What is an affine space?

Discussion in 'Physics & Math' started by arfa brane, Dec 5, 2020.

  1. arfa brane call me arf Valued Senior Member

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    Is it a Euclidean space, but without distances and angles; or without any functions that transform distances or angles, neither being defined?

    So there's no inner product, but apparently in an affine space, parallel lines still have the same 'structure' as in Euclidean space. So parallelograms are defined and there are volume-preserving functions. Distance isn't defined, but affine distance is (how, though?).

    What affine distance actually means, mathematically, is an object which is "like distance", but has nowhere to define a unit of distance.

    Or should I be saying, correct me if I'm wrong about this, but . . .
     
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  3. mathman Valued Senior Member

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  5. arfa brane call me arf Valued Senior Member

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    --https://en.wikipedia.org/wiki/Affine_geometry

    Aha so, relations between points and lines; both exist mathematically in any real dimension (degree of freedom) one or greater.
    Polynomials: I've seen one or two of these, I know you can define a ring of polynomials over a field. Everyone's seen a polynomial in one variable, x, and how to write a closed form expression (ok maybe not everyone's seen that) for a polynomial.

    But say, for something like \( x^2 - 1 \), you might get a question about this in high school math. You might be able to sketch this as a function of x, or assign f(x) to a y axis. This isn't a polynomial, it's a projection; the polynomial is a function f(x), not a function f(x,y). The sketch of the polynomial-as-a-function is really a g(x,f(x)).

    There are two roots for this expression, it can be written (x + 1)(x - 1). This shows that the original expression (which plots a parabola in \( \mathbb R^2\)) is the product of two degree-1 polynomials, or two 'straight lines'. If you add two straight lines, you get another straight line.
     
    Last edited: Dec 10, 2020
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  7. Write4U Valued Senior Member

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    set of straight lines, parabola.......
     
  8. arfa brane call me arf Valued Senior Member

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    Which is it? You get a parabola by multiplying lines, I can't see multiplication in your diagram.
    The parabola isn't there, but a subset of tangent lines to a parabola is.

    To be a ring, a set of polynomials needs addition and multiplication; usually you write k[x] for the polynomial ring over a field, k, in one (independent) variable x. That x is independent also means it doesn't need to be an element of the field; you can solve polynomials in x where x is a pizza topping, or a breed of sheep, it can be anything.

    In one variable, you get a general form: \( a_0x^0 + a_1x^1 + a_2x^2 + . . . + \; a_nx^n \) with closed form \( \sum_{i=0}^n a_ix^i \)
     
    Last edited: Dec 11, 2020
  9. Write4U Valued Senior Member

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    I did not mention multiplication. You said "add straight lines".
    If you say so. I merely demonstrated that by adding straight lines in a particular fashion, a parabola emerges from all the points where the straight lines come in contact with each other. I seem to remember something that what emerges is not a true parabola, but a variation. A hyperbolic paraboloid?

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    https://thebridgetcd.files.wordpress.com/2014/09/fig1-hyperbolic-paraboloid.jpg
     
  10. arfa brane call me arf Valued Senior Member

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    7,159
    Adding lines to a diagram isn't the same as mathematical addition of two lines; let's say you have a pair of lines, ax + b and cx + d, their sum is (a + c)x + (b + d) and is one line, not two or more.
     
  11. Write4U Valued Senior Member

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    I accept your mathematics.
    What I should like to know is by what equation the hyperbolic paraboloid emerges from 2 sets of straight lines? (as illustrated)

    Obviously there must be an equation that explains the shown illustration?
     
    Last edited: Dec 11, 2020
  12. QuarkHead Remedial Math Student Valued Senior Member

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    OK.Returning to the OP "What is an Affine Space?"

    First note that, for the vector \(v\) the transformation \(Tv=v+k\) where the vector \(k\) is a fixed vector is called a translation.

    Next suppose a linear transformation \(Lv=w\). Then an affine transformation is just \(Av=Lv+k = w+k\). Simple.

    But recall that the set of all linear transformations is a group, as is the set if all translations (this is easily proved -try it!). So that the set of all affine transformations is also a group: \(\mathcal{A}=\mathcal{L} \oplus \mathcal{T}\). In fact it is nt hard to prove directly that the set of all affine transformations is a group. Again, try it

    So an affine space is a vector space invariant under the affine group. c'est tout

    Similarly, affine geometry is that geometry invariant under the affine group. It has some strange properties to those brought up with Euclidean geometry
     
  13. arfa brane call me arf Valued Senior Member

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    Hopeful question: Is a polynomial ring invariant under this affine group?
     
  14. QuarkHead Remedial Math Student Valued Senior Member

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    No, in fact the question is without meaning. Why? Because transformations, aka operators act on a vector space, and that alone.

    To see why (the following is intended as an elucidation, not a proof.)

    Any vector space has, by definition a distinguished subspace that consists of those vectors that are not a linear combination of any others in the same subspace. and again by definition any other vector in the total space is such a linear combination.

    We can think of a transformation in 2 ways - hold the basis fixed and transform the other vectors relative to it. Or hold the other vectors fixed and transform the basis. It turns out the result is the same either way.

    But in each case, the basis vectors enter the picture. But rings do not have a basis, so the term "transformation" (aka operator) cannot apply to them
     
  15. arfa brane call me arf Valued Senior Member

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    7,159
    But isn't a set of polynomials a vector space? For instance, a textbook I have tells me:
    "The set \( \mathbb Z_p [x] \) of polynomials with coefficients from \( \mathbb Z_p\) is a vector space over \( \mathbb Z_p\), where p is prime.

    Or rather, k[x] for a field k is a vector space if x is also in k. This extends to any number of independent variables, [x,y], [x,y,z] . . .

    Even more, erm, confusing is this question from a lecture handout which is about affine geometry:

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    Last edited: Dec 19, 2020
  16. QuarkHead Remedial Math Student Valued Senior Member

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    Well yes, but you should be clear as to the question you are asking.

    First you asked about transformations of polynomial rings, and I gave the most complete answer that I could. Now you asking about transformations on a set of polynomials regarded as a vector space. Different questions.

    Look, a polynomial \(p(x)\) can only be regarded as a vector if and only if a) the variable \(x\) is Real and b) the coefficients are Real or Complex numbers.

    In either case the set of all such polynomials form an infinite-dimensional vector space, I leave it to you to see if this set is subject to affine transformations,
     
  17. arfa brane call me arf Valued Senior Member

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    It seems that Purdue Uni takes the approach that an affine space is the same as a set of polynomials (possibly finite) over a field (possibly a commutative ring).

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    --https://www.math.purdue.edu/~arapura/preprints/BasicAG1.pdf

    footnote #1 says:

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    And the ring of polynomials, defined as a set, is the usual sum of products; the ith coefficient is the product of n other coefficients. So if say, n = 1, you just have \( k[x] = \mathbb A_k \).

    I also assume that "\(ev_a(f)\)" means evaluate f at a.
     
    Last edited: Dec 21, 2020
  18. QuarkHead Remedial Math Student Valued Senior Member

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    arfa brane, you seem to spend your days searching the internet for things that you think will wrong-foot me.

    Never mind that you don't understand what you find. Sorry, I am not playing that game, and I am more sorry I made a genuine and honest attempt to help you.

    Have a good Xmas
     
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  19. arfa brane call me arf Valued Senior Member

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    Sad. That's possibly a reflection back onto you and something I call, "who cares?".
    Pathetic. Seriously.
     
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  20. arfa brane call me arf Valued Senior Member

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    7,159
    The questions I have, or to some extent had, are what are the differences between vector spaces and affine spaces, and if distance isn't defined in an affine space, what's an affine distance? If parallel lines are translation invariant, what do you call the distance between two such lines, e.g. (x - a) and (x + a), where a is constant?
     
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  21. arfa brane call me arf Valued Senior Member

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    7,159
    Another zinger:

    The Euclidean plane is the affine plane with a metric (a way to to define a unit interval (0,1) and angles between lines).
     
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  22. arfa brane call me arf Valued Senior Member

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    Hokay. Now I'm looking at the differences between vector spaces and modules.

    And a curious observation that a group is a mathematical object (a set with a binary operation on pairs from the set), which allows you to find solutions to equations.
    The equations might (or not) just be polynomials, and the solutions are obtained by setting one or more polynomials to arbitrary values; if so, then the values can be 0.

    This is trivially true because you can subtract one side of an equation from the other; this is equivalent to subtracting one side, an expression, from itself which is arithmetically = 0. 0 is the additive identity because adding 0 to an expression doesn't change the value of the expression. Thank goodness.

    So for instance the integers, \( \mathbb Z \) are a group under addition; this is something I've seen notated as \( (\mathbb Z, \;+)\) or \( \mathbb Z^+ \).

    The symmetries of this group let you solve addition problems in which you invoke the use of addition of an inverse, e.g. -1 is the inverse of 1, in \( \mathbb Z^+ \).

    Apparently modules are generalizations of vector spaces. So a vector space over a field generalizes to a module over a ring. Polynomials can form rings which are vector spaces if the field is the real or complex numbers, or if the independent variables and coefficients are both real.

    So presumably there's a proof somewhere that only those kinds of polynomial rings are also vector spaces. Ahem.

    Do I understand what a vector is? Why should you care?
     
    Last edited: Jan 30, 2021
  23. arfa brane call me arf Valued Senior Member

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    I think the word "affine" when used in mathematics has a somewhat mangled history; it does seem to mean "related", but related how? It does also seem to mean "linear", almost interchangeably.

    I found a good-looking explo, at stackexchange.
    https://english.stackexchange.com/q...the-word-affine-in-the-context-of-mathematics

    And I think the original Latin meaning of affinis was familial, it was about family connections.
     

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