Show that \(S O(3)\) is embedded in \(R^{9}\), and at the same time, that \(S O(3)\) is a manifold.

To show that \(SO(3)\) is embedded in \(R^9\), we shall start by showing that there exists an injective function from \(SO(3)\) to \(R^9\). An injective function maps distinct elements of the domain to distinct elements of the codomain, without any repetitions. Let's consider the following function, \(F: SO(3) \to R^9\), which maps a matrix in \(SO(3)\) to a vector in \(R^9\): $$ F(A) = (a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}, a_{31}, a_{32}, a_{33}) $$ where, \(A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \in SO(3)\). As the function \(F\) maps a matrix in \(SO(3)\) to a unique vector in \(R^9\), it is injective.

To show that the image of the embedding function F in step 1 is a closed subset of \(R^9\), we must show that the set of all orthogonal matrices satisfying the condition \(det(A) = 1\) is closed in \(R^9\). Let's consider the set: $$ SO(3) = \{ A \in R^{3 \times 3} \mid A^TA = I, det(A) = 1 \}. $$ For \(A \in SO(3)\), we observe that: 1. \(A^TA = I\) can be rewritten as a system of equations in terms of the entries of the matrix A: $$ a_{11}^2 + a_{21}^2 + a_{31}^2 = 1 \\ a_{12}^2 + a_{22}^2 + a_{32}^2 = 1 \\ a_{13}^2 + a_{23}^2 + a_{33}^2 = 1 \\ a_{11}a_{12} + a_{21}a_{22} + a_{31}a_{32} = 0 \\ a_{11}a_{13} + a_{21}a_{23} + a_{31}a_{33} = 0 \\ a_{12}a_{13} + a_{22}a_{23} + a_{32}a_{33} = 0 $$ These are linear and quadratic equations, which are closed under addition in \(R^9\). 2. \(det(A) = 1\) can be written in terms of the entries of the matrix A: $$ a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) = 1 $$ This is also a quadratic equation in terms of the elements of A, which is closed under addition in \(R^9\). The set of solutions that satisfies these conditions simultaneously is a closed subset of \(R^9\). Therefore, the image of the embedding function \(F\) is a closed subset of \(R^9\).

To show that \(SO(3)\) is a topological space, we need to show that it inherits a topology from \(R^9\). Since \(SO(3)\) is embedded in \(R^9\), the subset topology induced by \(R^9\) on \(SO(3)\) is the topology of the space. A set endowed with an induced topology is a topological space. Therefore, \(SO(3)\) is a topological space.

For \(SO(3)\) to be a smooth manifold, it must have a smooth structure with a compatible smooth atlas. We will use the standard matrix representation of elements in \(SO(3)\), which are parameterized by three variables, usually the Euler angles \((\phi, \theta, \psi)\). We can choose a smooth atlas, which is a collection of charts with smooth transition maps, for \(SO(3)\) by considering the standard parameterization of these elements. For example, consider a chart \(\mathcal{U}_{\phi,\theta,\psi} \subset SO(3)\), and a mapping \(\Phi: \mathcal{U}_{\phi,\theta,\psi} \to R^3\), defined by: $$ \Phi(A) = (\phi, \theta, \psi). $$ The transition map between two charts \(\mathcal{U}_{\phi_1,\theta_1,\psi_1}\) and \(\mathcal{U}_{\phi_2,\theta_2,\psi_2}\) is given by the change of coordinates, which consists of the composition of the inverse of one chart with another chart. As these mappings involve standard transformations, such as rotation and translation, the transition maps are smooth, and these charts form a smooth atlas on \(SO(3)\). Since \(SO(3)\) is a topological space with a smooth structure, it is a smooth manifold. Therefore, we have shown that \(SO(3)\) is embedded in \(R^9\), and it is a manifold.