Right negative coefficients are allowed. I haven't been thinking clearly about the problem. What I meant is there is no way of getting negative coefficients if they all start out positive.
This is what is now obvious in retrospect. For those coordinates where the coefficients a, b, and c are all positive, then the square has all positive coefficients:
Ie., if x = a +bi +cj then x
2 = 1(a
2 +2bc) +i( c
2+2ab) +j(b
2+2ac), which remain positive.
And they will always be positive when squaring over many iterations, so all orbits of x are positive..
This is unlike the complex plane where, when squared, the real part is a
2 -b
2 (potentially becoming negative even though both coefficients start out positive.)
If at least one coefficient in C3 is negative, then upon iterated squaring, negative values can eventually mix into all coefficients. In this case orbits can perhaps wander almost anywhere in the space.
Now I think I have answered my question:
The top two images are that of C3 sliced through the i and j axes respectively. These shapes are sharply truncated in the upper right because the coefficients are all too positive, and cause divergence at the sharp boundary.
Now the unanswered question is why does the remainder of the images look so close to the M-set? The orbits all wander around in 3 dimensions and still give the same results as the M-set. My conjecture is that a projection of the 3-D orbits onto the 2-D planes are the same as the M-set orbits. Otherwise it is too coincidental. To prove that is going to be an interesting exercise.
An even harder problem is why does C11 have cross-sections that look like the M-set? See
http://www.insidetheoutbox.net/recmaps/recmap-index.php?genre=c11
Dilber MC Theme by HarzeM