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August 16, 2024

Bernoulli Numbers and the Harmonic Oscillator

Posted by John Baez

I keep wanting to understand Bernoulli numbers more deeply, and people keep telling me stuff that’s fancy when I want to understand things simply. But let me try again.

The Bernoulli numbers can be defined like this:

xe x1=B 0+B 1x+B 2x 22!+B 3x 33!+ \frac{x}{e^x - 1} = B_0 + B_1 x + B_2 \frac{x^2}{2!} + B_3 \frac{x^3}{3!} + \cdots

and if you grind them out, you get

B 0 = 1 B 1 = 12 B 2 = 16 B 3 = 0 B 4 = 130 \begin{array}{lcr} B_0 &=& 1 \\ B_1 &=& -\frac{1}{2} \\ B_2 &=& \frac{1}{6} \\ B_3 &=& 0 \\ B_4 &=& -\frac{1}{30} \end{array}

and so on. The pattern is quite strange.

Bernoulli numbers are connected to hundreds of interesting things. For example if you want to figure out a sum like

1 10++1000 10 1^{10} + \cdots + 1000^{10}

you can use Bernoulli numbers — indeed Jakob Bernoulli boasted

It took me less than half of a quarter of an hour to find that the tenth powers of the first 1000 numbers being added together will yield the sum 91,409,924,241,424,243,424,241,924,242,500.

For some more mysterious appearances of the Bernoulli numbers, see:

But where the hell did this function x/(e x1)x/(e^x - 1) come from?

If DD means derivative:

(Df)(x)=f(x) (D f)(x) = f'(x)

then e D1e^D - 1 is a so-called ‘difference operator’:

((e D1)f)(x)=f(x+1)f(x) ((e^D - 1)f)(x) = f(x+1) - f(x)

which you can show using the Taylor series for ff if ff is an entire function. So D/(e D1)D/(e^D - 1) is about derivatives versus differences, and its inverse is about integrals versus sums. This lets you reduce sums like the one above to integrals… if you know your Bernoulli numbers. For details try this:

But x/(e x1)x/(e^x - 1) also shows up when you compute the expected energy of a quantum harmonic oscillator in thermal equilibrium!

Let’s work in units where Planck’s constant and Boltzmann’s constant are 11. Say we have a quantum harmonic oscillator whose allowed energies are 0,1,2,3,0, 1, 2, 3, \dots etcetera. (Sometimes people add 12\frac{1}{2} to each of these numbers, but let’s not.) If we compute this oscillator’s average or ‘expected’ energy at temperature TT, and divide it by TT, we get

xe x1 \frac{x}{e^x - 1}

where

x=1T x = \frac{1}{T}

So the quantum harmonic oscillator secretly knows about Bernoulli numbers.

What does this fact really mean??? I don’t know. I once read a book called Triangle of Thought about a conversation between Alain Connes and two other mathematicians, and he vaguely alluded to a fact of this sort, and said it was important.

I forget exactly what Connes said, so I imagine it’s something about how the Todd class in algebraic topology, which is usually defined using the function x/(1e x)x/(1 - e^{-x}) (whose power series also gives Bernoulli numbers), can be understood using the harmonic oscillator—perhaps because it appears in the Riemann–Roch theorem, which can probably be proved using ideas from quantum field theory (since it’s a special case of the Atiyah–Singer index theorem, which has a quantum proof). But all this erudite stuff is probably a complicated spinoff of the basic ideas, and I’d like to understand the basic ideas first.

By the way, here you can see a calculation of the expected energy of a quantum harmonic oscillator:

And here’s a little sanity check which is rather revealing. I said the expected energy divided by temperature is

xe x1 \frac{x}{e^x - 1}

where x=1/Tx = 1/T. But in the limit T+T \to +\infty the quantum harmonic oscillator should reduce to the classical harmonic oscillator. For that, the expected energy divided by temperature is just 1, since the oscillator has 2 degrees of freedom (position and momentum), and the equipartition theorem, which holds for classical systems with quadratic Hamiltonians, says we should get 1/2 times the number of degrees of freedom. And indeed, it works just as we’d expect:

lim x0xe x1=1 \lim_{x \to 0} \frac{x}{e^x - 1} = 1

This limit is also, by definition, the zeroth Bernoulli number. So the zeroth Bernoulli number is telling us the energy over temperature of a quantum harmonic oscillator in the high-temperature limit. The rest of the Bernoulli numbers are telling us all the ‘low-temperature corrections’ to the oscillator’s energy over temperature:

xe x1=B 0+B 1x+B 2x 22!+B 3x 33!+ \frac{x}{e^x - 1} = B_0 + B_1 x + B_2 \frac{x^2}{2!} + B_3 \frac{x^3}{3!} + \cdots

where x=1/Tx = 1/T.

I hope that if I think about it a bit harder, I’ll see that we’re using an analogy:

derivative operator DD : difference operator e D1e^D - 1 ::
classical harmonic oscillator : quantum harmonic oscillator

and that Bernoulli numbers arise from comparing the derivative and the difference operator — or comparing the classical harmonic oscillator with its continuous energy spectrum, to the quantum oscillator, with its discretely spaced energy spectrum — or more poetically, the continuous and the discrete!

Posted at August 16, 2024 2:41 PM UTC

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Re: Bernoulli Numbers and the Harmonic Oscillator

Misc. comments:

You might be interested in “Interactions between Lie theory and algebraic geometry” by Shilin Yu, tying together the GRR and the BCH formula.

The Todd op (mod signs) and its inverse are usually presented as differential ops, but, in a new MathOverflow question (not well received), I present the Todd op and its inverse also in terms purely of the finite difference op, which makes a direct connection to the Helmut Hasse formula for the Hurwitz zeta function and the Bernoulli function that Milnor presented in one of his papers. (The finite diff rep actually allows one of his divergent series derived from the diff op rep and related to the Harer-Zagier formula to be derived as a convergent series, but it is the asymptotic series that is desired.)

The Hurwitz ζ(s,z)\zeta(s,z) is related to the Bernoulli function via the derivative B s(z)= zζ(s,z)=sζ(s+1,z)B_s(z)=-\partial_z \zeta(-s,z) = -s\zeta(-s+1,z), which evaluates as the Bernoulli polynomial B n(x)B_n(x) for a positive integer n=sn=s. Alternatively, a Mellin transform gives the same result. The Hurwitz zeta and the Barnes zeta function (which can be expanded in terms of the Hurwtiz zeta) are related to the Casimir effect, Bose-Eisntein condensation, and solutions for the Schrodinger equation of atoms in a harmonic oscillator potential in “Basic zeta functions and some applications in physics” by Klaus Kirsten.

Posted by: Tom Copeland on August 16, 2024 9:09 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

Shilin Yu’s article Interactions between Lie theory and algebraic geometry is interesting, but it’s a personal research statement, so it doesn’t try to explain the connection Kontsevich discovered between the Duflo isomorphism and the Hochschild–Kostant–Rosenberg isomorphism, much less what I’m really interested in: why both these isomorphisms involve an expression that resembles the inverse of the Todd class, namely the ‘Duflo element’

J(T)=det(1e TT) J(T) = \det\left(\frac{1- e^{-T}}{T}\right)

of a linear transformation TT, or in fact actually the square root of this Duflo element. A much more expository account is here:

This is really nice. However, it still starts by introducing the formula for the Duflo element or its close relative

det(sinh(T/2)T/2) \det\left(\frac{\mathrm{sinh}(T/2)}{T/2}\right)

by fiat rather than by motivating it. It then explains a lot of other interesting math before explaining Kontsevich’s work.

Right now I’m trying to go down to the basic ideas, not up into the applications. So what I’d really enjoy is someone’s simple explanation of why you’d come up an expression like

det(1e TT) \det\left(\frac{1- e^{-T}}{T}\right)

or

x1e x \frac{x}{1 - e^{-x}}

if you didn’t already know about it. There are probably a number of good answers. I gave two in my post here.

I enjoyed this one:

He says the only power series f(x)f(x) for which f(0)=1 f(0) = 1 and the coefficient of x nx^n in f(x) n+1f(x)^{n+1} is always 11 is

f(x)=x1e x f(x) = \frac{x}{1 - e^{-x}}

And he explains why this funny condition on coefficients is necessary for the Todd class of P n\mathbb{C}\mathrm{P}^n to evaluate to 11 on the fundamental class of P n\mathbb{C}\mathrm{P}^n.

However, this still feels a bit ‘fancy’ to me. So I would like to connect this to the fact I presented here: at temperature TT, the ratio of the expected energy of the quantum harmonic oscillator with energy levels 0,1,2,0,1,2,\dots to the expected energy of the classical harmonic oscillator with the same frequency is

f(x)=x1e x f(x) = \frac{x}{1 - e^{-x}}

where x=1/Tx = 1/T.

I guess one concrete way forward is to think about the meaning of the powers of f(x)f(x), and their coefficients, from a physics viewpoint.

Posted by: John Baez on August 17, 2024 10:00 AM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

In the book Concrete Mathematics they hand-wave the Euler summation formula based on the operator equation you mentioned elsewhere,

Δ=exp(D)1\Delta = \exp(D) - 1

where D=ddxD = \frac{d}{d x}. Inverting, we get

=1exp(D)1=1D(Dexp(D)1)=1D+\sum = \frac{1}{\exp(D)-1} = \frac{1}{D} (\frac{D}{\exp(D)-1}) = \frac{1}{D} + \ldots

the first term of which is \int. That is, a sum equals an integral plus Bernoulli correction terms. Anyway that’s how I motivate x/(1exp(x))x/(1-\exp(-x)) to people.

I feel like there should be a generalization in which one compares two complex-oriented cohomology theories, involving a ratio of something involving their formal group laws, where the x/(1exp(x))x/(1-\exp(-x)) case is about HH vs. KK. More specifically, let xx be a weight for a torus action on a line LL, and consider the class of 0L\vec 0 \in L as an element of equivariant cohomology/KK-theory. I would write those as xx, 1exp(x)1-\exp(-x) respectively, though they live in different enough rings that it’s tricky to think about dividing one by the other.

Posted by: Allen Knutson on August 17, 2024 4:04 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

Allen wrote:

I feel like there should be a generalization in which one compares two complex-oriented cohomology theories, involving a ratio of something involving their formal group laws…

That reminds me of this paper that Jack Morava pointed me to when I told him I was struggling to understand Bernoulli numbers:

He defines generalized Bernoulli numbers for any formal group FF over a commutative algebra AA over \mathbb{Q}. Apparently there’s a unique isomorphism of formal groups

log F:FG a \log_F: F \to G_a

where G aG_a is the additive formal group of AA, and this has an inverse

exp F:G aF \exp_F: G_a \to F

Then he defines ‘Bernoulli numbers’ B n(F)AB_n(F) \in A by

Texp F(T)= n=0 B n(F)n!T n \frac{T}{\exp_F(T)} = \sum_{n = 0}^\infty \frac{B_n(F)}{n!} T^n

When F=G aF = G_a these Bernoulli numbers are all zero except for B 0(F)=1B_0(F) = 1, since exp F\exp_F is the identity. When F=G mF = G_m is the multiplicative formal group of AA we get the usual Bernoulli numbers since then exp F(T)=1e T\exp_F(T) = 1 - e^{-T}.

So this sounds less general than what you’re dreaming of, but still hard for me to understand! For example, I don’t even see why exp G m(T)=1e T\exp_{G_m}(T) = 1 - e^{-T}. But it sounds like what I need is sitting in front of me. I think I need to just buckle down, learn some stuff and think about it.

Posted by: John Baez on August 17, 2024 5:19 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

A convention in the FG literature is to write the completion of a nice group law FF with identity element ee in terms of a coordinate near the identity, for example

(0x)+ 𝔾 a(0y)=0(x+y)=0x+ 𝔾^ ay, (0-x)+_{\mathbb{G}_a}(0-y) = 0-(x+y) = 0-x+_{\hat{\mathbb{G}}_a}y,

while

(1x)+ 𝔾 m(1y)=1(x+yxy)=1x+ 𝔾^ my, (1-x)+_{\mathbb{G}_m}(1-y) = 1-(x+y-x y) = 1-x+_{\hat{\mathbb{G}}_m}y ,

and similarly

log 𝔾^ m(x)=log 𝔾 m(1x). - \log_{\hat{\mathbb{G}}_m}(x) = \log_{\mathbb {G}_m}(1-x) .

With this convention, Miller’s account of the generalized Hirzebruch genus may be exactly what Allen K is asking for. The issue is that the logarithm of a formal group law is a priori only defined over \mathbb{Q}, e.g.

log 𝔾^ m(x)= n1x nn. \log_{\hat{\mathbb{G}}_m}(x) = -\sum_{n \geq 1} \frac{x^n}{n} .

Posted by: jack morava on August 17, 2024 9:22 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

Thanks to John for help posting the note above. In my haste I forgot to say that (good multiplicative) cohomology theories over \mathbb{Q}-algebras are, like cats in the dark, all isomorphic, so the ratio in the Miller’s formula makes sense, cf eg

https://ncatlab.org/nlab/show/Chern-Dold+character

Posted by: jack morava on August 17, 2024 10:05 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

There are other interesting number arrays and associated combinatorics related to the deformed Todd operator variant of 11te x\frac{1}{1-te^x} (see OEIS A131758 for a ref). A Taylor series expansion about x=0x=0 gives rational functions with the Eulerian polynomials (see OEIS A008292 & A173018) in the variable tt in the numerator, and, when these are expanded as a power series, the powers of the integers appear; e.g., for x 4/4!x^4/4! the coefficient is (t 4+11t 3+11t 2+t)/(1t) 5=t+2 4t 2+3 4t 3+5 4t 4+(t^4+11t^3+11t^2+ t)/ (1-t)^5 = t + 2^4 t^2 + 3^4 t^3 + 5^4 t^4 + \cdots. The Eulerian polynomials are related to a slew of other number arrays and functions important in analysis and number theory as noted in the OEIS entries above, including, of course, the Bose-Einstein and Fermi-Dirac distributions.

Posted by: Tom Copeland on August 16, 2024 10:00 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

The definition of the Bernoulli numbers that sticks in my head is the one in Pierre Cartier’s paper Mathemagics: they are defined by the equation

B n=(B+1) n B^n = (B + 1)^n

for n2n \geq 2, and by B 0=1B^0 = 1. Then you change superscripts to subscripts.

This means the following. We start with B 0=B 0=1B_0 = B^0 = 1. Then the equation above with n=2n = 2 says

B 2=B 2+2B 1+1. B^2 = B^2 + 2B^1 + 1.

Cancelling gives

0=2B 1+1 0 = 2B^1 + 1

and so B 1=B 1=1/2B_1 = B^1 = -1/2.

Next, take n=3n = 3. The equation above says

B 3=B 3+3B 2+3B 1+1. B^3 = B^3 + 3B^2 + 3B^1 + 1.

Cancelling and changing superscripts to subscripts gives

0=3B 2+3B 1+1. 0 = 3B_2 + 3B_1 + 1.

Then substituting in the value B 1=1/2B_1 = -1/2 that we just computed gives B 2=1/6B_2 = 1/6.

“And so on”. Clearly this is formal nonsense, but it’s formal nonsense that works.

Posted by: Tom Leinster on August 17, 2024 3:53 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

This is very pretty, and also quite practical if you’re wanting to compute some Bernoulli numbers. It seems memorable, yet in fact I’d forgotten it—even though 21 years ago I gave a homework problem where I led students through how to derive it from

xe x1= n=0 B nx nn! \frac{x}{e^x - 1} = \sum_{n = 0}^\infty B_n \frac{x^n}{n!}

(It’s problem 8 here.)

Is this stuff about acting like subscripts are powers the ‘umbral calculus’? I can never remember what the umbral calculus is, though I remember Rota trying to rescue it from the shadow of disrepute.

Posted by: John Baez on August 17, 2024 5:34 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

Alas, Pierre Cartier died on Saturday, August 17th, 2024.

Anyone who never met him should watch this interview of him, in French with English subtitles. It really gets across his personality. For some of his thoughts related to “mathemagics”—that is, math using mysterious formal manipulations—go to 48:32. Great expression on his face when he describes how Bourbaki’s negative attitude to mathematical physics led them to neglect two of Weyl’s most important books. Then he turns to Boole’s algebraic approach to logic, then Dirac, then Heaviside….

Posted by: John Baez on August 19, 2024 8:56 AM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

Alas. He was a great man, with far more influence on the history of math than many another more well-known figure. His account of Grothendieck (as a person) is for example a non-mathematical example of the depth and lucidity of his thinking.

He had a sense of humor and a sense of irony. This really hurts.

Posted by: jack morava on August 19, 2024 12:29 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

Oh, that’s a great loss. I met him a few times at the IHES and conferences, and he seemed like one of those guys who would go on forever, in unbelievably robust health. His knowledge was amazing, and although I didn’t know him well, I had the strong impression that he loved to share it: he was always holding forth, in French or English, always with a little audience listening with rapt attention. The interview John links gives a flavour of that.

Posted by: Tom Leinster on August 19, 2024 1:23 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

It is interesting that those formulae show up that way. When one derives the formulae for the sum of consecutive powers using the binomial theorem and the fact that the difference operator and summation operator are roughly inverses, this kind of patterns shows up there.

I tend to think that the Bernoulli numbers are encoding something deep about the difference between the discrete and continuous. You see this with the shift map being the exponential of the derivative and the Euler -Maclaurin summation.

Posted by: Ana N Mouse on August 22, 2024 2:44 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

Let’s look at some really simple umbral Sheffer calculus (courtesy of Blissard–the gnarly stuff was used by Sylvester and Cayley in their explorations of invariant theory). Let the umbral substitution op and its action be given by a differential op as

(a.) n=S xa.x n=e a. x=0x n(a.)^n = S_{x \to a.} x^n = e^{a.\partial_{x=0}}x^n

= k0(a.) k x=0 kk!x n= k0a k x=0 kk!x n=a n. = \sum_{k \geq 0} (a.)^k \frac{\partial_{x=0}^k}{k!} x^n = \sum_{k \geq 0} a_k \frac{\partial_{x=0}^k}{k!} x^n = a_n.

Then we can easily rigorously derive a fundamental identity of the finite difference calculus

(a.) n=a n=S xa.x n=e a. x=0x n=e a. x=0(1(1x)) n(a.)^n = a_n = S_{x \to a.} x^n = e^{a.\partial_{x=0}}x^n = e^{a.\partial_{x=0}}(1-(1-x))^n

=e a. x=0 k=0 n(1) k(nk) j=0 k(1) j(kj)x j = e^{a.\partial_{x=0}} \sum_{k=0}^n (-1)^k\binom{n}{k} \sum_{j=0}^k (-1)^j \binom{k}{j}x^j

= k=0 n(1) k(nk) j=0 k(1) j(kj)a j=(1(1a.)) n, =\sum_{k=0}^n (-1)^k\binom{n}{k} \sum_{j=0}^k (-1)^j \binom{k}{j} a_j= (1-(1-a.))^n,

and this suggests another useful rep for umbral substitution

a n=(a.) n=S xa.x n=e (1a.) x=1x n=(1(1a.)) na_n = (a.)^n = S_{x \to a.} x^n = e^{-(1-a.)\partial_{x=1}}x^n = (1-(1-a.))^n

that can lead to interpretation for complex ss of

(a.) s=a s=(1(1a.)) s=e (1a.) x=1x s(a.)^s = a_s = (1-(1-a.))^s =e^{-(1-a.)\partial_{x=1}}x^s

= k0(1) k(sk) j=0 k(1) j(kj)a j = \sum_{k \geq 0} (-1)^k\binom{s}{k} \sum_{j=0}^k (-1)^j \binom{k}{j} a_j

as a Newton series, which is intimately related to Mellin transform interpolation and its analytic continuation. (With a j=(x+j) αa_j = (x+j)^{\alpha} and s=b.s = b., the Bernoulli number umbra, the Newton series gives a variant of the Helmut Hasse formula for the Hurwitz zeta function, the Bernoulli function B α(x)=αζ(α+1,x)B_\alpha(x) = -\alpha \; \zeta(-\alpha+1,x) of Milnor.)

The sequence of Bernoulli polynomials is an iconic Appell-Sheffer polynomial sequence (ASPS). Given any sequence of real or complex numbers, say a na_n, with a 0=1a_0 =1, an ASPS can be defined by umbral translation, or binomial convolution, via

A n(x)=T x(x+a.)x n=(a.+x) n=e a. xx nA_n(x) = T_{x \to (x+a.)} x^n = (a.+x)^n = e^{a.\partial_x}x^n

= k=0 n(nk)a kx nk. =\sum_{k=0}^n \binom{n}{k} a_k x^{n-k} .

The Bernoulli polynomials and numbers B n(x)B_n(x) and b nb_n can be generated by

B n(x)=(b.+x) n=e b. xx n= xe x1x n,B_n(x) = (b. +x)^n = e^{b.\partial_x} x^n = \frac{\partial_x}{e^{\partial_x}-1} x^n,

so the associated e.g.f.s are

e B.(x)t=e (b.+x)t=e b. xe xt=e b.te xt=te t1e xt.e^{B.(x)t} = e^{(b.+x)t} = e^{b.\partial_x}e^{xt} =e^{b.t}e^{xt} = \frac{t}{e^t-1}e^{xt} .

Then

e B.(1)t=e (b.+1)t=e b.te t=te t1e t=te t1=e b.t,e^{-B.(1)t} = e^{-(b.+1)t} = e^{-b.t}e^{-t} = \frac{-t}{e^{-t}-1}e^{-t} =\frac{t}{e^{t}-1} = e^{b.t},

so

(B.(1)) n=(1) n(b.+1) n=(b.) n=b n,(-B.(1))^n = (-1)^n (b.+1)^n = (b.)^n = b_n,

the recursion relation noted by Cartier in Tom Leinster’s comments.

In these ways, the Bernoulli-Hirzebruch-Todd operator arises naturally in the umbral Sheffer operator calculus along with the associated inverse ops and the raising ops, which also involve the Bernoulli numbers, or more accurately the Riemann zeta numbers ζ(n)\zeta(-n).

Of course, there are direct formulas for the Bernoulli numbers involving the Stirling numbers of the second kind or the face counts of the permutahedra, which are easily derived from the umbral Sheffer calculus, and these you find in toric topology and soliton solutions of the KdV equation and associated quadratic Riccati equation. In fact, from my formulas in OEIS A008292 on the Eulerian polynomials, you can discern a relation of the Bernoulli numbers to a formal group law and generalized Hirzebruch-Todd classes as John discusses. Expand as a Taylor series the shifted reciprocal of the bivariate e.g.f. for the Eulerian polynomials A(x,a,b)= (e^(ax)-e^(bx))/(ae^(bx)-be^(ax)), that is , x * (ae^(b x)-be^(a x)) / (e^(a x)-e^(b x)), using Wolfram Alpha and you will find the Bernoulli numbers. See the links adjacent to the OEIS formulas for refs on the associated generalized cohomology theory and toric topology (related to the convex polytopes the associahedra, permutahedra, and stellahedra).

For a deeper investigation of the applications of umbral calculus to topology, see “Symbolic calculus: A 19th century approach to MU and BP” by Nigel Ray in Homotopy Theory: Proceedings of the Durham Symposium 1983, London Mmathematical Society Lecture Notes Series, 117.

See the two MSE questions “What’s umbral calculus about?” and “Question about “baffling” umbral calculus result” and the MO-Q “How are Sheffer polynomials related to Lie theory?” for more introductory info and links.

The relationships among the Newton series of the finite difference calculus; the compositional inverse pair h(x)=e x1h(x) = e^x-1 and h 1(x)=ln(1+x)h^{-1}(x)=\ln(1+x); the umbral inverse pair of binomial Sheffer Stirling polynomials of the second and first kinds, a.k.a. the Bell/Touchard/exponential and the falling factorial polynomials–core sequences in combinatorics; and the Appell Sheffer Bernoulli polynomials and their umbral inverse polynomials are perhaps most easily and elegantly revealed with the umbral Sheffer calculus. I can post a new short note on these in my web blog if you are interested. They reflect the relations among the basic Lie group operations of translation and multiplicative and compositional inversions.

Posted by: Tom Copeland on August 17, 2024 11:55 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

Thanks for the offer! Right now I’m trying to understand why the Bernoulli numbers are so important in topology. I think for that I need to understand in very simple terms how the relation between additive and multiplicative formal group laws gives rise to the Grothendieck–Hirzebruch–Riemann–Roch formula and the function x/(e x1)x/(e^x - 1). I believe for this I mainly need to understand this:

and the comments here by Allen Knutson and Jack Morava. I feel I’m pretty close.

Posted by: John Baez on August 25, 2024 11:27 AM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

You might find my new post helpful in delineating the particulars between a general formal group law, the Bernoulli numbers, the Hirzebruch criterion, and the K-P formula:

Umbral witchcraft with Bernoulli and Blissard.

Posted by: Tom Copeland on August 25, 2024 10:14 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

Yeah that was the link I posted before. The UI here is maddening.

Posted by: Steve Huntsman on August 26, 2024 1:58 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

Yes, I learned about the MathOverflow post Euler-Maclaurin formula and Riemann-Roch from your comment, and I thought more people would look at it if it came with a title. Sorry for not crediting you — it was very helpful.

You can use HTML or Markdown here, so it’s pretty flexible, but alas, you can cannot drop a URL here and have it automatically turn into a clickable link. So, typing

https://mathoverflow.net/questions/10667/

gives merely

https://mathoverflow.net/questions/10667/

while typing

[https://mathoverflow.net/questions/10667/](https://mathoverflow.net/questions/10667/)

gives

https://mathoverflow.net/questions/10667/

and typing

[Euler-Maclaurin formula and Riemann-Roch](https://mathoverflow.net/questions/10667/)

gives

Euler-Maclaurin formula and Riemann-Roch

which is probably what most people would enjoy seeing.

Posted by: John Baez on August 26, 2024 2:50 PM | Permalink | Reply to this

Euler-Maclaurin and Riemann-Roch

https://mathoverflow.net/questions/10667/

Posted by: Steve Huntsman on August 19, 2024 1:13 PM | Permalink | Reply to this

Re: Euler-Maclaurin and Riemann-Roch

This might be relevant or interesting:

M Ando, J Morava, A renormalized Riemann-Roch formula and the Thom isomorphism for the free loop space, in Topology, geometry, and algebra: interactions and new directions (Stanford, CA, 1999), Contemp. Math. 279, Amer. Math. Soc. 2001,

https://arxiv.org/abs/math/0101121

Posted by: jack morava on August 19, 2024 11:16 PM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

The question that now jumps out at me: why should Temperature be like Integrating? (Or coolness (see Baez on What is Entropy)) like Differentiating?)

Posted by: Jesse C. McKeown on August 25, 2024 11:34 AM | Permalink | Reply to this

Re: Bernoulli Numbers and the Harmonic Oscillator

Good question! I think it’s because multiplying by xx is differentiation conjugated by the Laplace transform, while multiplying by 1/x1/x is integration conjugated by the Laplace transform.

(In my post here, I used x=1/Tx = 1/T to stand for coolness. Talking to physicists, I’d instead write coolness as β=1/kT\beta = 1/k T. There are more important subtleties about boundary conditions:

(f)(s)=s(f)(s)f(0) \mathcal{L} (f') (s) = s \mathcal{L}(f) (s) - f(0)

but I think the basic idea is right.)

Posted by: John Baez on August 25, 2024 5:51 PM | Permalink | Reply to this

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