I have a small question about abstract algebra. In the book, the author writes, "every finite group may be represented by a diagram known as a Cayley diagram". My question is: it seems for me that Cayley tables/diagram is very strong in terms of definition, I took a couple of examples with one and two generators and it works well. So, my question is: Does he say "may be represented by a Cayley diagram" because it is conjecture and we don't have a proof or because it is really incomplete diagram
Suppose $f : M \to \Bbb R$ is a Morse function with distinct critical points. What's the cleanest proof that if $g$ is very close to $f$, then $g = \psi^{-1} f \varphi$ where $\psi$ is a diffeomorphism of $\Bbb R$ and $\varphi$ is a diffeomorphism of $M$?
Say $f$ has critical points $p_1, \cdots, p_n$. $g$ is immediately Morse, so $g$ has distinct critical points $q_1, \cdots, q_n$ s.t $q_i$ is close to $p_i$
$\text{Diff}(M)$ acts $n$-transitively on $M$, so apply a diffeomorphism that takes $q_i$ to $p_i$
Then we have made $f$ and $g$ look the same near the common critical points $p_1, \cdots, p_n$, as $-x_1^2 + \cdots - x_{k_i}^2 + x_{k_i + 1}^2 + \cdots + x_n^2$, by Morse lemma, where $k_i$ is the index of $p_i$
All we have to do is apply a diffeomorphism of $M$ fixing $p_1, \cdots, p_n$ and a diffeomorphism of $\Bbb R$ to make $f = g$ globally
That should follow because $f, g$ are both submersive restricted to $M \setminus \{p_1, \cdots, p_n\}$, and submersions are structurally stable. What's a simple argument for that, now?
$f : M \to N$ be a submersion and $g$ be very close to $f$. Let $X$, $Y$ be vector fields on $N$ and lift $X$ up to $\widetilde{X}$ in $M$ by the submersion. We want $f(\varphi^t_{\widetilde{X}}(x)) = \varphi^t_Y(g(x))$ for some $t$. Maybe we can solve for this.
@BalarkaSen before proving that a vector bundle on $X \times I$ is the same as a vector bundle on $X$ they now asserted this to be a special case of $G$-principal bundles
@Thorgott Perfect I'll have a look at those books and see how it goes. But in simple terms, is the following sufficient: Fubini's theorem has three "segments"or "parts" (1) for simple rectangular regions where the function doesn't blow up to infinity anywhere [Fubini's theorem for rectangular regions (2) for simple, general regions where the function doesn't blow up to infinity anywhere [Fubini's theorem for general regions]
@Thorgott (3) for unbounded regions or for regions in which the function DOES blow up to infinity [Fubini's theorem for improper regions]. The condition for all three of these is that the integral has to be absolutely convergent (i.e. the double integral of the absolute value of the function must be finite).
If the integral is not absolutely convergent (like those conditionally convergent improper Riemann integrals I was talking about), then you can no longer use Lebesgue theory and it's best to not use Fubini's theorem (though of course, as you mentioned, you can do things like check for HK integrability and consult that version of Fubini's theorem for absolute certainty).
Is this a complete and accurate enough description/overview at an introductory level to the subject?
I'm looking at a definition of the Fisher Information matrix through an integral where you integrate w.r.t. x. I'd like to apply it to the logistic regression model where the response variable $y \in \{0,1\}$. Do I need to integrate w.r.t. x and y or can I just w.l.o.g. assume $y_i = 1$ and only integrate w.r.t. x?
vector bundle yes --- fiber bundles with fiber $\Bbb R^n$ are classified by $B\text{Homeo}(\Bbb R^n)$, so you're asking for spaces $X$ so that the map $[X, BO(n)] \to [X, B\text{Homeo}(\Bbb R^n)]$ has non-trivial kernel; in particular you have examples over spheres because these two groups are not homotopy equivalent
@MikeMiller the fundamental group of a wedge of two manifolds is always the free product of the fundamental groups, right? This is because any point on a manifold admits a neighbourhood homeomorphic to R^n and thus is a retract onto the point. Then, you just use van Kampen.
For example, there will soon be some major changes to the regulations of how various exams related to hinting permits are handled. And I have been the one in chanrge of planning how to implement these in the syatem
Where on a larger project, that would probably have been the job of a senior or even managing architect, and I would just have been told what to do afterwards
Some of the tasks are very challenging, some are rather boring, but most are just right
Planning a major change like this is on the challenging side. But I am aiming to become a senior architect at some point, so it is good to try it out
Just glad they decided to bring in an actual senior architect for the upcoming transition to digital hunting permits. That would have been way over my head to do right.
@anakhro I think he will mainly be on for the first part where the overall architecture needs to be set up. But there is still a bunch of stuff being worked out (such as whether they can actually get the budget approved from the relevant branch of government and so on)
Though it sounded like the approval was almost guaranteed, since they want the digital hunting permit to be the spearhead of digitizing various permits and IDs (digital driver's licence was delayed significantly recently, so it is not expected to be ready for a few years yet)
I think fishing licences are already digital. But there is basically no security on those anyway, whereas the hunting licence is also the permit for owning a shotgun
Can I ask for your thoughts on answering (as opposed to voting to close) problems that are clearly homework? Some time ago I posted an answer to a question regarding propositional logic and told the user where he/she/whatever could find additional resources on the matter. I got some upvotes and then 5 days later it was closed.
So if I have a $S^2\vee M$ for another 2-manifold $M$, the universal cover of this \vee-sum is the universal cover of $M$ with a sphere glued to each lift of the shared point (from the \vee-sum), right?
Today I was studying for a qualifying exam, and I came up with the following question;
Is there a simple description in terms of the subspaces universal covers for the universal cover of a wedge product?
This question came about after calculating universal covers of the wedge of spheres ($\...
If I have m smooth 1-forms on an n-dimensional manifold which are pointwise linearly independent, then can I get (n-m) more smooth 1-forms, so that I have n pointwise linearly independent smooth 1-forms?
Okay. For simplicity, assume m=n-1. So that I only need to obtain 1 1-form. At each point these (n-1) span a hyperplane in the cotangent space. I choose a vector outside this hyperplane.
I am trying to prove Cartan's lemma: $\omega_1, \dots, \omega_p$ are pointwise linearly independent 1-forms. And $\theta_1, \dots, \theta_p$ are 1-forms such that $\sum_{i=1}^p \theta_i \wedge \omega_i = 0$. Then $\theta_i = \sum_{j=1}^p A_{ij} \omega_j$ such that $A_{ij} = A_{ji}$.
On a chart I can extend $\omega_i$ to a basis, right?
@VJ123 I think there's only really one version and that's the general one; the boundedness has little to do with what makes the theorem work. (1) and (2) are more like corollaries of Fubini's theorem together with the easy fact that bounded measurable functions on bounded sets are integrable. But of course, none of what you said is wrong and if structuring like that helps you think about it, then I'm in no position to object.
Another way to phrase my question: without assuming the axioms of foundation or replacement, is there a formula $\phi(x)$ that is true if and only if $x$ is a stage/level of the von Neumann hierarchy? An example of a formula that is true iff $x$ is an ordinal would be: $x$ is well-founded and transitive, and every element of $x$ is transitive.
Basically I'm interested in seeing whether, given an ordinal $\alpha$, one can pick out the stage $V_\alpha$ corresponding to that ordinal (i.e. the minimum stage $x$ such that $\alpha \subseteq x$).
Oh it looks like Zuhair added an answer that might be relevant. I have to read that.
@TedShifrin Hey Ted. I think you're an expert on geometry, hopefully, also projective geometry. If yes, could you please have a look at this question: https://math.stackexchange.com/q/3607262/168764?
@nbro: They are not “straight” in standard models of projective geometry. They are projective lines. Mathematicians do not use the word “line” for general curves.
Your setup is confusing. I assume you're working inside the projective space $\Bbb P^2$, in which case the word "parabola" is strange terminology because parabola, ellipse, circle are all projectively the same thing.
Exactly, so the explanation that a homography cannot be convert a line to a parabola is because a line can be represented by a vector, while a parabola cannot?
Proof is easy, a projective transformation of $\Bbb P^2$ comes from a linear transformation of $\Bbb R^3$, and a line $\ell \subset \Bbb P^2$ is associated to plane $P \subset \Bbb R^3$ passing through the origin. Under a linear transformation of $\Bbb R^3$, $P$ will map to a plane through the origin, which will be associated to a line in $\Bbb P^2$
So image of $\ell$ under the projective transformation is a line.
Well, I thought you knew how to answer it, because I assumed you had more knowledge than me
Anyway, let me try to give you a visual example
Consider the following picture of a stadium
If I am able to upload it...
You can see that lines that are straight in the real world have become parabolas (or, in any case, another type of curve)
This means that, if I had a scheme of a football pitch, like the following one
There's no homography that will be able to map this scheme to the picture of the stadium above, right? Because any homography will convert any of the straight-lines of the scheme to other straight-lines, but the picture of the stadium above now contains parabolas
For given $\infty \ge k_1 \ge k_2 \ge 0$, are there non-trivial $C^{k_1}$ vector bundles over a "sufficiently nice" space that are trivial as a $C^{k_2}$ fibre bundle? How nice can the space be and still admit counter-examples?
An $n$-plane bundle, aka a fiber bundle with $\Bbb R^n$ fibers, is said to be $C^k$ if the transition functions are $C^k$-diffeomorphisms of the fibers. A vector bundle is a smooth $n$-plane bundle by that token
So what's the point of the adjective "$C^{k_1}$-"?
There seems to be something lost in translation here. A "$C^k$ fiber bundle" should be whatever structure you need for the total space to be $C^k$ and the projection map to be $C^k$
Sure, for a vector bundle each $\varphi_{ij}(x) \in GL_m$ is a smooth map, but that's not the right condition --- it's that $\varphi_{ij}: U_i \cap U_j \to GL_m$ is a smooth map
Sure, I call something a smooth fiber bundle only if it's a fiber bundle between smooth manifolds with projection map also smooth. It just didn't make sense in Leaky's context for me
If you want call them "$C^{r,s}$-bundles, where $r \leq s$, and you demand that the diffeomorphisms are fiberwise $C^s$ and that the transition maps $U_i \cap U_j \times F \to F$ are $C^r$
And then $\text{Bun}_{r,s}(M) \cong \text{Bun}_{r',s'}(M)$ so long as $s,s' \geq 1$ or $s = s' = 0$
Take $M$ be the exotic $7$-sphere. Then you have two bundles over $S^7$, given by $TS^7$ and $TM$. In the $C^{0, 0}$ sense these are isomorphic, but what else can you say?.
Yeah I'm sure what you really want to do is prove that if $E \to C$ is a bundle of Hilbert spaces, then $GL(E)$ is a fiber bundle of Hilbert Lie groups, I'm definitely not just conning you into listening
Today I was studying for a qualifying exam, and I came up with the following question;
Is there a simple description in terms of the subspaces universal covers for the universal cover of a wedge product?
This question came about after calculating universal covers of the wedge of spheres ($\...
For given $\infty \ge k_1 \ge k_2 \ge 0$, are there non-trivial $C^{k_1}$ vector bundles over a "sufficiently nice" space that are trivial as a $C^{k_2}$ fibre bundle? How nice can the space be and still admit counter-examples?
