Hyper Dimensional Philosophizing

[Fredrik]

Oppressed, I just have to say that I think it's just amazing how you can ignore everything that people say to you. You have started this discussion in three different forums, and you've been getting lots of comments from at least fifteen people who are far more knowledgeable about these things than you. In these discussions, you have consistently avoided to answer questions, and consistently ignored everything that proves you wrong. Well, at least you're consistent.

Your refusal to answer question isn't just preventing these discussions from making progress, it's also very very rude to the people who have gone through a lot of effort to try to help you understand these things, which are very basic to some of us.

I'm sure you have noticed that I'm starting to lose my patience with you. The biggest reason for that is that you aren't answering questions. It also doesn't help that you often repeat the same false claims immediately after they've been proven wrong, without so much as a comment about the argument that proved them wrong in the first place. Sometimes I wonder if you even read our posts, but you obviously do, since you keep finding things to misinterpret. That makes me wonder if you stopped being interested in a serious discussion a long time ago, and have decided to just be a troll.

 

[Ziggurat]

I'm sure you have noticed that I'm starting to lose my patience with you. The biggest reason for that is that you aren't answering questions. It also doesn't help that you often repeat the same false claims immediately after they've been proven wrong, without so much as a comment about the argument that proved them wrong in the first place. Sometimes I wonder if you even read our posts, but you obviously do, since you keep finding things to misinterpret. That makes me wonder if you stopped being interested in a serious discussion a long time ago, and have decided to just be a troll.

I reached that conclusion after he posted nonsense about a "conciousness" dimension. You've shown far more tolerance and patience than he deserves, and a lot more than I was able to muster. Good effort, but it's time to give up on him.

 

[Molinaro]

Hey that was my nonsense! And realy.. I didn't expect anyone to think I was being serious.

 

[Ziggurat]

Hey that was my nonsense!

My bad. I guess I stopped paying close attention after it became clear Oppressed was never really going to answer questions.

 

[Oppressed]

Wow,

I feel like I am living in two different realties.

One reality is where I have gone through 6 years of study during which most of that time I was studying math, science and physics in one form or another and I excelled in my classes. During this time I never had any particular problem being understood by my professors or fellow students over such fundamental concepts as dimension, Euclidian space or what basis means in vector space instead of what dimension means in vector space. Even today, I have met with two professors I consider friends and both have no problem understanding what I am talking about.

The other reality is on these forums, where it seems like not a single person understands what I am talking about and there seems to be a general consensus among posters on the forum that I do not understand the basic topics under discussion, such fundamental concepts as dimension, Euclidian space or what basis means in vector space instead of what dimension means in vector space. In this forum reality I should fail math, science and physics studies according to my lack of basic knowledge of the fundamental concepts involved.

There are a number of pretty decent online resources that explain about the fact that Euclidian space is flat, uniform and that the dimensions are perpendicular to each other. And yet I can not appear to get a single person on this forum to understand or agree with that. From the view point of posters on this forum I must considered a fool who does know what he is talking about, certainly not an experienced professional who is a expert in his field.

But if I go talk to people I know who are familiar with the subject, like Professor Greene and Professor Lee, I have no problem talking about the subject and no disagreement with them over the meanings being discussed. As I've worked over the past 10 years I've become a senior experience professional considered an expert in my field.

The point of Euclidian space being flat and uniform is that, as an example, if you have a two dimensional Euclidian space that the distance between two points P = (p1, p2) and Q = (q1,q2) is equal to the squareroot( (p1-q1)^2 + (p2-q2)^2 ) which is the same as you would measure with a real physical ruler. If the dimensions were not perpendicular then this would not be true for all points in the plane.

This is true for Euclidean space of n dimensions. This is very fundamental and basic so I do not understand how you do not understand this. In Euclidean space of n dimensions, the distance between two points P = (p1, p2,…,pn) and Q = (q1,q2,…,qn) is equal to the squareroot( (p1-q1)^2 + (p2-q2)^2 + … + (pn-qn)^2 ). If the dimensions in this Euclidean space were not perpendicular, this would not be true.

If you take a flat plane, which you can represent with a piece of paper, you can use a ruler and draw one line on the paper. It does not matter where on the plane you draw the first line or at what angle. Once you have drawn the line you have defined a reference line that lies in one dimension of the plane.

Pick a point in the middle of the line and draw a second line so that it is 45 degrees in the counterclockwise direction of whatever side of the first line you decide is positive. This line now serves as a second axis on the Euclidian plane. But while it does lie in a single dimension of the plane, because it is not perpendicular to the first line, the two axes do not represent the two dimensions of the Euclidean plane. Instead, they represent a warped space.

If you use these two axes to define two points along these two axes P(0,0) and Q(1,1) the distance between the two points will NOT be squareroot( (p1-q1)^2 + (p2-q2)^2 ) = squareroot( (0-1)^2 + (0-1)^2 ) = squareroot(2) = approx 1.41. Instead, it measures out with a ruler as approx 1.98.

This is limiting the two dimensions of the Euclidian space to being dimensions of length which I agreed to state as a condition to avoid the argument of whether or not something other than length can be used for the two dimensions of this Euclidean plane. But, given the two dimensions are lengths, the two dimensions need to be perpendicular.

I can not believe that no one here understands that.

This is a very fundamental concept about Euclidean space and understanding Euclidean space is very fundamental at early stages of math which just about every posting her indicates they have not only studied but moved on to higher levels of math.

Well, guess we are all back at a point of considering further discussion pointless.

 

[Ziggurat]

There are a number of pretty decent online resources that explain about the fact that Euclidian space is flat, uniform and that the dimensions are perpendicular to each other. And yet I can not appear to get a single person on this forum to understand or agree with that.

That's because you've still got it wrong. Dimensions aren't defined by orthogonality, but by linear independence. Perpendicular basis vectors are linearly independent, but you need not have orthogonality in order to have linear independence. In fact, it's possible to construct spaces of whatever dimensionality you want in which the concept of orthogonality isn't even defined. In spaces where orthogonality is defined, it is often convenient to use basis vectors which are perpendicular, but that is never a requirement. And the basis vectors themselves don't define the dimensionality anyways. The maximum number of linearly independent vectors does. There's an infinite number of such sets, most of which are not orthogonal, but the number of vectors in such a set is always the same. It is the number of such vectors, NOT the vectors themselves, which defines the dimensionality of the space.

From the view point of posters on this forum I must considered a fool who does know what he is talking about, certainly not an experienced professional who is a expert in his field.

Who gives a damn? You're still sloppy with your definitions, and you blame others for that sloppiness. The fact that most of the time you can get away with being sloppy, or that people you talk to face to face can still figure out what you mean (and possibly refrain from correcting you out of a desire to be polite), doesn't change that.

The point of Euclidian space being flat and uniform is that, as an example, if you have a two dimensional Euclidian space that the distance between two points P = (p1, p2) and Q = (q1,q2) is equal to the squareroot( (p1-q1)^2 + (p2-q2)^2 ) which is the same as you would measure with a real physical ruler. If the dimensions were not perpendicular then this would not be true for all points in the plane.

Dimensions are not vectors. They cannot be perpendicular. Orthogonality is defined for vectors, not for dimensions.

Pick a point in the middle of the line and draw a second line so that it is 45 degrees in the counterclockwise direction of whatever side of the first line you decide is positive. This line now serves as a second axis on the Euclidian plane. But while it does lie in a single dimension of the plane, because it is not perpendicular to the first line, the two axes do not represent the two dimensions of the Euclidean plane. Instead, they represent a warped space.

No, they represent a space with a non-Euclidean metric. "Warped" doesn't have a precise definition here.

If you use these two axes to define two points along these two axes P(0,0) and Q(1,1) the distance between the two points will NOT be squareroot( (p1-q1)^2 + (p2-q2)^2 ) = squareroot( (0-1)^2 + (0-1)^2 ) = squareroot(2) = approx 1.41. Instead, it measures out with a ruler as approx 1.98.

So what? As I said, that just means you have to use a non-Euclidean metric. That's easy to do, if you know what you're doing.

This is limiting the two dimensions of the Euclidian space to being dimensions of length which I agreed to state as a condition to avoid the argument of whether or not something other than length can be used for the two dimensions of this Euclidean plane. But, given the two dimensions are lengths, the two dimensions need to be perpendicular.

Once again: dimensions cannot be perpendicular, because they are not vectors. And your basis vectors do NOT need to be perpendicular. It is merely often convenient to choose them so that they are.

 

[Fredrik]

The point of Euclidian space being flat and uniform is that, as an example, if you have a two dimensional Euclidian space that the distance between two points P = (p1, p2) and Q = (q1,q2) is equal to the squareroot( (p1-q1)^2 + (p2-q2)^2 ) which is the same as you would measure with a real physical ruler. If the dimensions were not perpendicular then this would not be true for all points in the plane.

Ziggurat said this three times in his reply, and I said it myself in #326 and probably other places, but I have a feeling we may have to say it again: "Perpendicular" is a concept that isn't used about dimensions. Vectors can be perpendicular, but dimensions can't, because no angle is defined between them.

Your argument doesn't make sense and your conclusion is false.

It's possible that the argument would make sense (but still be wrong) if we knew that you were using a non-standard definition of the word dimension, and knew that definition, but you have only implied that you are using a non-standard definition by introducing the term "(type a) dimension". You still haven't told us what that definition is, even though I have asked you to do that in every post I've written lately. Why haven't you by the way?

This line now serves as a second axis on the Euclidian plane.
the two axes do not represent the two dimensions of the Euclidean plane. Instead, they represent a warped space.

No, they don't. Unless we introduce some new postulate, the lines only represent two one-dimensional subspaces of the Euclidean plane. If we postulate that those two lines are in fact orthogonal (even though we drew them at a 45 degree angle), then we can think of this picture as representing a non-euclidean metric (or rather a class of non-euclidean metrics). A manifold endowed with such a metric can still be flat though.

If you use these two axes to define two points along these two axes P(0,0) and Q(1,1) the distance between the two points will NOT be squareroot( (p1-q1)^2 + (p2-q2)^2 ) = squareroot( (0-1)^2 + (0-1)^2 ) = squareroot(2) = approx 1.41. Instead, it measures out with a ruler as approx 1.98.

This only means that it's more difficult to calculate things when you use a non-orthogonal basis. That's why the orthonormal bases are so convenient. The exact answer is sqrt(2+sqrt(2)) by the way, which is approximately 1.85.

This is limiting the two dimensions of the Euclidian space to being dimensions of length...

You have neither defined your "(type a) dimension", nor what it means for a (type a) dimension to be "of length".

Oh yeah, I've been meaning to ask...How do you define the concept "(type a) dimension"?

 

[Fredrik]

Even though some parts of your argument didn't make sense (the parts where you used concepts that weren't defined), I think I know what you were trying to say, and what your mistake is. I have of course told you most of this before, and you completely ignored it then, just like you ignore everything that's important, but I'm going to try one more time.

Let {E1,E2} be the standard orthonormal basis.
Let {F1,F2} be some other (any other) orthonormal basis.
Let {G1,G2} be the non-orthogonal basis defined by G1=(1,0) and G2=1/sqrt(2)*(1,1).

Any vector V can be expressed as either a*E1+b*E2, c*F1+d*F2 or e*G1+f*G2. You are aware of the fact that the norm |V| takes a particularly simple form when it's expressed in terms the components of V in an orthonormal basis:

|V| = sqrt(a²+b²) = sqrt(c²+d²)

You are also aware that |V| takes a more complicated form when it's expressed in terms of the components of V in a basis that isn't orthonormal, for example:

|V| = sqrt(e²+f²+sqrt(2)*e*f)

Your mistake is to believe that the definition of the Euclidean norm states that the norm of a vector "has the simple form, period".

If that had been true, then the norm of e*G1+f*G2 would have been sqrt(e²+f²) while the actual length of that vector had been sqrt(e²+f²+sqrt(2)*e*f). This would have made orthonormal bases "special" in the sense that they are the only ones that make the norm of an arbitrary vector equal to the actual length of that vector. And that would have made it natural to associate the dimensions of the vector space with the basis vectors of an orthonormal basis, or with the one-dimensional subspaces they are members of, or with...something like that.

This seems to be exactly what you are arguing above, except you aren't saying "if that had been true". You are implicitly assuming that it is true.

However, it isn't true. That's not how the norm is defined. The definition is that the norm takes that simple form when it's expressed in terms of the components of the vector in an orthonormal basis. And your whole argument falls with that.