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Theorem 2.14 (One-Parameter Subgroups) If $A(\cdot)$ is a one-parameter subgroup of $\text{GL}(n;\mathbb{C})$, there exists a unique $n\times n$ complex matrix $X$ such that $A(t)=\mathrm{e}^{tX}$

In the proof concerning existence, it is mentioned that $U$ is a open set in $\text{GL}(n,\mathbb{C})$ ($U:=\exp(B_{\epsilon/2}$) with $B_{\epsilon/2}$ the ball of radius $\epsilon/2$ around the origin in $M_{n}(\mathbb{C}),\, \epsilon/2<\log(2))$. The continuity of $A$ guarantees that there exists $t_0>0$ such that $A(t)\in U$ for all $t$ with $|t|\leq t_0$.

Question 1. Why $U$ is open?

My attempt. It is necessary to prove that for all $\mathrm{e}^{B}\in U$, exists $r>0$ such that $B(\mathrm{e}^{B};r)\subset U$. Let $Y\in B(\mathrm{e}^{B};r)$ then $\left\|Y-\mathrm{e}^B\right\|<r$, so $\left\|Y\right\|\leq \left\|\mathrm{e}^{B}\right\|+r\leq \mathrm{e}^{\left\|B\right\|}+r\leq \mathrm{e}^{\epsilon/2}+r\leq \mathrm{e}^{\log(2)}+r=2+r$

Question 2. From what I understand, if there exists $t_0>0$ then by continuity, (being close to $t_0$) we have that $A(t)\in U$ for all $|t|\leq t_0$. But, Why there exists $t_0>0$ such that $A(t_0)\in U$?

eraldcoil
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    If you view $M = GL(n; \mathbb{C})$ as a Lie group and $M_n(\mathbb{C})=T_pM$ as its Lie algebra (which I believe you're doing), then the first thing that comes to mind for me is that the exponential map is a diffeomorphism of some small neighborhood of $0 \in T_p M$ into $M$. From Prop 2.9 in Do Carmo: Given $p \in M$, there exists $\epsilon>0$ such that $exp_p: B_\epsilon(0) \subset T_pM \rightarrow M$ is a diffeomorphism of $B_\epsilon(0)$ onto an open subset of $M$. – locally trivial Apr 27 '22 at 17:35
  • Do Carmo's Riemannian Geometry? – eraldcoil Apr 27 '22 at 22:53
  • Yes, do Carmo's Riemannian Geometry. – locally trivial Apr 27 '22 at 22:55

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