## Formal group laws and binomial Sheffer sequences

Given a compositional inverse pair $f(x)$ and $f^{-1}(x)$, i.e.,

$f(f^{-1}(x)) = x$,

with $f(x) = e^{a. x}$ with $a_0 = 0$$a_1 = 1$, and $(a.)^n = a_n$,  construct the binomial Sheffer sequence $p_n(t)$ with the exponential generating function

$e^{x p.(t)} = e^{t f^{-1}(x)}$.

Then the associated formal group law (FGL) may be expressed as

$FGL(x,y) = f[f^{-1}(x)+f^{-1}(y)] = e^{a.[f^{-1}(x)+f^{-1}(y)]}$

$= e^{a. D_t}|_{t=0} \;\; e^{t[f^{-1}(x)+f^{-1}(y)]}$

$= f(D_t)|_{t=0} \;\; e^{t[f^{-1}(x)+f^{-1}(y)]}$

$= f(D_t)|_{t=0} \;\; e^{t f^{-1}(x)} e^{tf^{-1}(y)}$

$= \sum_{j \geq 0 } \; \; \sum_{k \geq 0 } \; \; \frac{x^j}{j!} \frac{y^k}{k!} \; f(D_t)|_{t=0} \;\; p_j(t)p_k(t).$

The last operator factor may be expressed several ways:

$f(D_t)|_{t=0} \;\; p_j(t)p_k(t)$

$= e^{a. D_t}|_{t=0} \;\; p_j(t)p_k(t) = p_j(a.)p_k(a.) = p_j(t)p_k(t)|_{t^n=a_n}$

$= \sum_{m \geq 1 } \; \; \;\; a_m \frac{D_t^m}{m!} |_{t=0} \; p_j(t)p_k(t)$

$= \sum_{m \geq 1 } \; \; \;\; a_m \; \sum_{q=0}^m \; [\frac{D_t^{m-q}}{(m-q)!} p_j(t)] \; [\frac{D_t^{q}}{q!} p_k(t)] |_{t=0}$

$= \sum_{m \geq 1 } \; \; \;\; a_m \; \sum_{q \geq 0 } \; p_{j,m-q} \; p_{k,q}$

where $p_0(t) = 1$ and $p_n(t) = \sum_{m=1}^n \; \; p_{n,m} t^m$.

The product of the Sheffer polynomials may be written, as discussed by J. Taylor in his thesis “Formal group laws and hypergraph colorings“, in terms of linear combinations of the Sheffer polynomials as

$p_j(t)p_k(t) = \sum_{n \leq (j+k)} \; c^n_{j,k} \; p_n(t)$.

Then noting that $f(D_t) = L$ is the lowering operator for the Sheffer polynomials, defined by $L \; p_n(t) = n \; p_{n-1}(t)$, and that the raising operator, defined by $R \; p_n(t) = p_{n+1}(t)$, is $R = t \; 1/[df(D_t)/dD_t]$ (see Sec. 3: A Walk with Bruno and Blissard, pages 12 and 13, of my pdf “Mathemagical Forests” for derivations of these relations),

$f(D_t) \; p_j(t)p_k(t) = L \; p_j(t)p_k(t) = L \; (R^j 1) (R^k 1)$

$= \sum_{n \leq (j+k)} \; c^n_{j,k} \; L \; p_n(t)$

$= \sum_{n \leq (j+k)} \; c^n_{j,k} \; n \; p_{n-1}(t)$

$= \sum_{n \leq (j+k)} \; c^n_{j,k} \; n \; R^{n-1}1$.

Evaluated at the origin,

$f(D_t) |_{t=0} \; p_j(t)p_k(t) = \sum_{n \leq (j+k)} \; c^n_{j,k} \; n \; p_{n-1}(0) = \; c^1_{j,k} = p_j(a.)p_k(a.),$

where the umbral evaluation is carried out only after multiplying the polynomials together treating the umbral variable as a regular variable.

A good inverse pair for a sanity check is $f(x) = e^x - 1$ and $f^{-1}(x) = ln(1+x)$ for which the Sheffer polynomials are the falling factorials or Stirling polynomials of the first kind, $p_n(t) = t!/(t-n)!$. For example,

$f(D_t)|_{t=0} \;\; e^{t f^{-1}(x)} e^{tf^{-1}(y)} = (e^{D_t} -1) |_{t=0} \;\; (1+x)^t (1+y)^t$

$= [(1+x)(1+y)]^{t+1}-[(1+x)(1+y)]^{t} |_{t=0} = (1+x)(1+y)-1= x+y+xy.$

Note also that with $g(y) = 1 / (df^{-1}(y)/dy)$ and  $(x,y) = (f^{-1}(y),f(x))$,

$e^{u \; g(y)D_y}y = e^{u \; d/df^{-1}(y)}y = e^{u \; d/dx}f(x) = f(x+u) = f(f^{-1}(y)+u)= H(y,u),$

so

$H(0,x) = f(x) = \sum_{n>0} \; [g(y)D_y]^n y |_{y=0} \; \frac{x^n}{n!},$

and clearly if the coefficients of the formal Taylor series expansion of $g(x)$ are non-negative integers, then so are the coefficients of $f(x)$. These coefficients are given in OEIS A145271.

My 2014 formulas in the OEIS entry for the Eulerian numbers A008292 give an FGL for $g(x) = (1+x_1x)(1+x_2x)$. An extrapolated $g(x) = (1+x_1x)(1+x_2x)(1+x_3x) \cdots$, which can be simply expressed as a polynomial, or infinite series, in the elementary symmetric polynomials/functions, gives the same $f^{-1}$ in terms of the complete homogeneous symmetric polynomials/functions.

Explicitly, expressing $g(x)$ in terms of the elementary symmetric polynomials $e_n$ and the complete homogeneous symmetric polynomials $h_n$,

$g(x) = (1+x_1x)(1+x_2x)(1+x_3x) \cdots= 1 + e_1(x_1,x_2,..) x + e_2(x_1,x_2,..)x^2+ \cdots$

$= E(x) = 1 / H(-x) = 1 / [1-h_1(x_1,x_2,..)x+h_2(x_1,x_2,..)x^2- \cdots].$

Then

$f^{-1}(x) = x-h_1x^2/2+h_2x^3/3 - \cdots = -ln[1-u.x]$ with $u_n = (-1)^{n-1}h_{n-1},$

so $p_n(t) = St1_n(u_1,u_2,u_3,...,u_n;t)=St1_n(1,-h_1,h_2,...,(-1)^{n-1}h_{n-1};t),$

$= St1_n(t,-h_1t,h_2t,...,(-1)^{n-1}h_{n-1}t;1),$

the Stirling partition polynomials of the first kind, a.k.a., the cycle index polynomials for the symmetric groups, described in A036039. See the link there to my pdf “Lagrange a la Lah” (particularly, pages 4 and 23 of Part I) for the first few polynomials and their characterization with $t=1$ as an Appell sequence in the indeterminate $u_1$ with the lowering and raising operators $L =d/du_1$ and $R = u./(1-u.L) = u_1 + u_2L+u_3L^2+\cdots.$

A related sequence of Appell polynomials in $z$ can be devised such that

$p_n(t) = w_n(z;t) |_{z=t},$ where $w_n(z;t) = R^n 1$ and

$R = z + t [H(-D_z)-1] = z+ t[\frac{1}{g(D_z)}-1] = z +t[f^{-1}(D_z)'-1]$

$= z - th_1D_z+th_2D_z^2-th_3D_z^3- \cdots = z + [e^{c.D_z}-1]$

with $c_n = (-1)^n n! h_nt$ for $n >0$ and $c_0=1$. Giving for the first few,

$w_0 = 1$

$w_1(z;t) = z$

$w_2(z;t) = z^2 - h_1t$

$w_3(z;t) = z^3-3h_1tz+2h_2t$

$w_4(z;t) = z^4 -6h_1tz^2+8h_2tz+3h_1^2 t^2-6h_3 t.$

Using the notation in Lagrange a la Lah,

$w_n(z;t) = St1_n(z,-h_1t,h_2t,-h_3t,...,(-1)^{n-1}h_{n-1}t;1).$

Then

$p_0 = 1$

$p_1(t) = t$

$p_2(t) = t^2 - h_1t$

$p_3(t) = t^3-3h_1t^2+2h_2t$

$p_4(t) = w_4(t;t) = t^4 -6h_1t^3+(8h_2+3h_1^2)t^2-6h_3t.$

Note that $D_t \; p_n(t) = (D_t + D_z) w_n(z;t) |_{z=t}.$

An umbral recursion relation follows from the last expression for the raising op:

$w_n(z;t) = zw_{n-1}(z;t)+w_{n-1}(z+c.;t)-w_{n-1}(z;t)$

$= zw_{n-1}(z;t)+(w.(z;t)+c.)^{n-1}-w_{n-1}(z;t).$

The OEIS entry A263633 can be used to express the complete symmetric polynomials in terms of the elementary symmetric polynomials or vice versa. For easy reference for spot checks,

$h_1=e_1,$

$h_2 = e_1^2-e_2,$ and

$h_3 = e_1^3 - 2 e_1e_2+e_3,$

and the same holds for $e_n$ and $h_n$ interchanged.

Since $D_z w_{n}(z;t) = n \; w_{n-1}(z;t)$, for $q>j>k$,

$p_j(t)p_k(t) = w_{j}(z;t) \; k! \; D^{j-k}_z \; \frac{w_{j}(z;t)}{j!} |_{z=t}.$

$= [j! \; D^{q-j}_z \; \frac{w_q(z;t)}{q!}] [ k! \; D^{q-k}_z \; \frac{w_q(z;t)}{q!} ]|_{z=t}.$

$D^m_t \; p_j(t) p_k(t) |_{t=0} = (D_{t_1}+D_{t_2})^m \; p_j(t_1)p_k(t_2) |_{t_1=t_2=0}$

$= (D_t +D_z)^m \; w_j(z;t) w_k(z;t) |_{z=t=o}$

$= (D_{t_1} +D_{z_1}+D_{t_2} +D_{z_2})^m \; w_j(z_1;t_1) w_k(z_2;t_2) |_{z_1=z_2=t_1=t_2=0}$

$= (D_{t_1} +D_{z_1}+D_{t_2} +D_{z_2})^m \; [j! \; D^{q-j}_{z_1} \; \frac{w_q(z_1;t_1)}{q!}] [ k! \; D^{q-k}_{z_2} \; \frac{w_{q}(z_2;t_2)}{q!} ]|_{z_1=z_2=t_1=t_2=0}$

Returning to the connection coefficients, we have

$p_1(t)p_1(t) = h_1p_1(t)+p_2(t)=e_1p_1(t)+p_2(t),$

$p_2(t)p_1(t)= 2(h_1^2-h_2)p_1(t)+2h_1p_2(t)+p_3(t)$

$= 2e_2p_1(t)+2e_1p_2(t)+p_3(t),$ and

$p_2(t)p_2(t)= 2(5h_1^3+3h_3-8h_2h_1) p_1(t)+2(5h_1^2-4h_2)p_2(t)+4h_1p_3(t)+p_4(t)$

$= 2(2e_1e_2+3e_3)p_1(t)+2(e_1^2+4e_2)p_2(t)+4e_1p_3(t)+p_4(t),$

so

$c^1_{1,1}=h_1 = e_1,$

$c^1_{1,2}=c^1_{2,1}=2(h_1^2-h_2) = 2e_2,$ and

$c^1_{2,2}= 2(5h_1^3+3h_3-8h_2h_1)= 2(2 e_1 e_2+3e_3),$

giving for the general FGL

$FGL(x,y) = x+y+c^1_{1,1}xy+\frac{c^1_{1,2}}{2!}(x^2y+y^2x)+\frac{c^1_{2,2}}{2!2!}x^2y^2+\cdots.$

This agrees with the particular

$FGL(x,y)= \frac{x+y+(a+b)xy}{1-(ab)xy}$

$= x+y+(a+b)xy+(ab)(x^2y+y^2x)+(a^2b+b^2a)x^2y^2+\cdots$

$= x+y+e_1xy+e_2(x^2y+y^2x)+e_1e_2x^2y^2+\cdots$

for the bivariate generating function of the Eulerian numbers A008292, for which $e_n = 0$ for $n > 2$. (Here $a$ and $b$ are regular variables, not umbral variables. Note that a dot as a subscript is used to flag umbral variables in my notation.)

Spot checking with previous formulas, we have

$c^1_{j,k} = p_j(a.) p_k(a.) = w_j(a.;a.) w_k(a.;a.),$

and

$c^1_{1,2} = p_1(a.) p_2(a.) = (a.)(a.^2-h_1a.)=a.^3-h_1a.^2=a_3-h_1a_2$

$= (e_1^2+2e_2)-h_1e_1= 2e_2$

since $h_1=e_1$ and

$f(x) = e^{a.x}= x + e_1 \frac{x^2}{2!} + (e_1^2+2e_2) \frac{x^3}{3!}+ (e_1^3 + 8 e_1 e_2 + 6 e_3) \frac{x^4}{4!} + \cdots .$

The coefficients of this last expression follow from A145271 with $g_n=g^{(n)}(0)=n!e_n$ (cf. also A190015).

Note also, for $n >1$, $p_n(a.)=0$ since

$f(D_t) p_n(t) |_{t=0}= L p_n(t) |_{t=0}=n\; p_{n-1}(0)$

$= e^{a.D_t} p_n(t) |_{t=0} = p_n(a.),$

giving

$a_n = -\sum_{k>0}^{n-1} p_{n,k}a_k.$

For consistency between the operator formalism for umbral substitution and the direct replacement of an argument with the umbral variable, we must evaluate $p_0(t) = t^0=1$ to $p_0(a.) = (a.)^0=a_0=0=f(D_t)|_{t=0}p_0(t)=e^{a.D_t}|_{t=0}p_0(t)$. Then, since $p_1(a.)=a_1=1$,

$e^{xp.(a.)}=e^{a.f^{-1}(x)}=f(f^{-1}(x))=x.$

The coefficients of the FGL remain invariant with a  scaling of the indeterminates, i.e., with

$f^{-1}(x) = -ln(1-u.x) = x/(1-v.x) = e^{w.x}-w_0.$

Then $(-1)^{n-1}h_{n-1}=u_n = nv_n=(n-1)!w_n$ and the associated $p_n$ are the refined Stirling polynomials of the first kind A036039, the refined Lah polynomials A130561, and the refined Bell polynomials A036040, respectively. The associated inversion partition polynomials for obtaining $f(x)=e^{b.x}$ in terms of the same indeterminates are given by A133932, A133437, and A134685, respectively. Alternatively, A134264A248120, or A248927 can be used for the inversion in terms of the coefficients of $x/f^{-1}(x)$, for which calculation A133314 and A263633 are useful. Ultimately,

$c^1_{j,k} = p_j(b.) p_k(b.)$ and $p_n(b.)=0$ except for $p_1(b.)=b_1=1$.

Then $p_n(b.)= \delta_{n-1}$ can be used to give a recursion relation for each set of inversion polynomials in terms of the coefficients of the associated binomial Sheffer sequence and the lower order partition polynomials.

For example, consider A133437, the inversion formula for power series, or o.g.f.s, involving the refined face polynomials of the Stasheff associahedra, and A130561, the refined Lah polynomials, related to the elementary Schur polynomials. Given

$f^{-1}(x) = x - (c_2 x^2 + b_3 x^3 + \cdots),$

the inversion polynomials of A133437 give

$f(x)= e^{a.x} = x + 2c_2 \frac{x^2}{2!}+(12c_2^2+6c_3)\frac{x^3}{3!}+(120c_2^3+ 120 c_2 c_3 + 24 c_4)\frac{x^4}{4!} + \cdots.$

$= x + c_1 x^2+(2c_2^2+c_3)x^3+(5c_2^3+ 5 c_2 c_3 + c_4)x^4+ \cdots.$

The associated binomial Sheffer polynomials as given by A130561 are the refined Lah polynomials

$e^{tf^{-1}(x)}=e^{xp.(t)}= 1 + tx + (t^2-2c_2t) \frac{x^2}{2!} + (t^3-6c_2t^2-6c_3t) \frac{x^3}{3!} + \cdots.$

Then, e.g.,

$p_3(a.) = -6c_3a_1-6c_2a_2+a_3=0,$ or, equivalently,

$a_3 = 6c_3a_1+6c_2a_2=6c_3+6c_2(2c_2) = 6c_3+12c_2^2.$

And,

$p_2(a.)p_2(a.) = (t^2-2c_2t)^2||_{t^n=a_n}= a_4-4c_2a_3+4c_2^2a_2$

$=(120c_2^3+120c_2c_3+24 c_4)-4c_2(12c_2^2+6c_3)+4c_2(2c_2)$

$=80 c_2^3+96c_2c_3+24c_4=80 (h_1/2)^3+96(h_1/2)(-h_2/3)+24h_3/4$

$=2(5h_1^3-8h_1h_2+3h_4)=c^1_{2,2},$

agreeing with the earlier result.

As another example, consider A134264, enumerating non-crossing partitions of polygons, giving the inversion partition polynomials in terms of the coefficients of the reciprocal

$h(x) = 1 + h_1 x + h_2 x^2 + \cdots = x/f^{-1}(x) = 1/[1-c_2 x - c_3 x^2 - c_4 x^3 - \cdots ].$

(The indeterminates here are not the complete homogeneous symmetric polynomials .)

Then

$f(x) = e^{a.x} = x + h_1x^2 + (h_2+h_1^2) x^3 + (h_3 + 3h_1h_2+h_1^3) x^4 + \cdots ,$

and, either from the e.g.f. $e^{tf^{-1}(x)}= e^{xp.(t)}$ or using the raising op

$R = t \cdot 1/f'(D_t) = t - 2h_1tD_t + (h_1^2-3h_2)tD_t^2-4h_3tD_t^3+ \cdots,$

$p_0 = 1$

$p_1(t) = t$

$p_2(t) = t^2 - 2h_1t$

$p_3(t) = t^3-6h_1t^2+6(h_1^2-h_2)t$

$p_4(t) = t^4 -12h_1t^3+12(3h_1^2-2h_2)t^2-24(h_1^3-2h_1h_2+h_3)t.$

And consistently, $p_4(a.)=0$ gives

$a_4 = 12h_1a_3-12(3h_1^2-2h_2)a_2+24(h_1^3-2h_1h_2+h_3)a_1$

$= 72h_1(h_2+h_1^2)-24(3h_1^2-2h_2)h_1+24(h_1^3-2h_1h_2+h_3)$

$= 4!(h_1^3+3h_1h_2+h_3).$