What is the equation of the orthogonal group (as a variety/manifold)? I have been studying some elementary Lie theory recently, so I have been thinking about matrix groups as manifolds. Most simple examples of manifolds that we learn in high school or college even are solutions to algebraic equations (for example, conic sections) i.e. algebraic varieties.
This led to the thought (which is also my question): 

Is the orthogonal group also a manifold which is the solution to some multivariable polynomial equation? If so, what is it?

Proposed answer:
If $X$ denotes the $n \times n$ real-valued matrix whose $ij$th entry is the monomial/variable $x_{ij}$.
Then orthogonal group is the zero set of the system of polynomial equations (in $n^2$ variables):
$$X^T X - I_n = 0$$
Is this really the answer? It seems too dumb to be possible. Then again, I have never really thought of an entire collection of matrices as the set of solutions to a system of multilinear equations before.
Check: this implies that the determinants of these matrices are $\pm 1$ by the multiplicative property of determinants.
Check: for the case $n=1$ we have $$x^2 -1 = 0 \implies x = \pm 1$$ which is O(1) as necessary/expected.
Check: for the case $n=2$ we have $$\begin{bmatrix} x_1 & x_3 \\ x_2 & x_4 \end{bmatrix}  \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 0 \implies \begin{array}{c}  x_1^2 + x_3^2 = 1 \\ x_1x_2 + x_3x_4 = 0 \\ x_2 x_1 + x_4 x_3 = 0 \\ x_2^2 + x_4^2 = 1 \end{array}$$ I know that O(2) has two path components, each of which is diffeomorphic to SO(2), which is just the unit circle, and the solution to this system of equations does suggest the disjoint union of two unit circles because of the equations $x_1^2 + x_3^2 =1$ and $x_2^2 + x_4^2=1$; however I am not sure how to interpret the auxiliary condition that $$x_1 x_2 = -x_3x_4$$ does this guarantee that the circles are disjoint somehow? Otherwise I am not sure if this equation is giving me the correct answer for the case $n=2$.
 A: $f(X)=X^TX-I=0$ is an implicit algebraic equation of $O(n)$ (there are $n(n+1)/2$ distinct equations in the $n^2$ real unknowns $(x_{i,j})_{i,j}$). Consequently  $O(n)$ is a real algebraic variety of dimension $\geq n^2-n(n+1)/2=n(n-1)/2$. Since $O(n)$ is a group, to obtain its dimension, it suffices to consider the kernel of the derivative of $f$ in $I$; $Df_I:H\rightarrow H^T+H$, $\ker(Df_I)=SK_n$  is the set of skew symmetric matrices, a vector space of dimension $n(n-1)/2$; conclusion: $dim(O(n))=n(n-1)/2$. 
From a topological point of view, $O(n)$ is complicated; it has $2$ connected components $SO(n),O^-(n)$ that are algebraically diffeomorphic ($X\in O^-(n)\rightarrow Xdiag(-1,1,\cdots,1)\in SO(n)$); note that $SO(n)$ is a group and $O^-(n)$ is not. In the sequel we consider only $SO(n)$. Note that $SO(n)$ is not simply connected; yet $SO(2)=S^1$ ($S^1$ can be covered by (uncoiled in the form of) the simply connected $\mathbb{R}$ of dimension $1$); $SO(3)=RP^3$, the real projective space (it can be covered by the simply connected $S^3$ of dimension $3$); $SO(4)$ can be covered by the simply connected $S^3\times S^3$ of dimension $6$. 
Removing a part of $SO(n)$, we can trivialize it. Let $U=\{X\in SO(n)|X+I$ is singular $\}$. Then the Cayley transform 
https://en.wikipedia.org/wiki/Cayley_transform
$K\in SK_n\rightarrow (I-K)(I+K)^{-1}\in SO(n)\setminus U$ is an algebraic  parametrization of $SO(n)\setminus U$. In other words, $SO(n)\setminus U$ is flat. That is complicated is the reattachment of $U$ over $SK_n$.
