Understanding difference between a distinguished boundary and 'normal' boundary in several complex variables I am reading through Tasty Bits of Several Complex Variables and I come across the term distinguished boundary. It seems a distinguished boundary is different from a normal boundary as the author explains the latter has fewer dimensions than the latter. What is also confusing me is the image of the bidisk where I can't tell the difference between the distinguished boundary and the normal boundary?
Is it that the distinguished boundary is illustrated as the bold lines/edges of the image and while the boundary cannot be fully represented on a 2D paper? Does the boundary include the shaded grey area whilst the distinguished boundary does not? Also the union of inside and outside of the donut confuses more.
I feel i'm being short sighted here, can someone correct my vision?

 A: The diagram is an attempt to depict in schematic form some aspects of the fully 4D configuration by keeping track only of the absolute values of the two complex coordinates. (This of course suppresses the angular coordinate of $z_1= r_1 e^{i \theta_1}$ and likewise for $z_2=r_2 e^{i \theta_2}$.) (This trick is motivated by the happy fact that all the relevant pieces of the puzzle can be described by equations involving only absolute values.) The schematic diagram is usually called an absolute space picture.
The ordered pair $(z_1, z_2)$ is represented schematically by $(|z_1|, |z_2|)$.
The distinguished boundary is just the Cartesian product of two circles, $S^1\times S^1$ = $\{ |z_1|=1 \} \times \{ |z_2| =1\}$
which in absolute terms is simply depicted by $(1,1)$. Note  that is the corner point of the dark square in absolute space.
In contrast, the normal or ordinary boundary is the union of two parts, one of which is $|z_1|<1 $ crossed with $|z_2|=1$. In absolute space this part is depicted by the upper dark line segment.  The other part is the vertical dark line segment.
