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Thread: Euclidean geometry...subgroups

  1. #1
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    Euclidean geometry...subgroups

    Let E*(2) be the set of maps t:R2->R2 x->Ux+a with a in R2 and U an orthogonal matrix of determinant 1. Show that E*(2) is a subgroup of the group of Euclidean transformations and determine which of the isometries are in E*(2).

    I'm not really sure where to even start. Thanks.
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  2. #2
    Member HappyJoe's Avatar
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    By "Euclidean transformation", do you mean a map $\displaystyle T\colon\mathbb{R}^2 \rightarrow \mathbb{R}^2$, such that $\displaystyle T(x) = Ax + b$, where $\displaystyle A$ is an orthogonal matrix, and $\displaystyle b$ is some vector?

    If so, to show that $\displaystyle E^*(2)$ is a subgroup of the group of Euclidean transformations, you need to show that $\displaystyle E^*(2)$ is closed under composition of maps, under inverses and that it contains the neutral element.

    Closed under composition:

    Take $\displaystyle t_1,t_2\in E^*(2)$, and say $\displaystyle t_1(x) = O_1x+b_1$ and $\displaystyle t_2(x) = O_2x+b_2$, where $\displaystyle O_1$ and $\displaystyle O_2$ are orthogonal matrices of determinant 1. Then:

    $\displaystyle (t_1\circ t_2)(x) = t_1(t_2(x)) = t_1(O_2x+b_2) = O_1(O_2x+b_2) + b_1 = (O_1O_2)x + (O_1b_2+b_1),$

    and provided that $\displaystyle O_1O_2$ is an orthogonal matrix of determinant 1, this composition is indeed in $\displaystyle E^*(2)$. But this is true, as you may try to show.

    Closed under inverses.

    Let $\displaystyle t\in E^*(2)$. We need to prove that the inverse $\displaystyle t^{-1}$ is also an element of $\displaystyle E^*(2).$ Suppose $\displaystyle t$ is of the form $\displaystyle t(x)=Ox+b$, where $\displaystyle O$ is orthogonal of determinant 1. Then the inverse of $\displaystyle t$ is $\displaystyle t^{-1}(x) = O^{-1}x - O^{-1}b$. Since $\displaystyle O^{-1}$ is again an orthogonal matrix of determinant 1 (which you may check), the inverse is an element of $\displaystyle E^*(2)$ as needed.

    The neutral element is there.

    The neutral element is the map $\displaystyle T$ given by $\displaystyle T(x)=Ix = x$, where $\displaystyle I$ is the identity matrix. This is certainly an element of $\displaystyle S^*(2)$.

    As for the question "Determine which of the isometries are in $\displaystyle E^*(2)$", it depends on your definition of isometry. Are you isometries linear or are they just supposed to preserve the distance in $\displaystyle \mathbb{R}^2$?

    All of the Euclidean transformations preserve the distance in $\displaystyle \mathbb{R}^2$, and so does $\displaystyle E^*(2)$ in particular. If you are thinking in geometric terms, then $\displaystyle E^*(2)$ consists of the maps, which are rotations followed by a translation.
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