The Complexity Zoo points out in the entry on EXP that if L = P then PSPACE = EXP.  Since NPSPACE = PSPACE by Savitch, as far as I can tell the underlying padding argument extends to show that $$(\text{NL} = \text{P}) \Rightarrow (\text{PSPACE} = \text{EXP}).$$  We also know that L $\subseteq$ NL $\subseteq$ NC $\subseteq$ P via Ruzzo's resource-bounded alternating hierarchy.If NC = P, does it follow that PSPACE = EXP?A different interpretation of the question, in the spirit of Richard Lipton: is it more likely that some problems in P cannot be parallelized, than that no exponential-time procedure requires more than polynomial space?I would also be interested in other surprising consequences of NC = P (the more unlikely the better).Edit: Ryan's answer leads to a further question: what is the weakest hypothesis that is known to guarantee PSPACE = EXP?W. Savitch. Relationships between nondeterministic and deterministic tape complexities, Journal of Computer and System Sciences 4(2):177-192, 1970. W. L. Ruzzo. On uniform circuit complexity, Journal of Computer and System Sciences 22(3):365-383, 1971.Edit (2014): updated old Zoo link and added links for all other classes.
NC = P consequences?
cc.complexity theory;conditional results
Yes. $NC$ can be seen as the class of languages recognized by alternating Turing machines that use $O(\log n)$ space and $(\log n)^{O(1)}$ time. (This was first proved by Ruzzo.) $P$ is the class where alternating Turing machines use $O(\log n)$ space but can take up to $n^{O(1)}$ time. For brevity let's call these classes $ATISP[(\log n)^{O(1)},\log n] = NC$ and $ASPACE[O(\log n)] = P$. Suppose the two classes are equal. Replacing the $n$ with $2^n$ in the above (i.e., applying standard translation lemmas), one obtains $TIME[2^{O(n)}] = ASPACE[O(n)] = ATISP[n^{O(1)}, n] \subseteq ATIME[n^{O(1)}] = PSPACE$. If $TIME[2^{O(n)}] \subseteq PSPACE$ then $EXP = PSPACE$ as well, since there are $EXP$-complete languages in $TIME[2^{O(n)}]$.Edit: Although the above answer is perhaps more educational, here's a simpler argument: $EXP = PSPACE$ already follows from $P$ is contained in polylog space and standard translation. Note $P$ is contained in polylog space is a much weaker hypothesis than $NC = P$. More details: Since $NC$ circuit families have depth $(\log n)^c$ for some constant, every such circuit family can be evaluated in $O((\log n)^c)$ space. Hence $NC \subseteq \bigcup_{c > 0} SPACE[(\log n)^c]$. So $P = NC$ implies $P \subseteq \bigcup_{c > 0} SPACE[(\log n)^c]$. Applying translation (replacing $n$ with $2^n$) implies $TIME[2^{O(n)}] \subseteq PSPACE$. The existence of an $EXP$-complete language in $TIME[2^{O(n)}]$ finishes the argument.Update: Addressing Andreas' additional question, I believe it should be possible to prove something like: $EXP=PSPACE$ iff for all $c$, every polynomially sparse language in $n^{O(\log^c n)}$ time is solvable in polylog space. (Being polynomially sparse means that there are at most $poly(n)$ strings of length $n$ in the language, for all $n$.) If true, the proof would probably go along the lines of Hartmanis, Immerman, and Sewelson's proof that $NE = E$ iff every polynomially sparse language in $NP$ is contained in $P$. (Note, $n^{O(\log^c n)}$ time in polylog space is still enough to imply $PSPACE=EXP$.)