Now let's bypass Hirzebruch's criterion above and connect the Bernoullis, using their basic operational definition rather than their e.g.f., to the Todd genus and more through formal group laws and associated Lie ops.

First, define the Bernoulli polynomials as the Appell sequence, $(B.(0)+x)^n=B_n(x)$, such that, $$f(B.(x+1))-f(B.(x))={f}'(x)$$ when convergent. Action on $f(x)=e^{xt}$ gives the e.g.f. since $$e^{B.(x+1)t}-e^{B(x)t}=t\;e^{xt}$$ implies $$e^{B.(x)t}(e^t-1)=t\;e^{xt}$$ and $$e^{B.(x)t}=\frac{t}{e^t-1}e^{xt}.$$

Action on $f(x)=-ln(1-xt)$ gives

$$-ln\left [1-B.(x+1)t \right ]+ln\left [1-B.(x)t \right ]= ln\left [ \frac{1-B.(x)t}{1-(B.(x)+1)t} \right ]=\frac{d\left [ -ln(1-xt) \right ]}{dx}=\frac{t}{1-xt}\;,$$

an iconic o.g.f., and using the special linear fractional (Mobius) transformation $L(t,x)=\frac{t}{1+xt}$, whose inverse in $t$ is $L(t,-x)$, this can be expressed succinctly as

$$ln\left [ 1+L[t,-(B.(x)+1)] \right ]=ln\left [ \frac{1-B.(x)t}{1-(B.(x)+1)t} \right ]=L(t,-x)=\frac{t}{1-xt},$$

with compositional inverse in $t$

$$L[e^t-1,B.(x)+1]=\frac{e^t-1}{1+(B.(x)+1)(e^t-1)}=L(t,x)=\frac{t}{1+xt}.$$

Together they comprise the formal group law

$$FGL(y,z;x)=L[L(y,-x)+L(z,-x),x]=\frac{y+z-2x\;yz}{1-x^2yz},$$

which (according to Lenart and Zainoulline) corresponds to the Euler characteristic.

For $x=0$, $(B.(0))^n=B_n$ are the Bernoulli numbers, $(B.(0)+1)^n=(B.(1))^n=(-B.(0))^n=(-1)^nB_n$, and the FGL specializes to

$$FGL(y,z;0)=y+z,$$

the fundamental additive FGL associated with the infinitesimal generator $d/dt=D_t$ and the iterated op $(D_t)^n$ with action of translation $exp(x\;D_t)f(t)=f(t+x)$.

For $x=-1$, this specializes to

$$FGL(y,z;-1)=\frac{y+z+2yz}{1-yz},$$

a fundamental FGL associated to the infinitesimal generator $(1+t)^2\frac{d}{dt}$, and, with a shift in coordinates, to the iterated op $(t^2D_t)^n=t^{2n}Lah(:tD_t:)$, related to the Lah polynomials, with the action $exp(x\;t^2D_t)f(t)=f(t/(1-xt))$, the special linear fractional transformation.

More generally for indeterminates $a_n$ with $a_0=1$, action on $f(a.,t)=-ln(1-a.\;t)$

gives (precisely when convergent and formally usefully otherwise)

$$ln\left [ \frac{1-B.(a.)t}{1-(B.(a.)+1)t} \right ]=\frac{t}{1-a.t}=\sum_{n \ge 0} a_n t^{n+1},$$

an o.g.f. for the series, which itself can be extended as an Appell sequence defined by the o.g.f.

$$\boldsymbol{O}_A(t,x)=\frac{t}{1-(a.+x)t}=\sum_{n \ge 0} (a.+x)^n t^{n+1}=\sum_{n \ge 0} A_n(x) t^{n+1}.$$

Letting $a_n=\bar{B}.(x)$, the umbral compositional inverse for the Bernoulli polynomials, i.e., $B_n(\bar{B}.(x))=x^n=\bar{B}_n(B.(x))$, we get

$$\boldsymbol{O}_\bar{B}(t,x)=ln\left [ \frac{1-B.(\bar{B}.(x))t}{1-(B.(\bar{B}.(x))+1)t} \right ]=ln\left [ \frac{1-xt}{1-(x+1)t} \right ]$$

$$=\sum_{n \ge 0} \frac{(x+1)^{n+1}-x^{n+1}}{n+1} t^{n+1},$$

and, consistently,

$$\bar{B}_n(B.(x))=\frac{(B.(x)+1)^{n+1}-(B.(x))^{n+1}}{n+1}=\frac{d}{dx}\;\frac{x^{n+1}}{n+1}=x^n.$$

The compositional inverse, through the same substitution above, is

$$\boldsymbol{O}_{\bar{B}}^{(-1)}(t,x)=\frac{e^t-1}{1+(x+1)(e^t-1)}.$$

Together they comprise the formal group law

$$FGL(y,z;x)=\boldsymbol{O}_{\bar{B}}^{(-1)}[\boldsymbol{O}_\bar{B}(y,x)+\boldsymbol{O}_\bar{B}(z,x),x]=\frac{y+z-(1+2x)\;y z}{1-x(1+x)\;yz}.$$

For $x=-1$, this specializes to

$$FGL(y,z;-1)=y+z+yz,$$ the multiplicative FGL associated with the Todd genus and the infinitesimal generator $(1+t)\frac{d}{dt}$ at the identity related by a coordinate shift to the iterated op. $(t\frac{d}{dt})^n=\phi_n(:tD_t:)$, the Bell or Stirling polynomials of the second kind, with action $exp(x\;tD_t)f(t)=f(e^xt)$, a dilation.

For $x=-1/2$,

$$FGL(y,z;-1/2)=\frac{y+z}{1+\;yz/4},$$

the Lorentz FGL, related to the Atiyah-Singer signature (Lenart and Zainoulline, "Towards generalized cohomology Schubert calculus via formal root polynomials").

So the Bernoullis are taking us on a tour through some basic formal groups, associated genera, and the conformal Lie algebra $sl_2(C)$ and associated group, the conformal global subgroup of the Virasoro algebra.

(There's a relation to elliptic functions, and more, as discussed by L $ Z. For infinite (and finite) matrix reps related to the Pascal matrix, see my mini-arxiv at my website--the notes on Infinigens.)