Challenge Math Challenge - November 2018

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fresh_42

Mentor
2018 Award
I guess when you originally set the question you didn't want to give any clues away that the irreducible components are organised as $\{1 \}$, $\{x , y \}$, and $\{ x^2 , xy , y^2 \}$ as that was question 16 b).

I'll be adjusting the calculations I did before (by adjusting $\varphi$) to get the appropriate homomorphism later when I'm less preoccupied.
I actually calculated from behind: took some irreps of $\mathfrak{sl}_2$, wrote them as polynomials, and changed the basis to $\mathfrak{su}_2$ and finally choose the Pauli-matrices ($\sigma_j$) instead of proper Lie algebra vectors $(i\sigma_j$), mixed the basis vectors, in order - yes - not to give away the solution for free, et voilà: the standard example in every book about Lie algebras became a standard example of what physicists use instead. I thought this is an interesting (and simple) example to demonstrate what physicists call "ladder-operators".

TGVF

I think I found solution for Basics 4:
Let the bug altitude in the evening of day i be noted $z_i$ (in meters).
The tree height in the evening of day n is $L_0 +ir$ with r being the daily rate of tree growth (0.20 m/day).
Then the relative bug position w.r.t. tree top in the evening of day i is $\lambda _i = \frac {z_i} {L_0+ir}$
On the next day morning (day i+1), the bug altitude becomes: $z_i + d = \lambda _i *(L_0+ir) +d$, with d being the nighty rate of climb (0.10 m/night)
On the next day evening (day i+1), the relative bug position will be the same as in the morning: $\lambda _{i+1} = \frac {z_i +d} {L_0+ir} = \lambda _i +\frac {d} {L_0+ir}$. Obviously, its absolute altitude will be higher to uniform tree growth.
The bug will reach the tree top as soon as $\lambda _{i} \geq 1$. Although this can take a long time, we can be sure the bug will succeed because if we consider i forming a continuum (changing it from integer to real), then the relative position is a real function with logarithmic growth: $\lambda(x)= \int \frac {d} {L_0+(x-1)r} \, dx = \frac {d} {r} Ln(1+ \frac {(x-1)r} {L_0})$ which is increasing with x without upper bound.
The number of days for reaching the tree top is obtained by solving the equation $\lambda(x)= 1$. That is: $x= 1+ \frac {L_0}{r}*(e^{\frac {r}{d}} -1)$.
Numerically: $x=3,195.53$ days that is for an integer number of days: 3,196 days. However, the precise solution obtained by discrete summation is only 3,192 days. The actual result is the lower Darboux stepwise sum, as compared to the continuous logarithmic integral which is consequently too large.

lpetrich

I'll post proofs of parts a and b of fresh_42's proof outline for Problem 18 in post #125.
(a): I had earlier proved that for every positive integer $n > 1$, $|n|$ is at most 1. If for all such $n$, $|n| = 1$, then that is the trivial case. So to be nontrivial, there must be some such $n$ where $|n| < 1$.

Since the positive integers are bounded from below, there must be some minimum value of $n$ that satisfies $|n| < 1$, and since $|1| = 1$, this minimum value must be greater than 1.

(b): To prove that this minimum $n$ must be a prime number, I consider the case of composite $n$, $n = ab$, where both $a$ and $b$ are greater than 1 and less than $n$. If $n$ is this minimum $n$, then $|a|$ and $|b|$ must both equal 1. But $|n| = |ab| = |a| |b| = 1$, which contradicts this premise. Therefore, this minimum $n$ must be a prime number.

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lpetrich

I continue to be stumped by parts c and d of fresh_42's proof outline in post #125, even with his hints in post #139. In particular, I find that the inequality of part c is a very weak constraint, not much stronger than the triangle inequality in the original definition of the value function. It appears to be too weak to prove anything in part d.

fresh_42

Mentor
2018 Award
I continue to be stumped by parts c and d of fresh_42's proof outline in post #125, even with his hints in post #139. In particular, I find that the inequality of part c is a very weak constraint, not much stronger than the triangle inequality in the original definition of the value function. It appears to be too weak to prove anything in part d.
You know that there is a minimal prime with $|p|<1$ and that all natural numbers have $|n| \leq 1\,.$ The goal is to find out which numbers are of exactly norm $1\,.$

lpetrich

You know that there is a minimal prime with $|p|<1$ and that all natural numbers have $|n| \leq 1\,.$ The goal is to find out which numbers are of exactly norm $1\,.$
I can easily prove that for all $n$ with prime factors less than $p$, $|n| = 1$. But if $n$ has prime factors larger than $p$ and is not divisible by $p$, then I am stuck. Expressing $n$ in the form $kp+m$ with $0 < m < p$, I find
$|kp + m| \leq \max(|kp|,|m|) = \max(|kp|,1) = 1$,
and that does not add anything to the non-Archimedeanness of the value function.

fresh_42

Mentor
2018 Award
I can easily prove that for all $n$ with prime factors less than $p$, $|n| = 1$. But if $n$ has prime factors larger than $p$ and is not divisible by $p$, then I am stuck. Expressing $n$ in the form $kp+m$ with $0 < m < p$, I find
$|kp + m| \leq \max(|kp|,|m|) = \max(|kp|,1) = 1$,
and that does not add anything to the non-Archimedeanness of the value function.
Have I missed that you have already shown the maximum formula, because you use it?

Anyway, if you have that all numbers are of norm one, which are coprime to $p$, then you also know all others for which $|n|<1$ and even their norm in dependence of $p$. With that, the rest is only a bit technical stuff.

lpetrich

Have I missed that you have already shown the maximum formula, because you use it?
I don't know what you have in mind, but I must note that $|kp+m| \leq 1$ is not equivalent to $|kp+m| = 1$. Or is there something that I was missing somewhere?

fresh_42

Mentor
2018 Award
I don't know what you have in mind, but I must note that $|kp+m| \leq 1$ is not equivalent to $|kp+m| = 1$. Or is there something that I was missing somewhere?
We know $|p| < 1$ and $|n| \leq 1$ and assume for the moment that $|a+b| \leq \operatorname{max}\{\,|a|\, , \,|b|\,\}\,.$

Next we show that $|a+b|=\operatorname{max}\{\,|a|,|b|\,\}$:
Let $|a|<|b|\,.$ Then $|a|<|b|=|(a+b)-a| \leq \operatorname{max}\{\,|a+b|,|a|\,\} =|a+b| \leq \operatorname{max}\{\,|a|,|b|\,\} = |b|\,.$

Thus we have $|m|=1$ for all $m$ which are coprime to $p$. All other numbers are of the form $n=p^r\cdot m$ with $|n|=|p|^r$.

lpetrich

But if $|a + b| = \max(|a|,|b|)$ for all integers a and b, then |0| = |1 - 1| = max(|1|,|-1|) = max(1,1) = 1, which is contrary to our definition of this value function or norm. So there must be some flaw in the proof of this statement.

fresh_42

Mentor
2018 Award
But if $|a + b| = \max(|a|,|b|)$ for all integers a and b, then |0| = |1 - 1| = max(|1|,|-1|) = max(1,1) = 1, which is contrary to our definition of this value function or norm. So there must be some flaw in the proof of this statement.
Yes. I had to use $|a| \neq |b|$ for otherwise I couldn't have concluded, that $|a|<\max\{\,|a+b|,|a|\,\}=|a+b|\,.$
In your post #146 where it is needed, we have $|kp| =|k|\cdot |p| \leq |p| \lt |m|\,.$

lpetrich

So with $|a+b| = \max(|a|,|b|)$ if $b \neq a$, I can do all the positive integers. For n relatively prime to p, the smallest number where |p| < 1, I get n = kp + m, where m is nonzero and less than p:
$|n| = |kp+m| = \max(|kp|,|m|) = \max(|kp|,1) = 1$.

With that case solved, consider n having p as a prime factor: $n = k p^m$ for k is relatively prime to p. Then,
$|n| = |k p^m| = |k||p|^m = |p|^m$.

For fraction $x = n_1/n_2$ where $n_1 = k_1 p^{m_1}$ and likewise for $n_2$, then $|x| = |p|^{m_1 - m_2}$.

fresh_42

Mentor
2018 Award
So with $|a+b| = \max(|a|,|b|)$ if $b \neq a$, I can do all the positive integers. For n relatively prime to p, the smallest number where |p| < 1, I get n = kp + m, where m is nonzero and less than p:
$|n| = |kp+m| = \max(|kp|,|m|) = \max(|kp|,1) = 1$.

With that case solved, consider n having p as a prime factor: $n = k p^m$ for k is relatively prime to p. Then,
$|n| = |k p^m| = |k||p|^m = |p|^m$.

For fraction $x = n_1/n_2$ where $n_1 = k_1 p^{m_1}$ and likewise for $n_2$, then $|x| = |p|^{m_1 - m_2}$.
That doesn't define the norm. In the end the definition should work for any rational number including powers of $p$, $0$ and without self references.
Plus you still must show $|a+b| \leq \max\{\,|a|,|b|\,\}\,.$

julian

Gold Member
Solution to problem 16.

The actual calculations are in the attached pdf file. I just outline in this post what was done.

We define a basis for $\mathfrak{su}(2,\mathbb{C})$ as:

$u_1 = i \sigma_1 = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} , \quad u_2 = i \sigma_2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} , \quad u_3 = i \sigma_3 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$

with brackets:

$[ u_1 , u_2 ] = - 2 u_3 , \quad [ u_2 , u_3 ] = - 2 u_1 , \quad [ u_3 , u_1 ] = - 2 u_2 .$

We compute $[(\alpha_1 , \alpha_2 , \alpha_3) , (\alpha_1' , \alpha_2' , \alpha_3')]$ defined by $[(\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3) , (\alpha_1' u_1 + \alpha_2' u_2 + \alpha_3' u_3)]$. We easily find:

$[(\alpha_1 , \alpha_2 , \alpha_3) , (\alpha_1' , \alpha_2' , \alpha_3')] =$
$= 2 (\alpha_3 \alpha_2' - \alpha_2' \alpha_3) u_1 + 2 (\alpha_1 \alpha_3' - \alpha_3 \alpha_1') u_2 + 2 (\alpha_2 \alpha_1' - \alpha_1 \alpha_2') u_3 .$

We define an adjusted $\varphi$ by:

\begin{align*}
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3) &= \varphi (\alpha_1 (i \sigma_1) + \alpha_2 (i \sigma_2) + \alpha_3 (i \sigma_3)) \\
&= \varphi( (i \alpha_1) \sigma_1 + ( i\alpha_2) \sigma_2+ (i \alpha_3) \sigma_3) \\
\end{align*}

Then from

\begin{align*}
\varphi(\alpha_1 \sigma_1 +\alpha_2 \sigma_2+\alpha_3 \sigma_3)&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)= \\
&= x(- i \alpha_1 a_3 + \alpha_2 a_3 - \alpha_3 a_1 )+\\
&+ x^2(2 i \alpha_1 a_5 + 2 \alpha_2 a_5 + 2 \alpha_3 a_2 )+\\
&+ y(i \alpha_1 a_1 + \alpha_2 a_1 + \alpha_3 a_3 )+\\
&+ y^2(2 i \alpha_1 a_5 -2 \alpha_2 a_5 -2 \alpha_3 a_4 )+\\
&+ xy(- i \alpha_1 a_2 - i \alpha_1 a_4 +\alpha_2 a_2 - \alpha_2 a_4 )
\end{align*}

we have:

\begin{align*}
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy) = \\
= \varphi( (i \alpha_1) \sigma_1 + ( i\alpha_2) \sigma_2+ (i \alpha_3) \sigma_3)&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)= \\
&= x(\alpha_1 a_3 +i\alpha_2 a_3 - i\alpha_3 a_1 )+\\
&+ x^2(-2\alpha_1 a_5 +2 i\alpha_2 a_5 + 2i\alpha_3 a_2 )+\\
&+ y(- \alpha_1 a_1 + i \alpha_2 a_1 +i\alpha_3 a_3 )+\\
&+ y^2(-2\alpha_1 a_5 -2i\alpha_2 a_5 -2i\alpha_3 a_4 )+\\
&+ xy(\alpha_1 a_2 +\alpha_1 a_4 +i\alpha_2 a_2 -i\alpha_2 a_4 ) \qquad Eq (1)
\end{align*}

Part 16 a) Is done in pdf file where I prove

$[\tilde{\varphi} (\alpha_1 , \alpha_2 , \alpha_3) , \tilde{\varphi} (\alpha_1' , \alpha_2' , \alpha_3')] = \tilde{\varphi} ([(\alpha_1 , \alpha_2 , \alpha_3) , (\alpha_1' , \alpha_2' , \alpha_3')])$

when applied to $1,x,x^2,y,y^2$, and $xy$ in turn.

16 b) Components that are transformed into linear combinations of themselves under repeated application of $\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 +\alpha_3 u_3)$ form an invariant subspace. Irreducible components are an invariant subspace which cannot be separated into smaller invariant subspaces. From Eq 1 we have:

\begin{align*}
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.1 = 0\\
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.x= - x i \alpha_3 + y (- \alpha_1 + i \alpha_2) \\
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.x^2 = 2 x^2 i \alpha_3 + xy (\alpha_1 + i \alpha_2) \\
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.y = x (- \alpha_1 + i \alpha_2) + y i \alpha_3 \\
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.y^2 = - 2 y^2 i \alpha_3 - xy (- \alpha_1 + i \alpha_2) \\
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.xy = 2 x^2 (- \alpha_1 + i \alpha_2) - 2 y^2 (\alpha_1 + i \alpha_2) .
\end{align*}

It is then obvious that the irreducible components are:

$\{ 1 \}$
$\{ x , y \}$
$\{ y^2 , xy , x^2 \} .$

16 c) The corresponding vectors of maximum weight are

$1 ,$
$y ,$
$x^2$

respectively - see the pdf file.

p.s. Hello @fresh_42. I've done the calculation with the adjusted $\varphi$ and still think there is a slight typo in the question. In the original question you wrote:

\begin{align*}
\varphi(\alpha_1 \sigma_1 +\alpha_2 \sigma_2+\alpha_3 \sigma_3)&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)= \\
&= \dots\\
&+ y(-i \alpha_1 a_1 - \alpha_2 a_1 +i\alpha_3 a_3 )+\\
&+ \dots
\end{align*}

but I think it should say:

\begin{align*}
\varphi(\alpha_1 \sigma_1 +\alpha_2 \sigma_2+\alpha_3 \sigma_3)&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)= \\
&= \dots\\
&+ y(i \alpha_1 a_1 + \alpha_2 a_1 +i\alpha_3 a_3 )+\\
&+ \dots
\end{align*}

Could you check my calculations in the pdf.

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fresh_42

Mentor
2018 Award
@julian
Great job, Julian!
Btw.: The Heisenberg algebra question in 15. is far easier with less computations. Only a bit thinking about centers is needed.

You are right, there is a sign error in the setup of $\varphi$ somewhere.
I took the basis and the representation theorem from
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/
which I now checked again, and which turned out to be correct.

I thought I had taken $x$ as the eigenvector of minimal weight $-1$ and $y$ for the maximal eigenvector. But with this setting my definition of $\varphi$ doesn't match up. Guess I deserved all these calculations to check again:

Pauli - matrices $\sigma_j \, \longrightarrow$ skew-Hermitian version $i \cdot \sigma_j \,\longrightarrow$ basis $\langle U,V,W\,|\,[U,V]=W\; , \;[V,W]=U\; , \;[W,U]=V \rangle$ because this is easiest to memorize $\longrightarrow$ standard basis $\langle H,X,Y\,|\,[H,X]=2X\; , \;[H,Y]=-2Y\; , \;[X,Y]=H \rangle$ in order to have the example $\mathfrak{sl}_2$ from the textbooks with CSA $\langle H \rangle \,\longrightarrow$ and all the way back to the Pauli matrices to make the question "physics" compatible

At least I haven't chosen a weird basis in $\mathbb{C}x \oplus \mathbb{C}y$ and something like $3x-5y$ the maximal vector.

$\mathfrak{su}_2.\mathbb{C}=\{\,0\,\}$
\begin{align*}
(0,1) &\stackrel{X}{\longrightarrow} (1,0)\stackrel{X}{\longrightarrow} (0,0)\\
(1,0)&\stackrel{Y}{\longrightarrow}(0,1)\stackrel{Y}{\longrightarrow}(0,0)
\end{align*}
\begin{align*}
(0,0,1)&\stackrel{X}{\longrightarrow}(0,-i,0)\stackrel{X}{\longrightarrow}(2,0,0)\stackrel{X}{\longrightarrow}(0,0,0)\\
(1,0,0)&\stackrel{Y}{\longrightarrow} (0,-i,0) \stackrel{Y}{\longrightarrow} (0,0,2)\stackrel{Y}{\longrightarrow}(0,0,0)
\end{align*}

julian

Gold Member
A partial go a problem 21:

The deck transformations of the covering space $\tilde{X}$ constitute a group of homeomorphisms of that covering space (where the group operation is the usual operation of composition of homeomorphisms).

Definitions:

Definition of a homeomorphism:

A function $h: X \rightarrow Y$ between two topological spaces is a homeomorphism if it has the following properties:

$h$ is a bijection (one-to-one and onto),
$h$ is continuous,
the inverse function $h^{-1}$ is continuous ($h$ is an open mapping).

Proving the Group property of $\mathcal{D} (p)$:

Closure under composition of two deck transformations:

Say $p \circ h_1 = p$ and $p \circ h_2 = p$ then we wish to prove that $p \circ (h_1 \circ h_2) = p$. For any $\tilde{x} \in \tilde{X}$ we have

\begin{align*}
[p \circ (h_1 \circ h_2)] (\tilde{x}) &= p \big( h_1 \big( h_2 (\tilde{x}) \big) \big) \\
& = [p \circ h_1] \big( h_2 (\tilde{x}) \big) \\
& = p \big( h_2 (\tilde{x}) \big) \qquad \qquad \text{as } p \circ h_1 = p \\
& = [p \circ h_2] (\tilde{x}) \\
\end{align*}

hence $p \circ (h_1 \circ h_2) = p$.

Identity Homeomorphism is a deck transformation:

The identity map $id_{\tilde{X}} : \tilde{X} \rightarrow \tilde{X}$ is a homeomorphism and we obviously have $p \circ id_{\tilde{X}} = p$.

Inverse of a deck transformation is a deck transformation:

Say $p \circ h = p$. For any $\tilde{x} \in \tilde{X}$ we have:

\begin{align*}
p (\tilde{x}) &= [p \circ (h \circ h^{-1})] (\tilde{x}) \\
&= [p \circ h] \big( h^{-1} (\tilde{x}) \big) \\
& = p \big( h^{-1} (\tilde{x}) \big) \qquad \qquad \qquad \text{as } p \circ h = p \\
& = [p \circ h^{-1}] (\tilde{x}) \\
\end{align*}

hence $p \circ h^{-1} = p$.

Associativity of deck transformations:

Suppose that $h_1 : \tilde{X} \rightarrow \tilde{X}$, $h_2 : \tilde{X} \rightarrow \tilde{X}$, and $h_3 : \tilde{X} \rightarrow \tilde{X}$ are deck transformations. Then $p \circ [(h_3 \circ h_2) \circ h_1] = p \circ [h_3 \circ (h_2 \circ h_3)]$. Proof. For any $\tilde{x} \in \tilde{X}$ we have

\begin{align*}
[ p \circ [(h_3 \circ h_2) \circ h_1 ] ] (\tilde{x}) & = p \big( ( h_3 \circ h_2 ) \big( h_1 (\tilde{x}) \big) \big) \\
& = p \big( h_3 \big( h_2 \big( h_1 (\tilde{x}) \big) \big) \big) \\
& = p \big( h_3 \big( (h_2 \circ h_1) (\tilde{x}) \big) \big) \\
& = [ p \circ [h_3 \circ (h_2 \circ h_1)] ] (\tilde{x}) . \\
\end{align*}

Proof that $h (\tilde{x}) = \tilde{x}$ for $\tilde{x} \in \tilde{X}$ implies $h = id_{\tilde{X}}$:

First we prove that if $p \circ g = p \circ h$, where $g$ and $h$ are deck transformations, and $g (\tilde{x}) = h(\tilde{x})$ for some point $\tilde{x} \in \tilde{X}$ then $g = h$. It then easily follows that if $h (\tilde{x}) = \tilde{x}$ for some point $\tilde{x} \in \tilde{X}$ then $h = id_{\tilde{X}}$.

We prove that $A := \{ \tilde{x} \in \tilde{X} : g (\tilde{x}) = h (\tilde{x}) \}$ is both open and closed, thereby proving that $A = \tilde{X}$ because of the connectedness of $\tilde{X}$.

Both $g : \tilde{X} \rightarrow \tilde{X}$ and $h : \tilde{X} \rightarrow \tilde{X}$ are continuous maps as they are homeomorphisms. Let $\tilde{x} \in \tilde{X}$. There exists an open neighborhood $U \in X$ containing the point $p (g(\tilde{x}))$ with $p^{-1} (U)$ a disjoint union of open sets $V_\iota$ each of which is homeomorphically mapped onto $U$ by $p$ (for $\tilde{U} \in V_\iota$ we write $p | \tilde{U} : \tilde{U} \rightarrow U$ is a homeomorphism). One of these open sets contains $g (\tilde{x})$, denote it by $\tilde{U}$. Also one of these open sets contains $h (\tilde{x})$ (this is because $p \circ h = p \circ g$) Let us denote this open set by $\tilde{V}$. Let $N_{\tilde{x}} = g^{-1} (\tilde{U}) \cap h^{-1} (\tilde{V})$. Then $N_{\tilde{x}}$ is an open set in $\tilde{X}$ containing $\tilde{x}$ (open because $g$ and $h$ are continuous functions as they are homeomorphisms).

Consider the case when $\tilde{x} \in A$. Then $g (\tilde{x}) = h (\tilde{x})$, and therefore $\tilde{V} = \tilde{U}$. It follows from this that both $g$ and $h$ map the open set $N_{\tilde{x}}$ in to $\tilde{U}$. We now use that $p \circ g = p \circ h$ and that $p | \tilde{U} : \tilde{U} \rightarrow U$ is a homeomorphism to prove that $g | N_{\tilde{x}} = h | N_{\tilde{x}}$. Take an arbitrary point $\tilde{x}' \in N_{\tilde{x}}$ other than the point $\tilde{x}$. Suppose that $g (\tilde{x}') \not= h (\tilde{x}')$, but we have $(p \circ g) (\tilde{x}') = (p \circ h) (\tilde{x}')$. This says two different points get mapped to the same point by $p | \tilde{U}$, but that contradicts that $p | \tilde{U}$ is injective. As such we must have that $g (\tilde{x}') = h (\tilde{x}')$ for all $\tilde{x}' \in N_{\tilde{x}}$. Thus $N_{\tilde{x}} \subset A$. We have thus shown that for each $\tilde{x} \in A$ there exists an open set set $N_{\tilde{x}}$ such that $\tilde{x} \in N_{\tilde{x}}$ and $N_{\tilde{x}} \subset A$. Thus $A$ is open.

Next we show that the set $\tilde{X} / A$ is open as well. So consider the case $\tilde{x} \in \tilde{X} / A$. In this case $\tilde{U} \cap \tilde{V} = \emptyset$ since $g (\tilde{x}) \not= h (\tilde{x})$ (having both $g (\tilde{x})$ and $h (\tilde{x})$ in $\tilde{U}$ together with $(p \circ g) (\tilde{x}) = (p \circ h) (\tilde{x})$ is in contradiction with the injectivity of $p | \tilde{U}$). Again define $N_{\tilde{x}} = g^{-1} (\tilde{U}) \cap h^{-1} (\tilde{V})$. But $g (N_{\tilde{x}}) \subset \tilde{U}$ and $h (N_{\tilde{x}}) \subset \tilde{V}$. Therefore $g (\tilde{x}') \not= h (\tilde{x}')$ for $\tilde{x}' \in N_{\tilde{x}}$, and thus $N_{\tilde{x}} \subset \tilde{X} / A$. We have then shown that for each $\tilde{x} \in \tilde{X} / A$ there exists an open set $N_{\tilde{x}}$ such that $\tilde{x} \in N_{\tilde{x}}$ and $N_{\tilde{x}} \subset \tilde{X} / A$. Thus $\tilde{X} / A$ is open.

The subset $A$ of $\tilde{X}$ is therefore both open and closed. As $A$ was assumed to be non-empty, we deduce that $A = \tilde{X}$, because $\tilde{X}$ is connected. Thus $g = h$, which was the required result

The last step is to note that a function such that $h (\tilde{x}) = \tilde{x}$ for some point $\tilde{x} \in \tilde{X}$ and the identity function $id_{\tilde{X}}$ (satisfying $h (\tilde{x}) = id_{\tilde{X}} (\tilde{x})$ for this point $\tilde{x} \in \tilde{X}$) means that $h = id_{\tilde{X}}$.

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fresh_42

Mentor
2018 Award
A partial go a problem 21: ....
What do you mean by partial? It looks fine to me!

Only a few minor remarks:
• Subsequent application of functions is associative. Done. And $p\circ (h\circ g^{-1})=p$ does the other group properties in one step.
• For the second part, you could have started as you did with the pair $(h,g)$ but then switched to $g=1$. Just because it is less to write.
• $A$ is closed can be done shorter by using the fact that diagonals in Hausdorff spaces are closed. This is even equivalent to $T_2$.

julian

Gold Member
What do you mean by partial? It looks fine to me!

Only a few minor remarks:
• Subsequent application of functions is associative. Done. And $p\circ (h\circ g^{-1})=p$ does the other group properties in one step.
• For the second part, you could have started as you did with the pair $(h,g)$ but then switched to $g=1$. Just because it is less to write.
• $A$ is closed can be done shorter by using the fact that diagonals in Hausdorff spaces are closed. This is even equivalent to $T_2$.
I'm still looking into the issue of when homeomorphims form a group. I have proven that the general result that the composition of homeomorphisms is a homoemorphism and a result about an inverse function being a homeomorphism.

But I think there are subtle issues relating to the topology. For example the Identity Homeomorphism:

The identity map $id_X : X \rightarrow X$ is obviously bijective. However, the identity mapping $id_X : (X ,\tau) \rightarrow (X , \tau')$ is continuous if and only if $\tau' \subseteq \tau$. Thus if we topologize $X$ in such a way that this inclusion is proper the identity mapping in this direction will be continuous while its inverse (also the identity) will not. We will only have a continuous inverse if $\tau = \tau'$.

Does this mean that in order to have a group like structure do we have to consider only homeomorphims that preserve the topology $\tau$?

I have some intuition that deck transformation would preserve the topology of $\tilde{X}$ but haven't formed a rigorous proof yet.

fresh_42

Mentor
2018 Award
Does this mean that in order to have a group like structure do we have to consider only homeomorphims that preserve the topology?
I don't quite understand. The definition was for homeomorphisms, so the question is obsolete. Whether $p\circ h = p$ forms a group if $h$ isn't homeomorph is another issue. I don't think so, since the bijection might be sufficient for the algebraic properties, but not for the topological. And isn't the whole issue of homeomorphisms, that they preserve the topologies?
I have some intuition that deck transformation would preserve the topology of $\tilde{X}$ but haven't formed a rigorous proof yet.
It's defined that way.

julian

Gold Member
I was trying to prove the continuity of the inverse of $h^{-1}$ by showing $( h^{-1} )^{-1} = h$ using bijective properties and that $h$ is continuous. Now I realise you can and should prove the continuity of the inverse $h^{-1}$ by topological arguments. Cleared up now.

"Math Challenge - November 2018"

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