# Degree of a commutator in a hyperalgebra or enveloping algebra

Consider a semisimple algebraic group $G$ over an algebraically closed field of arbitrary characteristic and let $\bar U(G)$ denote its hyperalgebra (ie, the restricted Hopf dual of the coordinate ring of $G$). In characteristic 0, $\bar U(G)$ is just the enveloping algebra of Lie($G$) but this is not the case in positive characteristic. There is a standard degree filtration $\bar U_{\leq n}(G)$ on $\bar U(G)$. For an element $X \in \bar U(G)$ let us define the degree $\textrm{deg}(X)$ of $X$ to be the minimum $n$ such that $X \in \bar U_{\leq n}(G)$.

It is a straightforward verification that for $X$ of degree n and $Y$ of degree m we have $XY - YX \in \bar U_{\leq n + m -1}(G)$ (cf for example Jantzen's Representations of Algebraic Groups). What I am wondering is when we achieve the maximum degree; eg, when is it the case that $\textrm{deg}(XY - YX) = n + m -1$?

I would not be surprised if this always holds in characteristic 0, ie when we are just considering the enveloping algebra. However, it is clear that this equality does not always hold in positive characteristic since there are zero divisors in $\bar U(G)$ in that case. Nevertheless, I would be happy if there was a weaker statement that could be made, perhaps something like: $\textrm{deg}(XY - YX) = n + m -1$ when $X,Y$ are basis element monomials (with respect to some ordering of Lie($G$)) such that $XY \neq 0$.

EDIT: As Bruce Westbury points out below, a lot of what I wrote above is silly and I've struck it out. Perhaps my question still has merit though.

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Isn't $X=Y$ a counterexample to both your proposals? –  Bruce Westbury Mar 29 '13 at 21:23
Oh yikes, yes. That's a good point. I obviously haven't thought this through well enough. –  Chuck Hague Mar 29 '13 at 21:31
The associated graded is a Poisson algebra and the Poisson bracket is graded there. If $\sigma(X)$ denotes the symbol of $X$, then I suppose your question is when ${\sigma(X), \sigma(Y)}\neq 0$. I'm not sure there's more you can say. –  Reimundo Heluani Mar 30 '13 at 22:09