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added to the Properties-section at Hopf algebra a brief remark on their interpretation as 3-vector spaces.
Are they a particular kind of 3-vector space? I mean is the Hopf algebra construction a categorification of a particular kind of 2-vector space?
And does the 3-dimensionality of a Hopf algebra show itself somehow?
I do not understand. Page 27 of Lurie et al. is giving an example of a Hopf algebra (group algebra for a finite group) which provides such a structure. I do not see a claim that all Hopf algebras, especially infinite dimensional provide something like that. Namely coalgebras and dual algebras correspond only in finite dimensional situation. Second I do not see why would Hopf having anything to do with this – I mean bialgebra requirement in finite dimensional situation does what can be done, I do not see why would antipode do anything here. Not to mention Hopf algebras in more general tensor categories, where the situation would be even worse.
I write out more details when I have a minute. But it’s straightforward to work it out.
coming back to the discussion of Hopf algebras as (bases for) 3-vector spaces:
I see that this is secretly discussed in
What they call a sesquiunital sesquialgebra in their def. 2.5 is precisely a (basis for a) 3-vector space: an algebra object internal to 2-vector spaces (with basis), hence internal to algebras+bimodules.
Have to run now. Maybe more later.
added brief mentioning and pointer to the literature in Tannaka duality for Hopf algebras
the entry Hopf algebra could do with some editing.
I started by polishing and expanding the Idea-section a bit. But have to interrupt now.
I have added cross-links between Hopf algebra and cogroup.
These and all related entries (e.g. Hopf algebroid, Steenrod algebra, etc. ) still could do with much editing.
I am sorry but I do not see a point of the sentence
In particular a co-group in rings is a Hopf algebra; a fact highlighted by Haynes Miller in the context of discussion of dual Steenrod algebras, see (Ravenel 86, appendix A) for review.
I mean why to mention Haynes Miller, what is so deep about mentioning somebody years after this point of view gave the whole branches of mathematics ? Also the message is not clear. Where is Miller, in Ravanel’s book, and what is the usage of this information ? To say that it was widely used point of view since early 1960s, see for example that M. Kac in early 1960-s started a huge subject of what is now called Kac algebras motivated by this commutative picture and then going beyond into nocommutative. Or Bergmann who had extensive work on cogroups in associative algebra setup, both commutative and noncommutative in 1970s. I. Bernstein had shown in late 1950s, I think, that unlike in commutative algebras, cogroups in unital associative algebras are very rare, that is why people need less categorical regular thing like Hopf algebras, as cogroups in associative algebras are so few. Algebraic geometers of course had thought of the affine group schemes (that is group objects in affine schemes) as spectra of cogroups in commutative algebras by the very definition of the subject from around 1960.
Steenrod algebra is a major historical motivation for Hopf algebras and this has been also used almost at the very beginning of the subject.
Or maybe you just want to say that you learned this from Miller so you want to keep the record in $n$Lab ? (I can not see its universal meaning, so it is probably particular and of less usage for others, unless it has some additional info which is yet not revealed here in the article)
The Ravenel’s book asserts that the term Hopf algebroid (for commutative Hopf algebroid) is due Miller. I believe this. But the observation that the cogroups in commutative rings are commutative Hopf rings and cogroups in commutative algebras are commutative Hopf algebras has been widely used way of thinking before Miller and is widely documented (perhaps not in homotopy theory, but in algebra and algebraic geometry).
In any case I will add commutative to all points in the article as it is not true without saying “commutative”: noncommutative Hopf algebras are not cogroups in any category.
I changed the critical part in the entry into:
In particular, a co-group in the category of (unital) commutative rings is a commutative Hopf ring and a cogroup in the category of (unital) commutative $k$-algebras is a commutative Hopf $k$-algebra; a fact highlighted in homotopy theory by Haynes Miller (in view of his generalization to commutative Hopf algebroids as cogroupoids in commutative algebra) in the context of discussion of dual Steenrod algebras, see (Ravenel 86, appendix A) for review.
Discussion of commutative Hopf algebras as cogroups is in
Here is the Israel Berstein’s (not Joseph Bernstein!) article showing that cogroups in associative algebras are extremely few (unlike Hopf algebras), and are basically free as algebras (the context is also algebraic topology):
This fact is observed in bigger generality by
Okay, thanks. I thought I had added the “commutative” qualifier where necessary, but thanks for catching places where I missed it.
Under “Examples” I added the line
the ordinary homology of an H-space (for instance a based loop space) is a Hopf algebra via its Pontrjagin ring-structure
added pointer to
(this used to be referenced only at W. S. Wilson and without the pdf link, so I completed the item and copied it to here)
added pointer to:
I have added mentioning (here) of the notion of involutive Hopf algebra (and will give this its own little page, for ease of hyperlinking).
Then in the proposition (here) that the antipoide is an anti-homomorphism I have added the statement that, hence, involutive Hopf algebras are star-algebras.
Also added hyperlinking to anti-homomorphism.
Following the language of star algebra maybe it is better to say algebra anti-involution as the antipode is an antihomomorphism. I don’t care, just suggestion so maybe resulting in a bit more uniform convention. Set theoretically it is just an involution, but in algebra world there is a distinction.
Okay, sure, anti-involution.
All calculations with Sweedler notation would expand immensely if one does not give some thinking on its use in first couple of instances of usage. Here for example, the second line taken for $h_{(1)}\otimes g_{(1)}$ instead of $h\otimes g$ is multiplied (and summed over dummy indices) with $S g_{(2)} S h_{(2)}$ and then just the renaming of indices using coassociativity gives 3rd line. Just spend one hour practicing Sweedler-assisted computations (once in your lifetime) and you will automatically observe such things on the fly for the rest of your life. (Easily observed) rule of a thumb is that the notation is bilinear so you can “contract” it with following leg if the coassociativity holds. I think it is more useful to figure out this rule yourself (why the step is legal and not out of nowhere) when you see it for the first time than to read the verbose explanation. Every reference using Sweedler notation in practice will freely contract with the next “leg”.
added pointer to:
added pointer to:
adding Noam Chomsky’s recent papers on Hopf algebras in generative linguistics:
Matilde Marcolli, Noam Chomsky, Robert Berwick, Mathematical Structure of Syntactic Merge (arXiv:2305.18278)
Matilde Marcolli, Robert Berwick, Noam Chomsky, Old and New Minimalism: a Hopf algebra comparison (arXiv:2306.10270)
Anonymouse
removing old query box.
Mike, can you do something with these notes that I took at some point as a grad student? I don't know this stuff very well, which is why I don't incorporate them into the text, but at least I cleaned up the formatting a bit so that you can if you like it. —Toby
One can make a group into a Hopf algebra in at least $2$ very different ways. Both ways have a discrete version and a smooth version.
Given a (finite, discrete) group $G$ and a ground ring (field?) $K$, then the group ring $K[G]$ is a cocommutative Hopf algebra, with $M(g_0,g_1) = g_0 g_1$, $I = 1$, $D(g) = g \otimes g$, $E(g) = 1$, and the nifty Hopf antipodal operator $S(g) = g^{-1}$. Notice that the coalgebra operations $D,E$ depend only on $Set|G|$.
Given a (finite, discrete) group $G$ and a ground ring (field?) $K$, then the function ring $Fun(G,K)$ is a commutative Hopf algebra, with $M(f_0,f_1)(g) = f_0(g)f_1(g)$, $I(g) = 1$, $D(f)(g,h) = f(g h)$, $E(f) = f(1)$, and the nifty Hopf antipodal operator $S(f)(g) = f(g^{-1})$. Notice that the algebra operations $M,I$ depend only on $Set|G|$.
Given a (simply connected) Lie group $G$ and the complex (real?) field $K$, then the universal enveloping algebra $U(G)$ is a cocommutative Hopf algebra, with $M(\mathbf{g}_0,\mathbf{g}_1) = \mathbf{g}_0 \mathbf{g}_1$, $I = 1$, $D(\mathbf{g}) = \mathbf{g} \otimes 1 + 1 \otimes \mathbf{g}$, $E(\mathbf{g}) = 0$, and the nifty Hopf antipodal operator $S(\mathbf{g}) = -\mathbf{g}$. Notice that the coalgebra operation $D,E$ depend only on $K Vect|\mathfrak{g}|$.
Given a (compact) Lie group $G$ and the complex (real?) field $K$, then the algebraic function ring $Anal(G)$ is a cocommutative Hopf algebra, with $M(f_0,f_1)(g) = f_0(g) f_1(g)$, $I(g) = 1$, $D(f)(g,h) = f(g h)$, $E(f) = f(1)$, and the nifty Hopf antipodal operator $S(f)(g) = f(g^{-1})$. Notice that the algebra operations $M,I$ depend only on $Anal Man|G|$.
Anonymouse
have hyperlinked Matilde Marcolli
Corrected the incorrect claim “Both Hopf algebras and Frobenius algebras are examples of bialgebras”
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