# The Lie Algebra of $O(n)$ is the set of $n \times n$ skew-symmetric matrices

I'm trying to show that the Lie Algebra for $O(n)$ is the set of $n \times n$ skew-symmetric matrices. Here is what I have so far.

Since $O(n)$ is the union of two disjoint subsets, the matrices with determinant of $1$ ($SO(n)$) and determinant $-1$, then the Lie Algebra of $O(n)$ is the same as the Lie Algebra for $SO(n)$ because of course, $I \in SO(n)$. I'll denote the set of skew-symmetric matrices by $Sk(n)$ (is there a convention for denoting this set?) and the Lie Algebra of $O(n)$ with $\mathfrak{so}(n)$.

A skew-symmetric matrix is such that $-M = M^\top$. Then $\exp(-M) = [\exp(M)]^{-1}$. So then, $\exp(M^\top) = [\exp(M)]^{-1}$. But the LHS equals $[\exp(M)]^\top$. Thus, $\exp(m) [\exp(M)]^\top = I$ which implies that $\exp(M) \in O(n)$. Consider $A \in Sk(n)$. We know that $\exp(A) \in O(n)$. In fact, it is not hard to show that $\exp(tA) \in O(n)$ as well. Consider the curve $\alpha(t) = \exp(tA)$. Now, $\alpha(0) = I$, $\alpha'(t) = A \exp(tA)$, and $\alpha'(0) = A$. So we've shown that $Sk(n) \subseteq \mathfrak{so}(n)$.

However, I'm having trouble showing the other direction: $\mathfrak{so}(n) \subseteq SK(n)$. I don't think this direction should require much theory beyond the definitions. Hints are very welcome.

Thank you.

Well let's see. Say $$\gamma:(-1,1)\to O(n)$$is differentiable and $\gamma(0)=I$; we need to show that $$\gamma'(0)+\gamma'(0)^T=0.$$
We know that $\gamma(t)\gamma(t)^T=I$. Differentiating that gives $$\gamma'(0)\gamma(0)^T+\gamma(0)\gamma'(0)^T=0.$$Oh. We're done, because $\gamma(0)=I$.
Here is another approach: let $f:M_n(R)\rightarrow M_n(R)$ defined by $f(X)=XX^T$, it is submersion in a neighborhood of $f^{-1}(I)$ and $O(n)$ is the submanifold defined by $f^{1}(I)$. If $X\in O(n)$ the tangent space $T_XO(n)$ of $X$ is the set of matrices $M$ such that $df_X(M)=0$. This is equivalent to saying that $MX^T+XM^T=0$. In particular, if $X=I$ then $M+M^T=0$.
• How can you talk about a tangent space in a matrix $X$? Is $O(n)$ the manifold in this case, $X$ a "point" of the manifold and $f^{-1}(0)=O(n)$? – Algebear Nov 5 '18 at 20:50