I came across the following property :

Let $\mathfrak{g}$ be a Lie algebra over a ring $k$ without zero divisors,
$\mathcal{U}=\mathcal{U}(\mathfrak{g})$ be its enveloping algebra. As such, $\mathcal{U}$ is a Hopf algebra and $\epsilon$, its counit, is the only character of $\mathcal{U}\rightarrow k$ which vanishes on $\mathfrak{g}$.

Set $\mathcal{U}_+=ker(\epsilon)$. We build the following filtrations ($N\geq 0$) $$ \mathcal{U}_N=\mathcal{U}_+^{N}=\underbrace{\mathcal{U}_+.\ldots.\mathcal{U}_+}_{N\mbox{ times}}\qquad (1) $$ (in fact $\mathcal{U}_0=\mathcal{U}, \mathcal{U}_{N+1}=\mathcal{U}.\mathcal{U}_N$) and, for $N\geq -1$ $$ \mathcal{U}^*_N=\mathcal{U}_{N+1}^\perp=\{f\in \mathcal{U}^*|(\forall u\in \mathcal{U}_{N+1})(f(u)=0)\}\qquad (2) $$ the first one is decreasing and the second one increasing (in particular $\mathcal{U}^*_{-1}=\{0\},\ \mathcal{U}^*_{0}=k.\epsilon$).

One shows easily that, for $p,q\geq 0$ (with $\diamond$ as the convolution product) $$ \mathcal{U}^*_p\diamond \mathcal{U}^*_q\subset \mathcal{U}^*_{p+q} $$ so that $\mathcal{U}^*_\infty=\cup_{n\geq 0}\mathcal{U}^*_n$ is a convolution subalgebra of $\mathcal{U}^*$.

Now, we can state the

Theorem A : The set of characters of $(\mathcal{U},.,1_{\mathcal{U}})$ is linearly free w.r.t. $\mathcal{U}^*_\infty$.

Remarks i) $\mathcal{U}^*_\infty$ is a commutative $k$-algebra.

ii) The title comes from the fact that, with $(k\langle X\rangle,conc,1)$ (non commutative polynomials), $k$ a $\mathbb{Q}$-algebra (without zero divisors) and one of the usual comultiplications (with $\Delta_+$ cocommutative and nilpotent) one takes $\mathfrak{g}$ as the space of primitive elements, $\mathcal{U}^*=k\langle\langle X\rangle\rangle$ (series) and $\mathcal{U}^*_\infty=k\langle X\rangle$.

As I am not a specialist, my questions are the following

Q1) Are the filtrations (1) and (2) known ? if yes, what is their name ?

Q2) Is the theorem above known ? if yes reference(s) would be appreciated.

Late edit I was asked by colleagues to provide a written proof which I happily insert there.

Proof (of theorem A) Let $\Xi(\mathcal{U})$ be the set of characters of $(\mathcal{U},.,1_{\mathcal{U}})$. For non zero $\beta\in \mathcal{U}^*_\infty$, we set $\deg(\beta)=p$ as being the unique index $p$ such that $\beta \in \mathcal{U}^*_p$ and $\beta \notin \mathcal{U}^*_{p-1}$. We consider linear relations between characters of the form $$ \sum_{i\in I} \beta_i\diamond f_i=0\ ;\ \beta_i\in \mathcal{U}^*_\infty\setminus \{0\}\mbox{ and } f_i\in \Xi(\mathcal{U}) \qquad (3) $$ Either all of them are trivial ($I=\emptyset$), or there are non trivial ones ($I\not=\emptyset$) among those we choose one with $\sum_{i\in I}\deg(\beta_i)$ minimal. WLOG we can consider that $(\exists i_0\in I)(f_{i_0}=\epsilon)$ (all characters being invertible we can multiply (3) by $f_{i_0}^{-1}$ for the law $\diamond$). Then the choosen relation becomes $$ \beta_{i_0}+\sum_{i\in I\setminus \{i_0\}} \beta_i\diamond f_i=0\qquad (4) $$ we now use the shift representation of $\mathcal{U}$ in $\mathcal{U}^*$ defined, for $\varphi\in \mathcal{U}^*,\ u,m\in \mathcal{U}$ by $$ \langle\varphi\triangleleft u| m\rangle=\langle\varphi| um\rangle $$ Remark that $\mathfrak{g}$ acts on $\mathcal{U}^*$ by derivations i.e. for $a,b\in \mathcal{U}^*$, $g\in \mathfrak{g}$ $$ (a\diamond b)\triangleleft g=(a\triangleleft g)\diamond b+ a\diamond (b\triangleleft g) $$ Observing that $I=\{i_0\}$ is impossible, we pick $i_1\in I\setminus \{i_0\}$ and $g\in \mathfrak{g}$ such that $\langle f_{i_1}|g\rangle\not=0$ (this is possible because $f_{i_1}\not=\epsilon$). We now shift (4) on the right by $g$ and get $$ \beta_{i_0}\triangleleft g+\sum_{i\in I\setminus \{i_0\}} (\beta_i\triangleleft g+ \beta_i\langle f_i|g\rangle)\diamond f_i=0 \qquad (5) $$ Now by $\deg(\beta_{i_0}\triangleleft g)< \deg(\beta_{i_0})$, $\deg(\beta_i\triangleleft g+ \beta_i\langle f_i|g\rangle)\leq \deg(\beta_i)$ and $(\beta_{i_1}\triangleleft g+ \beta_{i_1}\langle f_{i_1}|g\rangle)\not=0$ we get a contradiction w.r.t. minimality.

Remarks i) For $\mathfrak{g}$ simple, we have $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ and all the filtrations are stationary.
ii) The conclusion does not hold true if $k$ has zero divisors as the characters are no longer independent even w.r.t. $k$. For example with $X\not=\emptyset$ an alphabet, $\alpha.\beta=0$ with $\alpha,\beta\in k\setminus \{0\}$, we have for all $x\in X$ $$ \beta.(\alpha x)^*-\beta.\epsilon=0 $$

  • 1
    Interesting! My intuition says that this should follow from known results at least when $k$ is a field of characteristic $0$. Isn't $\mathcal{U}^\ast_\infty$ a primitively generated Hopf subalgebra of the graded dual of $\mathcal{U}$, whereas the characters are grouplike elements in the completion of said graded dual? I think grouplike elements should be linearly independent over a primitively generated Hopf subalgebra under suitably general conditions. – darij grinberg Sep 11 at 16:18
  • Actually, of course, $\mathcal{U}^\ast_\infty$ is not primitively generated, but it's a connected graded Hopf algebra, and that alone might suffice... – darij grinberg Sep 11 at 16:38
  • (+1) for interaction. I have two remarks (a) the coproduct $\Delta$ need not be graded for the theorem to be true (b) even in this case $\mathcal{U}^*_\infty$ is not cocommutative in general. Is it a problem ? – Duchamp Gérard H. E. Sep 11 at 16:46
  • 1
    Actually, I think I need the Hopf algebra to be commutative (or at least the subalgebra to be part of its center?), and $\mathcal{U}^\ast_\infty$ fits that bill perfectly. But I need to think about whether my approach requires characteristic $0$. I take it that yours doesn't? – darij grinberg Sep 11 at 17:01
  • Ok, I understand what you have in mind. But, yes, as the ring mustn't have zero divisors, characteristic must be prime or 0. But is this theorem known (even in a particular case), do you have a name to suggest ? – Duchamp Gérard H. E. Sep 11 at 18:00

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