Subspace topology of matrices I have been given homework in which the $2\times 2$ matrices of determinant $1$ are equipped with the subspace topology of $\mathbb{R}^4$. However, $\mathbb{R}^4$ is a space of 4-tuples, while 2x2 matrices are not n-tuples. How do I get the open sets of this topology?
How is this set of matrices a subset of $\mathbb{R}^4$?
 A: Any matrix 
$$\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}\in\mbox{Mat}^{2\times 2}(\mathbb{Z})$$
can be interpreted as the touple $(a,b,c,d)\in\mathbb{R}^4$, so $\mbox{Mat}^{2\times 2}(\mathbb{Z})$ can be thought of a subspace of $\mathbb{R}^4$ and be given the subspace topology.
A: The space of $2 \times 2$ matrices is identified with $\mathbb{R}^{2^2}$ via
$$\left(\begin{matrix}a&b\\c&d\end{matrix}\right) \longmapsto \left(\begin{matrix}a\\b\\c\\d\end{matrix}\right)$$
We deduce the topology from this identification. So your subset of matrices is identified with $\{(a,b,c,d) \in \mathbb{R}^4, \, ad -bc = 1\} \subset \mathbb{R}^4$.
Similarly, the space of $n \times n$ matrices is identified with $\mathbb{R}^{n^2}$ via $(a_{i,j}) \mapsto (a_{j+n(i-1)})$.
A: If $A=(a_{ij})$ and $B=(b_{ij})$ then the inherited metric will be: $$d(A,B)=\sqrt{\sum_{i,j} (a_{ij}-b_{ij})^2}$$
no matter whether you associate $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ with $(a,b,c,d)$, $(a,c,b,d)$ or any other re-ordering of the values.
