@TedShifrin I'm still trying to figure out why the vertical line and the horizontal plane, being orthogonal in R^3, turns out to be not orthogonal in R^4...
What are the chances that if I plug in a random vector into a quadratic form with 8 or 9 negative eigenvalues about the same size and 1 positive eigenvalue like 100,000 times smaller than any of the negative eigenvalues of getting a positive answer?
Ya they look reasonably like $E_8$. They are unimodular (which I can prove) and clearly odd. Maybe there's an obvious reason why they aren't diagonalizable and then I would have the most elaborate proof of a small positive eigenvalue in a matrix I could imagine.
The machine I am typing into tells me the have moderately valued negative eigenvalues and one extremely small positive eigenvalue. The things look reasonably enough like E_8 (to me) that I realized that I should maybe be surprised when the squares I was checking were positive. I am wondering if I can use some topology to deduce the existence of the positive eigenvalue (i.e. Donaldson + non-diag would imply a positve eigenvalue).
You say an eigenvector is a vector with a property, and then there's "and then scaling." You mean, I guess, that the vector is invariant under the procedure of first multiplying and second scaling, but, of course, it is not invariant, unless you scale by the reciprocal of the eigenvalue. And then you're in trouble if the eigenvalue is 0.
There is prize in a box. The prize has a value of a positive integer between 1 and N and you are given N. To win the prize, you have to guess its value. Your goal is to do it in as few
guesses as possible; however, among those guesses, you may only make at most g guesses that are too high. The
v...
Hi, why is this true? Given any $\epsilon > 0$, there are partitions $Q, R$ such that $$U(f; Q) < U(f) + \epsilon/2, \quad L(f; R) > L(f) - \epsilon/2$$
If there is no gravity on the moon, how could this flag be flapping in the wind? (see link)
http://www.stumbleupon.com/su/2wD6eg/hea-www.harvard.edu/~fine/images/desktops/Armstrong.jpg
Seeing vote counts and making tags seen rather cool. Where the latter is something you probably wouldn't ever get to use.(since the community always veto's it)
Terminology question: We write $\text{spec}(A)$ for the set of all prime ideals in a ring $A$, and we write $\text{spm}(A)$ for the set of all maximal ideals of $A$. What does $\text{spm}$ stand for? I imagine $\text{spec}$ is for 'spectrum'?
@I'mmostlyjustanidiot I think algebraic geometry needs to be learned with the classic approach first, to have some idea of why things are done.
@BalarkaSen The difference is that the functor of points starts with a scheme. This functor is part of the definition of a scheme (in the functorial point of view)
@BalarkaSen Near the singularity, we replace each point with a pair $(x,s)$ where $x$ is the point itself and $s$ is the slope at the point seen as an element in the projective line
(this is of course not quite right, but it is the idea)
so for one things this only makes any sense for 1-dimenional varieties of course, but these are easier to picture it for
Hmm, I remember making a nice picture of this at some point, starting with the usual cusp and making some program draw the result in 3d. But I have no idea where that might be now
I think $r$ is coming from a splitting field, in any case the argument I've found is that because $m \in \mathbb{Z}/ p\mathbb{Z}^\times \subset K^\times$ we have that $m\in K^X$ but this seems weird to me
The thing prof told me was that you can also do blow-up on complex submanifolds, but now you might not disingularize the variety as for 0-dimensional submanifolds. The question is whether doing finitely many blow-up on a certain submanifold here it's singular removes all singularities.
@FrankScience I disagree. Starting with the basics of algebraic curves can be a good way to get an idea of why any of those other things are interesting.
I have heard many people recommending to study about the complex analytic part of the story before algebraic geometry. Some Riemann surface theory, say.
So I believe you're a student in algebraic geometry? How come you're studying topology too, then? I know they don't contradict each other, but I'd say that's unusual (maybe not true, though, I don't know many grad students in algebraic geometry).
I have just learned the definition of Krull dimension for noetherian rings. I am still confused on some concepts, for example, can we pick $\langle 3,5,7,11,13,17,\cdots\rangle$ as an infinitely generated ideal of $\Bbb Z$(I.e. making $\Bbb Z$ non noetherian).
@FrankScience i aked a question how to show that $f$ is closed and not open: $$\begin{array}{ccc} f:[0,2] & \longrightarrow &[0,2] \\ x& \longmapsto & f(x) \\ \end{array},\; f(x)=\left\{\begin{array}{cc} 0& x\in [0,1], \\ x-1 & x\in [1,2]. \end{array} \right.$$
The point is that in United States, maybe the strength of students are much more diverse than in China, say. The top students are very very excellent, while others are good at anything else, not necessarily at math even for math majors.
Confused. It is written that given a 2-disk inside an manifold of dimension $\geq$ 5, one can find another disk with the same boundary which is embedded by Whitney's theorem. I can believe this. But it says for dimension $4$ this does not work, which I can't digest. I mean, it works for dimension $3$ (given that the boundary has a collar neighborhood), right? That's precisely what the Dehn's lemma says.
@Vrouvrou After reading some posts of you in stackexchange, it's very surprising that you asked about this. You indeed know Sobolev spaces, etc, but get stuck in this one...
@TedShifrin / @MikeMiller I was thinking about how nonstandard analysis allows us to combine finite and infinitesimal values together, and how elements of Lie algebras were considered "infinitesimal elements" of a Lie group back in the day. Say $G$ is a Lie group with Lie algebra $\frak g$ and adjoint action $\rho:G\to{\rm GL}({\frak g})$. Form the free product ${\cal U}(G):=G*({\frak g},+)$ modulo the relation $g\cdot V\cdot g^{-1}=\rho(g)V$.
This doesn't use the information of the Lie bracket unfortunately, but it allows us to differentiate e.g. $a(t)\cdot V\cdot a(t)^{-1}$ wrt $t$ at $t=0$ and get $[A,V]$ (the usual Lie bracket in $\frak g$, not just the commutator bracket in the tensor algebra). If we differentiate e.g. $a(t)Vb(t)Wc(t)$ at $t=0$ we get $AVW+VBW+VWC$. It seems like the "Lie algebra" of ${\cal U}(G)$ is the the universal enveloping algebra ${\cal U}({\frak g})$. Is this a thing?
If there is no gravity on the moon, how could this flag be flapping in the wind? (see link)
http://www.stumbleupon.com/su/2wD6eg/hea-www.harvard.edu/~fine/images/desktops/Armstrong.jpg
