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Relating Two-Dimensional Vectors and Complex Numbers

  1. Dec 23, 2008 #1
    My aim is to interpret two-dimension vector quotients. I will be using this as the analogue to what a quaternion is.

    Ok here goes:
    *only considering vectors in two-dimensions*
    We define a vector to be a directed line from A to B.
    We define equality between two vectors if we can use translations to bring the two starting points and two final points to coincide.

    Remark 1)
    Any vector can be defined by two parameters, its tensor and its unit vector.
    The unit vector has length 1 and direction.
    Direction can be specified by the angle which it makes with the x-axis, restricted to 0 and 2pi and measured anticlockwise from the x-axis
    Thus, to identify a vector, all we need is an tensor and an angle.

    Remark 2)
    We take two vectors, a and b. We wish to look at what is meant by the quotient of a and b.
    q = a/b, that is q * b = a.
    q being a quantity such that when the vector b is left multiplied by q, it becomes a.
    Geometrically, q changes b by 1) scaling its length 2) changing its direction.
    The scaling length is what is meant by a tensor.
    The change in direction is precisely a angle which is turned through, anti clockwise.

    Claim:
    q is a vector as it has precisely tensor and angle.



    What I am stuck on now is explicitly relating this to complex numbers.
    A complex number can be expressed as re^ix where r is a positive real number and x is a number between 0 and 2pi. Thus, it is much like length and angle.
    I can see that there is a bijection between complex numbers and vectors.
    I can also see that with the bijection, addition and subtraction hold. So there is a isomorphism between vectors and complex numbers.
    But how on earth do I begin to relate the multiplication of complex numbers to vectors?
    Is it as simple as saying, define multiplication like that which we get some complex number multiplication. But how do I go about showing that this is a good definition.... look at distributivity, etc? (not communitivity?)

    Thanks in advance

    Joe V
     
  2. jcsd
  3. Dec 23, 2008 #2

    HallsofIvy

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    What do YOU mean by "tensor" here and in the first line?

     
  4. Dec 23, 2008 #3
    By tensor I mean a positive real number which acts as a "scale factor". It simply changes the length of the vector.

    A interesting point to note here is that 0 is *not* a tensor as this would change the direction, though the limit exists.
     
  5. Dec 23, 2008 #4

    HallsofIvy

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    You are assigning a specific direction to the 0 vector then? What direction?
     
  6. Dec 23, 2008 #5

    jambaugh

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    You mean "length" not "tensor". A vector is a particular type of tensor (rank 1). A matrix is a rank 2 tensor, and so on.
    Do you intend that q be a vector? There are also matrix operators which can multiply a vector to yield a vector. With regard to your attempt, you should first define a product between vectors and then look for an identity 1 such that 1*x = x for any x. Then you look for inverses xy = 1 -> y = 1/x. You then have quotients as products of inverses.
    Yes and so your product should multiply lengths and add angles (modulo 2pi).

    For complex numbers you take the units 1 and i as your basis:
    [tex](x,y)=x+iy[/tex]
    Thence
    [tex] 1 = (1,0)[/tex]
    [tex](x,y)(u,v) = (xu-yv,xv+yu)[/tex]

    And you'll find then that:
    [tex]\frac{1}{(x,y)} = \frac{(x,-y)}{x^2 + y^2}[/tex]

    There is no other way to define in two dimensions what you are trying to define which is a division algebra. These have been well mapped out.
    Real numbers (dim=1)
    Complex numbers (dim=2),
    Hamilton's quaternions (dim =4) (non-commutative).
    Octonions (dim = 8) (non-commutative and non-associative).

    Here's another way to look at it. Polynomials are also vectors (in and infinite dimensional space) in that you can add them and multiply by scalars. You also have a product on polynomials which yields polynomials. Now the degrees of a product will add so if you want to restrict yourself to a finite dimensional space you need to restrict the degree via some identity on the variable. Call the variable i and use the identity [itex] i^2+1=0[/tex] and you can work with only first degree polynomials x+iy. This defines the complex number system.

    Trying other identities will yield other algebras but you will have something call zero divisors and not all elements (if any) will have inverses in the system so you won't get a division algebra. Note that a real algebra is also a vector space but with an additional multiplication. So this is what you are constructing.

    I suggest you take a look at the various wikipedia articles on algebras, division algebras, rings, fields, quotient rings, et cetera.
     
  7. Dec 23, 2008 #6
    Thank you for the reply. This has helped me quiet alot, and has changed how I plan to continue my essay.

    EDIT: What i meant is definatly a tensor. The tensor you refer to is a modern definition. By tensor I mean a signless number which is a stretch factor rather than length.
    I am suggesting that a vector is made of a stretch vector and unit vector with direction rather than a length and direction.

    What i want q to be is some operation that will turn b into a, regardless of what vectors b and a were. What other options could there be except another vector? If we were to say it was a matrix, surely it would be some more complication version, but essentually to same, as the vector. Like if q needed to turn b by pi/2 and strect it by 2, it could be the rotation matrix doubled, but surely this would just be the more complicated version of the vector.
    In fact, this is interesting in itself.
    So essentially a rotation stretch matrix is the q I am looking for. It is just another representation of the same thing. Oh err this is much more fun than spending time with the family :)

    Oh thank you very much, darn got a party to go to 2nite. I can see myself thinking about this instead of socialising :P
     
  8. Dec 23, 2008 #7

    jambaugh

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    Words mean things. Do not use the word tensor. You should use the word "scalar" as you are talking about rescaling the vector without changing its direction. The term tensor historically came from the theory of elasticity in more than one dimension. The original concept of a tensor was the operator which could stretch vectors but not uniformly and also could shear vectors. These will change the direction of an arbitrary vector and only rescales in particular directions (eigen-vectors).

    Consider if you draw a horizontal vector, a vertical vector and an oblique vector on a rubber sheet. Then apply "tension" in one direction, say left to right, you will stretch the horizontal vector, do nothing to the vertical vector, and stretch the horizontal component of the oblique vector thereby changing its direction.

    The mathematical representation of this transformation is a 2x2 matrix (rank 2 tensor):

    [tex]\left[{\array{cc}{ S & 0 & 0 & 1 &} }\right][/tex]

    [edit:] Let me add that the term tensor has a precise meaning and you shouldn't try to redefine it... especially as there is a term already defined for what you mean i.e. scalar and the current definition of tensor is intimately connected with the subject of vectors and vector spaces. You could say use the word "pointer" instead of "vector" or "flabugalar" for the number 2. Changing common terms will however only add to confusion and miscommunication. Learn and use the standard terms and their definitions and not only will you be understood better, you will later better understand.

    Believe me I sympathize with the desire to redefine terms. I grate at the use of the "super-" qualifier in the current context of mathematical physics when it has a better older use (dual to "sub-") in mathematics and I would love to get everyone to change how they use the term. But this can also be seen from the opposite side, namely the persons invoking "super-algebras" and "super-symmetry" should have at the time chosen a different qualifier in light of the potential confusion. (Although in the case of "super-symmetry" it can also be understood in both contexts.)
     
    Last edited: Dec 23, 2008
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