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.

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    $\begingroup$ Welcome to MSE! If you show what you have tried so far, you would get better help. $\endgroup$ – Ehsan M. Kermani May 6 '13 at 11:18
  • $\begingroup$ I get a vibe that a suitable theorem has been given in your course material, and the hints for question (i) are leading up to that. Since I don't know the theorem in question, I'm unsure what to suggest. I would approach it differently: the first row of $X$ can be any unit vector, i.e. a point on $S^{n-1}$. The second row can be any unit vector orthogonal to the first row, i.e. a point on $S^{n-2}$. Et cetera. This leads to a fibration and strongly suggests that the dimension is $(n-1)+(n-2)+\cdots+2+1=n(n-1)/2$. $\endgroup$ – Jyrki Lahtonen May 6 '13 at 16:10

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.

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    $\begingroup$ I could not understand how you get the final answer of the directional derivative.....could u please explain it in details? $\endgroup$ – Secretly Sep 30 '18 at 22:13

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