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I am going through Phil Hanlon's paper and on page 127, right after the first paragraph, "It is well known that.." which boils down to the following identity:

$$ \prod_{i=0}^{n-1}(\beta-i) = \sum_{\sigma \in S_n}\beta^{c(\sigma)} $$

where $c(\sigma)$ is the number of cycles (of any length) in a permutation.

I suspect this is not a difficult thing to prove, but I have not been able to find any literature that deals directly with this identity. Any help?

From the context, I think $ \beta $ is positive and strictly less than 1.

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  • $\begingroup$ Are you missing a sign somewhere? If $\beta=1$ then the LHS is zero but the RHS is $n!$. EDIT: for a fixed $n$, both sides are polynomials in $\beta$, so if they are equal for $0<\beta<1$ then they are equal for $\beta=1$ too. $\endgroup$
    – Linus Hamilton
    Commented Jul 24, 2016 at 19:46
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    $\begingroup$ I edited the question, I'm pretty sure that from the context, beta is positive strictly less than 1. $\endgroup$
    – I.K
    Commented Jul 24, 2016 at 19:51
  • $\begingroup$ Looking at the paper, I think it's $\sum_{i=0}^{n-1} (1-i\alpha) = \prod_{\sigma \in S_n} \alpha^{n-c(\sigma)}$. $\endgroup$
    – Brian Hopkins
    Commented Jul 24, 2016 at 19:51
  • $\begingroup$ Yes. I rewrote it. I used $ \beta = \alpha^{-1} $ Then it's the same equality, isn't it? $\endgroup$
    – I.K
    Commented Jul 24, 2016 at 19:53
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    $\begingroup$ I wonder if Hanlon is missing a $\text{sgn}(\sigma)$ term in the summation. Working out the $S_3$ example he has running through the article gives $2\alpha^2 - 3 \alpha + 1$ for the product but $2\alpha^2 + 3\alpha + 1$ for the sum (which is $\beta^3-3\beta^2+2\beta$ vs. $\beta^3 + 3\beta^2 + 2\beta$ in your formulation). $\endgroup$
    – Brian Hopkins
    Commented Jul 24, 2016 at 20:11

2 Answers 2

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It seems that the correct formula is actually

$$ \prod_{i=0}^{n-1}(\beta+i) = \sum_{\sigma \in S_n}\beta^{c(\sigma)} $$

And this has a quite nice proof over on math.SE here.

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  • $\begingroup$ Thank you. I'm going through it by hand for n=1,2,3.. and it does seem that it should be + and not - $\endgroup$
    – I.K
    Commented Jul 24, 2016 at 20:15
  • $\begingroup$ Thank you very much! This also clears up some later applications in the paper. This is just the result in Stanley V1 p. 27 (Prop. 1.3.7) $\endgroup$
    – I.K
    Commented Jul 24, 2016 at 20:37
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Since I can't see immediately how Hanlon uses the equation, I guess the question is which "typo" was made. Linus's equation is correct, but I also think $$ \prod_{i=0}^{n-1} (\beta - i) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \beta^{c(\sigma)} $$ is true (and probably follows easily from Stanley's third proof referenced on the math.SE problem).

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  • $\begingroup$ I see that now. From the context, it's clear he means the usual: $$ \prod_{i=0}^{n-1}(\beta+i) = \sum_{\sigma \in S_n}\beta^{c(\sigma)} $$ as Linus noted. Thanks everyone for the quick help! $\endgroup$
    – I.K
    Commented Jul 24, 2016 at 20:44

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