Recently, I am studing an interesting coalgebra which was called the infinitesimal bialgebra by Joni and Rota. It can be regarded as an algebraic framework for the calculus of divided differences. See Aguiar's paper http://pi.math.cornell.edu/~maguiar/cor.pdf for more details.

I want to know how does this coalgebra act as an algebraic framework for the calculus of divided differences. Then I go to see Joni and Rota paper https://onlinelibrary.wiley.com/doi/abs/10.1002/sapm197961293

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I am very confused about $\Delta(pq)=\Delta p(q\otimes 1)+(1\otimes p)\Delta(q)$.

My question is how to understand this strange coproduct and how to check it satisfies above equation. It seems that $\Delta(pq)=\Delta p(q\otimes 1)+(1\otimes p)\Delta(q)$ is not the same as the one studied by Aguiar.


The definition of the comultiplication (you should probably not use the word "coproduct", it leads to too much confusion) you're interested in is $$(\Delta p)(x,y) = \frac{p(x)-p(y)}{x-y}\in K[x,y]$$ where of course we see $K[x,y]$ as canonically isomorphic to $K[x]\otimes_K K[x]$ (so $x$ corresponds to the first tensor factor, and $y$ to the second). This means that if $p\in K[x]$, we see $p\otimes 1\in K[x,y]$ as $p(x)$, and $1\otimes p\in K[x,y]$ as $p(y)$.

Then the article mention that this comultiplication happens to satisfy the formula $$\Delta(pq) = \Delta(p)\cdot (q\otimes 1) + (1\otimes p)\cdot \Delta(q)$$ which just means, using the variables notations: $$\Delta(pq)(x,y) = \Delta(p)(x,y)\cdot q(x) + p(y)\cdot \Delta(q)(x,y)$$ or in other words $$\frac{(pq)(x)-(pq)(y)}{x-y} = \frac{p(x)-p(y)}{x-y}\cdot q(x) + p(y)\cdot \frac{q(x)-q(y)}{x-y}$$ which is a completely elementary computation.

The coassociativity is a little trickier since it involves three tensor factors. The key point is to understand what $\Delta\otimes Id$ actually means. Once we can write the correct formula to check, actually checking it is trivial.

The notations can be really confusing, so to be as clear as possible I will use the following conventions: let $A=K[t]$ be our initial algebra. I will use the variable $t$ when I talk about an element of $A$. Then $A\otimes_k A$ is a polynomial algebra in two variables, and I will use $K[u,v]$ in this case. Finally, $A\otimes_k A\otimes_k A$ is a polynomial algebra in three variables, and I will use $K[x,y,z]$. Of course the variable names don't have actual meaning, but I think it clarifies things if we don't use the same variables.

For instance, if $p(t)\in A$, then $\Delta p = \frac{p(u)-p(v)}{u-v}\in A\otimes_k A = K[u,v]$.

When we write $$\Delta\otimes Id: A\otimes_k A\to A\otimes_k A \otimes_k A,$$ the left-side copy of $A$ in the domain is transformed into the left and middle copy of $A$ in the codomain, and the right-side copy of $A$ in the domain stays on the right in the codomain. This means that is we write $$\Delta\otimes Id: K[u,v]\to K[x,y,z],$$ the variable $u$ will "become" $x$ and $y$, and $v$ will "become" $z$.

Explicitly, take $q(u,v)\in K[u,v]$. Then when we apply $\Delta\otimes Id$ to $q$, we see it as a polynomial in $u$, and we see $v$ as a non-variable which just becomes $z$. This gives: $$(\Delta\otimes Id)(q) = \frac{q(x,z)-q(y,z)}{x-y}\in K[x,y,z].$$ Similarly, when we apply $Id\otimes \Delta$, we see $q$ as a polynomial in $v$, $u$ becomes $x$ and $v$ becomes $y$ and $z$: $$(Id\otimes \Delta)(q) = \frac{q(x,y)-q(x,z)}{y-z}\in K[x,y,z].$$

If we start from $p(t)\in A$ and apply that to $$q(u,v)=\Delta(p) = \frac{p(u)-p(v)}{u-v}$$ we find $$(\Delta\otimes Id)(\Delta p) = \frac{\frac{p(x)-p(z)}{x-z}-\frac{p(y)-p(z)}{y-z}}{x-y} $$ and $$(Id\otimes \Delta)(\Delta p) = \frac{\frac{p(x)-p(y)}{x-y}-\frac{p(x)-p(z)}{x-z}}{y-z}.$$

You can check that those two things are really equal!

  • $\begingroup$ Perfect answer. Thank you very much. I try to verify the coassociativity. $(I \otimes \Delta )\Delta p(x)=\frac{-1\otimes \Delta(p)}{x-y}=\frac{\Delta p(y)}{y-x}$. On the other hand, $(\Delta \otimes I) \Delta p(x)=\frac{\Delta p(x)}{x-y}$. As desired. Is it right? $\endgroup$
    – Daisy
    Apr 2 '20 at 19:51
  • $\begingroup$ No, for coassociativity, you have to show that $\frac{\frac{p(x)-p(z)}{x-z}-\frac{p(y)-p(z)}{y-z}}{x-y}$ is equal to $\frac{\frac{p(x)-p(y)}{x-y}-\frac{p(x)-p(z)}{x-z}}{y-z}$. $\endgroup$ Apr 5 '20 at 17:24
  • $\begingroup$ Thank you for your comment. But I still can't understand you. I use the fact that $\Delta(1)=0$. But it seems that you are right and this coassociativity is precisely what I want. Can you give me some details? $\endgroup$
    – Daisy
    Apr 6 '20 at 1:42
  • 1
    $\begingroup$ It's a little tricky, so I added a full edit to explain as best as I could. $\endgroup$ Apr 6 '20 at 8:06
  • $\begingroup$ Thank you veru much. Well done. $\endgroup$
    – Daisy
    Apr 7 '20 at 0:06

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