What are the consequences of $L = \oplus L$? Shiva Kintali has just announced a (cool!) result that graph isomorphism for bounded treewidth graphs of width $\geq 4$ is $\oplus L$-hard.  Informally, my question is, "How hard is that?"

We know that nonuniformly $NL \subseteq \oplus L$, see the answers to this question.  We also know that it is unlikely that $\oplus L = P$, see the answers to this question.  How surprising would it be if $L=\oplus L$?  I have heard many people say that $L=NL$ would not be shocking the way $P=NP$ would.


  What are the consequences of $L=\oplus L$?


Definition: $\oplus L$ is the set of languages recognized by a non-deterministic Turing machine which can only distinguish between an even number or odd number of "acceptance" paths (rather than a zero or non-zero number of acceptance paths), and which is further restricted to work in logarithmic space.
Wigderson proved that $NL/poly \subseteq \oplus L/poly$. By standard arguments, $L = \oplus L$ would imply $L/poly = NL/poly$. (Take a machine $M$ in $NL/poly$. It has an equivalent machine $M&#39;$ in $\oplus L/poly$. Take the $\oplus L$ language of instance-advice pairs $S = \{(x,a)~|~M&#39;(x,a)~\textrm{accepts}\}$. If this language is in $L$, then by hardcoding the appropriate advice $a$ we get an $L/poly$ machine equivalent to $M$.) 

I think that would be surprising: nondeterministic branching programs would be equivalent to deterministic branching programs (up to polynomial factors).