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Thread: Cross product vs Tensor product

  1. #11
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by BuleriaChk View Post
    And I thought John Gabriel had a problem with slope (not to mention curvature). At least he (almost) recognized two dimensions.... (well, ok, sort of, anyway)...

    But I certainly see where the issue of disagreement lies... whether more than one dimension is admissible in the context of a mathematical discussion.... (so there is a religious faith that Cartesian coordinates are not somehow "real mathematics", and yet one can mention the (Cartesian) mapping without having to worry about ...

    Are you SURE you are not John Gabriel... ? You sure sound like him
    I am sure I am not him, I am smarter, more intelligent and more knowledagble than you both combined.

    How about you define what "dimension" even mean and then show us it is always meaningful.

    Remember, there are spaces such that the cartesian product of two of the same sapces is isomorphic to the original space. As such it is meaningless.

  2. #12
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by emperorzelos View Post
    Remember, there are spaces such that the cartesian product of two of the same sapces is isomorphic to the original space. As such it is meaningless.
    Only in a single dimension. e.g. (not a vector sum) and (not a vector product).

    "Cartesian product" actually refers to independent sets (real number lines with independent metrics, which are actually vector spaces). Such "isomorphic" operations work on scalars in a single dimension, not vectors).

    i.e., in the space (where the 's refer to scalars in the same dimension), not

    or

    or (whichever is greater in magnitude for the individual vectors under the operation; i.e., the projection of the smaller vector onto the larger).


    That is the difference between arithmetic sums and products and direct (vector) sums and vector products ("dot" and "cross" products).
    Last edited by BuleriaChk; 12-21-2016 at 02:57 PM.
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  3. #13
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by BuleriaChk View Post
    Only in a single dimension. e.g. (not a vector sum) and (not a vector product).

    "Cartesian product" actually refers to independent sets (real number lines with independent metrics, which are actually vector spaces). Such "isomorphic" operations work on scalars in a single dimension, not vectors).

    i.e., in the space (where the 's refer to scalars in the same dimension), not

    or

    or (whichever is greater in magnitude for the individual vectors under the operation; i.e., the projection of the smaller vector onto the larger).


    That is the difference between arithmetic sums and products and direct (vector) sums and vector products ("dot" and "cross" products).
    What a great abuse of notation and concepts. You clearly do not know what a scalar is, vectors, modules or anything. First of if you are dealing with just a ring using the additional is completely superflous, it adds nothing but confusion.

    Define what a scalar is mathematically.

    Define a vector space and a module and what seperates them.

  4. #14
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by emperorzelos View Post
    .

    Define a vector space and a module and what seperates them.
    You say "module"; do you mean "modulus"?

    (I Googled "module number theory" and only came up with descriptions of courses.....

    ------------------------------------------------------------------------

    Note: the concept of "modulo" does arise, since Pythagorean triples can be characterized terms of modulii. Their lowest common denominator, as I (tried to) explain to John Gabriel some time ago.

    Modulo operation
    (Wiki article)

    Is that what you are referring to? (if so, thanks for the reminder - I had forgotten about that, and it is relevant to clarifying my proof for Fermat's theorem to number theorists, to convince them that there is a second dimension).

    To the reader -------------------------------
    (as noted below, I'm a little new to the concept of modulus, which is, indeed relevant to Fermat's proof and my characterization of it in terms of vectors and the relativistic unit circle; so the following definition may not be correct yet and have to be revised. I'm crunching on it. Looks like it might have something to do with computer programming. Are we SURE John Gabriel is not responding in this thread? ...

    The issue is now how the concept of modulus applies to vector spaces (particularly Cartesian coordinates, but also radial coordinates), if at all (I think).........

    Time will tell...

    So - off the top of my head (from dim memory, long ago), and tentatively speaking for now (I'll get better):
    -------------------------------

    The division of an integer by an another (if it is a Pythagorean triple, mod 0 ?

    i.e. mod 0 if the triple is not Pythagorean ?)

    (uh, oh, there is a fraction.... warning - a second dimension may be necessary...

    So mod 0 if the triple is not Pythagorean?

    (omg, another fraction.. and there may be irrationals involved. Maybe even circles..

    i.e., where what about c = a + b and where a,b,c are integers, but not Pythagorean triples?

    (Note: I was writing tex on the fly, and have deleted a number of phrases here; I will return when I finally get the tex right offline....)

    Nevertheless, the plot thickens...

    (I'm also a little new to the language of modulus, but it is an interesting perspective in relation to division of integers, and certainly relevant to my proof of Fermat's Theorem, particularly in reference to Lorentz rotation as an integer generator in two dimensions and the language of vectors in contrast to the single dimension integer generation (i.e., counting on a single number line) from Frege, Russell, and constructivists in general, which in turn has deep relevance to Quantum Field Theory. In the end, it will turn out that particle count cannot be conserved in two dimensions, except for relativistic Pythagorean energies (n = 2) by virtue of both the relativistic unit circle and the Binomial Theorem
    Last edited by BuleriaChk; 12-22-2016 at 02:06 AM.
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  5. #15
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    Default Re: Cross product vs Tensor product

    Here is what I think is going on with Modulus vs. vector space:

    Fermat's Theorem and Modulus

    That is, modular integer division simply verifies if the result is an integer or not; if it is not an integer (mod 0), it is a fraction, and therefore not in the set to begin with.

    However, it does reside in the set of fractions, of which the general form is z = f(x,y) = y/x where x = 0 means that z is undefined; x =0 is the midpoint of all possible metrics in that dimension, and so is a point, not a line - the ratio does not exist for x-0=x if x= 0.

    This is true for all elements in the set of fractions, integers or not (i.e., including ratios of irrationals and transcendentals (e.g. where itself is the ratio of the circumference C to the diameter D (a one-dimensional metric) of a two-dimensional circle, and is the length (metric) diagonal of a two dimensional unit square, for independent variables x and y.

    (a "module" may be a different thing entirely, though)...
    Last edited by BuleriaChk; 12-22-2016 at 04:19 AM.
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  6. #16
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by BuleriaChk View Post
    You say "module"; do you mean "modulus"?

    (I Googled "module number theory" and only came up with descriptions of courses.....

    ------------------------------------------------------------------------

    Note: the concept of "modulo" does arise, since Pythagorean triples can be characterized terms of modulii. Their lowest common denominator, as I (tried to) explain to John Gabriel some time ago.

    Modulo operation
    (Wiki article)

    Is that what you are referring to? (if so, thanks for the reminder - I had forgotten about that, and it is relevant to clarifying my proof for Fermat's theorem to number theorists, to convince them that there is a second dimension).

    To the reader -------------------------------
    (as noted below, I'm a little new to the concept of modulus, which is, indeed relevant to Fermat's proof and my characterization of it in terms of vectors and the relativistic unit circle; so the following definition may not be correct yet and have to be revised. I'm crunching on it. Looks like it might have something to do with computer programming. Are we SURE John Gabriel is not responding in this thread? ...

    The issue is now how the concept of modulus applies to vector spaces (particularly Cartesian coordinates, but also radial coordinates), if at all (I think).........

    Time will tell...

    So - off the top of my head (from dim memory, long ago), and tentatively speaking for now (I'll get better):
    -------------------------------

    The division of an integer by an another (if it is a Pythagorean triple, mod 0 ?

    i.e. mod 0 if the triple is not Pythagorean ?)

    (uh, oh, there is a fraction.... warning - a second dimension may be necessary...

    So mod 0 if the triple is not Pythagorean?

    (omg, another fraction.. and there may be irrationals involved. Maybe even circles..

    i.e., where what about c = a + b and where a,b,c are integers, but not Pythagorean triples?

    (Note: I was writing tex on the fly, and have deleted a number of phrases here; I will return when I finally get the tex right offline....)

    Nevertheless, the plot thickens...

    (I'm also a little new to the language of modulus, but it is an interesting perspective in relation to division of integers, and certainly relevant to my proof of Fermat's Theorem, particularly in reference to Lorentz rotation as an integer generator in two dimensions and the language of vectors in contrast to the single dimension integer generation (i.e., counting on a single number line) from Frege, Russell, and constructivists in general, which in turn has deep relevance to Quantum Field Theory. In the end, it will turn out that particle count cannot be conserved in two dimensions, except for relativistic Pythagorean energies (n = 2) by virtue of both the relativistic unit circle and the Binomial Theorem
    BAHAHAHA! You do not know what a module is! Yeah thank you for demonstrating my superiority to you
    https://en.wikipedia.org/wiki/Module_(mathematics)

    Wiki

  7. #17
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by emperorzelos View Post
    BAHAHAHA! You do not know what a module is! Yeah thank you for demonstrating my superiority to you
    https://en.wikipedia.org/wiki/Module_(mathematics)

    Wiki
    There are lots of things I have never heard of, particularly in number theory. Of course, you seem to have never heard of Descartes....

    Anyway, thanks for the link, I'll read further...

    I suspect the answer is that module operations apply within a single number line (dimension) (i.e., within a single vector space) as global, with internal dimensions (geometric objects) determining the nature of scalar operations in the coefficients within the global vector space.

    (an area or a volume (or hyper-volume) can be characterized within the coefficient of the global vector, the division over a ring includes two dimensions for general functions within that vector space). So two dimensional "scalar" vector space multiplying the unit vector of another - a vector space within a vector space.

    e.g.,
    or

    (graphing the function still requires two dimensions within the "module".

    pretty easy answer.... (they just describe operations of scalar coefficients multiplying a global unit vector).

    (Shows the lengths number theorists will go to in trying to ignore Descartes...

    For my proof of Fermat's theorem, the question is moot anyway, since the Binomial Expansion holds for all (positive) real numbers....

    (This does relate to GTR, howevever, where the tensor analysis and trnsformations is performed on coefficients of the metric and stress-energy tensors (i.e., a tensor transformation on the metric tensor), so it is an interesting point, and confirms my analysis - that if the coordinate space is related to the energy-momentum space, it is because is interpreted as a coordinate density (or a mass multiplier) in STR, and is only constant (inertial) if is preserved in transformation. (the metric tensor, whatever it is, is invariant for a particular universe).

    If the metric and the stress-energy tensors are diagonalized with constant coefficients, there are no further ("spin") perturbations - i.e., the characterization of a black hole.

    Vector spaces within vector spaces within vector spaces..... ad infinitem
    Fractions over (and under) fractions over fractions.... ad infinitem

    Little fleas have smaller fleas
    upon their backs to bite 'em
    and smaller fleas have smaller fleas
    and so, ad infinitem

    "And it's turtles, all the way down..." - Number theorist to Bertrand Russell ..

    Mandelbrot, where are you?.....
    Last edited by BuleriaChk; 12-22-2016 at 02:27 PM.
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
    Ignore List -The Peanut Gallery.

  8. #18
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by BuleriaChk View Post
    There are lots of things I have never heard of, particularly in number theory. Of course, you seem to have never heard of Descartes....

    Anyway, thanks for the link, I'll read further...

    I suspect the answer is that module operations apply within a single number line (dimension) (i.e., within a single vector space) as global, with internal dimensions (geometric objects) determining the nature of scalar operations in the coefficients within the global vector space.

    (an area or a volume (or hyper-volume) can be characterized within the coefficient of the global vector, the division over a ring includes two dimensions for general functions within that vector space). So two dimensional "scalar" vector space multiplying the unit vector of another - a vector space within a vector space.

    e.g.,
    or

    (graphing the function still requires two dimensions within the "module".

    pretty easy answer.... (they just describe operations of scalar coefficients multiplying a global unit vector).

    (Shows the lengths number theorists will go to in trying to ignore Descartes...

    For my proof of Fermat's theorem, the question is moot anyway, since the Binomial Expansion holds for all (positive) real numbers....

    (This does relate to GTR, howevever, where the tensor analysis and trnsformations is performed on coefficients of the metric and stress-energy tensors (i.e., a tensor transformation on the metric tensor), so it is an interesting point, and confirms my analysis - that if the coordinate space is related to the energy-momentum space, it is because is interpreted as a coordinate density (or a mass multiplier) in STR, and is only constant (inertial) if is preserved in transformation. (the metric tensor, whatever it is, is invariant for a particular universe).

    If the metric and the stress-energy tensors are diagonalized with constant coefficients, there are no further ("spin") perturbations - i.e., the characterization of a black hole.

    Vector spaces within vector spaces within vector spaces..... ad infinitem
    Fractions over (and under) fractions over fractions.... ad infinitem

    Little fleas have smaller fleas
    upon their backs to bite 'em
    and smaller fleas have smaller fleas
    and so, ad infinitem

    "And it's turtles, all the way down..." - Number theorist to Bertrand Russell ..

    Mandelbrot, where are you?.....
    First of, modules are not number theory.

    Second, modules do not have a dimension necciserily.

  9. #19
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by emperorzelos View Post
    First of, modules are not number theory.

    Second, modules do not have a dimension necciserily.
    Ok, they're abstract algebra. I'm not sure which is worse, number theory or abstract algebra

    Finally, modules do not necessarily have a BASIS. (From the Wiki article) Those that don't are maybe affine vectors as the scalar coefficient of another vector? (I'm not even sure they have a metric, but I'm going to avoid researching it to find out).

    In any case I have zero interest in them personally, since my proof of Fermat's theorem does not depend on them, and plain ol' functions as coefficients of vectors are fine with me.

    "Twas brillig, and the slithy toves,
    Did gyre and gimble in the wabe...

    But did the momraths really outgrabe?..

    Anyway, I am outta here.
    Good luck in finding someone who is interested in that sort of thing....
    (or who knows enough to at least read the Wiki article on it...
    Last edited by BuleriaChk; 12-23-2016 at 01:48 AM.
    _______________________________________
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
    Ignore List -The Peanut Gallery.

  10. #20
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    Default Re: Cross product vs Tensor product

    Quote Originally Posted by emperorzelos View Post
    First of, modules are not number theory.

    Second, modules do not have a dimension necciserily.
    The subject got me to thinking this morning, however.

    The reason Fermat's expression can be related to the Binomial Theorem is becauset the integers a and b are independent. That is, with the exception of Pythagorean triples, is not an integer. In fact if the ring is restricted to positive integers, there is not even an additive inverse that can be applied (b > a implies that a-b < 0, and so is not a positive integer).

    That means that with the exception of Pythagorean triples, the coefficients each vector are a ring without division, which is why fractions consisting only of individual number lines as coefficients of vectors are independent.

    E.g. For fractions such as z = y/x, x and y are generally independent (which it is why z is the slope of a straight line in two dimensions), but not fractions such as z = f(x,y)/g(x,y). So they are also independent for , which is why they can form the bases of linear systems consisting of polynomials with constant coefficients.
    Last edited by BuleriaChk; 12-23-2016 at 11:26 AM.
    _______________________________________
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
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