6
$\begingroup$

Given a coalgebra $C$, can there exist more than one algebra structure on $C$ giving it the structure of a bialgebra? I will also ask the same question for Hopf algebras.

$\endgroup$
1
  • 3
    $\begingroup$ by PBW, universal enveloping algebra of a Lie algebra is isomorphic to a symmetric coalgebra (in char zero at least). So the coalgebra structure depends only on dimension, and from the bialgebra structure you can reconstruct the Lie algebra you started with $\endgroup$ Sep 17 at 19:55
8
$\begingroup$

Yes: if $k$ is a field of characteristic 2, let $C$ be the coalgebra over $k$ spanned by 1, $x$, $y$, and $z$ with $x$ and $y$ primitive, $\Delta z = z \otimes 1 + 1 \otimes z + x \otimes y + y \otimes x$ — in the algebra structure, $z$ is going to equal $xy = yx$, so $\Delta z$ has to equal $(\Delta x)(\Delta y)$. Then put algebra structures on this as follows:

  • $xy=yx$
  • $y^2 = 0$
  • either $x^2 = 0$ or $x^2 = y$

This gives two graded (with $\deg x = 1$, $\deg y = 2$) connected cocommutative bialgebras, and so by a theorem of Milnor and Moore, there is a unique antipode making such a bialgebra into a Hopf algebra.

$\endgroup$
3
  • $\begingroup$ I can believe it, but is it obvious those two algebras are non-isomorphic? $\endgroup$
    – LSpice
    Sep 17 at 21:57
  • 2
    $\begingroup$ One of the algebras ($x^2=0$) has all nontrivial nilpotents of order two, while the other has a nilpotent of order 4 ($x^2 = y \neq 0$, $x^3=xy=z$, $x^4=y^2=0$). $\endgroup$ Sep 17 at 23:10
  • $\begingroup$ Right, one is $k[x,y]/(x^2, y^2)$ while the other is $k[x]/(x^4)$. $\endgroup$ Sep 18 at 16:37
6
$\begingroup$

By dualizing, you are looking for an algebra $A$ such that there are two different coproducts $\Delta_1, \Delta_2: A \to A\otimes A$ that make $A$ into a bialgebra resp. Hopf algebra.

As an example let $p$ be prime, $k$ a field of char. $p$ and let $A=k[x_1,...,x_n]/(x_1^p,...,x_n^p)$ be the truncated polynomial algebra.

$A$ is the group algebra of the elementary abelian group $(\mathbb{Z}/p)^n=\langle g_1,...,g_n\rangle$ via $x_i=g_i-1$. Thus $A$ is a Hopf with coproduct $\Delta_1(x_i)=x_i\otimes 1 + 1\otimes x_i + x_i\otimes x_i$ and antipode $S_1(x_i)=g_i^{-1}-1=(-x_i)+\cdots +(-x_i)^{p-1}$.

$A$ is also the restricted enveloping algebra of $k^n$ considered as trivial restricted $p$-Lie algebra. As such $A$ is a Hopf algebra with coproduct $\Delta_2(x_i)=x_i \otimes 1 + 1\otimes x_i$ and antipode $S_2(x_i)=-x_i$.

These Hopf algebras where used by Avrunin and Scott in their proof of Carlson's conjecture on the rank variety in group cohomology. For a reference see section 4 of Carlson, Iyengar: Hopf algebra structures and tensor products for group algebras

$\endgroup$
2
  • $\begingroup$ Note that the Hopf algebras are not isomorphic because the space of primitives in the group Hopf algebra is trivial while it contains the $x_i$ in the other case. $\endgroup$
    – tj_
    Sep 17 at 23:58
  • $\begingroup$ Why the downvote? $\endgroup$
    – tj_
    Sep 17 at 23:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.