13
$\begingroup$

Given a field $k$ of characteristic $0$. For every $k$-bialgebra $A$, let $\mathrm{Prim} A$ denote the $k$-vector subspace of $A$ consisting of all primitive elements of $A$.

What conditions can we put on two $k$-bialgebras $A$ and $B$ to ensure that $\mathrm{Prim}\left(A\otimes B\right) = k\otimes \left(\mathrm{Prim}A\right) + \left(\mathrm{Prim} B\right)\otimes k$ ?

I haven't given this much thought, but I am not good at constructing counterexamples and it seems pointless to try proving anything here before having an "upper bound" on how far we can go. The only results I know about is that $k\otimes \left(\mathrm{Prim}A\right) + \left(\mathrm{Prim} B\right) \otimes k \subseteq \mathrm{Prim}\left(A\otimes B\right)$ always holds (for trivial reasons), and that if $A$ and $B$ are two connected graded cocommutative bialgebras, then $\mathrm{Prim}\left(A\otimes B\right) = k\otimes \left(\mathrm{Prim}A\right) + \left(\mathrm{Prim} B\right)\otimes k$ (as a consequence of Cartier-Milnor-Moore and Poincaré-Birkhoff-Witt).

It sounds rather natural to assume $A$ and $B$ to be cocommutative (after all, $\mathrm{Prim} A$ is always $=\mathrm{Prim}\left(A^c\right)$, where $A^c$ the greatest cocommutative sub-bialgebra of $A$), but I am not sure whether we can WLOG assume this to be so (maybe $\left(A\otimes B\right)^c$ is greater than $A^c\otimes B^c$ ?).

$\endgroup$
0

3 Answers 3

10
$\begingroup$

Well first when we restrict to the case when $A, B$ are filtered (see Bourbaki for example), in this case $\log_{*}$ always converges at $Id$ (as $Id=e+I_+$, $e$ being the unit for the convolution, it suffices to remark that $I_+^{*N}(h)=0$ for $N=N(h)$ large enough).

Now, in the general case, you can adapt the following computation to the algebra $H(A\otimes B)$ generated by the primitive elements of $A\otimes B$ where the series of $\log_{*_{12}}$ always converges.

For clarity, I note $A=A_1,B=A_2$ and and $e_i=1_{A_i}\circ \epsilon_i$.

Then $$ \log_{*_{12}}(I_1\otimes I_2)=\log_{*_{12}}((I_1\otimes e_2)*_{12}(e_1\otimes I_2))= $$ $$ \log_{*_{12}}(I_1\otimes e_2)+\log_{*_{12}}(e_1\otimes I_2) $$
as the two terms $(I_1\otimes e_2), (e_1\otimes I_2)$ commute. Now $\log_{*_{12}}(I_1\otimes e_2)=\log_{*_{1}}(I_1)\otimes e_2$ and $\log_{*_{12}}(e_1\otimes I_2)=e_1\otimes\log_{*_{2}}(I_2)$.

Which, in view of $\left(\mathrm{Prim}A_1\right) \otimes k + k \otimes \left(\mathrm{Prim} A_2\right) \subseteq \mathrm{Prim}\left(A_1\otimes A_2\right)$, proves that $Prim(A_1\otimes A_2)=Prim(A_1)\otimes k+k\otimes Prim(A_2)$.

Which does your job.

Addition : To answer your first question, $I_1$ and $e_2$ are morphisms of bialgebras so $I_1\otimes e_2$ maps $Prim(A_1\otimes A_2)$ into $Prim(A_1\otimes A_2)$ and then $H$ into $H$ (in fact the image of $H(A_1\otimes A_2)$ is a subbialgebra of $H(A_1)\otimes k.1_{A_2}$).

To answer the second point. For a bialgebra let us denote $I^+=Id-e$ (the complement projector of $e$) and $H(?)$ the subalgebra generated by the primitive elements. One has, with the morphism of bialgebras $$ (I_1\otimes e_2) : H(A_1\otimes A_2) \rightarrow H(A_1)\otimes k.1_{A_2} $$ the intertwining $$ (I_1^+\otimes e_2)\circ (I_1\otimes e_2)=(I_1\otimes e_2)\circ (I_1\otimes I_2)^+ $$ so, using series, we get $$ (\log_{*_1}(I_1)\otimes e_2)\circ (I_1\otimes e_2)=(I_1\otimes e_2)\circ \log_{*_{12}}(I_1\otimes I_2) $$ This is because, as a general principle, the intertwining intertwines the convolution. Let, $$ \begin{matrix} A & \stackrel{\varphi}{\longrightarrow} & B \cr \downarrow && \downarrow \cr A & \stackrel{\varphi}{\longrightarrow} & B \end{matrix} $$ with $\varphi$ a morphism of bialgebras and the down arrows $f,g$ such that $g\varphi=\varphi f$. Then, if the bialgebras are generated by primitive elements, if $f(1_A)=0,g(1_B)=0$ and if $S\in k[[x]]$ is a series, we have $S(g)\varphi=\varphi S(f)$. This is not difficult and argued in details (in particular the notion of summability and substitution) in my paper.

In conclusion, I think that $$ Prim(A_1\otimes A_2)=Prim(A_1)\otimes k.1_{A_2}+k.1_{A_1}\otimes Prim(A_2) $$ is true in full generality. One even does not have to suppose that $k$ is a field, only $\mathbb{Q}\subseteq k$ seems to be needed.

Do not hesitate to question and comment if something is unclear or wrong.

Regards

$\endgroup$
22
  • $\begingroup$ This is a nice argument handling a multitude of cases; thanks a lot. I am not sure about its generality, though. How do we know that $I_1\otimes e_2$ is well-defined as an endomorphism of the subalgebra of $A\otimes B$ generated by the primitives? How do we know that $\log_{*_1}\left(I_1\right) \otimes e_2$ projects onto $Prim\left(A_1\right)\otimes k$ (without having $A_1$ cocommutative to begin with)? $\endgroup$ Feb 22, 2013 at 18:11
  • $\begingroup$ The idea is to restrict to the algebra $H$ generated by the primitive elements of $A\otimes B$ (I do not now is this has a name in general). This algebra is automativally cocommutative. For details, you can have a look at my recent paper hal.archives-ouvertes.fr/… In order to know the degree of generality, the point is to examine in detail the "domains of convergence" of the elements of the aforementioned computation. For example, $\log_{∗_1}(I_1)\otimes e_2$ projects onto $Prim(A_1)\otimes k$, from $H$. $\endgroup$ Feb 22, 2013 at 18:40
  • $\begingroup$ Ah!! Very nice proof. (You forgot an $I_1$ in $(\log_{*_1}\otimes e_2)\circ (I_1\otimes e_2)=(I_1\otimes e_2)\circ \log_{*_{12}}(I_1\otimes I_2)$.) The paper you are talking about, is it hal.archives-ouvertes.fr/hal-00793118 ? Because the link doesn't work for me. $\endgroup$ Feb 23, 2013 at 5:38
  • $\begingroup$ (Also, the statement "in fact the image of [...]") is slightly wrong.) $\endgroup$ Feb 23, 2013 at 5:41
  • $\begingroup$ Yes it is this paper. As we aim at coefficients which are rings of function spaces, we had to revisit the theorem of Cartier-Milnor-Moore without PBW (because it is not assumed that Prim(B) has a linear basis). It turns out that only the fact that the ring contains the rationals is needed. $\endgroup$ Feb 23, 2013 at 8:29
11
$\begingroup$

Hi Darij, Hi Gérard,

Here is an elementary proof of $Prim(A \otimes B)=Prim(A)\otimes 1_B+1_A \otimes Prim(B)$, using the counities $\epsilon_A$ and $\epsilon_B$.

Let $X$ be a primitive element of $A \otimes B$. It can be written as :

(1) $X=\lambda 1_A \otimes 1_B+x\otimes 1_B+1_A \otimes y+ \sum x_i \otimes y_i$,

with $\epsilon_A(x)=\epsilon_B(y)=\epsilon_A(x_i)=\epsilon_B(y_i)=0$. Then $\Delta(X)=\lambda 1_A \otimes 1_B\otimes 1_A \otimes 1_B+x^{(1)}\otimes 1_B \otimes x^{(2)}\otimes 1_B+1_A \otimes y^{(1)}\otimes 1_A\otimes y^{(2)}$ $+\sum x_i^{(1)}\otimes y_i^{(1)}\otimes x_i^{(2)}\otimes y_i^{(2)}$ (we are using Sweedler notation, with $z^{(1)} \otimes z^{(2)}$ standing for $\Delta(z)$). Compared with $\Delta(X)=X\otimes 1_A \otimes 1_B+1_A \otimes 1_B \otimes X$, this becomes

(2) $\lambda 1_A \otimes 1_B\otimes 1_A \otimes 1_B+x^{(1)}\otimes 1_B \otimes x^{(2)}\otimes 1_B+1_A \otimes y^{(1)}\otimes 1_A\otimes y^{(2)}$ $+\sum x_i^{(1)}\otimes y_i^{(1)}\otimes x_i^{(2)}\otimes y_i^{(2)}$ $=X\otimes 1_A \otimes 1_B+1_A \otimes 1_B \otimes X$.

Applying $Id \otimes \epsilon_B \otimes \epsilon_A\otimes Id$ gives: $\lambda 1_A \otimes 1_B+x \otimes 1_B+1_A \otimes y+\sum x_i \otimes y_i$ $=\lambda 1_A \otimes 1_B+x \otimes 1_B+\lambda 1_A\otimes 1_B+1_A \otimes y.$ So $\lambda=0$ and $\sum x_i \otimes y_i=0$ (since $\epsilon_A \otimes \epsilon_B$ annihilates all terms but the $\lambda 1_A \otimes 1_B$ ones). Hence (2) simplifies to

(3) $x^{(1)}\otimes 1_B \otimes x^{(2)}\otimes 1_B+1_A \otimes y^{(1)}\otimes 1_A\otimes y^{(2)}$ $=X\otimes 1_A \otimes 1_B+1_A \otimes 1_B \otimes X$,

and (1) simplifies to $X = x \otimes 1_B + 1_A \otimes y$.

Applying $\epsilon_A\otimes Id \otimes \epsilon_A \otimes Id$ to (3) gives: $y^{(1)}\otimes y^{(2)}=y\otimes 1_B+1_B \otimes y$. So $y$ is primitive. Applying $Id \otimes \epsilon_B\otimes Id \otimes \epsilon_B$ to (3) gives: $x^{(1)}\otimes x^{(2)}= x\otimes 1_A+1_A\otimes x$. So $x$ is primitive. Finally, $X\in Prim(A)\otimes 1_B+1_A\otimes Prim(B)$.

$\endgroup$
2
  • 1
    $\begingroup$ Hi Loïc, welcome to MO and thanks for this beautiful and astonishingly elementary proof! (Sorry I can't accept this, since I don't want to unaccept Gérard's theoretical perspective.) I took the liberty to edit it to make it (IMHO) clearer and more structured. I hope I didn't adulterate anything. If you don't like my edits, feel free to go on mathoverflow.net/revisions/126362/list , scroll down to version 1 and click "rollback". $\endgroup$ Apr 3, 2013 at 16:50
  • 1
    $\begingroup$ Hi Loïc and Darij, I like very much Loïc's proof, not only it is elementary but it proves the claim whatever the characteristic. As a matter of fact, my machinery was devoted to prove CQMM without PBW (which is useful when one has $\Q$-algebras) and not really aimed to Darij's question. Cheers $\endgroup$ Apr 4, 2013 at 16:37
4
$\begingroup$

There is a short proof in Proposition 2.12 of my paper http://arxiv.org/pdf/1502.02150v1 which works over an arbitrary base ring.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.