Mathematics - 2022-11-27

8:21 AM

Suppose that in R^2, I take a loop A based at (0,0). Suppose that there is a hole at (0,1). Inside region of A does not contain (0,0) inside it. Now there is another loop B at (0,0) whose inside contains (0,1). Then intuitively there is no path homotopy between A and B as continuous deformation would mean crossing (0,1) once.

9:45 AM

Why $(1-\zeta^2)^{-1/2} = i(\zeta^2-1)^{-1/2}$ when $\zeta>1$? $(1-\zeta^2)^{-1/2} = (1-\zeta)^{-1/2}(1+\zeta)^{-1/2}$.
\begin{align*}
(1-\zeta)^{-1/2} & = e^{-{1\over 2}\log(1-\zeta)}\\
& = e^{-{1\over 2}(\ln |1-\zeta|+i\arg(1-\zeta))}\\
& = e^{-{1\over 2}(\ln|\zeta-1|+i\arg(\zeta-1)+i\pi)}\\
& = e^{-{\pi i\over 2}}e^{-{1\over 2}(\ln|\zeta-1|+i\arg(\zeta-1))}\\
& = -i(\zeta-1)^{-1/2}.\\
\end{align*}

10:55 AM

The fundamental groups of convex subspaces of R^n are the identity groups as convex subspaces are path connected.

the conclusion sounds right to me. there's a cheap generalization to star-shaped subsets of R^n. i wonder about the justification.

Ah, that's an exercise problem. I'll get there soon.

S^1 is path connected and its fundamental group is nontrivial, of course. the thing about convexity is it lets you homotope loops to constant loops in a particularly simple way.

and for a star shaped set, at least for one basepoint you can do that, and then use path connectedness to get it for other basepoints.

yeah. linearly

some say rectilinearly

i don't do that, if only because i want the number of meanings of the word 'linearly' and its variants to be minimized.

11:11 AM

Leslie, for star convex, is the following justification correct?: Suppose A is star convex. Then there is a point p in A such every line segment connecting p to other points in A lies completely inside A. So take any x in A, connect it to p by a line segment. So we have a path in A from p to x. It follows that fundamental group of A at x is isomorphic to as that at p.

QED.

"It follows that..." is because of the theorem that "If f is a path in X from a to b, then $\pi_1(X,a)$ is isomorphic to $\pi_1(X,b)$."

if i were in a class i would annotate that with pin citations to the results i am using (e.g. the theorem you cite).

also, if the idea is to show that pi_1 is trivial, i'd include something about the straight line homotopies you can use at p.

Oh right. I'll use that now.

so what's on your plate right now? algebraic topology, elsewhere it seems ring/module theory, what else?

Constant map c:I--> A, defined by c(t)= p for all t in I is a path. I want to show that path homotopy class of c, i.e., [c] is the only element in $\pi_1(A, p)$. This step requires straight lines as you said.

Leslie, we have algebraic topology next semester. This semester, we have Topology + introduction to algebraic topology, measure theory, abstract algebra, linear algebra, calculus of several variables.

Q: Can any harmonic function on $\{z:0<|z|<1\}$ be extended to $z=0$?

10

After looking at this question for quite some time, I've asked a couple of other students, and they also couldn't seem to come up with an answer. This is from an old qualifying exam at our university. Let $u$ be a harmonic function bounded on the set $\{z:0 < |z| < 1\}$. Can it always be defin...

11:20 AM

koro: that seems like fun. with the exception of measure theory and calculus of several variables (puke emoji)

I didn't know you liked general topology

suppose that [f] is in $\pi_1(A, p)$. Consider F(x,t):= (1-t)f(x)+ t p. (this lies in A as A is star shaped). This F is path homotopy between f and c. So [f]=[c]. This concludes the proof.

oh, i'll puke at that too. good point.

Leslie: I find several variables calculus also very interesting. It's just that there is so much syllabus and very less time to cover.

Next semester we also have complex analysis and number theory.

complex analysis can be frighteningly difficult.

it was the one class in grad school where not only was i lost, but i thought, 'i can't understand how anyone could understand this.'

11:31 AM

I had complex analysis during my UG also.

well, i hope for your sake it is just a redo of that.

let's see.

@Koro UG means undergraduate? You're a graduate student?

I have a doubt

Umm so, there is this question which has been bugging me for a while.. f(x)= x-(1/x).. we can see its derivative is positve in its domain that means function is strictly increasing but still it is not injective why?

it's injective on some domains, but not others.

what domain are you considering f as a function on?

note that for a function defined only on some open subset of R, f'(x) > 0 for all x does not actually imply that f is strictly increasing. it implies only that if is increasing on any connected subset of the domain of f.

11:47 AM

I'm taking R-{0} as domain

the attempt at a proof that f is strictly increasing goes, okay, consider a < b. then f(b) - f(a) is by the mean value theorem f'(c) (b - a) for some c in (a,b). that requires f to be defined on (a,b).

in particular it doesn't work if the domain of f doesn't include the interval (a,b).

so here, f(-1) = 0 and f(1) = 0 but there is no c in between with f(1) - f(-1) = f'(c) (1 - (-1)).

as one example.

@onepotatotwopotato yes UG means undegraduate. I'm doing masters now.

Ok I got it, So when does it helps to check if derivative of a function is increasing to find whether a function is injective?

if the domain is an interval, knowing that the function is increasing will tell you that it is injective.

i should rephrase that. if the domain of a function is an interval, the positivity of the derivative will tell you that the function is injective.

any strictly increasing function whose domain is a subset of R is injective. but when the domain is not connected, you can't deduce "strictly increasing" from a derivative condition.

Raffallo

12:03 PM

anyone is able to help me with this?
https://math.stackexchange.com/questions/4584985/estimate-variable-using-curve-formula

Thanks a lot you were a big help :)

Q: $\int_{D^2}f^*\omega = 0$ for any smooth map $f$ that maps $\partial D^2\to S^1$

12:49 PM

1

Consider the differential $2$-from $\omega = {dx\wedge dy\over x^2+y^2}$ on $X =\Bbb R^2-\{0\}$. Prove that, for the unit disk $D^2$ and for any smooth map $f:D^2\to X$ which sends the boundray circle to the unit circle $S^1$, $$\int_{D^2}f^*(\omega) =0.$$ Since $df^*(\omega) = f^*(d\omega) =0$...

differential-geometry solution-verification smooth-manifolds

Suppose that p is point of $S^1$. Why is $(S^1\times \{p\})\cup (\{p\}\times S^1)$ a figure 8 ?

I thought it was not a very difficult question but now I would say it's not very easy (to me).

I don't see how it's a union of two circles touching at one point.

Why is either of the sets in the union is a subset of the circle in the first place?

@Koro Embed that into a torus $S^1\times S^1$.

why? sorry, I am new to this.

if looks like union of a circle in xy plane with a circle in yz plane that is passing through the diameter of the first circle.

In no way, it looks like 8.

12:56 PM

@Koro I guess you have a copy of Munkres topology. See section 53 (covering space) example 5.

yes, that's where I am coming from.

"B_0 is a union of two circles " why?

B_0 is the set that I wrote above.

1:21 PM

$S^1\times\{p\}$ is a circle, $\{p\}\times S^1$ is a circle, their intersection is $\{p\}\times\{p\}=\{(p,p)\}$

thus, it's two circles intersecting in a point

I understood that now. :-)

if you picture $S^1\times S^1$ as torus like it has been suggested, this is the union of a longitudinal and meridian circle

Q: Bijective fiber bundle map is a bundle isomorphism proof attempt

I had asked this in General Topology room and got the answer.

ILikeMathematics

> A function of degree $n$ has at most $n - 1$ extrema.

This is wrong, right? It could have at most $n + 1$ extrema. It could have $n - 1$ inner extrema, but there is the possibility of the domain not being $\mathbb{R}$ which gives two more edge extrema.

monoidaltransform

1:39 PM

0

We are working in the continuous category. Definition 1: A bundle map between two fiber bundles $\pi_1:E_1\rightarrow B$, $\pi_2:E_2\rightarrow B$ is a continuous map $f:E_1\rightarrow E_2$ such that $\pi_2\circ f = \pi_1$ Definition 2: A bundle map $f:E_1\rightarrow E_2$ between two fiber bundle...

differential-geometry solution-verification fiber-bundles

this is not true

ever space is a fiber bundle over the one-point space in a unique way and then this would imply that continuous bijections are homeomorphisms

2:04 PM

I know there is a branch cut of $\sqrt{1-z^2}$ that makes the function analytic on $\Bbb C-[-1,1]$. Is there a branch cut that makes the function analytic on $\Bbb C-((-\infty,-1]\cup [1,\infty))$?

copper.hat

2:31 PM

The downvote dribble continues.

user123234

2:42 PM

is there a way to show that a state is recurrent if I have the transition matrix and a measure which is invariant under the transition matrix?

3:02 PM

$S^1$ is not simply connected.

Because its fundamental group at (1,0) is isomorphic to $\mathbb Z$.

But that's strange. I remember a discussion here a few days ago about $S^n$ being simply connected.

I suppose the discussion was for $n\ge 2$.

Q: How do I show that all the states are recurrent in this Markov chain?

yes, precisely

the $n$-sphere has connectivity $n-1$, which is $\ge1$ precisely when $n\ge2$

user123234

1

Let me consider a Markov chain with transition matrix given by $$P_{0,1} =1, P_{i,i-1}=1-p, P_{i,i+1}=p$$for $i\geq 1$ and $p\in (0,1)$. I need to prove that for $p<\frac{1}{2}$ all states are recurrent. We have got the hint from our Prof that we need to find a measure which is invariant for th...

probability probability-theory markov-chains stochastic-calculus

my last question referrs to this one here.

3:30 PM

thank @Thorgott.

4:25 PM

Suppose that Y is path connected. [I,Y]= the set of homotopy classes of maps from I into Y. It is to be shown that [I,Y] is singleton.

I tried to do it like this: Fix a b in Y. Let c: I--> Y be the constant map mapping I to b. So [c] is in [I,Y]. Now let f:I--> Y be continuous. By path connectedness, for every x in I, there exists a path $p_x$ such that $p_x(0)=b, p_x(1)=f(x)$. Now consider $F:I\times I\to Y$ defined by $F(x,t)= p_x(t)$. $F(x,0)= b, F(x,1)=f(x)$. It only remains to now show that F is continuous.

But it seems that F is not continuous. How to fix this?

Tempting to say that for any open set U in Y: $F^{-1}(U)= \cup_{x\in I} I\times p_x^{-1}(U)$.

But there is a trouble proving that RHS is contained in LHS, which makes me want to believe that F is not continuous.

you have chosen the paths $p_x$ completely independently of one another

there's no reason why this should yield something continuous in $x$

But if the equality were true, then F^{-1} will indeed be open.

another good indicator this cannot work is that you actually haven't used any property of the interval $I$

you could repeat your argument verbatim with any space $X$ in place of $I$ and surely not all homotopy sets $[X,Y]$ are singletons

I think you're right.

Then how to prove this?

I would do this in steps

first, show that any map I -> Y is homotopic to a constant map (this is where you have to use the particular structure of the space I). then, show that any two constant maps are homotopic (this only uses that Y is path-connected).

4:41 PM

the first step is what I tried to do.

@Koro this equality is wrong, to be clear. the correct equality is $F^{-1}(U)=\bigcup_{x\in I}\{x\}\times p_x^{-1}(U)$

and drawing a picture should clarify that just because the "slices" of a set are open doesn't mean the set is open

I agree.