Regularity assumption in the simple trace formula In the simple trace formula of Deligne Kazhdan one assumes that the test function is supported at the elliptic regular elements at one place and is a supercusp form at another place.  Why can't one just assume that it is supported on elliptic elements (not necessarily regular)?
More specifically, I don't see how the regularity assumption is used in the proofs of the simple trace formula.  I understand that one can get around it using Arthur's work.  
 A: If you don't make that assumption, you also will get "one-dimensional" representations. For example, consider $G=GL(2)$ and $F$ global function field, assume at two place pseudo matrix coefficient for the Steinberg representation and let the remaining local test functions be the characteristic function of $F_v^\times GL_2(o_v)$. This gives you then $(-1)^2$ for the the trivial representation. The trace formula is still called simple then (see Gelbart-Jacquet), not really more complicated than the one of Flicker-Kazdhan, but you get for each automorphic 1-dim'l representation $\mu \circ \det$ for $\mu^2 =\chi$ one extra contribution.
A: Probably the OP has figured this out by now, but for posterity let me explain (in the general setting of simple trace formulas, rather than the specific Deligne-Kazhdan case):
A trace formula $I(f) = J(f)$ is, crudely, an identity of a geometric distribution $I(f)$ with a spectral distribution $J(f)$.  Here $f$ is a test function that will be varied and ideally both $I(f) = \prod I(f_v)$ and $J(f) = \prod J(f_v)$ are factorizable.  
In simple trace formulas, one typically chooses $f_{v_1}$ at some place $v_1$.  with some regular or elliptic support so that globally only regular or elliptic terms contribute to the geometric distribution $I(f)$.  At another place $v_2$, one chooses $f_{v_2}$ to be a supercusp form so that $J(f)$ globally only involves only cuspidal representations.  These simplifications do two major things for you:


*

*They simplify convergence issues of $I(f)$ and $J(f)$.

*They make comparisons between trace formulas on different groups easier (you only need to compare regular or elliptic geometric terms, and then you don't need to analyze continuous spectrum to deduce an equality of cuspidal spectra). 
As I recall, a simple trace formula is explained in Gelbart's Lectures on the Arthur-Selberg Trace Formula.
