# Independence of characters with respect to polynomials

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 $$x\in X$$ $$\beta.(\alpha x)^*-\beta.\epsilon=0$$

• 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
• 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