# Validity of the Following Calculation (Generators of $SO(3)$)

I was reading the following page from the Physics from Symmetry by Jakop Schwichtenberg, I didn't quite understand the calculation.

Basically what he does is the following:

$$J_1$$ is one of the basis element for the generators of the group $$SO(3)$$ and is given by \begin{align} J_1 = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{pmatrix} \end{align} He is trying to show that, $$e^{\theta J_{1}}$$ will produce a well known rotation matrix in 3D as in the bottom of the page. So far so good. What I don't understand is that he defines the lower right $$2\times 2$$ matrix $$j_1$$ in $$J_1$$ and uses this $$2\times 2$$ matrix to show that

\begin{align} e^{\theta j_{1}} = \begin{pmatrix} cos(\theta) & -sin(\theta)\\ sin(\theta) & cos (\theta) \end{pmatrix} \end{align}

But how does this helps us? How does he conclude the final result from it?

• That looks like a very thorough explanation. What's the problem with it? Commented Aug 2, 2020 at 19:58
• Then I am certainly missing something. Let me rephrase: how did he go from (3.68) to (3.69). Just because a submatrix of the matrix equals to (3.68) when exponantiated, can we say that the submatrix of the result (when the original matrix exponantiated) will also equal to (3.68). Basically, I don't understand how 3.68 helps us. Also, what does "using $e^0 = 1$ for the upper left component" means?
– Etg
Commented Aug 2, 2020 at 21:18
• If we ignore the lower $2\times 2$ matrix, $I=\pmatrix{1&0&0\\0&*&*\\0&*&*}$ and $J_1^n=\pmatrix{0&0&0\\0&*&*\\0&*&*}$ for $n\ge1$ giving $\exp(\theta J_1)=\pmatrix{1&0&0\\0&*&*\\0&*&*}$. Commented Aug 3, 2020 at 4:52
• Is this trivial? Because I'm still unable to see how this holds.
– Etg
Commented Aug 3, 2020 at 9:32