If $|y_m| to 0$ then there exists $k_m$ such that $k_m |y_m| to t$

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Lemma 3.43
Suppose $B_m$ are elements of $G$ and that $B_m to I$.
Let $Y_m = log B_m$, which is defined for all sufficiently large $m$.
Suppose that $Y_m$ is nonzero for all $m$ and that $Y_m/|Y_m| to Y in operatornameM_n(mathbbC)$.
Then $Y$ is in $mathfrakg$.



Proof.
For any $t in mathbbR$, we have $(t/|Y_m|)Y_m to t Y$.
Note that since $B_m to I$, we have $|Y_m| to 0$.
Thus, we can find integers $k_m$ such that $k_m |Y_m| to t$.
We have, then,
$$
e^k_m Y_m
= exp
left[
(k_m |Y_m|) fracY_mY_m
right]
to e^tY.
$$



(Original image here.)




The group $G$ is a Lie group of matrices. This was taken from Brian Hall. It’s reallly just analysis question. But I’m not entirely sure why the $k_m$’s exists




real-analysis




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    up vote
    2
    down vote

    favorite













    Lemma 3.43
    Suppose $B_m$ are elements of $G$ and that $B_m to I$.
    Let $Y_m = log B_m$, which is defined for all sufficiently large $m$.
    Suppose that $Y_m$ is nonzero for all $m$ and that $Y_m/|Y_m| to Y in operatornameM_n(mathbbC)$.
    Then $Y$ is in $mathfrakg$.



    Proof.
    For any $t in mathbbR$, we have $(t/|Y_m|)Y_m to t Y$.
    Note that since $B_m to I$, we have $|Y_m| to 0$.
    Thus, we can find integers $k_m$ such that $k_m |Y_m| to t$.
    We have, then,
    $$
    e^k_m Y_m
    = exp
    left[
    (k_m |Y_m|) fracY_mY_m
    right]
    to e^tY.
    $$



    (Original image here.)




    The group $G$ is a Lie group of matrices. This was taken from Brian Hall. It’s reallly just analysis question. But I’m not entirely sure why the $k_m$’s exists




    real-analysis




    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite












      Lemma 3.43
      Suppose $B_m$ are elements of $G$ and that $B_m to I$.
      Let $Y_m = log B_m$, which is defined for all sufficiently large $m$.
      Suppose that $Y_m$ is nonzero for all $m$ and that $Y_m/|Y_m| to Y in operatornameM_n(mathbbC)$.
      Then $Y$ is in $mathfrakg$.



      Proof.
      For any $t in mathbbR$, we have $(t/|Y_m|)Y_m to t Y$.
      Note that since $B_m to I$, we have $|Y_m| to 0$.
      Thus, we can find integers $k_m$ such that $k_m |Y_m| to t$.
      We have, then,
      $$
      e^k_m Y_m
      = exp
      left[
      (k_m |Y_m|) fracY_mY_m
      right]
      to e^tY.
      $$



      (Original image here.)




      The group $G$ is a Lie group of matrices. This was taken from Brian Hall. It’s reallly just analysis question. But I’m not entirely sure why the $k_m$’s exists




      real-analysis




      share|cite|improve this question















      Lemma 3.43
      Suppose $B_m$ are elements of $G$ and that $B_m to I$.
      Let $Y_m = log B_m$, which is defined for all sufficiently large $m$.
      Suppose that $Y_m$ is nonzero for all $m$ and that $Y_m/|Y_m| to Y in operatornameM_n(mathbbC)$.
      Then $Y$ is in $mathfrakg$.



      Proof.
      For any $t in mathbbR$, we have $(t/|Y_m|)Y_m to t Y$.
      Note that since $B_m to I$, we have $|Y_m| to 0$.
      Thus, we can find integers $k_m$ such that $k_m |Y_m| to t$.
      We have, then,
      $$
      e^k_m Y_m
      = exp
      left[
      (k_m |Y_m|) fracY_mY_m
      right]
      to e^tY.
      $$



      (Original image here.)




      The group $G$ is a Lie group of matrices. This was taken from Brian Hall. It’s reallly just analysis question. But I’m not entirely sure why the $k_m$’s exists





      real-analysis





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      edited Aug 23 at 10:37









      Jendrik Stelzner

      7,57221037




      7,57221037










      asked Aug 22 at 4:55









      Hawk

      5,27393699




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          You see, $Y_m neq 0$ for all large $m$ so $|Y_m| neq 0$. Just set $k_m = t/|Y_m|$.



          Then $k_m|Y_m| = t$.






          share|cite|improve this answer




















          • Ah okay thanks.
            – Hawk
            Aug 22 at 5:24










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          up vote
          5
          down vote



          accepted










          You see, $Y_m neq 0$ for all large $m$ so $|Y_m| neq 0$. Just set $k_m = t/|Y_m|$.



          Then $k_m|Y_m| = t$.






          share|cite|improve this answer




















          • Ah okay thanks.
            – Hawk
            Aug 22 at 5:24














          up vote
          5
          down vote



          accepted










          You see, $Y_m neq 0$ for all large $m$ so $|Y_m| neq 0$. Just set $k_m = t/|Y_m|$.



          Then $k_m|Y_m| = t$.






          share|cite|improve this answer




















          • Ah okay thanks.
            – Hawk
            Aug 22 at 5:24












          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          You see, $Y_m neq 0$ for all large $m$ so $|Y_m| neq 0$. Just set $k_m = t/|Y_m|$.



          Then $k_m|Y_m| = t$.






          share|cite|improve this answer












          You see, $Y_m neq 0$ for all large $m$ so $|Y_m| neq 0$. Just set $k_m = t/|Y_m|$.



          Then $k_m|Y_m| = t$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 22 at 5:13









          4-ier

          5989




          5989











          • Ah okay thanks.
            – Hawk
            Aug 22 at 5:24
















          • Ah okay thanks.
            – Hawk
            Aug 22 at 5:24















          Ah okay thanks.
          – Hawk
          Aug 22 at 5:24




          Ah okay thanks.
          – Hawk
          Aug 22 at 5:24

















           

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