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Thread: Affine structure, Lambda_k

  1. #1
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    Question Affine structure, Lambda_k

    I am reading "Applied Differential Geometry" by William Burke. While I always had no trouble in standard math classes (diff. eq. and calc.) modern math (diff. geom.) is another story. This level of abstractness eludes me. So here is my question:

    The author writes:
    "The parametrization along the line passing through two points in an affine space is not unique." -p. 14.

    This is perfectly clear to me. The scale is arbitrary as well as where I start and stop on the line. (It's similar to arc length along a curve. E.g. I can use feet or inches, cm, or m along the curve. Also, where I start on the curve is completely arbitrary.)

    He continues:
    "If we single out the parametrization that runs from zero to one between two points, then the structure of the affine space A is given by a map

    $\displaystyle \Lambda_{.} : A A R \rightarrow A; (a,b,k) \mapsto \Lambda_{k}(a,b)$

    with conditions

    $\displaystyle \Lambda_{0} = a, \Lambda_{1} = b$"

    Now there is a figure which shows four points on a straight line. Point 'a' is labeled $\displaystyle \Lambda_{0}$, and point 'b' is labeled $\displaystyle \Lambda_{1}$. This makes sense according to the above definition of the map. However, the other two points, $\displaystyle \Lambda_{\frac{1}{2}}$ and $\displaystyle \Lambda_{2}}$ do not work.

    I think I understand them correctly - that they are two other (arbitrary) scalings for the parametrization of the line. But seeing that in detail completely - that is applying them - eludes me.
    Last edited by livingdog; Jul 6th 2010 at 09:53 AM.
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  2. #2
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    The image is here: Applied differential geometry - Google Books

    I don't understand what your question is. Are you thinking of a specific application?

    An affine space is just a linear space without distinguished origin. If you chuse an origin then you can write explicitly $\displaystyle \Lambda : (a,b,t) \mapsto (1-t)a + tb$.
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  3. #3
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    Quote Originally Posted by maddas View Post
    The image is here: Applied differential geometry - Google Books

    I don't understand what your question is. Are you thinking of a specific application?


    An affine space is just a linear space without distinguished origin. If you chuse an origin then you can write explicitly $\displaystyle \Lambda : (a,b,t) \mapsto (1-t)a + tb$.
    EDIT: let me restate this (and it's my fault I cannot be more clear through my confusion)

    There is a figure (thx to maddas for posting it) which shows four points on a straight line.
    Point 'a' is labeled $\displaystyle \Lambda_{0}$, and point 'b' is labeled $\displaystyle \Lambda_{1}$.
    This makes sense according to the above definition of the map.

    However, I do not know how to apply (use?) the other two scalings: $\displaystyle \Lambda_{\frac{1}{2}}$, and $\displaystyle \Lambda_{2}}$.

    I think I understand them conceptually - that they are just two other arbitrary scalings for the parametrization of the line. But seeing that in detail - i.e. applying them to the line - completely eludes me. So how does one apply $\displaystyle \Lambda_{\frac{1}{2}}$, and $\displaystyle \Lambda_{2}}$?

    If I crank out the first one, then $\displaystyle \Lambda_{\frac{1}{2}}(a,b) = \frac{1}{2}(b + a)$

    and if I crank out the second one, then $\displaystyle \Lambda_{2}(a,b) = 2b - a$.

    Where does that get me? Are the $\displaystyle \Lambda_{k}$ just points on the line - and not scalings??

    I apologize b/c I don't think I asked a ask... ...maybe this is my question: WHAT THE HEC IS$\displaystyle \Lambda_{k}$???

    ...oh wait a minute, I think I see it now. Let me try to explain this.

    $\displaystyle \Lambda_{k}$ chooses the start and end point of the scale - the 'k' factor - that one is using for the line. Thus the entire line can be located using these different scales, $\displaystyle \Lambda_{1}$ for scaling in units of '1', $\displaystyle \Lambda_{0.75}$ for scaling in units of 0.75, etc., where k $\displaystyle \in$ R.

    See the attached image: Affine structure, Lambda_k-untitled.jpg

    I apologize if I am still too thick to understand.
    Last edited by livingdog; Jul 6th 2010 at 09:55 AM.
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  4. #4
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    Errrr, let me write $\displaystyle \Lambda_k(a,b)$ as $\displaystyle \Lambda(a,b,k)$. Then fixing a and b, $\displaystyle \Lambda(a,b,\cdot):\mathbb{R} \to A$ is a parameterized (constant speed, I assume) line in the affine space. For a given value of k, $\displaystyle \Lambda(a,b,k)$ is a point on the line. A (constant speed) reparamterization of the line is a map $\displaystyle \gamma : \mathbb{R} \to A, t \mapsto \Lambda_k(a,b,c_1+c_2t)$ for constants c_1 and c_2. Intuitively, if you start at a at t=0 and you get to b at time t=1, then $\displaystyle \Lambda(a,b,k)$ is where you are at time t=k...
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  5. #5
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    Quote Originally Posted by maddas View Post
    Errrr, let me write $\displaystyle \Lambda_k(a,b)$ as $\displaystyle \Lambda(a,b,k)$. Then fixing a and b, $\displaystyle \Lambda(a,b,\cdot):\mathbb{R} \to A$ is a parameterized (constant speed, I assume) line in the affine space. For a given value of k, $\displaystyle \Lambda(a,b,k)$ is a point on the line. A (constant speed) reparamterization of the line is a map $\displaystyle \gamma : \mathbb{R} \to A, t \mapsto \Lambda_k(a,b,c_1+c_2t)$ for constants c_1 and c_2. Intuitively, if you start at a at t=0 and you get to b at time t=1, then $\displaystyle \Lambda(a,b,k)$ is where you are at time t=k...
    I think I see now:
    1. take a bare line
    2. attach, and fix, two points: 'a' and 'b'
    3. to locate any point on the line, use $\displaystyle \Lambda_k$


    So e.g. $\displaystyle \Lambda_\frac{1}{2}(a,b)$ locates the $\displaystyle \frac{1}{2}$ point between 'a' and 'b'; $\displaystyle \Lambda_{2}(a,b)$ locates the $\displaystyle 2$ point using 'a' and 'b' as the scale of the line.

    If that's it then I got it.

    Sorry for being so dense. But abstract math was never my forte'. Beyond linear algebra I get bogged down not being able to "see" the concepts without many questions/answers - as my posts in this thread demonstrates.

    Thank you very much for your explanations and patience with me!
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