perhaps one thing - while $f$ intuitively does map each prime ideal $x$ to an element in its residue field, it's slightly awkward if you think about, say $\operatorname{Spec} \mathbb{Z}$
because by the above you'd like to say $f$ maps $\operatorname{Spec} \mathbb{Z} $ to some product of fields thing
@Adeek If $X$ is affine, then $X \cong \operatorname{Spec} \mathcal{O}(X)$ is always true (no assumptions!)
So I think another way to 'say' what $f$ really is, is the following - and let's assume $X$ is a scheme over $k$ for some field $k$ for simplicity for now to help with intuition: 1) Because $X$ is a scheme over $k$, $\mathcal{O}(X)$ is a $k$-algebra. This comes from the map on sheaves induced by the map $X\rightarrow \operatorname{Spec} k$ 2) Giving an element $f\in \mathcal{O}(X)$ is equivalent to giving a $k$-algebra homomorphism $k[t] \rightarrow \mathcal{O}(X)$ mapping $t\mapsto f$. 3) This induces a map $X\rightarrow \mathbb{A}^1_k$ (I'm not sure if you know this - but in general there…
But anyway this is pretty tangential to what you were saying/asking i think lol
geometry of schemes, algebraic geometry, 3264 and all that,
although he has this book called moduli of curves which i found quite hard to follow-- but maybe if i pick it up again now i might find it more accessible
for what's it worth right now im learning stuff like gromov witten theory and some other things which are apparently also inspired by physics so i also get to pretend that im touching some physics-y stuff
I started actually in physics but because of family death and going bat shit crazy and attending finals drunk lol for some period I only did math. I wanted to do both math,physics, and computer science honours.
I want to find weak, non trivial, continuous, solutions of $$\Delta u - \lambda u = 0$$ for a square domain in $\mathbb{R}^N$, $N \ge 2$, under periodic boundary conditions, and under an added constraint that, the weak solutions $u$ should take given values, at a given finite set of points in the...
@Fargle ah I see. How to interprete 1+i geometrically then? I have a triangle with vertex 0,2,1+i which I though it was similar to the triangle with vertex 0,1,i
I'd like to think mathematicians are smart enough to understand the lack of value in a symbolic medal, and whoever travels there is not poor enough to try. But I have not underestimated people's stupidity since 2 years ago.
@mercio Yeah, and maybe after that you could have a look at this math.stackexchange.com/questions/2211278/… ? The current stage of which is a bit of a mystery to me. There was Bellissards critics and then there was a reply from Bender&co to Bellissard. After that I have not heard any news on the subject.
@Rudi_Birnbaum I recall reading something along the lines of the approach being not really anything new once everything was made precise. It was something along the lines of needing to show some operator was self-adjoint or something like that, but proving that was already pretty clearly equivalent to RH by stuff well-known to experts
(I may well be misremembering some details though)
@TobiasKildetoft yes, this is some aspect of it as I also have understood it. Then Bender seems currently to focus on "PT-invariance" (which the BBM operator is), rather than self-adjontness. As interestingly PT-invariant operators also have real eigenvalues. As far as I know now the key question is if they have exclusively such. But I don't know about the current state of affairs of that.
Is this true or false?
Let $g:[0,1] \to \Bbb R$ be a function continuous on $[0,1]$ and differentiable on $(0,1)$, such that $g'$ is a function of $g$, i.e. for every $x, y \in (0,1)$, if $g(x) = g(y)$ then $g'(x) = g'(y)$. Then $g$ is monotonic.
I posted my proof as a self-answer, and wond...
Consider the movement of any point $(x,y)$ on this semicircle (e.g. $(0,1)$ if you have a unit circle at $t=0$). Its position after a rotation with angular frequency $\omega$ and time $t$ should be given by $(x,\cos(\omega t)y)$, which is just due to the parametric description of a circle using the trigonometric functions.
Note that this describes the movement of ALL points on the semicircle, i.e. the $x$-position obviously never changes and the "height" or $y$-position varies as described above.
So the area really just changes by the factor $\cos(\omega t)$.
Let $K$ be your non-archimedean local field, $\mathcal O_K$ be the ring of integers, $\mathfrak p$ be the maximal ideal of the ring, and $k := \mathcal O_K / \mathfrak p$ be the residue field.
If $k$ is a finite field and $|k| = q$, then we assemble $\overline k$ this way: by the classificatio...
Hi @all again! Lunch newspaper reading left me with a mysterious quote from Scholze: "Die ganzen Zahlen seien Funktionen in einem dreidimensionalen Raum, erklärt Scholze, und die Primzahlen seien in gewisser Weise Knoten darin. "Aber das darf man nicht ganz wörtlich nehmen." Who can possible make some sense of that?
Let $M$ be a complete Riemannian manifold and let $S_1,S_2 \subset M$ be totally geodesic submanifolds which are closed as subspaces of $M$.
Question: Is $S_1 \cap S_2$ a submanifold of $M$? Or is it at least locally path connected?
Notice that if we drop the condition that $S_1,S_2$ are totall...
@lattice Thanks, yeah it helps. I found it helpful to picture turning the semicircle sideways and looking at it from the edge on. Then if you pick some point, say the peak, and follow it's path, it will trace out a circle. The coordinates of this point are $(-\sin(\omega t), \cos(\omega t))$ still looking from the edge on. Now turning the picture back, only the second coordinate is the one changing the height and this is $\cos(\omega t)$.
@rschwieb Im currently learning it for a project Im working on. Any good resources to learn? Currently working through a four hour video as a crash course
Luckily most of the coding is gonna be in matlab but Ive been told to try and introduce python
How does the equation $da=\sigma R^2\sin \theta d\theta d\phi$ come?
$\sigma$ is the surface charge density of the given sphere $x^2+y^2+z^2=R^2.$
**My attempt**
$da=\sqrt{1+z_x^2+z_y^2}dx dy$. Upper Hemisphere is $z=\sqrt{R^2-x^2-y^2}$. Hence we get $da=\frac{R}{z}dx dy$. We have $x=R\sin\theta \cos\phi$ and $y=R\sin\theta \sin\phi$. Using Jacobian, $dxdy=R^2\sin\theta \cos\theta.$ Am I doing the logically correct step?
I couldn't find the area of interest during my course. Now I am hard to find my own area of interest. currently I am trying to study Electrodynamics. :)
For a surface element imagine moving in the phi coordinate. This needs an element of $R\sin\theta d\phi$, the phi coordinate requires an $Rd\theta$. And that gives you a $da$ when you multiply them together.
For most simple coordinate systems the jacobian is a waste of time. If you want to be a good physicist you need simply imagine a lot of things.
Working out the area and volume elements is much easier by visualisation than the jacobian for simple cases like this
So when you move around a sphere they're arcs and not lines like in standard coordinates
Its hard without drawing it out
The reason it said can be seen from the diagram is because most physicists have came across this before with a visual representation. Really it should have shown how.
I have a variational problem of the form
$$
E(u) = \int_{\Omega} F(x,y,u,u_x,u_y)dxdy
$$
which leads me
$$
\frac{dE}{du} = \int_{\Omega} \left( \frac{\partial F}{\partial u} - \frac{d}{dx}\frac{\partial F}{\partial u_x} - \frac{d}{dy}\frac{\partial F}{\partial u_y} \right)h dxdy + \int_{\Gamma...
Problem: Suppose that $X$ is a complete metric space without isolated points and $\{F_n\}_{n=1}^\infty$ is a collection of hollow closed sets. Show that $X - \bigcup_{n=1}^\infty F_n$ is dense and uncountable...I've already shown density but I am having trouble showing uncountability. A hint would be appreciated.
suppose $Y$ is countable, then $X=Y \cup \bigcup_{n=1}^\infty F_n= \bigcup_{y \in Y} \{y\} \cup \bigcup_{n=1}^\infty F_n$ which is a countable union of nowhere dense set (singletons are closed and hollow, as $X$ doesn't have isolated points)
@AlessandroCodenotti wouldn't you need separability for that as well?
I think you can just say that a complete metric space without isolated point has cardinality at least $\mathfrak c$ and if it is also separable, then it has cardinality exactly $\mathfrak c$
@AlessandroCodenotti anyway, I think your approach works, too, if you want to do it in a more complicated way: $Y$ is $G_\delta$ and hence completely metrizable. As $Y$ is dense and $X$ is T1, any isolated point in $Y$ must also be isolated in $X$, so $Y$ is a completely-metrizable space without isolated points, thus it has cardinality at least $\mathfrak c$
can anyone think of a good notion of Hausdorff distance between multisets in a fixed metric space $X$?
it should be true that the spectrum of a family of Fredholm operators depends continuously on an operator, but you'd have to pay attention to multiplicity
i think the best multiset distance i can think of is to restrict to subsets of fixed cardinality and take $\text{inf}_\phi \text{sup}_x d(x, \phi(x))$ as $\phi: A \to B$ varies over all bijections
(to see that if $X$ is T1 and $Y$ is a dense subspace, then any isolated point in $Y$ is also isolated in $X$, let $y \in Y$ be an isolated point, then there exists an open set $U \subset X$ such that $Y \cap U = \{y\}$. As $X$ is T1, $X \setminus \{y\}$ is open, so we get that $U \cap (X \setminus \{y\}) \cap Y= \varnothing$, which implies that $U \cap (X \setminus \{y\}) = \varnothing$ as $U \cap (X \setminus \{y\})$ is open and $Y$ is dense, this means that $U=\{x\}$, so $x$ is isolated in $X$)
Is this true or false?
Let $g:[0,1] \to \Bbb R$ be a function continuous on $[0,1]$ and differentiable on $(0,1)$, such that $g'$ is a function of $g$, i.e. for every $x, y \in (0,1)$, if $g(x) = g(y)$ then $g'(x) = g'(y)$. Then $g$ is monotonic.
I posted my proof as a self-answer, and wond...
Reminds me of the take-home final I had in graduate real analysis in undergraduate school, @Mathein. The professor (who also wrote the book) couldn't do two of the five questions, as I recall.
@TedShifrin yeah the thing here is that all who tried to do them are some of the best algebra/number theory students I know here in Heidelberg and we all have taken the prereq courses. And nobody was able to solve a single problem
@TedShifrin I'm not sure, I asked to withdraw my participation so that the course wouldn't show up as failed in the transcript and the teacher agreed to that, so I won't show up when the problems are discussed
I can't believe they'd fail the whole class, @Mathein, if students were working hard. Of course, I don't believe in having the grade be based just on one exam, either.
Ah, this is like what Antonios was going through with a visitor teaching undergraduate algebra last spring. ... This is why I believe in teaching mentoring. :P
Well, we saw Balarka a few weeks ago. But I haven't seen him in over a week, I think. I haven't been around as much, though. Demonark says he's started college.
LOL, well, it's odd that he wouldn't have discussed homotopy invariance before making this claim. He's ordinarily extremely careful. But I don't know this book at all.
When he taught out of my undergrad diff geo text, he was supremely picky, even when I made it clear I wasn't writing it to be ultimately pedantic about analytic issues.
Side note: this answer made me think about the metric structure of the universal cover of $\Bbb R^2 \setminus \{0\}$, which was fun. Exercise: Show that the completion of the universal cover adds precisely one point
Is this true or false?
Let $g:[0,1] \to \Bbb R$ be a function continuous on $[0,1]$ and differentiable on $(0,1)$, such that $g'$ is a function of $g$, i.e. for every $x, y \in (0,1)$, if $g(x) = g(y)$ then $g'(x) = g'(y)$. Then $g$ is monotonic.
I posted my proof as a self-answer, and wond...
Let $M$ be a complete Riemannian manifold and let $S_1,S_2 \subset M$ be totally geodesic submanifolds which are closed as subspaces of $M$.
Question: Is $S_1 \cap S_2$ a submanifold of $M$? Or is it at least locally path connected?
Notice that if we drop the condition that $S_1,S_2$ are totall...
I have a variational problem of the form
$$
E(u) = \int_{\Omega} F(x,y,u,u_x,u_y)dxdy
$$
which leads me
$$
\frac{dE}{du} = \int_{\Omega} \left( \frac{\partial F}{\partial u} - \frac{d}{dx}\frac{\partial F}{\partial u_x} - \frac{d}{dy}\frac{\partial F}{\partial u_y} \right)h dxdy + \int_{\Gamma...
