Coalgebras and the Leibniz Rule for Higher-Order Derivatives

Algebra and CoalgebraCoalgebra and Leibniz rule for higher-order derivativesBinomial CoalgebraLeibniz Law for higher-order derivatives

 

Algebra and Coalgebra

Definition.

An $R$$R$-algebra $(R,\mu ,\eta )$$(R,\mu,\eta)$ is a monoidal object in $R$$R$-$\mathrm{Mod}$$\mathrm{Mod}$.

That is, an $R$$R$ module $A$$A$, a linear map $\mu :A\otimes A\to A$$\mu:A\otimes A\to A$, called multiplication and the unit map $\eta :R\to A$$\eta:R\to A$.

It satisfies following diagrams

Associative Law:

Unit Law:

 

Definition. A co-$R$$R$-algebra $(\mathcal{C},\mathrm{\Delta },ϵ)$$(\mathcal C,\Delta,\epsilon)$ is a comonoidal object in $R$$R$-$\mathrm{Mod}$$\mathrm{Mod}$.

That is, an $R$$R$-module $\mathcal{C}$$\mathcal C$, a comultiplication $\mathrm{\Delta }:\mathcal{C}\to \mathcal{C}\otimes \mathcal{C}$$\Delta:\mathcal C\to\mathcal C\otimes\mathcal C$, and a counit map $ϵ:\mathcal{C}\to R$$\epsilon:\mathcal C\to R$. Then reverse all the arrows in the above diagrams, i.e.

Coassociative

i.e.

$$(\mathrm{\Delta }\otimes \mathrm{id})\circ \mathrm{\Delta }=(\mathrm{id}\otimes \mathrm{\Delta })\circ \mathrm{\Delta }.$$

Counit

The counit axioms are captured by two triangles:

i.e.

$$(\epsilon \otimes \mathrm{id})\circ \mathrm{\Delta }={\mathrm{id}}_C\text{and}(\mathrm{id}\otimes \epsilon )\circ \mathrm{\Delta }={\mathrm{id}}_C$$

Coalgebra and Leibniz rule for higher-order derivatives

Binomial Coalgebra

Consider the ring of polynomials $K[X]$$K[X]$. View it as a vector space over $K$$K$.

Define $\mathrm{\Delta }:K[X]\to K[X]{\otimes }_{K}K[X]$$\Delta:K[X]\to K[X]\otimes_K K[X]$ and $ϵ:K[X]\to K$$\epsilon:K[X]\to K$ by:

$$\begin{array}{c}\mathrm{\Delta }(1)=1\otimes 1,\mathrm{\Delta }(X^n)=(1\otimes X+X\otimes 1{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})X^k\otimes X^{n-k}=\mathrm{\Delta }(X{)}^n\end{array}$$

$$ϵ=e{v}_0,ϵ(f(X))=f(0)$$

It is called binomial coalgebra.

Now let us check it forms a coalgebra. Since $\mathrm{\Delta }$$\Delta$ is a $K$$K$-alg homomorphism and $K[X]$$K[X]$ is polynomial ring, we only need to check the value at $X$$X$ for $(\mathrm{\Delta }\otimes \mathrm{id})\circ \mathrm{\Delta }=(\mathrm{id}\otimes \mathrm{\Delta })\circ \mathrm{\Delta }.$$(\Delta\otimes \mathrm{id})\circ\Delta = (\mathrm{id}\otimes \Delta)\circ\Delta.$

$$(\mathrm{\Delta }\otimes \mathrm{id})\circ \mathrm{\Delta }(X)=(\mathrm{\Delta }\otimes \mathrm{id})(1\otimes X+X\otimes 1)=\mathrm{\Delta }(1)\otimes X+\mathrm{\Delta }(X)\otimes 1$$

$$\mathrm{\Delta }(1)\otimes X+\mathrm{\Delta }(X)\otimes 1=1\otimes 1\otimes X+1\otimes X\otimes 1+X\otimes 1\otimes 1$$

and

$$(\mathrm{id}\otimes \mathrm{\Delta })\circ \mathrm{\Delta }(X)=(\mathrm{id}\otimes \mathrm{\Delta })(1\otimes X+X\otimes 1)=1\otimes \mathrm{\Delta }(X)+X\otimes \mathrm{\Delta }(1)$$

$$1\otimes \mathrm{\Delta }(X)+X\otimes \mathrm{\Delta }(1)=1\otimes 1\otimes X+1\otimes X\otimes 1+X\otimes 1\otimes 1$$

Hence we have the identity $(\mathrm{\Delta }\otimes \mathrm{id})\circ \mathrm{\Delta }=(\mathrm{id}\otimes \mathrm{\Delta })\circ \mathrm{\Delta }.$$(\Delta\otimes \mathrm{id})\circ\Delta = (\mathrm{id}\otimes \Delta)\circ\Delta.$

For

$$(\epsilon \otimes \mathrm{id})\circ \mathrm{\Delta }={\mathrm{id}}_C\text{and}(\mathrm{id}\otimes \epsilon )\circ \mathrm{\Delta }={\mathrm{id}}_C$$

Consider

$$(\epsilon \otimes \mathrm{id})\circ \mathrm{\Delta }(X)=(\epsilon \otimes \mathrm{id})(1\otimes X)+(\epsilon \otimes \mathrm{id})(X\otimes 1)=X$$

By symmetry we have $(\mathrm{id}\otimes \epsilon )\circ \mathrm{\Delta }={\mathrm{id}}_C$$(\mathrm{id}\otimes \varepsilon)\circ\Delta = \mathrm{id}_C$. Hence it is a coalgebra.

Leibniz Law for higher-order derivatives

Now let us consider the Leibniz Law via this structure.

Consider $(C^{\mathrm{\infty }}(\mathbb{R}),\mu ,\eta )$$(C^{\infty}(\mathbb R),\mu,\eta)$ and $(\mathbb{R}[D],\mathrm{\Delta },ϵ)$$(\mathbb R[D],\Delta,\epsilon)$. Then $\mathrm{\Delta }(D)=1\otimes D+D\otimes 1$$\Delta(D)=1\otimes D+D\otimes 1$.

$$\mathrm{\Delta }(D^n)(f\otimes g)=\mathrm{\Delta }(D{)}^n(f\otimes g)=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}\otimes g^{(n-k)}$$

The Leibniz law tells us that

$$D\circ \mu =\mu \circ \mathrm{\Delta }(D)$$

Lemma. Let $T:X\to Y$$T:X\to Y$ be a morphism and $S\in \mathrm{End}(Y),U\in \mathrm{End}(X)$$S\in\mathrm{End}(Y),U\in\mathrm{End}(X)$. If $ST=TU$$ST=TU$, then $S^{n}T=T{U}^n$$S^nT=TU^n$.

Proof.

We use mathematical induction here.

Assume that $S^{n-1}T=T{U}^{n-1}$$S^{n-1}T=TU^{n-1}$, then

$$S^{n}T=S^{n-1}(ST)=(S^{n-1}T)U=T{U}^n◻$$

Hence we have

$$D^n\circ \mu =\mu \circ \mathrm{\Delta }(D{)}^n$$

Input $f\otimes g$$f\otimes g$ we get

$$D^n\circ \mu (f\otimes g)=\mu \circ (\mathrm{\Delta }(D){)}^n(f\otimes g)=\mu (\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(f^{(k)}\otimes g^{(n-k)}))$$

That is

$$D^n(fg)=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}{g}^{(n-k)}◻$$

In general, let $(R,\mu ,\eta ,d)$$(R,\mu,\eta,d)$ be a differential ring and $K_d=\{r\in R|d(r)=0\}$$K_d=\{r\in R|d(r)=0\}$ be the constant ring with respect to $d$$d$.

Then we could consider the coalgebra of $(K_d[d],\mathrm{\Delta },ϵ)$$(K_d[d],\Delta,\epsilon)$. Again, $\mathrm{\Delta }(d)=1\otimes d+d\otimes 1$$\Delta(d)=1\otimes d+d\otimes 1$...We prove the Leibniz rule for higher-order derivatives for arbitrary differential ring.

The binomial coalgebra shows why the Leibniz Law looks like the binomial theorem.