What are the consequences of $\mathsf{L}^2 \subseteq \mathsf{P}$?

Question

We know that $\mathsf{L} \subseteq \mathsf{NL} \subseteq \mathsf{P}$ and that $\mathsf{L} \subseteq \mathsf{NL} \subseteq \mathsf{L}^2 \subseteq $ $\mathsf{polyL}$, where $\mathsf{L}^2 = \mathsf{DSPACE}(\log^2 n)$. We also know that $\mathsf{polyL} \neq \mathsf{P}$ because the latter has complete problems under logarithmic space many-one reductions while the former does not (due to the space hierarchy theorem). In order to understand the relationship between $\mathsf{polyL}$ and $\mathsf{P}$, it may help to first understand the relationship between $\mathsf{L}^2$ and $\mathsf{P}$.

What are the consequences of $\mathsf{L}^2 \subseteq \mathsf{P}$?

What about the stronger $\mathsf{L}^{k} \subseteq \mathsf{P}$ for $k>2$, or the weaker $\mathsf{L}^{1 + \epsilon} \subseteq \mathsf{P}$ for $\epsilon > 0$?

Answer 1

The following is an obvious consequence: $\mathsf{L}^{1+\epsilon} \subseteq \mathsf{P}$ would imply $\mathsf{L} \subsetneq \mathsf{P}$ and therefore $\mathsf{L} \neq \mathsf{P}$.

By the space hierarchy theorem, $\forall \epsilon > 0: \mathsf{L} \subsetneq \mathsf{L}^{1+\epsilon}$ . If $\mathsf{L}^{1+\epsilon} \subseteq \mathsf{P}$ then $\mathsf{L} \subsetneq \mathsf{L}^{1+\epsilon} \subseteq \mathsf{P}$.

Answer 2

$ \newcommand{\DSPACE}{\mathsf{DSPACE}} \newcommand{\L}{\mathsf{L}} \newcommand{\P}{\mathsf{P}} \newcommand{\DTIME}{\mathsf{DTIME}} $

$\L^2 \subseteq \P$ would refute the Exponential Time Hypothesis.

If $\L^2 \subseteq \P$ then by a padding argument $\DSPACE(n) \subseteq \DTIME(2^{O(\sqrt n)})$. This means that the satisfiability problem $\mathsf{SAT} \in \DSPACE(n)$ can be decided in $2^{o(n)}$ steps, refuting the Exponential Time Hypothesis.

More generally, $\DSPACE(\log^{k} n) \subseteq \P$ for $k\geq1$ implies $\mathsf{SAT} \in \DSPACE(n) \subseteq \DTIME(2^{O(n^{\frac{1}{k}})})$.

(This answer is expanded from a comment by @MichaelWehar.)

Answer 3

Group isomorphism (with groups given as multiplication tables) would be in P. Lipton, Snyder, and Zalcstein showed this problem is in $\mathsf{L}^2$, but it is still open whether it is in P. The best current upper bound is $n^{O(\log n)}$-time, and because it reduces to graph isomorphism, stands as a significant obstacle to putting graph iso into P.

Makes me wonder what other natural and important problems this would apply to: that is, in $\mathsf{L}^2$ but with their best known time upper bound quasi-polynomial.

Answer 4

Claim: If $L^k \subseteq P$ for some $k > 2$, then $P \neq \log(CFL)$ and $P \neq NL$.

Suppose that $L^k \subseteq P$ for some $k > 2$.

From "Memory bounds for recognition of context-free and context-sensitive languages", we know that $CFL \subseteq DSPACE(\log^2(n))$. By the space hierarchy theorem, we know that $DSPACE(\log^2(n)) \subsetneq DSPACE(\log^k(n))$.

Therefore, we get $\log(CFL) \subseteq DSPACE(\log^2(n)) \subsetneq DSPACE(\log^k(n)) \subseteq P$.

Also, by Savitch's Theorem, we know that $NL \subseteq L^2$. Therefore, we get $NL \subseteq DSPACE(\log^2(n)) \subsetneq DSPACE(\log^k(n)) \subseteq P$.