Proof. We will show by induction on k that for each $1\le k\le n$ , $\widetilde {f}_1$ and $\widetilde {f}_2$ coincide on $\mathbb {R}^m \times (\mathbb {R} \cup i\mathbb {R})^k \times \mathbb {R}^{n-k}$ . First, take $k=1$ . Let the domains of $\widetilde {f}_1$ and $\widetilde {f}_2$ be $\mathbb {R}^m \times \Omega _1$ and $\mathbb {R}^m \times \Omega _2$ . Fix $x_1,\ldots ,x_m, y_2,\ldots ,y_n\in \mathbb {R}$ . Let

$$\begin{align*}\Omega:=\{z\in \mathbb{C}: (x_1,\ldots,x_m, z, y_2,\ldots,y_n)\in \Omega_1\cap \Omega_2\}. \end{align*}$$

Since $\Omega _1\cap \Omega _2$ contains $(\mathbb {R}\cup i\mathbb {R})^n$ , it follows that $\Omega $ contains $\mathbb {R} \cup i\mathbb {R}$ . Since $\mathbb {R} \cup i\mathbb {R}$ is a connected set, $\mathbb {R} \cup i\mathbb {R}$ must be a subset of one of the connected components $\Omega _0$ of $\Omega $ . Since $\Omega _1\cap \Omega _2$ is open, it follows that $\Omega $ is open. Thus, every connected component of $\Omega $ is open. In particular, $\Omega _0$ is open.

Define $h_1, h_2:\Omega _0\to \mathbb {C}$ as

$$\begin{align*}h_1(z) := \widetilde{f}_1(x_1,\ldots,x_m, z, y_2,\ldots,y_n), \ \ \ h_2(z) := \widetilde{f}_2(x_1,\ldots,x_m, z, y_2,\ldots,y_n). \end{align*}$$

Then $h_1$ and $h_2$ are holomorphic on the open connected set $\Omega _0$ , and coincide on the real line due to the hypothesis that $\widetilde {f}_1=\widetilde {f}_2=f$ on $\mathbb {R}^m \times (\mathbb {R}\cup i\mathbb {R})^n$ . Thus, $h_1(z)=h_2(z)$ for all $z\in \Omega _0$ . In particular, $h_1(iy)=h_2(iy)$ for all $y\in \mathbb {R}$ . This proves the claim for $k=1$ .

Next, suppose that we have proved the claim up to $k-1$ , for some $k\ge 2$ . Fix some $x_1,\ldots ,x_m, y_1,\ldots , y_{k-1}, y_{k+1},\ldots ,y_n\in \mathbb {R}$ , and define

$$\begin{align*}\Omega:=\{z\in \mathbb{C}: (x_1,\ldots,x_m, y_1,\ldots, y_{k-1}, z, y_{k+1}\ldots,y_n)\in \Omega_1\cap \Omega_2\}. \end{align*}$$

Let $\Omega _0$ be the connected component of $\Omega $ that contains $\mathbb {R}\cup i\mathbb {R}$ . Define $h_1, h_2:\Omega _0\to \mathbb {C}$ as

$$ \begin{align*} h_1(z) &:= \widetilde{f}_1(x_1,\ldots,x_m, y_1,\ldots, y_{k-1},z, y_{k+1},\ldots,y_n), \\ h_2(z) &:= \widetilde{f}_2(x_1,\ldots,x_m, y_1,\ldots, y_{k-1},z, y_{k+1},\ldots,y_n). \end{align*} $$

As before, $\Omega _0$ is an open connected set, $h_1$ and $h_2$ are holomorphic on $\Omega _0$ , and coincide on $\mathbb {R}$ . Thus, $h_1=h_2$ on $i\mathbb {R}$ . This completes the induction step. The case $k=n$ proves the lemma.

Proof. Let $\widetilde {f}$ be the analytic continuation of f on the domain $\mathbb {R}^m \times \Omega $ , satisfying the required criteria. We will prove by induction on k that for any $1\le k\le n$ and any $x_1,\ldots ,x_m, y_1,\ldots ,y_n\in \mathbb {R}$ ,

(2.4.2) $$ \begin{align} \sum_{s_1,\ldots,s_k\in \{-1,1\}} \widetilde{f}(x_1,\ldots,x_m, i s_1 y_1,\ldots, i s_k y_k, y_{k+1},\ldots,y_n)\in \mathbb{R}. \end{align} $$

First, take $k=1$ . Fix $x_1,\ldots ,x_m, y_2,\ldots ,y_n\in \mathbb {R}$ and let

$$\begin{align*}\Omega_*:=\{z\in \mathbb{C}: (x_1,\ldots,x_m, z, y_2,\ldots,y_n)\in \Omega\}. \end{align*}$$

Define $h:\Omega _* \to \mathbb {C}$ as

$$\begin{align*}h(z) :=\widetilde{f}(x_1,\ldots,x_m, z, y_2,\ldots,y_n). \end{align*}$$

As in the proof of Lemma 2.4.1, $\Omega _*$ has an open connected component $\Omega _0$ containing $\mathbb {R} \cup i\mathbb {R}$ . Let $\Omega _0^+ := \{z\in \Omega _0: \operatorname {Im}(z)>0\}$ be the intersection of $\Omega _0$ with the open upper half-plane. Clearly, $\Omega _0^+$ is open. We claim that $\Omega _0^+$ is connected. To see this, suppose not. Since a subset of Euclidean space is connected if and only if it is path connected, this implies that $\Omega _0^+$ has multiple nonempty path connected components. It is easy to see that each component is open. Thus there exist disjoint nonempty open sets $A,B$ whose union is $\Omega _0^+$ , such that no point in A is connected to a point in B by a continuous path that lies entirely in $\Omega _0^+$ .

Since $\Omega _0$ is open and contains $\mathbb {R}$ , we deduce that for each $x\in \mathbb {R}$ , there is an open disk $B_x$ with center x such that $B_x\subseteq \Omega _0$ . Let $B_x^+ := B_x\cap \Omega _0^+$ , so that $B_x^+$ is also the intersection of $B_x$ and the open upper half-plane. Thus, $B_x^+$ is open and connected, but is also the union of the disjoint open sets $B_x^+\cap A$ and $B_x^+\cap B$ . Therefore, one of these two sets must be empty. In other words, either $B_x^+ \subseteq A$ or $B_x^+\subseteq B$ . Let

$$\begin{align*}U := \bigcup_{x: B_x^+\subseteq A} (B_x\cap \mathbb{R}), \ \ \ V := \bigcup_{x: B_x^+\subseteq B} (B_x\cap \mathbb{R}). \end{align*}$$

Since U and V are unions of open intervals, they are open subsets of $\mathbb {R}$ . By the above argument, they are disjoint, and their union is $\mathbb {R}$ . Since $\mathbb {R}$ is connected, this implies that one of U and V must be empty. Suppose that V is empty, so that no $B_x^+$ intersects B.

Take any $x\in B$ and $y\in A$ . Since $\Omega _0$ is connected and hence path connected, there is a continuous curve $\gamma :[0,1]\to \Omega _0$ such that $\gamma (0)=x$ and $\gamma (1)=y$ . By the construction of A and B, we know that this path cannot lie entirely in $\Omega _0^+$ . Let $t:= \inf \{s: \gamma (s)\notin \Omega _0^+\}$ . By the continuity of $\gamma $ , we deduce that $\operatorname {Im}(\gamma (t)) = 0$ , and hence, $w:=\gamma (t)\in \mathbb {R}$ . Let $B_w$ be as above, so that $B_w^+\subseteq A$ . Then, again by the continuity of $\gamma $ , there exists $s\in [0,t)$ that is so close to t that $\gamma (s)\in B_w$ . Then the restriction of $\gamma $ to the interval $[0,s]$ is a continuous curve, fully contained in $\Omega _0^+$ , that connects a point in A to a point in B. This is a contradiction that proves our claim that $\Omega _0^+$ is connected.

Let $\widetilde {\Omega }$ be the union of $\Omega _0^+$ , its reflection across the real line, and $\mathbb {R}$ . It is easy to see that $\widetilde {\Omega }$ is open, by checking that each point in $\widetilde {\Omega }$ belongs to an open disk that is contained in $\widetilde {\Omega }$ . Note that h is holomorphic in $\Omega _0^+$ , h real-valued on $\mathbb {R}$ , and each $x\in \mathbb {R}$ is contained in an open disk $B_x$ such that the intersection of $B_x$ and the open upper half-plane is contained in $\Omega _0^+$ . Thus, by the generalized form of the Schwarz reflection principle [Reference Rudin51, Theorem 11.14], h has an extension to a holomorphic function $\widetilde {h}$ on $\widetilde {\Omega }$ , defined as

$$\begin{align*}\widetilde{h}(z) := \begin{cases} h(z) &\text{ if } z\in \Omega_0^+\cup \mathbb{R},\\ \overline{h(\overline{z})} &\text{ if } \overline{z}\in \Omega_0^+. \end{cases} \end{align*}$$

Note that $\widetilde {\Omega }$ and $\Omega _0$ are open sets that contain $i\mathbb {R}$ . Thus, for each $ix\in i \mathbb {R}$ , there is an open disk $B_{ix}$ centered at $ix$ such that $B_{ix}\subseteq \widetilde {\Omega }\cap \Omega _0$ . Let W be the union of these disks. Since $i \mathbb {R}$ is path connected, it follows that W is path connected, and hence, connected. Clearly, W is open. Now, note that both h and $\widetilde {h}$ are holomorphic on W, and coincide on the upper half of the imaginary axis. Therefore, $h=\widetilde {h}$ everywhere on W. In particular, for any $ix$ on the upper half of the imaginary axis,

$$\begin{align*}h(-ix) = \widetilde{h}(-ix) = \overline{h(ix)}. \end{align*}$$

This is the same as saying that any $y_1\in \mathbb {R}$ ,

$$\begin{align*}\widetilde{f}(x_1,\ldots,x_m,-iy_1,y_2,\ldots,y_n) = \overline{\widetilde{f}(x_1,\ldots,x_m,i y_1, y_2,\ldots,y_n)}. \end{align*}$$

This proves the claim (2.4.2) for $k=1$ . Next, suppose that it holds for $k-1$ . Fix $x_1,\ldots ,x_m,y_1,\ldots ,y_{k-1}, y_{k+1},\ldots ,y_n\in \mathbb {R}$ , and define

$$\begin{align*}\Omega_*:=\{z\in \mathbb{C}: (x_1,\ldots,x_m, y_1,\ldots, y_{k-1}, z, y_{k+1},\ldots,y_n)\in \Omega\}. \end{align*}$$

Define $h:\Omega _* \to \mathbb {C}$ as

$$\begin{align*}h(z) := \sum_{s_1,\ldots,s_{k-1}\in \{-1,1\}} \widetilde{f}(x_1,\ldots,x_m, i s_1y_1,\ldots,is_{k-1}y_{k-1}, z, y_{k+1},\ldots,y_n). \end{align*}$$

Note that $\Omega _*$ is open, $\Omega _*$ contains $\mathbb {R}\cup i\mathbb {R}$ , h is holomorphic on $\Omega _*$ , and by the induction hypothesis, h real-valued on $\mathbb {R}$ . Proceeding exactly as before, we deduce that $h(-ix) = \overline {h(ix)}$ for any $x\in \mathbb {R}$ . Thus,

$$ \begin{align*} \sum_{s_1,\ldots,s_{k-1}\in \{-1,1\}} &\widetilde{f}(x_1,\ldots,x_m, i s_1y_1,\ldots,is_{k-1}y_{k-1}, -iy_k, y_{k+1},\ldots,y_n) \\ &= \overline{\sum_{s_1,\ldots,s_{k-1}\in \{-1,1\}} \widetilde{f}(x_1,\ldots,x_m, i s_1y_1,\ldots,is_{k-1}y_{k-1}, iy_k, y_{k+1},\ldots,y_n)} \end{align*} $$

for any $x_1,\ldots ,x_m, y_1,\ldots ,y_n\in \mathbb {R}$ . This completes the induction step.

Let $W_1,\ldots ,W_{m+n}$ be i.i.d. $N(0,1)$ random variables. Since the standard Gaussian distribution is symmetric around zero, we have that for any $s_1,\ldots ,s_n\in \{-1,1\}$ ,

$$ \begin{align*} \mathbb{E}(\widetilde{f}(W_1,\ldots,W_m, i W_{m+1},\ldots, iW_{m+n})) =\mathbb{E}(\widetilde{f}(W_1,\ldots,W_m, is_1W_{m+1},\ldots,i s_n W_{m+n})). \end{align*} $$

This shows that

$$ \begin{align*} &\mathbb{E}(\widetilde{f}(W_1,\ldots,W_m, i W_{m+1},\ldots, iW_{m+n}))\\ &= \frac{1}{2^n}\sum_{s_1,\ldots,s_n\in \{-1,1\}}\mathbb{E}(\widetilde{f}(W_1,\ldots,W_m, is_1W_{m+1},\ldots,i s_n W_{m+n}))\\ &= \frac{1}{2^n}\mathbb{E}\bigg(\sum_{s_1,\ldots,s_n\in \{-1,1\}}\widetilde{f}(W_1,\ldots,W_m, is_1W_{m+1},\ldots,i s_n W_{m+n})\bigg). \end{align*} $$

By equation (2.4.2) for $k=n$ , the right side is real. This completes the proof.

Proof. Since $\partial _{m+j}f \in \mathcal {F}_{m,n}$ , it has an analytic continuation h to $\mathbb {R}^m \times \Omega $ , where $\Omega $ is an open subset of $\mathbb {C}^n$ that contains $(\mathbb {R}\cup i\mathbb {R})^n$ . Define $\widetilde {f}: \mathbb {R}^m \times \Omega \to \mathbb {C}$ as

$$ \begin{align*} \widetilde{f}(x_1,\ldots,x_m, z_1,\ldots,z_n) &:= f(x_1,\ldots,x_m, z_1,\ldots,z_{j-1},0,z_{j+1},\ldots,z_n) \\ &\qquad + \int_0^{1} z_j h(x_1,\ldots, x_m, z_1,\ldots,z_{j-1}, tz_j, z_{j+1},\ldots,z_n) dt. \end{align*} $$

Then $\widetilde {f}$ is an analytic function on $\mathbb {R}^m \times \Omega $ , and is equal to f on $\mathbb {R}^{m+n}$ . Thus, the function

$$\begin{align*}\widetilde{g}(x_1,\ldots,x_m, z_1,\ldots,z_n) := z_j \widetilde{f}(x_1,\ldots,x_m, z_1,\ldots,z_n) \end{align*}$$

is the analytic continuation of g to $\mathbb {R}^m \times \Omega $ . So by Lemma 2.4.1 and the fact that $g\in \mathcal {F}_{m,n}$ , we conclude that $\mathbb {E}|\widetilde {g}(W)|<\infty $ and $\mathbb {E}(g(Z)) = \mathbb {E}(\widetilde {g}(W))$ , where

$$\begin{align*}W := (W_1,\ldots,W_m, iW_{m+1},\ldots,iW_{m+n}), \end{align*}$$

and $W_1,\ldots ,W_{m+n}$ are i.i.d. $N(0,1)$ random variables. Thus,

$$\begin{align*}\mathbb{E}(Y_j f(Z)) = \mathbb{E}(g(Z)) = \mathbb{E}(\widetilde{g}(W)) = \mathbb{E}(iW_{m+j} \widetilde{f}(W)). \end{align*}$$

On the other hand, $\mathbb {E}(\partial _{m+j} f(Z)) =\mathbb {E}(h(W))$ and $\mathbb {E}|h(W)|<\infty $ . By the definition of $\widetilde {f}$ , we have that $\partial _{m+j} \widetilde {f} = h$ . Thus, by the usual integration by parts formula for Gaussian random variables, we get

$$ \begin{align*} \mathbb{E}(iW_{m+j}\widetilde{f}(W))= i^2\mathbb{E}(\partial_{m+j} \widetilde{f}(W)) = - \mathbb{E}(h(W)) = -\mathbb{E}(\partial_{m+j} f(Z)). \end{align*} $$

Combining the last two displays completes the proof.

Proof. Take any $t\in [0,T)$ and $x\in \mathbb {R}^n$ . Define $g:\mathbb {C}^n \to \mathbb {C}$ as

$$\begin{align*}g(z) := h(x+\sqrt{2t} z). \end{align*}$$

If $t=0$ , then g is just the constant $h(x)$ . Thus,

$$\begin{align*}f(0,x) = h(x). \end{align*}$$

Next, suppose that $t>0$ . By the Cauchy–Schwarz inequality, we have that for any $z\in \mathbb {C}^n$ and any $\alpha> 0$ ,

$$ \begin{align*} |x+\sqrt{2t} z|^2 &\le (|x|+\sqrt{2t} |z|)^2\\ &= (\alpha^{-1}\alpha |x|+ \sqrt{2t}|z|)^2\\ &\le (\alpha^{-2}+1) (\alpha^2 |x|^2 + 2 t|z|^2). \end{align*} $$

Let us now choose $\alpha $ so large that $\gamma := (\alpha ^{-2}+1) t < T$ (which is possible since $t< T$ ), and let $\beta := 1+\alpha ^2$ . Then the above inequality and the hypothesis of the theorem show that there is a finite constant C such that for all $z\in \mathbb {C}^n$ ,

$$ \begin{align*} |g(z)| &= |h(x+\sqrt{2t} z)|\\ &\le C \exp\bigg(\frac{1}{4T}|x+\sqrt{2t}z|^2\bigg)\\ &\le C\exp\bigg(\frac{\beta}{4T}|x|^2 + \frac{2\gamma}{4T}|z|^2\bigg). \end{align*} $$

Since $\gamma <T$ , this proves that $\mathbb {E}|g(iZ)|< \infty $ , where $Z\sim N_{n,0}$ . Thus, $f(t,x) = \mathbb {E}(g(X))$ is well-defined.

Next, fix some $(t,x)\in [0,T)\times \mathbb {R}^n$ . The above inequality shows that there is a small enough ball B around $(t,x)$ , and some $\beta>0$ and $\gamma \in (0,T)$ such that for all $(s,y)\in B$ with $s\ge 0$ , and all $z\in \mathbb {C}^n$ ,

(2.7.1) $$ \begin{align} |h(y + \sqrt{2s}z)|\le C\exp\bigg(\frac{\beta}{4T}|x|^2 + \frac{\gamma}{2T}|z|^2\bigg). \end{align} $$

By the dominated convergence theorem, this implies that for any sequence $(s_n,y_n)\to (t,x)$ , we have

$$ \begin{align*} f(s_n,y_n) &= \mathbb{E}(h(y_n + i\sqrt{2s_n} Z)) \to \mathbb{E}(h(x+ i \sqrt{2t} Z)) = f(t,x). \end{align*} $$

Thus, f is continuous on $[0,T)\times \mathbb {R}^n$ . Since we have already observed that $f(0,x)=h(x)$ , it only remains to show that f satisfies the backward heat equation in $(0,T)\times \mathbb {R}^n$ .

For this, take any $(t,x)\in (0,T)\times \mathbb {R}^n$ and let B be a ball as above, with the radius r of the ball so small that the time coordinate is bigger than $\frac {1}{2}t$ everywhere in the ball. Then note that for any $(s,y)\in B$ and $z\in \mathbb {C}^n$ ,

(2.7.2) $$ \begin{align} \partial_s h(y + \sqrt{2s}z)= \frac{1}{\sqrt{2s}} \sum_{j=1}^n z_j\partial_j h(y+\sqrt{2s} z), \end{align} $$

where $\partial _s$ denotes partial derivative with respect to the parameter s, and $\partial _j h$ is the partial derivative of h in coordinate j. Let $B'$ denote the ball of radius $\frac {1}{2}r$ with center $(t,x)$ . Since h is holomorphic, Cauchy’s integral formula for partial derivatives of a holomorphic function in several complex variables gives that for any $w_1,\ldots ,w_n \in \mathbb {C}$ and $r_1,\ldots ,r_n>0$ ,

$$ \begin{align*} &\partial_j h(w_1,\ldots,w_n) \\ &= \frac{1}{(2\pi i)^n} \oint_{C(w_1, r_1)}\cdots\oint_{C(w_n, r_n)}\frac{h(u_1,\ldots,u_n)}{(u_j-w_j)\prod_{k=1}^n(u_k-w_k)} du_n \cdots du_1, \end{align*} $$

where $C(w_j, r_j)$ is the circular contour of radius $r_j$ centered at $w_j$ , traversed in the counterclockwise direction. From this formula, and the inequality (2.7.1), it follows that there is a constant $C_1$ such that for any $(s,y)\in B'$ and $1\le j\le n$ ,

(2.7.3) $$ \begin{align} |\partial_j h(y+\sqrt{2s} z)| &\le C_1\exp\bigg(\frac{\beta}{4T}|x|^2 + \frac{\gamma}{2T}|z|^2\bigg). \end{align} $$

By the identity (2.7.2), this allows us to apply the dominated convergence theorem and conclude that

$$ \begin{align*} \partial_t f(t,x) &= \mathbb{E}(\partial_t h(x + i\sqrt{2t} Z))= \frac{1}{\sqrt{2t}}\sum_{j=1}^n \mathbb{E}(iZ_j \partial_j h(x+i \sqrt{2t} Z)). \end{align*} $$

Again by the inequality (2.7.3), we can apply usual Gaussian integration by parts to get

$$\begin{align*}\mathbb{E}(iZ_j \partial_j h(x+i \sqrt{2t} Z)) = - \mathbb{E}(\partial_j^2 h(x+i\sqrt{2t}Z)). \end{align*}$$

Combining the last two displays completes the proof.

Proof. From the given condition on g, it is easy to see that if $b< (2c)^{-1/2}$ , then the functions $g(bx)$ and $g(bx)e^{-F(bx)/b^2}$ are in $\mathcal {F}_{0,1}$ . Let us henceforth assume that $b< (2c)^{-1/2}$ . Let $X\sim N(0,-1)$ and $Y\sim N(0,1)$ . Then note that

$$ \begin{align*} \mathbb{E}(e^{-F(bX)/b^2}) &= \mathbb{E}(e^{-F(ibY)/b^2}) = \mathbb{E}(e^{-iabY/b^2}) = e^{-a^2/2b^2} = e^{H(a)/b^2}, \end{align*} $$

which proves the second claim of the proposition. Next, note that

(2.8.3) $$ \begin{align} \mathbb{E}(g(bX) e^{-F(bX)/b^2}) &= \mathbb{E}(g(ibY) e^{-iabY/b^2})\notag \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty g(iby)e^{-iay/b} e^{-y^2/2} dy. \end{align} $$

To evaluate the above integral, we will change the contour of integration. Since $|g(z)|=O(e^{c|z|^2})$ as $|z|\to \infty $ and $b< (2c)^{-1/2}$ , we can change the contour of integration in (2.8.3) from $\mathbb {R}$ to $\mathbb {R} + iL$ for any $L\in \mathbb {R}$ without affecting the value of the integral. Then, taking $L = -a/b$ , we get

$$ \begin{align*} \mathbb{E}(g(bX) e^{-F(bX)/b^2}) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty g(ib(y-ia/b))e^{-ia(y-ia/b)/b} e^{-(y-ia/b)^2/2} dy\\ &= \frac{e^{-a^2/2b^2} }{\sqrt{2\pi}}\int_{-\infty}^\infty g(iby + a) e^{-y^2/2} dy. \end{align*} $$

Thus,

$$\begin{align*}\langle g\rangle_b = \frac{1 }{\sqrt{2\pi}}\int_{-\infty}^\infty g(iby + a) e^{-y^2/2} dy, \end{align*}$$

which tends to $g(a)$ as $b\to 0$ , by a simple application of the dominated convergence theorem and the growth rate of g obtained above.

Proof. Let $Z\sim N(0,1)$ . Then note that

(2.8.5) $$ \begin{align} \mathbb{E}(e^{-F(bX)/b^2}) &= \mathbb{E}(e^{-e^{i\alpha bZ}/b^2})\notag \\ &= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \exp\bigg(-\frac{1}{b^2} e^{i\alpha b x} - \frac{1}{2}x^2\bigg)dx. \end{align} $$

Take any $\beta \in \mathbb {R}$ . From the form of the integrand, it is not hard to justify that the contour of integration can be shifted to the line parallel to the real line that passes through $-i\beta /b$ . Thus,

$$ \begin{align*} \mathbb{E}(e^{-F(bX)/b^2}) &= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \exp\bigg(-\frac{1}{b^2} e^{i\alpha b (x-i\beta/b)} - \frac{1}{2}(x-i\beta/b)^2\bigg)dx\\ &= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \exp\bigg(-\frac{e^{\alpha \beta} }{b^2} e^{i\alpha bx} - \frac{1}{2}x^2 + \frac{i\beta}{b} x + \frac{\beta^2}{2b^2}\bigg)dx. \end{align*} $$

Let us now choose $\beta =x_0$ . Then for any x, we have that as $b\to 0$ ,

$$ \begin{align*} &\frac{e^{\alpha \beta}}{b^2} -\frac{e^{\alpha \beta} }{b^2} e^{i\alpha bx} - \frac{1}{2}x^2 + \frac{i\beta}{b} x \\ &= \frac{e^{\alpha \beta}}{b^2} -\frac{e^{\alpha \beta} }{b^2} \bigg(1+ i\alpha bx -\frac{1}{2}\alpha^2 b^2 x^2 + o(b^2)\bigg) - \frac{1}{2}x^2 + \frac{i\beta}{b} x\\ &= -\frac{1}{2}(1-\alpha^2e^{\alpha \beta}) x^2 + o(1) = -\frac{1}{2}(1-\alpha \beta)x^2 + o(1). \end{align*} $$

Moreover, for any x,

$$ \begin{align*} \bigg|\exp\bigg(\frac{e^{\alpha \beta}}{b^2} -\frac{e^{\alpha \beta} }{b^2} e^{i\alpha bx} - \frac{1}{2}x^2 + \frac{i\beta}{b} x\bigg)\bigg| &= \exp\bigg(\operatorname{Re}\bigg(\frac{e^{\alpha \beta}}{b^2} -\frac{e^{\alpha \beta} }{b^2} e^{i\alpha bx} - \frac{1}{2}x^2 + \frac{i\beta}{b} x\bigg)\bigg)\\ &= \exp\bigg(\frac{e^{\alpha \beta}}{b^2}(1-\cos(\alpha bx)) - \frac{1}{2}x^2\bigg)\\ &\le \exp\bigg(\frac{e^{\alpha \beta}}{2b^2}\alpha^2 b^2 x^2 - \frac{1}{2}x^2\bigg) \\ &= \exp\bigg(-\frac{1}{2}(1-\alpha \beta) x^2\bigg). \end{align*} $$

Since $\alpha \beta =\alpha x_0< 1$ , this allows us to apply to dominated convergence theorem to conclude that

$$ \begin{align*} &\lim_{b\to 0} \exp\bigg(\frac{e^{\alpha \beta}}{b^2} -\frac{\beta^2}{2b^2}\bigg) \mathbb{E}(e^{-F(bX)/b^2}) \\ &= \lim_{b\to 0}\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \exp\bigg(\frac{e^{\alpha \beta}}{b^2} -\frac{e^{\alpha \beta} }{b^2} e^{i\alpha bx} - \frac{1}{2}x^2 + \frac{i\beta}{b} x \bigg) dx\\ &= \int_{-\infty}^\infty \lim_{b\to 0} \frac{1}{\sqrt{2\pi}} \exp\bigg(\frac{e^{\alpha \beta}}{b^2} -\frac{e^{\alpha \beta} }{b^2} e^{i\alpha bx} - \frac{1}{2}x^2 + \frac{i\beta}{b} x \bigg) dx\\ &= \int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}} \exp\bigg(-\frac{1}{2}(1-\alpha \beta) x^2\bigg)dx= \frac{1}{\sqrt{1-\alpha \beta}}. \end{align*} $$

This proves the claim about the limiting behavior of the partition function. Next, let $g(x) := f(x) e^{-F(x)/b^2}$ . Then by the given conditions on f and b, we have

$$ \begin{align*} \mathbb{E} (g(bX)) &= \mathbb{E}(f(ibZ) e^{-e^{i\alpha bZ}/b^2})\\ &= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} f(ibx) \exp\bigg(-\frac{1}{b^2} e^{i\alpha b x} - \frac{1}{2}x^2\bigg)dx. \end{align*} $$

Shifting the contour exactly as before (with $\beta =x_0$ ), we get

$$ \begin{align*} \mathbb{E}(g(bX)) &= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} f(ib(x-i\beta/b))\exp\bigg(-\frac{1}{b^2} e^{i\alpha b (x-i\beta/b)} - \frac{1}{2}(x-i\beta/b)^2\bigg)dx. \end{align*} $$

Proceeding exactly as before, this leads to

$$ \begin{align*} \lim_{b\to 0} \exp\bigg(\frac{e^{\alpha \beta}}{b^2} -\frac{\beta^2}{2b^2}\bigg) \mathbb{E}(g(bX)) &= \int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}} f(\beta) \exp\bigg(-\frac{1}{2}(1-\alpha \beta) x^2\bigg)dx\\ &= \frac{f(\beta)}{\sqrt{1-\alpha \beta}}. \end{align*} $$

Since $\mathbb {E}(f(X_b)) = \mathbb {E}(g(bX))/\mathbb {E}(e^{-F(bX)/b^2})$ , this completes the proof.

Proof. Let $Z = (Z_{lm})_{0\le l\le L,\, -l\le m\le l}$ be a collection of i.i.d. $N(0,1)$ random variables, and define the Gaussian field

(3.1.1) $$ \begin{align} Z_{\lambda,L}(x) := \sum_{l=1}^L \sum_{m=-l}^l \lambda^{l/2}\sqrt{\frac{2\pi}{l(l+1)}} Z_{lm} Y_{lm}(x) \end{align} $$

on $S^2$ , and the Gaussian random variable

$$\begin{align*}E := \frac{Z_{00}}{\sqrt{8\pi \epsilon}}. \end{align*}$$

Then note that $\mathbb {E}(Z_{\lambda , L}(x) Z_{\lambda ,L}(y)) = G_{\lambda ,L}(x,y)$ , where $G_{\lambda , L}$ is the regularized Green’s function defined in equation (1.4.7). Noticing that

$$\begin{align*}1+\sum_{l=1}^L (2l+1) = L^2 + 1, \end{align*}$$

let us index the components of points in $\mathbb {C}^{L^2+1}$ as $z = (z_{lm})_{0\le l\le L, \, -l\le m\le l}$ . Define the function $f:\mathbb {C}^{L^2+1}\times S^2 \to \mathbb {C}$ as

(3.1.2) $$ \begin{align} f(z, y) := \sum_{l=1}^L \sum_{m=-l}^l \lambda^{l/2}\sqrt{\frac{2\pi}{l(l+1)}} z_{lm} Y_{lm}(y), \end{align} $$

and the function $c:\mathbb {C}^{L^2+1}\to \mathbb {C}$ as

$$\begin{align*}c(z) := \frac{z_{00}}{\sqrt{8\pi\epsilon}}. \end{align*}$$

Then define $h:\mathbb {C}^{L^2+1}\to \mathbb {C}$ as

$$ \begin{align*} h(z) &:=\bigg(\prod_{j=1}^ke^{\chi\alpha_j(b-\alpha_j)} g(\sigma(x_j))^{-\Delta_\alpha} e^{2\alpha_j f(z,x_j) + 2\alpha_j^2G_{\lambda, L}(x_j,x_j)}\bigg)\\ &\qquad \cdot \exp\bigg(-2wbc(z) -\mu e^{2bc(z)} \int_{S^2}e^{2bf(z,y)+2b^2 G_{\lambda, L}(y,y)}da(y) \bigg). \end{align*} $$

Then it is clear that the regularized correlation function defined in equation (1.4.9) can be expressed as $\mathbb {E}(h(X))$ , where X is a wrong sign standard Gaussian random vector in $\mathbb {R}^{L^2+1}$ . It is easy to check that h is analytic on $\mathbb {C}^{L^2+1}$ and $\mathbb {E}|h(i Z)| <\infty $ . Thus, the restriction of h to $\mathbb {R}^{L^2+1}$ is in $\mathcal {F}_{0,L^2+1}$ , and by our definition of wrong sign expectations, we have

$$\begin{align*}\mathbb{E}(h(X)) = \mathbb{E}(h(iZ)). \end{align*}$$

Now note that h can be expanded as

$$ \begin{align*} h(z) &:=\bigg(\prod_{j=1}^ke^{\chi\alpha_j(b-\alpha_j)} g(\sigma(x_j))^{-\Delta_\alpha} e^{2\alpha_j f(z,x_j) + 2\alpha_j^2G_{\lambda, L}(x_j,x_j)}\bigg)\\ &\qquad \cdot \sum_{n=0}^\infty\frac{(-\mu)^n e^{2(n-w)bc(z)}}{n!}\bigg( \int_{S^2}e^{2bf(z,y)+2b^2 G_{\lambda, L}(y,y)}da(y) \bigg)^n. \end{align*} $$

This gives

$$ \begin{align*} h(iZ) &:=\bigg(\prod_{j=1}^ke^{\chi\alpha_j (b-\alpha_j)} g(\sigma(x_j))^{-\Delta_\alpha} e^{2i\alpha_j Z_{\lambda, L}(x_j) + 2\alpha_j^2G_{\lambda, L}(x_j,x_j)}\bigg)\\ &\qquad \cdot \sum_{n=0}^\infty\frac{(-\mu)^n e^{2i(n-w)b E}}{n!}\bigg( \int_{S^2}e^{2ibZ_{\lambda,L}(y)+2b^2 G_{\lambda, L}(y,y)}da(y) \bigg)^n. \end{align*} $$

We will show later, in Lemma 3.2.3, that $G_{\lambda ,L}$ is a bounded function. Using this, it is easy to justify that while evaluating $\mathbb {E}(h(iZ))$ , we can take the expectation inside the above infinite sum. A simple calculation completes the proof.

Proof. If we take the formula for $C_{\epsilon , \lambda , L}$ from Theorem 3.1.1 and send $\epsilon $ to zero, only the term corresponding to $n = w$ survives; all other terms tend to zero. The absolute convergence of the series allows us to apply the dominated convergence theorem to obtain the claimed result.

Proof. It is well known that the real spherical harmonics satisfy the addition theorem [Reference Müller46, Theorem 2]

$$\begin{align*}P_l(x\cdot y) = \frac{4\pi}{2l + 1}\sum_{m=-l}^l Y_{lm}(x)Y_{lm}(y), \end{align*}$$

This gives

$$ \begin{align*} G_{\lambda, L}(x,y) &= \sum_{l=1}^L \sum_{m=-l}^l \lambda^l \frac{2\pi}{l(l+1)} Y_{lm}(x)Y_{lm}(y) \\ &= \sum_{l=1}^L \lambda^l \frac{2l+1}{2l(l+1)} P_l(x\cdot y), \end{align*} $$

proving the first claim. Next, recall the well-known fact [Reference Szegö56, Equation (7.21.1)] that

(3.2.1) $$ \begin{align} |P_l(x)|\le 1 \text{ for all } l\ge 0 \text{ and } x\in [-1,1]. \end{align} $$

Using this bound in the above expression proves the second and third claims of the lemma.

Proof. This is a simple consequence of Lemma 3.2.2, Lemma 3.2.3, and the dominated convergence theorem.

Proof. The generating function of Legendre polynomials is given by [Reference Szegö56, Equation (4.7.23) with $\lambda =1/2$ ]:

(3.2.2) $$ \begin{align} \sum_{l=0}^\infty t^l P_l(x) = \frac{1}{\sqrt{1-2xt + t^2}}, \end{align} $$

where the series converges absolutely when $|x|\le 1$ and $|t|<1$ . Using the bound (3.2.1), we can integrate both sides with respect to t from $0$ to $\lambda $ , and move the integral inside the sum to get

$$ \begin{align*} \sum_{l=0}^\infty \frac{\lambda^{l+1}P_l(x)}{l+1} = \int_0^\lambda \frac{1}{\sqrt{1-2xt + t^2}}dt\ge 0. \end{align*} $$

Consequently,

(3.2.3) $$ \begin{align} \sum_{l=1}^\infty \frac{\lambda^{l}P_l(x)}{l+1} &\ge -1. \end{align} $$

Next, note that by equation (3.2.2), we have that for any $t\in (0,1)$ ,

(3.2.4) $$ \begin{align} \sum_{l=1}^\infty t^{l-1} P_l(x) &= \frac{1}{t}\bigg(\frac{1}{\sqrt{1-2xt + t^2}} - 1\bigg). \end{align} $$

Now, for any $y\in \mathbb {R}$ ,

$$ \begin{align*} (1-y)\bigg(1+\frac{y}{2}\bigg)^2 &= (1-y)\bigg(1+y+\frac{ y^2}{4}\bigg) \\ &= 1-\frac{3y^2}{4} - \frac{y^3}{4} = 1 -\frac{3y^2 + y^3}{4}. \end{align*} $$

This shows, in particular, that for all $y\ge -3$ ,

$$\begin{align*}(1-y)\bigg(1+\frac{y}{2}\bigg)^2 \le 1. \end{align*}$$

Thus, for $y> -3$ , we can take square root of both sides to get

$$\begin{align*}\frac{1}{\sqrt{1-y}} \ge 1+\frac{y}{2}. \end{align*}$$

Take any $x,t$ such that $|x|\le 1$ and $|t|< 1$ . Then applying the above inequality with $y = 2xt - t^2> -3$ , we get

$$\begin{align*}\frac{1}{\sqrt{1-2xt+t^2}} \ge 1+xt - \frac{t^2}{2}. \end{align*}$$

Using this in equation (3.2.4), and integrating t from $0$ to $\lambda $ , we get

(3.2.5) $$ \begin{align} \sum_{l=1}^\infty \frac{\lambda^l P_l(x)}{l} &\ge \int_0^\lambda \bigg(x - \frac{t}{2}\bigg) dt =x\lambda - \frac{\lambda^2}{4} \ge -\frac{5}{4}. \end{align} $$

Combining equations (3.2.3) and (3.2.5), we get

$$ \begin{align*} \sum_{l=1}^\infty \lambda^l \frac{2l+1}{2l(l+1)} P_l(x) &= \sum_{l=1}^\infty \frac{\lambda^l P_l(x)}{2(l+1)} + \sum_{l=1}^\infty \frac{\lambda^l P_l(x)}{2l}\\ &\ge -\frac{1}{2} - \frac{5}{8} = -\frac{9}{8}. \end{align*} $$

By the formula for $G_\lambda $ from Lemma 3.2.3, this completes the proof.

Proof. By [Reference Szegö56, Theorem 7.3.3],

$$ \begin{align*} |P_l(\cos \theta)| \le \sqrt{\frac{2}{\pi l \sin \theta}} \end{align*} $$

for all $\theta \in (0,\pi )$ . This allows us to show, via the dominated convergence theorem, that for any distinct $x,y\in S^2$ ,

(3.2.6) $$ \begin{align} G(x,y) := \lim_{\lambda \uparrow 1} G_\lambda(x,y) = \sum_{l=1}^\infty \frac{2l+1}{2l(l+1)}P_l(x\cdot y), \end{align} $$

where the series on the right converges absolutely. By the addition theorem for spherical harmonics, it follows that this is indeed the Green’s function for the spherical Laplacian. It remains to prove the explicit formula for G. Recall that the Legendre polynomials form an orthonormal basis of $L^2_{\mathbb {R}}([-1,1])$ , and any $f\in L^2_{\mathbb {R}}([-1,1])$ admits an expansion

$$\begin{align*}f(x) = \sum_{l=0}^\infty a_l P_l(x), \end{align*}$$

where

$$\begin{align*}a_l = \frac{2l+1}{2}\int_{-1}^1 f(x) P_l(x) dx, \end{align*}$$

and the sum converges in $L^2$ [Reference Newman and Rudin47]. Let $f(x) = \ln (1-x)$ and $a_l$ be as above. Recall that for $l\ge 1$ , $P_l$ satisfies Legendre’s differential equation

$$\begin{align*}(1-x^2)P_l"(x) -2xP_l'(x) + l(l+1)P_l(x)=0. \end{align*}$$

Thus,

$$ \begin{align*} a_l &= \frac{2l+1}{2}\int_{-1}^1\ln(1-x) P_l(x) dx\\ &= -\frac{2l+1}{2l(l+1)}\int_{-1}^1((1-x^2) \ln(1-x)P_l"(x)-2x\ln(1-x)P_l'(x)) dx. \end{align*} $$

Now, integration by parts (and the fact that $P_l$ is a polynomial) shows that

$$ \begin{align*} \int_{-1}^1(1-x^2) \ln(1-x)P_l"(x)dx &= -\int_{-1}^1 P_l'(x) \frac{d}{dx}((1-x^2)\ln (1-x)) dx\\ &= -\int_{-1}^1P_l'(x) (-2x\ln (1-x) -(1+x)) dx. \end{align*} $$

Plugging this into the previous display, we get

$$ \begin{align*} a_l &= -\frac{2l+1}{2l(l+1)}\int_{-1}^1(1+x)P_l'(x)dx. \end{align*} $$

Applying integration by parts once again, we have

$$ \begin{align*} \int_{-1}^1(1+x)P_l'(x)dx &= 2P_l(1) - \int_{-1}^1 P_l(x) dx = 2. \end{align*} $$

Thus, for any $l\ge 1$ ,

$$\begin{align*}a_l = -\frac{2l+1}{l(l+1)}. \end{align*}$$

By direct calculation,

$$\begin{align*}a_0 = \frac{1}{2}\int_{-1}^1 \ln(1-x) dx = \ln 2 -1. \end{align*}$$

Thus, by the formula (3.2.6), we get

$$ \begin{align*} G(x,y) &= \frac{1}{2}\sum_{l=1}^\infty a_l P_l(x\cdot y) = -\frac{1}{2}(\ln(1-x\cdot y) + 1 - \ln 2). \end{align*} $$

But

$$\begin{align*}\|x-y\|^2 = \|x\|^2+\|y\|^2-2x\cdot y = 2(1-x\cdot y). \end{align*}$$

Plugging this into the previous display shows that it is equal to

$$ \begin{align*} -\frac{1}{2}(\ln (\|x-y\|^2/2) + 1- \ln 2) = -\ln \|x-y\|-\frac{1}{2} + \ln 2. \end{align*} $$

Since the series in equation (3.2.6) is absolutely convergent, this completes the proof.

Proof. By the bound (3.2.1), it is easy to justify that $G_\lambda (x,y)$ can be differentiated with respect to $\lambda $ by moving the derivative inside the infinite sum, giving

$$ \begin{align*} \frac{\partial }{\partial \lambda}(\lambda G_\lambda(x,y) ) &= \sum_{l=1}^\infty \frac{\partial}{\partial \lambda} \bigg(\lambda^{l+1}\frac{2l+1}{2l(l+1)} P_l(x\cdot y)\bigg)\\ &= \sum_{l=1}^\infty \lambda^{l}\frac{2l+1}{2l} P_l(x\cdot y)\\ &= \sum_{l=1}^\infty \lambda^{l}P_l(x\cdot y) + \sum_{l=1}^\infty\frac{ \lambda^{l}}{2l} P_l(x\cdot y). \end{align*} $$

By equation (3.2.2) and the inequality (3.2.5), the above quantity is bounded below by $-1-\frac {5}{8}$ . This proves the claim.

Proof of Theorem 3.2.1.

By Lemma 3.2.4 and Lemma 3.2.6, we only need to show that as we take $\lambda \uparrow 1$ , the $G_\lambda $ ’s in the integral displayed in the statement of Lemma 3.2.4 can be replaced by G. We already know that $G_\lambda $ converges to G pointwise. So it remains to verify the condition needed to apply the dominated convergence theorem. Recall that the integrand is

(3.2.7) $$ \begin{align} \bigg(\prod_{j=1}^k \prod_{l=1}^{w} e^{-4b\alpha_jG_\lambda(x_j, y_l)}\bigg) \bigg(\prod_{1\le l<l'\le w}e^{-4b^2G_\lambda(y_l, y_{l'})} \bigg). \end{align} $$

Let S be the set of j such that $\operatorname {Re}(\alpha _j) < 0$ . Take any $j\notin S$ . Then by Lemma 3.2.5,

$$\begin{align*}|e^{-4b\alpha_jG_\lambda(x_j, y_l)}|\le e^{\frac{9}{2}b\operatorname{Re}(\alpha_j)}. \end{align*}$$

Similarly, for all $l,l'$ ,

$$\begin{align*}e^{-4b^2G_\lambda(y_l, y_{l'})} \le e^{\frac{9}{2}b^2}. \end{align*}$$

Next, take any $j\in S$ . Then by Lemma 3.2.7,

$$ \begin{align*} |e^{-4b\alpha_jG_\lambda(x_j, y_l)}| &\le e^{-4b\operatorname{Re}(\alpha_j)(\frac{13}{8\lambda} + \frac{1}{\lambda}G(x_j,y_l))}. \end{align*} $$

Now recall that by assumption, $\operatorname {Re}(\alpha _j)> -1/2b$ for each j. This implies that for each $j\in S$ , $-4b\operatorname {Re}(\alpha _j )< 2$ . Consequently, we can find $\lambda _0\in (0,1)$ so close to $1$ that

$$\begin{align*}\kappa := \max_{j\in S} \frac{-4b\operatorname{Re}(\alpha_j)}{\lambda_0} < 2. \end{align*}$$

Thus, if $\lambda \in (\lambda _0, 1)$ , then for each $j\in S$ ,

$$\begin{align*}e^{-4b\operatorname{Re}(\alpha_j)G_\lambda(x_j, y_l)} \le e^{\kappa(\frac{13}{8} + G(x_j, y_l))}. \end{align*}$$

Combining all of the above, we see that if $\lambda \in (\lambda _0,1)$ , then the (nonnegative) quantity displayed in equation (3.2.7) is bounded above by

(3.2.8) $$ \begin{align} C \prod_{j\in S} \prod_{l=1}^w e^{\kappa G(x_j, y_l)}, \end{align} $$

where C has no dependence on $y_1,\ldots ,y_{w}$ . Since $\kappa < 2$ and $x_1,\ldots ,x_k$ are distinct, the above function has a finite integral over $(S^2)^{w}$ . This allows us to apply the dominated convergence theorem to complete the proof of Theorem 3.2.1. Observe that the $(-1)^w$ term can be written as $e^{-i\pi w}$ since w is an integer.

Theorem Restatement of Theorem 1.3.1.

Suppose that $k\ge 3$ , $\operatorname {Re}(\alpha _j)> -1/2b$ for each j, and the parameter $w := (Q-\sum _{j=1}^k \alpha _j)/b$ is a positive integer. Let $x_1,\ldots ,x_k$ be distinct points on $S^2$ , none of which are the north pole $e_3 = (0,0,1)$ . Let $z_1,\ldots ,z_k$ be their stereographic projections on $\mathbb {C}$ . Then

$$ \begin{align*} &C(\alpha_1,\ldots,\alpha_k; z_1,\ldots,z_k;b;\mu) = \widetilde{C}(\alpha_1,\ldots,\alpha_k; x_1,\ldots,x_k;b;\mu)\\ &= \frac{e^{-i\pi w}\mu^{w}}{w!} (4/e)^{1-1/b^2}\prod_{1\le j<j'\le k} |z_j - z_{j'}|^{4\alpha_j \alpha_{j'}}\\ &\qquad \qquad \cdot \int_{\mathbb{C}^w} \bigg(\prod_{j=1}^k \prod_{l=1}^{w} |z_j-t_l|^{4b\alpha_j}\bigg)\bigg(\prod_{1\le l<l'\le w}|t_l-t_{l'}|^{4b^2} \bigg) d^2t_1\cdots d^2t_w. \end{align*} $$

Proof. Take any $x=(x_1,x_2,x_3)\in S^2\setminus \{e_3\}$ , and let $u=u_1+i u_2\in \mathbb {C}$ be its stereographic projection on  $\mathbb {C}$ . Then the following relations hold:

$$\begin{align*}u =\frac{x_1+i x_2}{1-x_3}, \ \ x = \frac{1}{1+|u|^2}(2u_1, 2u_2, |u|^2-1). \end{align*}$$

Thus, if $x,y\in S^2\setminus \{e_3\}$ and $u,v$ are the stereographic projections of $x,y$ , then

$$ \begin{align*} |u-v|^2 &= \bigg(\frac{x_1}{1-x_3} - \frac{y_1}{1-y_3}\bigg)^2 + \bigg(\frac{x_2}{1-x_3} - \frac{y_2}{1-y_3}\bigg)^2\\ &= \frac{x_1^2(1-y_3)^2 + y_1^2(1-x_3)^2-2x_1y_1(1-x_3)(1-y_3)}{(1-x_3)^2(1-y_3)^2}\\ &\qquad + \frac{x_2^2(1-y_3)^2 + y_2^2(1-x_3)^2-2x_2y_2(1-x_3)(1-y_3)}{(1-x_3)^2(1-y_3)^2}. \end{align*} $$

Since $\|x\|^2=\|y\|^2=1$ , we have

$$ \begin{align*} &x_1^2(1-y_3)^2 + y_1^2(1-x_3)^2 + x_2^2(1-y_3)^2 + y_2^2(1-x_3)^2 \\ &= (1-x_3^2)(1-y_3)^2 + (1-y_3^2)(1-x_3)^2\\ &= (1-x_3)(1-y_3)((1+x_3)(1-y_3)+ (1+y_3)(1-x_3))\\ &= 2(1-x_3)(1-y_3)(1-x_3y_3). \end{align*} $$

Using this in the previous display, we get

$$\begin{align*}|u-v|^2 = \frac{2(1-x\cdot y)}{(1-x_3)(1-y_3)} = \frac{\|x-y\|^2}{(1-x_3)(1-y_3)}. \end{align*}$$

Now note that

$$ \begin{align*} 1+|u|^2 &= 1+\frac{x_1^2+x_2^2}{(1-x_3)^2} \\ &=1+ \frac{1-x_3^2}{(1-x_3)^2}= \frac{2-2x_3}{(1-x_3)^2} = \frac{2}{1-x_3}. \end{align*} $$

Similarly,

$$\begin{align*}1+|v|^2 = \frac{2}{1-y_3}. \end{align*}$$

Combining the last three displays completes the proof.

Proof of Theorem 1.3.1.

Note that by Lemma 3.3.1, for any $x,y\in S^2\setminus \{e_3\}$ ,

(3.3.1) $$ \begin{align} G(x,y) &= -\ln |\sigma(x)-\sigma(y)| + \frac{1}{2}\ln (1+|\sigma(x)|^2) \notag \\ &\qquad \qquad +\frac{1}{2}\ln (1+|\sigma(y)|^2) -\frac{1}{2}. \end{align} $$

Take any $y_1,\ldots , y_w\in S^2\setminus \{e_3\}$ . Then the above identity shows that

$$ \begin{align*} -4b^2 \sum_{1\le l<l'\le w} G(y_l, y_{l'}) &= 4b^2 \sum_{1\le l<l'\le w} \ln |\sigma(y_l)-\sigma(y_{l'})|\\ &\qquad - 2b^2(w-1)\sum_{l=1}^w \ln (1+|\sigma(y_l)|^2) + b^2w(w-1). \end{align*} $$

Similarly,

$$ \begin{align*} -4b\sum_{j=1}^k \sum_{l=1}^w\alpha_j G(x_j, y_l) &= 4b\sum_{j=1}^k \sum_{l=1}^w \alpha_j \ln |\sigma(x_j)-\sigma(y_l)|\\ &\qquad - 2bw\sum_{j=1}^k \alpha_j \ln(1+|\sigma(x_j)|^2) \\ &\qquad - 2b\bigg(\sum_{j=1}^k\alpha_j\bigg)\sum_{l=1}^w\ln (1+|\sigma(y_l)|^2) +2bw\sum_{j=1}^k\alpha_j. \end{align*} $$

Now note that the coefficients of each $\ln (1+|\sigma (y_l)|^2)$ from the above two displays add up to

$$ \begin{align*} -2b^2(w-1) - 2b\sum_{j=1}^k \alpha_j &= -2b\bigg(Q-\sum_{j=1}^k \alpha_j\bigg) +2b^2 -2b\sum_{j=1}^k \alpha_j\\ &= -2bQ + 2b^2 = 2. \end{align*} $$

Lastly, note that the constant terms add up to

$$ \begin{align*} b^2w(w-1) + 2bw\sum_{j=1}^k\alpha_j &= b^2w^2 - b^2w + 2bw(Q-bw) \\ &=-b^2w^2 -b^2 w + 2bwQ\\ &= -b^2 w^2 + b^2 w -2w. \end{align*} $$

Thus, combining the last four displays, we get

$$ \begin{align*} &-4b^2 \sum_{1\le l<l'\le w} G(y_l, y_{l'}) -4b\sum_{j=1}^k \sum_{l=1}^w\alpha_j G(x_j, y_l)\\ &= 4b^2 \sum_{1\le l<l'\le w} \ln |\sigma(y_l)-\sigma(y_{l'})| + 4b\sum_{j=1}^k \sum_{l=1}^w \alpha_j \ln |\sigma(x_j)-\sigma(y_l)|\\ &\qquad +2 \sum_{l=1}^w \ln (1+|\sigma(y_l)|^2) - 2bw\sum_{j=1}^k\alpha_j \ln(1+|\sigma(x_j)|^2)\\ &\qquad -b^2 w^2 + b^2 w -2w. \end{align*} $$

It is well known that for any $f:S^2\to \mathbb {C}$ that is integrable with respect to the area measure on $S^2$ , we have

(3.3.2) $$ \begin{align} \int_{S^2} f(y) da(y) = \int_{\mathbb{C}} f(\sigma^{-1}(z)) \frac{4}{(1+|z|^2)^2} d^2z. \end{align} $$

Combining this with the previous display, and recalling the $u_j := P(x_j)$ , we get

$$ \begin{align*} &\int_{(S^2)^{w}}\bigg(\prod_{j=1}^k \prod_{l=1}^{w} e^{-4b\alpha_jG(x_j, y_l)}\bigg) \bigg(\prod_{1\le l<l'\le w}e^{-4b^2G(y_l, y_{l'})} \bigg) da(y_1)\cdots da(y_{w})\\ &= 4^we^{-b^2 w^2 + b^2 w -2w}\bigg(\prod_{j=1}^k (1+|z_j|^2)^{-2bw\alpha_j}\bigg)\\ &\qquad \qquad \cdot \int_{\mathbb{C}^w} \bigg(\prod_{j=1}^k \prod_{l=1}^{w} |z_j-t_l|^{4b\alpha_j}\bigg)\bigg(\prod_{1\le l<l'\le w}\|t_l-t_{l'}\|^{4b^2} \bigg) d^2t_1\cdots d^2t_w. \end{align*} $$

Again, by equation (3.3.1),

$$ \begin{align*} &\prod_{1\le j < j'\le k} e^{-4\alpha_j\alpha_{j'}G(x_j, x_{j'})} = \prod_{1\le j<j'\le k} \bigg(\frac{\sqrt{e}|z_j - z_{j'}|}{\sqrt{(1+|z_j|^2) (1+|z_{j'}|^2)}}\bigg)^{4\alpha_j \alpha_{j'}}\\ &= e^{2\sum_{1\le j<j'\le k} \alpha_j \alpha_{j'}} \prod_{j=1}^k (1+|z_j|^2)^{-2\alpha_j \sum_{j'\ne j} \alpha_{j'}}\prod_{1\le j<j'\le k} |z_j -z_{j'}|^{4\alpha_j \alpha_{j'}}. \end{align*} $$

Next, note that

$$ \begin{align*} \prod_{j=1}^k g(\sigma(x_j))^{-\Delta_{\alpha_j}} &= 4^{-\sum_{j=1}^k \Delta_{\alpha_j}} \prod_{j=1}^k (1+|z_j|^2)^{2\Delta_{\alpha_j}} \end{align*} $$

We have to now multiply the expressions from the last three displays. Upon multiplying them together, the exponent of $1+|z_j|^2$ becomes

$$ \begin{align*} -2bw\alpha_j -2\alpha_j\sum_{j'\ne j }\alpha_{j'}+2\Delta_{\alpha_j} &= -2\bigg(Q-\sum_{j'=1}^k \alpha_j\bigg) \alpha_j -2\alpha_j\sum_{j'\ne j }\alpha_{j'}+2\alpha_j(Q-\alpha_j)\\ &= 0. \end{align*} $$

The exponent of $4$ becomes

$$ \begin{align*} w - \sum_{j=1}^k \Delta_{\alpha_j} &= 1 - \frac{1}{b^2} - \frac{1}{b}\sum_{j=1}^k\alpha_j - \sum_{j=1}^k \alpha_j \bigg(b-\frac{1}{b} - \alpha_j\bigg)\\ &= 1-\frac{1}{b^2} -b \sum_{j=1}^k \alpha_j +\sum_{j=1}^k \alpha_j^2. \end{align*} $$

Lastly, the exponent of e is

$$ \begin{align*} &-b^2w^2 +b^2w - 2w + 2\sum_{1\le j<j'\le k} \alpha_j \alpha_j' \\ &= -\bigg(Q-\sum_{j=1}^k\alpha_j\bigg)^2 + \bigg(b-\frac{2}{b}\bigg)\bigg(Q-\sum_{j=1}^k\alpha_j\bigg)+ 2\sum_{1\le j<j'\le k} \alpha_j \alpha_j'\\ &= -Q^2 +\bigg(2Q- b + \frac{2}{b}\bigg)\sum_{j=1}^k\alpha_j + Q\bigg(b-\frac{2}{b}\bigg) - \sum_{j=1}^k\alpha_j^2\\ &= \frac{1}{b^2}-1 + b \sum_{j=1}^k \alpha_j - \sum_{j=1}^k \alpha_j^2. \end{align*} $$

Note that this is exactly the negative of the exponent of $4$ . Thus, the constant factor in the combined expression is equal to

$$ \begin{align*} \exp\bigg(\bigg(1-\frac{1}{b^2} -b \sum_{j=1}^k \alpha_j +\sum_{j=1}^k \alpha_j^2\bigg)(\ln 4 - 1)\bigg) = e^{\chi(1-1/b^2)} e^{-\chi\sum_{j=1}^k \alpha_j (b-\alpha_j)}. \end{align*} $$

Combining the above calculations yields the claimed result.

Proof. Since $\alpha _j> -1/2b$ for $j=1,2,3$ and w is a positive integer, Theorem 1.3.1 gives

$$ \begin{align*} &C(\alpha_1,\alpha_2,\alpha_3;0,1,z_3;b;\mu) \\ &= \frac{e^{-i\pi w}\mu^{w}}{w!} (4/e)^{1-1/b^2} |z_3|^{4\alpha_1\alpha_3} |1-z_3|^{4\alpha_2 \alpha_3} \\ &\qquad \cdot \int_{\mathbb{C}^w} \bigg(\prod_{l=1}^w |t_l|^{4b\alpha_1}|1-t_l|^{4b\alpha_2}|z_3 - t_l|^{4b \alpha_3}\bigg)\bigg(\prod_{1\le l<l'\le w}|t_l-t_{l'}|^{4b^2} \bigg) d^2t_1\cdots d^2t_w. \end{align*} $$

The powers of the terms involving $z_3$ add up to

$$ \begin{align*} 4\alpha_1\alpha_3 + 4\alpha_2 \alpha_3 + 4bw\alpha_3 &= 4(\alpha_1+\alpha_2 + (Q-\alpha_1-\alpha_2-\alpha_3)) \alpha_3\\ &= 4\alpha_3(Q-\alpha_3) = 4\Delta_{\alpha_3}. \end{align*} $$

This shows that it suffices to prove the following:

(3.4.1) $$ \begin{align} &\lim_{|z_3|\to \infty} \int_{\mathbb{C}^w} \bigg(\prod_{l=1}^w |t_l|^{4b\alpha_1}|1-t_l|^{4b\alpha_2}\frac{|z_3 - t_l|^{4b \alpha_3}}{|z_3|^{4b\alpha_3}}\bigg)\bigg(\prod_{1\le l<l'\le w}|t_l-t_{l'}|^{4b^2} \bigg) d^2t_1\cdots d^2t_w\notag \\ &= \int_{\mathbb{C}^w} \bigg(\prod_{l=1}^w |t_l|^{4b\alpha_1}|1-t_l|^{4b\alpha_2}\bigg)\bigg(\prod_{1\le l<l'\le w}|t_l-t_{l'}|^{4b^2} \bigg) d^2t_1\cdots d^2t_w. \end{align} $$

Let $\nu $ be the probability measure on $\mathbb {C}$ with density function $\frac {1}{4\pi }g(z)$ with respect to Lebesgue measure, where g is the round metric. Then note that the integral in the first line of equation (3.4.1) can be written as

$$\begin{align*}\int_{\mathbb{C}^w} f_{z_3}(t_1,\ldots,t_w) d\nu(t_1)\cdots d\nu(t_w), \end{align*}$$

where

$$ \begin{align*} f_{z_3}(t_1,\ldots,t_w) := \bigg(\prod_{l=1}^w \frac{4\pi}{g(t_l)}\bigg) \bigg(\prod_{l=1}^w |t_l|^{4b\alpha_1}|1-t_l|^{4b\alpha_2}\frac{|z_3 - t_l|^{4b \alpha_3}}{|z_3|^{4b\alpha_3}}\bigg)\bigg(\prod_{1\le l<l'\le w}|t_l-t_{l'}|^{4b^2} \bigg). \end{align*} $$

The integral in the second line in equation (3.4.1) can be written as

$$\begin{align*}\int_{\mathbb{C}^w} f(t_1,\ldots,t_w) d\nu(t_1)\cdots d\nu(t_w), \end{align*}$$

where

$$ \begin{align*} f(t_1,\ldots,t_w) := \bigg(\prod_{l=1}^w \frac{4\pi }{g(t_l)}\bigg) \bigg(\prod_{l=1}^w |t_l|^{4b\alpha_1}|1-t_l|^{4b\alpha_2}\bigg)\bigg(\prod_{1\le l<l'\le w}|t_l-t_{l'}|^{4b^2} \bigg). \end{align*} $$

Clearly, $f_{z_3}\to f$ pointwise as $|z_3|\to \infty $ . We need to show that the integral of $f_{z_3}$ with respect to $\nu ^w$ converges to the integral of f with respect to $\nu ^w$ . Since $\nu ^w$ is a probability measure on $\mathbb {C}^w$ , well-known results from probability theory imply that it is sufficient to prove that for some $\delta>0$ ,

(3.4.2) $$ \begin{align} \limsup_{|z_3|\to \infty} \int_{\mathbb{C}^w} |f_{z_3}(t_1,\ldots,t_w)|^{1+\delta} d\nu(t_1)\cdots d\nu(t_w) <\infty. \end{align} $$

Let us fix some $\delta $ , to be chosen later. By the change of variable formula from equation (3.3.2), we have

$$ \begin{align*} &\int_{\mathbb{C}^w} |f_{z_3}(t_1,\ldots,t_w)|^{1+\delta} d\nu(t_1)\cdots d\nu(t_w) \\ &= \frac{1}{(4\pi)^w}\int_{(S^2)^w} |f_{z_3}(\sigma(y_1),\ldots,\sigma(y_w))|^{1+\delta} da(y_1)\cdots da(y_w). \end{align*} $$

Now, recall that by Lemma 3.3.1,

$$\begin{align*}\|x-y\| = g(\sigma(x))^{1/4} g(\sigma(y))^{1/4} |\sigma(x)-\sigma(y)| \end{align*}$$

for all $x,y\in S^2\setminus \{e_3\}$ . Let $u := \sigma ^{-1}(z_3)$ . Then note that by the above identity, and the fact that

$$\begin{align*}-\frac{1}{4}(4b\alpha_1 + 4b\alpha_2 + 4b\alpha_3 + 4b^2(w-1))= 1, \end{align*}$$

we get

$$ \begin{align*} &f_{z_3}(\sigma(y_1),\ldots,\sigma(y_w)) \\ &= \bigg(\prod_{l=1}^w \frac{4\pi}{g(\sigma(y_l))}\bigg) \bigg(\prod_{l=1}^w |\sigma(-e_3) - \sigma(y_l)|^{4b\alpha_1}|\sigma(e_1)-\sigma(y_l)|^{4b\alpha_2}\frac{|\sigma(u) - \sigma(y_l)|^{4b \alpha_3}}{|\sigma(u)-\sigma(-e_3)|^{4b\alpha_3}}\bigg)\\ &\qquad \cdot \bigg(\prod_{1\le l<l'\le w}|\sigma(y_l)-\sigma(y_{l'})|^{4b^2} \bigg)\\ &= (4\pi)^w g(\sigma(-e_3))^{-bw(\alpha_1-\alpha_3)} g(\sigma(e_1))^{-bw\alpha_2}\\ &\qquad \cdot\bigg(\prod_{l=1}^w \|e_3+ y_l\|^{4b\alpha_1}\|e_1-y_l\|^{4b\alpha_2}\frac{\|u - y_l\|^{4b \alpha_3}}{\|u+e_3\|^{4b\alpha_3}}\bigg) \bigg(\prod_{1\le l<l'\le w}\|y_l-y_{l'}\|^{4b^2} \bigg). \end{align*} $$

From this, it is clear that if u is in the upper hemisphere, then

$$\begin{align*}|f_{z_3}(\sigma(y_1),\ldots,\sigma(y_w))| \le K \prod_{l=1}^w \|e_3+ y_l\|^{4b\operatorname{Re}(\alpha_1)}\|e_1-y_l\|^{4b\operatorname{Re}(\alpha_2)}\|u - y_l\|^{4b\operatorname{Re}( \alpha_3)}, \end{align*}$$

where K is a constant that does not depend on $y_1,\ldots ,y_l$ . Since $4b\operatorname {Re}(\alpha _j )> -2$ for $j=1,2,3$ , it is easy to see from the above bound that the condition (3.4.2) holds for sufficiently small $\delta $ . This completes the proof of the lemma.

Proof of Theorem 3.4.1.

Integrals of the type appearing in Lemma 3.4.2 are sometimes called complex Selberg integrals. They can be exactly evaluated using formulas given by Dotsenko and Fateev [Reference Dotsenko and Fateev20] and Aomoto [Reference Aomoto5]. Let $\gamma (x) := \Gamma (x)/\Gamma (1-x)$ , where $\Gamma $ is the classical Gamma function. Then by [Reference Dotsenko and Fateev20, Equation (B.9)], the integral in Lemma 3.4.2 is equal to

(3.4.3) $$ \begin{align} w! \pi^w \gamma(b^2)^{-w} \prod_{j=1}^w \gamma(b^2j)\prod_{r=1}^3\prod_{j=1}^{w}\gamma(1+2b\alpha_r + (j-1)b^2). \end{align} $$

Aomoto [Reference Aomoto5] gives precise conditions on the parameters under which the above formula holds. In our setting, the conditions are that

$$ \begin{align*} &2b\operatorname{Re}(\alpha_1)> -1, \ \ 2b\operatorname{Re}(\alpha_2 ) > -1, \ \ 2b\operatorname{Re}(\alpha_1+\alpha_2)< -1, \\ &2(w-1)b^2 + 2b\operatorname{Re}(\alpha_1+\alpha_2) < -1. \end{align*} $$

Let us quickly verify that these conditions hold under the assumptions of the theorem. The first two inequalities are part of the assumptions. For the third inequality, note that since $w\ge 1$ and $\operatorname {Re}(\alpha _3)> -1/2b$ , we have

$$\begin{align*}\operatorname{Re}(\alpha_1 + \alpha_2) \le Q-\operatorname{Re}(\alpha_3)-b = -\frac{1}{b} - \operatorname{Re}(\alpha_3) < -\frac{1}{2b}. \end{align*}$$

Finally, for the fourth inequality, note that

$$\begin{align*}2(w-1)b^2 + 2b\operatorname{Re}(\alpha_1+\alpha_2) = 2b(Q -\operatorname{Re}(\alpha_3) - b) = -2 - 2b\operatorname{Re}(\alpha_3)< -1, \end{align*}$$

where the last inequality holds because $\operatorname {Re}(\alpha _3)> -1/2b$ . This proves that all of Aomoto’s conditions hold, and therefore the formula (3.4.3) is valid.

Now, it is known [Reference Giribet24] that the $\Upsilon _b$ functions satisfies

$$\begin{align*}\Upsilon_b(x+b) = \gamma(bx) b^{1-2bx}\Upsilon_b(x) \end{align*}$$

for all $x\in \mathbb {C}$ . It is also known that the zeros of $\Upsilon _b$ are the points $mb + nb^{-1}$ , where $m,n$ are either both positive integers, or both nonpositive integers [Reference Harlow, Maltz and Witten32]. Thus, if $x\in \mathbb {C}$ is not one of the zeros, then for any positive integer n, iterating the above identity gives

(3.4.4) $$ \begin{align} \Upsilon_b(x+nb) &= \Upsilon_b(x) b^{n - 2b\sum_{j=0}^{n-1} (x+jb)}\prod_{j=0}^{n-1} \gamma(b(x+jb))\notag\\ &= \Upsilon_b(x) b^{n - 2bnx - b^2 n(n-1)}\prod_{j=0}^{n-1} \gamma(b(x+jb)). \end{align} $$

In particular, since b is not a zero of $\Upsilon _b$ (because $b\ne mb + nb^{-1}$ for any positive integers $m,n$ ), this shows that

$$ \begin{align*} \prod_{j=1}^n \gamma(j b^2) = b^{b^2n^2 + b^2n - n} \frac{\Upsilon_b(b+nb)}{\Upsilon_b(b)}. \end{align*} $$

Next, we claim that $2\alpha _1+1/b$ is a not a zero of $\Upsilon _b$ . To see this, first note that since $\operatorname {Re}(\alpha _1)> -1/2b$ , we have that $2\operatorname {Re}(\alpha _1) + 1/b> 0$ . So, if $2\alpha _1+1/b$ is a zero of $\Upsilon _b$ , it must be of the form $mb + nb^{-1}$ for some positive integers $m,n$ . Consequently, $2\alpha _1 = mb + (n-1)b^{-1}\ge 0$ . But since $\operatorname {Re}(\alpha _2), \operatorname {Re}(\alpha _3)> -1/2b$ and $w \ge 1$ , we have

$$\begin{align*}\operatorname{Re}(\alpha_1) = Q- \operatorname{Re}(\alpha_2+\alpha_3) -bw < Q+ \frac{1}{2b}+\frac{1}{2b} - b = 0. \end{align*}$$

This shows that $\Upsilon _b(2\alpha _1 + 1/b)\ne 0$ . Similarly, $\Upsilon _b(2\alpha _r + 1/b)\ne 0$ for $r=2,3$ . Thus, by equation (3.4.4),

$$ \begin{align*} &\prod_{j=1}^w \gamma(b^2j)\prod_{r=1}^3\prod_{j=1}^{w}\gamma(1+2b\alpha_r + (j-1)b^2) \\ &= \prod_{j=1}^w \gamma(b^2j)\prod_{r=1}^3\prod_{j=1}^{w}\gamma(b(2\alpha_r + 1/b + (j-1)b)) \\ &=b^{b^2w^2 + b^2w - w} \frac{\Upsilon_b(b+wb)}{\Upsilon_b(b)}\prod_{r=1}^3\bigg(\frac{\Upsilon_b(2\alpha_r + 1/b + wb)}{\Upsilon_b(2\alpha_r + 1/b)} b^{-w+2bw(2\alpha_r +1/b) + b^2w(w-1)}\bigg)\\ &= b^{2 b^2w -2w } \frac{\Upsilon_b(b+wb)}{\Upsilon_b(b)}\prod_{r=1}^3\frac{\Upsilon_b(2\alpha_r + 1/b + wb)}{\Upsilon_b(2\alpha_r + 1/b)}. \end{align*} $$

The proof is completed by using the relation $\Upsilon _b(x) =\Upsilon _b(b+1/b - x)$ in the denominator (see [Reference Giribet24]), and the fact that $wb = Q - \sum _{j=1}^3 \alpha _j$ , and finally, by noting that

$$ \begin{align*} \gamma(b^2)^{-w} &= \frac{\Gamma(1-b^2)^w}{\Gamma(b^2)^w} =\frac{\Gamma(-b^2)^w(-b^2)^w}{\Gamma(b^2+1)^w (b^2)^{-w}} = \gamma(-b^2)^w (-1)^w b^{4w}, \end{align*} $$

where we used the relation $\Gamma (x+1)=x\Gamma (x)$ . Lastly, to see that $\Upsilon _b(b- 2\alpha _j)\ne 0$ for $j=1,2,3$ , recall that $\operatorname {Re}(\alpha _j)>-1/2b$ for each j. Thus, $\operatorname {Re}(b - 2\alpha _j) < 2b$ . On the other hand, since $\operatorname {Re}(w)\ge 1$ , we have that for each j,

$$ \begin{align*} \operatorname{Re}(\alpha_j) &= \operatorname{Re}(Q-bw) - \sum_{1\le j'\le 3, \, j'\ne j}\operatorname{Re}(\alpha_{j'})\\ &< Q - b + \frac{2}{2b} = 0, \end{align*} $$

which gives $\operatorname {Re}(b - 2\alpha _j)> b$ . Since $\operatorname {Re}(b-2\alpha _j)$ is positive, it cannot be that $b-2\alpha _j = m b + n b^{-1}$ for two nonpositive integers m and n. On other other hand, since $b\in (0,1)$ and $\operatorname {Re}(b - 2\alpha _j)<2b$ , it cannot be that $b-2\alpha _j = m b + n b^{-1}$ for some positive integers m and n. Thus, $b-2\alpha _j$ cannot be a zero of $\Upsilon _b$ .

Proof. Let f be the function displayed in equation (4.1.1). It is clear that f is holomorphic on $\mathbb {C}\setminus \{-d/c\}$ , with derivative

(4.1.2) $$ \begin{align} f'(z) = \frac{ad-bc}{(cz+d)^2} = \frac{1}{(cz+d)^2}. \end{align} $$

Now, we can write f as a function from $\mathbb {R}^2\setminus \{(-d/c,0)\}$ into $\mathbb {R}^2$ , given by

$$\begin{align*}f(x,y) = (u(x,y), v(x,y)), \end{align*}$$

where u and v are the real and imaginary parts of f. By the Cauchy–Riemann equations,

$$\begin{align*}\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \ \ \ \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}. \end{align*}$$

Thus, the determinant of the Jacobian of f (as a map from $\mathbb {R}^2\setminus \{(-d/c,0)\}$ into $\mathbb {R}^2$ ) is given by

$$ \begin{align*} \det J_f(x,y) &= \frac{\partial u}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\frac{\partial v}{\partial x}= \bigg(\frac{\partial u}{\partial x}\bigg)^2 +\bigg( \frac{\partial v}{\partial x}\bigg)^2. \end{align*} $$

But, since f is holomorphic, we have

$$\begin{align*}f'(x+iy) = \frac{\partial u}{\partial x} +i \frac{\partial v}{\partial x}. \end{align*}$$

Thus, by equation (4.1.2),

$$\begin{align*}\det J_f(x,y) = |f'(x+iy)|^2 = \frac{1}{|c(x+iy) + d|^4}. \end{align*}$$

Lastly, note that f is a bijection of $\widehat {\mathbb {C}}$ onto itself. By these facts, and the formula from Theorem 1.3.1, we get

$$ \begin{align*} &C(\alpha_1,\ldots,\alpha_k; f(z_1),\ldots,f(z_k);b;\mu) \\ &= \frac{e^{-i\pi w}\mu^{w}}{w!} (4/e)^{1-1/b^2}\prod_{1\le j<j'\le k} |f(z_j) - f(z_{j'})|^{4\alpha_j \alpha_{j'}}\\ &\qquad \qquad \cdot \int_{\mathbb{C}^w} \bigg(\prod_{j=1}^k \prod_{l=1}^{w} |f(z_j)-t_l|^{4b\alpha_j}\bigg)\bigg(\prod_{1\le l<l'\le w}|t_l-t_{l'}|^{4b^2} \bigg) d^2t_1\cdots d^2t_w\\ &= \frac{e^{-i\pi w}\mu^{w}}{w!} (4/e)^{1-1/b^2}\prod_{1\le j<j'\le k} |f(z_j) - f(z_{j'})|^{4\alpha_j \alpha_{j'}}\int_{\mathbb{C}^w} \bigg(\prod_{j=1}^k \prod_{l=1}^{w} |f(z_j)-f(s_l)|^{4b\alpha_j}\bigg)\\ &\qquad \qquad \cdot \bigg(\prod_{1\le l<l'\le w}|f(s_l)-f(s_{l'})|^{4b^2} \bigg) \prod_{l=1}^w \frac{1}{|cs_l + d|^4} d^2s_1\cdots d^2s_w. \end{align*} $$

The function f also has the interesting property that for any $z,z'\in \mathbb {C} \setminus \{-d/c\}$ ,

$$ \begin{align*} f(z) - f(z') &= \frac{az+b}{cz+d} - \frac{az' + b}{cz' + d}\\ &= \frac{(az+b)(cz'+d) - (az' + b)(cz+d)}{(cz+d)(cz'+d)}\\ &= \frac{(ad-bc)(z-z')}{(cz+d)(cz'+d)} = \frac{z-z'}{(cz+d)(cz'+d)}. \end{align*} $$

If we use this identity in the previous display, each $|f(z_j)-f(z_{j'})|$ is replaced by

$$\begin{align*}\frac{|z_j - z_{j'}|}{|cz_j+d||cz_{j'} + d|}, \end{align*}$$

each $|f(z_j)-f(s_l)|$ is replaced by

$$\begin{align*}\frac{|z_j - s_l|}{|cz_j+d||cs_l + d|}, \end{align*}$$

and each $|f(s_l)-f(s_{l'})|$ is replaced by

$$\begin{align*}\frac{|s_l - s_{l'}|}{|cs_l+d||cs_{l'} + d|}. \end{align*}$$

In the resulting expression, the powers of $|c s_l + d|$ add up to

$$ \begin{align*} -4b\sum_{j=1}^k\alpha_j -4b^2(w-1) -4 &= -4b\sum_{j=1}^k\alpha_j - 4b\bigg(Q- \sum_{j=1}^k \alpha_j\bigg) + 4b^2 - 4\\ &= -4bQ + 4b^2 - 4 = 0. \end{align*} $$

Similarly, the powers of $|cz_j + d|$ add up to

$$ \begin{align*} -4\alpha_j\sum_{j'\ne j} \alpha_{j'} - 4bw\alpha_j &= -4\alpha_j\sum_{j'\ne j} \alpha_{j'} - 4\bigg(Q- \sum_{j=1}^k \alpha_j\bigg) \alpha_j \\ &= -4Q\alpha_j +4\alpha_j^2 = -4\Delta_{\alpha_j}. \end{align*} $$

This completes the proof.

Proof. Take any distinct $z_1, z_2,z_3,z_4\in \mathbb {C}$ . Define $f:\mathbb {C}\setminus \{z_4\}\to \mathbb {C}$ as

$$\begin{align*}f(z) := \frac{(z-z_1)(z_2-z_4)}{(z-z_4)(z_2-z_1)}. \end{align*}$$

Note that f is an $\mathrm {SL}(2,\mathbb {C})$ -transform, since f can be expressed as $(az+b)/(cz+d)$ , where

$$\begin{align*}a = \frac{z_2 - z_4}{u}, \ \ b = -\frac{z_1(z_2-z_4)}{u}, \ \ c = \frac{z_2-z_1}{u}, \ \ d = -\frac{z_4(z_2-z_1)}{u}, \end{align*}$$

where u is any square-root of $(z_1-z_4)(z_2-z_4)(z_2-z_1)$ . Note that

$$\begin{align*}f(z_1) = 0, \ \ f(z_2) = 1, \ \ f(z_3) = \frac{(z_3-z_1)(z_2-z_4)}{(z_3-z_4)(z_2-z_1)}. \end{align*}$$

Note also that

$$ \begin{align*} |f'(z_1)| &= \frac{|(z_1-z_4)(z_2-z_4)(z_2-z_1)|}{|(z_2-z_1)(z_1-z_4)|^2}= \frac{|z_2-z_4|}{|(z_1-z_2)(z_1-z_4)|},\\ |f'(z_2)| &= \frac{|(z_1-z_4)(z_2-z_4)(z_2-z_1)|}{|(z_2-z_1)(z_2-z_4)|^2}= \frac{|z_1-z_4|}{|(z_1-z_2)(z_2-z_4)|},\\ |f'(z_3)| &= \frac{|(z_1-z_4)(z_2-z_4)(z_2-z_1)|}{|(z_2-z_1)(z_3-z_4)|^2}= \frac{|(z_2-z_4)(z_1-z_4)|}{|(z_1-z_2)(z_3-z_4)^2|}. \end{align*} $$

Now, by Theorem 4.1.1,

$$ \begin{align*} C(\alpha_1,\alpha_2,\alpha_3; z_1,z_2,z_3;b;\mu) &= \frac{C(\alpha_1,\alpha_2,\alpha_3; 0,1,f(z_3);b;\mu)}{|f'(z_1)|^{2\Delta_{\alpha_1}} |f'(z_2)|^{2\Delta_{\alpha_2}} |f'(z_3)|^{2\Delta_{\alpha_3}} }. \end{align*} $$

Also, $|f(z_3)|\to \infty $ as $z_4 \to z_3$ . Thus, taking $z_4\to z_3$ and applying Theorem 3.4.1, we get

$$ \begin{align*} &C(\alpha_1,\alpha_2,\alpha_3; z_1,z_2,z_3;b;\mu) \\ &= C(\alpha_1,\alpha_2,\alpha_3; b;\mu)\lim_{z_4\to z_3} \frac{|f(z_3)|^{4\Delta_{\alpha_3}}}{|f'(z_1)|^{2\Delta_{\alpha_1}} |f'(z_2)|^{2\Delta_{\alpha_2}} |f'(z_3)|^{2\Delta_{\alpha_3}} }\\ &= C(\alpha_1,\alpha_2,\alpha_3; b;\mu) |z_1-z_2|^{2\Delta_{\alpha_1}+2\Delta_{\alpha_2}-2\Delta_{\alpha_3}}|z_2 - z_3|^{-2\Delta_{\alpha_1}+2\Delta_{\alpha_2}+2\Delta_{\alpha_3}}\\ &\qquad \qquad \cdot |z_1-z_3|^{ 2\Delta_{\alpha_1}-2\Delta_{\alpha_2}+2\Delta_{\alpha_3}}. \end{align*} $$

This completes the proof.

Proof. Suppose, without loss of generality, that we take $\alpha _1 \to -1/2b$ while keeping $\operatorname {Re}(\alpha _j)>-1/2b$ for each j and w fixed at a positive integer value. Then $\Upsilon _b(b - 2\alpha _1) \to 0$ . So we only have to show that the numerator in the timelike DOZZ formula does not tend to zero.

Let $(\alpha _1,\alpha _2,\alpha _3)$ be the limiting value, so that $\alpha _1=-1/2b$ and $\operatorname {Re}(\alpha _2),\operatorname {Re}(\alpha _3)\ge -1/2b$ . The terms appearing in the numerator that can potentially tend to zero are $\Upsilon _b(bw+b)$ and $\Upsilon _b(2\alpha _j + bw + 1/b)$ for $j=1,2,3$ . We will now show that these are all nonzero.

As noted in [Reference Harlow, Maltz and Witten32], the zeros of $\Upsilon _b$ consist of the points $mb + nb^{-1}$ where either both $m,n$ are positive integers, or both $m,n$ are nonpositive integers. Since $b^2$ is irrational and w is a positive integer, it follows that $bw+b$ can attain neither of these forms, and is therefore not a zero of $\Upsilon _b$ .

Next, note that $2\alpha _1+bw+1/b = bw$ is not a zero of $\Upsilon _b$ since $b^2$ is irrational. Suppose that $2\alpha _2+bw+1/b$ is a zero of $\Upsilon _b$ . Since $\operatorname {Re}(\alpha _2) \ge -1/2b$ , we have $2\operatorname {Re}(\alpha _2) + bw+1/b \ge bw>0$ , which means (again, by the irrationality of $b^2$ ) that it must be of the form $mb + nb^{-1}$ for some positive integers $m,n$ . Thus,