Anyone seen these conclusions about the Riemann xi function or see any errors here?

With $\xi(s)$ the entire Landau Riemann xi function defined by the Hadamard product representation

$$\xi(s) = (1/2) \prod_{\rho_m} [1 - \frac{s}{\rho_m}]$$

where the product ranges over the nontrivial zeros of the Riemann zeta function, let

$$\vartheta(z) = 2\xi(\frac{1}{1-z}) $$

$$= \Xi(\frac{1}{1-z}) $$

$$= \prod_{\rho_m}[1- \frac{1}{\rho_m} \frac{1}{(1-z)}].$$

Let $u = 1/(1-z)$, then $z = (u-1)/u$ and

$$\vartheta(\frac{u-1}{u}) = \Xi(u)$$

$$= \Xi(1-u) = \vartheta(\frac{u}{u-1}),$$


$$\vartheta(z) = \vartheta(\frac{1}{z})$$ $$ = \Xi[\frac{1}{1-(1/z)}]= \Xi[\frac{z}{z-1}]$$

$$= \prod_{\rho_k} [1 - \frac{1}{\rho_k} \frac{z}{z-1} ].$$

(Edit June 10: Corrected transcription error. At last moment I changed notation from $\Omega$ in my notes to $\Xi$ but missed the change in this last set of equalities since I read them in my mind as the same Hadamard product. Sorry for the confusion.)

We then have the two equivalent expressions

$$\vartheta(z) = \prod_{\rho_m}[1- \frac{1}{\rho_m} \frac{1}{(1-z)}] $$

$$ = \prod_{\rho_k} [1 - \frac{1}{\rho_k} \frac{z}{z-1} ].$$

The zeros of the last product are given implicitly by

$$1 = \frac{1}{\rho_k} \frac{z_k}{z_k-1},$$


$$z_k= \rho_k/(\rho_k-1)$$

and the other product gives a null factor when

$$1 = \frac{1}{\rho_m}\frac{1}{1-z_k},$$


$$z_k = (\rho_m-1)/\rho_m.$$

Each zero is paired with its complex conjugate so that both equations are simultaneously satisfied only if $\rho_k = \bar{\rho}_m$ giving

$$\frac{\rho_m-1}{\rho_m} = \frac{\bar{\rho}_m}{\bar{\rho}_m-1}$$


$$(\rho_m-1)(\bar{\rho}_m-1) = \rho_m \bar{\rho}_m$$

and, therefore, that

$$Real(\rho_m) = 1/2.$$

Edit to address some comments:

If $\vartheta$ were an entire function with real zeros only, e.g., a polynomial with real zeros only, or the reciprocal of the gamma function written as it's Weierstrass factorization, the result would not apply since the necessary pairing that each zero's complex conjugate is also a zero, which is distinct but related to $\rho = 1 - \bar{\rho}$ since $ 3 + i \alpha$ and $3 - i\alpha$ are a conjugate pair that do not satisfy that relation, nor does the symmetry relation $f(s) = f(1-s)$ apply. The symmetry relation and factorization are indeed enough (explicit values for the zeros are not needed) to ensure the critical line--that's the point. Check that $\vartheta(z) = \vartheta(1/z)$ is a severe restriction that when evaluated at $z =0$ giving $\vartheta(0)) = \vartheta(\infty) = 1 = \prod_{\rho_k}[1+(1/\rho_k]$ enforces that $\rho_m = 1/2 + i\alpha$ and $ \rho_k = 1/2 - i\alpha$ are indeed solns.

(See, I believe Bump and others on the local Riemann hypothesis of other functions with critical lines.)

The two conditions of the general form of the factorization and the symmetry are necessary and sufficient to prove the critical line hypothesis without resort to calculating the actual values of the imaginary parts of the zeros (regardless of how interesting they are and the associations to the PNT) is the crux of the matter. (Anticlimactic, if true.)

Edit to address a comment:

From "Remark on Dirichlet Series Satisfying Functional Equations" by Eugenio P. Balanzario:

"By a theorem of H. Hamburger (see [6], page 31) the zeta function of Riemann is determined by its functional equation (FE). Hence, if we want to produce other Dirichlet series satisfying a functional equation, then it is necessary to change (FE) somehow."

[6] Titchmarsh, E. C. The theory of the Riemann zeta function, Oxford, 1986.

I should say, since I use this in the formal analysis, that the FE, the Hadamard factorization, and the fact that if $\rho$ is a zero then so is its complex conjugate are necessary (but not necessarily independent) conditions. I do not assume that $\rho = 1 - \bar{\rho}$, that is to be determined from the conditions.

Edit Jun 11;

Prompted by the initial responses, I thought more carefully about my assumptions and concluded before looking at the extended and revised responses that I indeed made a mistake by omitting potential solutions of the form of multiple factors, say $$(1-\frac{1}{p_k}\frac{z}{z-1}) (1- \frac{1}{p_m}\frac{z}{z-1}) = 0,$$ so, in effect, tantamount to assuming the RH is true, as LI did, to arrive at his criteria. Thanks to the responders for the input, the attention, and effort to show me the errors of my way--I learned some things along the way. I'll look over your last responses to see if I can learn some more.

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    $\begingroup$ note that leaving aside questions of convergence as pretty much all those products are conditional convergent, we know that $y_k=1-\rho_k$ is a root for each $\rho_k$ and obviously $\frac{y_k-1}{y_k}=\frac{\rho_k}{\rho_k-1}$ so that is the required pairing not that $\rho_k=\bar \rho_m$ which happens precisely when $\Re \rho =1/2$ so the above is the tautology $\Re z =1/2$ iff $\bar z =1-z$ which is true but not particularly helpful $\endgroup$
    – Conrad
    Jun 9, 2020 at 11:02
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    $\begingroup$ note also that the manipulations above have nothing to do with zeta or xi, only with the symmetry $f(s)=f(1-s)$ so you can pick any entire function of small enough order that satisfies that (so we have a representation without say exponential terms) and conclude that its roots are on the critical line and that is manifestly untrue as there are lots of counterxamples $\endgroup$
    – Conrad
    Jun 9, 2020 at 11:09
  • $\begingroup$ @Conrad, well that's interesting that it has nothing to do with the Riemann xi and zeta since it is precisely the notation and function defined in Li's classic "The Positivity of a Sequence of Numbers and the Riemann Hypothesis" and explored by Coffey in "Relations and positivity results for the derivatives of the Riemann ξ function" $\endgroup$ Jun 9, 2020 at 15:12
  • 1
    $\begingroup$ I meant in the sense that the manipulation above applies to any function that has a symmetry like the zeta and which is nice enough so it is factorable (even conditionally only), so one cannot conclude anything particular about the zeroes from it - now if one adds other conditions etc, who knows, but my remark was strictly about the formal manipulation above $\endgroup$
    – Conrad
    Jun 9, 2020 at 15:26
  • $\begingroup$ For some conditions that the symmetry condition alone imposes, see math.stackexchange.com/questions/28737/… $\endgroup$ Jun 9, 2020 at 17:05

2 Answers 2


There is the following error: Given a zero $\rho$ of $\xi(s)$, by the functional equation $\xi(s)=\xi(1-s)$, also $1-\rho$ is a zero of $\xi(s)$. Because $\xi$ is a real${}^*$ function, $\overline{\xi(s)}=\xi(\bar s)$, then also $\bar\rho$ is a zero of $\xi(s)$. The error in the deduction above is the implicit assumption that these two different operations of producing a new zero from a known one, yield the same result: $\bar\rho = 1-\rho$. But there is no reason for that. Instead, the four zeros $\rho$, $1-\rho$, $\bar\rho$, and $1-\bar\rho$ can all be mutually different.

${}^*$(Edit) This shall mean here: A function $f:\Omega\subset\mathbb{C}\to\mathbb{C}$ is real, if its domain $\Omega$ is invariant under complex conjugation, i.e. $\overline{\Omega} = \Omega$, and if it has the property $\overline{f(\bar{z})} = f(z)$ for each $z\in\Omega$ (where $z\mapsto\bar{z}$ denotes complex conjugation in both the domain and range of $f$). If we define the complex conjugate $\bar f$ of the function $f$ (regardless whether it is real or not) by $\bar f:\overline{\Omega}\to\mathbb{C}, z\mapsto \overline{f(\bar{z})}$, then $f$ is real if $\bar f = f$.

(Edit) Some more detailed explanation: The functional equation $\vartheta(z^{-1}) = \vartheta(z)$ is only a reformulation of $\xi(s)=\xi(1-s)$. So, the corresponding respective pairings $(z, z')$ and $(\rho, \rho')$ of zeros of $\vartheta$ and $\xi$ with $z'= z^{-1}$ and $\rho'=1-\rho$ are essentially the same:

With $z=(\rho-1)/\rho$ and $z'=(\rho'-1)/\rho'$ we have

$$z= z'^{-1}\Leftrightarrow\frac{(\rho-1)}{\rho} = \frac{\rho'}{(\rho'-1)} \Leftrightarrow\rho=1-\rho'.$$

On the other hand, there is the pairing $(\rho,\bar\rho)$ of zeros of the Riemann zeta function $\zeta$ with their complex conjugates, which comes from the fact that $\zeta$ (besides $\xi$) is real in the sense defined above, i. e. $\overline{\zeta(s)}=\zeta(\bar s)$, which is easily seen from its defining sum in the half plane ${\rm Re}\,s>1$ and its unique analytic continuation.

Now, in the sentence justifiying the equation in the Question post $$\frac{\rho_m−1}{\rho_m}=\frac{\bar\rho_m}{\bar\rho_m−1}$$ both pairings were implicitly assumed to be identical. Instead, $\vartheta(z^{-1}) = \vartheta(z)$ implies for each zero $\rho$ of $\xi$ only the existence of a zero $\rho'$ with $$\frac{\rho−1}{\rho}=\frac{\rho'}{\rho'−1}.$$ But there is no a priori reason why $\rho'$ should be $\bar\rho$.

  • $\begingroup$ First, $\xi(s)$ is not a real function. See Mathworld where you can see separate plots of the real and imaginary parts over a region and a Taylor series in a complex variable. But more importantly your line of reasoning does not reflect mine. I'll add more detail in the question in the morning to hopefully convince you and others of that, but nothing to contradict what I have already written. $\endgroup$ Jun 11, 2020 at 8:42
  • $\begingroup$ @Tom, I added the definition of my use of "real function". In this sense, $\xi$ is a real function, which can also be seen from the (conditionally convergent) product representation you gave, because in this product each zero $\rho$ is paired with its complex conjugate $\bar\rho$, which leads to factors $(1-s/\rho)(1-s/\bar\rho)=1-s (1/\rho + 1/\bar\rho) + s^2/|\rho|^2$ where $s$ has only real coefficients. Regarding your reasoning, I tried to exhibit just the core of your argument, because neither the product representation nor the introduction of $\vartheta$ seem to be essential. $\endgroup$
    – Uwe
    Jun 11, 2020 at 10:51

This is an extended comment based on my original comments and edits in the OP - the three conditions for an entire function $F$ mentioned in the OP:

1.symmetry (symmetrized functional equation) : $F(1-s)=F(s)$

2.$F$ real (various equivalent definitions being: conjugate invariant, $F(\bar z)=\bar F(z)$, real values on the real line, real Taylor coefficients at real values etc)

3.$F$ has a (conditional) Hadamard factorization of the given type (in terms of zeroes only with no exponential factors)

are very general and there are fairly arbitrary functions that satisfy them.

Basically one needs a set of complex numbers $\rho_k=\sigma_k+it_k, \sigma_k \ge \frac{1}{2}, t_k \ge 0, \rho_k \ne 1$ with a density condition of the type: $A_R$ is the number of $\rho_k, |\rho_k| \le R$, then the cardinality of $A(R)$ is $O(R^{2-\delta})$ for some fixed $\delta >0$ (which implies by standard stuff that their convergence exponent which is obviously the same as the convergent exponent of the set obtained from them under $s \to 1-s, s \to \bar s$ is at most $2-\delta$ so the sum of $1/|\rho_k|^2$ converges) and then one can construct a function like that by taking for each $\rho_k$ the polynomial (omitting repeated factors if any)

$P_k(z)=(1-\frac{z}{\rho_k})(1-\frac{z}{1-\rho_k})(1-\frac{z}{\bar \rho_k})(1-\frac{z}{1-\bar \rho_k})$

which is generally of degree $4$ but it can be of degree $2$ if either $\sigma_k=1/2$ or $t_k=0$ and even of degree $1$ if for some $k$'s, we have $\rho_k=1/2$

Obviously $P_k(1-z)=P_k(z)$ and $P_k$ is real (and it trivially satisfies Hadamard factorization of order zero) while if $|z| \le M$ and $|\rho_k|$ large, one clearly has $|\log P_k(z)| \le C_M/|\rho_k|^2$ (since $|(1-\frac{z}{\rho_k})(1-\frac{z}{1-\rho_k})| \le 1+C_M/|\rho_k|^2) $ which means that under the mentioned density hypothesis on the $|\rho_k|$, we get an entire function $F(z)=\Pi_{k}P_k(z)$ where the product is actually normally convergent and which has zeroes precisely at $\rho_k$ and its transforms under $s \to 1-s, s \to \bar s$, while it has the required (conditional as we need to group the zeroes as noted like in the usual $\xi$ case for that matter) Hadamard factorization

This being said, obviously most $F$ as above won't be symmetrizations of Dirichlet series like for $\xi$ or the symmetrizations of $L_{\chi}$ functions (all those have extra properties which are quite stringent - for example since the symmetrization involves the $\Gamma$ function which is of order $1$ and maximal type (in the extended Nevanlinna sense for meromorphic functions), all those have order $1$ and maximal type too as their zeroes satisfy the usual Riemann-von Mangoldt density $O(R\log R)$ and even among symmetrizations of Dirichlet series, we have such that come from Dirichlet series that do not have multiplicative coefficients so they are without Euler products

(eg the symmetrization of the Davenport-Heilbronn function is one such - for reference, I include the definition of both:

$f(s)= 1/1^s+\tan \theta/2^s-\tan \theta/3^s-1/4^s+1/6^s+\tan \theta/7^s...$, where $\theta$ is the unique angle in $(0, \pi/2)$ for which $\tan 2\theta =\frac{\sqrt 5 -1}{2}$

$F(s)=(\frac{5}{\pi})^{s/2}\Gamma(\frac{1+s}{2})f(s), F(s)=F(1-s)$

and a very detailed analysis of its zeroes is in the Karatsuba-Voronin RZ bible starting in chapter 6, page 212)


In particular, the OP's hypothesis that those three conditions by themselves allow one to draw conclusions about the zeroes is not correct

  • $\begingroup$ Also, and more immediately, Landau's $\zeta(2s)\zeta(2s-1)$... $\endgroup$ Jun 11, 2020 at 18:29

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