I have been reading a paper lately. It said that, "consider the subset $S=\{R\in SO(3):2<tr[R]\le3\}$. Since $−1 ≤ tr[R] ≤ 3$ on $SO(3)$, we express the set $S=\{R\in SO(3):2<tr[R]\le3\}$= $SO(3) \cap tr^{-1}\{(2,4)\}$ where $tr^{−1}(B)$ denotes the preimage of $B$ under the trace map. Now as $tr^{-1}\{(2,4)\}$ is an open set so $\{R\in SO(3):2<tr[R]\le3\}$ is a submanifold of $\mathbb{R}^{3\times 3}$."
However, I cannot understand why $\{R\in SO(3):2<tr[R]\le3\}$ is a submanifold of $\mathbb{R}^{3\times 3}$. Thanks a lot for your help.
For each $S\in SO(3)$, $\operatorname{tr}S\leqslant3$, and therefore given an element $S\in SO(3)$, asserting that $2<\operatorname{tr}S\leqslant3$ is the same thing as asserting that $2<\operatorname{tr}S<4$. On the other hand, $SO(3)\cap\operatorname{tr}^{-1}\bigl((2,4)\bigr)$ is an open subset of $SO(3)$, and therefore a submanifold. In other words, $SO(3)\cap\operatorname{tr}^{-1}\bigl((2,3]\bigr)$ is a submanifold of $SO(3)$.