I have heard of that object in the context of computing $\pi_4 S^2$ (I think there is a framed torus in $S^4$ that gives the nontrivial element of that homotopy group, but I do not know which one).
define a map $\Bbb{Q}(x) \to \Bbb Q$ as follows: Write $f \in \Bbb{Q}$ as $f=\frac{g}{h}$ with $a,b \in \Bbb{Z}[x]$, then send that to $\mathrm{lc}(g)/\mathrm{lc}(h)$ where $\mathrm{lc}$ is leading coefficient. Your ring is the inverse image of $\Bbb Z$ under that map
It's preimage of a circle by the Hopf map, yeah. But the preimage of a point by Hopf compose Sigma Hopf is preimage of a circle in S^3 by Sigma Hopf : S^4 --> S^3...
Writing it down properly would probably require us to figure out how the Hopf foliation and the singular foliation on S^3 by spheres (w/ north and south pole) interacts.
The thing that confuses me is when they say framed links spit out a number in Z/2 I don't get how to say a framing is 0. I guess it bounds a framed disc
SO(2) I meant, if you're in a 3-manifold. Normal fibers are 2-dimensional.
Uh, Z/2
@MikeMiller Is this the first in the "Examples" section? They're computing the framings of the generator circlesin the surface embedding in some 4 (or more) dimensional manifold?
@MikeMiller I get it. If a surface $\Sigma \subset M^4$ is framed nullcobordant, then the generating circles $a_k, b_k \subset \Sigma$ generating $H^1(\Sigma)$ are also framed nullcobordant. We're exploiting that.
Hm, not quite is it. The frames over the circles are 3-frames, while the frames over the surface are 2-frames.
But I guess I can just do the cross product construction to get a 3-frame from the 2-frame over the circle
Isn't that the claim that's being made? Maybe I don't understand. $\Phi(\Sigma)$ seems to measure if there are a pair of circles in $\Sigma$ intersecting at a point and both circles having an odd-twist frame. And the claim that is being made is that it detects if $\Sigma$ is framed nullcobordant in $M$.
Ah, am I flipping implications? Having a generator set of circles which are framed nullcobordant implies the surface is framed nullcobordant?
That might be true, by extending sections of the frame bundle over the 2-skeleton using obstruction theory etc.
(Because if a circle is framed nullcobordant it has the "trivial non-twisty 3-framing" on it)
Is it true that $\lim\limits_{h\to0}\dfrac{f(a+h)-f(a)-mh}{h}=0$ iff $\lim\limits_{h\to0}\dfrac{f(a+h)-f(a)-mh}{|h|}=0$? I think it is correct, I need an expert opinion.
@Silent The forward implication is definitely true. If the first limit is zero, its negative is zero, and from those two facts you can derive that the second limit is zero as well.
@Fargle So, the other implication must be true, too, right? because, second expression says (i think) $\lim\limits_{h\to0^-}\dfrac{f(a+h)-f(a)-mh}{-h}=0$ which implies $\lim\limits_{h\to0^-}\dfrac{f(a+h)-f(a)-mh}{h}=0$, and it also says $\lim\limits_{h\to0^+}\dfrac{f(a+h)-f(a)-mh}{h}=0$. These imply $\lim\limits_{h\to0}\dfrac{f(a+h)-f(a)-mh}{h}=0$.
I'm running an epidemic model using a gillespie algorithm, an counting the number of infecteds. However, when I solve the corresponding master equations and calculate the mean number of infecteds, I get a different shape, and even a different endemic state.
Is there a reason this could be the case (such as me calculating the mean infecteds)?
I'm doing this for a household model, and it all works fine when the external rate of infection is constant. It's only when I change this external rate to be proportional to the number of infecteds in the population at each time step that the two disagree with each other.
why can you find an extension such that $f$ and $g$ split into linear factors (I guess that's what you mean by enough roots)? That's usually only proved for fields in a first course on algebra
What is meant by learning math historically (NOT learning math history only, but learning math with a historical development perspective)? I've seen some sources say that to learn a math topic X, you need to look at the historical development of the topic X and go over the famous questions by you...
it's not actually about writing the email first , but more about figuring out whom to write an email, as you have to do it as fast possible and that means finding out which person to write an email before the course is announced
Problem: Let $R$ be any commutative ring with $1$, let $V$ be an $R$-module and let $x_1,...,x_n \in V$. Assume that for some matrix $A \in M_n(R)$, $A(x_1,...,x_n)^T = 0$. Prove that (det A)xi = 0 for all i...I could use a hint.
Isn't the proof basically: $0 = adj(A) A (x_1,...,x_n)^T = (\det A) I(x_1,..,x_n)^T = \det A (x_1,...,x_n)^T$, which implies $(\det A)x_i = 0$ for each $i$. Why does $V$ need to be a free $R$-module?
Define $\|\cdot\|\colon V \rightarrow \mathbb{R} $ by $\|a\| = \sum 2^n |a_n| $, where $V$ denotes the vector space of all sequences $a=(a_1,a_2,a_3,\dotsc)$ of real numbers such that $\sum 2^n |a_n| $ converges.
Which of the following are true?
1) $V$ contains only the sequence $(0,0,\...
Question
Given a "discontinuous differential equation"(see example) is the below technique (see conjecture) known for "integration"? Is it possible to make any of this more rigorous? (for example what is the set of $f$ for which the conjecture holds true)
Example
Consider a non-continuous dif...
Oh that will be gr8 ... though I would totally understand if he finds any parts of it too dense and wants to talk to me or something ... Thank you soooo much @rschwieb
@MoreAnonymous I know that there are integrals which yield values for functions which are highly discontinuous, but I've never heard of one that does it for all such functions. I think you might need to be more specific about what special properties the class of functions which are discontinuous are expected to have.
@MoreAnonymous Yeah... I don't mean properties of the integral, I mean properties of the function being integrated. It seems like that collection needs to be constrained sommehow, even if the members are wild
Ah you mean about $\lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) $ ? $d_r$ seems to have no constraints (as far as I know)
Possibly dumb question: how many subfields does $\Bbb{R}$ contain? I just proved that any subfield of $\Bbb{R}$ must contain $\Bbb{Q}$, which seems to imply that there couldn't be very many, although this is but a mere suspicion.
There are also some fun non-field extensions in a similar vein as the complex numbers: $\Bbb R[x]/(x^2)$ and $\Bbb R[x]/(x^2 - 1)$, respectively the dual numbers and the split-complex numbers.
"The campus houses the KTIS-AM(Faith Network) and KTIS-FM(Life Network) radio stations, broadcasting contemporary Christian music and programming to the Twin Cities area." (from wikipedia)
oh hey, it's that one Christian radio channel I always skip past on my way to the classical radio channel
hey, in regards to probability, if i have a mutually exclusive events and it happens because they can't occur at the same time, is there a proper name for a situation in converse?
The following image is a Penrose Diagram. After doing some research I am curious, has anyone tried applying Penrose Diagrams to Tensor Networks, where each intersection between space and time curves denotes a node/tensor? In the diagram the interior nodes/tensors would all be 4-tensors since each...
Hey I have a question about why a certain double-integral is wrong: I want to find the area of a rectangle in the first quadrant bounded by the lines $x=0$, $x=a$, $y=0$, $y=b$
This is what I tried to do: $2 \cdot \int_{0}^{\sqrt{a^2+b^2}}\int_{0}^{arccos(a/r)} 1\cdot r \cdot dr \cdot d \theta$
Did I enter the latex wrong or do I just not have the extension I need to see it...?
Anyways, I tried to just integrate over the first triangle (then multiply by 2 for area of rectangle) by defining $\theta$ in terms of $r$ and then performing a double integral over the region... what did I do wrong?
The reason I integrated $\theta$ from $0$ to $arccos(a/r)$ is because the line $x=a$ in polar coordinates is $x=a \rightarrow a=r \cdot cos(\theta) \rightarrow \theta = arccos(a/r)$
But evaluating this integral gives me imaginary solutions!
Define $\|\cdot\|\colon V \rightarrow \mathbb{R} $ by $\|a\| = \sum 2^n |a_n| $, where $V$ denotes the vector space of all sequences $a=(a_1,a_2,a_3,\dotsc)$ of real numbers such that $\sum 2^n |a_n| $ converges.
Which of the following are true?
1) $V$ contains only the sequence $(0,0,\...
Question
Given a "discontinuous differential equation"(see example) is the below technique (see conjecture) known for "integration"? Is it possible to make any of this more rigorous? (for example what is the set of $f$ for which the conjecture holds true)
Example
Consider a non-continuous dif...
Isn't any extension of $\mathbb{R}$ either $\mathbb{R}$ or $\mathbb{C}$ ?
