4. If three angles $\alpha, \beta, \gamma$ form an arithmetic sequence with a common difference of $\frac{\pi}{3}$, then $\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \gamma \cdot \tan \alpha$ $\qquad$
4. -3 .

According to the problem, $\alpha=\beta-\frac{\pi}{3}, \gamma=\beta+\frac{\pi}{3}$. Therefore, $\tan \alpha=\frac{\tan \beta-\sqrt{3}}{1+\sqrt{3} \tan \beta}, \tan \gamma=\frac{\tan \beta+\sqrt{3}}{1-\sqrt{3} \tan \beta}$.
Then, $\tan \alpha \cdot \tan \beta=\frac{\tan ^{2} \beta-\sqrt{3} \tan \beta}{1+\sqrt{3} \tan \beta}$, $\tan \beta \cdot \tan \gamma=\frac{\tan ^{2} \beta+\sqrt{3} \tan \beta}{1-\sqrt{3} \tan \beta}$, $\tan \gamma \cdot \tan \alpha=\frac{\tan ^{2} \beta-3}{1-3 \tan ^{2} \beta}$.
Thus, $\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \gamma \cdot \tan \alpha$ $=\frac{9 \tan ^{2} \beta-3}{1-3 \tan ^{2} \beta}=-3$.