@Knight In terms for these exams, I think the general idea is doing more practice questions is the best policy.
The next thing to do after practicing a lot is doing actual practice exams. That is, you take an exam of which you have never taken before, and you do a "mock exam": pretend you are taking the exam, time limit and everything.
And see how you do.
That way, you get a feeling for progress.
"Am I better able to do this exam now that I have done all those practice questions?"
I like some elegant work. I don’t have any desire to become like Von Neumann, Einstein or Gauss. All I want to do is that I should be know for my elegant idea, no matter in how small group but the idea should be real
First, I want to make it clear to you that you can both do your own learning and research on your own, while studying at school stuff that you have to study.
If I have a system of two equations and two unknowns and I believe there is only one solution, how do I prove or disprove that there is a unique solution
If I have a system of two equations and two unknowns and I believe there is only one solution, how do I prove or disprove that there is a unique solution?
Below is a problem that I am working on. I believe that my incomplete solution is correct as far as it goes. I would like to know
if my solution is incorrect. I plan to solve the system of two equations by using a computer program such as SciLab. What I am wondering is, does this system of equati...
If you want to do research, you should find a way to do research.
If you aren't going to try to get into a college program to do so, then first you need to find a job which you can sustain yourself with while you do your own learning/reading.
But if you are going to try to get into a college program, you should at least try to do well on the entrance exams, since that's the first step.
Mind you, you don't have to get into the top university/college to do research.
I am joking people. There is a trope among gamers, that a game is literally unplayable if it misses some trivial detail from real-life. I was hoping run around with that trope. Turns out I can't.
@BalarkaSen Hatcher has another exercise about covering space actions. If $G$ acts on $M$ as a covering space action via orientation preserving homemorphisms, then $M/G$ is orientable if $M$ is.
I assume that the point is that we are inducing the orientation along the projection $p\colon M\to M/G$
And you get the local consistency from the orientation preserving homemorphisms $g\colon M\to M$
I wouldn't describe the orientation as a map $\mu : M \to \{\pm 1\}$ because there's no canonical identification of the generators of $H_n(M, M-x)$ togather a-priori. I would describe it as a collection $\{\mu_x\}$ of generators of $H_n(M, M-x)$ for each $x \in M$.
I mean ok, the understandable idea is that some quantities are homotopy invariant, but what kind of information are those quantities encoding? Where do they come from?
But yes, essentially you want to give me a collection $\{\nu_x\}$ of generators of $H_n(M/G, M/G - x)$ from there which satisfies the local consistency. Thinking of it in the map way, given a map $\mu : M \to \{\pm 1\}$ you want a map $\nu : M/G \to \{\pm 1\}$. When do you have such an induced map? What do you know about $\mu$?
I'm asking because apparently Yu proved the conjecture for manifolds whose $\pi_1$ has finite asymptotic dimension (so for example for all manifolds with hyperbolic $\pi_1$), and that really surprised people and got them interested in asymptotic dimension, which Gromov had defined years earlier but had been kind of ignored
@BalarkaSen so what I am guessing is that $\nu_x := p_*\mu_x$, and this is well-defined because if $Gx = Gy$, there exists a $g\in G\,.\, gx=y$, and since $g_*$ preserves the generators, $\nu_x = p_*\mu_x = p_*g_*\mu_x = p_*\mu_y =\nu_y$.
@anakhro Right, the "map" $\mu : M \to \{\pm 1\}$ is $G$-equivariant since $G$ acts by orientation-preserving homeomorphisms on $M$, so descends down to a map $\nu : M/G \to \{\pm 1\}$.
And that is your orientation on $M/G$. You should write everything down explicitly once, of course, instead of philosophizing; if $x \in M$ then you have an isomorphism $H_n(M, M - x) \to H_n(M/G, M/G - Gx)$ because $M \to M/G$ is a local homeomorphism near $x$ (it's a covering map). The image of $\mu_x$ you call $\nu_{Gx}$.
And then $\{\nu_{Gx}\}$ is your candidate orientation for $M/G$. You need to check local compatibility, which you can, etc.
$\nu_{Gx}$ only depends on the class $Gx$ and not on $x$ because $G$ acts transitively on the fiber over $Gx$, so you have well-definedness; essentially what you wrote.
Namely, if $x$ and $y$ such that $Gx = Gy = p$, there is a commutative triangle involving $H_n(M, M - x) \to H_n(M/G, M/G - p) \leftarrow H_n(M, M - y)$ and $H_n(M, M - x) \to H_n(M, M - y)$ where the last map is induced by the deck transformation permuting $x$ and $y$ on the fiber over $p$ in $M \to M/G$.
Therefore image of $\mu_x$ and $\mu_y$ are the same.
@BalarkaSen Hatcher has another exercise about covering space actions. If $G$ acts on $M$ as a covering space action via orientation preserving homemorphisms, then $M/G$ is orientable if $M$ is.
Off-topic again: Is any cobordism invariant class a polynomial on the Pontryagin classes? Something like this was true but I forget what is the precise statement
Given $p \in M$, $M$ a connected manifold, there is a map $\Omega_p M \to \Bbb Z/2$ sending $\gamma$ to the neutral element iff $\gamma^*TM$ is an orientable bundle over the circle. Then we have the following properties. 1) This is a group map: if you compose two loops, then $(\gamma_1 \ast \gamma_2)^* TM$ is orientable iff $\gamma_1^*TM$ and $\gamma_2^*TM$ are either both oriented or both unorientable. This requires a little understanding of bundles over the circle. In particular, the map on $\pi_0$ is a homomorphism $w: \pi_1 M \to \Bbb Z/2$.
Unfortunately this isn't the right way to think about orientations to be well-behaved for group quotients. The issue is that non-null loops downstairs lift upstairs to paths between different points. So the above description of orientability (it's a manifold for which loops don't flip orientation) isn't the right phrasing
But it's close to one!
Instead we're going to have to thicken this up so that it allows paths between different points --- instead of the fundamental group, this is one of those rare places where it's smart to talk about the fundamental groupoid
$\Pi(M)$ is a category whose objects are the points of $M$, and $$\text{Mor}(x,y) = \pi_0 \text{Map}([0,1],0,1) \to (X,x,y),$$ aka, the set of paths from $x$ to $y$ up to homotopy rel endpoints.
The composition operation just being concatenation of paths; the fact that we make this up to homotopy rel endpoints means that this is a category (composition is actually associative on the nose, and you have inverses).
What is the role of $\pm 1$ here, as we vary between points? Well, $\pm 1$ is really $\pi_0 GL(n)$, right? We check whether a self-map of $T_p M$ is homotopic to the identity ornot.
Set $O(M)$ to be the category whose objects are the points of $M$, and $$\text{Mor}_O(x,y) = \pi_0 \text{Iso}(T_x M, T_y M).$$
Claim: there is a functor $\Pi(M) \to O(M)$, which is the identity on objects. It's a little irritating to write down. The point is that $\gamma^*TM$ is trivial over $[0,1]$, so that there are sections $e_1(t), \cdots, e_n(t)$ over the interval which restrict to a basis for $\gamma^*TM$ at each point. Then the map $T_x M \to T_y M$ sends $e_i(0)$ to $e_i(1)$.
All of the environments you define have an optional parameter, if you have the environment thm for theorems and you want a named theorem you can do \begin{thm}[the name of your theorem goes here] the statement goes here \end{thm}, ok?
Now using a \cite{reference} inside the name of a theorem works fine. Using \cite[optional parameter]{reference}inside the name of a theorem breaks things and cannot be compiled
So you need to wrap the \cite with a \protect because of black wizardry
This is the effect I wanted to obtain, rather than only having the Eng89 reference
Another funny one I stumbled upon is that in a figure environment you need to put the \caption before the \label or you can't actually reference that label elsewhere in the document
Dually, orientation is determined by triviality of the determinant bundle over $M$, which is a real line bundle. Give $M$ a CW-structure and look at the $1$-skeleton; it suffices to understand if the bundle is trivial over the 1-skeleton or not to determine global triviality of a line bundle.
orientations are determined by elements of $H^1(M;\Bbb Z/2)$, in other words
that's really 2-skeletal since you need to know the relations in $\pi_1 M$ in the end
If $G$ acts on $M$, then we also have $G$ acting on $\Pi(M), O(M)$. On $\Pi(M)$ the action is obvious; on $O(M)$, if we have a homotopy class of isomorphism $T_x M \to T_y M$, it is sent to the composite $$T_{gx} M \xrightarrow{g^{-1}} T_x M \to T_y M \xrightarrow{g} T_{gy} M$$
And the functor $F: \Pi(M) \to O(M)$ is equivariant under this group action; $F(gx) = gF(x)$ and $F(g[x \to y]) = gF(x \to y)$
If $M/G$ is to be orientable, then necessarily $M$ must be as well. So suppose $M$ is orientable. Why must $G$ act by orientation preserving transformations?
Suppose $g: T_x M \to T_{gx} M$ is not orientation-preserving. In the above language, this means that $g_* \in O(M)(x,gx)$ disagrees with $F(x \to gx)$.
But that means precisely that if we have a loop $\gamma$ in $M/G$ which is induced from a loop $x \to gx$ in $M$, then the induced map $F(\gamma) \in O(M)([x],[x])$ is orientation-reversing, the non-identity element.
I hate the language I used here. This is a straightforward idea that I have obfuscated.
The point of this stupid language is: so long as the map $w: \pi_1 M \to \Bbb Z/2$ is trivial, then "pushing along a path" lets you take an orientation at any one $T_x M$ and take it to an orientation at any other $T_y M$.
That gives you the global orientation. If $M/G$ is not orientable, then that means that the orientation on some two $T_x M$ and $T_{gx} M$ don't agree, where we're comparing them using the fact above.
And that means that $G$ acted by transformations which didn't preserve the orientation
@BalarkaSen a video showing police brutality on August 31 (every Hongkonger remembers this day) is now being circulated in India as a video showing Chinese police arresting COVID patients?
Boom live, a company in India to verify videos, has issued the following statement: FALSE: चीन में संदिग्ध कोविड -19 मरीज़ों पर पुलिस की कार्यवाही दिखाने वाला वीडियो फ़र्ज़ी है
Correct, let $B$ be an evenly covered open set in $M/G$ containing $x$ and $y$ and consider a slice $\widetilde{B}$ above it containing two fibers $\widetilde{x}$ and $\widetilde{x}$ above $x$ and $y$ in $M$
Then draw the commutative diagram interpolating between $H_n(M/G, M/G - x) \to H_n(M/G, M/G - B) \leftarrow H_n(M/G, M/G - y)$ and $H_n(M, M - \widetilde{x}) \to H_n(M, M - \widetilde{x}) \leftarrow H_n(M, M - \widetilde{y})$.
You should try practicing these arguments more, anakhro.
Actually, it is not a very difficult problem. We just need a stronger form of Bertrand's postulate, that for every positive integer $n ≥ 8$ there is a prime between $n$ and $n·3/2$. This fact has been known for a long time, for example based on a theorem of Jitsuro Nagura that if $n ≥ 25$ then th...
Can someone help to check my solution? I make mistakes now and then, but recently some trolls have been downvoting some of my posts, so I'm not sure whether this is one of them or not. Thanks!
By the way, in case anyone is interested in the coronavirus and haven't seen these yet:
@BalarkaSen a problem I am seeing is that I see a lot of these things for the first time, and then they don't show up again---or if they do show up again, I don't notice that they are there again. Perhaps this is indicative that I am moving too quickly through Hatcher and I am not taking my time to relate things back to what I saw before.
Or simply that I am not actively thinking as much as I should be!
Probably a mixture of these and more.
Granted I regret not going through Hatcher last summer when you told me to.
What's an example of a parametric family of distributions that is closed under translation, scaling, and maximum? This family, for example, is closed under maximum.
This would correspond to the underlying CDFs being closed under translation, scaling, and pointwise product.
