# Direct calculation of the tangent space of $SO(3)$

Let $SO(3)$={$RR^T=I$, $det(R)=1$}, I need to show that a base of the tangent space in the identity is given by: $$E_i=\frac{d}{dt}\exp(tL_i)|_{t=0}$$ where $$L_1= \left(\begin{matrix} 0& 1& 0\\ -1& 0& 0\\ 0& 0& 0\end{matrix}\right), L_2= \left(\begin{matrix} 0& 0& -1\\ 0& 0& 0\\ 1& 0& 0\end{matrix}\right), L_3= \left(\begin{matrix} 0& 0& 0\\ 0& 0& 1\\ 0& -1& 0\end{matrix}\right).$$ I would like to prove it directly without involving higher knowledge on Lie Algebras and Lie Groups relations. How would you proceed?

Edit: I would like to do this as an example before introducing any notion of Lie Algebras and tangent spaces of Lie Groups.

## 3 Answers

Hint:By differentiating $XX^T$, we obtain that the tangent space is given by $RX+XR^T=0$ take $X=Id$ you have $R+R^T=0$the matrices $L_i$ are antisymmetric and are a base of the tangent space since they are linearly independent and the dimension of $SO(3)$ is 3.

• how do you demonstrate the differentiation that brings you to the tangent space equation? – Dac0 Mar 29 '16 at 16:04
• The map which send $X$ to $XX^T$ is the composition of $X\rightarrow (X,X)$ and $(X,Y)\rightarrow XY^T$. The first map is linear and the second bilinear. – Tsemo Aristide Mar 29 '16 at 16:08
• I think I didn't get your point – Dac0 Mar 29 '16 at 16:17
• You have to compose the derivatives. A linear map is equal to its derivative. And the derivativeof the bilinear msp $b$ at $(x,y)$ is $b(u,y)+b(x,v)$ where $(u,v)$ is tangent to $(x,y)$. – Tsemo Aristide Mar 29 '16 at 16:21

We can observe that the group $SO(3)$ is a connected component of the group $O(3)$ so the Lie algebra (i.e. the tangent space) is $\mathcal{o}(3)$ which is the tangent space at the identity element of the $O(3)$ group. So after this consideration we want to find a path $\gamma(t) \subset O(3)$ such that $\gamma(0)= I_3$ ; it will then follow that $\{\dot \gamma(0) \}\ = T_e O(3)$. If $\gamma(t)=A(t)$ then by definition must be $A^T (t) A(t) = I_3$ for all $t$ and in particular the derivative at $t=0$ of the previous relation must be zero. You can compute easily the derivative and notice that if $\gamma(t)=A(t) \subset O(3)$ then $\dot A(t)_{t=0} \in T_e O(3)$ is a skew symmetric matrix. We should now prove that all skew symmetric matrices of dimension 3 belongs to the tangent space at the origin of $O(3)$. This can be done by choosing the classical exponentiation of a matrix, I mean choosing this path $\gamma(s)=exp(sA)$ where $A$ is a skew symmetric matrix then you can obtain with one line computation this result $exp(sA)^T exp(sA)= exp(sA^T + sA)=1$ notice that in this second passage we don't need to apply the Backer-Campbell-Hausdorff formula since $[A^T,A]=0$. So we can conclude that the vectors of the tangent space of $O(3)$ are all and only the skew symmetric matrices. What is the dimension of that tangent space? in general observing the independent components of a generic element of $T_e O(n)$ we can observe that the dimension is $(n^2 -n)/2$ so for the 3-dimensional case the dimension is 3. Which means that the 3 skew-symmetric matrices you wrote above forms a basis of the $T_e O(3)$

(1) $\exp\ (A):= \sum_{i=0}^\infty \frac{A^i}{i!}$

Note that $c_i(t):=\exp\ (tL_i)$ is a curve in $SO(3)$ : $$c_i(t)c_i(t)^T = \exp\ tL_i(\exp\ tL_i)^T=\exp\ tL_i \exp\ tL_i^T = \exp\ t(L_i+ L_i^T) = I$$

Clearly $\frac{d}{dt}\bigg|_{t=0} c_i(t) =L_i \in T_ISO(3)$ Since ${\rm dim}\ SO(3)=3$, then three vectors may be a basis

(2) As far as I know to show that without calculating $\exp\ L_i$, we need some about Lie group. That is $c_1(t)=c_2(t)$ may happen. That is we need an inner product $g$ on $T_I SO(3)$ to distinguish $L_i$ : Define $$g (A,B):= {\rm trace}\ AA^T,\ A\in T_I SO(3)$$

From $g$, $L_i$ are different. Lastly, we must check that $g$ is $SO(3)$-invariant : For $h\in SO(3)$, define $$g_h (dh\ L_i,dh\ L_j) := g_I(L_i,L_j)$$ where $dh\ L_i:= \frac{d}{dt} h\exp\ (tL_i)$