Dominations under oracles which is closed under complement? Edited at 2010/11/29:

As John Watrous have mentioned, the class $\mathsf{C^O}$ may be not well-defined. After reading some early posts, I try to restate my question in an unambiguous way.

Let $\mathsf{O}$ be a complexity class that is closed under complement, i.e. $\mathsf{O} = \mathsf{coO}$. Also we assume that the logspace, $\mathsf{L}$, is a subset of $\mathsf{O}$.


  When does the equality $\mathsf{L^O} = \mathsf{O}$ hold?


We define $\mathsf{L^O}$ as languages accepted by logspace oracle machines with an $\mathsf{O}$ oracle, where queries are written on a separated oracle tape not restricted to the logspace bound, and after each query the tape is automatically erased.

We know that $\mathsf{NL} = \mathsf{coNL}$ by Immerman-Szelepcsényi Theorem, and we have $\mathsf{L^{NL}} = \mathsf{NL}$.
Before the era of Reingold, when nobody knows whether $\mathsf{SL} = \mathsf{L}$, Nisan and Ta-Shma have proved that $\mathsf{SL}$ is closed under complement. They also show that $\mathsf{L^{SL}} = \mathsf{SL}$ in the paper.

In the paper "Directed Planar Reachability Is in Unambiguous Log-Space" by Bourke, Tewari and Vinodchandran, they gave a claim in corollary 4.3 that $\mathsf{L^{UL \cap coUL}} = \mathsf{UL \cap coUL}$. Clearly $\mathsf{UL \cap coUL}$ is closed under complement, but is this equality holds so trivially?

Do we have any easy conditions to decide if $\mathsf{L^O}$ and $\mathsf{O}$ are in fact the same? For easy conditions it means we only have to check some properties about $\mathsf{O}$, then we can decide if they are equal, without using definitions of the classes to prove the inclusion $\mathsf{L^O} \subseteq \mathsf{O}$. 

Another related question would be:


  Do we have any oracle $\mathsf{O}$ such that $\mathsf{L^O} \neq \mathsf{O}$?
For a given choice of complexity classes $\mathsf{C}$ and $\mathsf{O}$, by which we mean sets of languages and nothing else, the class $\mathsf{C}^{\mathsf{O}}$ is not well-defined: we need the definition of $\mathsf{C}$ to determine how oracle queries are defined.  Sometimes even then there could be ambiguities that need to be resolved, or the definition might not provide a reasonable notion of an oracle query at all.  (Attaching oracles to space-bounded complexity classes is one example in the first category, where the relativized classes we obtain are very sensitive to the exact capabilities of the oracle tape and whether we consider that it contributes to our space bounds.)

So, I don't think you can possibly say anything about $\mathsf{C}^{\mathsf{O}}$ without using the definitions of the classes, because if you don't use the definition of $\mathsf{C}$ you aren't working with a well-defined mathematical object.  

If I have misunderstood your question, please clarify what sort of properties of $\mathsf{C}$ and $\mathsf{O}$ you permit to be used.