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