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Somebody more experienced than I am please check the formatting for the references section. Do I need to put a link to the reference somewhere near the beginning?

Schepler 22:48, 18 November 2006 (UTC)Reply

I'm wondering whether it would be worth it to indicate exactly what the PD structure on ${\displaystyle (S^{\cdot }M){\check {~}}}$ is, or whether it is a bit too complex and would obscure things.

If I included it, it would go something like:

Addition is just the normal pointwise addition of functions. For multiplication, given ${\displaystyle \phi ,\psi :S^{\cdot }M\to A}$, their product ${\displaystyle \phi \psi :S^{\cdot }M\to A}$ is defined so that for ${\displaystyle x_{1},\ldots ,x_{n}\in M}$,

${\displaystyle (\phi \psi )(x_{1}\cdots x_{n})=\sum _{S\subseteq \{1,2,\ldots ,n\}}\phi \left(\prod _{i\in S}x_{i}\right)\psi \left(\prod _{j\notin S}x_{j}\right).}$

The set I of functions ${\displaystyle \phi }$ such that ${\displaystyle \phi (1)=0}$ can easily be seen to be an ideal with respect to this ring structure. Then defining ${\displaystyle \gamma _{m}\phi :S^{\cdot }M\to A}$ such that

${\displaystyle (\gamma _{m}\phi )(x_{1}\cdots x_{n})=\sum _{\pi \in P_{n,m}}\prod _{S\in \pi }\phi \left(\prod _{i\in S}x_{i}\right)}$

gives a divided power structure on I. Here ${\displaystyle P_{n,m}}$ denotes the set of (unordered) partitions of ${\displaystyle \{1,2,\ldots ,n\}}$ into m parts.

(Note that by definition, ${\displaystyle (\phi ^{m})(x_{1}\cdots x_{n})}$ is equal to the corresponding sum where ${\displaystyle \pi }$ ranges over ordered partitions of ${\displaystyle \{1,2,\ldots ,n\}}$ into m parts, thus making the above definition of the PD structure a natural one.)

Daniel Schepler 15:20, 21 November 2006 (UTC)Reply

The article on tensor algebra now gives lots of explicit detail on how to construct Hopf algebras from generic tensor products (exterior product, symmetric product, etc). At the bottom is a short section that briefly mentions the divided-power Hopf algebra. What you (or someone) would need to do is to go through the same steps, but now for the divided power algebra, and to then replace the tensor by the symmetric product, and thus gain the insight you are looking for ... It probably would not be a bad idea to expand this article to do all this. 67.198.37.16 (talk) 17:54, 10 March 2018 (UTC)Reply