PLAIN ENGLISH FIRST
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You wrote an entry sequence in dots and arrows. This turns it into a
machine you can run. It walks down into nested voxels, comes back to
zero, then CROSSES THROUGH zero to a root that lives on the far side,
and opens the first frame. Along the way it proves two smaller things:
a linear step can address a space that is astronomically large, and
two kinds of motion (surface + depth) do not interfere.

THE SPINE  (read left to right)
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  ~> 0 ~~> . ~~> .. ~~> ... ~~> ..... ~~> 0 ~~> [0 ~> -1] ~~> r0 ~~> >[1] : begin
   enter    burrow the ladder (dots = depth)    return   THE CROSSING   root   frame 1
   origin                                        to 0    (through floor) below  opens
                                                                          zero

THE THREE FINDINGS
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[LIT]  THE LADDER ~~> depth n addresses a field of size (2n)^(2^n).
       n=2 -> 3 digits · n=4 -> 15 digits · n=6 -> 70 digits.
       You only ever TOUCH a window of size 1+n (linear).
       ~~> a linear reach inside a doubly-exponential space. The clamp holds.

[LIT]  THE COMMUTATOR ~~> MOVE and BURROW commute. Run both orders,
       the difference is 0, over all 343 tested input triples.
       ~~> "surface and depth are orthogonal" is now MEASURED, not asserted.

[LIT]  THE CROSSING ~~> zero is not a floor. +z (depth) and -z (root)
       both exist; the root sits at z=-1, reached by passing through 0.
       ~~> frame 1 opens BELOW zero. The space is signed.

[AMBER] the -1 |-> r0 topology (root reached from below) is your design
        choice, not derived from anything. Labelled, not proven.

THE KNOWN-BAD  (an instrument whose good news is silence needs one)
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A planted void: a "spilling" burrow that SHOULD break orthogonality and
MUST make the commutator non-zero. If it ever reads zero, the instrument
is blind. It reads 12 vs 11 ~~> non-zero ~~> the instrument can see a fault.

WHAT THIS DOES NOT DO
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- does not run your real compiler (no src/ available) — this is the
  entry MATH, verified in Node, not the I13-H1.1 execution authority.
- does not decide whether (2n)^(2^n) is the RIGHT field law — it tests
  the shape you gave, it does not derive it.
- the sign topology is measured as PRESENT, not justified as CORRECT.