Show that an orthogonal group is a $ \frac{n(n−1)}2 $-dim. $ C^{\infty} $-Manifold and find its tangent space The orthogonal group is defined as (with group structure inherited from $n\times n$ matrices)
$$O(n) := \{X\in \mathbb{R}^{n\times n} : X^\text{t}X=I_n\}.$$
(i) Show that $O(n)$ is an $\frac{n(n-1)}{2}$-dimensional $C^\infty$-manifold in the space of $n\times n$ matrices. 
Hints. Exhibit $O(n)$ as the preimage of $0$ of the function $\phi$ from $n\times n$ matrices to the symmetric $n\times n$ matrices given by $X\mapsto X^\text{t}X − I_n$. (Note that the target space of $\phi$ is very important in order to satisfy the maximal rank condition.)
Then show that the equation $\phi'(A)H=S$ has a solution $H$ for each $A\in O(n)$ and each symmetric $n\times n$ matrix $S$. You will also need to compute the dimension of the space of symmetric $n\times n$ matrices.
(ii) Show that the tangent space $T_{I_n}O(n)$ at the identity is the space of antisymmetric matrices.
 A: They've already given you the smooth map $\phi(X) = X^{t}X - I_n,$ and $O(n) = \phi^{-1} (0),$ so it is automatically a manifold (if the rank is constant). To see the dimension, you need to calculate the nullity of $\phi^{\prime}.$ See below for a full solution.

 The point here is that $\phi$ takes $n\times n$ matrices to symmetric matrices since $(X^{t}X- I)^t = X^{t}(X^{t})^{t} - I = X^t X - I.$ Take the directional derivative $\phi_v (X) = \lim_{h\to 0} \frac{(X+hv)^t(X+hv) - I - X^t X + I}{h} = X^t v + v^t X = (X^t v) + (X^t v)^t.$ That is, $\phi^{\prime}(X)$ takes $v$ to $(X^t v)+(X^t v)^t.$ Then we note that this is symmetric, and $\phi^{\prime}$ is onto the space of symmetric matrices since $X$ is invertible. The space of symmetric matrices has dimension $n(n+1)/2$ (you can do this yourself), whence the dimension of $O(n)$ is $n^2 - n(n+1)/2 = n(n-1)/2.$ 

To calculate the tangent space at $I$, you just want to look at $\phi^{\prime} (I)$ and calculate its null space.
A: To show that the map $p\colon X\mapsto X^{t} \cdot X$ from $M_n(\mathbb{R})$ to $\mathcal{Sym}_n(\mathbb{R})$ is a submersion at $X= I$, it is enough to produce a section around $p(I) = I$, that is a map $s\colon U\to M_n(\mathbb{R})$ smooth, $s(I) = I$, and $p\circ s = Id$ on $U$ a neighborhood of $I$ in $\mathcal{Sym}_n(\mathbb{R})$. Take $s$ to be the square root, that is
$$s(I + \delta) = \sqrt{I+ \delta} = \sum_{k\ge 0} \binom{1/2}{k} \delta^k$$
Now to produce a section around any $X \in O(n, \mathbb{R})$, use the fact that $O(n)$ is a subgroup.
