# Fundamental polygon

A First Course in String Theory

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2.5 Constructing simple orbifolds

(b) Consider a torus $\displaystyle{T^2}$, presented as the $\displaystyle{(x,y)}$ plane with the identifications $\displaystyle{x \sim x + 2}$ and $\displaystyle{y \sim y+2}$. Choose $\displaystyle{-1 < x, y, \le 1}$ as the fundamental domain. The orbifold $\displaystyle{T^2/\mathbb{Z}_2}$ is defined by imposing the $\displaystyle{\mathbb{Z}_2}$ identification $\displaystyle{(x,y) \sim (-x,-y)}$.

Prove that there are four points on the torus that are left fixed by the $\displaystyle{\mathbb{Z}_2}$ transformation. Show that the orbifold $\displaystyle{T^2/\mathbb{Z}_2}$ is topologically a two-dimensional sphere, naturally presented as a square pillowcase with seamed edges.

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To find the fixed points, we consider the cases when $\displaystyle{-x = x + 2m}$ and $\displaystyle{-y = y + 2n}$, where $\displaystyle{m,n \in \mathbb{Z}}$. Since the length of the interval is only 2, we can consider only the cases when $\displaystyle{m,n = 0, 1}$. Then the only solutions are

$\displaystyle{(0,0)}$
$\displaystyle{(0,1)}$
$\displaystyle{(1,0)}$
$\displaystyle{(1,1)}$

— Me@2021-01-17 04:14:44 PM

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— Wikipedia on Surface (topology)

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The formula for this topology is $\displaystyle{ABB^{-1}A^{-1}ABB^{-1}A^{-1} = ABB^{-1}A^{-1} }$, which is a sphere.

— Me@2021-01-29 06:10:46 PM

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# Cosmic computer

Consistent histories, 9

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There is a cosmic computer there

which is responsible to make sure that

quantum mechanics (laws) will always give consistent measurement results,

such as the ones of the EPR entangled pairs.

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NO. That is wrong.

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Instead, quantum mechanics itself is THAT cosmic computer that renders all the measurement results consistent.

— Me@2021-01-27 3:54 PM

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# Feeling

feeling ~ receiving data non-sequentially

— Me@2017-06-03 02:53:16 PM

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feeling

~ receiving data in parallel

~ receiving data at once

— Me@2021-01-27 08:19:33 PM

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For example, you would not feel that it is actually that the air molecules keep hitting you. You would not feel the speed and force of each individual particle. Instead, you have overall feelings of “pressure” and “temperature”.

Actually, you do not feel the temperature. Instead, you feel “cold”, “cool”, “warm”, or “hot”, etc.

— Me@2021-01-27 08:19:33 PM

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feeling

~ turning data into information

~ statistics in real time

— Me@2021-01-27 08:47:39 PM

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# 機遇創生論, 2.2

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（安：我剛才都是想到「大種子論」。）

（安：但「種子大論」比較奇怪。）

（安：這個比較難，因為，一個就講人生的某些方面，而另一個則講人的某些性質。換句話說，一個是人生理論，另一個則是人心理論。我暫時看不出，它們如何可以，自然地合體。）

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（安：等等。你剛才那兩句，互相矛盾。你第一句的意思是，一個是「心靈作業系統」，另一個即是「心靈應用程式」。「大種子論」歸類成「應用程式」。但是，在第二句，你又把「大種子論」歸類成「系統程式」（的一部分）。）

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（安：我們原本的「心靈作業系統」，似乎不是講這些東西。

— Me@2021-01-23 09:17:24 PM

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# 1990s, 6

— Photomyne colorization

— Me@2021-01-25 03:49:32 PM

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# Superposition always exists, 2.2.1

Decoherence and the Collapse, 2.1 | Quantum decoherence 7.2.1

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But wait! Doesn’t this mean that the “consciousness causes collapse” theory is wrong? The spin bit was apparently able to cause collapse all by itself, so assuming that it isn’t a conscious system, it looks like consciousness isn’t necessary for collapse! Theory disproved!

No. As you might be expecting, things are not this simple. For one thing, notice that this ALSO would prove as false any other theory of wave function collapse that doesn’t allow single bits to cause collapse (including anything about complex systems or macroscopic systems or complex information processing). We should be suspicious of any simple argument that claims to conclusively prove a significant proportion of experts wrong.

To see what’s going on here, let’s look at what happens if we don’t assume that the spin bit causes the wave function to collapse. Instead, we’ll just model it as becoming fully entangled with the path of the particle, so that the state evolution over time looks like the following:

$\displaystyle{|O, \uparrow \rangle \to \frac{1}{\sqrt{2}} |A, \downarrow \rangle + \frac{1}{\sqrt{2}} |B, \uparrow \rangle \to \frac{1}{\sqrt{2}}\sum_i \left( \alpha_i | i, \downarrow \rangle + \beta_i |i, \uparrow \rangle \right) = | \Psi \rangle}$

The interference has vanished, even though we never assumed that the wave function collapsed!

And all that’s necessary for that is environmental decoherence, which is exactly what we had with the single spin bit!

A particle can be in a superposition of multiple states but still act as if it has collapsed!

You might be tempted to say at this point: “Well, then all the different theories of wave function collapse are empirically equivalent! At least, the set of theories that say ‘wave function collapse = total decoherence + other necessary conditions possibly’. Since total decoherence removes all interference effects, the results of all experiments will be indistinguishable from the results predicted by saying that the wave function collapsed at some point!”

But hold on! This is forgetting a crucial fact: decoherence is reversible, while wave function collapse is not!!!

Now the two branches of the wave function have “recohered,” meaning that what we’ll observe is back to the interference pattern!

— Decoherence is not wave function collapse

— MARCH 17, 2019

— SQUARISHBRACKET

— Rising Entropy

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Decoherence is not wave function collapse

In case the original link does not work, use the Internet Archive version:

https://web.archive.org/web/20210124095054/https://risingentropy.com/decoherence-is-not-wave-function-collapse/

— Me@2021-01-24 07:14:50 PM

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A particle can be in a superposition of …

Note that it is not that the particle is in a superposition. Instead, it is that the system is in a superposition.

— Me@2021-01-24 07:16:49 PM

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2021.01.25 Monday ACHK

# Superposition always exists, 2.2.2

Decoherence and the Collapse, 2.2 | Quantum decoherence 7.2.2

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superposition ~ indistinguishability

superposition state ~ logically indistinguishable states (forming one SINGLE quantum state)

logically indistinguishable ~ indistinguishable by definition ~ indistinguishable due to “the experiment setup is without detector” part of the definition

By the Leibniz’s Law (Identity of indiscernibles), logically indistinguishable cases are actually the same one SINGLE case, represented by one SINGLE quantum state.

Classically, there are no such logically indistinguishable cases because classically, all particles are distinguishable. So the probability distribution in the newly invented non-classical state should be completely different from any probability distributions provided by classical physics. Such cases of a new kind are called quantum states.

A quantum state’s probability distribution can be calculated from its wave function.

“Why that single quantum state is represented by a superposition of eigenstates and why its wave function is governed by the Schrödinger equation” is ANOTHER set of questions, whose correct answers may or may not be found in the Wikipedia article Theoretical and experimental justification for the Schrödinger equation.

Superpositions always exist. Logically indistinguishable cases are always there. You just trade some logically indistinguishable cases with some other logically indistinguishable cases.

The “superpositions” are superpositions in definition, in language, in logic, in calculation, and in mathematics, but not in physical reality, not in physical spacetime.

— Me@2021-01-24 09:29:13 PM

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# … is in a superposition

Quantum decoherence 5.3.2 | Wheeler’s delayed choice experiment 1.2

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For example, in the double-slit experiment, if no detector is installed, the system is in a quantum superposition state.

It is not that each individual photon is in a superposition. Instead, it is that the system of the whole experimental setup is in a superposition.

— Me@2021-01-23 12:57 AM

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[guess]

However, “what ‘the whole experimental setup‘ is” is not 100% objective. In other words, it is a little bit subjective.

“The whole experimental setup”, although largely objective, is partially defined with respect to an observer.

— Me@2021-01-23 12:58 AM

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So quantum probability/indistinguishability effect is partly observer-dependent, although the subjectivity is just tiny compared with that of the classical probability in a mixed state.

— Me@2021-01-23 12:59 AM

[guess]

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# Consistent histories, origin

The square root of the probability, 6

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There is no wave function collapse.

For example, in the double-slit experiment, with-detector and without-detector are actually two different physics systems. Different experimental setups provide different probability distributions, encoded in the wave functions. So different experimental setups result in different wave functions.

That is the key to understanding strange quantum phenomena such as EPR. A classical system has consistent results is no magic.

You create either a system with a detector or a system without a detector. With a detector, it will have only distinguishable-at-least-in-definition states, aka classical states. A system with only classical states is a classical system. Then, why so shocked when a classical system has consistent results?

Quantum mechanics is “strange”, but not “that strange”. It is not so strange that it is unexplainable.

— Me@2021-01-20 07:11 PM

— Me@2021-01-22 08:48 AM

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# 2004, 2

— Me@2021-01-22 07:24:57 PM

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# Ex 1.15 Central force motion

Structure and Interpretation of Classical Mechanics

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Find Lagrangians for central force motion in three dimensions in rectangular coordinates and in spherical coordinates. First, find the Lagrangians analytically, then check the results with the computer by generalizing the programs that we have presented.

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The Lagrangians in rectangular coordinates:

\displaystyle{ \begin{aligned} &L (t; x, y, x; v_x, v_y, v_z) \\ &= \frac{1}{2} m \left( v_x^2 + v_y^2 + v_z^2 \right) - U \left( \sqrt{x^2 + y^2 + z^2} \right) \\ \end{aligned}}

The Lagrangians in spherical coordinates:

\displaystyle{ \begin{aligned} &L (t; r, \theta, \phi; \dot r, \dot \theta, \dot \phi) \\ &= \frac{1}{2} m r^2 \dot \phi^2 (\sin \theta)^2 + \frac{1}{2} m \dot \theta^2 r^2 + \frac{1}{2} m \dot r^2 - U (r) \\ &= \frac{1}{2} m \left( \dot r^2 + r^2 \dot \theta^2 + r^2 (\sin^2 \theta) \dot \phi^2 \right) - U (r) \\ \end{aligned}}

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(define ((F->C F) local)
(->local (time local)
(F local)
(+ (((partial 0) F) local)
(* (((partial 1) F) local)
(velocity local)))))

(define (p->r local)
(let ((polar-tuple (coordinate local)))
(let ((r (ref polar-tuple 0))
(theta (ref polar-tuple 1))
(phi (ref polar-tuple 2)))
(let ((x (* r (sin theta) (cos phi)))
(y (* r (sin theta) (sin phi)))
(z (* r (cos theta))))
(up x y z)))))

(define ((L-central-rectangular m U) local)
(let ((q (coordinate local))
(v (velocity local)))
(- (* 1/2 m (square v))
(U (sqrt (square q))))))

(define (L-central-polar m U)
(compose (L-central-rectangular m U) (F->C p->r)))

(show-expression
((L-central-polar 'm (literal-function 'U))
(->local 't
(up 'r 'theta 'phi)



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— Me@2021-01-22 02:59:11 PM

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# Quantum Computing, 2.2

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2021.01.21 Thursday ACHK

# Fallacy of infinity thinking, prequel

Why should you NOT murder one innocent person in order to save millions of people?

(Note: This question is NOT the same as “Could we give up one innocent person’s life in order to save millions of people’s?”)

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Could we murder one innocent person in order to save millions of people?

— Me@2017-06-20 01:04:56 PM

— Me@2021-01-19 06:15:54 PM

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NO.

If we can murder one innocent person in order to save all other people, then no one is safe after all, because anyone could be THAT innocent person, being sacrificed at any time.

(In the situation that we cannot save all the people at the same time, which person or which group has higher or lower priorities depends on context. There is no universal answer.)

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Instead, if we protect each person’s life, then all the people’s lives are protected.

— Me@2021-01-19 06:01:24 PM

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If you start with protecting each one, then every one person and thus the whole society will be protected.

If you start with protecting the whole society at all costs, then no one will be safe, because anyone could be that cost; any innocent person could be sacrificed at any time.

— Me@2021-01-20 6:48 AM

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# 鐵達尼極限 2.2

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Enjoy everything, need nothing, …

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… especially for human relationships.

– Conversations with God

（那某些原因，其實就是「基因」。「基因」只顧自己生命的延續，企圖不斷複製自己，即使犧牲「基因載體」的利益、幸福，甚至生命，也在所不惜。「基因載體」者，人或其他生物也。）

— Me@2021-01-03 04:18:55 PM

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# 1985

— Photomyne colorization

— Me@2021-01-18 10:14:30 PM

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# Problem 2.5b1

A First Course in String Theory

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2.5 Constructing simple orbifolds

~~~

The wikipedia page “Fundamental polygon”, specifically the subsection entitled “group generators”, has a serious mathematical error. You cannot derive a presentation for the fundamental group from the fundamental polygon using the side labels in the manner described on that page (and which you have copied), unless all of the vertices of the polygon are identified to the same point. In the picture you provided and which can be seen on that page, one opposite pair of vertices of the square is identified to one point on the sphere, the other opposite pair of vertices is identified to a different point on the sphere.

There is still a way to derive a presentation for the fundamental group from a fundamental polygon, but it is not the way described on the wikipedia page. In the sphere example of your question, you have to ignore one of the two letters $\displaystyle{A}$, $\displaystyle{B}$, keeping only the other letter. For example, ignoring $\displaystyle{A}$ and keeping $\displaystyle{B}$, you get a presentation $\displaystyle{ \langle B \mid B B^{-1} = 1 \rangle }$, which is a presentation of the trivial group. The way you tell which to ignore and which to keep is by taking the quotient of the boundary of the polygon which is a graph with vertices and edges, choosing a maximal tree in that graph, ignoring all edge labels in the maximal tree, and keeping all edge labels not in the maximal tree.

On that wikipedia page, the Klein bottle and the torus examples are correct and you do not have to ignore any edge labels: all vertices are identified to a single point and the maximal tree is just a point. The sphere and the projective plane examples are incorrect: the four vertices are identified to two separate points, the maximal tree has one edge, and you have to ignore one edge label. The example of a hexagon fundamental domain for the torus is also incorrect: the six vertices are identified to two separate points, the maximal tree has one edge, and you have to ignore one edge label.

edited Jul 23 ’14 at 17:17

answered Jul 23 ’14 at 17:11

Lee Mosher

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yes, i thought that the fundamental polygon is this quotient space. – user159356 Jul 23 ’14 at 17:28

That’s backward: in your example, the sphere is the quotient space of the fundamental polygon, not the other way around. – Lee Mosher Jul 23 ’14 at 17:30

— Mathematics Stack Exchange

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2021.01.18 Monday ACHK

# Summing over histories, 2

The square root of the probability, 5

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If there is more than one way to achieve the present state, present == sum over all possible pasts, with weightings.

— Me@2011.06.26

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This is false for a physical state. This is only true for wave functions, which are NOT probabilities.

Wave functions are used for calculating probabilities; but they are not themselves probabilities.

Wave functions are quantum states, but not physical states.

Wave functions are logical and mathematical, but not physical.

A physical state is something observable, something can be measured, at least in principle.

A physical state is something that exists in spacetime, a wave function is not.

— Me@2021-01-16 06:12:08 PM

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# Zooming out, 2

In this morning’s dream, I could hold my lucid dream without feeling a headache. I felt comfortable as awake.

I was in a city and listening to an Inception-style music.

I tried to fold the city as in Inception so that I could have walked on a vertical land.

However, it did not work. The city got folded together, not as an L shape, [?]but as an L with a horizontal line head shape.[?]

Then I got myself waking up by flying-falling downwards.

— Me@2011.08.03

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# 機遇創生論, 2.1

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「緣份攻略」都不行，因為感覺有點怪。

（安：那就不如叫做「緣份理論」。）

「理論」很空泛。不應把「理論」，視為名字的一部分。

（安：不如叫做「超級種子理論」，或者「廣義種子理論」？）

（安：無錯，那相當「拗口」。)

（安：我剛才都是想到「大種子論」。）

（安：但「種子大論」比較奇怪。）

— Me@2021-01-13 04:52:02 PM

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