A confusion in finding the tangent space of $O(n)$ group.

Question

The question and the solution is given below:

But I could not understand the last 2 paragraphs, why the kernel of $df_{A}$ is as described and why the dimension of the subspace is n choose 2, could anyone explain this for me please?

Answer 1

The author computes that $$ df_A(B) = (A^tB) + (A^tB)^t. $$ The kernel of $df_A$ is the literally the set of all matrices $B$ so that $df_A(B) = 0$ (by $0$ I mean the zero matrix). Hence $B$ is in the kernel of $df_A$ if and only if $$ (A^tB) + (A^tB)^t =0 \Longrightarrow (A^tB)^t = -A^tB. $$ This is exactly the definition for $A^tB$ being anti-symmetric. ($X$ is anti-symmetric if $X^t = -X$, so take $X=A^tB$.)

However, we have assumed that $A \in O(n)$ and so $A^t=A^{-1}$. So far we have $$ \mathrm{ker}(df_A) = \{B : A^{-1}B \text{ is anti-symmetric}\}. $$ Next the author pulls a translation trick, which might be where you got confused. Note that $$ \{B : A^{-1}B \text{ is anti-symmetric}\} = \{AC : C \text{ is anti-symmetric}\} $$ is true by an easy (but omitted) argument. If $B$ belongs to the left-hand set then $A^{-1}B$ is antisymmetric, and then with $C=A^{-1}B$ we get $AC = AA^{-1}B=B$ belongs to the right. Reverse this line of thought to get containment in the other direction.

Now for the dimension count. As $A$ is invertible, multiplication by $A$ yields the map $C \mapsto AC$ which is a bijection from the subspace of all anti-symmetric matrices to the kernel of $df_A$. Hence the dimension of the kernel of $df_A$ is equal to the dimension of the spaces of all anti-symmetric $n \times n$ matrices. But if a matrix $X = (x_{ij})$ is anti-symmetric, you must have $x_{ji} = -x_{ij}$. Thus you only have freedom of choice on the lower triangular half of $X$, excluding the diagonal entries as they must be $0$ by anti-symmetry. Thus, you only have free choices on the entries $$ x_{21} $$ $$ x_{31}, x_{32} $$ and so on down to $$ x_{n1}, x_{n2}, \ldots, x_{n,n-1} $$ How many free parameters is this? Add them up by rows: $$ 1 + 2 + 3 + \cdots +n-1 $$ which is well-known to equal $\frac{(n-1)n}{2} = \binom{n}{2}$.