Mixing
Is there any complex number $a$ solving $exp (X)=1+aX$ for given square matrix $X$?
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I am doing exercises in Rossmann's book on Lie groups. Exercise 1.2.12 goes like this:
$X in M_n (mathbb{C})$, $L$ is a subspace of $M_n (mathbb{C})$, s.t. $[X, Y]in L$ for $Yin L$. Prove that $exp (-X)exp (X+Y)in 1+L $.
There is a hint that we consider derivative of $exp (-X)exp (X+tY)$. That is
$$frac{I-e^{-mathrm{ad}X}}{mathrm{ad}X}Y,$$
which 'seems' to sit in $L$.
However, if you let $L=mathbb{C} X$, one claims that it contradicts for $exp (X)$ may not be expressed as $1+aX$.
My question is whether the claim is true or not. Any help is appreciated.
lie-groups lie-algebras matrix-exponential
asked Oct 28 at 7:45
Kirby Lee
1919
add a comment |
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2
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I am doing exercises in Rossmann's book on Lie groups. Exercise 1.2.12 goes like this:
$X in M_n (mathbb{C})$, $L$ is a subspace of $M_n (mathbb{C})$, s.t. $[X, Y]in L$ for $Yin L$. Prove that $exp (-X)exp (X+Y)in 1+L $.
There is a hint that we consider derivative of $exp (-X)exp (X+tY)$. That is
$$frac{I-e^{-mathrm{ad}X}}{mathrm{ad}X}Y,$$
which 'seems' to sit in $L$.
However, if you let $L=mathbb{C} X$, one claims that it contradicts for $exp (X)$ may not be expressed as $1+aX$.
My question is whether the claim is true or not. Any help is appreciated.
lie-groups lie-algebras matrix-exponential
asked Oct 28 at 7:45
Kirby Lee
1919
2
The $ 2times 2$ diagonal matrix with diagonal entries $1$ and $2$ is a counterexample to the statement in the title.
– Kavi Rama Murthy
Oct 28 at 12:19
@KaviRamaMurthy Thank you so much. It is brief and clear!
– Kirby Lee
Oct 28 at 23:09
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am doing exercises in Rossmann's book on Lie groups. Exercise 1.2.12 goes like this:
$X in M_n (mathbb{C})$, $L$ is a subspace of $M_n (mathbb{C})$, s.t. $[X, Y]in L$ for $Yin L$. Prove that $exp (-X)exp (X+Y)in 1+L $.
There is a hint that we consider derivative of $exp (-X)exp (X+tY)$. That is
$$frac{I-e^{-mathrm{ad}X}}{mathrm{ad}X}Y,$$
which 'seems' to sit in $L$.
However, if you let $L=mathbb{C} X$, one claims that it contradicts for $exp (X)$ may not be expressed as $1+aX$.
My question is whether the claim is true or not. Any help is appreciated.
lie-groups lie-algebras matrix-exponential
asked Oct 28 at 7:45
Kirby Lee
1919
I am doing exercises in Rossmann's book on Lie groups. Exercise 1.2.12 goes like this:
$X in M_n (mathbb{C})$, $L$ is a subspace of $M_n (mathbb{C})$, s.t. $[X, Y]in L$ for $Yin L$. Prove that $exp (-X)exp (X+Y)in 1+L $.
There is a hint that we consider derivative of $exp (-X)exp (X+tY)$. That is
$$frac{I-e^{-mathrm{ad}X}}{mathrm{ad}X}Y,$$
which 'seems' to sit in $L$.
However, if you let $L=mathbb{C} X$, one claims that it contradicts for $exp (X)$ may not be expressed as $1+aX$.
My question is whether the claim is true or not. Any help is appreciated.
lie-groups lie-algebras matrix-exponential
lie-groups lie-algebras matrix-exponential
asked Oct 28 at 7:45
Kirby Lee
1919
asked Oct 28 at 7:45
Kirby Lee
1919
asked Oct 28 at 7:45
Kirby Lee
1919
asked Oct 28 at 7:45
Kirby Lee
1919
asked Oct 28 at 7:45
Kirby Lee
1919
1919
2
The $ 2times 2$ diagonal matrix with diagonal entries $1$ and $2$ is a counterexample to the statement in the title.
– Kavi Rama Murthy
Oct 28 at 12:19
@KaviRamaMurthy Thank you so much. It is brief and clear!
– Kirby Lee
Oct 28 at 23:09
add a comment |
2
The $ 2times 2$ diagonal matrix with diagonal entries $1$ and $2$ is a counterexample to the statement in the title.
– Kavi Rama Murthy
Oct 28 at 12:19
@KaviRamaMurthy Thank you so much. It is brief and clear!
– Kirby Lee
Oct 28 at 23:09
2
2
The $ 2times 2$ diagonal matrix with diagonal entries $1$ and $2$ is a counterexample to the statement in the title.
– Kavi Rama Murthy
Oct 28 at 12:19
The $ 2times 2$ diagonal matrix with diagonal entries $1$ and $2$ is a counterexample to the statement in the title.
– Kavi Rama Murthy
Oct 28 at 12:19
@KaviRamaMurthy Thank you so much. It is brief and clear!
– Kirby Lee
Oct 28 at 23:09
@KaviRamaMurthy Thank you so much. It is brief and clear!
– Kirby Lee
Oct 28 at 23:09
add a comment |
1 Answer
1
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1
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I would like to note that you incorrectly used Theorem 5 in the book. I restate it here (with minor changes to make things clearer and less confusing).
Theorem 5 [section 1.2, page 15, in Lie Groups: An Introduction Through Linear Groups by Wulf Rossman]. Let $Z:Bbb{R}to M_n(Bbb{C})$ be a differentiable matrix function. Then,
$$frac{d}{dtau}e^{Z(tau)}=e^{Z(tau)}left(frac{1-e^{-operatorname{ad}Z(tau)}}{operatorname{ad}Z(tau)}right)frac{d}{dtau}Z(tau).$$
Here, $$left(frac{1-e^{-operatorname{ad}Z(tau)}}{operatorname{ad}Z(tau)}right)=sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}Z(tau)big)^k.$$
Let's look at your counter-example. I will illustrate here how to correctly apply the theorem above. For $L=Bbb{C}X$, we set, WLOG, $Y=X$. So,
begin{align}frac{d}{dt}e^{-X}e^{X+tY}&=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y\&=e^{-X}e^{(1+t)X}left(sum_{k=0}^inftyfrac{(-1)^k(1+t)^k}{(k+1)!}(operatorname{ad}X)^kright)X
\&=e^{-X}e^{(1+t)X}X=Xe^{tX}.end{align}
That is, for $L=Bbb C X$ and $Y=X$,
$$e^{-X}e^{X+tY}=e^{tX}.$$
In particular, $e^{-X}e^{X+Y}=e^X=e^Y$ when $Y=X$. This is not in the form $I+aX$ in general just as Kavi Rama Murthy's comment shows, unless we have a stronger condition on $L$ such as $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$.
In the section below, I verify that the assumption that $L$ is a subspace of $M_n(mathbb{C})$ with the properties that $[X,T]in L$ for all $Tin L$ and that $L$ is closed under left multiplication by $X$ and elements of $L$ itself, then the claim holds. Without this extra assumption, we already see that the problem is wrong.
Modified Problem Statement. Fix $Xin M_n(mathbb{C})$. Let $L$ be a subspace of $M_n(mathbb{C})$ with the properties that $[X,T]in L$ for all $Tin L$, and that $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$, then $$e^{-X}e^{X+Y}in I+L$$ for all $Yin L$.
From the theorem, you should take $Z(t)=X+tY$. That is,
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}frac{d}{dt}e^{Z(t)}=e^{-X}Biggl(e^{Z(t)}left(frac{1-e^{-operatorname{ad}Z(t)}}{operatorname{ad}Z(t)}right)frac{d}{dt}Z(t)Biggr).$$
Since $frac{d}{dt}Z(t)=Y$, we get
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y.$$
Because $operatorname{ad}X(Y)=[X,Y]in L$ and $operatorname{ad}Y(Y)=[Y,Y]=0in L$, we obtain $$big(operatorname{ad}X+toperatorname{ad}Ybig)^kYin L.$$ This proves that
$$left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y in L,.$$ Now, because $e^{X+tY}Tin L$ for all $Tin L$ because $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$, we see also that $e^{-X}e^{X+tY}Tin L$ as well. This means
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Yin L.$$
This proves that
$$e^{-X}e^{X+tau Y}-I=int_0^taufrac{d}{dt}e^{-X}e^{X+tY} dt in L,$$
and the conclusion follows.
answered Oct 28 at 12:50
Zvi
3,205222
Thank you for your detailed calculation. But I am most doubtful whether we can claim that a formal series is in L with each term in L. In arbitrary vector space it might not allowed to sum up infinite vectors. And what's your opinion to Murthy's comment above? I thought it is solid.
– Kirby Lee
Oct 28 at 23:16
Sorry I have checked it again. Say in your notation $Z'(t)=Z(t)v(t)$. You proved $vin L$. How to show that $Zv$ is still in $L$?
– Kirby Lee
2 days ago
I don't have any $v$ in my notation. What is $v$? Is it $Y$? But then $Y$ does not depend on $t$.
– Zvi
yesterday
My fault. I mean $frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y$. The infinite sum is in $L$. There is an $e^{X+tY}$ in the formula. For $t=0$ it is fine. How about general $t$?
– Kirby Lee
22 hours ago
1
Thanks for this. I am okay with the fact that the sum is in $L$. But my question is that, just on the left of the sum, there is an $e^{-X}e^{X+tY}$, which is not identity when $tne 0$. How to show that it keeps the sum in $L$?
– Kirby Lee
5 hours ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I would like to note that you incorrectly used Theorem 5 in the book. I restate it here (with minor changes to make things clearer and less confusing).
Theorem 5 [section 1.2, page 15, in Lie Groups: An Introduction Through Linear Groups by Wulf Rossman]. Let $Z:Bbb{R}to M_n(Bbb{C})$ be a differentiable matrix function. Then,
$$frac{d}{dtau}e^{Z(tau)}=e^{Z(tau)}left(frac{1-e^{-operatorname{ad}Z(tau)}}{operatorname{ad}Z(tau)}right)frac{d}{dtau}Z(tau).$$
Here, $$left(frac{1-e^{-operatorname{ad}Z(tau)}}{operatorname{ad}Z(tau)}right)=sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}Z(tau)big)^k.$$
Let's look at your counter-example. I will illustrate here how to correctly apply the theorem above. For $L=Bbb{C}X$, we set, WLOG, $Y=X$. So,
begin{align}frac{d}{dt}e^{-X}e^{X+tY}&=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y\&=e^{-X}e^{(1+t)X}left(sum_{k=0}^inftyfrac{(-1)^k(1+t)^k}{(k+1)!}(operatorname{ad}X)^kright)X
\&=e^{-X}e^{(1+t)X}X=Xe^{tX}.end{align}
That is, for $L=Bbb C X$ and $Y=X$,
$$e^{-X}e^{X+tY}=e^{tX}.$$
In particular, $e^{-X}e^{X+Y}=e^X=e^Y$ when $Y=X$. This is not in the form $I+aX$ in general just as Kavi Rama Murthy's comment shows, unless we have a stronger condition on $L$ such as $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$.
In the section below, I verify that the assumption that $L$ is a subspace of $M_n(mathbb{C})$ with the properties that $[X,T]in L$ for all $Tin L$ and that $L$ is closed under left multiplication by $X$ and elements of $L$ itself, then the claim holds. Without this extra assumption, we already see that the problem is wrong.
Modified Problem Statement. Fix $Xin M_n(mathbb{C})$. Let $L$ be a subspace of $M_n(mathbb{C})$ with the properties that $[X,T]in L$ for all $Tin L$, and that $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$, then $$e^{-X}e^{X+Y}in I+L$$ for all $Yin L$.
From the theorem, you should take $Z(t)=X+tY$. That is,
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}frac{d}{dt}e^{Z(t)}=e^{-X}Biggl(e^{Z(t)}left(frac{1-e^{-operatorname{ad}Z(t)}}{operatorname{ad}Z(t)}right)frac{d}{dt}Z(t)Biggr).$$
Since $frac{d}{dt}Z(t)=Y$, we get
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y.$$
Because $operatorname{ad}X(Y)=[X,Y]in L$ and $operatorname{ad}Y(Y)=[Y,Y]=0in L$, we obtain $$big(operatorname{ad}X+toperatorname{ad}Ybig)^kYin L.$$ This proves that
$$left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y in L,.$$ Now, because $e^{X+tY}Tin L$ for all $Tin L$ because $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$, we see also that $e^{-X}e^{X+tY}Tin L$ as well. This means
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Yin L.$$
This proves that
$$e^{-X}e^{X+tau Y}-I=int_0^taufrac{d}{dt}e^{-X}e^{X+tY} dt in L,$$
and the conclusion follows.
answered Oct 28 at 12:50
Zvi
3,205222
Thank you for your detailed calculation. But I am most doubtful whether we can claim that a formal series is in L with each term in L. In arbitrary vector space it might not allowed to sum up infinite vectors. And what's your opinion to Murthy's comment above? I thought it is solid.
– Kirby Lee
Oct 28 at 23:16
Sorry I have checked it again. Say in your notation $Z'(t)=Z(t)v(t)$. You proved $vin L$. How to show that $Zv$ is still in $L$?
– Kirby Lee
2 days ago
I don't have any $v$ in my notation. What is $v$? Is it $Y$? But then $Y$ does not depend on $t$.
– Zvi
yesterday
My fault. I mean $frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y$. The infinite sum is in $L$. There is an $e^{X+tY}$ in the formula. For $t=0$ it is fine. How about general $t$?
– Kirby Lee
22 hours ago
1
Thanks for this. I am okay with the fact that the sum is in $L$. But my question is that, just on the left of the sum, there is an $e^{-X}e^{X+tY}$, which is not identity when $tne 0$. How to show that it keeps the sum in $L$?
– Kirby Lee
5 hours ago
|
show 2 more comments
up vote
1
down vote
I would like to note that you incorrectly used Theorem 5 in the book. I restate it here (with minor changes to make things clearer and less confusing).
Theorem 5 [section 1.2, page 15, in Lie Groups: An Introduction Through Linear Groups by Wulf Rossman]. Let $Z:Bbb{R}to M_n(Bbb{C})$ be a differentiable matrix function. Then,
$$frac{d}{dtau}e^{Z(tau)}=e^{Z(tau)}left(frac{1-e^{-operatorname{ad}Z(tau)}}{operatorname{ad}Z(tau)}right)frac{d}{dtau}Z(tau).$$
Here, $$left(frac{1-e^{-operatorname{ad}Z(tau)}}{operatorname{ad}Z(tau)}right)=sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}Z(tau)big)^k.$$
Let's look at your counter-example. I will illustrate here how to correctly apply the theorem above. For $L=Bbb{C}X$, we set, WLOG, $Y=X$. So,
begin{align}frac{d}{dt}e^{-X}e^{X+tY}&=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y\&=e^{-X}e^{(1+t)X}left(sum_{k=0}^inftyfrac{(-1)^k(1+t)^k}{(k+1)!}(operatorname{ad}X)^kright)X
\&=e^{-X}e^{(1+t)X}X=Xe^{tX}.end{align}
That is, for $L=Bbb C X$ and $Y=X$,
$$e^{-X}e^{X+tY}=e^{tX}.$$
In particular, $e^{-X}e^{X+Y}=e^X=e^Y$ when $Y=X$. This is not in the form $I+aX$ in general just as Kavi Rama Murthy's comment shows, unless we have a stronger condition on $L$ such as $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$.
In the section below, I verify that the assumption that $L$ is a subspace of $M_n(mathbb{C})$ with the properties that $[X,T]in L$ for all $Tin L$ and that $L$ is closed under left multiplication by $X$ and elements of $L$ itself, then the claim holds. Without this extra assumption, we already see that the problem is wrong.
Modified Problem Statement. Fix $Xin M_n(mathbb{C})$. Let $L$ be a subspace of $M_n(mathbb{C})$ with the properties that $[X,T]in L$ for all $Tin L$, and that $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$, then $$e^{-X}e^{X+Y}in I+L$$ for all $Yin L$.
From the theorem, you should take $Z(t)=X+tY$. That is,
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}frac{d}{dt}e^{Z(t)}=e^{-X}Biggl(e^{Z(t)}left(frac{1-e^{-operatorname{ad}Z(t)}}{operatorname{ad}Z(t)}right)frac{d}{dt}Z(t)Biggr).$$
Since $frac{d}{dt}Z(t)=Y$, we get
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y.$$
Because $operatorname{ad}X(Y)=[X,Y]in L$ and $operatorname{ad}Y(Y)=[Y,Y]=0in L$, we obtain $$big(operatorname{ad}X+toperatorname{ad}Ybig)^kYin L.$$ This proves that
$$left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y in L,.$$ Now, because $e^{X+tY}Tin L$ for all $Tin L$ because $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$, we see also that $e^{-X}e^{X+tY}Tin L$ as well. This means
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Yin L.$$
This proves that
$$e^{-X}e^{X+tau Y}-I=int_0^taufrac{d}{dt}e^{-X}e^{X+tY} dt in L,$$
and the conclusion follows.
answered Oct 28 at 12:50
Zvi
3,205222
Thank you for your detailed calculation. But I am most doubtful whether we can claim that a formal series is in L with each term in L. In arbitrary vector space it might not allowed to sum up infinite vectors. And what's your opinion to Murthy's comment above? I thought it is solid.
– Kirby Lee
Oct 28 at 23:16
Sorry I have checked it again. Say in your notation $Z'(t)=Z(t)v(t)$. You proved $vin L$. How to show that $Zv$ is still in $L$?
– Kirby Lee
2 days ago
I don't have any $v$ in my notation. What is $v$? Is it $Y$? But then $Y$ does not depend on $t$.
– Zvi
yesterday
My fault. I mean $frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y$. The infinite sum is in $L$. There is an $e^{X+tY}$ in the formula. For $t=0$ it is fine. How about general $t$?
– Kirby Lee
22 hours ago
1
Thanks for this. I am okay with the fact that the sum is in $L$. But my question is that, just on the left of the sum, there is an $e^{-X}e^{X+tY}$, which is not identity when $tne 0$. How to show that it keeps the sum in $L$?
– Kirby Lee
5 hours ago
|
show 2 more comments
up vote
1
down vote
up vote
1
down vote
I would like to note that you incorrectly used Theorem 5 in the book. I restate it here (with minor changes to make things clearer and less confusing).
Theorem 5 [section 1.2, page 15, in Lie Groups: An Introduction Through Linear Groups by Wulf Rossman]. Let $Z:Bbb{R}to M_n(Bbb{C})$ be a differentiable matrix function. Then,
$$frac{d}{dtau}e^{Z(tau)}=e^{Z(tau)}left(frac{1-e^{-operatorname{ad}Z(tau)}}{operatorname{ad}Z(tau)}right)frac{d}{dtau}Z(tau).$$
Here, $$left(frac{1-e^{-operatorname{ad}Z(tau)}}{operatorname{ad}Z(tau)}right)=sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}Z(tau)big)^k.$$
Let's look at your counter-example. I will illustrate here how to correctly apply the theorem above. For $L=Bbb{C}X$, we set, WLOG, $Y=X$. So,
begin{align}frac{d}{dt}e^{-X}e^{X+tY}&=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y\&=e^{-X}e^{(1+t)X}left(sum_{k=0}^inftyfrac{(-1)^k(1+t)^k}{(k+1)!}(operatorname{ad}X)^kright)X
\&=e^{-X}e^{(1+t)X}X=Xe^{tX}.end{align}
That is, for $L=Bbb C X$ and $Y=X$,
$$e^{-X}e^{X+tY}=e^{tX}.$$
In particular, $e^{-X}e^{X+Y}=e^X=e^Y$ when $Y=X$. This is not in the form $I+aX$ in general just as Kavi Rama Murthy's comment shows, unless we have a stronger condition on $L$ such as $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$.
In the section below, I verify that the assumption that $L$ is a subspace of $M_n(mathbb{C})$ with the properties that $[X,T]in L$ for all $Tin L$ and that $L$ is closed under left multiplication by $X$ and elements of $L$ itself, then the claim holds. Without this extra assumption, we already see that the problem is wrong.
Modified Problem Statement. Fix $Xin M_n(mathbb{C})$. Let $L$ be a subspace of $M_n(mathbb{C})$ with the properties that $[X,T]in L$ for all $Tin L$, and that $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$, then $$e^{-X}e^{X+Y}in I+L$$ for all $Yin L$.
From the theorem, you should take $Z(t)=X+tY$. That is,
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}frac{d}{dt}e^{Z(t)}=e^{-X}Biggl(e^{Z(t)}left(frac{1-e^{-operatorname{ad}Z(t)}}{operatorname{ad}Z(t)}right)frac{d}{dt}Z(t)Biggr).$$
Since $frac{d}{dt}Z(t)=Y$, we get
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y.$$
Because $operatorname{ad}X(Y)=[X,Y]in L$ and $operatorname{ad}Y(Y)=[Y,Y]=0in L$, we obtain $$big(operatorname{ad}X+toperatorname{ad}Ybig)^kYin L.$$ This proves that
$$left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y in L,.$$ Now, because $e^{X+tY}Tin L$ for all $Tin L$ because $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$, we see also that $e^{-X}e^{X+tY}Tin L$ as well. This means
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Yin L.$$
This proves that
$$e^{-X}e^{X+tau Y}-I=int_0^taufrac{d}{dt}e^{-X}e^{X+tY} dt in L,$$
and the conclusion follows.
answered Oct 28 at 12:50
Zvi
3,205222
I would like to note that you incorrectly used Theorem 5 in the book. I restate it here (with minor changes to make things clearer and less confusing).
Theorem 5 [section 1.2, page 15, in Lie Groups: An Introduction Through Linear Groups by Wulf Rossman]. Let $Z:Bbb{R}to M_n(Bbb{C})$ be a differentiable matrix function. Then,
$$frac{d}{dtau}e^{Z(tau)}=e^{Z(tau)}left(frac{1-e^{-operatorname{ad}Z(tau)}}{operatorname{ad}Z(tau)}right)frac{d}{dtau}Z(tau).$$
Here, $$left(frac{1-e^{-operatorname{ad}Z(tau)}}{operatorname{ad}Z(tau)}right)=sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}Z(tau)big)^k.$$
Let's look at your counter-example. I will illustrate here how to correctly apply the theorem above. For $L=Bbb{C}X$, we set, WLOG, $Y=X$. So,
begin{align}frac{d}{dt}e^{-X}e^{X+tY}&=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y\&=e^{-X}e^{(1+t)X}left(sum_{k=0}^inftyfrac{(-1)^k(1+t)^k}{(k+1)!}(operatorname{ad}X)^kright)X
\&=e^{-X}e^{(1+t)X}X=Xe^{tX}.end{align}
That is, for $L=Bbb C X$ and $Y=X$,
$$e^{-X}e^{X+tY}=e^{tX}.$$
In particular, $e^{-X}e^{X+Y}=e^X=e^Y$ when $Y=X$. This is not in the form $I+aX$ in general just as Kavi Rama Murthy's comment shows, unless we have a stronger condition on $L$ such as $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$.
In the section below, I verify that the assumption that $L$ is a subspace of $M_n(mathbb{C})$ with the properties that $[X,T]in L$ for all $Tin L$ and that $L$ is closed under left multiplication by $X$ and elements of $L$ itself, then the claim holds. Without this extra assumption, we already see that the problem is wrong.
Modified Problem Statement. Fix $Xin M_n(mathbb{C})$. Let $L$ be a subspace of $M_n(mathbb{C})$ with the properties that $[X,T]in L$ for all $Tin L$, and that $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$, then $$e^{-X}e^{X+Y}in I+L$$ for all $Yin L$.
From the theorem, you should take $Z(t)=X+tY$. That is,
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}frac{d}{dt}e^{Z(t)}=e^{-X}Biggl(e^{Z(t)}left(frac{1-e^{-operatorname{ad}Z(t)}}{operatorname{ad}Z(t)}right)frac{d}{dt}Z(t)Biggr).$$
Since $frac{d}{dt}Z(t)=Y$, we get
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y.$$
Because $operatorname{ad}X(Y)=[X,Y]in L$ and $operatorname{ad}Y(Y)=[Y,Y]=0in L$, we obtain $$big(operatorname{ad}X+toperatorname{ad}Ybig)^kYin L.$$ This proves that
$$left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y in L,.$$ Now, because $e^{X+tY}Tin L$ for all $Tin L$ because $L$ is closed under left multiplication by any element of $tilde{L}=L+mathbb{C}X$, we see also that $e^{-X}e^{X+tY}Tin L$ as well. This means
$$frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Yin L.$$
This proves that
$$e^{-X}e^{X+tau Y}-I=int_0^taufrac{d}{dt}e^{-X}e^{X+tY} dt in L,$$
and the conclusion follows.
answered Oct 28 at 12:50
Zvi
3,205222
edited 3 hours ago
answered Oct 28 at 12:50
Zvi
3,205222
answered Oct 28 at 12:50
Zvi
3,205222
answered Oct 28 at 12:50
Zvi
3,205222
3,205222
Thank you for your detailed calculation. But I am most doubtful whether we can claim that a formal series is in L with each term in L. In arbitrary vector space it might not allowed to sum up infinite vectors. And what's your opinion to Murthy's comment above? I thought it is solid.
– Kirby Lee
Oct 28 at 23:16
Sorry I have checked it again. Say in your notation $Z'(t)=Z(t)v(t)$. You proved $vin L$. How to show that $Zv$ is still in $L$?
– Kirby Lee
2 days ago
I don't have any $v$ in my notation. What is $v$? Is it $Y$? But then $Y$ does not depend on $t$.
– Zvi
yesterday
My fault. I mean $frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y$. The infinite sum is in $L$. There is an $e^{X+tY}$ in the formula. For $t=0$ it is fine. How about general $t$?
– Kirby Lee
22 hours ago
1
Thanks for this. I am okay with the fact that the sum is in $L$. But my question is that, just on the left of the sum, there is an $e^{-X}e^{X+tY}$, which is not identity when $tne 0$. How to show that it keeps the sum in $L$?
– Kirby Lee
5 hours ago
|
show 2 more comments
Thank you for your detailed calculation. But I am most doubtful whether we can claim that a formal series is in L with each term in L. In arbitrary vector space it might not allowed to sum up infinite vectors. And what's your opinion to Murthy's comment above? I thought it is solid.
– Kirby Lee
Oct 28 at 23:16
Sorry I have checked it again. Say in your notation $Z'(t)=Z(t)v(t)$. You proved $vin L$. How to show that $Zv$ is still in $L$?
– Kirby Lee
2 days ago
I don't have any $v$ in my notation. What is $v$? Is it $Y$? But then $Y$ does not depend on $t$.
– Zvi
yesterday
My fault. I mean $frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y$. The infinite sum is in $L$. There is an $e^{X+tY}$ in the formula. For $t=0$ it is fine. How about general $t$?
– Kirby Lee
22 hours ago
1
Thanks for this. I am okay with the fact that the sum is in $L$. But my question is that, just on the left of the sum, there is an $e^{-X}e^{X+tY}$, which is not identity when $tne 0$. How to show that it keeps the sum in $L$?
– Kirby Lee
5 hours ago
Thank you for your detailed calculation. But I am most doubtful whether we can claim that a formal series is in L with each term in L. In arbitrary vector space it might not allowed to sum up infinite vectors. And what's your opinion to Murthy's comment above? I thought it is solid.
– Kirby Lee
Oct 28 at 23:16
Thank you for your detailed calculation. But I am most doubtful whether we can claim that a formal series is in L with each term in L. In arbitrary vector space it might not allowed to sum up infinite vectors. And what's your opinion to Murthy's comment above? I thought it is solid.
– Kirby Lee
Oct 28 at 23:16
Sorry I have checked it again. Say in your notation $Z'(t)=Z(t)v(t)$. You proved $vin L$. How to show that $Zv$ is still in $L$?
– Kirby Lee
2 days ago
Sorry I have checked it again. Say in your notation $Z'(t)=Z(t)v(t)$. You proved $vin L$. How to show that $Zv$ is still in $L$?
– Kirby Lee
2 days ago
I don't have any $v$ in my notation. What is $v$? Is it $Y$? But then $Y$ does not depend on $t$.
– Zvi
yesterday
I don't have any $v$ in my notation. What is $v$? Is it $Y$? But then $Y$ does not depend on $t$.
– Zvi
yesterday
My fault. I mean $frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y$. The infinite sum is in $L$. There is an $e^{X+tY}$ in the formula. For $t=0$ it is fine. How about general $t$?
– Kirby Lee
22 hours ago
My fault. I mean $frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}left(sum_{k=0}^inftyfrac{(-1)^k}{(k+1)!}big(operatorname{ad}X+toperatorname{ad}Ybig)^kright)Y$. The infinite sum is in $L$. There is an $e^{X+tY}$ in the formula. For $t=0$ it is fine. How about general $t$?
– Kirby Lee
22 hours ago
1
1
Thanks for this. I am okay with the fact that the sum is in $L$. But my question is that, just on the left of the sum, there is an $e^{-X}e^{X+tY}$, which is not identity when $tne 0$. How to show that it keeps the sum in $L$?
– Kirby Lee
5 hours ago
Thanks for this. I am okay with the fact that the sum is in $L$. But my question is that, just on the left of the sum, there is an $e^{-X}e^{X+tY}$, which is not identity when $tne 0$. How to show that it keeps the sum in $L$?
– Kirby Lee
5 hours ago
|
show 2 more comments
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The $ 2times 2$ diagonal matrix with diagonal entries $1$ and $2$ is a counterexample to the statement in the title.
– Kavi Rama Murthy
Oct 28 at 12:19
@KaviRamaMurthy Thank you so much. It is brief and clear!
– Kirby Lee
Oct 28 at 23:09