Proof of Logarithm map formulae from $SO(3)$ to $mathfrak {so}(3)$












2












$begingroup$


According to exponential map, there also exist a logarithm map
$$log:SO(3) to mathfrak {so}(3).$$
Suppose a vector $t inmathfrak {so}(3)$ and $t=|t|w$, according to exponential map
$$R = cos|t|I+(1-cos |t|)ww^T+sin |t|w^{land}.$$
So the inverse(i.e. from rotation matrix $R$ to vector $t$) is
$$|t|=cos^{-1}left(frac{operatorname{Tr}(R)-1}{2}right)tag 1$$
$$w = frac{1}{2sin(|t|)}
begin{bmatrix} r_{32}-r_{23} \ r_{13}-r_{31} \ r_{21}-r_{12}
end{bmatrix} tag 2$$
where $r_{ij}$ is elements of $R$. I can easily derive (1) while stuck in (2), can anyone help me out?






linear-algebra lie-groups lie-algebras rotations







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$endgroup$












  • $begingroup$
    WP not enough?
    $endgroup$
    – Cosmas Zachos
    Dec 10 '18 at 23:14


















2












$begingroup$


According to exponential map, there also exist a logarithm map
$$log:SO(3) to mathfrak {so}(3).$$
Suppose a vector $t inmathfrak {so}(3)$ and $t=|t|w$, according to exponential map
$$R = cos|t|I+(1-cos |t|)ww^T+sin |t|w^{land}.$$
So the inverse(i.e. from rotation matrix $R$ to vector $t$) is
$$|t|=cos^{-1}left(frac{operatorname{Tr}(R)-1}{2}right)tag 1$$
$$w = frac{1}{2sin(|t|)}
begin{bmatrix} r_{32}-r_{23} \ r_{13}-r_{31} \ r_{21}-r_{12}
end{bmatrix} tag 2$$
where $r_{ij}$ is elements of $R$. I can easily derive (1) while stuck in (2), can anyone help me out?






linear-algebra lie-groups lie-algebras rotations







share|cite|improve this question











$endgroup$












  • $begingroup$
    WP not enough?
    $endgroup$
    – Cosmas Zachos
    Dec 10 '18 at 23:14
















2












2








2





$begingroup$


According to exponential map, there also exist a logarithm map
$$log:SO(3) to mathfrak {so}(3).$$
Suppose a vector $t inmathfrak {so}(3)$ and $t=|t|w$, according to exponential map
$$R = cos|t|I+(1-cos |t|)ww^T+sin |t|w^{land}.$$
So the inverse(i.e. from rotation matrix $R$ to vector $t$) is
$$|t|=cos^{-1}left(frac{operatorname{Tr}(R)-1}{2}right)tag 1$$
$$w = frac{1}{2sin(|t|)}
begin{bmatrix} r_{32}-r_{23} \ r_{13}-r_{31} \ r_{21}-r_{12}
end{bmatrix} tag 2$$
where $r_{ij}$ is elements of $R$. I can easily derive (1) while stuck in (2), can anyone help me out?






linear-algebra lie-groups lie-algebras rotations







share|cite|improve this question











$endgroup$




According to exponential map, there also exist a logarithm map
$$log:SO(3) to mathfrak {so}(3).$$
Suppose a vector $t inmathfrak {so}(3)$ and $t=|t|w$, according to exponential map
$$R = cos|t|I+(1-cos |t|)ww^T+sin |t|w^{land}.$$
So the inverse(i.e. from rotation matrix $R$ to vector $t$) is
$$|t|=cos^{-1}left(frac{operatorname{Tr}(R)-1}{2}right)tag 1$$
$$w = frac{1}{2sin(|t|)}
begin{bmatrix} r_{32}-r_{23} \ r_{13}-r_{31} \ r_{21}-r_{12}
end{bmatrix} tag 2$$
where $r_{ij}$ is elements of $R$. I can easily derive (1) while stuck in (2), can anyone help me out?






linear-algebra lie-groups lie-algebras rotations




linear-algebra lie-groups lie-algebras rotations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 14:28









amWhy

192k28225439




192k28225439










asked Dec 9 '18 at 4:16









FinleyFinley

375213




375213












  • $begingroup$
    WP not enough?
    $endgroup$
    – Cosmas Zachos
    Dec 10 '18 at 23:14




















  • $begingroup$
    WP not enough?
    $endgroup$
    – Cosmas Zachos
    Dec 10 '18 at 23:14


















$begingroup$
WP not enough?
$endgroup$
– Cosmas Zachos
Dec 10 '18 at 23:14






$begingroup$
WP not enough?
$endgroup$
– Cosmas Zachos
Dec 10 '18 at 23:14












1 Answer
1






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oldest

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1












$begingroup$

This is a standard formula accompanying the Rodrigues rotation formula, obvious if only you use the right notation! The important thing is is to streamline your orthogonal matrix R. Its logarithm is the antisymmetric matrix K, defined by the linear cross product at the end of your expansion of R,
$$
Kequiv
left[begin{array}{ccc}
0 & -t_3 & t_2 \
t_3 & 0 & -t_1 \
-t_2 & t_1 & 0
end{array}right] ~,
$$
so that $| t|^2= -frac{operatorname {Tr} (K^2)}{2}equiv | K|^2$; hence $K=|t| hat K$ with $| hat K|=1$ .



Now the crucial thing is to re-express the Rodrigues formula in this notation,
$$
R=(R^T)^{-1}= e^K= {mathbb 1} + sin |t| ~ hat K + (1-cos|t|) hat K ^2 .
$$
Note how different the coefficient of the identity looks, contrasted to your vector expression: it is not a mistake (read up on the point in WP.... well-wishers frequently "correct" it! Of course, this expression follows directly from the characteristic equation for K, not Rodrigues' original argument).



So, tracing this expression, and the fact that K is traceless since antisymmetric, your (1) follows.



Now antisymmetrize R,
$$
frac{R-R^T}{2}= sin|t| ~ hat K ,
$$
since the other two terms are symmetric matrices. Convert your "unit" antisymmetric matrix to your unit vector w and you have (2).






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    1 Answer
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    1 Answer
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    active

    oldest

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    1












    $begingroup$

    This is a standard formula accompanying the Rodrigues rotation formula, obvious if only you use the right notation! The important thing is is to streamline your orthogonal matrix R. Its logarithm is the antisymmetric matrix K, defined by the linear cross product at the end of your expansion of R,
    $$
    Kequiv
    left[begin{array}{ccc}
    0 & -t_3 & t_2 \
    t_3 & 0 & -t_1 \
    -t_2 & t_1 & 0
    end{array}right] ~,
    $$
    so that $| t|^2= -frac{operatorname {Tr} (K^2)}{2}equiv | K|^2$; hence $K=|t| hat K$ with $| hat K|=1$ .



    Now the crucial thing is to re-express the Rodrigues formula in this notation,
    $$
    R=(R^T)^{-1}= e^K= {mathbb 1} + sin |t| ~ hat K + (1-cos|t|) hat K ^2 .
    $$
    Note how different the coefficient of the identity looks, contrasted to your vector expression: it is not a mistake (read up on the point in WP.... well-wishers frequently "correct" it! Of course, this expression follows directly from the characteristic equation for K, not Rodrigues' original argument).



    So, tracing this expression, and the fact that K is traceless since antisymmetric, your (1) follows.



    Now antisymmetrize R,
    $$
    frac{R-R^T}{2}= sin|t| ~ hat K ,
    $$
    since the other two terms are symmetric matrices. Convert your "unit" antisymmetric matrix to your unit vector w and you have (2).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      This is a standard formula accompanying the Rodrigues rotation formula, obvious if only you use the right notation! The important thing is is to streamline your orthogonal matrix R. Its logarithm is the antisymmetric matrix K, defined by the linear cross product at the end of your expansion of R,
      $$
      Kequiv
      left[begin{array}{ccc}
      0 & -t_3 & t_2 \
      t_3 & 0 & -t_1 \
      -t_2 & t_1 & 0
      end{array}right] ~,
      $$
      so that $| t|^2= -frac{operatorname {Tr} (K^2)}{2}equiv | K|^2$; hence $K=|t| hat K$ with $| hat K|=1$ .



      Now the crucial thing is to re-express the Rodrigues formula in this notation,
      $$
      R=(R^T)^{-1}= e^K= {mathbb 1} + sin |t| ~ hat K + (1-cos|t|) hat K ^2 .
      $$
      Note how different the coefficient of the identity looks, contrasted to your vector expression: it is not a mistake (read up on the point in WP.... well-wishers frequently "correct" it! Of course, this expression follows directly from the characteristic equation for K, not Rodrigues' original argument).



      So, tracing this expression, and the fact that K is traceless since antisymmetric, your (1) follows.



      Now antisymmetrize R,
      $$
      frac{R-R^T}{2}= sin|t| ~ hat K ,
      $$
      since the other two terms are symmetric matrices. Convert your "unit" antisymmetric matrix to your unit vector w and you have (2).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        This is a standard formula accompanying the Rodrigues rotation formula, obvious if only you use the right notation! The important thing is is to streamline your orthogonal matrix R. Its logarithm is the antisymmetric matrix K, defined by the linear cross product at the end of your expansion of R,
        $$
        Kequiv
        left[begin{array}{ccc}
        0 & -t_3 & t_2 \
        t_3 & 0 & -t_1 \
        -t_2 & t_1 & 0
        end{array}right] ~,
        $$
        so that $| t|^2= -frac{operatorname {Tr} (K^2)}{2}equiv | K|^2$; hence $K=|t| hat K$ with $| hat K|=1$ .



        Now the crucial thing is to re-express the Rodrigues formula in this notation,
        $$
        R=(R^T)^{-1}= e^K= {mathbb 1} + sin |t| ~ hat K + (1-cos|t|) hat K ^2 .
        $$
        Note how different the coefficient of the identity looks, contrasted to your vector expression: it is not a mistake (read up on the point in WP.... well-wishers frequently "correct" it! Of course, this expression follows directly from the characteristic equation for K, not Rodrigues' original argument).



        So, tracing this expression, and the fact that K is traceless since antisymmetric, your (1) follows.



        Now antisymmetrize R,
        $$
        frac{R-R^T}{2}= sin|t| ~ hat K ,
        $$
        since the other two terms are symmetric matrices. Convert your "unit" antisymmetric matrix to your unit vector w and you have (2).






        share|cite|improve this answer











        $endgroup$



        This is a standard formula accompanying the Rodrigues rotation formula, obvious if only you use the right notation! The important thing is is to streamline your orthogonal matrix R. Its logarithm is the antisymmetric matrix K, defined by the linear cross product at the end of your expansion of R,
        $$
        Kequiv
        left[begin{array}{ccc}
        0 & -t_3 & t_2 \
        t_3 & 0 & -t_1 \
        -t_2 & t_1 & 0
        end{array}right] ~,
        $$
        so that $| t|^2= -frac{operatorname {Tr} (K^2)}{2}equiv | K|^2$; hence $K=|t| hat K$ with $| hat K|=1$ .



        Now the crucial thing is to re-express the Rodrigues formula in this notation,
        $$
        R=(R^T)^{-1}= e^K= {mathbb 1} + sin |t| ~ hat K + (1-cos|t|) hat K ^2 .
        $$
        Note how different the coefficient of the identity looks, contrasted to your vector expression: it is not a mistake (read up on the point in WP.... well-wishers frequently "correct" it! Of course, this expression follows directly from the characteristic equation for K, not Rodrigues' original argument).



        So, tracing this expression, and the fact that K is traceless since antisymmetric, your (1) follows.



        Now antisymmetrize R,
        $$
        frac{R-R^T}{2}= sin|t| ~ hat K ,
        $$
        since the other two terms are symmetric matrices. Convert your "unit" antisymmetric matrix to your unit vector w and you have (2).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 19:36

























        answered Dec 11 '18 at 0:50









        Cosmas ZachosCosmas Zachos

        1,582520




        1,582520






























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