**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.)



Lower bound on the degree of a product of elements in a hyperalgebra/enveloping algebra
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 $f(a,b) = a + b - \max(a,b)$.


