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 $A^c\otimes B^c$ ?).


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.


  • $\begingroup$ 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)? $\endgroup$ – darij grinberg Feb 22 '13 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 $H$. $\endgroup$ – Duchamp Gérard H. E. Feb 22 '13 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. $\endgroup$ – darij grinberg Feb 23 '13 at 5:38
  • $\begingroup$ (Also, the statement "in fact the image of [...]") is slightly wrong.) $\endgroup$ – darij grinberg Feb 23 '13 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. $\endgroup$ – Duchamp Gérard H. E. Feb 23 '13 at 8:29

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, $X\in Prim(A)\otimes 1_B+1_A\otimes Prim(B)$.

  • 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". $\endgroup$ – darij grinberg Apr 3 '13 at 16:50
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    $\begingroup$ 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 $\endgroup$ – Duchamp Gérard H. E. Apr 4 '13 at 16:37

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.


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