@BalarkaSen Please don't use offensive algebra notation, the modern geometric and politically correct term is fuzzy point (or point with order 1 fuzziness if you want to be very precise)
I just read a proof where a Ramsey like property is established by picking a colouring and then doing some magic with its orbit under a group action in the space of all colourings, my mind is blown
the nilradical is always nilpotent in an Artinian ring
that's what we agreed to above
since prime<=>maximal in an Artinian ring and maximals are comaximal, the product of the finitely many maximal ideals is the nilradical (and also the Jacobson radical)
What is this group law $(e^{-a},e^{-\frac{1}{a}}) ⭐️ (e^{-b},e^{-\frac{1}{b}})=(e^{-ab},e^{-\frac{1}{ab}})$ isomorphic to? I think it's isomorphic to the multiplicative group of real numbers?
The mulitplicative group of real numbers being denoted by $G=(\Bbb R^+, ⭐️)$
So if the above group law gives a way to add "points" on $xy=1,$ What's the correct way to express the addition of all points on $xy=1$ to all points on $xy=4$ for example?
Oh scratch that. I could write: $1/x \oplus 4/x=5/x$ which can be generalized to $a/x \oplus b/x=(a+b)/x$
so the former group law maps points "along" the curve, while this latter group law maps curves to other curves in the family, and the groups are isomorphic
Arclength of the boundary of $\Omega$ is the boundary measure so you wish to estimate $(\mu(B_\varepsilon(\Omega)) - \mu(\Omega))/\epsilon$ where $B_\epsilon(-)$ is the $\epsilon$-enlargement of the domain
I think the manifestation of Pythagoras is $\mu(A + B)^{1/2} \leq \mu(A)^{1/2} + \mu(B)^{1/2}$ for any two subsets $A, B \subset \Bbb R^2$
shouldn't the integral on the RHS of the equality here imgur.com/a/T7nMwmD be with respect to $E$? Since $\mu \restriction E$ is a measure on $\mathcal{F}_E$ which is a $\sigma$-algebra on $E$, not on $X$ necessarily?
I don't know, it's folklore to me. You can learn Ricci flows from Toppings' notes; I think you want to flow along the ODE $d\gamma_t/dt = -\kappa \mathbf{N}$ where $\kappa$ is the curvature
you have to show (1) every curve goes extinct in finite time under this flow (2) there are no neck pinches
then some calculus will tell you that asymptotically every curve becomes the round circle
@Astyx Note that there are two key ideas in the Fourier analytic proof, one is the Wirtinger inequality $\|f\|_{L^2} \leq \|f'\|_{L^2}$ which falls out of Plancharel's formula, and the other is the trick of replacing arclength by energy of a curve
Wirtinger is actually not a Fourier analytic fact, that's a 1D phenomenon. In general you have Poincare inequality, $\|f\|_{L^p} \leq C \|\nabla f\|_{L^p}$
and these are useful in highdim isoperineq
I accidentally figured out how to get it for $p = 2$ yesterday or something
the $\int \|y'\| = \int 1 = \int \|y'\|^2$ trick is wild though
I'll probably stick to the path-connectedness of $\mathrm{GL}(n,\mathbb{C})$. last thing I need to talk about is this fact relevance in the rest of math which... I don't know
You could try and show that there is a bijection between equivalence classes of irreducible admissible representations of $\operatorname{GL}_n(F)$ and equivalence classes of continuous Frobenius semisimple complex $n$-dimensional Weil-Deligne representations of the Weil group of $F$
@Thorgott I like "Verstehen Sie Deutsch?" ... "bisschen" ... "Sie sind Beschuldigter in einem Strafverfahren, ihnen wird Ladendiebstahl zur Last gelegt."
"Do you know algebra?" ... "a little" ... "There is a bijection between equivalence classes of irreducible admissible representations of GLn(F) and equivalence classes of continuous Frobenius semisimple complex n-dimensional Weil-Deligne representations of the Weil group of F"
