## Why do Bernoulli numbers arise everywhere?

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I have seen Bernoulli numbers many times, and sometimes very surprisingly. They appear in my textbook on complex analysis, in algebraic topology, and of course, number theory. Things like the criteria for regular primes, or their appearance in the Todd class, zeta value at even numbers looks really mysterious for me. (I remember in Milnor's notes about characteristic class there is something on homotopy group that has to do with Bernoulli numbers, too, but I don't recall precisely what that is. I think they also arise in higher K-theory.)

The list can go on forever. And the wikipedia page of Bernoulli number is already quite long.

My question is, why do they arise everywhere? Are they a natural thing to consider?

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p.s.----(maybe this should be asked in a separate question)

Also, I've been wondering why it is defined as the taylor coefficient of the particular function x/(e^x-1), was this function important? e.g. I could have taken the coefficient of the series that defines the L-genus, namely $\dfrac{\sqrt{z}}{\text{tanh}\sqrt{z}}$, which only amounts to change the Bernoulli numbers by some powers of 2 and some factorial. I guess many similar functions will give you the Bernoulli numbers up to some factor. Why it happen to be the function x/(e^x-1)?

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According to history of mathematics, at the beginning of Bernoulli's original idea, he did not begin defined as the Taylor coefficients of the function x/(e^x-1) at 0. – yaoxiao Apr 11 2011 at 2:56
That's a good point. So do you know how did he define it and what's his motivation? – 36min Apr 11 2011 at 4:19
Just a minor remark: the space of the areas you have mentioned might be simply connected. For example, the criteria for regular primes, the zeta values at even integers, and higher K-theory - these are all closely related areas. If you are willing to take on board everything we know and everything we conjecture to be true, then it follows that Bernoulli numbers appear in one of these if and only if they appear in all of them. Somewhat related is my answer here: mathoverflow.net/questions/45376/… – Alex Bartel Apr 11 2011 at 4:42

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@Alex, I take it that by "simply connected" you mean "connected in a simple way". – gowers Apr 11 2011 at 8:32
@Tim You are right, I guess I didn't mean simply connected, but rather a complete graph or something like that. – Alex Bartel Apr 11 2011 at 12:54
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## 6 Answers

I don't know of a universal theory of all places where Bernoulli numbers arise, but Euler-Maclaurin summation explains many of their more down-to-earth occurrences.

The heuristic explanation (due to Lagrange) is as follows. The first difference operator defined by $\Delta f(n) = f(n+1)-f(n)$ and summation are inverses, in the same sense in which differentiation and integration are inverses. This just amounts to a telescoping series: $\sum_{a \le i < b} \Delta f(i) = f(b) - f(a)$.

Now by Taylor's theorem, $f(n+1) = \sum_{k \ge 0} f^{(k)}(n)/k!$ (under suitable hypotheses, of course). If we let $D$ denote the differentation operator defined by $Df = f'$, and $S$ denote the shift operator defined by $Sf(n) = f(n+1)$, then Taylor's theorem tells us that $S = e^D$. Thus, because $\Delta = S-1$, we have $\Delta = e^D - 1$.

Now summing amounts to inverting $\Delta$, or equivalently applying $(e^D-1)^{-1}$. If we expand this in terms of powers of $D$, the coefficients are Bernoulli numbers (divided by factorials). Because of the singularity at "$D=0$", the initial term involves antidifferentiation $D^{-1}$, i.e., integration. Thus, we have expanded a sum as an integral plus correction terms involving higher derivatives, with Bernoulli number coefficients.

Specifically, $$\sum_{a \le i < b} f(i) = \int_a^b f(x) \, dx + \sum_{k \ge 1} \frac{B_k}{k!} (f^{(k-1)}(b) - f^{(k-1)}(a)).$$ (Subtracting the values at $b$ and $a$ just amounts to the analogue of turning an indefinite integral into a definite integral.)

This equation isn't literally true in general: the infinite sum usually won't converge and there's a missing error term. However, it is true when $f$ is a polynomial, and one can bootstrap from this case to the general one using the Peano kernel trick.

So from this perspective, the reason why $t/(e^t-1)$ is a natural generating function to consider is that we sometimes want to invert $e^t-1$ (the factor of $t$ is just to make it holomorphic), and the most important reason I know of to invert it is that we want to invert $\Delta = e^D-1$.

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If I recall correctly, the infinite sum converges to the correct answer also for polynomials times exponentials. – Allen Knutson Apr 11 2011 at 3:42
You can do this in higher dimensions too, integrating over polytopes. – Steve Huntsman Apr 11 2011 at 12:44

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See also this question: mathoverflow.net/questions/10667 – aorq Apr 12 2011 at 0:34
Also umbrally $p(B(x)+1)-p(B(x))={p}'(x)$ where $B(x)$ are the Bernoulli polynomials and $p(x)$ any polynomial, so you might expect them to pop up in approximation and linearization problems. – Tom Copeland Sep 18 at 2:56

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The main reason I know for the appearance of Bernoulli numbers is the one Henry Cohn already explained: we'd like to invert the difference operator $e^D - 1$, so we'd like to expand $1/(e^D - 1)$ as a Taylor series. But $1/(e^x - 1)$ doesn't have a Taylor series, because it has a pole at the origin. It has a perfectly nice Laurent series, but just to make things more obscure people prefer to discuss the Taylor series of $x/(e^x - 1)$. And the coefficients of this are called Bernoulli numbers.

I understand how Bernoulli numbers are used to compute $\sum_{i=1}^n i^k$ and how they show up in formulas for the Riemann zeta function.

However, Alain Connes loses me here:

• Alain Connes, Andre Lichnerowicz and Marcel Paul Schutzenberger, A Triangle of Thoughts, AMS, Providence, 2000.

He points out that if $H$ is the Hamiltonian for some sort of particle in a box and $\beta$ is the inverse temperature,

$$1/(1 - e^{-\beta H}) = 1 + e^{-\beta H} + e^{-2 \beta H} + \cdots$$

is the operator you take the trace of to get the partition function for a collection of an arbitrary number of particles of this sort. And he claims that pondering this explains all the appearances of $x/(1 - e^x)$ and the Bernoulli numbers in topology!

Does anyone understand that? I imagine he's hinting at some relation between characteristic classes, the heat equation, the Laplacian on differential forms, and things like that. But I've never understood how these pieces are supposed to fit together.

And here's something that remains more mysterious to me. The paper by Kervaire and Milnor has a cool formula for the order of the group of smooth structures on the (4n-1)-sphere for n > 1. It's:

$$2^{2n-4} (2^{2n-1} - 1) P(4n-1) B(n) a(n) / n$$

where:

$P(k)$ is the order of the $k$th stable homotopy group of spheres

$B(k)$ is the $k$th Bernoulli number, in the sequence 1/6, 1/30, 1/42, 1/30, 5/66, 691/2730, 7/6, ...

$a(k)$ is 1 or 2 according to whether k is even or odd

How do the Bernoulli numbers weasel their way into this game?

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I think I have a rough idea of how Bernoulli numbers get into exotic spheres. There's this "J-homomoprhism" Z x BO --> Pic(S), which sends a (stable) real vector bundle to the (stable) spherical fibration given by one-point compactifying the fibers. The Spivak normal fibration over S^n gives a canonical map S^n --> Pic(S), and finding manifold structures on S^n is tied to lifting this map along J (so saying that the normal fibration actually came from a normal vector bundle). This lifting problem comes down to understanding what J does on homotopy groups. This in turn can be understood ... – Dustin Clausen Apr 11 2011 at 16:19

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in terms of Pontryagn-Thom theory: a class in the kernel of J on pi_n is represented by an "exotic" null-bordism of S^n. These correspond to almost parallelizable manifolds (manifolds parallelizable away from a fixed point), and once you've got one of those most of your characteristic classes vanish, and (for instance) the integrality of the A-hat genus will tell you something about just one particular Bernoulli number. Well OK, but what does this have to do with Euler-Maclaurin summation?? :) – Dustin Clausen Apr 11 2011 at 16:23
John discussed the relevant passage of 'A Triangle of Thoughts' at greater length here golem.ph.utexas.edu/category/2008/02/… – David Corfield Apr 12 2011 at 8:27
Also Harer and Zagier have shown that the orbifold Euler characteristic of the moduli space $\mathcal M_{g,n}$ is $(-1)^{n-1}\frac{(2g+n-3)!}{(2g-2)!} \zeta(1-2g)$. – Tom Copeland Sep 18 at 2:46

In algebraic topology one key point is as follows. The complex $K$-theory spectrum has homotopy groups $KU_{\ast}=\mathbb{Z}[u,u^{-1}]$, with $u$ in degree two. This ring maps in an obvious way to $\pi_{\ast}(H\wedge KU)$, and it is not hard to calculate that the resulting map $\mathbb{Z}[u,u^{-1}]\to\pi_{\ast}(H\wedge KU)$ induces an isomorphism $\mathbb{Q}[u,u^{-1}]\to\pi_{\ast}(H\wedge KU)$. The ring $(H\wedge KU)^0(\mathbb{C}P^\infty)$ can be described as $\mathbb{Q}[[ux]]$ or as $\mathbb{Q}[[y]]$, where $x$ comes from $H^2(\mathbb{C}P^\infty)$ and $y$ comes from $KU^0(\mathbb{C}P^\infty)$. Specifically, if we let $L$ denote the tautological line bundle, then $y$ can be taken to be the $K$-theory class of the virtual bundle $L-1$. It then works out that $y=e^x-1$, so $x/y$ is the Bernoulli series. The Bernoulli numbers occur as coefficients of $x^k/k!$ rather than $x^k$ itself, which suggests that one should work with $\Omega S^3$ rather then $\mathbb{C}P^\infty$: there is a canonical map $\Omega S^3\to\mathbb{C}P^\infty$ using which we can identify $H^*(\Omega S^3)$ with the ring of all series of the form $\sum_ka_kx^k/k!$ with $a_k\in\mathbb{Z}$. All this is of course linked with Adams's treatment of the $J$-homomorphism and the $e$-invariant. However, I think that much of this is still mysterious, at least to me. I think there are some missing ingredients involving the relationship between $R$ and $gl_1(R)$ for various $E_\infty$ ring spectra $R$, particularly those related to surgery and the $J$-homomorphism. There is a lot of literature about this kind of thing from the 1970s but I have not managed to extract the answers that I wanted.

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I remember this question coming up one tea time when I was sitting next to Frank Adams, and he said the reasons for the occurrence of Bernoulli numbers in topology were "trivial". Makes me wish I'd asked "compared to what?" – Charles Matthews Apr 11 2011 at 14:44

I have been told (by J.H. Conway) that Bernoulli numbers were first discovered by Faulhauber. See the http://en.wikipedia.org/wiki/Faulhaber's_formula wikipedia article for details. This reference hardly answers your question, but one possible characterization is that they are useful in summing nth powers. That fact alone indicates that their ubiquity is quite natural to expect.

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Another way where they show up is in Lie theory. If you want to compute the derivative of the exponential map (of a Lie group) you encounter the function $x/(e^x - 1)$ quite inevitably. The already posted questions can partly be viewed as incarnations of this. THis results also in the appearence of the Bernoulli numbers in the BCH series, which is of course of fundamental importance far beyond the usage in Lie algebra theory...

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One of ways for understanding the power of Bernoulli numbers and your question is that Bernoulli numbers grow faster than any geometric series.see http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2023%20Bernoulli%20numbers.pdf

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