Is there a connected Lie group $G \subset GL_n(\mathbb{C})$ such that $G \cap B_1(I)$ is not connected? This came up while trying to solve this problem, which has been previously posted but not correctly answered:
Exercise 11, chapter 2 in Lie Groups, Lie Algebras, and Representations: An Elementary Introduction
The question is, given a matrix Lie group $G \subset GL_n(\mathbb{C})$ and an element $A \in G$ with $||A - I|| < 1$ (so that $\log A$ is defined by convergent power series), is $\log A \in \mathfrak{g}$ necessarily?
My thoughts: If $G \cap B_1(I)$ is connected, then we can take a path $\gamma: [0,1] \to G \cap B_1(I)$ from $I$ to $A$, and then $\log \gamma$ is a well defined and analytic path from $0$ to $\log A$ which lies in $\mathfrak{g}$ for sufficiently small $t$, consequently for all $t$.
But does such a path always exist? Is $G \cap B_1(I)$ always connected?
Edit: A comment pointed out an easy example (roots of unity) with  $G$ disconnected, but does there exist one with $G$ connected?
Thanks
 A: Both answers are negative. Consider a subgroup of $G$ of $U(2)$ (unitary $2\times 2$ matrices) consisting of matrices $\left (\begin{matrix}e^{it}&0\\0&e^{i a t}\end{matrix}\right)$. The Lie algebra $\mathfrak{g}$ is consists of matrices 
$\left (\begin{matrix}it &0\\0&i a t\end{matrix}\right)$.  If $a$ is irrational then for every neighborhood $V$ of $I$ the set $V\cap G$ will be disconnected. If $a$ is rational, there exists a neighborhood $V$ of $I$, and $V'$ of $O$ in $gl(n,\mathbb{C})$ so that $\exp$ maps $g\cap V'$ diffeomorphically to $V\cap G$ ( this is true for any closed subgroup of a Lie group). However, $B_1$ might be too large, in fact, if we fix $V$, then for some $a$ rational ( with large denominator) the intesection $G\cap V$ will be disconnected.
There is a reason that $\mathfrak{g}$ is given as $\{X\in gl(n, \mathbb{C})\ | \exp t X \in G \text{ for all } t \in \mathbb{R}\}$ (talking about real Lie groups in general). 
To better understand $G_a\subset U(1)\times U(1)\subset U(2)$, it's better to look at $U(1)\times U(1)$ as $\mathbb{R}^2/\mathbb{Z}^2$ and $G_a$ the image of a line through origin ( the line is $\mathfrak{g}$ ) inside 
$\mathbb{R}^2/\mathbb{Z}^2$.
ADDED: The  classical groups $G$ behave better. For instance, let $G=O(n,\mathbb{R})$. Let $X\in gl(n,\mathbb{R})$ such that $\exp X \in G$. When does this imply $X\in o(n,\mathbb{R})$, that is, when can we guarantee that $X$ is skew symmetric? 
Let's consider a similar problem. Assume that $\exp Y v= v$, for some linear map $Y$. 
Does this imply $Yv=0$? Yes, if $Y$ is such that $\frac{\exp Y-1}{Y}$ is invertible. 
Indeed, we have $(e^Y-1)v= (e^Y-1)/Y \cdot Y v$. 
Now, the roots of the function $z\mapsto (e^z-1)/z$ are $2k\pi i$, with $k\ne 0$. Therefore, 
the matrix $\frac{\exp Y-1}{Y}$ is invertible if and only if $Y$ does not have any eigenvalue
of form $2k\pi i$, $k\ne 0$. 
Consider now the representation of $GL(n,\mathbb{R})$ on  $\mathbb{R}^n\otimes \mathbb{R}^n$. As an action on matrices
$g$ takes a matrix $A$ to $g A g^t$. Now $O(n,\mathbb{R})$ is the fixator of the matrix $I$. Therefore,
if $\exp X$ fixes $I$ ( that is $\exp X \cdot \exp X ^t= I$), what can we say about $X$?
Note that the action of $gl(n,\mathbb{R})$ on matrices is 
$X\cdot A = XA + A X^t= L(X) (A)$.
 If $\lambda_i$'s are the eigenvalues of $X$, then $\lambda_i + \lambda_j$ are the eigenvalues of $L(X)$.
Therefore, if we assume that $|\lambda_i|<\pi$ for all $i$, then $L(X)$ does not have any eigenvalue of 
form $2k\pi i$. Conclusion: if $\exp (X)$ is orthogonal, $\rho(X)<\pi$, then $X$ skew symmetric.
This can be done for other classical groups. Note that $\det \exp X=1$ implies $Tr(X)=0$ in the real case, while
the complex case needs also the condition $|Tr(X)|<2\pi$. 
