User:Dom walden/Multivariate Analytic Combinatorics/Smooth Point via Surgery

Critical points
We attach a few conditions to the critical point $$\textbf{w}$$ and the function $$H$$ of this chapter. Later chapters relax or modify these conditions.

Squarefree
$$H$$ can be factored into $$H_1 H_2 \cdots H_n$$ where no factor has a power and no factor is repeated.

Minimality
The critical point $$\textbf{w}$$ is one of
 * Strictly minimal: the only critical point on its polytorus, i.e.
 * $$\mathcal{V} \cap T(\textbf{w}) = \{\textbf{w}\}$$


 * Finitely minimal: one of a finite number of critical point on its polytorus, i.e.
 * $$1 < \#(\mathcal{V} \cap T(\textbf{w})) < \infty$$


 * Torally minimal: one of an infinite number of critical point on its polytorus, i.e.
 * $$\#(\mathcal{V} \cap T(\textbf{w})) = \infty$$
 * and the the torality hypothesis is satisfied...

Smoothly varying
At least one of the partial derivatives $$\frac{\partial H}{\partial w_k}(\textbf{w}) \neq 0$$.

Formally, it means the gradient map of $$H$$ at $$\textbf{w}$$ is not equal to the zero vector.


 * $$\nabla H(\textbf{w}) \neq \textbf{0}$$

where:


 * $$\nabla f(p) = \begin{bmatrix}

\frac{\partial f}{\partial x_1}(p) \\ \vdots \\ \frac{\partial f}{\partial x_n}(p) \end{bmatrix}$$

Quadratically nondegenerate
In the proof, we make use of the Hessian matrix $$\mathcal{H}$$


 * $$\mathcal{H} = \left[ \frac{\partial^2 \phi(\textbf{w})}{\partial z_i \partial z_j} \right]$$

where $$\phi$$ is a function we will see later.

Quadratic nondegeneracy means the Hessian matrix is nonsingular, a fact we use in the proof.

One quadratically nondegenerate smooth point
Let $$F(\textbf{z}) = \frac{G(\textbf{z})}{H(\textbf{z})}$$ where $$H(z)$$ is squarefree and has a strictly minimal, smoothly varying and quadratically nondegenerate critical point $$\textbf{w} = (w_1, \cdots, w_d) \in \C_*^d$$ in the direction $$\textbf{r} = (r_1, \cdots, r_d) \in \N_*^d$$. Then,


 * $$a_{n\textbf{r}} \sim \textbf{w}^{-n\textbf{r}} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det \mathcal{H}}} \frac{G(\textbf{w})}{w_k \frac{\partial H}{\partial w_k}(\textbf{w})} (nr_k)^{(1-d)/2}$$

where $$\mathcal{H}$$ is the Hessian matrix of $$\phi(\theta)$$ at $$\textbf{w}$$.

Proof
By the multivariate Cauchy formula


 * $$a_{n\textbf{r}} = \frac{1}{(2\pi i)^d} \int_T F(\textbf{z}) \frac{d\textbf{z}}{\textbf{z}^{n\textbf{r}+1}}$$

To make use of the implicit function theorem, we choose one coordinate from the critical point for which $$\frac{\partial H}{\partial w_k}(\textbf{w}) \neq 0$$, possible because it is smooth. Call this coordinate $$w_k$$. Call the projection of the critical point $$\textbf{w}$$ into $$d-1$$ coordinates by $$\textbf{w}^\circ$$ and the projection of the direction $$\textbf{r}$$ into $$d-1$$ coordinates by $$\textbf{r}^\circ$$.

Rewrite the above Cauchy formula as an iterated integral, defining $$T^\circ$$ as the torus through $$\textbf{w}^\circ$$


 * $$a_{n\textbf{r}} = \frac{1}{(2\pi i)^d} \int_{T^\circ} ({\textbf{w}^\circ})^{-n\textbf{r}^\circ} \left( \int_{|w_k| = \rho - \delta_s} w_k^{-nr_k} F(\textbf{w}^\circ, w_k) \frac{dw_k}{w_k} \right) \frac{d\textbf{w}^\circ}{\textbf{w}^\circ}$$

By the implicit function theorem, there exists a neighbourhood $$\mathcal{N}$$ of $$\textbf{w}^\circ$$ such that $$H(\textbf{w}^\circ, z) = 0 \iff z = g(\textbf{w}^\circ)$$


 * $$I = \frac{1}{(2\pi i)^d} \int_{\mathcal{N}} ({\textbf{w}^\circ})^{-n\textbf{r}^\circ} \left( \int_{|w_k| = \rho - \delta_s} w_k^{-nr_k} F(\textbf{w}^\circ, w_k) \frac{dw_k}{w_k} \right) \frac{d\textbf{w}^\circ}{\textbf{w}^\circ}$$

and


 * $$I' = \frac{1}{(2\pi i)^d} \int_{\mathcal{N}} (\textbf{w}^\circ)^{-n\textbf{r}^\circ} \left( \int_{|w_k| = \rho + \delta_s} w_k^{-nr_k} F(\textbf{w}^\circ, w_k) \frac{dw_k}{w_k} \right) \frac{d\textbf{w}^\circ}{\textbf{w}^\circ}$$

As a result of the implicit function theorem, for any fixed $$\textbf{w}^\circ$$ there is a unique pole a $$z = g(\textbf{w}^\circ)$$ inside the annulus.

The difference between the two inner integrals of $$I$$ and $$I'$$ is that the latter has this pole inside and therefore by integration with residues


 * $$\int_{|w_k| = \rho + \delta_s} w_k^{-nr_k} F(\textbf{w}^\circ, w_k) \frac{dw_k}{w_k} - \int_{|w_k| = \rho - \delta_s} w_k^{-nr_k} F(\textbf{w}^\circ, w_k) \frac{dw_k}{w_k} = (2\pi i) Res\left( g(\textbf{w}^\circ)^{-nr_k} \frac{F(\textbf{w}^\circ, w_k)}{w_k}; w_k = g(\textbf{w}^\circ) \right)$$

therefore


 * $$\chi = I - I' = \frac{1}{(2\pi i)^{d-1}} \int_{\mathcal{N}} ({\textbf{w}^\circ})^{-n\textbf{r}^\circ} g(\textbf{w}^\circ)^{-nr_k} \psi(\textbf{w}^\circ) \frac{d\textbf{w}^\circ}{\textbf{w}^\circ}$$

where $$\psi(\textbf{w}^\circ) = Res\left( \frac{F(\textbf{w}^\circ, w_k)}{w_k}; w_k = g(\textbf{w}^\circ) \right)$$.

Because $$\textbf{w}$$ is a minimal point, the domain of convergence of the integral is greater than $$\rho$$ away from $$\textbf{w}$$.


 * $$a_{n\textbf{r}} - \chi = \frac{1}{(2\pi i)^{d-1}} \int_{T^\circ \setminus \mathcal{N}} ({\textbf{w}^\circ})^{-n\textbf{r}^\circ} g(\textbf{w}^\circ)^{-nr_k} \psi(\textbf{w}^\circ) \frac{d\textbf{w}^\circ}{\textbf{w}^\circ} < |\textbf{w}^{-n\textbf{r}}|$$

By changing variables $$z_j = \textbf{w}_j e^{i\theta_j} \quad (1 \leq j \leq d, j \neq k)$$. Let $$\mathcal{N'}$$ be the image of $$\mathcal{N}$$ under this change of variables. This is a neighbourhood of the origin in $$\R^{d-1}$$. $$g$$ and $$\psi$$ can be re-written after this change of variables


 * $$\phi(\theta) = log\frac{g(\textbf{w}^\circ) e^{i\theta}}{g(\textbf{w}^\circ)} + \frac{i}{r_k}\textbf{r}^\circ . \theta$$

Therefore, $$\chi$$ can be written


 * $$\chi = \frac{1}{(2\pi)^{d-1}} \textbf{w}^{-n\textbf{r}} \int_{\mathcal{N'}} e^{-nr_k \phi(\theta)} \psi(\textbf{w}^\circ e^{i\theta}) d\theta$$

By Fourier-Laplace integrals


 * $$\int_{\mathcal{N'}} e^{-nr_k \phi(\theta)} \psi(\textbf{w}^\circ e^{i\theta}) d\theta \sim \frac{ (2\pi)^{(d-1)/2}}{\sqrt{\det \mathcal{H}}} \psi(\textbf{w}^\circ e^{i0}) (nr_k)^{(1-d)/2} = \frac{(2\pi)^{(d-1)/2}}{\sqrt{\det \mathcal{H}}} \frac{G(\textbf{w})}{w_k \frac{\partial H}{\partial w_k}(\textbf{w})} (nr_k)^{(1-d)/2}$$

Therefore


 * $$\chi \sim \textbf{w}^{-n\textbf{r}} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det \mathcal{H}}} \frac{G(\textbf{w})}{w_k \frac{\partial H}{\partial w_k}(\textbf{w})} (nr_k)^{(1-d)/2}$$

Implicit function theorem
Theorem

If $$f(\textbf{z}^\circ, y)$$ is a holomorphic function at $$\textbf{w} \in \C^d$$ and $$\frac{\partial f}{\partial y}(\textbf{w}) \neq 0$$ then for $$\textbf{z}^\circ$$ in a neighbourhood of $$\textbf{w}^\circ$$ there is a unique holomorphic function $$g(\textbf{z}^\circ)$$ such that $$f(\textbf{z}^\circ, y) = 0 \iff y = g(\textbf{z}^\circ)$$.

Proof

The solution to $$f(\textbf{z}^\circ, g) = 0$$ exists(?) and can be differentiated with respect to $$\bar{z}_j (j = 2, \cdots, d)$$ in the form


 * $$\frac{\partial f}{\partial z_1}\frac{\partial g}{\partial \bar{z}_j} + \frac{\partial f}{\partial \bar{z}_1}\bar{\frac{\partial g}{\partial z_j}} + \frac{\partial f}{\partial \bar{z}_j} = 0$$

Because $$f$$ is holomorphic the Cauchy-Riemann conditions apply such that $$\frac{\partial f}{\partial \bar{z}_1} = \frac{\partial f}{\partial \bar{z}_j} = 0$$ leaving us with


 * $$\frac{\partial f}{\partial z_1}\frac{\partial g}{\partial \bar{z}_j} = 0$$

But, by the hypothesis $$\frac{\partial f}{\partial y}(\textbf{w}) \neq 0$$, therefore $$\frac{\partial g}{\partial \bar{z}_j} = 0.$$

Bear in mind that the neighbourhood $$\mathcal{N}$$ is a necessary condition otherwise we might have multiple poles...

Projection
Projection in this context is straight-forward. Simply remove the $$kth$$ coordinate from $$\textbf{w}$$ to give $$\textbf{w}^\circ = (w_1, \cdots, w_{k-1}, w_{k+1}, \cdots, w_d) \in \C_*^{d-1}$$. Similarly with $$\textbf{r}$$.

Iterated integral
An integral $$\int f(x, y)\,dx\,dy$$ can be re-written $$\int \left( \int f(x, y)\,dx \right) dy$$, where in the inner integral $$y$$ is kept constant.

Neighbourhood
We restrict all the coordinates to an arc containing $$w_i$$ for each $$w_i \in \textbf{w}^\circ$$, i.e. all coordinates except the $$kth$$.

Integration with residues
If $$f(z)$$ is an analytic function on and inside the contour $$C$$ except at a singularity $$z_0$$, then


 * $$\int_C f(z) dz = (2 \pi i) Res(f(z); z = z_0)$$

Fourier-Laplace integrals
Theorem

If $$A$$ and $$\phi$$ are complex-valued analytic functions on a compact neighbourhood $$\mathcal{N}$$ of the origin in $$\R^d$$, the real part of $$\phi$$ is nonnegative on $$\mathcal{N}$$ and vanishes only at the origin and the Hessian matrix $$\mathcal{H}$$ of $$\phi$$ at the origin is nonsingular, then


 * $$\int_\mathcal{N} A(z) e^{-\lambda \phi(z)} dz \sim A(0) \frac{(2\pi)^{d/2}}{\sqrt{\det \mathcal{H}}} \lambda^{-d/2}$$

Proof

By the complex Morse Lemma, there exists a change of variables $$\psi^{-1}$$ to move $$\mathcal{N}$$ to a neighbourhood of the origin in $$\C^d$$


 * $$\int_\mathcal{N} A(z) e^{-\lambda \phi(z)} dz = \int_{\psi^{-1} \mathcal{N}} A(\psi(y)) e^{-\lambda S(y)} (\det d \psi(y)) dy$$

where $$S(y) = \sum_{i=1}^d z_i^2$$.

Let $$C = \psi^{-1} \mathcal{N}$$ be a polydisk centred at the origin. Using the functions $$Id(z) = z$$ and $$\pi(z) = Re\{z\}$$ we construct a prism operator $$P$$ which has the property


 * $$\partial P(C) = C - Re\{C\} + P(\partial C)$$

Now, we can apply Stokes' formula


 * $$\int_{\partial P(C)} \omega = 0$$

which implies


 * $$\int_C \omega = \int_{Re\{C\}} \omega - \int_{P(\partial C)} \omega$$.

Therefore


 * $$\int_C A(\psi(y)) e^{-\lambda S(y)} (\det d \psi(y)) dy = \int_{Re\{C\}} A(\psi(y)) e^{-\lambda S(y)} (\det d \psi(y)) dy + O(e^{-\epsilon\lambda})$$

As proved in the complex Morse Lemma


 * $$\det d \psi(0) = \frac{2^{d/2}}{\sqrt{\det \mathcal{H}}}$$

By applying the Gaussian integral (like in the univariate saddle-point method) multiple times


 * $$\int_{\pi(C)} e^{-\lambda S(y)} dy = \prod_{j=1}^d \int_{-a}^a e^{-\lambda S(y)} dy \sim \prod_{j=1}^d \int_{-\infty}^\infty e^{-\lambda S(y)} dy = \prod_{j=1}^d \sqrt{\frac{\pi}{\lambda}} = \pi^{d/2} \lambda^{-d/2}$$

Complex Morse Lemma
Lemma

If $$\phi(x)$$ has vanishing gradient and nonsingular Hessian matrix $$\mathcal{H}$$ at the origin then there exists a change of variables $$x = \psi(y)$$ around $$x = y = 0$$ such that $$\phi(\psi(y)) = S(y) = \sum_{j=1}^d y_j^2$$ and $$(\det d\psi(0))^2 = \frac{2^d}{\det \mathcal{H}}$$.

 Proof

Prism operator
Between two functions $$f, g: X \to Y$$, a homotopy is a map $$F: X \times I \to Y$$ where $$F(z, 0) = f(z)$$ and $$F(z, 1) = g(z)$$.

Explain chain...

From a homotopy and a chain $$C$$, we can define the prism operator


 * $$P(C) = \sum_i (-1)^i F(C \times Id) | [v_0, \cdots, v_i, w_i, \cdots, w_n]$$

The prism operator satisfies the relation:


 * $$\partial P(C) = g_\#(C) - f_\#(C) - P(\partial C)$$

where $$f_\#$$ and $$g_\#$$ map the chains in $$X$$ to the chains in $$Y$$.

Stokes' formula
If $$M$$ is a complex manifold of dimension $$n$$, $$\omega$$ a holomorphic form of degree $$n$$ and $$\sigma$$ an $$(n+1)$$-dimensional chain


 * $$\int_{\partial \sigma} \omega = 0$$.