Roots of the derivative of the Riemann zeta function and of characteristic polynomials
Nonlinearity accepted (2010)
- DOI: 10.1088/0951-7715/23/10/014
- arXiv: 1002.0372
Available from
Eduardo Dueñez's profile on Mendeley.
or
Abstract
We investigate the horizontal distribution of zeros of the derivative of the Riemann zeta function and compare this to the radial distribution of zeros of the derivative of the characteristic polynomial of a random unitary matrix. Both cases show a surprising bimodal distribution which has yet to be explained. We show by example that the bimodality is a general phenomenon. For the unitary matrix case we prove a conjecture of Mezzadri concerning the leading order behavior, and we show that the same follows from the random matrix conjectures for the zeros of the zeta function.
Author-supplied keywords
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Eduardo Dueñez's profile on Mendeley.
Page 1
Roots of the derivative of the Riemann zeta function and of characteristic polynomials
ROOTS OF THE DERIVATIVE OF THE RIEMANN ZETA FUNCTION
AND OF CHARACTERISTIC POLYNOMIALS
EDUARDO DUE~NEZ, DAVID W. FARMER, SARA FROEHLICH, CHRIS HUGHES,
FRANCESCO MEZZADRI, AND TOAN PHAN
Abstract. We investigate the horizontal distribution of zeros of the derivative of the Rie-
mann zeta function and compare this to the radial distribution of zeros of the derivative of
the characteristic polynomial of a random unitary matrix. Both cases show a surprising bi-
modal distribution which has yet to be explained. We show by example that the bimodality
is a general phenomenon. For the unitary matrix case we prove a conjecture of Mezzadri
concerning the leading order behavior, and we show that the same follows from the random
matrix conjectures for the zeros of the zeta function.
1. Introduction
The zeros of the derivative 0(s) of the Riemann zeta-function are intimately connected
with the behavior of the zeros of (s) itself. Indeed, a theorem by Speiser [19] states that
the Riemann Hypothesis (RH) is equivalent to 0(s) having no zeros to the left of the crit-
ical line. Thus, understanding of the properties the zeros of 0(s) can provide important
tools and insight into the study of RH. After Speiser's article this idea was explored by
Berndt [1] and Spira [20], but not much progress was achieved until the work of Levinson
and Montgomery [11], who proved a quantitative re nement of Speiser's theorem. They
showed that (s) and 0(s) have essentially the same number of zeros to the left of the crit-
ical line = Re(s) = 12 , and proved that as T ! 1, where T is the height on the critical
line, a positive proportion of the zeros of 0(s) lie in the region
<
1
2
+ (1 + )
log log T
log T
; > 0: (1.1)
Consider the group of unitary matrices U(N) with probably distribution given by Haar
measure, which is the unique measure invariant under the left and right action of U(N) on
itself. Such a probability space is often known as the Circular Unitary Ensemble (CUE). Let
(z) be the characteristic polynomial of a matrix in the CUE. In recent years evidence has
been accumulated suggesting that, in the limit as T ! 1, the local statistical properties
of (s) can be modeled by the characteristic polynomials of matrices in the CUE where
N log(T=2). The connection between the Riemann zeta-function and characteristic
polynomials is extensive; examples include the distribution of the zeros of (s), its value
distribution and its moments. (For a series of review articles on the subject see [15] and
references therein.)
Assuming that random matrix theory (RMT) provides an accurate description of (s),
the horizontal distribution of the zeros of 0(s) in proximity of the critical line should be the
same as the radial distribution of the roots of 0(z) close to the unit circle. This idea was
rst developed by Mezzadri [13], who determined the distribution of the zeros of 0(z) that
Research supported by the American Institute of Mathematics and the National Science Foundation.
1
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AND OF CHARACTERISTIC POLYNOMIALS
EDUARDO DUE~NEZ, DAVID W. FARMER, SARA FROEHLICH, CHRIS HUGHES,
FRANCESCO MEZZADRI, AND TOAN PHAN
Abstract. We investigate the horizontal distribution of zeros of the derivative of the Rie-
mann zeta function and compare this to the radial distribution of zeros of the derivative of
the characteristic polynomial of a random unitary matrix. Both cases show a surprising bi-
modal distribution which has yet to be explained. We show by example that the bimodality
is a general phenomenon. For the unitary matrix case we prove a conjecture of Mezzadri
concerning the leading order behavior, and we show that the same follows from the random
matrix conjectures for the zeros of the zeta function.
1. Introduction
The zeros of the derivative 0(s) of the Riemann zeta-function are intimately connected
with the behavior of the zeros of (s) itself. Indeed, a theorem by Speiser [19] states that
the Riemann Hypothesis (RH) is equivalent to 0(s) having no zeros to the left of the crit-
ical line. Thus, understanding of the properties the zeros of 0(s) can provide important
tools and insight into the study of RH. After Speiser's article this idea was explored by
Berndt [1] and Spira [20], but not much progress was achieved until the work of Levinson
and Montgomery [11], who proved a quantitative re nement of Speiser's theorem. They
showed that (s) and 0(s) have essentially the same number of zeros to the left of the crit-
ical line = Re(s) = 12 , and proved that as T ! 1, where T is the height on the critical
line, a positive proportion of the zeros of 0(s) lie in the region
<
1
2
+ (1 + )
log log T
log T
; > 0: (1.1)
Consider the group of unitary matrices U(N) with probably distribution given by Haar
measure, which is the unique measure invariant under the left and right action of U(N) on
itself. Such a probability space is often known as the Circular Unitary Ensemble (CUE). Let
(z) be the characteristic polynomial of a matrix in the CUE. In recent years evidence has
been accumulated suggesting that, in the limit as T ! 1, the local statistical properties
of (s) can be modeled by the characteristic polynomials of matrices in the CUE where
N log(T=2). The connection between the Riemann zeta-function and characteristic
polynomials is extensive; examples include the distribution of the zeros of (s), its value
distribution and its moments. (For a series of review articles on the subject see [15] and
references therein.)
Assuming that random matrix theory (RMT) provides an accurate description of (s),
the horizontal distribution of the zeros of 0(s) in proximity of the critical line should be the
same as the radial distribution of the roots of 0(z) close to the unit circle. This idea was
rst developed by Mezzadri [13], who determined the distribution of the zeros of 0(z) that
Research supported by the American Institute of Mathematics and the National Science Foundation.
1
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X
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:1
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2.
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v1
[
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Page 2
2 DUE~NEZ, FARMER, FROEHLICH, HUGHES, MEZZADRI, AND PHAN
are very far from the unit circle and conjectured the leading order term of the distribution
very close to the unit circle. In this paper we prove his conjecture. We also perform an
analogous calculation for the Riemann zeta-function and conjecturally nd that the result
agrees with the RMT model. In addition, we do numerical computations in both cases
and nd a surprising feature in the distribution of zeros of the derivative, namely that the
probability distribution is bimodal.
2. The zeros of 0(s).
We have mentioned that the main motivation for studying the zeros of the derivative of
the Riemann zeta-function is its connection with RH. Indeed, Levinson and Montgomery's
result is the basis for Levinson's method [10], which Conrey [2] used to prove that at least
40% of the zeros of the zeta-function are on the line = 12 .
Levinson's method involves estimating a weighted average of the zeros of 0(s) to the left
of 12 + a= log T for some xed a > 0. Thus, zeros of
0(s) in the region 12 <
1
2 + a= log T
are an inherent loss in Levinson's method. It would be useful to understand the magnitude
of this loss. Alternatively, if we could nd a lower bound for the number of zeros of 0 in
this region we could improve the estimate for the number of zeros on the critical line.
0.5 1 1.5 2 2.5 3 3.5
1020
1025
1030
1035
1040
Figure 2.1. Zeros of (s) (dot), 0(s) (triangle), 00(s) (square), and ( 0=)0(s)
(star), with imaginary parts in the range 1015 < T < 1040.
Figure 2.1 gives a representative example of the location of zeros of 0(s) and the relation-
ship to the zeros of (s) and various other derivatives of the zeta-function. It illustrates that
zeros of 0(s) close to the critical-line correspond to closely spaced zeros of (s). We make
this statement precise in Section 6. Also, a zero of 0(s) seems to be \missing" when zeros
of are particularly far apart or when there are two successive large gaps. Indeed, there
are very far from the unit circle and conjectured the leading order term of the distribution
very close to the unit circle. In this paper we prove his conjecture. We also perform an
analogous calculation for the Riemann zeta-function and conjecturally nd that the result
agrees with the RMT model. In addition, we do numerical computations in both cases
and nd a surprising feature in the distribution of zeros of the derivative, namely that the
probability distribution is bimodal.
2. The zeros of 0(s).
We have mentioned that the main motivation for studying the zeros of the derivative of
the Riemann zeta-function is its connection with RH. Indeed, Levinson and Montgomery's
result is the basis for Levinson's method [10], which Conrey [2] used to prove that at least
40% of the zeros of the zeta-function are on the line = 12 .
Levinson's method involves estimating a weighted average of the zeros of 0(s) to the left
of 12 + a= log T for some xed a > 0. Thus, zeros of
0(s) in the region 12 <
1
2 + a= log T
are an inherent loss in Levinson's method. It would be useful to understand the magnitude
of this loss. Alternatively, if we could nd a lower bound for the number of zeros of 0 in
this region we could improve the estimate for the number of zeros on the critical line.
0.5 1 1.5 2 2.5 3 3.5
1020
1025
1030
1035
1040
Figure 2.1. Zeros of (s) (dot), 0(s) (triangle), 00(s) (square), and ( 0=)0(s)
(star), with imaginary parts in the range 1015 < T < 1040.
Figure 2.1 gives a representative example of the location of zeros of 0(s) and the relation-
ship to the zeros of (s) and various other derivatives of the zeta-function. It illustrates that
zeros of 0(s) close to the critical-line correspond to closely spaced zeros of (s). We make
this statement precise in Section 6. Also, a zero of 0(s) seems to be \missing" when zeros
of are particularly far apart or when there are two successive large gaps. Indeed, there
Page 3
ROOTS OF THE DERIVATIVE OF THE ZETA FUNCTION 3
can't be a zero of 0(s) between every pair of zeros of (s) because the density of zeros of
(s) is 12 log(T=2) while the density of zeros of
0(s) is 12 log(T=4). So on average there
is a \missing" zero of 0(s) in each T interval of width 2= log 2 9:06.
Conrey and Ghosh [4] and subsequently Guo [7] improved Levinson and Montgomery's
result (1.1) and showed that a positive proportion of the zeros of 0(s) are much closer to
the line = 12 . Indeed, for any xed a > 0, the region
1
2
a
log T
(2.1)
contains a positive proportion of the zeros. Soundararajan [18] made further progress and
introduced the functions
m (a) := lim inf
T!1
1
N1(T )
X
0 12+
a
log T
0<
0T
1; (2.2a)
m+(a) := lim sup
T!1
1
N1(T )
X
0 12+
a
log T
0<
0T
1; (2.2b)
where N1(T ) is the number of zeros 0+ i
0 of 0(s) with 0 <
0 T . Soundararajan proved
that m (a) > 0 for a > 2:6, and conjectured that
m(a) = m (a) = m+(a): (2.3)
He also conjectured that m(a) is continuous, that m(a) > 0 for all a > 0, and that m(a)! 1
as a ! 1. Zhang [23], Feng [6], Garaev and Yldrm [8], and Ki [9] proved re nements
of Soundararajan's results. In particular Feng [6] showed that m (a) > 0 for all a > 0
unconditionally of RH but assuming a conjecture on the frequency of small gaps between
consecutive critical zeros of (s).
3. Zeros of derivatives of polynomials and statement of results
Suppose f(z) is a polynomial with all zeros on the unit circle. (Eventually, f will be a
random polynomial obtained as the characteristic polynomial of a random unitary matrix.)
The Gauss-Lucas theorem assures that all the roots of f 0(z) lie on or inside the unit circle
(zeros of f 0 on the circle occur only if f has multiple zeros). If f(z) has two zeros which
are very close together, then f 0(z) will have a zero close by. (This is a consequence of the
continuous dependence of the zeros of f 0 on those of f . This dependence is actually piecewise
analytic as will be described below.) The speci c location of the nearby zero of f 0(z) will
depend primarily on how close those two zeros of f(z) are, and on the general position of
the remaining zeros of f(z). Thus, to leading order (with respect to the size of the gap),
the distribution of zeros of f 0(z) near jzj = 1 should largely depend on the distribution of
small gaps between zeros of f(z), that is, on the tail of the nearest-neighbor spacing of zeros
of f(z). We will make this idea precise and treat in detail the case where f(z) = (z) is the
characteristic polynomial of a CUE matrix (a matrix chosen from the unitary group U(N),
uniformly with respect to Haar measure).
Let z0 be a root of 0(z) and de ne the random variable
S := N(1 jz0j): (3.1)
can't be a zero of 0(s) between every pair of zeros of (s) because the density of zeros of
(s) is 12 log(T=2) while the density of zeros of
0(s) is 12 log(T=4). So on average there
is a \missing" zero of 0(s) in each T interval of width 2= log 2 9:06.
Conrey and Ghosh [4] and subsequently Guo [7] improved Levinson and Montgomery's
result (1.1) and showed that a positive proportion of the zeros of 0(s) are much closer to
the line = 12 . Indeed, for any xed a > 0, the region
1
2
a
log T
(2.1)
contains a positive proportion of the zeros. Soundararajan [18] made further progress and
introduced the functions
m (a) := lim inf
T!1
1
N1(T )
X
0 12+
a
log T
0<
0T
1; (2.2a)
m+(a) := lim sup
T!1
1
N1(T )
X
0 12+
a
log T
0<
0T
1; (2.2b)
where N1(T ) is the number of zeros 0+ i
0 of 0(s) with 0 <
0 T . Soundararajan proved
that m (a) > 0 for a > 2:6, and conjectured that
m(a) = m (a) = m+(a): (2.3)
He also conjectured that m(a) is continuous, that m(a) > 0 for all a > 0, and that m(a)! 1
as a ! 1. Zhang [23], Feng [6], Garaev and Yldrm [8], and Ki [9] proved re nements
of Soundararajan's results. In particular Feng [6] showed that m (a) > 0 for all a > 0
unconditionally of RH but assuming a conjecture on the frequency of small gaps between
consecutive critical zeros of (s).
3. Zeros of derivatives of polynomials and statement of results
Suppose f(z) is a polynomial with all zeros on the unit circle. (Eventually, f will be a
random polynomial obtained as the characteristic polynomial of a random unitary matrix.)
The Gauss-Lucas theorem assures that all the roots of f 0(z) lie on or inside the unit circle
(zeros of f 0 on the circle occur only if f has multiple zeros). If f(z) has two zeros which
are very close together, then f 0(z) will have a zero close by. (This is a consequence of the
continuous dependence of the zeros of f 0 on those of f . This dependence is actually piecewise
analytic as will be described below.) The speci c location of the nearby zero of f 0(z) will
depend primarily on how close those two zeros of f(z) are, and on the general position of
the remaining zeros of f(z). Thus, to leading order (with respect to the size of the gap),
the distribution of zeros of f 0(z) near jzj = 1 should largely depend on the distribution of
small gaps between zeros of f(z), that is, on the tail of the nearest-neighbor spacing of zeros
of f(z). We will make this idea precise and treat in detail the case where f(z) = (z) is the
characteristic polynomial of a CUE matrix (a matrix chosen from the unitary group U(N),
uniformly with respect to Haar measure).
Let z0 be a root of 0(z) and de ne the random variable
S := N(1 jz0j): (3.1)
Page 4
4 DUE~NEZ, FARMER, FROEHLICH, HUGHES, MEZZADRI, AND PHAN
Denote by Q(s;N) the probability density function (p.d.f.) of S. Mezzadri [13] showed that
the limit
Q(s) := lim
N!1
Q(s;N) (3.2)
exists, and proved that
Q(s;N)
1
s2
; N !1; s!1; (3.3)
with s = o (N). He also conjectured that
Q(s)
4
3
s1=2; s! 0: (3.4)
Formula (3.3) can be interpreted as the RMT counterpart of the Levinson-Montgomery
bound (1.1) for the roots of 0(s). The RMT model of the Riemann-zeta function is based on
the observation that the local correlations of the non-trivial zeros of (s) coincide with those
of the eigenvalues of matrices in the CUE. In order to make this correspondence quantitative,
the densities of the eigenvalues and of the zeros of (s) must be made (asymptotically) equal,
i.e.,
N
2
=
1
2
log
T
2
: (3.5)
It follows from (3.3) that the expected value of S does not exist|its p.d.f. does not decay
suciently rapidly. On the other hand, using (3.3), the average of the values of S not
exceeding N is
Z N
1
s
ds
s2
= logN; N !1: (3.6)
Recalling the relation (3.1) between S and jz0j, we conclude that a positive proportion of the
roots of 0(z) must lie within a distance from the unit circle bounded from above by
(1 + )
logN
N
(3.7)
from the unit circle. Because of (3.5), formula (3.7) corresponds to Levinson and Mont-
gomery's result (1.1).
Now consider the roots eit1 ; : : : ; eitN (with < ti ) of the characteristic polyno-
mial (z) of a random unitary matrix distributed with Haar measure. It is convenient for
our purposes to de ne
xj =
Ntj
2
; j = 1; : : : ; N; (3.8)
so that, on average, the distance between two consecutive xj's is one. The joint probability
density function (j.p.d.f.) of the eigenvalues is given in terms of x1; : : : ; xN as
p2(x1; : : : ; xN) :=
1
NNN !
Y
1j<kN
jeN(xk) eN(xj)j
2 ; (3.9)
where we have used the notation eN(x) := exp(2ix=N). Relabeling the indexes j =
1; : : : ; N , if necessary, we assume that
x1 : : : xN < x1 +N: (3.10)
Denote by Q(s;N) the probability density function (p.d.f.) of S. Mezzadri [13] showed that
the limit
Q(s) := lim
N!1
Q(s;N) (3.2)
exists, and proved that
Q(s;N)
1
s2
; N !1; s!1; (3.3)
with s = o (N). He also conjectured that
Q(s)
4
3
s1=2; s! 0: (3.4)
Formula (3.3) can be interpreted as the RMT counterpart of the Levinson-Montgomery
bound (1.1) for the roots of 0(s). The RMT model of the Riemann-zeta function is based on
the observation that the local correlations of the non-trivial zeros of (s) coincide with those
of the eigenvalues of matrices in the CUE. In order to make this correspondence quantitative,
the densities of the eigenvalues and of the zeros of (s) must be made (asymptotically) equal,
i.e.,
N
2
=
1
2
log
T
2
: (3.5)
It follows from (3.3) that the expected value of S does not exist|its p.d.f. does not decay
suciently rapidly. On the other hand, using (3.3), the average of the values of S not
exceeding N is
Z N
1
s
ds
s2
= logN; N !1: (3.6)
Recalling the relation (3.1) between S and jz0j, we conclude that a positive proportion of the
roots of 0(z) must lie within a distance from the unit circle bounded from above by
(1 + )
logN
N
(3.7)
from the unit circle. Because of (3.5), formula (3.7) corresponds to Levinson and Mont-
gomery's result (1.1).
Now consider the roots eit1 ; : : : ; eitN (with < ti ) of the characteristic polyno-
mial (z) of a random unitary matrix distributed with Haar measure. It is convenient for
our purposes to de ne
xj =
Ntj
2
; j = 1; : : : ; N; (3.8)
so that, on average, the distance between two consecutive xj's is one. The joint probability
density function (j.p.d.f.) of the eigenvalues is given in terms of x1; : : : ; xN as
p2(x1; : : : ; xN) :=
1
NNN !
Y
1j<kN
jeN(xk) eN(xj)j
2 ; (3.9)
where we have used the notation eN(x) := exp(2ix=N). Relabeling the indexes j =
1; : : : ; N , if necessary, we assume that
x1 : : : xN < x1 +N: (3.10)
Page 5
ROOTS OF THE DERIVATIVE OF THE ZETA FUNCTION 5
We also extend the sequence fxjg to be periodic by setting xN+1 = x1, xN+2 = x2, etc. Fix
integers n and j with 0 n N 2. Let us denote by p2(n; s) the probability density
function of xj+n+1 xj. Since the j.p.d.f. (3.9) is invariant under translations, which means
p2(x1 + ; : : : ; xN + ) = p2(x1; : : : ; xN) for all 2 R; (3.11)
it follows that p2(n; s) does not depend on j. If n = 0, p2(s) := p2(0; s) is known as the
spacing distribution. It has an asymptotic expansion in powers of s:
p2(s) =
1
3
1
3N2
2s2
2
45
1
9N2
+
1
15N4
4s4
+
1
315
2
135N2
+
1
45N4
2
189N6
6s6 +O(s7): (3.12)
This expansion follows from the pair correlation function for U(N) (see [15]),
sin2(y)
N2 sin2(y=N)
; (3.13)
and the fact that the pair correlation function and the nearest neighbor spacing for U(N)
agree to order 6.
To describe our main result, suppose that a root of (z) is degenerate (which means
xj+1 = xj) so that z0 = exp (2ixj=N) will also be a root of 0(z). Simple considerations
of continuity show that, if xj and xj+1 are slightly moved apart (say, while keeping the
remaining roots of f xed), then z0 will also move, but will still be close to the midpoint
of the segment joining exp(2ixj=N) to exp(2ixj+1=N). In Proposition 8.1 we make this
precise, showing that z0 stays close to that midpoint provided xj+1 xj < 1=. Henceforth
we assume that the rescaled distance
:= xj+1 xj (3.14)
is small.
By the translation invariance of the j.p.d.f. of the xj's, we may assume without loss of
generality that
xj+1 =
2
and xj =
2
: (3.15)
De ne
:= N(1 z0): (3.16)
In section 5 we shall show that
= b1
22 + b2
44 +O(6); as ! 0; (3.17)
where b1 and b2 are explicit functions of the zeros xk for k 6= j; j + 1.
By combining the distribution of given in (3.12) with information we will determine
about b1 and b2 in Section 5, we will prove
Theorem 3.1. Let (z) be the characteristic polynomial of a random matrix in U(N) dis-
tributed with respect to Haar measure. The distribution of = N(1 jz0j) arising from closely
spaced zeros of (z) is given by
4
3
s1=2
82
45
s3=2 +O
s5=2
; (3.18)
as N !1.
We also extend the sequence fxjg to be periodic by setting xN+1 = x1, xN+2 = x2, etc. Fix
integers n and j with 0 n N 2. Let us denote by p2(n; s) the probability density
function of xj+n+1 xj. Since the j.p.d.f. (3.9) is invariant under translations, which means
p2(x1 + ; : : : ; xN + ) = p2(x1; : : : ; xN) for all 2 R; (3.11)
it follows that p2(n; s) does not depend on j. If n = 0, p2(s) := p2(0; s) is known as the
spacing distribution. It has an asymptotic expansion in powers of s:
p2(s) =
1
3
1
3N2
2s2
2
45
1
9N2
+
1
15N4
4s4
+
1
315
2
135N2
+
1
45N4
2
189N6
6s6 +O(s7): (3.12)
This expansion follows from the pair correlation function for U(N) (see [15]),
sin2(y)
N2 sin2(y=N)
; (3.13)
and the fact that the pair correlation function and the nearest neighbor spacing for U(N)
agree to order 6.
To describe our main result, suppose that a root of (z) is degenerate (which means
xj+1 = xj) so that z0 = exp (2ixj=N) will also be a root of 0(z). Simple considerations
of continuity show that, if xj and xj+1 are slightly moved apart (say, while keeping the
remaining roots of f xed), then z0 will also move, but will still be close to the midpoint
of the segment joining exp(2ixj=N) to exp(2ixj+1=N). In Proposition 8.1 we make this
precise, showing that z0 stays close to that midpoint provided xj+1 xj < 1=. Henceforth
we assume that the rescaled distance
:= xj+1 xj (3.14)
is small.
By the translation invariance of the j.p.d.f. of the xj's, we may assume without loss of
generality that
xj+1 =
2
and xj =
2
: (3.15)
De ne
:= N(1 z0): (3.16)
In section 5 we shall show that
= b1
22 + b2
44 +O(6); as ! 0; (3.17)
where b1 and b2 are explicit functions of the zeros xk for k 6= j; j + 1.
By combining the distribution of given in (3.12) with information we will determine
about b1 and b2 in Section 5, we will prove
Theorem 3.1. Let (z) be the characteristic polynomial of a random matrix in U(N) dis-
tributed with respect to Haar measure. The distribution of = N(1 jz0j) arising from closely
spaced zeros of (z) is given by
4
3
s1=2
82
45
s3=2 +O
s5=2
; (3.18)
as N !1.
Page 6
6 DUE~NEZ, FARMER, FROEHLICH, HUGHES, MEZZADRI, AND PHAN
Note that Theorem 3.1 refers to the small values of that arise from closely spaced zeros
of the polynomial. The theorem does not account for all small values of . That distinction
is often missed, because examples such as shown in Figure 2.1 give the mistaken impression
that zeros of the derivative very close to the unit circle can only arise from closely spaced
zeros of the polynomial. Farmer and Ki [5] give examples of families of polynomials whose
(rescaled) zeros are bounded away from each other, but for which the density function of
vanishes like C s as s! 0. They also argue that any larger density of zeros of z0 near the unit
circle must arise from closely spaced zeros of the polynomial. Therefore we have the following
corollary of Theorem 3.1, which proves Mezzadri's conjecture about the distribution of jz0j.
Corollary 3.2. Let (z) be the characteristic polynomial of a random matrix in U(N)
distributed with respect to Haar measure, and let Q(s;N) be the p.d.f. of S = N(1 jz0j) for
z0 a root of 0(z). Then
Q(s) = lim
N!1
Q(s;N) =
4
3
s1=2 +O(s): (3.19)
Note that it remains an unsolved problem to show that Q(s) is a proper probability
distribution. That is, to show
R1
0 Q(s)ds = 1. This is the random matrix analogue of
Soundararajan's conjecture m(a)! 1 as a!1.
4. Comparison with data
We compare our formulas with numerical data.
We generated Haar-random matrices in U(N) using the simple algorithm described in [14].
Figure 4.1 shows the empirical distribution of the rescaled zeros of 0 for various size ma-
trices. Figure 4.2 shows the empirical cumulative distribution function Ip(x) =
R x
0 Q(s; 40)ds
for U(40) and a comparison with the tail of the empirical cumulative distribution function
with our results, showing good agreement.
We would like to know the underlying cause of the curious \second bump" in the dis-
tribution of zeros of derivatives. This seems to be a completely general phenomenon. In
Figure 4.3 we show the analogous distributions for characteristic polynomials of matrices
from COE(40) and for degree-40 polynomials whose roots are independently and uniformly
distributed on the unit circle. Both cases show the \second bump", although not quite at
the same location. In Figure 6.1 we nd a similar shape for the distribution of zeros of 0.
5. Proof of theorem 3.1
Suppose f(z) is a degree-N polynomial having all zeros on the unit circle, for which two
zeros z1, z2 are very close together. Then the derivative f 0(z) will have a zero close to the
midpoint (z1 + z2)=2. This follows because, if z1 = z2 (that is, if f(z) has a multiple root at
z1), then z1 is also a root of the derivative f 0(z), and the roots of f 0 are continuous functions
of the roots of f . By a rotation we can assume that
f(z) = F (z)(z e i=2)(z ei=2); (5.1)
where F (z) does not have any zeros eit with =2 t =2. The root of f 0 near 1 is the
root near 1 of
f 0
f
(z) =
F 0
F
(z) +
1
z e i=2
+
1
z ei=2
; (5.2)
and we denote that root by z0 = 1 .
Note that Theorem 3.1 refers to the small values of that arise from closely spaced zeros
of the polynomial. The theorem does not account for all small values of . That distinction
is often missed, because examples such as shown in Figure 2.1 give the mistaken impression
that zeros of the derivative very close to the unit circle can only arise from closely spaced
zeros of the polynomial. Farmer and Ki [5] give examples of families of polynomials whose
(rescaled) zeros are bounded away from each other, but for which the density function of
vanishes like C s as s! 0. They also argue that any larger density of zeros of z0 near the unit
circle must arise from closely spaced zeros of the polynomial. Therefore we have the following
corollary of Theorem 3.1, which proves Mezzadri's conjecture about the distribution of jz0j.
Corollary 3.2. Let (z) be the characteristic polynomial of a random matrix in U(N)
distributed with respect to Haar measure, and let Q(s;N) be the p.d.f. of S = N(1 jz0j) for
z0 a root of 0(z). Then
Q(s) = lim
N!1
Q(s;N) =
4
3
s1=2 +O(s): (3.19)
Note that it remains an unsolved problem to show that Q(s) is a proper probability
distribution. That is, to show
R1
0 Q(s)ds = 1. This is the random matrix analogue of
Soundararajan's conjecture m(a)! 1 as a!1.
4. Comparison with data
We compare our formulas with numerical data.
We generated Haar-random matrices in U(N) using the simple algorithm described in [14].
Figure 4.1 shows the empirical distribution of the rescaled zeros of 0 for various size ma-
trices. Figure 4.2 shows the empirical cumulative distribution function Ip(x) =
R x
0 Q(s; 40)ds
for U(40) and a comparison with the tail of the empirical cumulative distribution function
with our results, showing good agreement.
We would like to know the underlying cause of the curious \second bump" in the dis-
tribution of zeros of derivatives. This seems to be a completely general phenomenon. In
Figure 4.3 we show the analogous distributions for characteristic polynomials of matrices
from COE(40) and for degree-40 polynomials whose roots are independently and uniformly
distributed on the unit circle. Both cases show the \second bump", although not quite at
the same location. In Figure 6.1 we nd a similar shape for the distribution of zeros of 0.
5. Proof of theorem 3.1
Suppose f(z) is a degree-N polynomial having all zeros on the unit circle, for which two
zeros z1, z2 are very close together. Then the derivative f 0(z) will have a zero close to the
midpoint (z1 + z2)=2. This follows because, if z1 = z2 (that is, if f(z) has a multiple root at
z1), then z1 is also a root of the derivative f 0(z), and the roots of f 0 are continuous functions
of the roots of f . By a rotation we can assume that
f(z) = F (z)(z e i=2)(z ei=2); (5.1)
where F (z) does not have any zeros eit with =2 t =2. The root of f 0 near 1 is the
root near 1 of
f 0
f
(z) =
F 0
F
(z) +
1
z e i=2
+
1
z ei=2
; (5.2)
and we denote that root by z0 = 1 .
Page 9
ROOTS OF THE DERIVATIVE OF THE ZETA FUNCTION 9
We write z0 = 1 =N and wish to expand (1 =N eN(xn)) 1 as a Taylor series in .
This will be justi ed when jj < N j1 eN(xn)j. Hence, we have
F 0
F
1
N
= N
1X
j=0
Aj
j (5.8)
for suciently small, where
Aj = ( 1)
j 1
j!N j+1
F 0
F
(j)
(1)
=
1
N j+1
N 2X
n=1
1
(1 eN(xn))j+1
: (5.9)
We will see later that the prefactor of N is the right choice to make the coecients Aj
approximately bounded. Note that, for jj < 1 (equivalently, for jj < 2=N), any such
that jj < jj = N jj=(2) makes the expansion in (5.8) valid (independently of the exact
location of the zeros of F ), since
jj <
N
2
N j1 eN(=2)j N j1 eN(xn)j ; 1 n N 2: (5.10)
We also wish to expand the other terms in (5.2) as a series in . We have
1
z0 e i=2
+
1
z0 ei=2
=
2 2N 2 cos
N
2 2N +
2
N2 2
1 N
cos N
=
2 + 22N 1 112
44N 3 +O(6N 5)
2 + 22 22N 1 112
44N 2 + 112
44N 3 +O(6N 4)
: (5.11)
Combining this with equations (5.2) and (5.8), putting all terms over a common denominator,
and using the fact that f 0(z0) = 0, we have
0 =
1X
j=0
Aj
j
2 + 22 22N 1 112
44N 2 + 112
44N 3 +O(6N 4)
2 + 22N 1 112
44N 3 +O(6N 5): (5.12)
Note that a global factor of N canceled to give the above equation, suggesting that we have
chosen the correct scaling for .
Equation (5.12) is simply a more explicit and manageable form of the equation f 0(; z0) = 0
de ning z0 implicitly as a function of . Noting that f 0(0; 1) = 0, together with the functional
equation f 0(; z) = f 0( ; z), it follows that = () has an expansion in powers of 2, with
no constant term, of the form
= b1
22 + b2
44 +O(6): (5.13)
From (5.12), we obtain
0 =
A0 2b1 +
1
N
22
+
A1b1 + A0b
2
1 2b2
A0b1
N
A0
12N2
1
12N3
44 +O(6): (5.14)
We write z0 = 1 =N and wish to expand (1 =N eN(xn)) 1 as a Taylor series in .
This will be justi ed when jj < N j1 eN(xn)j. Hence, we have
F 0
F
1
N
= N
1X
j=0
Aj
j (5.8)
for suciently small, where
Aj = ( 1)
j 1
j!N j+1
F 0
F
(j)
(1)
=
1
N j+1
N 2X
n=1
1
(1 eN(xn))j+1
: (5.9)
We will see later that the prefactor of N is the right choice to make the coecients Aj
approximately bounded. Note that, for jj < 1 (equivalently, for jj < 2=N), any such
that jj < jj = N jj=(2) makes the expansion in (5.8) valid (independently of the exact
location of the zeros of F ), since
jj <
N
2
N j1 eN(=2)j N j1 eN(xn)j ; 1 n N 2: (5.10)
We also wish to expand the other terms in (5.2) as a series in . We have
1
z0 e i=2
+
1
z0 ei=2
=
2 2N 2 cos
N
2 2N +
2
N2 2
1 N
cos N
=
2 + 22N 1 112
44N 3 +O(6N 5)
2 + 22 22N 1 112
44N 2 + 112
44N 3 +O(6N 4)
: (5.11)
Combining this with equations (5.2) and (5.8), putting all terms over a common denominator,
and using the fact that f 0(z0) = 0, we have
0 =
1X
j=0
Aj
j
2 + 22 22N 1 112
44N 2 + 112
44N 3 +O(6N 4)
2 + 22N 1 112
44N 3 +O(6N 5): (5.12)
Note that a global factor of N canceled to give the above equation, suggesting that we have
chosen the correct scaling for .
Equation (5.12) is simply a more explicit and manageable form of the equation f 0(; z0) = 0
de ning z0 implicitly as a function of . Noting that f 0(0; 1) = 0, together with the functional
equation f 0(; z) = f 0( ; z), it follows that = () has an expansion in powers of 2, with
no constant term, of the form
= b1
22 + b2
44 +O(6): (5.13)
From (5.12), we obtain
0 =
A0 2b1 +
1
N
22
+
A1b1 + A0b
2
1 2b2
A0b1
N
A0
12N2
1
12N3
44 +O(6): (5.14)
Page 10
10 DUE~NEZ, FARMER, FROEHLICH, HUGHES, MEZZADRI, AND PHAN
Setting each term in (5.14) equal to 0 and solving for b1 and b2 we have the following:
Proposition 5.1. In the notation above, if 0 < 1= and N is suciently large then
= b122 + b244 +O(6) where
b1 =
A0
2
+
1
2N
(5.15)
b2 =
1
8
A30 + 2A0A1
+
A1
4N
A0
6N2
1
24N3
; (5.16)
with Aj given in (5.9).
Note that in this analysis we have treated F (and hence Aj) as being xed, in the sense
that we assume its zeros do not vary with . In the next section we will show that when
f(z) is the characteristic polynomial of a matrix drawn from the CUE, this can be justi ed
up to O(7).
Using (5.13) and (5.15), we can determine the distribution of from the distributions of
and the Aj. For small , this comes from the tail of the nearest-neighbor spacing.
5.2. Nearest-neighbor spacing. Proposition 5.1 provides a formula for = N(1 z0),
but what we really want is the distribution of := N(1 jz0j). So by (5.4) and (5.13), and
writing Bj = Re(bj), we have
= B1
22 +B2
44 +O
6 +
jj2
N
+
jj3
N2
(5.17)
= B1
22 +B2
44 +O
6 +
2
N
: (5.18)
The second line is a corollary of Proposition 8.1, because if < 1=.
Suppose for the moment that B1 and B2 were constants (instead of being random). We
would have = g() = B122 +B244 +ON(6) where is random with p.d.f. (3.12) given
by the nearest neighbor spacing of eigenvalues of unitary matrices. Then the distribution
function of would be given by
p2(g 1(s))
g0(g 1(s))
=
B 3=21
6
1
1
N2
s1=2
1
B 5=21
45
1
5
2N2
+
3
2N4
+
5B 7=21 B2
12
1
1
N2
!
s3=2
+ON(s
5=2): (5.19)
It turns out that B1 actually is a constant: in Section 5.3 we show that B1 = 14 . It is
fortunate that B1 is a constant, otherwise it could be dicult to determine the expected
value of quantities like B 3=21 .
The contribution of B2 takes a bit more work. If B2 was independent of then we could
just average over the possible contributions of B2 to (5.19). That is, in (5.19) replace B2 by
its expected value. This can be computed from the expected values of various combinations
of A0 and A1. But B2 is not independent of . However, it is independent of to leading
order. Our speci c concern is the distribution of the other roots when is very small.
This approximates the polynomial having a double zero. Since the next-nearest-neighbor
spacing of U(N) eigenvalues vanishes to order 7, the dependence of B2 on is only to order
Setting each term in (5.14) equal to 0 and solving for b1 and b2 we have the following:
Proposition 5.1. In the notation above, if 0 < 1= and N is suciently large then
= b122 + b244 +O(6) where
b1 =
A0
2
+
1
2N
(5.15)
b2 =
1
8
A30 + 2A0A1
+
A1
4N
A0
6N2
1
24N3
; (5.16)
with Aj given in (5.9).
Note that in this analysis we have treated F (and hence Aj) as being xed, in the sense
that we assume its zeros do not vary with . In the next section we will show that when
f(z) is the characteristic polynomial of a matrix drawn from the CUE, this can be justi ed
up to O(7).
Using (5.13) and (5.15), we can determine the distribution of from the distributions of
and the Aj. For small , this comes from the tail of the nearest-neighbor spacing.
5.2. Nearest-neighbor spacing. Proposition 5.1 provides a formula for = N(1 z0),
but what we really want is the distribution of := N(1 jz0j). So by (5.4) and (5.13), and
writing Bj = Re(bj), we have
= B1
22 +B2
44 +O
6 +
jj2
N
+
jj3
N2
(5.17)
= B1
22 +B2
44 +O
6 +
2
N
: (5.18)
The second line is a corollary of Proposition 8.1, because if < 1=.
Suppose for the moment that B1 and B2 were constants (instead of being random). We
would have = g() = B122 +B244 +ON(6) where is random with p.d.f. (3.12) given
by the nearest neighbor spacing of eigenvalues of unitary matrices. Then the distribution
function of would be given by
p2(g 1(s))
g0(g 1(s))
=
B 3=21
6
1
1
N2
s1=2
1
B 5=21
45
1
5
2N2
+
3
2N4
+
5B 7=21 B2
12
1
1
N2
!
s3=2
+ON(s
5=2): (5.19)
It turns out that B1 actually is a constant: in Section 5.3 we show that B1 = 14 . It is
fortunate that B1 is a constant, otherwise it could be dicult to determine the expected
value of quantities like B 3=21 .
The contribution of B2 takes a bit more work. If B2 was independent of then we could
just average over the possible contributions of B2 to (5.19). That is, in (5.19) replace B2 by
its expected value. This can be computed from the expected values of various combinations
of A0 and A1. But B2 is not independent of . However, it is independent of to leading
order. Our speci c concern is the distribution of the other roots when is very small.
This approximates the polynomial having a double zero. Since the next-nearest-neighbor
spacing of U(N) eigenvalues vanishes to order 7, the dependence of B2 on is only to order
Page 11
ROOTS OF THE DERIVATIVE OF THE ZETA FUNCTION 11
O(7). Thus, for the terms we are computing for the distribution of we can treat B2 as
independent of . The expected value of B2 is calculated in Section 5.3.
5.3. Expected value of Aj. We have
Aj :=
1
N j+1j!
F 0
F
(j)
(1)
= ( 1)jN j 1
N 2X
n=1
1
(1 eitn)j+1
; (5.20)
where t1; t2; : : : are the arguments of the zeros of F (z). Since
1
1 eit
=
1
2
+
i
2
cot
t
2
(5.21)
we see that
Re(A0) =
N 2
2N
; (5.22)
so
B1 = Re(b1) =
1
4
; (5.23)
as claimed. Note also that
hA0i =
1
2
1
N
; (5.24)
because the imaginary part of the summand is odd.
For Aj with j 1, we require a random matrix calculation. The sum in (5.20) is dominated
by the terms where eitn is close to 1, so one possibility is to determine the level densities of
the tn. We will nd the expected value of Aj by appealing to prior results on averages of
ratios of characteristic polynomials [3].
We assume that f(z) is the characteristic polynomial of a matrix chosen uniformly with
respect to Haar measure from the unitary group U(N). We restrict to those matrices which
have two eigenvalues very close to 1, and we wish to determine the joint distribution of the
remaining eigenvalues. This is very similar to the calculations of Due~nez [16] and Snaith [17]
for the orthogonal group SO(N).
First we restrict the measure on the entire ensemble to determine the measure on the
remaining eigenvalues. Haar measure on U(N) is given by
d = C
Y
1n<mN
eitn eitm
2 dt1 dtN : (5.25)
Here and following, C is a normalization constant which may vary from line to line, chosen so
that the measure has total mass 1. Restricting to those matrices which have one eigenvalue
at 1 is equivalent to rotating (changing variables) to move an eigenvalue to 1. So we can
also write the measure as
d1 = C
Y
1n<mN 1
eitn eitm
2
Y
1nN 1
eitn 1
2 dt1 dtN 1: (5.26)
The set of matrices which have a repeated eigenvalue at 1 has measure zero, so there is no
canonical way to restrict the measure. However, we are interested in the limiting case of two
O(7). Thus, for the terms we are computing for the distribution of we can treat B2 as
independent of . The expected value of B2 is calculated in Section 5.3.
5.3. Expected value of Aj. We have
Aj :=
1
N j+1j!
F 0
F
(j)
(1)
= ( 1)jN j 1
N 2X
n=1
1
(1 eitn)j+1
; (5.20)
where t1; t2; : : : are the arguments of the zeros of F (z). Since
1
1 eit
=
1
2
+
i
2
cot
t
2
(5.21)
we see that
Re(A0) =
N 2
2N
; (5.22)
so
B1 = Re(b1) =
1
4
; (5.23)
as claimed. Note also that
hA0i =
1
2
1
N
; (5.24)
because the imaginary part of the summand is odd.
For Aj with j 1, we require a random matrix calculation. The sum in (5.20) is dominated
by the terms where eitn is close to 1, so one possibility is to determine the level densities of
the tn. We will nd the expected value of Aj by appealing to prior results on averages of
ratios of characteristic polynomials [3].
We assume that f(z) is the characteristic polynomial of a matrix chosen uniformly with
respect to Haar measure from the unitary group U(N). We restrict to those matrices which
have two eigenvalues very close to 1, and we wish to determine the joint distribution of the
remaining eigenvalues. This is very similar to the calculations of Due~nez [16] and Snaith [17]
for the orthogonal group SO(N).
First we restrict the measure on the entire ensemble to determine the measure on the
remaining eigenvalues. Haar measure on U(N) is given by
d = C
Y
1n<mN
eitn eitm
2 dt1 dtN : (5.25)
Here and following, C is a normalization constant which may vary from line to line, chosen so
that the measure has total mass 1. Restricting to those matrices which have one eigenvalue
at 1 is equivalent to rotating (changing variables) to move an eigenvalue to 1. So we can
also write the measure as
d1 = C
Y
1n<mN 1
eitn eitm
2
Y
1nN 1
eitn 1
2 dt1 dtN 1: (5.26)
The set of matrices which have a repeated eigenvalue at 1 has measure zero, so there is no
canonical way to restrict the measure. However, we are interested in the limiting case of two
Page 12
12 DUE~NEZ, FARMER, FROEHLICH, HUGHES, MEZZADRI, AND PHAN
eigenvalues which are very close together, so we determine the measure by restricting the
measure (5.26) to have jtN 1j t, and then let t! 0. The resulting measure is
d2 = C
Y
1n<mN 2
eitn eitm
2
Y
1nN 2
eitn 1
4 dt1 dtN 2: (5.27)
Let U2(N 2) denote the ensemble of unitary matrices with joint eigenvalue measure 2.
Then if g = g(eit1 ; : : : ; eitn 2) we have
hgiU2(N 2) = CNhgj(1)j
4iU(N 2) (5.28)
where the right side is a Haar measure average, is the characteristic polynomial, and
CN = hj(1)j
4i 1U(N 2): (5.29)
In other words, an expectation involving repeated eigenvalues on U(N) is equivalent to an
expectation on U(N 2) with an extra factor of the 4th power of the characteristic polynomial.
This is the key observation for computing the expected values of the Aj because it reduces
it to the evaluation of known quantities.
Speci cally, let
G( 1; 2; 3; 4; 1; 2;
1;
2) =
(e 1)(e 2)(e 3)(e 4)(e 1)(e 2)
(e
1)(e
2)
: (5.30)
Theorem 4.1 of [3] provides an explicit formula for the expected value of G for the char-
acteristic polynomial of Haar distributed matrices on U(N 2). The formula is complicated
so we do not reproduce it here. This is sucient to determine the expected values of all the
quantities in (5.15). The calculation requires the assistance of a computer algebra package.
We now present the answers, which we determined with the help of Mathematica.
The normalization constant for the measure 2 is (the reciprocal of)
hG(0; 0; 0; 0; 0; 0; 0; 0)iU(N 2) = hj(1)j
4iU(N 2)
=
N4
12
N2
12
= C 1N ; (5.31)
eigenvalues which are very close together, so we determine the measure by restricting the
measure (5.26) to have jtN 1j t, and then let t! 0. The resulting measure is
d2 = C
Y
1n<mN 2
eitn eitm
2
Y
1nN 2
eitn 1
4 dt1 dtN 2: (5.27)
Let U2(N 2) denote the ensemble of unitary matrices with joint eigenvalue measure 2.
Then if g = g(eit1 ; : : : ; eitn 2) we have
hgiU2(N 2) = CNhgj(1)j
4iU(N 2) (5.28)
where the right side is a Haar measure average, is the characteristic polynomial, and
CN = hj(1)j
4i 1U(N 2): (5.29)
In other words, an expectation involving repeated eigenvalues on U(N) is equivalent to an
expectation on U(N 2) with an extra factor of the 4th power of the characteristic polynomial.
This is the key observation for computing the expected values of the Aj because it reduces
it to the evaluation of known quantities.
Speci cally, let
G( 1; 2; 3; 4; 1; 2;
1;
2) =
(e 1)(e 2)(e 3)(e 4)(e 1)(e 2)
(e
1)(e
2)
: (5.30)
Theorem 4.1 of [3] provides an explicit formula for the expected value of G for the char-
acteristic polynomial of Haar distributed matrices on U(N 2). The formula is complicated
so we do not reproduce it here. This is sucient to determine the expected values of all the
quantities in (5.15). The calculation requires the assistance of a computer algebra package.
We now present the answers, which we determined with the help of Mathematica.
The normalization constant for the measure 2 is (the reciprocal of)
hG(0; 0; 0; 0; 0; 0; 0; 0)iU(N 2) = hj(1)j
4iU(N 2)
=
N4
12
N2
12
= C 1N ; (5.31)
Page 13
ROOTS OF THE DERIVATIVE OF THE ZETA FUNCTION 13
say. As ! 0 we have
hA30i =
*
0
3
(1)
+
U2(N 2)
= CN
@3
@ 1; 2;
1
( 1; 2;
1)=(0;0;0)
hG( 1; 2; 0; 0; 0; 0;
1; 0)iU(N 2)
=
1
10
N3
7
10
N2 +
8
5
N
6
5
hA1i =
0
0
(1)
U2(N 2)
= CN
1 +
d
d 1
1=0
@
@ 1
1= 1
hG( 1; 0; 0; 0; 0; 0;
1; 0)iU(N 2)
=
1
15
N2
1
2
N +
11
15
hA0A1i =
0
(1)
0
0
(1)
U2(N 2)
= CN
1 +
d
d 1
1=0
@
@ 1
1= 1
@
@ 2
( 2;
2)=(0;0)
hG( 1; 2; 0; 0; 0; 0;
1;
2)iU(N 2)
=
1
30
N3
3
10
N2 +
13
15
N
4
5
: (5.32)
Thus,
hB2i = Rehb2i =
1
48
7
48
N 1 +O(N 2): (5.33)
Inserting this into (5.19) gives the expansion claimed in Theorem 3.1.
6. The Riemann zeta-function
We do analogous calculations for derivatives of the Riemann zeta-function and compare
our results with data.
We start with
0(s)
(s)
= b
1
s 1
1
2
0(12s+ 1)
(12s+ 1)
+
X
1
s
+
1
(6.1)
where b = log 2 1 12
. As in the polynomial case, we assume there are two very closely
spaced zeros of the -function and look for the nearby zero of 0. Suppose the closely spaced
zeros are
=
1
2
+ i(t 12) (6.2)
with
s0 =
1
2
+X + it (6.3)
say. As ! 0 we have
hA30i =
*
0
3
(1)
+
U2(N 2)
= CN
@3
@ 1; 2;
1
( 1; 2;
1)=(0;0;0)
hG( 1; 2; 0; 0; 0; 0;
1; 0)iU(N 2)
=
1
10
N3
7
10
N2 +
8
5
N
6
5
hA1i =
0
0
(1)
U2(N 2)
= CN
1 +
d
d 1
1=0
@
@ 1
1= 1
hG( 1; 0; 0; 0; 0; 0;
1; 0)iU(N 2)
=
1
15
N2
1
2
N +
11
15
hA0A1i =
0
(1)
0
0
(1)
U2(N 2)
= CN
1 +
d
d 1
1=0
@
@ 1
1= 1
@
@ 2
( 2;
2)=(0;0)
hG( 1; 2; 0; 0; 0; 0;
1;
2)iU(N 2)
=
1
30
N3
3
10
N2 +
13
15
N
4
5
: (5.32)
Thus,
hB2i = Rehb2i =
1
48
7
48
N 1 +O(N 2): (5.33)
Inserting this into (5.19) gives the expansion claimed in Theorem 3.1.
6. The Riemann zeta-function
We do analogous calculations for derivatives of the Riemann zeta-function and compare
our results with data.
We start with
0(s)
(s)
= b
1
s 1
1
2
0(12s+ 1)
(12s+ 1)
+
X
1
s
+
1
(6.1)
where b = log 2 1 12
. As in the polynomial case, we assume there are two very closely
spaced zeros of the -function and look for the nearby zero of 0. Suppose the closely spaced
zeros are
=
1
2
+ i(t 12) (6.2)
with
s0 =
1
2
+X + it (6.3)
Page 14
14 DUE~NEZ, FARMER, FROEHLICH, HUGHES, MEZZADRI, AND PHAN
a zero of 0.
Using
0
(s) = log(s) +O(1=s) (6.4)
we have
0 = b
1
2
log t+
1
s0
+
1
s0 +
+
X
6=
1
s0
+
1
; (6.5)
where
b = b+
1
2
log(2) i
4
+O(1=t): (6.6)
Note that
1
s0
+
1
s0 +
=
8X
4X2 + 2
(6.7)
and
1
s0
+
1
=
X i(t
)
X2 + (t
)2
+
1
2 i
1
4 +
2
; (6.8)
which has real part
X
X2 + (t
)2
+
1
2
1
4 +
2
: (6.9)
As in the polynomial case, we rescale:
X =x= log t
=2= log t: (6.10)
Note that this is analogous to the rescaling in the unitary case because N log(t=2). We
have
0 = b + I
1
2
log t+
2x log t
x2 + 22
+ x log t
X
1
x2 + 42
2
; (6.11)
where b is a real constant, I is purely imaginary, and the sum is over the rescaled zeros
= log t(t
)=2. We follow the same procedure as in the U(N) case. First multiply
through (6.11) by (x2 + 22)= log t and expand the nal summand as a series in x, giving
0 = 2x+ (x2 + 22)
1
2
+ x 1 x
3 2 +O(x
5)
+ smaller terms; (6.12)
where
j =
X
1
(42
2)j
: (6.13)
Note that (6.12) has the same form as (5.12). Now write x = 122 + 244 + O(6) and
gather terms to get
2 1
1
2
22 +
2 2
1
2
21 + 1
24 + smaller terms: (6.14)
Thus,
1
1
4
(6.15)
2
1
64
1
8
1; (6.16)
a zero of 0.
Using
0
(s) = log(s) +O(1=s) (6.4)
we have
0 = b
1
2
log t+
1
s0
+
1
s0 +
+
X
6=
1
s0
+
1
; (6.5)
where
b = b+
1
2
log(2) i
4
+O(1=t): (6.6)
Note that
1
s0
+
1
s0 +
=
8X
4X2 + 2
(6.7)
and
1
s0
+
1
=
X i(t
)
X2 + (t
)2
+
1
2 i
1
4 +
2
; (6.8)
which has real part
X
X2 + (t
)2
+
1
2
1
4 +
2
: (6.9)
As in the polynomial case, we rescale:
X =x= log t
=2= log t: (6.10)
Note that this is analogous to the rescaling in the unitary case because N log(t=2). We
have
0 = b + I
1
2
log t+
2x log t
x2 + 22
+ x log t
X
1
x2 + 42
2
; (6.11)
where b is a real constant, I is purely imaginary, and the sum is over the rescaled zeros
= log t(t
)=2. We follow the same procedure as in the U(N) case. First multiply
through (6.11) by (x2 + 22)= log t and expand the nal summand as a series in x, giving
0 = 2x+ (x2 + 22)
1
2
+ x 1 x
3 2 +O(x
5)
+ smaller terms; (6.12)
where
j =
X
1
(42
2)j
: (6.13)
Note that (6.12) has the same form as (5.12). Now write x = 122 + 244 + O(6) and
gather terms to get
2 1
1
2
22 +
2 2
1
2
21 + 1
24 + smaller terms: (6.14)
Thus,
1
1
4
(6.15)
2
1
64
1
8
1; (6.16)
Page 16
16 DUE~NEZ, FARMER, FROEHLICH, HUGHES, MEZZADRI, AND PHAN
The measure d(a;b) is Haar measure on U(M + a+ b) restricted to those matrices which
have a eigenvalues equal to 1 and b eigenvalues equal to 1. Note that W (a;b)1 is independent
of b.
Proof. For xed a; b > 12 de ne !(z) = jz 1j
2ajz + 1j2b. Let f ng1n=0 be the sequence in
C[x] uniquely determined by the following requirements:
(1) For all n 0, n is of degree n and has positive leading coecient.
(2) For all m;n 0,
h m; ni :=
1
2
Z
m(e
it) n(eit)w(e
it)dt = mn; (7.3)
where mn is the Kronecker delta.
Then f ng is the sequence of normalized orthogonal polynomials on the unit circle with
respect to the measure d(z) = (2) 1!(z)d`(z) (where d` is the arc-length element.)
Let
KM(z; w) :=
M 1X
n=0
n(z) n(w) (7.4)
be the projection kernel onto polynomials of degree less than M with respect to the inner
product (7.3). By the Gaudin-Mehta method, the probability measure (7.1), when regarded
as a measure on the unit circle, can be rewritten as
dk =
1
M !
det
1j;kM
(KM(zj; zk))
MY
j=1
d(zj) (7.5)
(note that the normalization constant CM;k is no longer needed). Then the 1-level measure
is W (M)1 (z)d(z), where
W (M)1 (z) = KM(z; z): (7.6)
(The normalization above is such that the total mass of the 1-level measure is equal to M .)
Let the \dual" of a polynomial (z) = cnzn + cn 1zn 1 + + c1z + c0 of degree n be
the polynomial
(z) = zn (z 1) = cn + cn 1 + + c1z
n 1 + c0z
n: (7.7)
Then we have the following formula of Szeg}o for the projection kernel ([21], Theorem 11.4.2):
KM(z; w) =
M(z)
M(w) M(z) M(w)
1 zw
: (7.8)
(This formula is analogous to the classical one of Christo el and Darboux for the projection
kernel of orthogonal polynomials on the line.)
In view of (7.6) and (7.8), in order to nd the rescaled limit of the 1-level measure as
M ! 1 it will suce to derive the asymptotic behavior of the orthogonal polynomials n
as n ! 1 near the point z = +1. Theorem 7.2 and formula (7.16) below are the key
ingredients, but rst we need to introduce some notation.
Denote by P (a;b)n the classical Jacobi polynomials: they are orthogonal in the interval
[ 1; 1] with respect to the measure
w(a;b)(x) := (1 x)a(1 + x)b (7.9)
The measure d(a;b) is Haar measure on U(M + a+ b) restricted to those matrices which
have a eigenvalues equal to 1 and b eigenvalues equal to 1. Note that W (a;b)1 is independent
of b.
Proof. For xed a; b > 12 de ne !(z) = jz 1j
2ajz + 1j2b. Let f ng1n=0 be the sequence in
C[x] uniquely determined by the following requirements:
(1) For all n 0, n is of degree n and has positive leading coecient.
(2) For all m;n 0,
h m; ni :=
1
2
Z
m(e
it) n(eit)w(e
it)dt = mn; (7.3)
where mn is the Kronecker delta.
Then f ng is the sequence of normalized orthogonal polynomials on the unit circle with
respect to the measure d(z) = (2) 1!(z)d`(z) (where d` is the arc-length element.)
Let
KM(z; w) :=
M 1X
n=0
n(z) n(w) (7.4)
be the projection kernel onto polynomials of degree less than M with respect to the inner
product (7.3). By the Gaudin-Mehta method, the probability measure (7.1), when regarded
as a measure on the unit circle, can be rewritten as
dk =
1
M !
det
1j;kM
(KM(zj; zk))
MY
j=1
d(zj) (7.5)
(note that the normalization constant CM;k is no longer needed). Then the 1-level measure
is W (M)1 (z)d(z), where
W (M)1 (z) = KM(z; z): (7.6)
(The normalization above is such that the total mass of the 1-level measure is equal to M .)
Let the \dual" of a polynomial (z) = cnzn + cn 1zn 1 + + c1z + c0 of degree n be
the polynomial
(z) = zn (z 1) = cn + cn 1 + + c1z
n 1 + c0z
n: (7.7)
Then we have the following formula of Szeg}o for the projection kernel ([21], Theorem 11.4.2):
KM(z; w) =
M(z)
M(w) M(z) M(w)
1 zw
: (7.8)
(This formula is analogous to the classical one of Christo el and Darboux for the projection
kernel of orthogonal polynomials on the line.)
In view of (7.6) and (7.8), in order to nd the rescaled limit of the 1-level measure as
M ! 1 it will suce to derive the asymptotic behavior of the orthogonal polynomials n
as n ! 1 near the point z = +1. Theorem 7.2 and formula (7.16) below are the key
ingredients, but rst we need to introduce some notation.
Denote by P (a;b)n the classical Jacobi polynomials: they are orthogonal in the interval
[ 1; 1] with respect to the measure
w(a;b)(x) := (1 x)a(1 + x)b (7.9)
Page 17
ROOTS OF THE DERIVATIVE OF THE ZETA FUNCTION 17
and are normalized as follows ([21], Equation 4.3.3):
h(a;b)n :=
Z 1
1
jP (a;b)n (x)j
2w(a;b)(x) dx =
2a+b+1
2n+ a+ b+ 1
(n+ a+ 1) (n+ b+ 1)
(n+ 1) (n+ a+ b+ 1)
: (7.10)
Let also
h+n = 2
a+bh(a+1=2;b+1=2)n ; h
n = 2
a+bh(a 1=2;b 1=2)n : (7.11)
Theorem 7.2. [21]
z n 2n(z) = AP
(a 1=2;b 1=2)
n
z + z 1
2
+B(z z 1)P (a+1=2;b+1=2)n 1
z + z 1
2
z n+1 2n 1(z) = CP
(a 1=2;b 1=2)
n
z + z 1
2
+D(z z 1)P (a+1=2;b+1=2)n 1
z + z 1
2
: (7.12)
Letting cn = (a+ b)=(n+ a+ b), we have
A =
s
2
1 + cn
h n
B =
1
2
s
2
1 cn
h+n 1
C =
s
2
1 cn
h n
D =
1
2
s
2
1 + cn
h+n 1
:
Equation (7.12) appears as (11.5.4) in Szeg}o's book (except for an obvious typograph-
ical mistake therein.) The constants A;B;C;D can be easily found using Szeg}o's equa-
tion (11.5.2) together with the fact that 2n 1 is a polynomial, which forces the coecient
of z n on the right-hand side of equation (7.12) to vanish. We omit the details.
A meaningful rescaling of the 1-level measure is achieved via the change of variables
t =
2
M
: (7.13)
Indeed, one nds that the limiting 1-level measure (as M !1) is W1()d, where
W1() = K1(; ); (7.14)
K1(; ) = lim
M!1
1
M
KM(e
2=M ; e2=M)
q
!(e2=M)!(e2=M): (7.15)
(Actually, all limiting local correlations can be expressed in terms of the limiting kernel K1,
not just the 1-level density.)
It remains to compute K1(; ). It suces to use equation (7.12) in formula (7.8) and an
asymptotic formula by Szeg}o's (see [21], equation (8.21.17)):
sin
t
2
a
cos
t
2
b
P (a;b)n (cos t) = N
a (n+ a+ 1)
n!
r
t
sin t
Ja(Nt) + t
a+2O(na); (7.16)
valid for xed a > 1, b 2 R, in the range 0 < t c=n for any xed constant c > 0, where
N = n+ (a+ b+ 1)=2 and Ja is a Bessel function of the rst kind.
A tedious but straightforward computation nally gives:
K1(; ) =
2
ei( )
p
Ja+1=2()Ja 1=2() Ja 1=2()Ja+1=2()
(7.17)
K1(; ) =
2
[Ja+1=2()
2 + Ja 1=2()
2] 2aJa+1=2()Ja 1=2()
: (7.18)
and are normalized as follows ([21], Equation 4.3.3):
h(a;b)n :=
Z 1
1
jP (a;b)n (x)j
2w(a;b)(x) dx =
2a+b+1
2n+ a+ b+ 1
(n+ a+ 1) (n+ b+ 1)
(n+ 1) (n+ a+ b+ 1)
: (7.10)
Let also
h+n = 2
a+bh(a+1=2;b+1=2)n ; h
n = 2
a+bh(a 1=2;b 1=2)n : (7.11)
Theorem 7.2. [21]
z n 2n(z) = AP
(a 1=2;b 1=2)
n
z + z 1
2
+B(z z 1)P (a+1=2;b+1=2)n 1
z + z 1
2
z n+1 2n 1(z) = CP
(a 1=2;b 1=2)
n
z + z 1
2
+D(z z 1)P (a+1=2;b+1=2)n 1
z + z 1
2
: (7.12)
Letting cn = (a+ b)=(n+ a+ b), we have
A =
s
2
1 + cn
h n
B =
1
2
s
2
1 cn
h+n 1
C =
s
2
1 cn
h n
D =
1
2
s
2
1 + cn
h+n 1
:
Equation (7.12) appears as (11.5.4) in Szeg}o's book (except for an obvious typograph-
ical mistake therein.) The constants A;B;C;D can be easily found using Szeg}o's equa-
tion (11.5.2) together with the fact that 2n 1 is a polynomial, which forces the coecient
of z n on the right-hand side of equation (7.12) to vanish. We omit the details.
A meaningful rescaling of the 1-level measure is achieved via the change of variables
t =
2
M
: (7.13)
Indeed, one nds that the limiting 1-level measure (as M !1) is W1()d, where
W1() = K1(; ); (7.14)
K1(; ) = lim
M!1
1
M
KM(e
2=M ; e2=M)
q
!(e2=M)!(e2=M): (7.15)
(Actually, all limiting local correlations can be expressed in terms of the limiting kernel K1,
not just the 1-level density.)
It remains to compute K1(; ). It suces to use equation (7.12) in formula (7.8) and an
asymptotic formula by Szeg}o's (see [21], equation (8.21.17)):
sin
t
2
a
cos
t
2
b
P (a;b)n (cos t) = N
a (n+ a+ 1)
n!
r
t
sin t
Ja(Nt) + t
a+2O(na); (7.16)
valid for xed a > 1, b 2 R, in the range 0 < t c=n for any xed constant c > 0, where
N = n+ (a+ b+ 1)=2 and Ja is a Bessel function of the rst kind.
A tedious but straightforward computation nally gives:
K1(; ) =
2
ei( )
p
Ja+1=2()Ja 1=2() Ja 1=2()Ja+1=2()
(7.17)
K1(; ) =
2
[Ja+1=2()
2 + Ja 1=2()
2] 2aJa+1=2()Ja 1=2()
: (7.18)
Page 18
18 DUE~NEZ, FARMER, FROEHLICH, HUGHES, MEZZADRI, AND PHAN
Remark. The kernel K1(; ) can be used to compute n-level correlations and spacing sta-
tistics through (matrix or operator) determinants (see [22] for an explanation and proof of
these applications). Incidentally, for the purposes of evaluating such determinants, the factor
ei( ) may be suppressed in (7.17) (this is tantamount to conjugating the corresponding
integral operator by a unitary transformation).
8. Appendix: The Domain of Analyticity of the Root of the Derivative
by Eduardo Due~nez
De ne as before e(x) := e2ix for x real. For any xed N 3 de ne eN(x) := e(x=N) =
e2ix=N . Consider N unit complex numbers
eN(0); eN( 0); eN(1); eN(2); : : : ; eN(N 2); (8.1)
where
0 0 j N 0; j = 1; 2; : : : ; N 2: (8.2)
(I. e., the open arc centered at 1 of the unit circle jzj = 1 having endpoints eN(0) contains
none of the remaining N 2 numbers). The inequalities (8.2) de ne a \pyramid" PN
contained in the cube [0; N ]N 1. For xed 0 < T N=2, denote by P(T )N the closed truncation
of PN at height T . Then P
(T )
N is de ned by the inequalities
0 j N 0; j = 1; 2; : : : ; N 2;
0 0 T: (8.3)
For notational convenience we will set = (1; 2; : : : ; N 2) and 0 = (0; ). Finally, let
f(z) = f(0; z) = (z eN(0))(z eN( 0))
N 2Y
j=1
(z eN(j)) (8.4)
be the monic polynomial of degree N with roots (8.1).
In this section we prove the following result.
Proposition 8.1. With the above notation, for every T < 1 there exists N(T ) such that for
all N N(T ) and for each 0 in P
(T )
N , the derivative f
0(z) = @@zf(0; z) of f has a unique
root z0 = &(0) in the open disk with diameter [eN( 0); eN(0)]. Moreover, & is an analytic
function of 0 in the interior of P
(T )
N .
We remark that, by the Gauss-Lucas theorem, the root z0 alluded to in Proposition 8.1
must lie on or to the left of the vertical diameter [eN( 0); eN(0)].
Proof. Fix T < 1= and consider N as a parameter (N 3) for the time being. Let
F (z) = F (; z) :=
N 2Y
j=1
(z eN(j)); (8.5)
so
f(z) = (z eN(0))(z eN( 0))F (z) (8.6)
and
f 0
f
(z) = g(z) + L(; z); (8.7)
Remark. The kernel K1(; ) can be used to compute n-level correlations and spacing sta-
tistics through (matrix or operator) determinants (see [22] for an explanation and proof of
these applications). Incidentally, for the purposes of evaluating such determinants, the factor
ei( ) may be suppressed in (7.17) (this is tantamount to conjugating the corresponding
integral operator by a unitary transformation).
8. Appendix: The Domain of Analyticity of the Root of the Derivative
by Eduardo Due~nez
De ne as before e(x) := e2ix for x real. For any xed N 3 de ne eN(x) := e(x=N) =
e2ix=N . Consider N unit complex numbers
eN(0); eN( 0); eN(1); eN(2); : : : ; eN(N 2); (8.1)
where
0 0 j N 0; j = 1; 2; : : : ; N 2: (8.2)
(I. e., the open arc centered at 1 of the unit circle jzj = 1 having endpoints eN(0) contains
none of the remaining N 2 numbers). The inequalities (8.2) de ne a \pyramid" PN
contained in the cube [0; N ]N 1. For xed 0 < T N=2, denote by P(T )N the closed truncation
of PN at height T . Then P
(T )
N is de ned by the inequalities
0 j N 0; j = 1; 2; : : : ; N 2;
0 0 T: (8.3)
For notational convenience we will set = (1; 2; : : : ; N 2) and 0 = (0; ). Finally, let
f(z) = f(0; z) = (z eN(0))(z eN( 0))
N 2Y
j=1
(z eN(j)) (8.4)
be the monic polynomial of degree N with roots (8.1).
In this section we prove the following result.
Proposition 8.1. With the above notation, for every T < 1 there exists N(T ) such that for
all N N(T ) and for each 0 in P
(T )
N , the derivative f
0(z) = @@zf(0; z) of f has a unique
root z0 = &(0) in the open disk with diameter [eN( 0); eN(0)]. Moreover, & is an analytic
function of 0 in the interior of P
(T )
N .
We remark that, by the Gauss-Lucas theorem, the root z0 alluded to in Proposition 8.1
must lie on or to the left of the vertical diameter [eN( 0); eN(0)].
Proof. Fix T < 1= and consider N as a parameter (N 3) for the time being. Let
F (z) = F (; z) :=
N 2Y
j=1
(z eN(j)); (8.5)
so
f(z) = (z eN(0))(z eN( 0))F (z) (8.6)
and
f 0
f
(z) = g(z) + L(; z); (8.7)
Page 19
ROOTS OF THE DERIVATIVE OF THE ZETA FUNCTION 19
where
g(z) :=
1
z eN(0)
+
1
z eN( 0)
; (8.8)
L(; z) :=
F 0
F
(z) =
N 2X
j=1
1
z eN(j)
: (8.9)
De ne cN(x) := cos(2x=N) and sN(x) := sin(2x=N), so eN(x) = cN(x) + isN(x).
Parametrize the boundary of the disk with diameter eN(0) (i. e., the disk jz cN(0)j
sN(0)) as:
z( ) = cN(0) + ie
i sN(0): (8.10)
Since roots of f occur at eN(0) whenever f has multiple roots there, it is best to work
instead with a slightly deformed contour C obtained as the boundary of the \twice bitten"
disk
jz cN(0)j sN(0)
jz eN(0)j
jz eN( 0)j :
missing tiny -neighborhoods of the points eN(0). We still assume that C is parametrized
by (8.10), except for in a small -neighborhood of any multiple of . (We will not write
down the exact parametrization of C for such values of since its precise form will not be
needed.)
Write C as a union of four pieces: C = Cd [ Cr [ Cu [ C` (down, right, up, left), where
Cd = fz( ) : + g;
Cr = fz( ) : + g;
Cu = fz( ) : g;
C` = fz( ) : g:
The assumed inequalities (8.3) ensure that there are no zeros of f anywhere on C (and, a
fortiori, no multiple zeros); hence, every zero of f 0 inside C is a zero of f 0=f with the same
multiplicity. It suces to ensure that that (for all suciently small ) the contour C encloses
exactly one zero of f 0=f . Note that, since f(0; z) has no zeros on or inside C, f 0=f has no
poles there either. By the Argument Principle, the claim in Proposition 8.1 regarding the
uniqueness of the zero z0 of f 0 is equivalent to showing that the image D of C under f 0=f
has index 1 about the origin for large N and all suciently small .
We let w( ) := f
0
f (z( )) and denote the images of the four pieces Cd; Cr; Cu; C` of C under
f 0=f by Dd;Dr;Du;D`.
The idea of the proof is very simple. As we shall show, the dominant part of the logarithmic
derivative f 0=f(z) (for z in C) is g(z). The image of C under g(z) is easy to describe explicitly;
indeed, it is a curve E having index 1 about the origin. The curve D can be regarded as a
perturbation of E . We show that, if H and are suciently small, then the remaining terms
1=(z eN(j)), j = 1; 2; : : : ; N 2, are small enough to keep the index of D equal to that
of E .
We denote by Ed; Er; Eu; E` the images under g of the respective parts of C ( gure 8.1).
where
g(z) :=
1
z eN(0)
+
1
z eN( 0)
; (8.8)
L(; z) :=
F 0
F
(z) =
N 2X
j=1
1
z eN(j)
: (8.9)
De ne cN(x) := cos(2x=N) and sN(x) := sin(2x=N), so eN(x) = cN(x) + isN(x).
Parametrize the boundary of the disk with diameter eN(0) (i. e., the disk jz cN(0)j
sN(0)) as:
z( ) = cN(0) + ie
i sN(0): (8.10)
Since roots of f occur at eN(0) whenever f has multiple roots there, it is best to work
instead with a slightly deformed contour C obtained as the boundary of the \twice bitten"
disk
jz cN(0)j sN(0)
jz eN(0)j
jz eN( 0)j :
missing tiny -neighborhoods of the points eN(0). We still assume that C is parametrized
by (8.10), except for in a small -neighborhood of any multiple of . (We will not write
down the exact parametrization of C for such values of since its precise form will not be
needed.)
Write C as a union of four pieces: C = Cd [ Cr [ Cu [ C` (down, right, up, left), where
Cd = fz( ) : + g;
Cr = fz( ) : + g;
Cu = fz( ) : g;
C` = fz( ) : g:
The assumed inequalities (8.3) ensure that there are no zeros of f anywhere on C (and, a
fortiori, no multiple zeros); hence, every zero of f 0 inside C is a zero of f 0=f with the same
multiplicity. It suces to ensure that that (for all suciently small ) the contour C encloses
exactly one zero of f 0=f . Note that, since f(0; z) has no zeros on or inside C, f 0=f has no
poles there either. By the Argument Principle, the claim in Proposition 8.1 regarding the
uniqueness of the zero z0 of f 0 is equivalent to showing that the image D of C under f 0=f
has index 1 about the origin for large N and all suciently small .
We let w( ) := f
0
f (z( )) and denote the images of the four pieces Cd; Cr; Cu; C` of C under
f 0=f by Dd;Dr;Du;D`.
The idea of the proof is very simple. As we shall show, the dominant part of the logarithmic
derivative f 0=f(z) (for z in C) is g(z). The image of C under g(z) is easy to describe explicitly;
indeed, it is a curve E having index 1 about the origin. The curve D can be regarded as a
perturbation of E . We show that, if H and are suciently small, then the remaining terms
1=(z eN(j)), j = 1; 2; : : : ; N 2, are small enough to keep the index of D equal to that
of E .
We denote by Ed; Er; Eu; E` the images under g of the respective parts of C ( gure 8.1).
Page 20
20 DUE~NEZ, FARMER, FROEHLICH, HUGHES, MEZZADRI, AND PHAN
A
C
D,F E B,H
G
Figure 8.1. The curve E .
Er is parametrized as
g(z( )) =
1
2sN(0) sin
; + : (8.11)
Thus, Er is a twice-traversed straight segment starting at the faraway point H =
g(z( +)) = 1=(2s(0) sin ), moving leftward to the pointA = g(z(0)) = 1=(2s(0))
and retracing itself back to B = g(z( )) (here B = H).
Eu is a large arc BCD
_ on the upper half-plane. (Eu is essentially a semicircle, for
small. This is easily seen from the fact that the dominant term in f 0=f(z) for z very
close to eN(0) is 1=(z eN(0)), namely an inversion with center eN(0), taking the
tiny (almost) semicircle Cu to a huge (almost) semicircle Eu.) Eu escapes any bounded
region of the plane as approaches zero.
E` and Ed are obtained from Er and Eu through central symmetry with respect to the
origin.
It is clear that E has index 1 about the origin for all suciently small .
For notational convenience, denote by ~A = w( =2); ~B = w( ); : : : ; ~H = w( + ) the
points on D analogous to A;B; : : : ; H on E .
We claim that if N is large enough and 0 2 P
(T )
N , then the index of D about the origin
is also 1 for all suciently small and all N N(T ). Note that is allowed to depend on
0, whereas the lower bound N(T ) for N depends only on T .
For notational simplicity, we set N 1 := 0. Write
f 0
f
(z) =
m
z eN(0)
+
N 1X
j=m
1
z eN(j)
| {z }
R(0; z)
;
where m 1 is the multiplicity of the zero eN(0) of f(z) and j, j = m; : : : ; N 1, are the
remaining zeros of f(z).
Let 0 > 0 be the minimum of the distances from eN(0) to the other eN(j), m
j N 1. Now let = 0=N . If jz eN(0)j = , then jz eN(j)j 0 (m j
A
C
D,F E B,H
G
Figure 8.1. The curve E .
Er is parametrized as
g(z( )) =
1
2sN(0) sin
; + : (8.11)
Thus, Er is a twice-traversed straight segment starting at the faraway point H =
g(z( +)) = 1=(2s(0) sin ), moving leftward to the pointA = g(z(0)) = 1=(2s(0))
and retracing itself back to B = g(z( )) (here B = H).
Eu is a large arc BCD
_ on the upper half-plane. (Eu is essentially a semicircle, for
small. This is easily seen from the fact that the dominant term in f 0=f(z) for z very
close to eN(0) is 1=(z eN(0)), namely an inversion with center eN(0), taking the
tiny (almost) semicircle Cu to a huge (almost) semicircle Eu.) Eu escapes any bounded
region of the plane as approaches zero.
E` and Ed are obtained from Er and Eu through central symmetry with respect to the
origin.
It is clear that E has index 1 about the origin for all suciently small .
For notational convenience, denote by ~A = w( =2); ~B = w( ); : : : ; ~H = w( + ) the
points on D analogous to A;B; : : : ; H on E .
We claim that if N is large enough and 0 2 P
(T )
N , then the index of D about the origin
is also 1 for all suciently small and all N N(T ). Note that is allowed to depend on
0, whereas the lower bound N(T ) for N depends only on T .
For notational simplicity, we set N 1 := 0. Write
f 0
f
(z) =
m
z eN(0)
+
N 1X
j=m
1
z eN(j)
| {z }
R(0; z)
;
where m 1 is the multiplicity of the zero eN(0) of f(z) and j, j = m; : : : ; N 1, are the
remaining zeros of f(z).
Let 0 > 0 be the minimum of the distances from eN(0) to the other eN(j), m
j N 1. Now let = 0=N . If jz eN(0)j = , then jz eN(j)j 0 (m j
Page 21
ROOTS OF THE DERIVATIVE OF THE ZETA FUNCTION 21
N 1), so jR(0; z)j (N m)=0. Moreover, m=(z eN(0)) maps Cu into (part of)
the upper half-circle S: jwj = m=, Rew > 0. Therefore, Du will be supported on the
N 1
0
-neighborhood T of S. Since T intersects the imaginary axis Rew = 0 on the interval
i
h
m
N m
0
; m +
N m
0
i
= i
h
m+N(m 1)
0
; N+m(N 1)0
i
, our claim that Du only intersects the
positive imaginary axis is proved. The claim that D` only intersects the negative imaginary
axis for suitably small is proved along identical lines.
In order to prove the claim that D has index 1 about the origin, it suces now to show
that Dr is contained in the right half-plane Re(w) > 0 and D` in the left half-plane Re(w) <
0. The truth of said statement for Dr is obvious, since each of the terms 1=(z eN(j))
(j = 0; : : : ; N 1) in f 0=f(z) (cf., equations (8.7){(8.9)) has positive real part for z on Cr.
Now note that
g(z( )) =
1
2sN(0) sin
; ; (8.12)
is real and positive; we anticipate it to be the dominant term of the logarithmic derivative
f 0=f(z). We will show that the real parts of each of the terms 1z( ) eN (j) (for
and j = 1; : : : ; N 2) are bounded above by a suitably small multiple of g(z( )).
Lemma 8.2. For all z with jzj < 1 we have
max
jj=1
Re
1
z
=
1 Re(z)
1 jzj2
: (8.13)
This lemma can be proved by rewriting the equation jj = 1 in the form
w +
z
1 jzj2
=
1
1 jzj2
;
in terms of the variable
w =
1
z
;
whence the result follows trivially.
Lemma 8.3. For N 3, 0 0 1, 0 < < and all we have
Re
1
z( ) eN( )
0
N
+ 22(7 + 4
p
3)
20
N2
g(z( )); (8.14)
First, since g(z( )) > 0 for 0 < < (cf., equation (8.12)), it suces to prove the
upper bound
0
N
+ 22(7 + 4
p
3)
20
N2
(8.15)
for the quantity
h(0; ; ) :=
1
g(z( ))
Re
1
z( ) eN( )
= sN(0) sin Re
1
z( ) eN( )
: (8.16)
By Lemma 8.2,
h(0; ; )
1 Re(z( ))
1 jz( )j2
sN(0) sin :
N 1), so jR(0; z)j (N m)=0. Moreover, m=(z eN(0)) maps Cu into (part of)
the upper half-circle S: jwj = m=, Rew > 0. Therefore, Du will be supported on the
N 1
0
-neighborhood T of S. Since T intersects the imaginary axis Rew = 0 on the interval
i
h
m
N m
0
; m +
N m
0
i
= i
h
m+N(m 1)
0
; N+m(N 1)0
i
, our claim that Du only intersects the
positive imaginary axis is proved. The claim that D` only intersects the negative imaginary
axis for suitably small is proved along identical lines.
In order to prove the claim that D has index 1 about the origin, it suces now to show
that Dr is contained in the right half-plane Re(w) > 0 and D` in the left half-plane Re(w) <
0. The truth of said statement for Dr is obvious, since each of the terms 1=(z eN(j))
(j = 0; : : : ; N 1) in f 0=f(z) (cf., equations (8.7){(8.9)) has positive real part for z on Cr.
Now note that
g(z( )) =
1
2sN(0) sin
; ; (8.12)
is real and positive; we anticipate it to be the dominant term of the logarithmic derivative
f 0=f(z). We will show that the real parts of each of the terms 1z( ) eN (j) (for
and j = 1; : : : ; N 2) are bounded above by a suitably small multiple of g(z( )).
Lemma 8.2. For all z with jzj < 1 we have
max
jj=1
Re
1
z
=
1 Re(z)
1 jzj2
: (8.13)
This lemma can be proved by rewriting the equation jj = 1 in the form
w +
z
1 jzj2
=
1
1 jzj2
;
in terms of the variable
w =
1
z
;
whence the result follows trivially.
Lemma 8.3. For N 3, 0 0 1, 0 < < and all we have
Re
1
z( ) eN( )
0
N
+ 22(7 + 4
p
3)
20
N2
g(z( )); (8.14)
First, since g(z( )) > 0 for 0 < < (cf., equation (8.12)), it suces to prove the
upper bound
0
N
+ 22(7 + 4
p
3)
20
N2
(8.15)
for the quantity
h(0; ; ) :=
1
g(z( ))
Re
1
z( ) eN( )
= sN(0) sin Re
1
z( ) eN( )
: (8.16)
By Lemma 8.2,
h(0; ; )
1 Re(z( ))
1 jz( )j2
sN(0) sin :
Page 22
22 DUE~NEZ, FARMER, FROEHLICH, HUGHES, MEZZADRI, AND PHAN
From Equation (8.10) it quickly follows that
1 jz( )j2 = sN(20) sin ; and
Re(z( )) = cN(0) sN(0) sin :
Thus,
h(0; ; )
1 cN(0) + sN(0) sin
sN(20) sin
sN(0) sin =
1 cN(0) + sN(0) sin
2cN(0)
=
1
2
tan
20
N
sin +
1
2
sec
20
N
1
1
2
tan
20
N
+
1
2
sec
20
N
1
: (8.17)
Let us now use the Taylor formul with remainder
j tan j
2
2
sup
j#j
j tan00 #j
j sec 1j
2
2
sup
j#j
j sec00 #j
with := 20=N . Both j tan00 #j and j sec00 #j are even functions of #, increasing with j#j, so
their respective suprema are bounded by j tan00(2=3)j = 14 and j sec00(2=3)j = 8
p
3 (since
# = = 20=N 2=3, by the assumptions 0 1 and N 3). The inequalities
tan + 72; and
sec 1 4
p
3 2;
follow immediately. These inequalities together with (8.17) complete the proof of Lemma 8.3
upon setting = 20=N .
To complete the proof that D` is contained in Rew < 0, it remains to show that
Re
f 0
f
(z( ))
> 0 for . (8.18)
But
Re
f 0
f
(z( ))
= g(z( ))
N 2X
j=1
Re
1
z( ) eN(j)
since g(z( )) > 0
g(z( ))
1 (N 2)
0
N
+ 22(7 + 4
p
3)
20
N2
by Lemma 8.3
= g(z( ))
1 0 + U
0
N
+ V
20
N
+W
20
N2
;
say, for suitable absolute constants U; V;W . As N !1, the last bracketed quantity above
has the limit 1 0. As long as j0j T < 1 , there will exist N(T ) such that said quantity
is positive for N N(T ). This completes the proof of (8.18) and of the uniqueness of the
From Equation (8.10) it quickly follows that
1 jz( )j2 = sN(20) sin ; and
Re(z( )) = cN(0) sN(0) sin :
Thus,
h(0; ; )
1 cN(0) + sN(0) sin
sN(20) sin
sN(0) sin =
1 cN(0) + sN(0) sin
2cN(0)
=
1
2
tan
20
N
sin +
1
2
sec
20
N
1
1
2
tan
20
N
+
1
2
sec
20
N
1
: (8.17)
Let us now use the Taylor formul with remainder
j tan j
2
2
sup
j#j
j tan00 #j
j sec 1j
2
2
sup
j#j
j sec00 #j
with := 20=N . Both j tan00 #j and j sec00 #j are even functions of #, increasing with j#j, so
their respective suprema are bounded by j tan00(2=3)j = 14 and j sec00(2=3)j = 8
p
3 (since
# = = 20=N 2=3, by the assumptions 0 1 and N 3). The inequalities
tan + 72; and
sec 1 4
p
3 2;
follow immediately. These inequalities together with (8.17) complete the proof of Lemma 8.3
upon setting = 20=N .
To complete the proof that D` is contained in Rew < 0, it remains to show that
Re
f 0
f
(z( ))
> 0 for . (8.18)
But
Re
f 0
f
(z( ))
= g(z( ))
N 2X
j=1
Re
1
z( ) eN(j)
since g(z( )) > 0
g(z( ))
1 (N 2)
0
N
+ 22(7 + 4
p
3)
20
N2
by Lemma 8.3
= g(z( ))
1 0 + U
0
N
+ V
20
N
+W
20
N2
;
say, for suitable absolute constants U; V;W . As N !1, the last bracketed quantity above
has the limit 1 0. As long as j0j T < 1 , there will exist N(T ) such that said quantity
is positive for N N(T ). This completes the proof of (8.18) and of the uniqueness of the
Page 23
ROOTS OF THE DERIVATIVE OF THE ZETA FUNCTION 23
desired root z0 = &(0) of f 0. The analyticity of & as a function of 0 follows from the joint
analyticity of the function f(0; z) in the variables 0 and z together with the formula
&(0) =
1
2i
I
C
zf 00(0; z)
f 0(0; z)
dz: (8.19)
While C depends on the choice of a xed , it is clear that can be chosen so that z0 = &(0)
is uniquely de ned by the integral (8.19) for all 0 in any desired compact subset of the
interior of P(T )N . This is enough to ensure that & is analytic in the whole interior of P
(T )
N and
concludes the proof of Proposition 8.1.
References
[1] Berndt, B. C., The number of zeros for (k)(s) J. London Math. Soc. (2) 2, 577{580 (1970).
[2] Conrey, J. B., More than two fths of the zeros of the Riemann zeta function are on the critical line J.
Reine Angew. Math. 399, 1{26 (1989).
[3] Conrey, J. B., Farmer, D. W., and Zirnbauer, M., Autocorrelation of rations of L-functions, Commun.
Number Theory Phys. 2, 593{636 (2008).
[4] Conrey, J. B. and Ghosh, A., Zeros of derivatives of the Riemann zeta-function near the critical line
Analytic Number Theory: Proc. Conf. in Honor of P. T. Bateman (Allenton Park, Ill., 1989) (Prog.
Math. vol 85) B. C. Berndt et. al. eds., Birkhauser Inc., Boston, pp. 95{110 (1990).
[5] Farmer, D. W. and Ki, H., Landau-Siegel zeros and zeros of the derivative of the Riemann zeta function,
preprint.
[6] Feng, S., A note on the zeros of the derivative of the Riemann zeta function near the critical line. Acta
Arith. 120, 59{68 (2005).
[7] Guo, C. R., On the zeros of (s) and 0(s) J. Number Theory 54, 206{210 (1995).
||. On the zeros of the derivative of the Riemann zeta function Proc. London Math. Soc. (3) 72,
28{62 (1996).
[8] Garaev, M. Z. and Yldrm, C. Y., On small distances between ordinates of zeros of (s) and 0(s).
Preprint: arXiv:math.NT/0610377
[9] Ki, H., The zeros of the derivative of the Riemann zeta function near the critical line.
Preprint: arXiv:math.NT/0701726
[10] Levinson, N., More than one third of zeros of Riemann's zeta function are on = 1=2 Advances in
Math. 13, 383{436 (1974).
[11] Levinson, N. and Montgomery, H. L., Zeros of the derivatives of the Riemann zeta-function Acta Math.
133, 49{65 (1974).
[12] Mehta, M. L., Random Matrices, second edition. Academic Press, San Diego, 1991.
[13] Mezzadri, F., Random matrix theory and the zeros of 0(s). J. Phys. A 36, 2945{2962 (2003).
[14] Mezzadri, F., How to generate random matrices from the classical compact groups. Notices Amer. Math.
Soc. 54, 592{604 (2007).
[15] Recent perspectives in random matrix theory and number theory. LMS Lecture Note Series, 322. Mez-
zadri, F. and Snaith, N. C. (Eds.) Cambridge University Press, Cambridge, 2005.
[16] Miller, S. J., Investigations of zeros near the central point of elliptic curve L-functions. With an appendix
by Eduardo Due~nez. Experiment. Math. 15, 257{279 (2006).
[17] Snaith, N. C., Derivatives of random matrix characteristic polynomials with applications to elliptic
curves. J. Phys. A 38, 10345{10360 (2005).
[18] Soundararajan, K., The horizontal distribution of the zeros of 0(s). Duke Math. J. 91, 33{59 (1998).
[19] Speiser, A., Geometrisches zur Riemannschen Zetafunktion Math. Ann. 110 514{21 (1934).
[20] Spira, R., Zero-free regions of (k)(s) J. London Math. Soc. 40, 677{682 (1965).
||. Zeros of 0(s) in the critical strip Proc. Amer. Math. Soc. 35, 59{60 (1972).
||. Zeros of 0(s) and the Riemann hypothesis Illinois J. Math. 17, 147{152 (1973).
[21] Szeg}o, G., Orthogonal Polynomials, American Mathematical Society, Providence, RI, 1939.
[22] Tracy, C. A. and Widom, H., Correlation Functions, Cluster Functions, and Spacing Distributions for
Random Matrices, J. Statist. Phys. 92, 809{835 (1998).
desired root z0 = &(0) of f 0. The analyticity of & as a function of 0 follows from the joint
analyticity of the function f(0; z) in the variables 0 and z together with the formula
&(0) =
1
2i
I
C
zf 00(0; z)
f 0(0; z)
dz: (8.19)
While C depends on the choice of a xed , it is clear that can be chosen so that z0 = &(0)
is uniquely de ned by the integral (8.19) for all 0 in any desired compact subset of the
interior of P(T )N . This is enough to ensure that & is analytic in the whole interior of P
(T )
N and
concludes the proof of Proposition 8.1.
References
[1] Berndt, B. C., The number of zeros for (k)(s) J. London Math. Soc. (2) 2, 577{580 (1970).
[2] Conrey, J. B., More than two fths of the zeros of the Riemann zeta function are on the critical line J.
Reine Angew. Math. 399, 1{26 (1989).
[3] Conrey, J. B., Farmer, D. W., and Zirnbauer, M., Autocorrelation of rations of L-functions, Commun.
Number Theory Phys. 2, 593{636 (2008).
[4] Conrey, J. B. and Ghosh, A., Zeros of derivatives of the Riemann zeta-function near the critical line
Analytic Number Theory: Proc. Conf. in Honor of P. T. Bateman (Allenton Park, Ill., 1989) (Prog.
Math. vol 85) B. C. Berndt et. al. eds., Birkhauser Inc., Boston, pp. 95{110 (1990).
[5] Farmer, D. W. and Ki, H., Landau-Siegel zeros and zeros of the derivative of the Riemann zeta function,
preprint.
[6] Feng, S., A note on the zeros of the derivative of the Riemann zeta function near the critical line. Acta
Arith. 120, 59{68 (2005).
[7] Guo, C. R., On the zeros of (s) and 0(s) J. Number Theory 54, 206{210 (1995).
||. On the zeros of the derivative of the Riemann zeta function Proc. London Math. Soc. (3) 72,
28{62 (1996).
[8] Garaev, M. Z. and Yldrm, C. Y., On small distances between ordinates of zeros of (s) and 0(s).
Preprint: arXiv:math.NT/0610377
[9] Ki, H., The zeros of the derivative of the Riemann zeta function near the critical line.
Preprint: arXiv:math.NT/0701726
[10] Levinson, N., More than one third of zeros of Riemann's zeta function are on = 1=2 Advances in
Math. 13, 383{436 (1974).
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Page 24
24 DUE~NEZ, FARMER, FROEHLICH, HUGHES, MEZZADRI, AND PHAN
[23] Zhang, Y., On the zeros of 0(s) near the critical line, Duke Math. J. 110, 555-572 (2001).
Department of Mathematics, The University of Texas At San Antonio
E-mail address: eduenez@math.utsa.edu
American Institute of Mathematics
E-mail address: farmer@aimath.org
Department of Mathematics and Statistics, McGill University
Department of Mathematics, University of York
E-mail address: ch540@york.ac.uk
Department of Mathematics, University of Bristol
E-mail address: f.mezzadri@bristol.ac.uk
Department of Economics, Northwestern University
[23] Zhang, Y., On the zeros of 0(s) near the critical line, Duke Math. J. 110, 555-572 (2001).
Department of Mathematics, The University of Texas At San Antonio
E-mail address: eduenez@math.utsa.edu
American Institute of Mathematics
E-mail address: farmer@aimath.org
Department of Mathematics and Statistics, McGill University
Department of Mathematics, University of York
E-mail address: ch540@york.ac.uk
Department of Mathematics, University of Bristol
E-mail address: f.mezzadri@bristol.ac.uk
Department of Economics, Northwestern University
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