/sci/ - Science & Math

>>12557410
Instantons are classified by the cohomology of certain sheaves on the moduli space of fields in the field theory.
>>12557436
>It's not at all clear why
It is clear. It's because these models are integrable and admit Lax pairs as operators on a Hilbert [math]D[/math]-module of a nCFT, which satisfy certain quantum flatness conditions. You can then begin to point-count stable curves in its moduli space via Gromov-Witten/quantum cohomology and reproduce the correlators/conserved quantities of your system.
>>12557970
I am now doing holography and HMS.

>>12507565
>Anybody here like them?
Sure.

>>11564366
You don't read algebra books anymore? What if you forgot something?

I'm using Aluffi, while reading some parts of Herstein, Aluffi has such a great and modern notation for everything, it's also so clear and he usually puts some little jokes in the book that makes me smile from time to time, heh.

>>11519032
Remember that the Bessels form an ONB on [math]L^2_r((0,\infty),rdr)[/math], namely [math]\langle J_n,J_m\rangle_r = \int_0^\infty dr rJ_m(r)J_n(n) = \delta_{nm}[/math] (up to normalization).

>>11503293
It sounds like you get an integrable tangent subbundle that you can polarize against, which is central in geometric quantization; it allows us to quantize a prequantum bundle. This is why a classification of Poisson structures is so sought after.
Perhaps this article https://aip.scitation.org/doi/abs/10.1063/1.529446 would be of interest to you.
>>11503409
中出しOKだし
>>11503838
Using Brown representability I can name as many as [math]|\pi_0 {\bf hSpec}|[/math] lol.

>>11490975
You're quite welcome sweetie.

>>11452402
>>11452419
Yep, it's pretty standard so it probably just doesn't have a name.
>the classic proof went something like this
That looks like the right idea, which is to identify the classifying space [math]BO(n)[/math] as the real flag Grassmannian [math]\operatorname{Gr}_n(\mathbb{R})[/math]. Each collection of [math]n+1[/math] non-planar points defines an [math]n[/math]-plane up to a frame, hence an element in [math]\operatorname{Gr}_n(\mathbb{R})[/math].
Also any non-symmorphicity (preventing the splitting into a semi-direct product) of the lattice is killed off in the continuum.

>>11426214
Conjugation is an inner automorphism. Specifically you need to prove that [math]f:G\rightarrow\mathbb{C}[/math] mapping [math]g \mapsto |g|[/math] is a class function, which doesn't follow as trivially as your post suggests.
>>11426564
Sure we can. QED proved that the fine-structure constant [math]\alpha = \frac{1}{137}[/math].

>>11419050
Yes. In general the non-linear Cauchy problem [eqn]\begin{cases} \Delta u = f(u,\cdot); & \text{in }\Omega \\ au+ b(n\cdot \nabla u) = 0; & \text{on } \partial\Omega\end{cases}[/eqn] is well-posed for [math]u\in H^1(\Omega)[/math] for [math]C^1[/math] boundary [math]\partial\Omega[/math], meaning that it has the usual Chauchy criterion for existence and uniqueness of solutions, at least for sufficiently small times (depending on how nice [math]f[/math] is; being Caratheodory is enough), as long as you propagate your boundary value from somewhere other than a hyperbolic critical point of the submanifold [math]N\subset H^1(\Omega)[/math] of functions satisfying the Cauchy boundary conditions.
Now "unique" here means that the tangent space has codimension one: this is a [math]functional[/math] condition, not a numerical one, in the sense that if you find two solutions [math]u,v[/math] solving the same problem then you have [math]v \in \operatorname{Span}_{H^1}u[/math], not [math]u=v[/math]. In the case of the harmonics, notice that the general solution [math]e^{ix} + e^{-ix}[/math] is a linear combination of [math]\sin x[/math] and [math]\cos x[/math].
>>11419118
[math]\sin(3x)[/math] satisfies [math]y'' + 9y = 0[/math] sweetie.
>>11419975
Energy is an eigenvalue of the Hamiltonian [math]H[/math]. Typically [math]H = K+V[/math] where [math]K[/math] is the kinetic energy and [math]V[/math] is the interaction; if you look at [math]fixed~V[/math]-slices of the energy landscape and find a local minimum at some slice, then that means the kinetic energy [math]K[/math] is minimized. [math]K[/math] is typically a symmetric bilinear form in the momenta [math]q[/math] (if there is [math]\operatorname{Isom}(X)[/math] symmetry on [math]H[/math]) so it means slower objects, but this is a classical description that is not exactly accurate. Better way is to think about the dispersion relation and the "group velocity" [math]\nabla_q K[/math].

>>11402053
Then I'm afraid your only option is to try to draw a bunch of graphs.
>>11402198
Depending on your function space, differential (as well as integral) operators can be represented as infinite matrices by passing to an ONB, which you can get from either Sturm-Liouville or [math]N[/math]-representation. For instance [math]\partial_x \cong [\delta_{n,n+1}]_{n\in\mathbb{N}}[/math] in the Fourier basis. Notice that these matrices are infinite dimensional, as are most Banach function spaces.
Cramer's rule and Gaussian elimination work for finite-dimensional matrices, but given sufficient regularity/invertibility (via Fredholm alternative, for instance) you can construct solutions via Cramer's rule by just plugging in the matrix entries. Gaussian elimination requires you to manipulate the rows/columns themselves, however, which you of course cannot do infinitely many times. Generally you cannot use induction (i.e. finitely many test cases) to prove convergence statements.

>>11378647
Since [math]SU(2)[/math] universally doubly covers [math]SO(3)[/math], their Lie algebras are isomorphic, so orbital algebra is the same as spin algebra. There's a difference in using the rank [math]l[/math] of the irrep as labels or the spin [math]j[/math], which are related by [math]l = 2j+1[/math]. In other words, spin-3 irreps have rank-7 Pauli matrices and I don't think anyone would ever bother writing them out explicitly. Besides, the decomposition of [math]{\bf 3}\otimes{\bf 2}[/math] literally contains every irrep up to [math]{\bf 5}[/math], as you have noticed, which makes computing the Clebsch-Gordans even more bothersome. It's reasonable that you wouldn't find any table for them because listing them all would take at least 5 pages.
>I am not sure if such manipulations are valid
Yes they are, as the decomposition is associative. Affine semisimple Lie's have representations rings that form Verlinde algebras over [math]\mathbb{C}[/math], which is unital and associative equipped with a fusion rule.
>one with l=3 and one with l=2
As the SCOP [math]\Delta[/math] itself decomposes into spherical harmonics on the Fermi surface, you can just start with first few terms of the direct sum since higher-wave contributions are suppressed. But in general [math]s,d,\cdots[/math]-waves are even while [math]p,f,\cdots[/math]-waves are odd.
If you really want every spin-wave contribution then, as you've said, you have to compute the CB's yourself.
Been a while since I did SC stuff but I'm sure there's a better way.

>>11377660
Your idea for the continuity of both [math]f,g[/math] being necessary for the continuity of [math]fg[/math] is correct. You essentially picked a random [math]a[/math] and applied the limit [math]x\rightarrow a[/math] (which you can pass into the function by continuity) to show that the multiple of the limit can exist only if both factors exist. The same can be done for the sum [math]f+g[/math].
Now just note that the indicator [math]\chi_W:X\rightarrow \mathbb{R}[/math] on any closed [math]W\subset X[/math] cannot be continuous. The reason is that the Hausdorffness of [math]\mathbb{R}[/math] guarantees the existence of an open [math]U\subset \mathbb{R}[/math] with [math]1\in U, 0\not\in U[/math], which has closed preimage [math]\chi_W^{-1}U = W[/math], contradicting the definition of continuity. With [math]\mathbb{Q}[/math] and [math]X=\mathbb{R}[/math] proves the discontinuity of [math]\chi_\mathbb{Q}[/math]

>>11355314
You flatter me sweetie, I just happen to also be doing a PhD in TQFT/topological ordering. For the general theory behind the classification problem, I personally recommend Bernevig's text on TI/TSCs and his set of notes https://arxiv.org/abs/1506.05805..

>>11331120
[math]u_n^2 - 2u_n +1 < \frac{1}{n} \iff (u_n-1)^2 < \frac{1}{n} \iff |u_n| < \frac{1}{\sqrt{n}}+1 \rightarrow 1[/math].

>>11316153
>copypaste their results
Thank you for subscribing to the Journal of Mathematical Physics, sweetie!

>>11313998
https://en.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem
>>11314610
I'm so proud of you for using Demos honey.

>>11307708
>functional analysis
Those are general operator algebra theory considerations. If you're on an infinite lattice, your Hilbert space [math]\mathcal{H} = \bigotimes_n \mathcal{H}_n[/math] is still infinite dimensional, and the domain problems crop up as soon as non-local things start propagating, like correlations in the long-range entangled phase. Not to mention difficulties with the continuum limit if you want an effective field theory.
In any case, your problem has been sufficiently answered I hope? Good luck out there.

>>11303098
Yep. My research heavily intersects his and Hopkins's result on the classification of invertible TOs so I know quite a bit about him.

>>11287930
Yes. If your system has a compact symmetry [math]G[/math] such that [math]A\mapsto A^g = g^{-1}Ag = A[/math] for all [math]g\in G[/math] then you can decompose [math]A \in \operatorname{Hom}_\mathbb{R}(V,W) \cong V^*\otimes W[/math] into diagonal representation blocks [math]A= \bigoplus_\pi A_\pi[/math], where [math]A_\pi = P_\pi AP_\pi[/math] transforms under the irrep [math]\pi[/math]. Given the orthogonal projection [math]P_\pi[/math] onto the subspace [math]V_\pi \subset V[/math] for which [math]A_\pi[/math] acts irreducibly, we can then first solve the partial problem [math]A_\pi x_\pi = P_\pi b[/math] for each [math]\pi[/math] then combine them to obtain the solution [math]x = \bigoplus_\pi x_\pi[/math] to the original equation. If [math]G[/math] is finite then you only have finitely many partial problems to solve.

>>11267336
I'll use this one since >>11267326 is here.

>>11266575
Read Nakahara.

>>11261160
Yeah that I completely agree with.

>>11245932
The objects are matrices themselves, with the semisimples being elements of the the natural basis [math]E_i[/math]. It's a special spherical 2-groupoid and the simplest realization of a SMFC.

>>11183933
No. If we let [math]\ddot{x} = F(x,\dot{x};t)[/math] where [math]F(x,\dot{x};t) = F_0(t) + {\bf a} \cdot \begin{pmatrix}x \\ \dot{x}\end{pmatrix}[/math] is a bilinear function (with the [math]\dot{x}[/math] term paying the role of drag) then [math]\frac{d}{dt}\begin{pmatrix}x \\ \dot{x}\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ a_1 & a_2 \end{pmatrix}\cdot\begin{pmatrix}x \\ \dot{x}\end{pmatrix} + \begin{pmatrix} 0 \\F_0(t)\end{pmatrix} \equiv M \cdot \begin{pmatrix}x\\ \dot{x}\end{pmatrix} + {\bf F}_0(t)[/math] then the Lyapunov exponents are determined by the eigenvalues of the matrix [math]M[/math], namely [math]a_1[/math]. In typical (linear) spring systems we have [math]a_1 = -k[/math], so this means an attractor appears near the fixed point [math]x^* = 0[/math].
Now we can generalize the above linearized argument to the general case where [math]{\bf a} = {\bf a}(t)[/math] is time-dependent. If the fixed point [math]x^* = 0[/math] is hyperbolic, then by Grobman-Hartman theorem the stability manifolds emanating from it in the full time-dependent problem is diffeomorphic to that of the linearized problem above. Hence the attractor is still there and solutions stay near each other.

Of course the situation can be very different if we consider general coupled springs, where [math]{\bf a}[/math] can now depend on [math]x,\dot{x}[/math]. This makes the problem very nonlinear and Lyapunov exponents can be super-log.