A quantum Grothendieck group? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T07:09:24Zhttp://mathoverflow.net/feeds/question/93016http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/93016/a-quantum-grothendieck-groupA quantum Grothendieck group?Ollie Margetts2012-04-03T17:12:45Z2012-04-03T23:16:22Z
<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#93020Answer by Bruce Westbury for A quantum Grothendieck group?Bruce Westbury2012-04-03T17:57:52Z2012-04-03T17:57:52Z<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#93050Answer by MTS for A quantum Grothendieck group?MTS2012-04-03T23:16:22Z2012-04-03T23:16:22Z<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>