Do stunted exponential series give projections of a cocommutative bialgebra on its coradical filtration? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T15:51:28Z http://mathoverflow.net/feeds/question/83654 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83654/do-stunted-exponential-series-give-projections-of-a-cocommutative-bialgebra-on-it Do stunted exponential series give projections of a cocommutative bialgebra on its coradical filtration? darij grinberg 2011-12-16T19:24:35Z 2012-08-07T19:55:25Z <p>Let $k$ be a field of characteristic $0$.</p> <p>Let $H$ be a cocommutative connected filtered bialgebra over $k$. ("Connected" means that the counit, restricted to the $0$-th part of the filtration, is an isomorphism. Since $H$ is a connected filtered bialgebra, it is automatically a Hopf algebra.)</p> <p>The convolution algebra $\mathcal L\left(H,H\right)$ of all $k$-linear maps $H\to H$ is local; its maximal ideal is the ideal $\mathfrak g\left(H,H\right)$ of all $k$-linear maps $H\to H$ sending $1$ to $0$. Denoting by $e$ the unity of the algebra $\mathcal L\left(H,H\right)$ (this $e$ is the map which sends every $x\in H$ to $\varepsilon\left(x\right)\cdot 1_H$), we can define an exponential mapping $\exp^{\ast}: \mathfrak g\left(H,H\right)\to e+\mathfrak g\left(H,H\right)$ and a logarithm mapping $\log^{\ast}: e+\mathfrak g\left(H,H\right)\to \mathfrak g\left(H,H\right)$ by the formulae</p> <p>$\exp^{\ast}\left(f\right)=\sum\limits_{i\geq 0}\dfrac{f^{\ast i}}{i!}$</p> <p>and</p> <p>$\log^{\ast}\left(e+f\right)=\sum\limits_{i\geq 1}\dfrac{\left(-1\right)^{i-1}}{i}f^{\ast i}$,</p> <p>where $f^{\ast i}$ denotes the $i$-th power of $f$ in the convolution algebra $\mathcal L\left(H,H\right)$. Both of these power series converge pointwise because $H$ is filtered. It is easy to see (and annoying to write up...) that $\exp^{\ast}\circ\log^{\ast}=\mathrm{id}$ and $\log^{\ast}\circ\exp^{\ast}=\mathrm{id}$.</p> <p>So far we have been doing general stuff, not using the cocommutativity of $H$, and not using the fact that $H=H$ either (i. e., we could just as well have done all the above in $\mathcal L\left(C,A\right)$ with $C$ being a connected filtered $k$-coalgebra, and $A$ a $k$-algebra). But in the case of cocommutative $H$, we have something nice:</p> <p><strong>Theorem (Patras, Reutenauer, Garsia?).</strong> The map $\log^{\ast}\mathrm{id}$ is a projection of $H$ onto the subspace consisting of all primitive elements of $H$.</p> <p>I can prove this using lots and lots of computations and an umbral-calculus trick (I can send the proof to anyone interested, but it's boredom guaranteed). The <a href="http://lacim.uqam.ca/~christo/idempotents.pdf" rel="nofollow">paper where I know this theorem from</a> gives three references I cannot get any quick use of ([R2] [GR] [P1] as referenced in the Example on page 4; two of these are Solomon descent algebra literature, and the third is a book by Reutenauer that I'll one day read from one end to the other, but not now...). There is a shortcut through finite-dimensionality and dualization, but I don't like this kind of approaches as they use the fact of $k$ being a field (whereas the Theorem holds for $k$ being any commutative ring with $1$).</p> <p>While I would be glad to hear a nicer proof of the Theorem, here is the</p> <p><strong>Actual question:</strong> For every $s\in\mathbb N$, we can define a mapping $\exp_s^{\ast}: \mathfrak g\left(H,H\right)\to e+\mathfrak g\left(H,H\right)$ by the formula</p> <p>$\exp_s^{\ast}\left(f\right)=\sum\limits_{i=0}^s \dfrac{f^{\ast i}}{i!}$.</p> <p>This is just the exponential series, stopped at $i=s$ and applied to $f$.</p> <p>Now, I conjecture that $\log^{\ast}\left(\exp_s^{\ast}\left(\mathrm{id}-e\right)\right)$ is a projection of $H$ onto the subspace $C^{i+}$, where $\left(C^{\ell}\right)_{\ell\geq 0}$ is the coradical filtration of the coalgebra $H$ (note that $C^0 = H^0 = k\cdot 1$, and that $C^1 = k\cdot 1 + \left(\text{subspace of primitive elements of }H\right)$), and $S^+$ denotes $S\cap \mathrm{Ker}\varepsilon$ for every subspace $S$ of $H$.</p> <p>The only evidence I have for this is that it's true for $s=0$ (obviously), for $s=1$ (this is the above Theorem) and for $s=\infty$ (not exactly a natural number, but a reasonable value to plug in the conjecture). Don't these data cry for a deeper meaning?...</p> http://mathoverflow.net/questions/83654/do-stunted-exponential-series-give-projections-of-a-cocommutative-bialgebra-on-it/83902#83902 Answer by darij grinberg for Do stunted exponential series give projections of a cocommutative bialgebra on its coradical filtration? darij grinberg 2011-12-19T21:50:36Z 2012-07-21T13:24:38Z <p>Sorry, people. The conjecture is false. For a counterexample, take $H=k\left[x\right]$ (the usual polynomial algebra with shuffle comultiplication) and $s=2$. The image of $x^4$ under $\log^{\ast}\left(\exp^{\ast}_2\left(\mathrm{id}-e\right)\right)$ will be $-2x^4$, and this is not in the second term of the coradical filtration.</p> <p>I was fooled by the fact that (in general, even if $H$ is not cocommutative!) the map $\log^{\ast}\left(\exp^{\ast}_s\left(\mathrm{id}-e\right)\right)$ restricts to the identity on $C^s$. This fact, though, is a triviality, since $C^s$ is a subcoalgebra of $C$ on which $\exp^{\ast}$ and $\exp^{\ast}_s$ are one and the same thing. (This argument doesn't always work over a ring, but this isn't hard to repair. Instead of restricting maps to $C^s$, consider maps modulo maps which vanish on $C^s$.)</p> <p>A projection from a vector space $V$ onto a subspace $U$ is characterized by two properties: that it restricts to the identity on $U$, and that it maps everything onto $U$. The first of these properties being satisfied, I got overly optimistic that the second one couldn't be that much false. Turned out that it is. (Also, I got overly optimistic by the fact that the conjecture holds for $s=2$ on the third graded component of the tensor Hopf algebra; but the fourth graded component gives a contradiction.)</p> <p>I am left wondering whether there are some reasonable algebraically-defined projections onto $C^s$ for $s\neq 0,1,\infty$. We shouldn't necessarily look for variations on the Eulerian idempotent; there might also be something similar to the Dynkin idempotent (or the Klyachko one, but I haven't even started to learn about that).</p> http://mathoverflow.net/questions/83654/do-stunted-exponential-series-give-projections-of-a-cocommutative-bialgebra-on-it/102809#102809 Answer by Duchamp Gérard H. E. for Do stunted exponential series give projections of a cocommutative bialgebra on its coradical filtration? Duchamp Gérard H. E. 2012-07-21T12:33:53Z 2012-08-07T19:55:25Z <blockquote> <blockquote> <p>While I would be glad to hear a nicer proof of the Theorem,</p> </blockquote> </blockquote> <p>Sorry for late answer. You can get a rather quick proof of this theorem by remarking that, in your situation, when H is cocommutative, the comultiplication is a morphism of bi-algebras. </p> http://mathoverflow.net/questions/83654/do-stunted-exponential-series-give-projections-of-a-cocommutative-bialgebra-on-it/102812#102812 Answer by Duchamp Gérard H. E. for Do stunted exponential series give projections of a cocommutative bialgebra on its coradical filtration? Duchamp Gérard H. E. 2012-07-21T13:02:22Z 2012-07-24T04:01:21Z <p>OK, The argument goes as follows : you do two or three commutative diagrams showing that morphism of bi-algebras commute with convolution (I can send you them) and end by </p> <p>\begin{eqnarray} &amp;&amp;log(I\otimes I)=log((I\otimes e)*(e\otimes I))=\cr &amp;&amp;log(I\otimes e)+log(e\otimes I)=log(I)\otimes e +e\otimes log(I) \end{eqnarray}</p> <p>This proves that log(I) maps INTO the set of primitives and to end the proof just remark that log(I)(f)=f when f is primitive because the series ends at the first term. </p> http://mathoverflow.net/questions/83654/do-stunted-exponential-series-give-projections-of-a-cocommutative-bialgebra-on-it/102813#102813 Answer by Duchamp Gérard H. E. for Do stunted exponential series give projections of a cocommutative bialgebra on its coradical filtration? Duchamp Gérard H. E. 2012-07-21T13:24:43Z 2012-07-21T13:31:56Z <p>Indeed ! But, you need to pass properly to the limit. You can use $log_s$ (the $s$ first terms of the expansion) and then remark that, when you apply both sides to a given $f\in H$, the limit is obtained by the polynomial $log_s$ with $s$ sufficiently large. </p> http://mathoverflow.net/questions/83654/do-stunted-exponential-series-give-projections-of-a-cocommutative-bialgebra-on-it/102815#102815 Answer by Duchamp Gérard H. E. for Do stunted exponential series give projections of a cocommutative bialgebra on its coradical filtration? Duchamp Gérard H. E. 2012-07-21T13:47:51Z 2012-07-21T13:47:51Z <p>OK, I did not apply the general fact you mention (which I do not use but can imagine true under certain conditions). I just used thet fact that, if $\phi : \mathcal{B}_1\rightarrow \mathcal{B}_2$ is a morphism of bialgebras, and $\phi\circ f_1=f_2\circ\phi\ ;\ \phi\circ g_1=g_2\circ\phi$ for $f_i,g_i\in End(\mathcal{B}_i)$, then $\phi\circ (f_1*g_1)=(f_2*g_2)\circ\phi$ and then, as $(I\otimes I)\circ\Delta=\Delta\circ I$, one has<br> $$log_s(I\otimes I)\circ\Delta=\Delta\circ log_s(I)$$ Hope that it helps. Tell me. </p> http://mathoverflow.net/questions/83654/do-stunted-exponential-series-give-projections-of-a-cocommutative-bialgebra-on-it/102816#102816 Answer by Duchamp Gérard H. E. for Do stunted exponential series give projections of a cocommutative bialgebra on its coradical filtration? Duchamp Gérard H. E. 2012-07-21T13:50:25Z 2012-07-21T13:50:25Z <p>OK, you have got it. I am glad that it could help. If you happen to come to Paris, please let me know, we can discuss these matters (I am often around the IHP of IHES areas). </p>