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
edited Aug 23 at 10:37
Jendrik Stelzner
7,57221037
asked Aug 22 at 4:55
Hawk
5,27393699
add a comment |Â
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
edited Aug 23 at 10:37
Jendrik Stelzner
7,57221037
asked Aug 22 at 4:55
Hawk
5,27393699
add a comment |Â
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
edited Aug 23 at 10:37
Jendrik Stelzner
7,57221037
asked Aug 22 at 4:55
Hawk
5,27393699
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
edited Aug 23 at 10:37
Jendrik Stelzner
7,57221037
asked Aug 22 at 4:55
Hawk
5,27393699
edited Aug 23 at 10:37
Jendrik Stelzner
7,57221037
edited Aug 23 at 10:37
Jendrik Stelzner
7,57221037
edited Aug 23 at 10:37
Jendrik Stelzner
7,57221037
7,57221037
asked Aug 22 at 4:55
Hawk
5,27393699
asked Aug 22 at 4:55
Hawk
5,27393699
asked Aug 22 at 4:55
Hawk
5,27393699
5,27393699
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
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 $k_m|Y_m| = t$.
answered Aug 22 at 5:13
4-ier
5989
Ah okay thanks.
– Hawk
Aug 22 at 5:24
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $k_m|Y_m| = t$.
answered Aug 22 at 5:13
4-ier
5989
Ah okay thanks.
– Hawk
Aug 22 at 5:24
add a comment |Â
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 $k_m|Y_m| = t$.
answered Aug 22 at 5:13
4-ier
5989
Ah okay thanks.
– Hawk
Aug 22 at 5:24
add a comment |Â
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 $k_m|Y_m| = t$.
answered Aug 22 at 5:13
4-ier
5989
You see, $Y_m neq 0$ for all large $m$ so $|Y_m| neq 0$. Just set $k_m = t/|Y_m|$.
Then $k_m|Y_m| = t$.
answered Aug 22 at 5:13
4-ier
5989
answered Aug 22 at 5:13
4-ier
5989
answered Aug 22 at 5:13
4-ier
5989
answered Aug 22 at 5:13
4-ier
5989
5989
Ah okay thanks.
– Hawk
Aug 22 at 5:24
add a comment |Â
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
add a comment |Â
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