Holomorphic Morse normalization and Gaussian contour

Because $H$ is complex symmetric and invertible, some vector $v$ has $v^{T}Hv\mathrm{\ne }0$: otherwise polarization would give $u^{T}Hv=0$ for every $u,v$, contradicting invertibility. Extend $v$ to a basis and use this as a complex linear coordinate change, so $H_{11}\mathrm{\ne }0$. From now on, compose each coordinate change with the preceding ones; $\mathrm{\Psi }$ denotes the final map from the original coordinates. The holomorphic equation ${\mathrm{\partial }}_{z_1}\varphi (z_1,z^{\mathrm{\prime }})=0$ has a unique holomorphic solution $z_1=\psi (z^{\mathrm{\prime }})$ near $0$ by [F3] and [F4], since ${\mathrm{\partial }}_{z_1}\varphi (0)=0$ and ${\mathrm{\partial }}_{z_1}^2\varphi (0)=H_{11}\mathrm{\ne }0$.

Set $t=z_1-\psi (z^{\mathrm{\prime }})$. Taylor's formula in $t$, with the linear term zero, gives $\varphi (\psi (z^{\mathrm{\prime }})+t,z^{\mathrm{\prime }})=\varphi (\psi (z^{\mathrm{\prime }}),z^{\mathrm{\prime }})+t^{2}a(t,z^{\mathrm{\prime }}),a(t,z^{\mathrm{\prime }})={\int }_0^1(1-s){\mathrm{\partial }}_{z_1}^2\varphi (\psi (z^{\mathrm{\prime }})+st,z^{\mathrm{\prime }})ds\mathrm{.}$ The integral is holomorphic by locally uniform convergence of the power series [F2]; $a(0,0)=H_{11}\mathrm{/}2\mathrm{\ne }0$. Shrink the polydisc so $a=a(0,0)(1+b)$ with $\mathrm{\mid }b\mathrm{\mid }<1$. The binomial series for $(1+b{)}^{1\mathrm{/}2}$ converges absolutely by [F2], so $w_1=t\sqrt{2a(t,z^{\mathrm{\prime }})}$ is holomorphic and has nonzero $t$-derivative at $0$.

By [F4], $(t,z^{\mathrm{\prime }})\mapsto (w_1,z^{\mathrm{\prime }})$ is biholomorphic near $0$, and the phase becomes $\varphi (\psi (z^{\mathrm{\prime }}),z^{\mathrm{\prime }})+w_1^2\mathrm{/}2$. The Hessian of the reduced phase in $z^{\mathrm{\prime }}$ is the Schur complement $H_{22}-H_{21}{H}_{11}^{-1}{H}_{12}$. It is invertible because the block determinant formula gives $\det H=H_{11}\det (H_{22}-H_{21}{H}_{11}^{-1}{H}_{12})$.

For real symmetric positive-definite $H$, [F8] gives $H=L{L}^{\ast }$ with $L$ lower triangular and positive diagonal. Since $H$ is real, $\bar{L}$ is another such factorization; uniqueness in [F8] gives $L=\bar{L}$, so $L$ is real and $H=L{L}^T$. On each ball $B_R$, the map $x\mapsto L^{T}x$ sends a compact Jordan set to a compact Jordan set; [F11] changes variables by $\mathrm{\mid }\det L{\mathrm{\mid }}^{-1}$. Letting $R\to \mathrm{\infty }$ and using [F12] gives the corresponding nonnegative improper-integral equality. Since $\det L>0$, [F9]–[F10] give $\det H=(\det L{)}^2$. Repeated compact-rectangle Fubini [F15] factors the standard Gaussian integral on $[-R,R{]}^n$ into one-dimensional integrals, and [F12] lets $R\to \mathrm{\infty }$. By [F13] and the substitution $t=x\mathrm{/}\sqrt{2}$ in [F14], each one-dimensional integral of $e^{-x^2\mathrm{/}2}$ is $\sqrt{2\pi }$. Finally, $z=\sqrt{\lambda }w$ scales the integral by ${\lambda }^{-n\mathrm{/}2}$, proving the formula for real positive-definite $H$ with the positive determinant root.

Repeating steps 1.1–2.1 on the reduced phase proves the normal form by induction on $n$, with the constant phase as the $n=0$ case. Because $\mathrm{\Psi }$ is the composite from the original coordinates, differentiating $\varphi (z)=\varphi (0)+\frac{1}{2}\mathrm{\Psi }(z{)}^T\mathrm{\Psi }(z)$ twice at $0$ gives the original Hessian identity $H=(D\mathrm{\Psi }(0){)}^{T}D\mathrm{\Psi }(0)$; taking determinants gives the asserted square relation. For a continuous parameter family, continuity of $\det H$ lets one restrict to a neighborhood whose determinant image lies in a disc avoiding zero, where a square-root branch can be fixed from its base value.

For complex symmetric $H$ with positive-definite real part, the Gaussian integral is holomorphic in its entries: on each compact subset of this matrix domain, the integrand and its entry derivatives are bounded by a polynomial times $e^{-c\mathrm{\mid }x{\mathrm{\mid }}^2}$ for some $c>0$, so [F6] permits differentiation under the integral. This domain is convex, hence connected, and $\det H\mathrm{\ne }0$ there because $Hv=0$ would imply $\mathrm{Re}(v^{\ast }Hv)=v^{\ast }(\mathrm{Re}H)v>0$ for $v\mathrm{\ne }0$. For $H_t=I+t(H-I)$, [F16] and determinant multiplicativity give $\frac{d}{dt}\det H_t=(\det H_t)\mathrm{tr}(H_t^{-1}(H-I))$. Thus $L(H)={\int }_0^1\mathrm{tr}(H_t^{-1}(H-I))dt$ depends holomorphically on $H$, satisfies $e^{L(H)}=\det H$, and is zero at $I$. When $H$ is real positive definite, its segment $H_t$ is real positive definite, so $L(H)$ is real and the branch $e^{L(H)\mathrm{/}2}$ is positive. The real Gaussian formula from step 3.2 and the complex integral formula agree on a real open set of positive-definite matrices. In a polydisc centered at any point of that set, apply the one-variable identity theorem successively in the independent entries of a complex symmetric matrix; then [F5] extends the identity from that polydisc to the connected matrix domain.

The holomorphic $n$-form in the statement is closed: in complex coordinates its exterior derivative has repeated holomorphic differentials and no antiholomorphic coefficients. The difference of two homologous admissible contours is the boundary of the swept chain. By [F7], its integral is the boundary integral on the truncation of that chain; the stated decay makes this term tend to zero as the truncation expands. Therefore the contour integrals agree.