A quantum Grothendieck group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T07:09:24Z http://mathoverflow.net/feeds/question/93016 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93016/a-quantum-grothendieck-group A quantum Grothendieck group? Ollie Margetts 2012-04-03T17:12:45Z 2012-04-03T23:16:22Z <p><strong>Question 1</strong>: Given a co-commutative bialgebra, does there exist a sort of Grothendick group type construction? Presumably this should take the form of a functor from co-commutative bialgebras to hopf algebras?</p> <p>My motivation: the finite particle vectors in the symmetric Fock space $\mathbb{C}\Omega\oplus \bigoplus_{n=1}^\infty H^{\vee n}$ have the natural structure of a graded bialgebra. Just set</p> <ol> <li>$m(f^{\otimes n},g^{\otimes m})=Sym(f^{\otimes n}\otimes g^{\otimes m})$,</li> <li>$\eta(\lambda)=\lambda\Omega$,</li> <li>$\Delta(f^{\otimes n})=\sum_{k=0}^n{ {n\choose k} f^{\otimes k}\otimes f^{\otimes n-k}}$, where $f^{\otimes 0}:=\Omega$, and</li> <li>$\epsilon(\cdot)=\langle \Omega,\cdot\rangle$.</li> </ol> <p>but it doesn't seem to have an antipode.</p> <p><strong>Question 2</strong>: Can we make a quantum group containing this bialgebra?</p> <p>Actually it occurs to me that it has probably been looked at, since a finite dimensional $H$ gives us an algebra isomorphic to $\mathbb{C}[x_1,\ldots,x_{\mathrm{dim} H}]$? I should say also that I'm neither an algebraist, nor a quantum groupie, so I'd appreciate any references/constructions readable by a non-expert!</p> http://mathoverflow.net/questions/93016/a-quantum-grothendieck-group/93020#93020 Answer by Bruce Westbury for A quantum Grothendieck group? Bruce Westbury 2012-04-03T17:57:52Z 2012-04-03T17:57:52Z <p>Connected graded bialgebras have an antipode (which is unique):</p> <p>The following book gives two formulae:</p> <p>MR2724388 Aguiar, Marcelo; Mahajan, Swapneel Monoidal functors, species and Hopf algebras. CRM Monograph Series, 29. American Mathematical Society, Providence, RI, 2010. lii+784 pp. ISBN: 978-0-8218-4776-3 </p> <p>These formulae are for the antipode of a connected graded bialgebra and are given in 2.3.3. These are the Takeuchi formula and the Milnor and Moore formula.</p> <p>The original references are</p> <p>MR0292876 (45 #1958) Takeuchi, Mitsuhiro Free Hopf algebras generated by coalgebras. J. Math. Soc. Japan 23 (1971), 561–582. </p> <p>MR0174052 (30 #4259) Milnor, John W.; Moore, John C. On the structure of Hopf algebras. Ann. of Math. (2) 81 1965 211–264. </p> http://mathoverflow.net/questions/93016/a-quantum-grothendieck-group/93050#93050 Answer by MTS for A quantum Grothendieck group? MTS 2012-04-03T23:16:22Z 2012-04-03T23:16:22Z <p>The forgetful functor from the category of Hopf algebras to the category of bialgebras has a left adjoint. This means that given a bialgebra $B$, there is a Hopf algebra $H(B)$ with a bialgebra morphism $\iota : B \to H(B)$ such that any bialgebra morphism from $B$ to a Hopf algebra $H$ factors through $\iota$ via a morphism of Hopf algebras. </p> <p>I do not know whether $H(B)$ is cocommutative if $B$ is, and I also do not know whether the morphism $\iota$ is always injective. For the latter question I sort of suspect the answer to be negative, just as a semigroup does not always inject into its Grothendieck group.</p> <p>For discussion of this and related issues, see <a href="http://arxiv.org/abs/0905.2613" rel="nofollow">http://arxiv.org/abs/0905.2613</a> (very short) and references therein, especially to the lecture notes of Bodo Pareigis available here: <a href="http://www.mathematik.uni-muenchen.de/~pareigis/pa_schft.html" rel="nofollow">http://www.mathematik.uni-muenchen.de/~pareigis/pa_schft.html</a></p>