4. Given $\boldsymbol{a}=\left(\cos \frac{2}{3} \pi, \sin \frac{2}{3} \pi\right), \overrightarrow{O A}=\boldsymbol{a}-\boldsymbol{b}, \overrightarrow{O B}=\boldsymbol{a}+\boldsymbol{b}$, if $\triangle O A B$ is an isosceles right triangle with $O$ as the right-angle vertex, then the area of $\triangle O A B$ is $\qquad$
4. 1 Detailed Explanation: Let vector $\boldsymbol{b}=(x, y)$, then $\left\{\begin{array}{l}(\boldsymbol{a}+\boldsymbol{b})(\boldsymbol{a}-\boldsymbol{b})=0 \\ |\boldsymbol{a}+\boldsymbol{b}|=|\boldsymbol{a}-\boldsymbol{b}|\end{array}\right.$,
$$
\begin{array}{l}
\text { i.e., }\left\{\begin{array} { l } 
{ ( x - \frac { 1 } { 2 } , y + \frac { \sqrt { 3 } } { 2 } ) \cdot ( - x - \frac { 1 } { 2 } , - y + \frac { \sqrt { 3 } } { 2 } ) = 0 } \\
{ ( x - \frac { 1 } { 2 } ) ^ { 2 } + ( y + \frac { \sqrt { 3 } } { 2 } ) ^ { 2 } = ( x + \frac { 1 } { 2 } ) ^ { 2 } + ( y - \frac { \sqrt { 3 } } { 2 } ) ^ { 2 } }
\end{array} , \text { i.e., } \left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
x= \pm \sqrt{3} y
\end{array}\right.\right. \\
\therefore \boldsymbol{b}=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \text { or }\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right), \therefore S_{\triangle A O B}=\frac{1}{2}|\boldsymbol{a}+\boldsymbol{b}||\boldsymbol{a}-\boldsymbol{b}|=1 .
\end{array}
$$