4
$\begingroup$

I am reading through the Dyer and Lashof's paper "Homology of Iterated Loop Spaces". They are quoting the following theorem:

If $A$ is a connected, free, associative, commutative, primitively generated mod $p$ Hopf algebra of finite type, then $A$ is isomorphic as a Hopf algebra to a tensor product of free monogenic Hopf algebras, i.e. to a tensor product of exterior and polynomial Hopf algebras each having one generator.

I can find respective theorems in Milnor-Moore's paper, but I cannot understand the "freeness" assumption, as it does not appear elsewhere.

So my questions are: 1) What "free Hopf algebra" might mean in this context? Is it free as an algebra?

2) Is it known that the only "free" Hopf algebras with one generator are external and polynomial ones?

$\endgroup$
9
$\begingroup$

1) Yes, free means free as an algebra. (I don't think it could possibly mean anything else. "Free Hopf algebras" and "free coalgebras" as left adjoints to the forgetful functor don't exist.)

2) is elementary, I think. Let $H$ be a connected free monogenic Hopf algebra. Let $x$ be a generator. Thanks to the connectivity assumption, $\deg x > 0$. The algebra $H$ is the free graded-commutative algebra on $x$, either $\Bbbk[x]$ if $\deg x$ is even or $p = 2$, or $\Lambda(x)$ otherwise. Then by counitality and degree arguments, $\Delta(x) = 1 \otimes x + x \otimes 1$, so $H$ is indeed either a polynomial or exterior Hopf algebra.

Note that I used the connectivity assumption above. If you don't assume that $H$ is connected (or at the very least locally conilpotent) then the result becomes false. Consider for example the Hopf algebra of dimension $2$, with basis $(1,x)$, both in degree zero, $x \cdot x = 1$, $\epsilon(x) = 1$ and $\Delta(x) = x \otimes x$. In other words, the group algebra of $\mathbb{Z}/2\mathbb{Z}$.

$\endgroup$
  • $\begingroup$ "free as an algebra" <-- as a commutative algebra, I hope. We algebraists need to do something about the proliferation of nested adjectives. $\endgroup$ – darij grinberg Mar 2 at 3:51
  • $\begingroup$ @darijgrinberg Yes, I meant free as a commutative algebra, sorry. $\endgroup$ – Najib Idrissi Mar 2 at 9:50
1
$\begingroup$

Without the freeness assumption, one also has the Hopf algebras $\mathbb Z/p[y]/(y^{p^n})$ with $y$ an even dimensional class. (This certainly is in Milnor-Moore.)

$\endgroup$

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.