The Hodge idea is that if you have $\mathcal{E}$ a vector bundle (viewed as a locally free sheaf) then you define a nondegenerate pairing $H^q(X,\mathcal{E})\otimes H^{n-q}(X,\mathcal{E}^*\otimes K_X)\to H^n(X,K_X)$.
the stuff in $H^q(X,\mathcal{E})$ can be represented by $(0,q)-$forms and the stuff in $H^{n-q}$ is represented by $(n,n-q)-$forms (using some Hodge theory ideas), then the pairing is just integrate
When $\mathcal{E}$ is holomorphic, there is a "Dolbeault resolution"
and this basically works because $\mathcal{E}$ is naturally a kernel of a suitable $\overline{\partial}$ operator defined on $\mathcal{E}-$valued forms.
I'm supposed to show the punctured projective plane is diffeomorphic to the Möbius strip and I'm trying to see how this works geometrically. Picture the projective plane as quotient of the sphere and puncture it at the north pole. Picture the Möbius strip as fiber bundle over the circle. The circle, which the Möbius strip is a bundle over, can be identified with the equator of that sphere, which is a copy of $\mathbb{R}\mathbb{P}^1\cong S^1$. The fibers of a point on the circle underlying the strip then corresponds to a pair of opposite meridians (which are antipodally identified). Traversi…
That looks more or less right. I would argue that it's a better idea to think of RP^2 as D^2 attached to S^1 by the map bd(D^2) = S^1 -> S^1, z -> z^2, and then puncture this D^2 at the center. This picture also translates into a formal proof now
Hmm, I find it hard to see there when exactly the twisting happens (it has something to do with the attachment, cause that corresponds to the antipodal identification, but I can't pinpoint it)
I guess I can make my picture even cleaner: Start by considering the sphere as-is, with both north and south pole removed. You can push this radially outward to obtain a $S^1\times[0,1]$. Quotienting by antipodals on $S^2$ and then puncturing at the north/south-pole equivalence class is exactly the same as just quotienting this cylinder by antipodals. Think of $S^1/\sim$ as a semi-circle (with two boundary points identified) instead and the cylinder as lying over that; the identifications on the fibers over the inner points of that semi-circle become trivial and the ones on the boundary poi…
I can't see why that should be true because I (rightly) shoved all topology out of my brain after realizing that I couldn't pour my coffee into a donut
The complex polynomial $x^n+a_{n-1}x^{n-1}+\dotsc+a_1x+a_0$ has, for $m\le n$, at least $m$ zeroes in the disc of radius $2\max_{0\le k\le m-1}|a_k|^{1/(n-k)}$ about the origin
if you have $k$ Gershgorin disks that are disjoint from the other $n-k$ Gershgorin disks, the union of those $k$ disks contains precisely $k$ of the Eigenvalues and the union of the other $n-k$ disks contains precisely the other $n-k$ (all with multiplicity, of course)
you prove this by essentially writing down a path from the matrix to its diagonal part (and the result holds trivially for diagonal matrices) and note that the Eigenvalues can't jump connected components cause they depend continuously on the entries
Wait am I being an idiot, can't you just argue (CP^1)^n/S_n -> CP^n, [z_0, ..., z_n] -> [e_1 : ... : e_n] is a homeomorphism because it's a continuous bijection and (CP^1)^n/S_n is compact.
Then you immediately get continuity of inverse and as a special case obtain continuous dependence of roots
hmm, that reminds me I once read that continuous dependence of roots on coefficients still holds if you allow the leading coefficient to vary continuously even when you allow it to vanish by compensating the disappearing zeroes as infinity on the Riemann sphere
yeah i think i want to identify CP^n with monic poly of degree n with complex coefficients upto scale and send [z_0, ..., z_n] to product (z + z_i) where by convention z + infty is zero
How to solve differential $ρ(\frac{∂u}{ ∂t} +(u·∇)u)−µu+∇p =f$ where there is initial condition$∇·u=0$. I know it is not allowed to ask Homework question but I don't know how to solve this one.
well, just saying something meaningful about existence and uniqueness of one particular PDE is one of the millenium problems, so I don't think you'll have much luck finding a general theorem
you do have a local existence and uniqueness theorem for certain PDEs with analytic coefficients, but I think that's about as general as it gets
So I will get 1 million dollars for doing homework?
I just searched millieum problem. Now I am wondering... How with the existence of professionals,supergenius...etc mathematician of you can there be a problem like this?
Shouldn't all math problems be solved by now?
May be Millieum problem are scams or some politician will silently kill you after you solve one.
Consider point A(5, 2) and variable points B(a, a) and C(b, 0). If the perimeter of triangle ABC is minimum, then find a, b. I know that for minimum length, path followed by light ray is to be traced but how to apply it here.
Any endomorphism of an algebraic (should've specified that) field extension $L/k$ is an automorphism. It's injective, because any non-zero field homomorphism is. Take an element in the image. It satisfies some polynomial over $k$. The endomorphism then induces an injective self-map on the roots of that polynomial, which are finitely many, so the self-map is a bijection and your element lies in the image.
Does a topological space have enough structure to define a curve on it? I'm a physics student watching a mathematical primer on diff. geometry for GR and we seem to be defining curves on a topological space and a topological space with an atlas. But these spaces don't have a notion of distance since they have no metric, how can we define and classify curves (continuity, differentiability etc.) before we have this structure?
I get that the charts map into Euclidean space in which these notions are defined, but we have to define the curve on the manifold before we map to the chart no?
but how in that map do we decide which points that are "adjacent" in $\Bbb{R}$ are close in the topological space?
I (think I) understand continuity, that doesn't make any reference to distances
I can kind of see how you could define a curve on a set that doesn't have a notion of distance, but at that point isn't not really a curve, may be I'm trying to impart too much physics on it
if you agree the topological notion of continuity generalizes the metric notion of continuity, then you should also agree the topological notion of a curve generalizes the metric notion of a curve
I'm not a physicist, but you can still think of it as a path that is traversed by some object: the interval gives you a time parameter and the image of the curve that is being traced is the path the object traverses. continuity just tells you the object doesn't randomly jump around in space
for what it's worth, every topological manifold is metrizable (meaning you can define a metric on it that induces the same topology). in fact, every topological manifold can be embedded into $\mathbb{R}^n$ for some large enough $n$, so, if you're trying to build intuition, there is not really a loss in just thinking about nice subsets of $\mathbb{R}^n$, e.g. the ones that you would intuitively call curves or surfaces in 3d space
Hi guys, small question in calculus. If one of the series in a sum of series is divergence and the other one is convergence, does it mean that the sum is divergence?
Given $$\int_{0}^{\infty} f(t) dt = \int_{0}^{\infty} g(t) dt$$ the most we can get is : $$ \int_{0}^{\infty} f(t) dt - \int_{0}^{\infty} g(t) dt = 0 $$ $$\lim_{x\to \infty} \int_{0}^{x} \left[f(t) -g(t) \right] dt = 0 $$ $$\lim_{x \to \infty} \frac{d}{dx} \int_{0}^{x} \left[f(t) -g(t)\right] $$
$$\lim_{x \to \infty} f(x) - g(x) =0$$
That is, $f$ and $g$ converge to each other as $x$ goes to infinity. But the problem is that someone is saying that it is wrong because I took the derivative without taking the limit, that is I didn’t justify the switching of limits. Can somebody help me here?
Suppose you have a double sequence $\displaystyle a_{nm}$. What are sufficient conditions for you to be able to say that $\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}} = \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}$? Bonus points for necessary and sufficient conditions.
For an...
no, you're trying to take the derivative of a limit of a function of $x$ as $x$ goes to infinity with respect to $x$, which doesn't make sense conceptually
@Thorgott I agree, but what if we see it like this this: that limit part is a number, and so it’s a constant and we took its derivative which is zero and hence things are consistent.
@Thorgott Do you know the proper formalism behind, say, constructing countably many iid random variables on $\Bbb R$?
I suppose it goes like this; let $\mu$ be any measure on $\Bbb R$. Then $\mu^n = \mu \times \cdots \times \mu$ defines a well-defined measure on $\Bbb R^n$ for any $n \in \Bbb N$, and since these measure spaces $(\Bbb R^n, \mu^n)$ are consistent, in the sense that they are compatible with the coordinate inclusions $\Bbb R^n \to \Bbb R^{n-1}$, there's a measure $\mu^{\Bbb N}$ on $\Bbb R^{\Bbb N}$ by some extension theorem which restrict to $\mu^n$ on $\Bbb R^n$.
I forget what this extension theorem is called, or the statement.
Can "countable" be improved? Can you do it on general spaces than $\Bbb R$? Polish spaces maybe?
Ok, I checked Durrett. He calls it the Kolmogorov extension theorem; if $\mu_n$ are probability measures on $(\Bbb R^n, \mathcal{B}^n)$ which are compatible in the sense that $\mu_{n+1}(A \times \Bbb R) = \mu_n(A)$, then there is a probability measure $\mu_\infty$ on $(\Bbb R^{\Bbb N}, \mathcal{B}^{\Bbb N})$ compatible with all these finite levels
He also mentions it's not true for arbitrary measurable spaces
@Knight Here's a concrete exercise for you. Give me a differentiable function $f$ for which $\lim_{x\to\infty} f(x) = 0$ but for which $\lim_{x\to\infty} f'(x)$ does not exist.
yeah, that's Kolmogorov, but you don't need that here
you can just construct this hands-on
define the sigma algebra on the countable product to be product by products of measurable sets from the respective spaces, all of which but finitely many are the entire space
define a premeasure on that by just making it the product of the individual measures and extend by Caratheodory
this relies crucially on the measures being probability measures, but should work for arbitrary products, I'm pretty sure
There must be something wrong in what you are saying
If there was an uncountable collection of iid nondegenerate random variables on $([0, 1], \mathcal{B}_{[0, 1]})$, the $L^2$ space on this measure space would not be separable.
A very concrete approach to construct a sequence of iid random variables with distribution $F$ (say, $F$ is the desired cumulative distribution function) is to proceed as follows.
It is enough to construct a sequence of iid uniform $[0,1]$ random variables $(X_i)_{i\ge 1}$, because then $(F^{\le...
Hi. I have a question : if domain of f(x) is -[2.2] then what's the domain of f(|x|+1) ?
Anonymous
Hello. I have a very silly question: consider the left group action of $G$ on itself. $a$ and $b$ are two group elements that stabilize $s \in G$, i.e., $as=bs$. Then can we always write $ass^{-1}=bss^{-1} \implies a = b$ by right multiplying by $s^{-1}$? That is, are $a$ and $b$ necessarily equal in this case?
Sir, i have problem in a particular set of problems. For ex- questions having complex equation and asking us to find number of value of x for which this equation hold true.
I am quoting a particular problem from my book- $log(4)+ \frac{2x+1}{2x}log(3)=log(\sqrt[x]3+27)$
