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Old 29th March 2017, 10:06 AM   #601
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Originally Posted by Dr.Sid View Post
Well yes, but there is no need in helping them, right ? Or spending precious little time we have left by participating in our own destruction .. go see a movie or something ..

Careful! If the AIs are evil enough, they might punish you for failing to help them get invented, by torturing a re-creation of you in a simulation.
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Old 29th March 2017, 01:06 PM   #602
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Originally Posted by ProgrammingGodJordan View Post
A lack of reading comprehension.
Of course a conference talk about a Euclidean superspace is about a Euclidean superspace! Non-anticommutative N=(1,1) Euclidean Superspace
An important key word in the title is "non-anticommutative" because it is that which makes the superspace Euclidean globally and locally !

Superspace
Quote:
"Superspace" is the coordinate space of a theory exhibiting supersymmetry. In such a formulation, along with ordinary space dimensions x, y, z, ..., there are also "anticommuting" dimensions whose coordinates are labeled in Grassmann numbers rather than real numbers. The ordinary space dimensions correspond to bosonic degrees of freedom, the anticommuting dimensions to fermionic degrees of freedom.
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Old 29th March 2017, 01:37 PM   #603
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A more complete explanation of how supermanifolds are not locally Euclidean

A more complete explanation of how supermanifolds are not locally Euclidean.
For a start Wikipedia states that supermanifolds are not locally Euclidean, even in its informal definition!
Supermanifold
Quote:
Informal definition
An informal definition is commonly used in physics textbooks and introductory lectures. It defines a supermanifold as a manifold with both bosonic and fermionic coordinates. Locally, it is composed of coordinate charts that make it look like a "flat", "Euclidean" superspace. These local coordinates are often denoted by
( x , θ , θ ¯ )
where x is the (real-number-valued) spacetime coordinate, and θ, and θ ¯ are Grassmann-valued spatial "directions".
the phrase "looks like" is not a mathematical definition!
The author uses the standard notation in English of putting double quotes around words that do not mean what they usually mean.

Later there are actual definitions as used in mathematics. The second definition is appropriate here since manifold learning is about manifolds.
Quote:
Concrete: as a smooth manifold
A different definition describes a supermanifold in a fashion that is similar to that of a smooth manifold, except that the model space Rp has been replaced by the model superspace Rcp × Raq.
The last 2 spaces are "the even and odd real subspaces of the one-dimensional space of Grassmann numbers"
Grassmann numbers are very different from normal numbers as in Euclidian spaces. For example everyone knows that the numbers on the real number line (a 1D Euclidean space) commute under addition and multiplication: a + b = b + a and ab = ba. Grassmann numbers anti-commute under multiplication so that θiθj = -θjθi (note the negative sign).

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Old 29th March 2017, 09:15 PM   #604
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Originally Posted by Reality Check View Post
A more complete explanation of how supermanifolds are not locally Euclidean.
For a start Wikipedia states that supermanifolds are not locally Euclidean, even in its informal definition!
Supermanifold

the phrase "looks like" is not a mathematical definition!
The author uses the standard notation in English of putting double quotes around words that do not mean what they usually mean.

Later there are actual definitions as used in mathematics. The second definition is appropriate here since manifold learning is about manifolds.

The last 2 spaces are "the even and odd real subspaces of the one-dimensional space of Grassmann numbers"
Grassmann numbers are very different from normal numbers as in Euclidian spaces. For example everyone knows that the numbers on the real number line (a 1D Euclidean space) commute under addition and multiplication: a + b = b + a and ab = ba. Grassmann numbers anti-commute under multiplication so that θiθj = -θjθi (note the negative sign).
(A)

The Grassmann numbers represent some direction sequence, from some real valued x, in ϕ(x,θ,θ_).

NOTE:

The hypothesis prescribes the clamping of the Grassmannian parameters in a particular regime, after which largely real numbers are usable... (In other words, some grasmannian bound properties are perhaps feasible, whence observations in deep neural models don't strictly require grassmann aligned numbers)

Feasible properties lay in the boundary of 'eta', or direct numerical simulations etc, at least for an initial 'trivial' example of reinforcement like learning in this paradigm.




Originally Posted by Reality Check View Post
A lack of reading comprehension.
Of course a conference talk about a Euclidean superspace is about a Euclidean superspace! Non-anticommutative N=(1,1) Euclidean Superspace
An important key word in the title is "non-anticommutative" because it is that which makes the superspace Euclidean globally and locally !

Superspace
(B)
Yes, your words to a small degree, do exhibit a lack of reading comprehension.
See the data above or below.



(C)


See 'euclidean supermanifold' via https://ncatlab.org/nlab/show/Euclidean+supermanifold.



(D)
FOOTNOTE:
Overall take away is that there is some regime in euclidean superspace, for which supermanifolds (some coordinate sequence of such), are feasible.
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Old 29th March 2017, 09:22 PM   #605
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Originally Posted by Dr.Sid View Post
Well yes, but there is no need in helping them, right ? Or spending precious little time we have left by participating in our own destruction .. go see a movie or something ..
The outcome is already positive for us, and may be positive for us in the long run.

For example, with advances in artificial cognitive machines, we are able to reduce errors in disease diagnosis.
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Old 30th March 2017, 01:25 PM   #606
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Originally Posted by ProgrammingGodJordan View Post
The Grassmann numbers ....
Shooting yourself in the foot by showing that you know the supermanifolds use Grassmann algebra!
More math word salad does not address my points.
A more complete explanation of how supermanifolds are not locally Euclidean
A basic point about supermanifolds is they are not actually Euclidean locally.
23 March 2017 ProgrammingGodJordan: Lies about Christopher Lu's code from Lu's Master's thesis (which does not contain his hypothesis).
24 March 2017 ProgrammingGodJordan: A valid hypothesis is not incoherent math word salad as I pointed out yesterday.

Your assertion is that supermanifolds are locally Euclidean but you have been citing the Wikipedia page on supermanifolds that says that you are wrong !.

That is a slightly broken web page stating that a subset of supermanifolds are labeled as Euclidean: Euclidean supermanifold
Quote:
A Euclidean supermanifold is a supermanifold that can be thought of as being equipped with a flat Riemannian metric.
Alternatively, it is a supermanifold for which the transition functions of an atlas are restricted to be elements of the super Euclidean group.
31 March 2017 ProgrammingGodJordan: A web page about a subset of supermanifolds does not state that all supermanifolds are locally Euclidean.

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Old 30th March 2017, 02:04 PM   #607
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Originally Posted by Reality Check View Post
A web page about a subset of supermanifolds does not state that all supermanifolds are locally Euclidean.
Regardless of what any web pages may say:

All Riemannian manifolds are locally Euclidean.

Locally Euclidean is not at all the same thing as Euclidean.

A torus, for example, is locally Euclidean: for every point of the torus, there is a neighborhood in which the local topology is diffeomorphic to an open subset of 2-dimensional Euclidean space. With a torus, however, there exist some closed curves that fail to divide the space into an outside and an inside. In 2-dimensional Euclidean spaces, that can't happen.

With regard to what ProgrammingGodJordan is going on about, a more relevant fact is that supermanifolds that aren't even Hausdorff can't be locally Euclidean, let alone Euclidean.
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Old 30th March 2017, 08:17 PM   #608
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(A)
Originally Posted by W.D.Clinger View Post
Regardless of what any web pages may say:
With regard to what ProgrammingGodJordan is going on about, a more relevant fact is that supermanifolds that aren't even Hausdorff can't be locally Euclidean, let alone Euclidean.
I see you've changed your stance on this argument.

Your initial argument was:

Originally Posted by W.D.Clinger
If the supermanifold were actually Euclidean, there would be no reason for Wikipedia to include the two words I highlighted.
ProgrammingGodJordan is telling us the supermanifold is Euclidean, when it isn't even Hausdorff.
I don't know why you attempted to pretend as if you had not made that initial post.

I particularly referred to supermanifolds, at the boundary of euclidean regimes.



(B)

Originally Posted by W.D.clinger
Regardless of what any web pages may say:

All Riemannian manifolds are locally Euclidean.

Locally Euclidean is not at all the same thing as Euclidean.

A torus, for example, is locally Euclidean: for every point of the torus, there is a neighborhood in which the local topology is diffeomorphic to an open subset of 2-dimensional Euclidean space. With a torus, however, there exist some closed curves that fail to divide the space into an outside and an inside. In 2-dimensional Euclidean spaces, that can't happen.

With regard to what ProgrammingGodJordan is going on about, a more relevant fact is that supermanifolds that aren't even Hausdorff can't be locally Euclidean, let alone Euclidean.
Also, supermanifolds are flat Riemannian metric. So your quote above is perhaps self-contradictory.

See https://ncatlab.org/nlab/show/Euclidean+supermanifold
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Old 30th March 2017, 08:22 PM   #609
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(A)

Originally Posted by Reality Check View Post
Shooting yourself in the foot by showing that you know the supermanifolds use Grassmann algebra!

Your assertion is that supermanifolds are locally Euclidean but you have been citing the Wikipedia page on supermanifolds that says that you are wrong !.

That is a slightly broken web page stating that a subset of supermanifolds are labeled as Euclidean: Euclidean supermanifold

31 March 2017 ProgrammingGodJordan: A web page about a subset of supermanifolds does not state that all supermanifolds are locally Euclidean.
Your "points" had long been demolished.

You failed to observe the wikipedia data, and so I directed you to another source which showed that supermanifolds may be euclidean, in toddler like, clear description.

What did you mean by broken link, is the link not functional for you?




(B)

Originally Posted by RealityCheck
The set of points in the neighborhood of any point in a supermanifold is NEVER EUCLIDEAN.
It is still clear that the following url disregards your nonsense quote above.
https://ncatlab.org/nlab/show/Euclidean+supermanifold
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Old 30th March 2017, 10:58 PM   #611
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Since memes appear to make an argument more compelling, I think gifs should do even more. Therefore I offer up a gif of an owl being given a hat, for the use of Reality Check or W.D.Clinger:



Use it wisely.
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Old 31st March 2017, 08:26 AM   #612
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Originally Posted by ProgrammingGodJordan View Post
I see you've changed your stance on this argument.
Not at all. I am also not surprised by your inability to understand what I have said.

Originally Posted by ProgrammingGodJordan View Post
I particularly referred to supermanifolds, at the boundary of euclidean regimes.
But you don't know what that means. You're just quoting some words you found on the web. For example:

Originally Posted by ProgrammingGodJordan View Post
Also, supermanifolds are flat Riemannian metric. So your quote above is perhaps self-contradictory.

See https://ncatlab.org/nlab/show/Euclidean+supermanifold
Your citation says "can be thought of as being equipped with a flat Riemannian metric." If what you were saying were true, those highlighted words would have been unnecessary.

Note also "is a submersion with flat Riemannian metric on the fibers." Not on the supermanifold itself, but on the fibers.

But you don't know what that means either.
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Old 31st March 2017, 09:44 AM   #613
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(A)
Originally Posted by W.D.Clinger View Post
Not at all. I am also not surprised by your inability to understand what I have said.
You first mentioned that no supermanifold can be euclidean. (source)

You then switched to "no supermanifold that that isn't even Hausdorff can be euclidean" (source), after I provided data (source) that contrasted your initial statement.

After the switch, you presented a new way to contradict yourself, by expressing that all Riemannian fabric were locally euclidean, but simultaneously expressing that supermanifolds (which are actually Riemannian (source)) are not locally euclidean.

Move on beyond that blunder, that can't be avoided.



(B)
Originally Posted by W.D.Clinger
Your citation says "can be thought of as being equipped with a flat Riemannian metric." If what you were saying were true, those highlighted words would have been unnecessary.

Note also "is a submersion with flat Riemannian metric on the fibers." Not on the supermanifold itself, but on the fibers.

But you don't know what that means either.
You appear to think "X thought of as being Riemann" renders X devoid of Riemann based on the source, yet in the same breadth you express that X is Riemannian on the fiber, also based on the source.

In the contradiction above, it is clearly seen that X is reimannian.
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Old 31st March 2017, 09:48 AM   #614
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Originally Posted by Reality Check View Post
A lie about my ""points" had long been demolished" when the truth is still that supermanifolds are not locally Euclidean.
.
(A)
Yes, the truth is, the following still holds:

Originally Posted by ProgrammingGodJordan
Originally Posted by RealityCheck
The set of points in the neighborhood of any point in a supermanifold is NEVER EUCLIDEAN.
It is still clear that the following url disregards your nonsense quote above.
https://ncatlab.org/nlab/show/Euclidean+supermanifold


(B)
Also, it is trivially observable that W.D.Clinger contradicted himself, by expressing that "all Riemannian fabric are locally euclidean", while in the same reply expressing that "no supermanifold is locally euclidean".

This is a blatant contradiction, because the source expresses that supermanifolds may be Riemannian.

So, I don't know why you select to follow his invalid path. (Your behaviour is unreal)
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Old 31st March 2017, 10:00 AM   #615
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Needs more cowbell. And stupid memes.
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Old 31st March 2017, 11:02 AM   #616
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Originally Posted by fagin View Post
Needs more cowbell. And stupid memes.
You're partially right.


Here goes:



Originally Posted by W.D.Clinger
Spaces that aren't even Hausrdoff can't be locally euclidean
Apart from the blunders of Clinger's pointed out in reply 613, this is another blunder, because locally euclidean spaces need not be Hausdorff.

WikiPedia: "The Hausdorff property is not a local one; so even though Euclidean space is Hausdorff, a locally Euclidean space need not be."
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Old 31st March 2017, 04:24 PM   #617
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Originally Posted by fagin View Post
Needs more cowbell. And stupid memes.
Here's more cowbell.

I have highlighted a few words. All other emphasis is as in the originals.
Originally Posted by Frank W Warner
1.3 Definitions A locally Euclidean space M of dimension d is a Hausdorff topological space M for which each point has a neighborhood homeomorphic to an open subset of Euclidean space Rd....

--- Frank W Warner. Foundations of Differentiable Manifolds and Lie Groups. Scott, Foresman and Company, 1971.

Originally Posted by Antoni A Kosinski
A differential manifold M of class Cr consists of a second countable Hausdorff space M and a Cr structure on it.

---Antoni A Kosinski. Differential Manifolds. Academic Press, 1993. Republished by Dover Publications, 2007.

Originally Posted by Jeffrey M Lee
Definition 1.12. An n-dimensional topological manifold is a paracompact Hausdorff topological space, say M, such that every point p ∈ M is contained in some open set Up that is homeomorphic to an open subset of the Euclidean space Rn. Thus we say that a topological manifold is "locally Euclidean". The integer n is referred to as the dimension of M, and we denote it by dim(M).

Note: At first it may seem that a locally Euclidean space must be Hausdorff, but this is not the case.

...snip...

Definition 1.33. A differentiable manifold of class Cr is a set M together with a specified Cr structure on M such that the topology induced by the Cr structure is Hausdorff and paracompact....

---Jeffrey M Lee. Manifolds and Differential Geometry. American Mathematical Society Graduate Studies in Mathematics, Volume 107, 2009.

As can be seen from the examples above, different authors use slightly different definitions. By Warner's definition, all locally Euclidean spaces are Hausdorff, but Lee notes that locally Euclidean spaces need not be Hausdorff.

All three authors agree, however, that locally Euclidean manifolds are Hausdorff. It's part of their definition.

Mathematicians are fond of generalization, so they sometimes consider what would happen if the definition of a manifold were relaxed to allow non-Hausdorff spaces. There is a Wikipedia article on non-Hausdorff manifolds, which starts out by acknowledging this fact:

Originally Posted by Wikipedia
In geometry and topology, it is a usual axiom of a manifold to be a Hausdorff space, that is "manifold" means "second countable Hausdorff manifold".

ProgrammingGodJordan is having trouble not only with mathematics, but also with the challenge of quoting other participants in this thread accurately. For example:

Originally Posted by ProgrammingGodJordan View Post
You first mentioned that no supermanifold can be euclidean.
Actually, I said the particular supermanifold you appeared to be citing (via Wikipedia) isn't Euclidean, because it isn't even Hausdorff.

Originally Posted by ProgrammingGodJordan View Post
You then switched to "no supermanifold that that isn't even Hausdorff can be euclidean"
That's true. Euclidean spaces are Hausdorff. Spaces that are not Hausdorff are therefore not Euclidean.

It's called logic. Try it sometime.

Originally Posted by ProgrammingGodJordan View Post
After the switch, you presented a new way to contradict yourself, by expressing that all Riemannian fabric were locally euclidean,
I have not used the word "fabric" in this discussion. I have never before heard someone use the phrase "Riemannian fabric".

A Google search on that two-word phrase turns up 5 results, with this thread as top hit. The other four hits include a site of which Google warns "This site may harm your computer", plus a "deconstructing God" blog whose most recent entry, dated Jury 6th, 2009, consists of gibberish under the title "Appreciating Michael Jackson".

Originally Posted by ProgrammingGodJordan View Post
but simultaneously expressing that supermanifolds (which are actually Riemannian (source)) are not locally euclidean.
No. You are quite confused. As I said to RealityCheck, all Riemannian manifolds are locally Euclidean.

Originally Posted by ProgrammingGodJordan View Post
You appear to think "X thought of as being Riemann" renders X devoid of Riemann based on the source, yet in the same breadth you express that X is Riemannian on the fiber, also based on the source.
The fibers can be Euclidean even if the manifold is not.

Originally Posted by ProgrammingGodJordan View Post
Also, it is trivially observable that W.D.Clinger contradicted himself, by expressing that "all Riemannian fabric are locally euclidean",
It is trivially observable that none of my previous posts within this thread have used the word "fabric".

Originally Posted by ProgrammingGodJordan View Post
while in the same reply expressing that "no supermanifold is locally euclidean".
It is trivially observable that I have never written the sequence of words within your quotes, in this thread or in any other.

Originally Posted by ProgrammingGodJordan View Post
This is a blatant contradiction, because the source expresses that supermanifolds may be Riemannian.

So, I don't know why you select to follow his invalid path. (Your behaviour is unreal)
Crackpots often invent quotations because it's easier to attack stupid stuff they've invented out of thin air than what others have actually written or said.

Last edited by W.D.Clinger; 31st March 2017 at 04:28 PM. Reason: added citation for Jeffrey M Lee
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Old 31st March 2017, 09:29 PM   #618
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Originally Posted by W.D.Clinger View Post
Here's more cowbell.

I have highlighted a few words. All other emphasis is as in the originals.






As can be seen from the examples above, different authors use slightly different definitions. By Warner's definition, all locally Euclidean spaces are Hausdorff, but Lee notes that locally Euclidean spaces need not be Hausdorff.

All three authors agree, however, that locally Euclidean manifolds are Hausdorff. It's part of their definition.

Mathematicians are fond of generalization, so they sometimes consider what would happen if the definition of a manifold were relaxed to allow non-Hausdorff spaces. There is a Wikipedia article on non-Hausdorff manifolds, which starts out by acknowledging this fact:




ProgrammingGodJordan is having trouble not only with mathematics, but also with the challenge of quoting other participants in this thread accurately. For example:


Actually, I said the particular supermanifold you appeared to be citing (via Wikipedia) isn't Euclidean, because it isn't even Hausdorff.


That's true. Euclidean spaces are Hausdorff. Spaces that are not Hausdorff are therefore not Euclidean.

It's called logic. Try it sometime.


I have not used the word "fabric" in this discussion. I have never before heard someone use the phrase "Riemannian fabric".

A Google search on that two-word phrase turns up 5 results, with this thread as top hit. The other four hits include a site of which Google warns "This site may harm your computer", plus a "deconstructing God" blog whose most recent entry, dated Jury 6th, 2009, consists of gibberish under the title "Appreciating Michael Jackson".


No. You are quite confused. As I said to RealityCheck, all Riemannian manifolds are locally Euclidean.


The fibers can be Euclidean even if the manifold is not.


It is trivially observable that none of my previous posts within this thread have used the word "fabric".


It is trivially observable that I have never written the sequence of words within your quotes, in this thread or in any other.


Crackpots often invent quotations because it's easier to attack stupid stuff they've invented out of thin air than what others have actually written or said.
Nonsense & redundancy above.





(A)
I don't recall stipulating any non-Hausdorff manifold.

This is why it appeared that you mentioned that all super-manifolds were not locally euclidean in nature. (source)



(B)
I still maintain my prior quote:
Originally Posted by ProgrammingGodJordan
Apart from the blunders of Clinger's pointed out in reply 613, this is another blunder, because locally euclidean spaces need not be Hausdorff.

WikiPedia: "The Hausdorff property is not a local one; so even though Euclidean space is Hausdorff, a locally Euclidean space need not be."
This means your quote from reply 577 is nonsense:

Originally Posted by W.D.Clinger
ProgrammingGodJordan is telling us the supermanifold is Euclidean, when it isn't even Hausdorff.


(C)
I think you are probably aware that my paraphrasing of your prior nonsensical expressions, does not perturb it to a nonsensical level. (it is nonsensical regardless)

In other words, I could have used your exact quote, it would still convey nonsense as I had notified with respect to the sources. (I have used your exact words in 'B' above.)



(D)
I still maintain the following:
Originally Posted by ProgrammingGodJordan
Originally Posted by W.D.Clinger
Your citation says "can be thought of as being equipped with a flat Riemannian metric." If what you were saying were true, those highlighted words would have been unnecessary.

Note also "is a submersion with flat Riemannian metric on the fibers." Not on the supermanifold itself, but on the fibers.

But you don't know what that means either.
You appear to think "X thought of as being Riemann" renders X devoid of Riemann, based on the source, yet in the same breadth you express that X is Riemannian on the fiber, also based on the source.

In the contradiction above, it is clearly seen that X is reimannian.


(E)
wrt topic;
Originally Posted by W.D.Clinger
The fibers can be Euclidean even if the manifold is not.
Can you provide an example of the above?
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Old 31st March 2017, 10:17 PM   #619
W.D.Clinger
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Originally Posted by ProgrammingGodJordan View Post
Originally Posted by W.D.Clinger
The fibers can be Euclidean even if the manifold is not.
Can you provide an example of the above?
Of course. If you look at any textbook on manifolds, many of the motivating examples of fiber bundles involve fibers that are Euclidean.

A cylinder is one of the simplest examples. It is, in fact, an example of what mathematicians refer to as a trivial bundle. The fibers of that bundle are Euclidean: each fiber is the real line, which is a 1-dimensional Euclidean space. The base space is a circle. Their product space is a cylinder, which is locally Euclidean but not Euclidean.

Another example: In the tangent bundle of a Riemannian manifold, the fibers are Euclidean.

But you didn't know any of that.
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Old 1st April 2017, 12:45 AM   #620
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Originally Posted by W.D.Clinger View Post
Of course. If you look at any textbook on manifolds, many of the motivating examples of fiber bundles involve fibers that are Euclidean.

A cylinder is one of the simplest examples. It is, in fact, an example of what mathematicians refer to as a trivial bundle. The fibers of that bundle are Euclidean: each fiber is the real line, which is a 1-dimensional Euclidean space. The base space is a circle. Their product space is a cylinder, which is locally Euclidean but not Euclidean.

Another example: In the tangent bundle of a Riemannian manifold, the fibers are Euclidean.

But you didn't know any of that.
(A)
Perhaps I did.

As you've just pointed out, both the manifold, and fiber sequence of said Rn are euclidean in nature.



(B)
Albeit, we see that based on your last quote, your prior quote (below), is not entirely true:

Originally Posted by W.D.Clinger
The fibers can be Euclidean even if the manifold is not.
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Last edited by ProgrammingGodJordan; 1st April 2017 at 12:51 AM.
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Old 1st April 2017, 04:29 AM   #621
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Originally Posted by ProgrammingGodJordan View Post
Perhaps I did.

But we know you didn't, because you are still making mistakes such as:

Originally Posted by ProgrammingGodJordan View Post
As you've just pointed out, both the manifold, and fiber sequence of said Rn are euclidean in nature.
That is not at all what I said.

Neither of the two examples I gave involves Rn.

For the two examples I gave, I said the fibers were Euclidean even though the manifold is not Euclidean.

The two examples I gave do of course involve locally Euclidean manifolds, because all Riemannian manifolds are locally Euclidean. You are having a great deal of trouble understanding that locally Euclidean and Euclidean are not the same thing.

Originally Posted by ProgrammingGodJordan View Post
Albeit, we see that based on your last quote, your prior quote (below), is not entirely true:
I said "The fibers can be Euclidean even if the manifold is not."

That is entirely true.

I gave examples, which you failed to understand even though you are pretending to understand words you learned at the University of Google.
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Old 1st April 2017, 01:47 PM   #622
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Originally Posted by W.D.Clinger View Post
But we know you didn't, because you are still making mistakes such as:


That is not at all what I said.

Neither of the two examples I gave involves Rn.

For the two examples I gave, I said the fibers were Euclidean even though the manifold is not Euclidean.

The two examples I gave do of course involve locally Euclidean manifolds, because all Riemannian manifolds are locally Euclidean. You are having a great deal of trouble understanding that locally Euclidean and Euclidean are not the same thing.


I said "The fibers can be Euclidean even if the manifold is not."

That is entirely true.

I gave examples, which you failed to understand even though you are pretending to understand words you learned at the University of Google.
(A)
Cylinders appear to have something to do with Rn, contrary to your expressions of the contrast.
https://en.wikipedia.org/wiki/Projective_plane


(B)
Locally euclidean or euclidean, both are degrees of the euclidean paradigm.
In simpler, toddler like words, at some point, both are euclidean.

So, the mistake is not mine.
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Old 1st April 2017, 04:54 PM   #623
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Originally Posted by ProgrammingGodJordan View Post
Cylinders appear to have something to do with Rn, contrary to your expressions of the contrast.
https://en.wikipedia.org/wiki/Projective_plane
Although cylinders are locally Euclidean, which does have something to do with R2, the word "cylinder" does not appear within that Wikipedia article.

Citing Wikipedia articles that don't support your claim would be a fine April Fool's joke, but doing so all year round might be noticed.

Originally Posted by ProgrammingGodJordan View Post
Locally euclidean or euclidean, both are degrees of the euclidean paradigm.
In simpler, toddler like words, at some point, both are euclidean.

So, the mistake is not mine.
Toddlers might not see anything wrong with your argument.
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Old 1st April 2017, 05:52 PM   #624
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Originally Posted by W.D.Clinger View Post
Although cylinders are locally Euclidean, which does have something to do with R2, the word "cylinder" does not appear within that Wikipedia article.

Citing Wikipedia articles that don't support your claim would be a fine April Fool's joke, but doing so all year round might be noticed.
(1)
It is optimal to see that you have corrected your prior blunder, where you nonsensically expressed: "cylinders don't involve Rn".


(2)
Oh, but the projected plane did have something to do with why I referenced it:

Manifold -> A finite cylinder may be constructed as a manifold by starting with a strip [0, 1] × [0, 1] and gluing a pair of opposite edges on the boundary by a suitable diffeomorphism. A projective plane may be obtained by gluing a sphere with a hole in it to a Möbius strip along their respective circular boundaries.



Originally Posted by W.D.Clinger View Post
Toddlers might not see anything wrong with your argument.
(3)
Perhaps one should be more like toddlers, such that one avoids seeing non-errors where possible, and avoid presenting nonsense such as cylinders don't involve Rn. etc.
That toddler would probably recognize that the above was not 'my argument'.
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