I'm personally of the perspective that the US has been this way. It's just more explicit now, and social media algorithms only increase the feedback loop.
Well, I guess what I'm particularly emphasizing are conspiracy theories that go around, and people willing to take such theories as facts that they will use to guide their decisions.
I was following from the sidelines. I was neither the one to write papers based on the conjecture (fortunately), nor the one to disprove it (unfortunately)
Suppose that $R$ is a commutative ring, $P$ is a projective $R$-module, and $g : Hom_{R}(P,P) \to M$ is a homomorphism, where $M$ is some $R$-module. Does $g$ induce a map from $P \to M$?
@Shiranai Where are you reading this? You're right that the conjugation action only makes sense when a left and right action are defined (and normally you're acting on the group itself)
On another note, how do you guys remember the stuff you read? Just re-read when you feel you need it, have some kind of system, or something like that?
When I'm reading something new, and that new thing depends on something I've previously learned, either I'll remember that previous thing in enough detail that I don't need to reread anything about it, or I don't, and I'll go reread about it. After a few rereads it'll tend to stick forever
It's also the natural approach to learning in general. When you're reading new material, and you don't know a word, you look up a reference, and then proceed. Whether it's new or something you used to know, you just go read about it again, not really a big deal forgetting stuff, we all do it.
when $\lim f(x)$ exists and is real, is it always possible to find upper and lower bounds for $f$? in other terms, in the given condition, is it always possible to show it with squeeze theorem?
Hmm...I was thinking that $\phi (f) = (\pi_1f \iota_{1}, \pi_2 f \iota_2)$ would be the isomorphism, where $\pi_1$ and $\pi_2$ are canonical projections and $\iota_1 : P \to P \oplus Q$ and $\iota_2 : Q \to P \oplus Q$ are the injections.
yeah, we have $\operatorname{End}_R(P\oplus Q)=\operatorname{Hom}_R(P\oplus Q,P\times Q)\cong\operatorname{Hom}_R(P,P\times Q)\oplus\operatorname{Hom}_R(Q,P\times Q)\cong\operatorname{Hom}_R(P,P)\oplus\operatorname{Hom}_R(P,Q)\oplus\operatorname{Hom}_R(Q,P)\oplus\operatorname{Hom}_R(Q,Q)$ naturally in both $P$ and $Q$
I'm having some calculus difficulties. I want to show something along the lines of $$\sqrt{\int_a^b\int_a^b\big\vert k(x,y)f(y)\big\vert^2dxdy}\leq\sqrt{\int_a^b\int_a^b\big\vert k(x,y)\big\vert^2dxdy}\sqrt{\int_a^b\int_a^b\big\vert f(y)\big\vert^2dxdy}$$ But I don't know if this necessarily true, and I think direction application of Cauchy-Schwarz should give me a different answer
@user193319 He wasn't answering the Hom_R(P,P) thingo, he was just explaining my matrix in regard to showing that End(P\oplus Q) isn't what you wrote above
@NeverEnoughTime I have the usual definition involving the certain diagrams commuting, I also have that $P$ is projective if and only if $P \oplus Q$ is free for some module $Q$, and I have that $P$ is projective if and only if short exact sequences where $P$ is the last nonzero term always split.
So you guys are saying that the isomorphism $End_{R}(P \oplus Q) \cong \operatorname{Hom}_R(P,P)\oplus\operatorname{Hom}_R(P,Q)\oplus\operatorname{Hom}_R(Q,P)\oplus\operatorname{Hom}_R(Q,Q)$ solves the problem?
I actually hadn't thought about $\operatorname{Hom}_R(P,P)$ being projective before, the functor $\operatorname{Hom}_R(\operatorname{Hom}_R(P,P),-)$ looks weird
The decomposition of End(P+Q) is not more complicated than the decomposition of matrices that NeverEnough gave. And showing that End(F) for F free isn't too hard
those are all easy facts if you know how $\operatorname{Hom}$ behaves with respect to products and coproducts, which is something one should know when doing commutative algebra as it's integral to pretty much anything you do there
the relevant facts are $\operatorname{Hom}_R(\bigoplus_{i\in I}M_i,N)\cong\prod_{i\in I}\operatorname{Hom}_R(M_i,N)$ and $\operatorname{Hom}_R(M,\prod_{i\in I}N_i)\cong\prod_{i\in I}\operatorname{Hom}_R(M,N_i)$
Oh, I checked your profile before for context about what you know, and you were doing first year algebra stuff over 4 years ago, so I assumed
By tutor I meant TA btw
Wait now I'm confused
You were studying intensely for the GRE in 2017, and worried you were losing time for grad subject prep, but you're a first year math undergrad? (and you say you're an undergraduate then, meaning in 2017)
Let $f : \Bbb Z^2 \to \Bbb R$ be a discrete harmonic function, that is average $f$ on neighbors of a vertex is the value of $f$ at the vertex. If $f$ is positive, then $f$ must be constant.
Here's the real puzzle. I wanted a proof along the following lines: Scale the ambient by $1/n$ to get a positive harmonic function on $(1/n \cdot \Bbb Z)^2$, and take $n \to \infty$. Then the ambient medium Gromov-Hausdorff converges to $\Bbb R^2$, and you get a positive harmonic function on $\Bbb R^2$.
There is no such nonconstant function so we are done.
But I think you need additional hypothesis on $f$ to do stuff like this
Here's one way to write it precisely, perhaps. Define $f_n : (1/n \cdot \Bbb Z)^2 \to \Bbb R$, $f_n(\alpha) = f(n \alpha)/n$.
Then in "the limit" we get a function $g : \Bbb Q^2 \to \Bbb R$, which is just $f_n$ restricted to $(1/n \cdot \Bbb Z)^2$
It seems the point is this $g$ is not necessarily continuous on $\Bbb Q^2$. It is clearly scale-invariant, $g(c x) = c g(x)$ for $c \in \Bbb Q$, by construct.
@BalarkaSen So if you had angular control, you'd get continuity of $g$. Eg, suppose $f$ was a sublinear (in the dumb sense...) harmonic function on $\Bbb Z^2$, $|f(x)| = o(\|x\|)$.
I'm not sure, but I think if you have a minimum inside the ball, then the function has to be constant in the ball. If you have a minimum on the edge, then you have to have an element that i lower than m-e where m is the minimum, and m+e is the minimum inside the ball
if the minimum is inside the ball (ie all its neighbours are in the ball) you get m = a+b+c+d, which leads to a==b=c=d=m, and propagates so that the function is constant in the entire ball except the corners
if the minimum in on the edge, , m is the sum of three in the ball and one out of the ball, meaning the one out of the ball has to compensate for the one inside the ball (not on the edge), otherwise you have a local minimum
Here's the only proof I know: Let $X_n$ be a random walk on $\Bbb Z^2$, let $Y_n = f(X_n)$. This is a martingale (check, by harmonicity of $f$). By martingale convergence theorem (hard!!), $Y_n$ converges and is independent of the initial condition, and the converging value is $f$. So it is constant.
So, I'm trying to show an inequality and I'm getting bogged down in the details
It is this: $$\sup\Bigg\{\sqrt{\int_a^b\big\vert\int_a^bk(x,y)f(y)dy\big\vert^2dx}:\|f(y)\|\leq 1\Bigg\}\leq\sqrt{\int_a^b\int_a^b\big\vert k(x,y)\big\vert^2dxdy}$$
It might be true that sufficiently slowly growing positive harmonic functions are all constant on (virtually) nilpotent groups, for example. If my proof idea works one could try this here.
Alright, I believe you (I have something similar to what you wrote. What I'm attempting is this): $$\sqrt{\int_a^b\big\vert\int_a^bk(x,y)f(y)dy\big\vert^2dx}\leq\sqrt{\int_a^b\int_a^b\big\vert k(x,y)f(y)\big\vert^2dxdy}\leq\sqrt{\int_a^b\int_a^b\big\vert k(x,y)\big\vert^2dxdy}\sqrt{\int_a^b\int_a^b\big\vert f(y)\big\vert^2dxdy}$$
@EdwardEvans yuuuup. I feel bad when I spend more than 80 euro a month on the stuff and that's not even close. Would've thought you grew it yourself or smth
and my recklessness meant I did no mathematics for a year and just spent all my money and then I moved to Heidelberg and had a course on L-functions from day one and just.. quit it after like 2 lectures 'cause I had no idea what was going on hahaha
But wait, wouldn't C-S be $$\vert\int_a^bk(x,y)f(y)dy\vert^2\leq\vert\int_a^bk(x,y)dy\vert^4$$ Because the LHS is $\vert\langle k(x,y),f(y)\rangle\vert^2$?
@Clarinetist i think my ill-formed question caused a little misunderstanding. let me rephrase it: let $\lim_{x\to a} f(x) = L$ with $a,L\in\mathbb{R}$ then i'm wondering if i can always show it with squeeze theorem? that is to construct two functions $g,h$ such that $\forall x\in\mathbb{R}, g(x)\leq f(x)\leq h(x)$
@TedShifrin that's also what my intuition says. and eventually, i want to use more the squeeze theorem, just for the sake of its beauty in my regard. i've found exercices in the website brilliant, and solved nearly all of them. do you have an exercise to apply squeeze theorem, for me, sir?
oh, ok. then $-1\leq \sin(1/x^l)\leq 1$ and $x^k\leq x^k\sin(1/x^l)\leq x^k$. and $\lim_{x\to 0}x^k=\lim_{x\to 0}-x^k=0$. thus the original limit is $0$.
I feel silly, but is there an easy reason for $\mathbb{R}^{\mathbb{N}}$ not being a manifold? It comes down to saying that $\mathbb{R}^{\mathbb{N}}$ doesn't embed in $\mathbb{R}^n$ for any $n$, which is definitely true, but I don't know if it's easy.
