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.
Super Artificial Intelligence, a naive approach
- Thread starter ProgrammingGodJordan
- Start date
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.
Reality Check
Penultimate Amazing
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
Reality Check
Penultimate Amazing
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
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.
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 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).
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ProgrammingGodJordan
Banned
(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.
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)
[IMGw=400]http://i.imgur.com/FqobtB7.jpg[/IMGw]
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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ProgrammingGodJordan
Banned
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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Reality Check
Penultimate Amazing
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
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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W.D.Clinger
Philosopher
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.
ProgrammingGodJordan
Banned
(A)
I see you've changed your stance on this argument.
Your initial argument was:
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)
[IMGw=300]http://i.imgur.com/sA6PAz9.jpg[/IMGw]
Also, supermanifolds are flat Riemannian metric. So your quote above is perhaps self-contradictory.
See https://ncatlab.org/nlab/show/Euclidean+supermanifold
I see you've changed your stance on this argument.
Your initial argument was:
W.D.Clinger said:
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)
[IMGw=300]http://i.imgur.com/sA6PAz9.jpg[/IMGw]
W.D.clinger said:
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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ProgrammingGodJordan
Banned
[IMGw=300]http://i.imgur.com/1jiSyAv.jpg[/IMGw]
(A)
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)
It is still clear that the following url disregards your nonsense quote above.
https://ncatlab.org/nlab/show/Euclidean+supermanifold
(A)
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)
RealityCheck said:
It is still clear that the following url disregards your nonsense quote above.
https://ncatlab.org/nlab/show/Euclidean+supermanifold
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Reality Check
Penultimate Amazing
A lie about my ""points" had long been demolished" when the truth is still that supermanifolds are not locally Euclidean.
27 March 2017: A basic point about supermanifolds is they are not actually Euclidean locally.
A more complete explanation of how supermanifolds are not locally Euclidean
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.
31 March 2017 ProgrammingGodJordan: A web page about a subset of supermanifolds does not state that all supermanifolds are locally Euclidean.
That working link is to a slightly broken web page where the equations are not formatted for me.
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Squeegee Beckenheim
Penultimate Amazing
- Joined
- Dec 29, 2010
- Messages
- 32,124
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.
Use it wisely.
W.D.Clinger
Philosopher
Not at all. I am also not surprised by your inability to understand what I have said.
But you don't know what that means. You're just quoting some words you found on the web. For example:
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.
ProgrammingGodJordan
Banned
[IMGw=300]http://i.imgur.com/JYrZOW4.jpg[/IMGw]
(A)
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)
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.
(A)
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)
W.D.Clinger said:
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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ProgrammingGodJordan
Banned
(A)
Yes, the truth is, the following still holds:
ProgrammingGodJordan said:
(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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fagin
Philosopher
Needs more cowbell. And stupid memes.
ProgrammingGodJordan
Banned
You're partially right.
Here goes:
[IMGw=300]http://i.imgur.com/FB63Njc.jpg[/IMGw]
W.D.Clinger said:
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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W.D.Clinger
Philosopher
Here's more cowbell.
I have highlighted a few words. All other emphasis is as in the originals.
Frank W Warner said:
Antoni A Kosinski said:
Jeffrey M Lee said:
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:
Wikipedia said:
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.
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ProgrammingGodJordan
Banned
Nonsense & redundancy above.
[IMGw=350]http://i.imgur.com/P9OKdTN.jpg[/IMGw]
(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:
ProgrammingGodJordan said:
This means your quote from reply 577 is nonsense:
W.D.Clinger said:
(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:
ProgrammingGodJordan said:
(E)
wrt topic;
W.D.Clinger said:
Can you provide an example of the above?
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W.D.Clinger
Philosopher
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.
ProgrammingGodJordan
Banned
(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:
W.D.Clinger said:
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W.D.Clinger
Philosopher
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.
ProgrammingGodJordan
Banned
(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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W.D.Clinger
Philosopher
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.
Toddlers might not see anything wrong with your argument.
ProgrammingGodJordan
Banned
(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.
(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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