Bernard d’Espagnat
Here I would like to mention one particular physicist and philosopher: Bernard d’Espagnat. In his paper “Problems in Objectizing”, d’Espagnat opts for “Mind-Independent Reality”, and, in particular, he also supports Hans Primas’s proposals which EEQT developed into a full-fledged mathematical model.
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But even d’Espagnat is somewhat pessimistic about the final results of such an approach. He ends his paper with the comment:
…although we cannot know Mind-Independent Reality, still we get, through physics, not fully deceitful glimpses of it. I feel Primas’s optimistic guess somehow strengthens my position in this matter.
Waves and Things: Information!
In EEQT we are dealing with two kinds of objects: material objects about the existence of which we have no doubts, and which are necessary if we want to tell engineers what to do; but also we have “wave functions” that are necessary for modeling quantum phenomena - two different categories of objects. These two kind of objects are coupled together – there is an exchange of information between them. That is why in our theory we have not only “dynamics” – exchange of forces, but also, what we called, “binamics” – exchange of “bits of information”. Therefore our theory can be considered as one step towards an introduction to the active role of information in physics.
I was already well-situated in my career as a scientist when I made the decision to follow this path so it was not so easy to attack me – it was enough to just ignore my work.
The Three Questions of John Archibald Wheeler
Here comes the important term that is rather new in physics: Information. It has become more popular since John Archibald Wheeler published his famous “It from Bit” article "Information, Physics, Quantum: The Search for Links", and asked three important questions:
How come existence?
How come the quantum?
How come the “one world” out of many observer-participants?
The simplest answer to the first question, “How come existence”, is, I would say, existence comes from “timeless information”.
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What does John Wheeler mean by his famous 'It from bit' idea?
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John Wheeler's "It from bit" idea is a concept that suggests that all physical reality is ultimately based on information. In other words, the physical world can be thought of as being made up of bits of information rather than physical particles. This idea is rooted in the field of quantum mechanics and suggests that the universe is fundamentally digital in nature.
Tell me more
P.S.2.
"When I speak, the reason says, “This will I say”; but God
takes the word out of my mouth and the lips say something else
at which reason trembles."
Sri Aurobindo
P.S.3. New version of my notes on the conformal group. Will add still more today.
"In EEQT we are dealing with two kinds of objects: material objects about the existence of which we have no doubts, and which are necessary if we want to tell engineers what to do; but also we have “wave functions” that are necessary for modeling quantum phenomena - two different categories of objects.".
ReplyDeleteRather, I would say that these are classes of objects, for categories require the identification of both objects and morphisms. I think this can be quickly rectified, however. What will be the morphisms in these categories?
The way I see it is this:
Material objects encompass all kinds of particles and systems that can be described by classical dynamical equations. The morphisms for these objects could include classical dynamical operations such as translations, rotations, as well as physical processes like collisions, thermodynamic phenomena, etc.
Wave functions, however, are described in the framework of quantum mechanics and represent the quantum states of a system. The morphisms for these objects could include quantum operations like evolution according to the Schrödinger equation, projections resulting from quantum measurements, as well as quantum transformations such as rotation and translation. Quantum observables would be morphisms too.
However, it's worth noting that one of the key aspects of EEQT is the merging of these two "categories" through processes like wave function collapse that affect both the quantum states and classical evolution. Thus, while we can think about these two categories separately, they are tightly interconnected in the context of EEQT.
So I would venture a question. Would it not be a good idea to create a quotient category given by a certain class of abstractions? For what is the transition between the two categories we have now, assuming that it can be imagined as I have described?
"Both multiplication and *-structure are too rigid to include information.".
DeleteAnd what does it require to include information? Why do you think this is currently impossible?
"And what does it require to include information? Why do you think this is currently impossible?"
DeleteA "feeling".
Quoting from : "Towards the theory of matter, geometry and information"
Delete"Continuing the analogy : in the same way as a gravitational field curves space-time [f6], the information field may curve the state space. May change the geometry of the space of quantum states. May enable the flow of information and of energy through new channels. Now quantum matter gets a worthy partner, just as the gravitational field was a worthy partner to classical matter. The same way as gravitational field is local[f7] in space-time, the information field is local in Hilbert space where " near" means " similar". The geometry of the information field must be, as we have said, a nonlinear geometry. Only in this way can we explain the stability of structures, such as the structure of life. With the phenomenon of life we can in this way, associate a topological invariant (a kind of a vortex) in the nonlinear field of information. Physical and chemical life processes would be then controlled by a quantum feedback between information and matter. And, when we speak about geometry, it must be noted that it must be more than a classical geometry such as is sufficient for the Einstein theory of gravitation. What is needed here is a kind of quantum geometry. Such a geometry is today only in statu nascendi."
"A "feeling".".
DeleteIt seems to me that the term 'feeling' used in this context is a metaphorical placeholder for a theoretical construct we have yet to fully comprehend and incorporate into our mathematical models.
When we say that both multiplication and *-structure are too rigid to include information, we're indicating that our traditional mathematical operations may not adequately encapsulate all facets of information transfer and processing. For instance, they might not be sufficiently flexible to capture dynamic, context-dependent processes that characterize complex information systems.
As for the 'feeling', it can be interpreted as our ability to intuitively understand and respond to these complex, often non-linear, dynamics in a way that is currently difficult to formalize mathematically. Our challenge, therefore, is to develop new mathematical tools or frameworks that can emulate this 'feeling'—the subtle, nuanced interplay of elements within an information system.
So isn't this 'feeling' simply the same as a semantic structure that we cannot describe at the moment? Doesn't this prompt us to mathematise philosophy? After all, we want to capture the 'feeling', and more specifically it seems to me that we just want to capture the sense. We are concerned with semantics. Time, too, has a semantic structure, but nobody cares about that because it's 'no longer physics, but philosophy, metaphysics'. Yes, but quantum mechanics used to be outside physics too. And I don't see any other way than combining separated disciplines. Otherwise we will go no further.
What I know is that mathematically EEQT is almost ok, and it does its job concerning physical applications of quantum theory. However its main category, the category of C* algebras is too restrictive. Both multiplication and *-structure are too rigid to include information. I have only glimpses of how it should be extended - a task for the future. And only then there will be something to guide us in the creation of a consistent philosophy.
Delete"What I know is that mathematically EEQT is almost ok, and it does its job concerning physical applications of quantum theory. However its main category, the category of C* algebras is too restrictive. Both multiplication and *-structure are too rigid to include information. I have only glimpses of how it should be extended - a task for the future. And only then there will be something to guide us in the creation of a consistent philosophy.".
DeleteOne possible avenue is to consider a new concept of "Semantic C*-algebras", essentially extending the traditional C*-algebras with additional structure that carries semantic information.
Formally, let's denote the category of C*-algebras as $\mathcal{A}$, and define a new category $\mathcal{A'}$ of "Semantic C*-algebras" as follows:
Objects in $\mathcal{A'}$ are pairs $(A, \sigma)$ where $A$ is a C*-algebra and $\sigma$ is a "semantic structure" on $A$.
Morphisms in $\mathcal{A'}$ are *-homomorphisms that preserve the semantic structure.
A semantic structure $\sigma$ could be thought of as a function that assigns to each element of the C*-algebra a "meaning" from some set of meanings $\mathcal{M}$, i.e. $\sigma : A \rightarrow \mathcal{M}$. The preservation of the semantic structure by a *-homomorphism $f$ could then mean that for all $a \in A$, $f(\sigma(a)) = \sigma(f(a))$.
While the specifics of what the set of meanings $\mathcal{M}$ and the semantic structure function $\sigma$ would be, and how they should interact with the algebraic operations in $A$, are open problems, they should be designed to capture the semantic aspects that we're interested in.
As an illustration, suppose we have two semantic C*-algebras $(A, \sigma_A)$ and $(B, \sigma_B)$ in $\mathcal{A'}$, and suppose that $\sigma_A(a) = \sigma_B(b)$ for some $a \in A$ and $b \in B$. Then, by the preservation of semantic structure, for any *-homomorphism $f: A \rightarrow B$ in $\mathcal{A'}$, we would have $\sigma_B(f(a)) = \sigma_B(b)$, meaning that $f$ preserves the "meaning" of $a$ from $A$ to $B$.
In terms of a theorem, the extension above guarantees that "semantics-preserving" morphisms indeed preserve semantics.
Well. I know why we don't deal with 'spirit'. Popper's verifiability and falsifiability. Well, yes, paranormal phenomena do not meet the criteria of repeatability of an experiment, therefore they do not exist, right? Well, electrons don't meet this criterion either, because repeatable are the probability distributions of the measurement, not the measurements themselves.
ReplyDeleteIn terms of the philosophy of science, I much prefer Kuhn's approach.
"P.S.3. New version of my notes on the conformal group. Will add still more today.".
ReplyDeleteIn (61) you probably forgot to close the bracket.
And also correct the \tilde{\gamma} above.
ReplyDeleteThank you. Corrected. The part with (61) has been removed. It does not belong to this section - it was put there by one of my many mistakes.
DeleteIn (53) you have not corrected \gamma.
DeleteFixed. Thank you.
DeleteBjab -> Ark
DeleteZdania na 4 stronie pod wzorem (17):
"It follows that the points of M ̃ which are not in the image of
M ̃..."
nie rozumiem - coś jest złe w tym zdaniu.
Fixed. Thank you.
DeleteBjab -> Ark
DeleteWiersz nad wzorem (18):
"dimension 4.." ->
dwie kropki
Wzor (18)
Czwórki powinny być w indeksie górnym
Obrazek w Proposition 1 ma jekieś dziwne literki przy strzałkach.
Fixed. Updated. Thank you.
DeleteBjab -> Ark
DeleteWzór (24) wydaje mi się błędny.
Pod wzorem (25) mamy:
"descends to its
action on
tildeM = PN ." ->
Niepotrzebne przejście do nowego wiersza i niezwinięty wyraz tildeM
(24) was not just wrong. It was nonsensical. tildeM also fixed. Thank you!
DeleteBjab -> Ark
DeleteW proposition 2 i w jej proofie nie zwinęło się tildeM.
tildeM nie zwinęło się też w Remark 1.
About Barbour:
ReplyDelete"His theory, called "Platonia" after the ancient philosopher Plato, proposes that reality consists of an infinite number of "nows", which are different configurations of the universe. Each of these "nows" is a static snapshot of the universe in a particular state. Our experience of time arises from comparing these different "nows" in our minds."
My wife favors this view. She says that time is an artificial creation, an illusion, and that one should think of time as the wheel of a slide projector. At any given point along the way, you are watching one particular slide. But all the rest of the slides are present on the wheel. The light that illuminates each slide is consciousness/perception. At the center of the wheel is the All.
"My wife favors this view. She says that time is an artificial creation, an illusion, and that one should think of time as the wheel of a slide projector. At any given point along the way, you are watching one particular slide. But all the rest of the slides are present on the wheel. The light that illuminates each slide is consciousness/perception. At the center of the wheel is the All.".
ReplyDeleteYes, this is an appropriate metaphor, however, the thing is that the devil is in the details. This needs to be worked out and described mathematically. It requires a special 'feel' and an extraordinarily subtle handling.
For this will start a new physics.
The consequence of properly capturing this hidden, final aspect of reality that has eluded us for millennia must be real on a principle such that it can be given to engineers or other biologists or technical physicists to do so, a structure describing the design of the Time Machine. Further challenges will no longer be posed to physics, but to metaphysics. The Time Machine is the final task of physics. Renormalisation.
ReplyDeleteBut then, if Time is an illusion, a "time machine" would necessarily be a matter of tweaking consciousness. Or some such thing.
Delete"But then, if Time is an illusion, a "time machine" would necessarily be a matter of tweaking consciousness. Or some such thing.".
DeleteOn the one hand yes, but on the other no. Things exist and events happen!
Tweaking consciousness is one thing. Mystical experiences are a separate category, these are not intersubjective. But there are more non-intersubjective things in physics, and there are also apparently objective issues. The issue is really subtle, and the devil this time hides in details I never dreamed of before.
The point is that this world of ideas is primary, but materialising an object from the world of ideas will require translating it into signs. In this case, I certainly have mathematical symbols in mind, but that is a lot of work. Nevertheless, it is the most fascinating task in the world.
@ M.S. That is what I work on from time to time. My wife inspires, I perspire.
ReplyDeleteUpdated the "Conformal group" pdf. Added at the end an exercise for readers.
ReplyDeleteWyraz tildeM się nie zwinął w trzech miejscach:
Delete1. Proposition 2
2. Proof of Proposition 2
3. Remark 1
we cane ->
we can
"Updated the "Conformal group" pdf. Added at the end an exercise for readers.".
DeleteThe problem is asking us to demonstrate that the action of the orthogonal group $O(4,2)$ on the compactified Minkowski space $\hat{M}$ is faithful. In other words, we need to show that if a group element $g$ in $O(4,2)$ acts as the identity on all points of $\hat{M}$, then $g$ must be the identity element of $O(4,2)$.
To begin, let's consider a matrix $A \in O(4,2)$. By definition, $A$ is orthogonal, which means that $A^T g A = g$ where $g$ is the Minkowski metric. Now, suppose that $A$ acts as the identity on all points in $\hat{M}$, i.e., $A.X = X$ for all $X \in \hat{M}$.
But the definition of the action of $O(4,2)$ on $\hat{M}$ is given by $A.X = AX$, which is just the usual matrix multiplication. So if $A.X = X$ for all $X$, this means that $AX = X$ for all $X$.
In particular, this must be true for the basis vectors of the space. Thus we have $A e_i = e_i$ for all $i = 1, ..., 6$, where $\{e_i\}$ is the standard basis of $\mathbb{R}^{4,2}$. But this precisely means that $A$ is the identity matrix in $O(4,2)$, as required.
Hence, the action of $O(4,2)$ on $\hat{M}$ is faithful.
Do you have any comments?
@M.S. "To begin, let's consider a matrix $A \in O(4,2)$. By definition, $A$ is orthogonal, which means that $A^T g A = g$ "
DeleteThat is an incorrect understanding.
"That is an incorrect understanding.".
DeleteWhat do you precisely mean? In the context of Lie groups, when we say $A \in O(4,2)$, we mean $A$ is an element of the orthogonal group $O(4,2)$.
The orthogonal group $O(4,2)$ is the group of matrices that preserve the quadratic form of a pseudo-Euclidean space of signature $(4,2)$, not necessarily the standard dot product in Euclidean space. This space has 4 positive dimensions and 2 negative dimensions.
The condition for a matrix $A$ to be in $O(4,2)$ is that it should satisfy the condition $A^T g A = g$, where $g$ is the metric tensor (which can be represented as a matrix) of the space, and $A^T$ is the transpose of $A$.
Is this wrong?
Perhaps Bjab will find out why it is incorrect. He pays attention to details.
Delete"Perhaps Bjab will find out why it is incorrect. He pays attention to details.".
DeleteThe statement "To begin, let's consider a matrix $A \in O(4,2)$. By definition, $A$ is orthogonal, which means that $A^T g A = g$ " is generally correct, but could be misleading if not understood properly.
The orthogonal group $O(4,2)$ is the group of matrices that preserve a specific bilinear form defined by the metric $g$ in a pseudo-Euclidean space of signature (4,2). So, the condition $A^T g A = g$ for a matrix $A$ to belong to $O(4,2)$ is correct.
However, what could potentially be misunderstood here is the meaning of "$A$ is orthogonal". In the context of $O(4,2)$, "$A$ is orthogonal" means "$A$ preserves the pseudo-Euclidean bilinear form defined by $g$", not that "$A$ preserves the standard Euclidean dot product". If one is accustomed to the terminology from a context where "orthogonal" refers exclusively to preserving the standard Euclidean dot product, then this could lead to confusion.
@Bjab
DeleteFixed. Updated. Thank you.
Bjab -> Ark
DeletePowyżej wzoru (29) jest coś takiego:
BR3,1
Powyżej wzoru (34) mamy coś takiego:
X5 − X5
Poniżej wzoru (34) mamy coś takiego:
written im ->
written in
@Bjab.
DeleteFixed. Updated. Thank you. Added another subsection and another exercise.
"Added another subsection and another exercise.".
Delete$\tau_+(M)$ and $\tau_-(M)$ are the images of Minkowski space $M$ under the maps $\tau_+$ and $\tau_-$ respectively.
By the definitions of $\tau_+$ and $\tau_-$, we can see that for a given $x \in M$:
$\tau_+(x) = (x, \frac{1}{2}(1 - q(x)), -\frac{1}{2}(1 + q(x)))$
and
$\tau_-(x) = (-x, -\frac{1}{2}(1 - q(x)), \frac{1}{2}(1 + q(x)))$
Let's consider the intersection $\tau_+(M) \cap \tau_-(M)$.
This is the set of all $X \in M˜$ such that $X = \tau_+(x) = \tau_-(y)$ for some $x,y \in M$.
This means that we need to solve the system of equations:
$(x, \frac{1}{2}(1 - q(x)), -\frac{1}{2}(1 + q(x))) = (-y, -\frac{1}{2}(1 - q(y)), \frac{1}{2}(1 + q(y)))$
This system is solved by $x = -y$ and $q(x) = q(y)$. Since $q$ is a quadratic form, it is invariant under sign changes, so $q(-x) = q(x)$.
Thus, we can see that the intersection $\tau_+(M) \cap \tau_-(M)$ is not empty - it includes at least the points of the form $\tau_+(x) = \tau_-(-x)$ for all $x \in M$.
Bjab -> Ark
DeleteW (38) za dużo literek T.
W (38) za dużo przecinków i literek s.
Czy w (36) nie przydałyby się dwie transpozycje w macierzy C?
DeleteFixed. Updated. But the number of commas in (38) seems to me ok. Thank you.
Delete@M.S.
Delete" (x, \frac{1}{2}(1 - q(x)), -\frac{1}{2}(1 + q(x))) =
(-y, -\frac{1}{2}(1 - q(y)), \frac{1}{2}(1 + q(y)))
This system is solved by $x = -y$"
x=-y does not solve the system.
Yes, you are correct. If $x = -y$, then we would have $\tau_+(x) = \tau_-(-x)$, but this would result in two points in $\tilde{M}$ that are negatives of each other, not the same point as required for them to be in the intersection $\tau_+(M) \cap \tau_-(M)$.
DeleteThis indicates that the intersection $\tau_+(M) \cap \tau_-(M)$ is indeed empty because there does not exist $x, y \in M$ such that $\tau_+(x) = \tau_-(y)$.
Therefore, the intersection of the images of $\tau_+(M)$ and $\tau_-(M)$ is empty.
Correction:
ReplyDelete"To begin, let's consider a matrix $A \in O(4,2)$. By definition, $A$ is orthogonal, which means that $A^T g A = g$" -> "To begin, let's consider a matrix $A \in O(4,2)$. By definition, $A$ preserves a quadratic form, which means that $A^T g A = g$"
@M.S.
Delete"$A^T g A = g$""
With the notation as in the paper that is incorrect.
q instead of g. Is that what you mean?
Delete@M.S. "$A^T g A = g$" has incorrect syntax. Mathematica would complain: "Tensors g and A" have incompatible shapes.
Delete"@M.S. "$A^T g A = g$" has incorrect syntax. Mathematica would complain: "Tensors g and A" have incompatible shapes.".
DeleteIn the context of the orthogonal group $O(4,2)$, the matrices involved are 6x6 matrices. If $A$ is a matrix in $O(4,2)$ and $g$ is a 6x6 matrix representing the metric tensor, then the multiplication operations in $A^T g A = g$ should be valid, since all the matrices involved have the same dimensions...
Maybe instead of "$A^T g A = g$", we should write "$A^T g_{ij} A = g_{ij}$" for all "i" and "j". This is a more precise way of stating that $A$ is orthogonal with respect to the metric $g$.
DeleteIf not in this way, how else do you see it?
g is 4x4. G is 6x6.
Deleteg is 4x4. G is 6x6.
DeleteOk, indeed. I marked as g what you have in formula (23) and in your paper this is G. You are right.
It was not only "marking". You wrote
Delete" where $g$ is the Minkowski metric"
" where $g$ is the Minkowski metric".
DeleteOk. Thus, the sentence should read:
"To begin, let's consider a matrix $A \in O(4,2)$. By definition, $A$ is orthogonal (in the sense of preserving the quadratic form), which means that $A^T G A = G$, where $G$ is the metric tensor of the pseudo-Euclidean space of signature $(4,2)$. (A is 6x6 and G is 6x6)"
Now it should be correct.
Latest exercise:
DeleteLet's consider the action of $\hat{\pi}$ on $M$ via $\tau^+$ and $\tau^-$.
$\hat{\pi} \circ \tau^+(M)$ denotes the set of all points in $N/ \approx$ that are the images of points in $M$ through the map $\tau^+$ followed by the projection $\hat{\pi}$.
Similarly, $\hat{\pi} \circ \tau^-(M)$ denotes the set of all points in $N/ \approx$ that are the images of points in $M$ through the map $\tau^-$ followed by the projection $\hat{\pi}$.
Given that $\hat{\pi}$ is a canonical projection, it maps points from $N$ to their equivalence classes in $N/ \approx$. Therefore, it may occur that different points from $N$ are mapped to the same equivalence class in $N/ \approx$.
Indeed, looking at the maps $\tau^+$ and $\tau^-$, we can see that for any point $x \in M$, $\tau^+(x) = -\tau^-(-x)$. This implies that these two points will be mapped to the same equivalence class by the projection $\hat{\pi}$, as they are opposite points in $N$ and thus belong to the same equivalence class.
From this, we infer that the intersection $(\hat{\pi} \circ \tau^+(M)) \cap (\hat{\pi} \circ \tau^-(M))$ is non-empty, as it contains at least one equivalence class for each point $x \in M$.
"But the definition of the action of $O(4,2)$ on $\hat{M}$ is given by $A.X = AX$"
DeleteThis is incorrect. This is the action on N. But \hat{M} is the set of equivalence classes of X, not the set of X's themselves.
PS.1. latest excercise:
DeleteThe canonical projection $\hat{\pi}$ is indeed acting on the space $N$, not on $M$. However, $M$ is embedded into $N$ through the mappings $\tau^+$ and $\tau^-$. So when we talk about $\hat{\pi} \circ \tau^+(M)$ and $\hat{\pi} \circ \tau^-(M)$, we are referring to the action of the projection $\hat{\pi$} on the images of points in $M$ under the mappings $\tau^+$ and $\tau^-$, which are in $N$.
"x \in M$, $\tau^+(x) = -\tau^-(-x)$. This implies that these two points will be mapped to the same equivalence class by the projection $\hat{\pi}$"
DeleteThis is incorrect. \hat{\pi} is defined through the equivalence relation \approx defined in Eq. (26).
Ok. I see Eq. (26).
DeleteSo, the equivalence relation $\approx$ is defined such that $X \approx X'$ if and only if $X = \lambda X'$, for $\lambda > 0$. In other words, two vectors are equivalent if they are proportional and $\lambda$ is positive. Under this relation, it's true that $[X] = [-X]$, because they're proportional ($\lambda = -1$), but $[[X]] \neq [[-X]]$, because $\lambda$ needs to be positive for $\approx$.
Given this, $\hat{\pi}(\tau^+(x))$ and $\hat{\pi}(\tau^-(-x))$ would indeed not be in the same equivalence class, because $\tau^+(x)$ and $\tau^-(-x)$ are negatives of each other. They would be equivalent under the equivalence relation $\sim$, but not under the relation $\approx$.
Therefore, $\hat{\pi} \circ \tau^+(M)$ and $\hat{\pi} \circ \tau^-(M)$ would indeed be disjoint under the equivalence relation $\approx$.
"This is a logically incorrect conclusion.".
DeleteOk, however, from the definition of the equivalence relation $\approx$, it is true that $\tau^+(x)$ and $\tau^-(-x)$ are not equivalent, because they are negatives of each other. This means that under the equivalence relation $\approx$, these points will not fall into the same equivalence class, and thus $\hat{\pi}(\tau^+(x))$ and $\hat{\pi}(\tau^-(-x))$ will represent different points in the quotient space $N / \approx$.
However, this does not necessarily mean that the image sets $\hat{\pi} \circ \tau^+(M)$ and $\hat{\pi} \circ \tau^-(M)$ are disjoint. It simply means that each point in $M$ will map to different points in the quotient space under the two mappings. There could still be other points in $M$ that map to the same point in the quotient space under $\tau^+$ and $\tau^-$, leading to an overlap of the image sets.
However, do I have enough information to definitively state whether these images will be disjoint or not? What other formulas from the paper should I use?
For a while I removed my last comment, because I got all confused myself.
Delete"For a while I removed my last comment, because I got all confused myself.".
DeleteIf so, I apologize for introducing confusion. Then you might want to go over it step by step the way you record difficult parts bar by bar in a music studio.
Hint: non-emptiness would require existence of x,y such that
Delete\tau_+ (x) = \lambda \tau_- (y)
for some \lambda>0.
"Hint: non-emptiness would require existence of x,y such that
Delete\tau_+ (x) = \lambda \tau_- (y)
for some \lambda>0.".
I know, I am just solving this system of equations and I think there is a contradiction and hence the intersection will be empty. I'll write it down explicitly right away.
Substituting the expressions for $\tau_+$ and $\tau_−$, we have:
Delete(x, 1/2(1 - q(x)), -1/2(1 + q(x))) = \lambda(-y, -1/2(1 - q(-y)), 1/2(1 + q(-y)))
Equating the corresponding components, we get:
x = -\lambda y
1/2(1 - q(x)) = -1/2(1 - q(-y))
-1/2(1 + q(x)) = 1/2(1 + q(-y))
From the first equation, we can see that x and y are proportional with a negative factor \lambda. However, in the subsequent equations, we encounter a contradiction. The second equation implies that q(x) = q(-y), while the third equation implies that q(x) = -q(-y). These two conditions cannot be simultaneously satisfied, indicating that there are no x and y that fulfill the equation $\tau_+(x) = \lambda\tau_−(y)$.
Therefore, based on the given information, we can conclude that the intersection from the excercise is empty.
"The second equation implies that q(x) = q(-y), while the third equation implies that q(x) = -q(-y). These two conditions cannot be simultaneously satisfied"
DeleteIncorrect conclusion. They can.
"Incorrect conclusion. They can.".
DeleteOk. It seems to me that I know.
The first equation states that q(x) = q(-y). This means that the value of function q for vector x is equal to the value of function q for vector -y. In the case of this quadratic function, this equality holds for any vector because the square of any number is the same regardless of its sign. Therefore, the first equality is always satisfied.
The second equation states that q(x) = -q(-y). This means that the value of function q for vector x is equal to the negation of the value of function q for vector -y. To achieve this, q(x) and q(-y) must have opposite signs. Since q(x) is a sum of squared numbers, and the values are always non-negative, in order to have opposite signs, one condition must be met: q(-y) must be negative, i.e., -(y_1)^2 - (y_2)^2 - (y_3)^2 + (y_4)^2 < 0.
In summary, these equations can be satisfied if q(-y) is negative. Otherwise, if q(-y) is non-negative, these two equalities cannot be simultaneously satisfied.
So the intersection will be empty only for negative q(-y).
Is this correct now?
"Since q(x) is a sum of squared numbers,"
DeleteThis is incorrect.
""Since q(x) is a sum of squared numbers,"
DeleteThis is incorrect.".
The sum of 3 squared numbers - 1 squared number.
@M.S., Bjab
ReplyDeleteThanks for you comments. Will addres them tomorrow.
Added new Exercise at the end. In fact this one is important, the previous one was just a preparation.
ReplyDeleteBjab -> Ark
ReplyDelete(41) jakieś BR
@Bjab
Delete"(41) jakieś BR".
To miało być zapewne \BR oznaczające zbiór liczb rzeczywistych.
Bjab -> Ark
ReplyDeleteNie wychodzi mi wzór (43)
Thanks. I will check it and provide a derivation.
Delete@Bjab
DeleteFormulated as a unnumbered Proposition and provided a detailed proof.
Thanks for telling me about the need of such a proof.
Bjab -> Ark
DeleteDziękuję.
Strasznie się namęczyłem szukając błędu u Ciebie lub u siebie. Nie mogłem znaleźć. W Twoim dowodzie nie ma błędu, ale i ja rozwijając lewą i prawą stronę wzoru (43) dochodziłem do wniosku że lewa nie jest taka sama jak prawa. Lewa strona bardzo różniła się wyglądem od prawej i pochopnie przyjąłem, że ich wartości się różnią.
No ale w końcu zrozumiałem.
(Po lewej stronie miałem składnik cosh - sinh a po prawej miałem składnik 1/(cosh + sinh).
Bjab -> Ark
Delete"Now cosh(α) = (e
α+e
−α
)/2, sinh(α) = (e
α−e
−α/2. Using ..." ->
brak nawiasu zamykającego w powyższym wierszu proofa.
Fixed. Updated. Thank you!
DeleteThere is still unsolved problem with faithful action of SO(4,2) on \hat{M}.
ReplyDelete