3
$\begingroup$

I seek a reference for the fact that "coefficients of the Hirzebruch $L$-polynomial have odd denominators". The coefficients are $$\frac{2^{2k}(2^{2k-1}-1)B_k}{(2k)!}$$ where $B_k$ is the Bernoulli number, but I cannot locate the appropriate divisibility property of $B_k$. Of course, $2^{2k-1}-1$ is odd, so it can be ignored.

$\endgroup$
2
  • $\begingroup$ The coefficient you wrote is only the leading coefficient of $L_k$ (i.e. the coefficient of $p_k$). Were you asking only for this or were you asking about the coefficients of all the terms? $\endgroup$ Commented May 16, 2018 at 19:04
  • $\begingroup$ @MichaelAlbanese: I cannot recall but I think the link below solved my problem. $\endgroup$ Commented May 16, 2018 at 20:22

2 Answers 2

4
$\begingroup$

This follows from the Clausen - von Staudt theorem. See http://www.bernoulli.org (structure of the denominator)

$\endgroup$
1
  • $\begingroup$ Thank you. One needs also a little combinatorial argument on how many powers of two are in $(2k)!$, but I think I got it. $\endgroup$ Commented Jun 4, 2012 at 23:24
8
$\begingroup$

The expression you wrote is the coefficient of $p_k$ in $L_k$, and as Igor Rivin's answer rightly points out, it follows from the known properties of Bernoulli numbers that the denominator is odd.

There are (complicated) expressions for the other coefficients in $L_k$, see Theorem 1 of Hirzebruch $L$-polynomials and multiple zeta values by Berglund & Bergström. Instead of trying to deduce that the denominators of these expressions are odd directly, we have the following (lemma 1.5.2) from Hirzebruch's Topological Methods in Algebraic Topology.

The polynomial $L_k$ can be written in a unique way as a polynomial with coprime integer coefficients, divided by a positive integer $\mu(L_k)$, where

$$\mu(L_k) = \prod q^{\left\lfloor\tfrac{2k}{q-1}\right\rfloor}$$

is a product over all primes $q$ with $3 \leq q \leq 2k + 1$.

It follows that the denominator of each coefficient in $L_k$ is odd.

Note, coprime coefficients does not mean pairwise coprime. For example,

$$L_3 = \frac{62p_3 - 13p_2p_1 + 2p_1^3}{3^3\cdot 5\cdot 7}$$

and clearly $62$ and $2$ are not coprime.

There is an analogous result for $A$ and $\hat{A}$ polynomials. The following is a remark is section 1.6:

The polynomial $A_k$ can be written in a unique way as a polynomial with coprime integer coefficients multiplied by $2^{\alpha(k)}/\mu(L_k)$.

Here $\alpha(k)$ is the number of non-zero terms in the dyadic expansion of $k$. Using the fact that $A_k = 2^{4k}\hat{A}_k$ (see the top of page 197), we obtain:

The polynomial $\hat{A}_k$ can be written in a unique way as a polynomial with coprime integer coefficients divided by $2^{4k-\alpha(k)}\mu(L_k)$.

There is also a similar result for Todd polynomials (lemma 1.7.3).

The polynomial $T_k$ can be written in a unique way as a polynomial with coprime integer coefficients, divided by a positive integer $\mu(T_k)$, where

$$\mu(T_k) = \prod q^{\left\lfloor\tfrac{k}{q-1}\right\rfloor}$$

is a product over all primes $q$ with $2 \leq q \leq k + 1$. Moreover $\mu(T_{2k+1}) = 2\mu(T_{2k}) = 2^{2k+1}\mu(L_k)$.

None of these results are proved in Hirzebruch's book. Instead a reference is given to the paper Cohomologie-Operationen und charakteristische Klassen by Atiyah and Hirzebruch. The proof for Todd polynomials is given in section 3.7 after which it is stated that one can prove similar results for the polynomials $L_k$ and $A_k$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .