Primitive elements of a tensor product of bialgebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:56:00Z http://mathoverflow.net/feeds/question/86230 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86230/primitive-elements-of-a-tensor-product-of-bialgebras Primitive elements of a tensor product of bialgebras darij grinberg 2012-01-20T16:35:11Z 2013-04-03T17:09:55Z <p>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 $A$.</p> <p>What conditions can we put on two $k$-bialgebras $A$ and $B$ to ensure that $\mathrm{Prim}\left(A\otimes B\right) = k\otimes \left(\mathrm{Prim}A\right) + \left(\mathrm{Prim} B\right)\otimes k$ ?</p> <p>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).</p> <p>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 $A^c\otimes B^c$ ?).</p> http://mathoverflow.net/questions/86230/primitive-elements-of-a-tensor-product-of-bialgebras/122606#122606 Answer by Duchamp Gérard H. E. for Primitive elements of a tensor product of bialgebras Duchamp Gérard H. E. 2013-02-22T06:16:29Z 2013-03-05T06:22:37Z <p>Dear Darij, </p> <p>Well first when we restrict to the case when $A, B$ are filtered (see Bourbaki for example), in this case $\log_{*}$ always converges at $Id$. </p> <p>Now, in the general case, it seems to me that 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.</p> <p>For clarity, I note $A=A_1,B=A_2$ and and $e_i=1_{A_i}\circ \epsilon_i$.</p> <p>Then $$\log_{*_{12}}(I_1\otimes I_2)=\log_{*_{12}}((I_1\otimes e_2)*_{12}(e_1\otimes I_2))=$$ $$\log_{*_{12}}(I_1\otimes e_2)+\log_{*_{12}}(e_1\otimes I_2)$$<br> 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 $\log_{*_{12}}(e_1\otimes I_2)=e_1\otimes\log_{*_{2}}(I_2)$.</p> <p>Which proves that $Prim(A_1\otimes A_2)=Prim(A_1)\otimes k+k\otimes Prim(A_2)$. </p> <p>This seems to do your job. I will check this more carefully in the train today. </p> <p>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 $H(A_1)\otimes k.1_{A_2}$). </p> <p>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, $$\begin{matrix} A &amp; \stackrel{\varphi}{\longrightarrow} &amp; B \cr \downarrow &amp;&amp; \downarrow \cr A &amp; \stackrel{\varphi}{\longrightarrow} &amp; B \end{matrix}$$ 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 within a series is correctly set there, I hope !) in my paper.</p> <p>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 $\mathbb{Q}\in k$ seems to be needed. </p> <p>Do not hesitate to question and comment if something is unclear or wrong. </p> <p>Regards</p> http://mathoverflow.net/questions/86230/primitive-elements-of-a-tensor-product-of-bialgebras/126362#126362 Answer by Loic Foissy for Primitive elements of a tensor product of bialgebras Loic Foissy 2013-04-03T08:27:22Z 2013-04-03T17:09:55Z <p>Hi Darij, Hi Gérard,</p> <p>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 $\epsilon_B$.</p> <p>Let $X$ be a primitive element of $A \otimes B$. It can be written as :</p> <p><strong>(1)</strong> $X=\lambda 1_A \otimes 1_B+x\otimes 1_B+1_A \otimes y+ \sum x_i \otimes y_i$,</p> <p>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</p> <p><strong>(2)</strong> $\lambda 1_A \otimes 1_B\otimes 1_A \otimes 1_B+x^{(1)}\otimes 1_B \otimes x^{(2)}\otimes 1_B+1_A \otimes y^{(1)}\otimes 1_A\otimes y^{(2)}$ $+\sum x_i^{(1)}\otimes y_i^{(1)}\otimes x_i^{(2)}\otimes y_i^{(2)}$ $=X\otimes 1_A \otimes 1_B+1_A \otimes 1_B \otimes X$.</p> <p>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 <strong>(2)</strong> simplifies to</p> <p><strong>(3)</strong> $x^{(1)}\otimes 1_B \otimes x^{(2)}\otimes 1_B+1_A \otimes y^{(1)}\otimes 1_A\otimes y^{(2)}$ $=X\otimes 1_A \otimes 1_B+1_A \otimes 1_B \otimes X$,</p> <p>and <strong>(1)</strong> simplifies to $X = x \otimes 1_B + 1_A \otimes y$.</p> <p>Applying $\epsilon_A\otimes Id \otimes \epsilon_A \otimes Id$ to <strong>(3)</strong> 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 <strong>(3)</strong> gives: $x^{(1)}\otimes x^{(2)}= x\otimes 1_A+1_A\otimes x$. So $x$ is primitive. Finally, $X\in Prim(A)\otimes 1_B+1_A\otimes Prim(B)$.</p>