@AlessandroCodenotti To my understanding the idea here is to consider each point of $\text{Spec} A$ as a map $\{pt\} \to \text{Spec} A$. In other words, a morphism $\text{Spec} F \to \text{Spec} A$ for some field $F$.
Those morphisms are in 1-1 correspondence with homomorphisms $A \to F$ by contravariance of the $\text{Spec}$ functor
This is the so-called "functor of points" viewpoint
Ah, I see, so in this case they explicitly construct the morphism. To each point $\wp \in \text{Spec} A$ they associate the morphism $A \to \text{Frac}(A/\wp)$.
https://math.stackexchange.com/questions/2558352/finding-uniform-convergance-over-a-sigma-finite-space someone can help? i can prove the "lemma" i wrote there , not sure if it is helpful
@Alessandro I suppose you are right that the "one-to-one correspondence" needs to be rigorously stated here. If $A \to F_1$ and $A \to F_2$ are two morphisms with the same kernel $\varphi$, then $A/\varphi \to F_1$ and $A/\varphi \to F_2$ are both isomorphisms to the image.
So I suspect the codomain of the correspondence are not quite homomorphisms of $A$ to fields, but upto the equivalence defined by "two such homomorphisms are considered to be the same if they have isomorphic images"?
I know the other complex section had very easy psets and at least one person emailed the professor and essentially asked him to make it harder. Others didn't ask the professor but commented on it to me at various times
I don't think my students had the latter complaint too often, Mathei. But I think it's a reasonable complaint. Just remember that a lot of your fellow students are struggling even with the "too easy" ones.
I mean ours weren't particularly hard either but I had less contact with the other students in this class, the only one I know didn't talk about psets but mentioned that we could've gone faster
If it's framed not as complaining but more as a request, wouldn't that potentially be appropriate? It's just input, and the professor in question would ideally decide whether to go with it or not based on class composition
Yes, @anon, I have taught how to make associated bundles from representations. In fact, I commented on a diff geo question involving that just yesterday or the day before.
geometrically, $A_4\times_{\Bbb Z_3}(\triangle)$ should be "almost" a tetrahedron, and similarly $S_4\times_{\Bbb Z_4}(\square)$ is "almost" a cube, etc. I suspect there should be a way to identify boundary pieces in a way to actually get the platonic solids that way, maybe by treating the $n$-gons as CW complexes and choosing an arbitrary (?) identification of edges and then "translating" that identification with the group action
@EricSilva it's not just an analogy. More generally, if you have a ring extension $R \subset S$, then you can turn any $R$-module $M$ into an $S$-module by looking at $S \otimes_R M$ (I will spare you the more general thing in terms of bimodules). For $\Bbb R \subset \Bbb C$, this gives complexification and for $k[H] \subset k[G]$ this gives induced representations
Kind of a trivial question, can we infer that $H^0_{DR}( \emptyset) =0$ from the sequence $\Omega^{-1}( \emptyset) \rightarrow \Omega^0( \emptyset) \rightarrow \Omega^1( \emptyset) \doublerightarrow 0 \rightarrow \Omega^0( \emptyset) \rightarrow 0$? My arguemtn says yes, but I want to check
suppose you wanted to describe the local ordering of a ferromagnet. then you'd say that at each point inside the ferrogmagnet there's a local magnetization field which says what direction the magnetic field points
if you're at a high temperature, that local field will fluctuate enough that there's not a net magnetization. so a magnet in high temperatures doesn't actually give a useful magnetic field
and those rods have a tendency to line up parallel to one another. so at low temperatures you'll tend to have a well-defined direction for those little rods
...except, it doesn't matter if those rods point up or down. parallel/antiparallel doesn't matter
so if you want to talk about the local order in a 3D nematic, you'd take a vector field in 3D space and mod out by reflections
Suppose I had some large ball $\Gamma$ of radius $\epsilon$, $\Gamma = B_{(\mathbb{R}^n, d)}(0, \epsilon)$ cetered at the origin and I shrunk it to the unit ball by $h : \Gamma \setminus\{0\} \to \mathbb{B}^n$ defined by $h(x) = \frac{x}{||x||}$, does $h$ have an inverse?
@Perturbative Actually I misunderstood what you were doing. No your function doesn't ahve an inverse becasue you're actually sending every point to the Sphere, not the ball. If you divide by the amgnitude, everyone now has magnitude 1.
You're collapsing a whole ray to a point on the sphere
Would you guys say you were always talented in the field of mathematics, or did you have to excessively study it and be molded by it before you really understood it.
@Dragneel In my primary school years I found mathematics very easy to learn and do. So that sounds like talent. But I was also exceedingly careless in my math, and I had to train tha tout of myself.
"What I fight against most in some sense, [when talking to the public,] is the kind of message, for example as put out by the film Good Will Hunting, that there is something you're born with and either you have it or you don't. That's really not the experience of mathematicians. We all find it difficult, it's not that we're any different from someone who struggles with maths problems in third grade.
It's really the same process. We're just prepared to handle that struggle on a much larger scale and we've built up resistance to those setbacks."
hmm, how odd. I very much disagree with the quote, and it certainly does not reflect my experience of mathematicians
they are not all some sort of savant, but they were certainly not the ones struggling with the math problems in third grade (unless they had teachers who decided to challenge them a lot extra)
Anyway, here is a good and challenging problem for representations of finite groups that I deemed too challenging by a huge amount for my students (even if I gave them some extra hints): Let $m$ be a natural number and assume that $m-1 = p^r$ is a prime power. Let $G$ be a group of order $n = mp^r$ and assume that $G$ does not have a normal Sylow $p$-subgroup and that all Sylow $p$-subgroups intersect trivially. Find an irreducible representation og $G$ with dimension $p^r$.
@MatheinBoulomenos Ahh, I see. So you don't know what will be covered yet?
well, we already know the amalgamation subrep is the trivial rep, so to verify the quotient is irreducible by schur orthogonality we need $\langle \chi,\chi\rangle=|G|^{-1}\sum |\chi(g)|^2$ to be $1+1=2$, but $\chi(g)={\rm Fix}(g)$ so this is $|G|^{-1}\sum {\rm Fix}(g)^2=|X^2/G|$ by Burnside's lemma, where $X$ is the set of $p$-sylows, so we need $G\curvearrowright X$ to be $2$-transitive
Hmm I need a little help with statistics :) I don't know how to answer the last question because I don't think I was given the Significance Level $\alpha$. https://i.gyazo.com/af9239d2255e9995b1b1efd60abba0a6.png
@anon Of course, if we just wanted an irreducible with dimension divisible by $p$ then we just needed that the Sylow $p$-subgroup was not normal and refer to Ito-Michler
@TobiasKildetoft Physicists write $\otimes$ when they mean $\times$ and vice versa. If you shot all of them, Ill be sad but itd probably also be good for my employment prospects
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@Semiclassical Hi, I have written an email to chief but I still can not ask a question on the main site. I did not understand absolutely what I had to do and how to unblock the situation.
