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Coalgebras and the Leibniz Rule for Higher-Order Derivatives

Coalgebra

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

Algebra and Coalgebra

Definition.

$R$ $(R,\mu,\eta)$ $R$ $\mathrm{Mod}$ .

$R$ $A$ $\mu:A\otimes A\to A$ $\eta:R\to A$ .

It satisfies following diagrams

Associative Law:

Unit Law:

Definition. $R$ $(\mathcal C,\Delta,\epsilon)$ $R$ $\mathrm{Mod}$ .

$R$ $\mathcal C$ $\Delta:\mathcal C\to\mathcal C\otimes\mathcal C$ $\epsilon:\mathcal C\to R$ . Then reverse all the arrows in the above diagrams, i.e.

Coassociative

i.e.

(Δ \otimes id) \circ Δ = (id \otimes Δ) \circ Δ .

Counit

The counit axioms are captured by two triangles:

i.e.

(ε \otimes id) \circ Δ = {id}_{C} and (id \otimes ε) \circ Δ = {id}_{C}

Coalgebra and Leibniz rule for higher-order derivatives

Binomial Coalgebra

$K[X]$ $K$ .

$\Delta:K[X]\to K[X]\otimes_K K[X]$ $\epsilon:K[X]\to K$ by:

\begin{matrix} Δ (1) = 1 \otimes 1, Δ (X^{n}) = (1 \otimes X + X \otimes 1)^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) X^{k} \otimes X^{n - k} = Δ (X)^{n} \end{matrix}

ϵ = e v_{0}, ϵ (f (X)) = f (0)

It is called binomial coalgebra.

$\Delta$ $K$ $K[X]$ $X$ $(\Delta\otimes \mathrm{id})\circ\Delta = (\mathrm{id}\otimes \Delta)\circ\Delta.$

(Δ \otimes id) \circ Δ (X) = (Δ \otimes id) (1 \otimes X + X \otimes 1) = Δ (1) \otimes X + Δ (X) \otimes 1

Δ (1) \otimes X + Δ (X) \otimes 1 = 1 \otimes 1 \otimes X + 1 \otimes X \otimes 1 + X \otimes 1 \otimes 1

and

(id \otimes Δ) \circ Δ (X) = (id \otimes Δ) (1 \otimes X + X \otimes 1) = 1 \otimes Δ (X) + X \otimes Δ (1)

1 \otimes Δ (X) + X \otimes Δ (1) = 1 \otimes 1 \otimes X + 1 \otimes X \otimes 1 + X \otimes 1 \otimes 1

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

For

(ε \otimes id) \circ Δ = {id}_{C} and (id \otimes ε) \circ Δ = {id}_{C}

Consider

(ε \otimes id) \circ Δ (X) = (ε \otimes id) (1 \otimes X) + (ε \otimes id) (X \otimes 1) = X

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

Leibniz Law for higher-order derivatives

Now let us consider the Leibniz Law via this structure.

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

Δ (D^{n}) (f \otimes g) = Δ (D)^{n} (f \otimes g) = \sum_{k = 0}^{n} (\binom{n}{k}) f^{(k)} \otimes g^{(n - k)}

The Leibniz law tells us that

D \circ μ = μ \circ Δ (D)

Lemma. $T:X\to Y$ $S\in\mathrm{End}(Y),U\in\mathrm{End}(X)$ $ST=TU$ $S^nT=TU^n$ .

Proof.

We use mathematical induction here.

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

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

Hence we have

D^{n} \circ μ = μ \circ Δ (D)^{n}

$f\otimes g$ we get

D^{n} \circ μ (f \otimes g) = μ \circ (Δ (D))^{n} (f \otimes g) = μ (\sum_{k = 0}^{n} (\binom{n}{k}) (f^{(k)} \otimes g^{(n - k)}))

That is

D^{n} (f g) = \sum_{k = 0}^{n} (\binom{n}{k}) f^{(k)} g^{(n - k)} ◻

$(R,\mu,\eta,d)$ $K_d=\{r\in R|d(r)=0\}$ $d$ .

$(K_d[d],\Delta,\epsilon)$ $\Delta(d)=1\otimes d+d\otimes 1$ ...We prove the Leibniz rule for higher-order derivatives for arbitrary differential ring.