@rschwieb Details: 1. Let $R$ be a non-discrete valuation domain. 2. Let $\mathfrak p$ be a prime ideal of $R$. 3. $R_{\mathfrak p}\supset R$ in the field of fractions of $R$. 4. $R_{\mathfrak p}$ is a valuation ring. 5. $R$ is Prufer. 6. $R$ is local; let $\mathfrak m$ be the maximal ideal. So $R_{\mathfrak m}\cong R$ is not a discrete valuation ring. 7. $R$ is not almost-dedekind.
Evolute means locus of all centre of curvature of points on the curve.
For Involute, tieing a string along the curve. Let P,Q be the points. When we move Q, we get involute right?
Attach a string to a point on a curve. Extend the string so that it is tangent to the curve at the point of attachment. Then wind the string up, keeping it always taut. The locus of points traced out by the end of the string is called the involute of the original curve
When can we say that a local Hamiltonian vector field is global. Can I say that its global when $H$ exists and is defined properly on the entire manifold?
So this book says that a locally Hamiltonian vector field with no zero points on a compact symplectic manifold cannot be globally Hamiltonian. How does one show this?
So, I have an associative binary operation $g:X\times X\to X$ and I want to characterize all unitary functions $f:X\to X$ such that $f\circ g$ is still associative. How would you approach that proof?
Hard for me to cook up an example but Kahler is a very strong restriction!
I can give you a couple basic things off the top of my head. (1) Spheres never admit symplectic structures, because you can take the kernel of the quadratic form of the symplectic 2-form and that'd give you a nonvanishing vector field which contradicts hairy ball. So look for better examples
(2) Symplectic manifold always has an almost complex structure (in fact the space of almost symplectic structures ~ space of almost complex structures), so if you strategy is to look for "symplectic but not complex" examples you have to be careful. I was thinking of CP2 # CP2 but I learnt from here that this is not even almost complex, so admits no symplectic structure actually
I can't help you more unfortunately, my knowledge of examples end here
So this book says that a locally Hamiltonian vector field with no zero points on a compact symplectic manifold cannot be globally Hamiltonian. How does one show this?
$X_H$ is called a Hamiltonian vector field on the manifold $(M , \omega)$ if for $X_H$ you define $i_{X) (\omega) = dH$ where $H$ is a smooth function on $M$
Umm I can't fix the latex, can you understand what I mean @BalarkaSen
@Albas I suppose locally Hamiltonian would just mean $\iota_X \omega$ is, around any neighborhood of a point, $df$ for some locally defined function $f$ on that neighborhood
That's the same as saying $i_X\omega$ is closed
If $X_H$ is a global Hamiltonian vector field corresponding to some $H : M \to \Bbb R$ and $M$ is compact, then $H$ has a critical point on $M$, which means $X_H$ vanishes somewhere. Actually, it has to vanish at two points.
@RyanUnger Oh I meant, what sort of mathematics are they doing in those three weeks? I saw a summer school that was three (may have been six) weeks long that was meant to go through derived algebraic geometry and algebraic K-theory for undergraduates, completely nuts
Surely depends on the student, but I personally find the simplest slice of modern algebraic geometry (eg, scheme theory) hard to understand - I think these sort of things require a lot of background on much more classical mathematics
Is there a way of deriving the formula $\dim (U+W) = \dim U + \dim W - \dim (U \cap W)$ from the rank nullity theorem or codimension equation? It feels like there should be one.
Historically things like scheme theory or stable homotopy theory or ... were developed as a context in which one can rephrase classical questions in algebraic geometry/algebraic topology/... in a way that decidedly gives better results and resolves classical questions
So I don't think it's possible to fully understand them without knowing concrete examples in which one can ask concrete things
@RyanUnger Analysis is a way of thinking :P. Modern algebra is about endlessly building a tower of abstraction. Analysis is about mastering latent inhibition
@BalarkaSen I have a vector field $X$ transverse to $S^2\subset\mathbb R^3$, and then I have an embedding $e$ of a sphere into a 3-manifold $M$. I compose this embedding with an orientation preserving diffeomorphism $f\colon e(S^2)\to S^2\subset\mathbb R^3$. The embedding $f\circ e$ is transverse to $X$, too, right?
I don't know what this book is saying then. So back to the drawing board.
@user681391 I would suppose $S\pitchfork X$ means (at the most basic level) that $T_pS + \text{span}\{X_p\} = T_p\mathbb R^3$, where $S$ is the embedded surface.
A point $a \in S $ is called an accumulation point or limit point of $S$ if every deleted $\delta$ neighborhood of $a$ contains points of $S$. Can we regarard these accumulation points $a$ as complements to $S$?
Although, I would like to get back at my previous answer to your two examples. Perhaps there can be infinitely many neighbourhoods if we take arbitrary point of either set with arbitrary small $\delta$
@user681391 it is a limit point, for $\delta = 1$. However not the kind of point you asked for. Its neighbourhood will overlap with some other point in $(-1;1)$
It has to be a limit point. But then again if it has to satisfy that"deleted $\delta$ neighborhood of $a$ contains points of $S$", then certainly "deleted neighborhood" implies that some point Is Not in $S$
$x\in S$ is a limit point of $S$ if for every $\delta>0$ the neighborhood has nontrivial intersection with $S$ after deleting $x$: $N_\delta(x)\cap (S-\{x\})\ne \emptyset$.
if $S\subset X$ is a subset (with $X$ a topological space, here $\Bbb R^1$), then $x\in X$ is a limit point of $S$ if for every $\delta>0$ the neighborhood has nontrivial intersection with $S$ after deleting $x$: $N_\delta(x)\cap (S-\{x\})\ne \emptyset$.
Can you find a neighborhood of $1$ that doesn't intersect $(-1,1)$?
$(-1,1)\cap (1-\delta,1+\delta)=(1-\delta,1)$ in which case for any $\delta>0$ this is a non-empty intersection, and thus $1$ is a limit point of $(-1,1)$
(well if $\delta\geq 2$ then $(-1,1)\cap (1-\delta,1+\delta)=(-1,1)$ but that stills gives nontrivial intersection)
For all $n \geq m$, $1/n \in Q$. We consider the worst possible scenario, i.e. none of $V_i$'s (other than $V_t$ ) contains more than one element of $T$. But, there can only be $m-1$ such open sets. So, we get a finite subcover.
@Misha.P Well if $\delta\geq 2$, then it is what you are saying (which I wrote above). But say $\delta =1/2$, then $(-1,1)\cap (1-\delta, 1+\delta) = (-1,1) \cap (1/2, 3/2) = (1/2,1)=(1-\delta,1)$ right?
@N.Maneesh I can't speak to the usefulness of these videos, but the speaker is regarded as something of a crank in some circles, and that may explain some habits of unclear explanation he may exhibit. Perhaps just find a different speaker?
officially it's called "trascendental number theory", apparently the aim is to prove Thue-Siegel-Roth, but it will also go through class numbers and stuff from algebraic number theory
Prove that there exist real numbers which are not algebraic:
Proof: $\mathbb{R}$ is perfect, so it is uncountable. [Since it is closed and every point of it is a limit point]. The set of algebraic numbers, say $A$ is countable. Therefore, $\mathbb{R} \setminus A$ must be uncountable.
I started teaching a brand new course out of it, @rschwieb, and after two months realized it was unusable in a serious way. Of course, lots of people use it and apparently love it, but the exercises are unpredictable, not enough middle-level problems, and the text has stuff that's interesting for pros but not — IMHO — appropriate for first-year college students. At any rate, I started writing the second half of my book during the fall, and then wrote the first half during the spring/summer
I have been writing up a test case today. The first steps can be summarized as 1. Create a new person. 2. Make that person dead (might require god mode)
@TobiasKildetoft the idea is this: let $C$ be a category with finitely many objects and $R$ be a ring, then $R[C]$ is defined as a free $R$-module on the set of all morphisms in $C$, with multiplication defined on the basis elements by $0$ if they are not composable and as composition if they are composable. Then one can show that $R[C]$-modules correspond to functors $C \to R\mathrm{-Mod}$, note that this implies that equivalent categories have Morita equivalent algebras. Take $X$ to be the category with $n$ elements and for each pair of objects, exactly one morphism, i.e. the trivial rela…
@Thorgott: You have Taylor's Theorem, of course. So you look at the cubic form, but I don't know classifications for cubic and higher as I know for quadratic.
Oh, I assumed @Thorgott was asking what to do if the second derivative test fails. It would be nice if one replied to questions like "What do you have in mind?" ...
@Ryan: I was out of the room and missed your pithy removed remark. Or is that a removed pithy remark?
Sorry about that. I was afk for a bit. Yeah, I wanted to know about the scenario where the second derivative test fails, but it seems the terms do grow too complicated. I also considered reducing the multi-dimensional case to the one-dimensional case, but composing with arbitrary smooth curves doesn't seem too fruitful. Thanks for the reply.
You have to be careful with such approaches. There's a classic example of a smooth function on $\Bbb R^2$ that has a minimum at the origin along every line through the origin but fails to have a local minimum at the origin.
I used to run into a number of my students/former students in chat here.
Something like $(y-x^2)(y-2x^2)$, yes. I think considering smooth curves rather than just straight lines should suffice, but that doesn't look like it would give rise to a practical test.
No, not every smooth curve. But you're never going to get any sort of criterion out of this. Ultimately, it has to be an algebraic criterion in terms of the cubic, quartic, ... forms.
