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Thread: Dimensions

  1. #1
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    Default Dimensions

    We have a moron here whom constantly go on about dimension but has no clue what it is really about so I will go through it here.

    First of we need to know what a vector space is versus a module.

    A left module over a ring is an abelian group with an action such that for and we have


    This is a module which is the general case, we can define the right module similarly by switching side but as it doesn't matter here we focus only on left. when our ring is a field then our module is a vector field.

    Now we have a generalization of dimension too, it is caleld the length of the module, let the following sequence of submodules be given



    and it is the maximum length of proper submodules, then the length of the module is . This generalization fits quite well with our intuitive understanding of dimensions for vector fields. Let us look at , we have then the chain

    as one of the longest possible chain of submodules. That gives us that it's length is . The reader should pay attention to that here we have not used basis, vectors or anything to define the dimension, or the general case of length, because it is quite superfluous and requires additional structure in many cases. The traditional idea of independent vectors is still valid as for vectors such that

    we have that there exists a vector field homomorphism, also known as a linear function, such that where is some field of our choice, most commonly the real numbers. As such we see that our old idea of dimensions is encompassed within the generalized idea. However one needs to pay attention to that the vector field commonly denoted does not contain a lot of the structure we normally associate with it, as we properly should write it as where hte last element indicates the scalar multiplication we are used to. Two things are lacking that we are used to, the concept of a norm in the vector field and the concept of orthogonality. Neither exist so our old call for orthonormal basis is meaningless, we cannot have it as neither concept makes sense.

    For that we need to introduce what is called an innerproduct, denoted . This innerproduct makes our vector field to an innerproduct space, which is a vector space with an inner product. The inner product makes it possible to claim that something is "orthogonal" to something else. This is done by saying that two vectors are orthogonal if . the ordinary innder product is the dot product, that is


    The inner product gives us also an induced normed, so all innerproduct spaces are normed spaces. That is , this makes it possible for us to normalize vectors such that we claim that a vector has in a sense a length of 1.

    This brings us to for example functions , this is a function from one set to another, is it meaningful to claim it has any form of dimension? No, it is not a module of any sort! It's just a set function. Alright, let's assume it is continuous, well it is not enough, still just a function! It is a subset of as I explained in my post about definition functions from set theoretical perspective. But now wait, we can impose a vector structure on , so what if we plot the function in ? That's not gonna help either, the thing to remember with the here is that we are just talking about a set, it has none of the structure required to make the concept of dimension meaningful.

    A function cannot have dimensionality in that sense as they are not a module of any sort. However we can have that module homomorphisms can give images that are modules, and we can make a vector field of functions, but talking about a dimension to a single function that is just a general one, is meaningless.

    So summa summarum
    A set has no meaningful way of claiming a dimension
    A module has a way to have a dimension, or length, but it is independent of what we are familiar with
    A vector space cannot have orthogonality or normal vectors as neither is defined explicitly for it.
    A normed space is a vector space where we can normalize vectors
    An inner product space can even make claims of orthogonality when it comes to vectors and here we can get a basis that are orthonormal.
    Functions and their images are not modules, only a specific subset of functions give an image that is a module, but the set of all continuous functiosn can be made into a vector space.

    Keep in mind that just because we project a graph on a 2D surface for our visual convinience does that not in anyway mean that the dimensionality of the surface we use, reflect anything about the dimension of the function, it makes no sense talking about dimensionality for a function.


  2. #2
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    Default Re: Dimensions

    Quote Originally Posted by emperorzelos View Post
    We have a moron here whom constantly go on about dimension but has no clue what it is really about so I will go through it here.

    First of we need to know what a vector space is versus a module.

    A left module over a ring is an abelian group with an action such that for and we have


    This is a module which is the general case, we can define the right module similarly by switching side but as it doesn't matter here we focus only on left. when our ring is a field then our module is a vector field.

    Now we have a generalization of dimension too, it is caleld the length of the module, let the following sequence of submodules be given



    and it is the maximum length of proper submodules, then the length of the module is . This generalization fits quite well with our intuitive understanding of dimensions for vector fields. Let us look at , we have then the chain

    as one of the longest possible chain of submodules. That gives us that it's length is . The reader should pay attention to that here we have not used basis, vectors or anything to define the dimension, or the general case of length, because it is quite superfluous and requires additional structure in many cases. The traditional idea of independent vectors is still valid as for vectors such that

    we have that there exists a vector field homomorphism, also known as a linear function, such that where is some field of our choice, most commonly the real numbers. As such we see that our old idea of dimensions is encompassed within the generalized idea. However one needs to pay attention to that the vector field commonly denoted does not contain a lot of the structure we normally associate with it, as we properly should write it as where hte last element indicates the scalar multiplication we are used to. Two things are lacking that we are used to, the concept of a norm in the vector field and the concept of orthogonality. Neither exist so our old call for orthonormal basis is meaningless, we cannot have it as neither concept makes sense.

    For that we need to introduce what is called an innerproduct, denoted . This innerproduct makes our vector field to an innerproduct space, which is a vector space with an inner product. The inner product makes it possible to claim that something is "orthogonal" to something else. This is done by saying that two vectors are orthogonal if . the ordinary innder product is the dot product, that is


    The inner product gives us also an induced normed, so all innerproduct spaces are normed spaces. That is , this makes it possible for us to normalize vectors such that we claim that a vector has in a sense a length of 1.

    This brings us to for example functions , this is a function from one set to another, is it meaningful to claim it has any form of dimension? No, it is not a module of any sort! It's just a set function. Alright, let's assume it is continuous, well it is not enough, still just a function! It is a subset of as I explained in my post about definition functions from set theoretical perspective. But now wait, we can impose a vector structure on , so what if we plot the function in ? That's not gonna help either, the thing to remember with the here is that we are just talking about a set, it has none of the structure required to make the concept of dimension meaningful.

    A function cannot have dimensionality in that sense as they are not a module of any sort. However we can have that module homomorphisms can give images that are modules, and we can make a vector field of functions, but talking about a dimension to a single function that is just a general one, is meaningless.

    So summa summarum
    A set has no meaningful way of claiming a dimension
    A module has a way to have a dimension, or length, but it is independent of what we are familiar with
    A vector space cannot have orthogonality or normal vectors as neither is defined explicitly for it.
    A normed space is a vector space where we can normalize vectors
    An inner product space can even make claims of orthogonality when it comes to vectors and here we can get a basis that are orthonormal.
    Functions and their images are not modules, only a specific subset of functions give an image that is a module, but the set of all continuous functiosn can be made into a vector space.

    Keep in mind that just because we project a graph on a 2D surface for our visual convinience does that not in anyway mean that the dimensionality of the surface we use, reflect anything about the dimension of the function, it makes no sense talking about dimensionality for a function.
    That's why I avoid modules. Proof of Fermat's theorem doesn't need them, only vector spaces.
    Fermat's theorem is lacking a multiplicative product. STR establishes the multiplicative product in each of the two bases for independent variables, and shows that the Binomial Expansion includes multiplicative products between independent variables, and so is complete.

    So abstract modules are completely irrelevant to proving Fermat's theorem; they are totally unnecessary if one admits functions according to the normal rules of arithmetic. Arithmetic alone cannot have multiple dimensions, and in the context of Fermat's theorem, such a reduction is a reduction ad absurdum.

    Just the declaration of the product ct is an example of multiplication not included in Fermat's expression.

    Modules in this sense are a solution looking for a problem... they are not even thought widgets... and even to create a thought widget, there has to be something there a rate of widget creation (call it c). And to declare an invariant one has to terminate the rate of widget creation by a scalar, t, so x = ct

    Modules are an attempt to characterize t without c as a rate of change; if c= 0, there is nothing except a tautology.

    A straight line has the equation y=Ax + b. For b=0, y=Ax. Even if A = 1, to declare two variables y=x, one needs two dimensions (x,y) unless the expression is a tautology. .

    Creating a basis in a single dimension for all possible scalar multiples is accomplished by

    (the foundation of the so-called "time dilation" equation, except it is a relation between an initial multiplicative state (ct) a change of state (vt') and a final state (ct) of widget creation, thought process or not.

    Dividing both sides by the final state (ct') yields the relativistic unit circle, thus defining sin and cos (if one multiplies by or a relation between squares geometrically, but abstractly it means that is independent of y = x, and the same is true of higher powers, which is why a basis of powers can form a linear system of polynomials with constant coefficients.

    To establish a second independent widget, a second circle is needed, where the unit basis is provided in the same manner (with independent field variables c,v,t, and t')

    The Binomial Expansion is then a function which can then be compared with Fermat's Expression , the latter of which has no multiplicative terms, and is thus the basis of a Presburger arithmetic.

    (Note that the constructivist approach does not include powers; which in one dimension is a tautology as compared with where A is a scaling factor (a slope) even if A = 1.

    Modules are just a complicated way of trying to ignore scalars by trying to provide a justification for powers in terms of counting (Peano's axions, group theory independent of dimension - distinguishable variables over two distinct fields).

    Modules are an attempt to characterize x=ct (or y=ct) without declaring a rate of change c (so even derivatives are irrelevant). again, c=0 means there is nothing there, not even a thought widget just a tautology as an expression .

    Regarding polynomials, independent variables, and Fermat's theorem vs. the Binomial Expansion, the concept of modules without dimension is an abstraction full of sound and fury and signifying nothing (except tautologies).

    Correction: vector spaces can be null or one-dimensional (I can't find where I stated the contrary; on the other hand I have used one-dimensional and null vectors throughout this thread). Doesn't change my argument here.
    Last edited by BuleriaChk; 03-21-2017 at 10:25 AM.
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  3. #3
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    Default Re: Dimensions

    stop with your idiocy of special relativity, it has no bearing on mathematics.

    And no, Modules are not trying to avoid scalars, they generalize them to NOT necciserily come from a field you ignoramus.

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