Csdrhrt

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.

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.