From Constant-Coefficient ODEs to the Hopf Algebra of Exponential Polynomials

From ODE Coalgebras to the Hopf Algebra of Exponential Polynomials

In the previous discussion, we considered a single constant-coefficient linear differential equation

$$p(D)f=0,D=\frac{d}{dt},$$

and observed that its solution space carries a natural coalgebra structure.

The algebraic starting point was the finite-dimensional quotient

$$A_p=\mathbb{C}[D]/(p(D)).$$

Its dual

$$A_p^{\ast }={\mathrm{Hom}}_{\mathbb{C}}(A_p,\mathbb{C})$$

is therefore a coalgebra. Under the usual identification with the solution space of $p(D)f=0$$p(D)f=0$, the comultiplication becomes

$$\mathrm{\Delta }f(s,t)=f(s+t),$$

and the counit is

$$\epsilon (f)=f(0).$$

For a single differential equation, this is where the story naturally stops. The solution space is generally not closed under multiplication.

But if we allow all constant-coefficient linear ODEs at once, a larger structure appears.

The total space of solutions

Let

$$\mathcal{E}=\{f:\mathbb{C}\to \mathbb{C}\mid p(D)f=0\text{ for some }0\ne p\in \mathbb{C}[x]\}.$$

If

$$p(x)=\prod _{i=1}^r(x-{\lambda }_i{)}^{m_i},$$

then the usual theory of constant-coefficient ODEs gives

$$\ker p(D)=\underset{i=1}{\overset{r}{⨁}}{e}^{{\lambda }_{i}t}\mathbb{C}[t{]}_{<m_i}.$$

Hence

$$\mathcal{E}=\{\sum _{i=1}^r{P}_i(t)e^{{\lambda }_{i}t}|P_i\in \mathbb{C}[t],{\lambda }_i\in \mathbb{C}\}.$$

Thus $\mathcal{E}$$\mathcal E$ is the space of exponential polynomials.

The individual solution spaces sit inside one another in a directed way. Indeed,

$$\ker p(D),\ker q(D)\subseteq \ker (pq)(D),$$

so that

$$\mathcal{E}=\underset{p\ne 0}{\underrightarrow{\lim }}\ker p(D).$$

Each $\ker p(D)$$\ker p(D)$ is a finite-dimensional coalgebra. Passing to their union reveals an additional operation: multiplication.

Multiplication relates different differential equations

A fixed solution space is usually not closed under pointwise multiplication.

For example,

$$(D-\lambda )f=0$$

has solution space

$$\mathbb{C}{e}^{\lambda t},$$

but

$$e^{\lambda t}{e}^{\lambda t}=e^{2\lambda t}$$

is generally not a solution of the same equation.

This obstruction disappears in $\mathcal{E}$$\mathcal E$. If

$$f(t)=P(t)e^{\lambda t},g(t)=Q(t)e^{\mu t},$$

then

$$f(t)g(t)=P(t)Q(t)e^{(\lambda +\mu )t},$$

which is again an exponential polynomial.

Thus $\mathcal{E}$$\mathcal E$ is a commutative algebra under pointwise multiplication,

$$m(f\otimes g)=fg,$$

with unit the constant function $1$$1$.

The multiplication usually moves us from one differential equation to another. Already for pure exponentials,

$$e^{\lambda t}{e}^{\mu t}=e^{(\lambda +\mu )t}.$$

Thus addition of characteristic roots is visible directly in multiplication of solutions.

Translation gives the coalgebra structure

The coalgebra structure from the finite-dimensional solution spaces extends to all of $\mathcal{E}$$\mathcal E$ by

$$\mathrm{\Delta }f(s,t)=f(s+t).$$

For this to define an algebraic comultiplication, one must check that $\mathrm{\Delta }f$$\Delta f$ lies in

$$\mathcal{E}\otimes \mathcal{E},$$

rather than merely being an arbitrary function of two variables.

It is enough to check this on the basic functions

$$t^n{e}^{\lambda t}.$$

Indeed,

$$(s+t{)}^n{e}^{\lambda (s+t)}=\sum _{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})s^j{e}^{\lambda s}{t}^{n-j}{e}^{\lambda t}.$$

Therefore

$$\mathrm{\Delta }(t^n{e}^{\lambda t})=\sum _{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})(t^j{e}^{\lambda t})\otimes (t^{n-j}{e}^{\lambda t}).$$

The counit is evaluation at the origin,

$$\epsilon (f)=f(0).$$

Coassociativity is simply associativity of addition:

$$((\mathrm{\Delta }\otimes 1)\mathrm{\Delta }f)(r,s,t)=f((r+s)+t),$$

while

$$((1\otimes \mathrm{\Delta })\mathrm{\Delta }f)(r,s,t)=f(r+(s+t)).$$

Hence

$$(\mathrm{\Delta }\otimes 1)\mathrm{\Delta }=(1\otimes \mathrm{\Delta })\mathrm{\Delta }.$$

Likewise,

$$f(0+t)=f(t)=f(t+0)$$

gives the counit identities.

The coalgebra structure is therefore translation written contravariantly on functions.

Time reversal gives the antipode

The additive group has inverse map

$$t⟼-t.$$

This induces

$$S(f)(t)=f(-t).$$

For example,

$$S(t)=-t,S(e^{\lambda t})=e^{-\lambda t}.$$

The antipode identity is just the equation

$$(-t)+t=0.$$

Indeed,

$$m(S\otimes 1)\mathrm{\Delta }(f)=\eta \epsilon (f),$$

and similarly

$$m(1\otimes S)\mathrm{\Delta }(f)=\eta \epsilon (f).$$

Thus $\mathcal{E}$$\mathcal E$ is a commutative and cocommutative Hopf algebra.

There is a useful geometric picture behind these formulas. Addition

$$+:\mathbb{C}\times \mathbb{C}\to \mathbb{C}$$

produces, by pullback of functions, the comultiplication

$$f(t)⟼f(s+t).$$

The identity element $0$$0$ gives the counit, and the inverse map $t\mapsto -t$$t\mapsto -t$ gives the antipode.

One should not, however, confuse $\mathcal{E}$$\mathcal E$ with the ordinary algebraic coordinate ring of the additive group. The coordinate ring of ${\mathbb{G}}_a$$\mathbb G_a$ is only

$$\mathbb{C}[t].$$

The algebra $\mathcal{E}$$\mathcal E$ is larger, since it also contains all exponential functions $e^{\lambda t}$$e^{\lambda t}$. It is naturally viewed as a Hopf algebra of representative functions on the additive group.

Exponential solutions are group-like

The simplest constant-coefficient ODE solutions have an immediate Hopf-algebraic interpretation.

Let

$$g_{\lambda }(t)=e^{\lambda t}.$$

Then

$$\mathrm{\Delta }(g_{\lambda })(s,t)=e^{\lambda (s+t)}=e^{\lambda s}{e}^{\lambda t},$$

so

$$\mathrm{\Delta }(g_{\lambda })=g_{\lambda }\otimes g_{\lambda }.$$

Also,

$$\epsilon (g_{\lambda })=1.$$

Thus $e^{\lambda t}$$e^{\lambda t}$ is group-like.

Moreover,

$$g_{\lambda }{g}_{\mu }=g_{\lambda +\mu },$$

and

$$S(g_{\lambda })=g_{-\lambda }.$$

The characteristic roots therefore appear inside $\mathcal{E}$$\mathcal E$ as a group of group-like elements.

The function $t$$t$ is primitive

The polynomial part of an ODE solution has a different but equally standard interpretation.

Since

$$\mathrm{\Delta }(t)(s,t)=s+t,$$

we have

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

Thus $t$$t$ is primitive.

It follows that

$$\mathrm{\Delta }(t^n)=(t\otimes 1+1\otimes t{)}^n=\sum _{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})t^j\otimes t^{n-j}.$$

Combining this with the group-like property of $e^{\lambda t}$$e^{\lambda t}$ gives

$$\mathrm{\Delta }(t^n{e}^{\lambda t})=\mathrm{\Delta }(t{)}^n\mathrm{\Delta }(e^{\lambda t}),$$

and therefore

$$\mathrm{\Delta }(t^n{e}^{\lambda t})=\sum _{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})t^j{e}^{\lambda t}\otimes t^{n-j}{e}^{\lambda t}.$$

Thus the familiar form

$$t^n{e}^{\lambda t}$$

of a generalized eigenfunction has a direct Hopf-algebraic interpretation: it is obtained by multiplying a power of a primitive element by a group-like element.

For example,

$$\ker (D-\lambda {)}^{n+1}=\mathrm{span}\{e^{\lambda t},t{e}^{\lambda t},\dots ,t^n{e}^{\lambda t}\}$$

is a finite-dimensional subcoalgebra of $\mathcal{E}$$\mathcal E$.

It is generally not a subalgebra. The Hopf algebra appears only after all characteristic roots and all polynomial degrees are allowed simultaneously.

The finite dual of the differential-operator algebra

There is another description of $\mathcal{E}$$\mathcal E$ which explains why this Hopf structure appears so naturally.

Consider

$$H=\mathbb{C}[D].$$

Its finite dual is

$$H^{\circ }=\{\phi \in H^{\ast }|\phi (I)=0\text{ for some finite-codimensional ideal }I\subseteq H\}.$$

Every nonzero ideal of $\mathbb{C}[D]$$\mathbb C[D]$ is generated by a nonzero polynomial, and

$${\dim }_{\mathbb{C}}\mathbb{C}[D]/(p(D))<\mathrm{\infty }.$$

Hence

$$H^{\circ }=\bigcup _{p\ne 0}{(\mathbb{C}[D]/(p(D)))}^{\ast }.$$

This is already suggestive: the finite dual is obtained by putting together the duals of exactly the same finite-dimensional quotients that produced the coalgebras attached to individual ODEs.

To make the relation explicit, associate to $\phi \in H^{\circ }$$\varphi\in H^\circ$ the function

$$f_{\phi }(t)=\sum _{n\ge 0}\phi (D^n)\frac{t^n}{n!}.$$

Suppose $\phi$$\varphi$ annihilates the ideal generated by

$$p(D)=a_0+a_{1}D+\cdots +a_m{D}^m.$$

Then for every $n\ge 0$$n\geq0$,

$$0=\phi (D^{n}p(D))=\sum _{j=0}^m{a}_j\phi (D^{n+j}).$$

On the other hand,

$$D^j{f}_{\phi }(t)=\sum _{n\ge 0}\phi (D^{n+j})\frac{t^n}{n!}.$$

Therefore

$$p(D)f_{\phi }=\sum _{n\ge 0}(\sum _{j=0}^m{a}_j\phi (D^{n+j}))\frac{t^n}{n!}=0.$$

Thus an element of the finite dual determines a solution of some constant-coefficient linear ODE.

Conversely, let $f$$f$ satisfy

$$p(D)f=0.$$

Define

$${\phi }_f(D^n)=f^{(n)}(0).$$

Then

$${\phi }_f(D^{n}p(D))=(D^{n}p(D)f)(0)=0$$

for every $n\ge 0$$n\geq0$.

Hence ${\phi }_f$$\varphi_f$ annihilates the ideal $(p(D))$$(p(D))$, so

$${\phi }_f\in H^{\circ }.$$

The two constructions are inverse to one another. Thus

$$\mathbb{C}[D{]}^{\circ }\cong \mathcal{E}.$$

The total space of constant-coefficient ODE solutions is therefore the finite dual of the algebra of constant-coefficient differential operators.

The Hopf structure on differential operators

The algebra $\mathbb{C}[D]$$\mathbb C[D]$ itself carries the standard Hopf structure

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

$$\epsilon (D)=0,$$

and

$$S(D)=-D.$$

The comultiplication is the Leibniz rule written algebraically.

Indeed,

$$D(fg)=(Df)g+f(Dg).$$

More generally,

$$\mathrm{\Delta }(D^n)=\sum _{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})D^j\otimes D^{n-j},$$

which is the higher Leibniz rule.

Since $\mathbb{C}[D]$$\mathbb C[D]$ is a Hopf algebra, its finite dual $\mathbb{C}[D{]}^{\circ }$$\mathbb C[D]^\circ$ is again a Hopf algebra. Under the identification

$$\mathbb{C}[D{]}^{\circ }\cong \mathcal{E},$$

the abstract dual Hopf structure becomes the elementary structure described above.

The multiplication on $\mathbb{C}[D]$$\mathbb C[D]$ dualizes to translation on $\mathcal{E}$$\mathcal E$.

For ${\phi }_f\in \mathbb{C}[D{]}^{\circ }$$\varphi_f\in\mathbb C[D]^\circ$,

$$\mathrm{\Delta }({\phi }_f)(D^m\otimes D^n)={\phi }_f(D^{m+n})=f^{(m+n)}(0).$$

These are exactly the mixed Taylor coefficients of

$$f(s+t).$$

Thus multiplication of differential operators becomes

$$\mathrm{\Delta }f(s,t)=f(s+t)$$

on the dual side.

Conversely, the coproduct on $\mathbb{C}[D]$$\mathbb C[D]$ dualizes to pointwise multiplication of solutions.

If ${\phi }_f$$\varphi_f$ and ${\phi }_g$$\varphi_g$ correspond to $f$$f$ and $g$$g$, then

$$({\phi }_f{\phi }_g)(D^n)=({\phi }_f\otimes {\phi }_g)\mathrm{\Delta }(D^n).$$

Using

$$\mathrm{\Delta }(D^n)=\sum _{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})D^j\otimes D^{n-j},$$

we obtain

$$({\phi }_f{\phi }_g)(D^n)=\sum _{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})f^{(j)}(0)g^{(n-j)}(0).$$

By the Leibniz rule, this is

$$(fg{)}^{(n)}(0).$$

Hence

$${\phi }_f{\phi }_g={\phi }_{fg}.$$

So pointwise multiplication of ODE solutions is precisely dual to the Leibniz coproduct on differential operators.