An important fact about the generalized von Mangoldt function

An important fact in analytic number theory, and especially the recent string of papers relating to the bounded gaps between primes, is the following: the generalized von Mangoldt function \Lambda_k(n) (defined below) vanishes when n has more than k distinct prime factors. I will give a proof of this fact here for those interested.

The generalized von Mangoldt function is defined as

\displaystyle \Lambda_k(n) = \sum_{d | n} \mu(d) \left(\log \frac{n}{d}\right)^k,

where \mu is the Mobius function. Unfortunately, like the usual von Mangoldt function (where k = 1), the generalized von Mangoldt function is not multiplicative. We will first treat the case when n is square-free. Write n = p_1 \cdots p_s, where s > k and the p_i are distinct primes. Then we see that

\displaystyle \Lambda_k(n) = (\log p_1 + \cdots + \log p_s)^k - (\log p_2 + \cdots + \log p_s)^k - (\log p_1 + \log p_3 + \cdots + \log p_s)^k - \cdots

\displaystyle + (-1)^{s-1} (\log p_1)^k + \cdots (-1)^{s-1} (\log p_s)^k.

Thus it suffices to check the following general polynomial identity

\displaystyle \Lambda_k(n) = (x_1 + \cdots + x_s)^k - (x_2 + \cdots + x_s)^k - (x_1 + x_3 + \cdots + x_s)^k - \cdots

\displaystyle + (-1)^{s-1} x_1^k + \cdots (-1)^{s-1} x_s^k = 0.

Here, the summands are all expressions of the form \displaystyle (-1)^{s - |S|}\left(\sum_{i \in S} x_i \right)^k for subsets S \subset \{1, \cdots, s\}.

To simplify the calculations, let us consider the coefficient of a single monomial x_1^{a_1} \cdots x_s^{a_s} that appears after we expand the above polynomial expression. Since s > k it follows that at most k of the a_i‘s are non-zero, and without loss of generality, assume that a_i is zero for i = k+1, \cdots, s. The only subsets S we need to consider are those such that \{1, \cdots, k\} \subset S. Each eligible subset S will contribute (-1)^{s - |S|} \binom{k}{a_1, \cdots, a_k} to the coefficient. There are \binom{s-k}{s - t} many eligible subsets S of size t. Whence, the coefficient is equal to

\displaystyle \binom{k}{a_1, \cdots, a_k} - (s-k)\binom{k}{a_1, \cdots, a_k} + \binom{s-k}{2} \binom{k}{a_1, \cdots, a_k} - \cdots

\displaystyle = \binom{k}{a_1, \cdots, a_k}\left(1 - (s-k) + \binom{s-k}{2} - \binom{s-k}{3} + \cdots \right) = 0,

by the binomial theorem and the fact that s - k > 0.

The general case is again done by verifying a general polynomial identity. For n = p_1^{b_1} \cdots p_s^{b_s}, where b_i \geq 1, write the linear form f(x_1, \cdots, x_s) = b_1 x_1 + \cdots b_s x_s. Then the vanishing of the generalized von Mangoldt function can be deduced from

\displaystyle f^k - (f - x_1)^k - \cdots - (f - x_s)^k + (f - x_1 - x_2)^k + \cdots + (f - x_{s-1} - x_s)^k - \cdots = 0.

The key is to match the powers of f in the above expression. For a fixed 0 \leq t \leq k, consider all terms where the exact power of f is t. We would have

\displaystyle f^t \left( (-1)^{k-t+1}(x_1^{k-t} + \cdots + x_s^{k-t}) + (-1)^{k-t+2}((x_1 + x_2)^{k-t} + \cdots + (x_{s-1} + x_s)^{k-t}) + \cdots \right).

By strong induction, the term in the parentheses would vanish according to the polynomial identity proved above. Since this is true for every 0 \leq t \leq k, we have the entire expression vanishing and we are done.

If anyone knows a shorter, cleaner proof I would be very happy to hear it.

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2 thoughts on “An important fact about the generalized von Mangoldt function

  1. David Tweedle

    Hi Stanley,
    I was reading one of your posts on facebook and eventually discovered this one.
    Very interesting blog!
    The polynomial (x_1 + \cdots + x_s)^k - (x_2 + \cdots + x_s)^k - \cdots
    is the coefficient of \lambda^k in the formal power series
    (\exp(\lambda x_1) -1)(\exp(\lambda x_2) -1)\cdots (\exp(\lambda x_s)-1)
    The coefficients of \lambda^k of the above power series vanish if k < s.
    I hope you find this interesting…
    -David Tweedle

    Reply

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