Fundamental Domains of Gamma and Zeta Functions

Theorem 2.1. The preimage by Γ of the real axis is formed with infinitely many unbounded curves (components). The components corresponding to the positive and to the negative real half axis alternate and do not cross each other. Some of them start however from the same poles of Γ.

Proof. The number Γ(x) is real for every real x and the graph of the function x → Γ(x) has the lines x = 0, x = −1, x = −2, … as vertical asymptotes [7].

Figure 1 can be found in most of the books of complex analysis serving as texts for graduate studies. We used the online document [7]. It shows the graph of the real function x → Γ(x), which can be used to draw some information about the complex function Γ.

The respective graph has local minima and maxima, which correspond to the points where Γ^′(z) = 0. All these points are on the real axis, namely, x₀ ∈ (1,2), and for every positive integer n, there is a unique x_n ∈ (−n, −n + 1) such that Γ^′(x_n) = 0. Indeed, let us denote

The sequence (Γ_n) converges uniformly on compact sets of ℂ∖A to Γ. It can be easily checked that for every n ∈ ℕ, the equation $$ is equivalent to an algebraic equation of degree n + 1 and has exactly n + 1 real solutions situated one in every interval (−k, −k + 1), k = 1,2, …, n and one in the interval (0, ∞). Therefore $$ cannot have nonreal solutions. Since $$ converges in turn uniformly on compact subsets of ℂ∖A to Γ^′, we infer that every interval (−n, −n + 1), n ∈ ℕ, contains exactly one solution x_n of the equation Γ^′(z) = 0, and there is one more solution x₀ ∈ (1,2). There are no other solutions of this equation. It is also obvious that

Based on this information, we can reveal the preimage by Γ of the real axis, denoted Γ⁻¹(ℝ). Since all x_n, n ≥ 0 are simple roots of Γ^′(z) = 0, in a neighborhood V_n of every x_n, Γ(z) has the form [1, page  133]

By the Big Picard Theorem, the preimage by Γ of Γ(x_n) is for every n a countable set of points. The formula (2.1) shows that $$, thus this set is of the form {z_n,k}∪$$, k = 0,1, 2, … having the unique accumulation point ∞. Suppose that z_n,0 is x_n. Then by (2.5), the preimage of a small interval (a_n, b_n) of the real axis centered at Γ(x_n) is the union of an interval (α_n, β_n)∋x_n of the real axis and another Jordan arc γ_−n orthogonal to the real axis at x_n (let n = 2 in [1, Figure  4.8]) and symmetric with respect to the real axis, as well as infinitely many other Jordan arcs passing each one through a z_n,k, respectively, $$, k ∈ ℕ. Simultaneous continuations [6] over the real axis of these preimages have as result the interval (−n, −n + 1) for (α_n, β_n). Indeed, −n and −n + 1 being poles for Γ, we have that for x ∈ (x_n, x_n+1), x − >−n and x − >−n + 1 if and only if Γ(x)−>±∞. The continuations have as result an unbounded curve crossing the real axis at x_n for γ_−n and infinitely many other unbounded curves passing each one through z_n,k, or through $$ for k ∈ ℕ. The unboundedness is guaranteed by the fact that the continuation is unlimited, since there are no poles of Γ off the real axis. We use the same notation γ_−n for the curves passing through x_n and γ_n,k, respectively, $$ for the others. We notice that these curves cannot intersect each other, since in such a point of intersection z₀ we would have Γ^′(z₀) = 0, which is excluded. Also, the curves γ_n,k, and $$, k ≠ 0 cannot intersect the real axis, for a similar reason. We call these curves components of Γ⁻¹(ℝ).

Since 0 is a lacunary value for Γ, there can be no continuity (except at the poles) between the preimage of the real positive half axis and negative half axis, which means that each one of these components is unbounded. Moreover, if we use two different colors, say black and red for the preimage of the negative, respectively, positive real half axis, then these colors must alternate, since minima and maxima for the real function Γ(x) are alternating. Hence the intervals (α_n, β_n) have alternating colors, which imply alternating colors for γ_−n. Indeed, due to the continuity of Γ, except at the poles, change of color can happen only there, which means that the color of γ_−n and that of (α_n, β_n) must agree. Thus, the colors of γ_−n are alternating. On the other hand, if a point travels on a small circle centered at origin in the w-plane (w = Γ(z)), it will meet alternatively the positive and the negative half axis, which implies alternation into the colors of γ_n,k. If x_n were multiple zeros of Γ^′ then more than one curve γ_−n of the same color would start from x_n violating this rule of color alternation. Thus, as previously stated, Γ^′ has only simple zeros.

We notice that, by Formula (2.1), if z = x + iy ∈ γ_n,k, or $$ then lim _x→−∞ Γ(z) = 0 and lim _x→+∞ Γ(z) = ∞. Therefore, when z describes any γ_n,k or $$ its image Γ(z) describes the positive or the negative real half axis, the correspondence z − >Γ(z) being bijective there. Also, if z ∈ γ_−n, then lim _x→−∞ Γ(z) = 0.

Theorem 2.2. The complex plane ℂ can be written as a disjoint union of sets bounded by the components of Γ⁻¹(ℝ) such that the interior of each one of them is a fundamental domain of Γ. These domains accumulate to infinity and only there. The function Γ extended to the boundary of each one of them maps them surjectively onto $$.

Proof. Let us introduce first some notations. We denote by Ω_−n the domain bounded by γ_−n−1 and γ_−n, n = 0,1, 2, …. Let Ω₁ be the domain from the upper half plane bounded by γ₀, the interval [x₀, +∞) and the second component of Γ⁻¹(ℝ) situated in the upper half plane and which does not intersect the real axis, Ω₂ be the domain bounded by this component and the fourth one and so forth. We denote by $$, n ∈ ℤ the domain symmetric to Ω_n with respect to the real axis. Obviously, for n ≤ 0, we have $$.

Let us notice that, by the conformal correspondence theorem, the image by Γ of every Ω_−n, n = 0,1, 2, … is the complex plane with a slit L_−n alongside the real axis, from Γ(x_n−1) to Γ(x_n), while the image of every Ω_n and of every $$, n ∈ ℕ is the complex plane with a slit L_n = [0, +∞) (the same for every n ∈ ℕ). It is obvious that the union of the closures of the domains Ω_n and $$, n ∈ ℤ, is the complex plane and if the common boundary of every adjacent couple of them is counted just once, we obtain a disjoint union. As proven in [2], for an arbitrary analytic function having the unique essential singularity at ∞, these domains accumulate to ∞ and only there in the sense that every neighborhood V of ∞ contains infinitely many domains Ω_n and $$ and any compact set in ℂ intersects only a finite number of these domains. Finally, since for every Ω_n and every w ≠ 0 there is $$ such that w = Γ(z) (z ∈ Ω_n if w is not on the corresponding slit and z ∈ ∂Ω_n if w is on the slit) we have that $$ is surjective. The same is true for every $$.

Theorem 3.1. The group G of cover transformations of (ℂ, Γ) has two generators: an involution and a transformation generating an infinite cyclic subgroup of G.

Proof. We notice that U_k are conformal mappings in ℂ∖E since the branch points x_n belong to E. For every k ∈ ℤ we have Γ∘U_k(z) = Γ(z), z ∈ ℂ∖E. Moreover, U_k(Ω_j∖E) = Ω_k+j∖E, $$.

Finally we define $$, j ∈ ℤ by

It can be easily seen that H is an involution and Γ∘H(z) = Γ(z), $$, $$, $$, and so forth. We notice also that

This shows in particular that U₁ generates an infinite cyclic subgroup. It is an elementary exercise to show that the group generated by U₁ and H is the group of covering transformations of (ℂ, Γ).

Theorem 5.1. Consecutive curves $$ and $$ form strips S_k which are infinite in both directions. The function ζ maps these strips (not necessarily bijectively) onto the complex plane with a slit alongside the interval [1, +∞) of the real axis.

Proof. Indeed, if two such curves met at a point s, one of the domains bounded by them would be mapped by ζ onto the complex plane with a slit alongside the real axis from 1 to ζ(s). Such a domain must contain a pole of ζ, which cannot happen.When a point s travels on $$ and $$ leaving the strip at left, ζ(s) moves on the real axis from 1 to ∞ and back.

Theorem 5.2. There are infinitely many points u with ζ(u) = 1.

Proof. Let us denote by u_k,j the points of S_k for which ζ(u_k,j) = 1, by Γ_k,j the components of ζ⁻¹(ℝ) containing u_k,j and by s_k,j the nontrivial zero of ζ situated on Γ_k,j. When lim _σ−>+∞ ζ(σ + it) = 1, σ + it ∈ Γ_k,j, we assign (by abuse!) the value ∞ to the respective u_k,j. We will see later that every S_k contains a unique Γ_k,j with this property. The monodromy theorem assures that there is a one to one correspondence between s_k,j (counted with multiplicities if they exist), u_k,j, and Γ_k,j. If s_k,j is a zero of order m, then m curves Γ_k,j cross at that zero making a star configuration. The color alternation rule is still respected. Since there are infinitely many nontrivial zeros of ζ, it follows that there are infinitely many points u_k,j.

Theorem 5.3. When the continuation takes place over the whole real axis, the components Γ_k,j are such that the branches corresponding to both the positive and the negative real half axis contain only points σ + it with σ < 0 for |σ| big enough.

Proof. Indeed, a point traveling in the same direction on a circle γ centered at the origin of the z-plane meets consecutively the positive and the negative real half axis. Thus the preimage of γ should meet consecutively the branches corresponding to the preimage of the positive and the negative real half axis. It can be easily seen that two components Γ_k,j can meet only at multiple nontrivial zeros of ζ (if they exist!) and no Γ_k,j can intersect any $$. Thus, those components of preimages of circles centered at the origin which cross a $$, will continue to cross alternatively red and black components of the preimage of the real axis. These last components are mapped by ζ either on the interval (−∞, 1), or on the whole real axis.

On the other hand, due to the continuity of ζ on $$, if a component of ζ⁻¹(γ) meets a $$, it should cross it and all $$, l ≥ 1 meeting consecutively the branches corresponding to the preimage of the positive and of the negative real half axis. Such an alternation is possible only if the previously stated condition on σ is fulfilled.

Theorem 5.4. For every k there is a unique component of ζ⁻¹(ℝ) situated in the strip S_k, say Γ_k,0, which is mapped bijectively by ζ onto (−∞, 1), that is, such that lim _σ−>+∞ζ(σ + it) = 1, and lim _σ−>−∞ζ(σ + it) = −∞, σ + it ∈ Γ_k,0.

Proof. The strip S_k is mapped by ζ onto the complex plane with a slit alongside the real axis from 1 to +∞. The mapping is not necessarily bijective. For every x₀ ∈ (1, +∞), there is $$ and $$ such that ζ(s_k) = ζ(s_k+1) = x₀. Let us connect s_k and s_k+1 by a Jordan arc η interior to S_k (except for its ends). Then ζ(η) is a closed curve C_η bounding a domain D or a Jordan arc travelled twice in opposite directions, in which case D = ∅. We need to show that C_η intersects again the real axis, in other words η intersects the preimage of (−∞, 1). Indeed, otherwise C_η would be contained either in the upper or in the lower half plane. Then ζ would map half of the strip S_k bounded by η and the branches of $$ and $$ corresponding to σ → +∞ onto $$ with a slit alongside the real axis from x₀ to 1. We can take x₀ big enough such that this half strip contains no zero of ζ, which makes impossible such a mapping.

Let us show that S_k cannot contain more than one component of the preimage of (−∞, 1). Indeed, if there were more, we could repeat the previous construction with two consecutive such components, taking s_k and s_k+1 with ζ(s_k) = ζ(s_k+1) > 0 and arrive again to a contradiction.

Theorem 5.5. Every strip S_k contains a unique unbounded component of the preimage of the unit disc.

Proof. Indeed, to see this it is enough to take the preimage of a ray making an angle α with the real half axis and let α → 0. A point s ∈ ζ⁻¹({z}) with |z| = 1, argz = α, tends to ∞ as α → 0 if and only if the corresponding component of the preimage of that ray tends to Γ_k,0 as α → 0, which happens if and only if the component of the preimage of the closed unit disc containing the point s is unbounded. The uniqueness of Γ_k,0 implies the uniqueness of such a component. Obviously, the respective unbounded component can contain besides s_k,0 some other nontrivial zeros of ζ, as it appears on the pictures in Figures 6 and 7. We notice that some of S_k contain bounded components of the preimage of the unit disc and some others do not contain such components, and this is another experimental fact. For example S₅,  S₇,  S₁₀ do contain one bounded component, while the others in this range do not. However, it looks like the existence of bounded components becomes a rule for k big enough, when there can be several bounded components of the preimage of the unit disc in every strip S_k (see Figure 7). We have found (see [8]) that the strip corresponding to t ∈ (10,008; 10,016) has three bounded components of the preimage of the unit circle, one of which contains two nontrivial zeros and the strip corresponding to t ∈ (1,000,002; 1,000,012) has six of them, one containing three nontrivial zeros.

Theorem 6.1. All the real zeros of ζ^′ are simple zeros and they alternate with the trivial zeros of ζ.

Proof. Indeed, since the color change can happen only at a zero of ζ (and at s = 1) and the trivial zeros of ζ are simple, if several branches of the preimage of the positive or of the negative half axis crossed the real axis at the same point or in different points between consecutive trivial zeros of ζ, the color alternation for those branches would be violated.

A way to envision this violation is to keep in mind that the function ζ is locally conformal, except at the branch points, hence the orthogonal net formed with the real axis and a family of circles centered at the origin of the z-plane is the image by ζ of an orthogonal net in which the components of the preimages of the negative and of the positive half axis must alternate when travelling in the same direction on each one of the components of the preimage of those circles.

Theorem 6.2. If S_k is a j_k-strip with j_k ≥ 2, then it contains at least one and at most j_k − 1 zeros (counted with multiplicities, if any) of the derivative ζ^′.

Proof. Let us notice first that S₁ is a 1 strip, thus the derivative ζ^′ cannot have any zero in S₁. This is attested by the fact that ζ maps conformally S₁ onto the complex plane with a slit alongside the interval (1, +∞) of the real axis; therefore there cannot be branch points of ζ in S₁. The strips S₂ and S₃ are 2 strips, S₄, S₅, and S₆ are 3 strips, and so forth. Let γ_ρ be a circle |z| = ρ for a small enough value of ρ, such that the preimage of γ_ρ consists of disjoint closed curves. If such a curve η_k,j is in the critical strip, then we can suppose that it contains a unique nontrivial zero s_k,j of ζ. Suppose first that s_k,j belongs to the unbounded component of the preimage of the unit disc. As ρ increases the corresponding curves η_k,j expand such that for some value of ρ,η_k,j will meet another curve of the same type at point v_k,h(j). Indeed, starting with S₂ there are at least two such curves and each one will cover, as ρ varies from 0 to 1, the whole component of the preimage of the unit disc. It is obvious that v_k,h(j) must be a branch point of ζ, due to the fact that ζ takes the same value in points situated on different curves η_k,j in every neighborhood of v_k,h(j). Since v_k,h(j) cannot be a multiple pole, we have necessarily that ζ^′(v_k,h(j)) = 0. All the zeros of ζ^′ we could visit were simple zeros. However, up to now, nothing allows us to say that this should always be the case. If more than two curves η_k,j touch at the same point v_k,h(j) for a ρ = ρ₀, then that point must be a multiple zero of ζ^′. Let us see what global mapping properties of ζ can be described in such a case. At a multiple zero v_k,h(j) of order m of ζ^′ the preimage of the segment of line γ from 1 to ζ(v_k,h(j)) produces a star configuration with m + 1 arcs converging to v_k,h(j). The simultaneous continuation of those arcs over γ must end up in points u_k,j with ζ(u_k,j) = 1, or at ∞, as lim _σ−>+∞ ζ(σ + it) = 1. The theorem of domain preservation (which says that an analytic function in an open and connected set maps that set onto another one of the same type) assures that the respective u_k,j are different. Thus, if m > 1, at least two of these arcs must turn to different points u_k,j with ζ(u_k,j) = 1. It is obvious that those u_k,j are consecutive on the preimage of the unit circle. Then the domains bounded by those arcs and the preimage of the unit circle is mapped conformally by ζ onto the unit disc with a slit alongside γ. Consequently, the domains bounded by the preimage of the interval [1, +∞) and those arcs are mapped conformally by ζ onto the complex plane with a slit alongside this interval followed by a slit alongside γ.

When ρ increases past ρ₀ the respective η_k,j   fuse into a unique closed curve. This last curve can meet for a ρ > ρ₀ another η_k,j or another curve obtained by fusion and so on until we obtain a curve turning around all the zeros of ζ contained in the respective unbounded component of the preimage of the unit disc. The fact that these curves must fuse and not simply intersect each other is obvious. Indeed, due to the continuity of ζ if η_k,j crossed each other, this would have happen at points as close as we wanted of v_k,h(j), which is impossible, since the zeros of ζ^′ are isolated points. Finally, since for a v in the preimage of the unit disc with ζ^′(v) = 0, the preimage of γ_ρ passing through v must contain at least two different components η_k,j, we conclude that the points v_k,h(j) are the only zeros of ζ^′ having images by ζ situated in the unit disc.

The preimage of any circle γ_ρ with ρ > 1 contains a unique unbounded component obtained by the fusion of all unbounded components when increasing ρ past 1. This component intersects every $$ since γ_ρ intersects the interval (1, +∞). As ρ increases, the respective component moves to the left covering unbounded domains in every S_k which are mapped by ζ onto a small quadrilateral situated in the neighborhood of the point z = 1 and exterior to the unit circle. The real axis divides this quadrilateral into two quadrilaterals, the preimages of which have unbounded components. The respective components are curvilinear 4n-gones with two unbounded sides, one corresponding to a segment of the real axis and the other one to an arc of the unit circle having both one end in z = 1. For bigger values of ρ (ρ = 2.5 is big enough!) these domains are the quadrilaterals which can be seen in Figures 6 and 7 on the left of the critical strip. The unbounded component of the preimage of γ_ρ touches bounded components of the preimage of γ_ρ for some values of ρ in zeros of ζ^′ which project by ζ outside the unit disc. Figures 6 and 7 do not give us an accurate description of the intermediate positions of this component at the right of critical strip, since the increment of ρ is too rough and the range of t is too small. Such a position can be seen in the supplementary pictures of [8] for t = 1,000,025. There the preimage of γ_ρ has a bounded component for ρ ≈ 3.5 and the unbounded component of γ_2.5 is on its right side.

A zero of ζ^′ in some strip S_k with image by ζ outside the unit disc cannot belong to components of preimages of γ_ρ with different values of ρ, due to the fact that ζ is a single valued function. The only way for a branch point of ζ to have the image on a $$ with ρ₀ > 1 is for it to be the touching point of two bounded components of the preimage of $$ or of a bounded and unbounded component of the preimage of $$. When increasing ρ past ρ₀ the two components fuse into a unique one, which is unbounded if the two components were not both bounded. To facilitate the counting of the zeros of ζ^′ in S_k, we form a full binary tree in which the leafs are the zeros of ζ and the internal vertices are the zeros of ζ^′,corresponding to the points where the curves η_k,j come into contact. If a zero of ζ^′ is multiple of order m, we can build the tree such that it generates m internal vertices. It follows easily by recursion that if the number of leafs is j_k, then the number of internal vertices is j_k − 1 and the conclusion of the theorem is obvious.

Conjecture 6. All the nonreal zeros of ζ^′ are situated in the right half plane.

Theorem 6.3. If S_k is a j_k-strip, then S_k is the disjoint union of exactly j_k  substrips, whose interiors are fundamental domains of ζ.

Proof. The only 1 strip in the upper half plane is S₁ and it has no branch point of ζ. It is by itself a fundamental domain of ζ and it is mapped conformally by ζ onto the complex plane with a slit alongside [1, +∞). As we have seen in Theorem 5.2, a j_k-strip S_k contains exactly j_k components Γ_k,j of ζ⁻¹(ℝ) and exactly j_k points u_k,j with ζ(u_k,j) = 1, and one of these points being considered (by abuse!) as ∞ (which is the only one for S₁). For k ≥ 2, there is a number h_k,1≤h_k ≤ j_k − 1 of branch points v_k,h(j) of ζ in S_k. If we connect all the points ζ(v_k,h(j)) to the point z = 1 by a segment of line γ_k,h(j) (which is the interval [0,1] when s_k,j    is a multiple zero of ζ) and follow the protocol of Theorem 6.2, we obtain j_k − 1 arcs or unbounded curves L_k,j belonging to S_k which are projected by ζ onto different γ_k,h(j). The arcs L_k,j connect consecutive points u_k,j via a point v_k,h(j), while the unbounded curves go from u_k,j to ∞ via a point v_k,h(j). For every v_k,h(j) at most one of the arcs L_k,j containing v_k,h(j) can be unbounded. There cannot be unbounded L_k,j containing a v_k,h(j) when this point is between two embraced curves Γ_k,j, as in the case where t ∈ (1,000,001; 1,000,002) appearing in supplementary pictures of [8]. By the conformal correspondence theorem, the sub-strips formed by consecutive arcs or unbounded curves L_k,j and consecutive components of ζ⁻¹{[1, +∞)} are mapped conformally by ζ onto the complex plane with a slit alongside [1, +∞) followed by at most three slits alongside intervals starting at z = 1 and ending in some ζ(v_k,h(j)). Every such sub-strip contains a unique s_k,j, if it is a simple zero of ζ, and if s_k,j is a multiple zero of order m then exactly m sub-strips meet at s_k,j. Thus the number of sub-strips of S_k is exactly j_k. If the joint boundary of every couple of adjacent sub-strips is counted just once, S_k is the disjoint union of these sub-strips, which proves completely the theorem.

Theorem 7.1. The group G of cover transformations of (ℂ, ζ) has two generators: an involution and a transformation generating an infinite cyclic subgroup of G.

Proof. In order to find the group of cover transformations of (ℂ, ζ) we need to rename the fundamental domains. We proceed in a way similar to what we did in the case of the function Gamma. Let us denote by σ_n the branch points of ζ situated on the negative real half axis counted in an increasing order of their module. Let Ω₀ be the domain bounded by the branches of the components of the preimage of the positive real half axis crossing the real axis at σ₁ (see Figure 4). It is mapped conformally by ζ onto the complex plane with a slit alongside the real axis from ζ(σ₁) to 1. Let Ω₋₁ be the domain bounded by the component of the preimage of the negative real half axis crossing the real axis in the s-plane at σ₂, by the boundary of Ω₀, by $$ and its symmetric with respect to the real axis. We notice that Ω₋₁ is mapped conformally by ζ onto the complex plane with a slit alongside the real axis complementary to the interval (ζ(σ₂), ζ(σ₁)), The domains Ω_−n, n ≥ 2 are those bounded by the branches of the preimage of the real axis crossing the real axis at σ_n, respectively σ_n+1. They are mapped conformally by ζ onto the complex plane with a slit alongside the complementary with respect to the real axis of the interval between ζ(σ_n) and ζ(σ_n+1). Finally the domains Ω_n, n ∈ ℕ are the former domains Ω_k,j counted starting from the positive real half axis and going up. In order to have consecutive subscripts for all adjacent domains, we take Ω₋₁ = Ω₁. All the domains Ω_n, n ∈ ℤ are fundamental domains of ζ. So are the domains $$ symmetric to them with respect to the real axis. For n ≤ 1, we have $$.

We define, as in the case of the function Gamma mappings U_k and H by the formulas similar to (3.1) and (3.3), where Γ is replaced by ζ. As in that case we need to perform slits into every Ω_j such that for every j and k, ∂Ω_j, and ∂Ω_j+k are mapped by ζ onto the same slits into the z-plane in order for these formulas to be applicable. We notice that the group generated by U₁ and H is the group G of covering transformations of (C, ζ). The group generated by U₁ is an infinite cyclic subgroup of G.