theorem to produce an ONB. See e.g. [math]N[/math]-representation theorem for tempered distributions.

>>11007311
I'm Canadian. While the competition is not exactly fierce, there's only a handful of institutions that has a dedicated mathematical physics research group.