AP = U{k>0} ATIME (n^k)

"polynomial time on an ATM" An ATM is nondeterministic.

APSPACE = U{k>0} ASPACE (n^k)

"polynomial space on an ATM".

AEXPTIME = U{k>0} ATIME (2^(n^k))

"exponential time on an ATM"

AP = PSPACE.  ("polynomial time on an ATM is the same class of problems that use polynomial space on a DTM.")

APSPACE = EXPTIME ("polynomial space on an ATM is the class of problems that use exponential time on a DTM.")

PH is the union of all sigmas, deltas, and pis in the polynomial hierarchy.

P = NP if and only if P = PH   (P = NP is equivalent to saying that the entire polynomial hierarchy can be solved in polynomial time.) ("complete collapse of the polynomial hierarchy").

If NP = co-NP   then NP = PH.  ("collapse to the second level".  Meaning P may not equal NP, but everything above that level is equal to each ther.)

PH is a subset of PSPACE.  (But is it a proper subset?  Does PH=PSPACE?)

If PH hierarchy has any complete problems, then there are only a finite number of levels in the polynomial hierarchy.

When we say a problem is NP-complete that means that ANY problem in NP can be "reduced" in polynomial time to the given problem.  And "reduce" that the input from any (NP) problem can be converted to input to the given problem in polynomial time such that the answer to both problems (accept,reject) is always the same.

Along the same lines, consider a class of problems in some vague polynomial system (like the hierarchy), call it K.  And a K-complete problem is one which a) the problem is in K and b) every problem in K can be reduced to it in polynomial time.  The input from every problem in K can be converted to valid for the given problem and both problems always give the same answer on their respective inputs.

So far, no PH-complete problem has ever been found.

If there were a PH-complete problem, call it X, then that problem would be in one of the sigmas or the pi's.  So for any problem in PH (at whatever level) there is a polynomial time reduction algorithm to X.  So, for any problem in PH, at whatever level, I can solve it by reducing to X (polynomial time) and then applying the X algorithm to get the answer.  Thus, my problem is solvable by an algorithm at the same level as X.  So, my problem is at X's level.