Vtyjkyuk

I was trying the following problem :

Let $operatornameO(n)$ be the group of $n times n$ orthogonal matrices. Show that if $gamma : (-1,1) to operatornameO(n)$ is a differentiable curve such that $gamma(0) = I_n$ , then $gamma'(0)$ is a skew-symmetric matrix.

What can be said if $gamma(0)$ is an arbitrary element $O in operatornameO(n)$?

Repeat the exercise replacing $operatornameO(n)$ by $operatornameSO(n)$.

I have no ideas regarding the problem and hence could not proceed.Thanks in advance for help.

If $M,N$ are smooth manifolds and $f colon M to N$ is a smooth map, then its pushforward at $p in M$ is $$f_p*colon T_pM rightarrow T_f(p)N :X_p mapsto fracddtf(alpha(t))biggrlvert_t=0$$
where $alpha$ is a curve on $M$ such that $alpha(0) = p$ and $alpha'(0) = X_p$.

In the case where $f$ is the curve $gamma$, then $M=(-1,1)$, $N=mathrmO(n)$, $T_pM = mathbbR$ and $T_f(p)N subseteq M(n,mathbbR)$ (the space of all $n times n$ matrices). So $gamma(t) in mathrmO(n)$ for any $t in (-1,1)$, hence $gamma(t)^T gamma(t) = mathrmid_n$ (where $^T$ denotes the transpose).
Check that its pushforward at the point $t$ is the derivative $gamma'(t)$. Since the identity map is constant, you can compute
$$fracddt gamma(t)^T gamma(t)biggrlvert_t=0 = 0 = gamma'(0)^Tgamma(0) +gamma(0)^Tgamma'(0).$$
Now $gamma(0) = mathrmid_n = gamma(0)^T$, then you get $gamma'(0)^T+gamma'(0) = 0$, that is $gamma'(0)^T = -gamma'(0)$. Thus $gamma'(0)$ is a skew-symmetric matrix in $M(n,mathbbR)$.

Now you should be able to complete the remaining points.