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Stream: theory: mathematics

Topic: Can we define a Hopf algebra as a group object?

Jack Jia (May 01 2025 at 17:45):

The antipode of a Hopf algebra is almost the same as the inverse of a group object.

I am wondering if a group object in $\mathsf{Alg}_{\mathbb{k}}^{\text{op}}$ is precisely a Hopf algebra over $\mathbb{k}$, where the comultiplication, counit, and antipode are given by the (opposites of) group multiplication, group unit, and the inverse, respectively.

[For clarity, I am defining the category $\mathsf{Alg}_{\mathbb{k}}$ to be the category of (unital) rings sliced under $\mathbb{k}$, I am taking for granted that the 2 definitions of $\mathbb{k}$-algebra (one as a monoid in $\mathbb{k}$-vector spaces, another as a ring that $\mathbb{k}$ maps into) coincide.]

I think all the diagrams that are supposed to commute translates between these, but the types of arrows don't quite check out. For example, the (opposite of) the inverse map is a map of $\mathbb{k}$-algebras, but the antipode of a Hopf algebra is only required to be $\mathbb{k}$-linear.

Mike Shulman (May 01 2025 at 18:43):

I'm not sure about the antipode, but a different point is that you do need to be talking about commutative rings in order that the tensor product in $\mathsf{Alg}_{\Bbbk}$ coincides with the cartesian product in $\mathsf{Alg}_{\Bbbk}^{\mathrm{op}}$. But maybe you're one of the people for whom all rings are commutative? (-:

Mike Shulman (May 01 2025 at 18:44):

That restriction does also mean that you only end up with commutative Hopf algebras this way. (I guess the non-commutative ones are "quantum groups".)

Jack Jia (May 01 2025 at 19:06):

@Mike Shulman

I think if all rings are commutative, those are already the affine group schemes and there's no real point in calling them Hopf algebras! Also thanks for pointing out the products don't really coincide!

Jack Jia (May 01 2025 at 19:09):

@Jean-Baptiste Vienney

I think maybe I didn't explain well: I meant that $\mathbb{k}$-algebras are monoids (not quite, because you will be taking $\mathbb{k}$ in the position of the terminal object in $\mathbb{k}$-vector spaces), not Hopf algebras.

Jean-Baptiste Vienney (May 01 2025 at 19:10):

Oh ok

Jean-Baptiste Vienney (May 01 2025 at 19:15):

I'm just confused :laughing:

Mike Shulman (May 01 2025 at 19:17):

Jack Jia said:

I think if all rings are commutative, those are already the affine group schemes and there's no real point in calling them Hopf algebras!

Well, if there's a difference between "commutative ring" and "affine scheme", there should be the same difference between "commutative Hopf algebra" and "affine group scheme", right?

Mike Shulman (May 01 2025 at 19:17):

(That is, they are the objects of a pair of contravariantly-equivalent categories.)

Jack Jia (May 01 2025 at 19:20):

Mike Shulman said:

Jack Jia said:

I think if all rings are commutative, those are already the affine group schemes and there's no real point in calling them Hopf algebras!

Well, if there's a difference between "commutative ring" and "affine scheme", there should be the same difference between "commutative Hopf algebra" and "affine group scheme", right?

Sorry, I am a bit confused - In this case, won't "commutative Hopf algebra" and "affine group scheme" both be on the "affine scheme" side?

David Egolf (May 01 2025 at 19:48):

On the off-chance you aren't already aware of this: the book "Monoidal Functors, Species and Hopf Algebras" talks about Hopf monoids internal to a braided monoidal category. I believe the nLab says that one can recover Hopf algebras as a special case.

Mike Shulman (May 01 2025 at 19:49):

Jack Jia said:

won't "commutative Hopf algebra" and "affine group scheme" both be on the "affine scheme" side?

I don't think so -- why would they be?

Mike Shulman (May 01 2025 at 19:49):

A homomorphism of commutative Hopf algebras is in particular a homomorphism of underlying commutative rings, which goes in the opposite direction from the map of affine (group) schemes.