Example 6 Given the hyperbola $2 y^{2}-2 x^{2}=1$, a line $l$ passing through point $F$ $(0,1)$ intersects the hyperbola at points $A$ and $B$. Find the range of $\overrightarrow{F A} \cdot \overrightarrow{F B}$.
If the slope of $l$ does not exist, its equation is $x=0$, and $l$ intersects the hyperbola at $A\left(0, \frac{\sqrt{2}}{2}\right), B\left(0,-\frac{\sqrt{2}}{2}\right)$,
$$
\begin{array}{c}
\overrightarrow{F A}=\left(0, \frac{\sqrt{2}}{2}-1\right), \overrightarrow{F B}=\left(0,-\frac{\sqrt{2}}{2}-1\right), \\
\therefore \overrightarrow{F A} \cdot \overrightarrow{F B}=\left(\frac{\sqrt{2}}{2}-1\right) \cdot\left(-\frac{\sqrt{2}}{2}-1\right)= \\
-\left(\frac{\sqrt{2}}{2}+1\right)\left(\frac{\sqrt{2}}{2}-1\right)=-\left[\left(\frac{\sqrt{2}}{2}\right)^{2}-1\right]=\frac{1}{2} .
\end{array}
$$

If the slope of $l$ exists, let its equation be $y=k x+1$, substituting into $2 y^{2}-2 x^{2}=1$ and simplifying yields
$2\left(k^{2}-1\right) x^{2}+4 k x+1=0$.
From $\left\{\begin{array}{l}k^{2}-1 \neq 0, \\ \Delta=16 k^{2}-8\left(k^{2}-1\right)>0,\end{array}\right.$ we get $k \neq \pm 1$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
then $\left(x_{1}, y_{1}-1\right) \cdot\left(x_{2}, y_{2}-1\right)=\left(x_{1}, k x_{1}\right) \cdot$
$\left(x_{2}, k x_{2}\right)=x_{1} x_{2}+k^{2} x_{1} x_{2}=\frac{k^{2}+1}{2\left(k^{2}-1\right)}=$
$$
\frac{1}{2}\left(1+\frac{2}{k^{2}-1}\right),
$$

When $-1 < k < 1$ and $k \neq 0$, $\frac{2}{k^{2}-1} < -2$, $\overrightarrow{F A} \cdot \overrightarrow{F B} < -\frac{1}{2}$; when $k > 1$ or $k < -1$, $\frac{2}{k^{2}-1} > 0$, $\overrightarrow{F A} \cdot \overrightarrow{F B} > \frac{1}{2}$.

In summary, $\overrightarrow{F A} \cdot \overrightarrow{F B} \in\left(-\infty,-\frac{1}{2}\right) \cup$ $\left[\frac{1}{2},+\infty\right)$.