# Lower bound on the degree of a product of elements in a hyperalgebra/enveloping algebra

Background: Fix a linear algebraic group $G$ over an algebraically closed field $k$ of arbitrary characteristic and let $B \subseteq G$ be a Borel subgroup with unipotent radical $N$. Let $\Delta^+$ denote the positive roots in a root system of a torus of $G$. Then we have the hyperalgebra $\bar U(N)$ of $N$ which is generated as a $k$-algebra by the divided-power elements $E_\beta^{(n)}$ for $\beta \in \Delta^+$ and $n \geq 0$. (I would also be happy to just consider the characteristic 0 case, where $\bar U(N)$ is just the enveloping algebra of Lie($N$) and is also generated by the non-divided power elements $E_\beta^n$).

Fix an ordering $\beta_1, \ldots, \beta_k$ of $\Delta^+$. Then the elements $E_{\beta_1}^{(n_1)} \cdots E_{\beta_k}^{(n_k)}$, for $n_1, \ldots, n_k \geq 0$, form a vector space basis of $\bar U(N)$. This gives a natural vector space grading on $\bar U(N)$ that I will denote by $\bar U(N)_n$ for $n \geq 0$. It is definitely not the case that $\bar U(N)_n$ is a multiplicative grading on $\bar U(N)$, as is easy to see. The following is, however, a well-known fact (by the PBW theorem): $$\bar U(N)_n \cdot \bar U(N)_m \subseteq \bar U(N)_{ \leq m+n } \,\,\, .$$ This gives an upper bound on the degree of a product of elements.

Question: I am wondering if anyone has considered the question of a lower bound on the degree of a product of elements in $\bar U(N)$. That is, I would like to see a fact on the following kind: $$\bar U(N)_n \cdot \bar U(N)_m \subseteq \bar U(N)_{ \geq f(m,n) } \,\,\, ,$$ where $f(m,n)$ is some reasonable function of $m$ and $n$ that hopefully gives a sharp bound.

(Note that whereas the PBW theorem gives an upper bound on degrees for any hyperalgebra/enveloping algebra, I expect that this lower bound, if it exists, will very much depend on the fact that $N$ is a unipotent algebraic group.)

-

I don't have a full answer, but the following calculation may be useful.

Consider the example $G = SL_3$ so that $N$ is the Heisenberg group on three generators and in characteristic zero, $\bar{U}(N)$ is the enveloping algebra of the three-dimensional Heisenberg Lie algebra. Thus

$\bar{U}(N) = k[x,z][y ; z \frac{d}{dx}]$

is a skew polynomial algebra. This notation means that every element in $\bar{U}(N)$ can be written uniquely in the form $\sum_{i=0}^r f_i y^i$ for some $f_i \in k[x,z]$ and $y$ satisfies the commutation relation

$ad(y)(f) := y f - f y = z \frac{df}{dx}$

for any $f \in k[x,z]$. Note that $z$ is a central element in $\bar{U}(N)$.

Now consider the element $y^a \cdot x^b$ for any $a,b \in \mathbb{N}$. Let us rewrite it in standard form as follows:

$y^a x^b = l_y^a(x^b) = (r_y + ad(y))^a(x^b) =$

$\left(\sum_{i=0}^a\binom{a}{i} r_y^{a-i} ad(y)^i \right)(x^b) = \sum_{i=0}^a \binom{a}{i} \frac{b!}{(b-i)!}z^i x^{b-i} y^{a-i}$.

Here $l_y$ and $r_y$ are operators of left-multiplication by $y$ and right-multiplication by $y$, respectively. Note that they commute, so $r_y$ and $ad(y) = l_y - r_y$ also commute, so the application of the binomial theorem is legal.

So in this case you can take $f(a,b) = a + b - \max(a,b)$.

-
Ah, that is very interesting! I wonder if this works in type A in general. –  Chuck Hague Sep 21 '11 at 15:42
I wonder, too. I guess the next step would be to try type $A_3$ and see what happens. Good luck! –  Konstantin Ardakov Sep 22 '11 at 8:46