[proofplan] The Kähler metric makes the Levi-Civita connection unitary on the holomorphic tangent bundle $T_X$ and therefore induces a Hermitian connection on the canonical bundle $K_X$. If the holonomy is contained in $SU(n)$, the standard determinant form is fixed by parallel transport, so it produces a globally defined parallel holomorphic volume form of constant norm. Conversely, a nowhere-vanishing holomorphic volume form of constant norm is parallel for the induced Chern connection on $K_X$; parallel transport must preserve it, and a unitary [linear map](/page/Linear%20Map) preserving a nonzero determinant form has determinant $1$. [/proofplan] [step:Use the Kähler metric to identify holonomy as a unitary determinant action] Fix $x\in X$. Let $T_{\mathbb R}X$ denote the real tangent bundle of $X$, and let $J:T_{\mathbb R}X\to T_{\mathbb R}X$ denote the complex structure endomorphism of $X$, characterized by multiplication by $i$ after passing to the holomorphic tangent bundle $T_X$. By the standard Kähler connection identity, the Levi-Civita connection $\nabla$ satisfies $\nabla J=0$ and preserves the Hermitian metric on $T_X$. Hence every element of $\operatorname{Hol}_x(\omega)$ acts as a unitary linear automorphism of the Hermitian [vector space](/page/Vector%20Space) $(T_X)_x$. The induced connection on \begin{align*} K_X=\det(T_X^*) \end{align*} has holonomy representation given by the inverse determinant character: if $A\in U((T_X)_x,\omega_x)$ is the holonomy action on $(T_X)_x$, then its action on $K_{X,x}$ is multiplication by $\det(A)^{-1}$. Therefore a nonzero element of $K_{X,x}$ is fixed by all holonomy transformations exactly when every holonomy transformation has determinant $1$. [guided] The first point is that Kähler holonomy is already unitary. Let $J:TX\to TX$ denote the complex structure of $X$. The Levi-Civita connection $\nabla$ preserves the Riemannian metric by definition, and the standard Kähler connection identity gives $\nabla J=0$. Therefore parallel transport preserves both the complex structure and the Hermitian metric induced by $\omega$ on $T_X$. Thus \begin{align*} \operatorname{Hol}_x(\omega)\subset U((T_X)_x,\omega_x). \end{align*} Now pass from the tangent bundle to the canonical bundle. The fibre of the canonical bundle is \begin{align*} K_{X,x}=\det((T_X)_x^*). \end{align*} If $A:(T_X)_x\to (T_X)_x$ is a unitary holonomy transformation and $\eta\in K_{X,x}$, then the induced action is \begin{align*} (A\cdot \eta)(v_1,\dots,v_n)=\eta(A^{-1}v_1,\dots,A^{-1}v_n) \end{align*} for $v_1,\dots,v_n\in (T_X)_x$. Since $\eta$ is an alternating $n$-linear form, this equals multiplication by $\det(A)^{-1}$. Hence $A$ fixes a nonzero volume form in $K_{X,x}$ if and only if $\det(A)=1$. Because $A$ is already unitary, this condition is exactly $A\in SU((T_X)_x,\omega_x)$. [/guided] [/step] [step:Construct a parallel holomorphic volume form from special unitary holonomy] Assume $\operatorname{Hol}_x(\omega)\subset SU((T_X)_x,\omega_x)$ for every $x\in X$. For each [connected component](/page/Connected%20Component) $C\subset X$, choose a point $x_C\in C$ and choose a nonzero element \begin{align*} \Omega_{x_C}\in K_{X,x_C}. \end{align*} For any point $y\in C$, define $\Omega_y\in K_{X,y}$ by parallel transporting $\Omega_{x_C}$ along a piecewise smooth path in $C$ from $x_C$ to $y$ using the connection induced by $\nabla$ on $K_X$. This definition is independent of the chosen path in $C$. Indeed, if two piecewise smooth paths from $x_C$ to $y$ are chosen, then following the first path and returning along the second gives a loop based at $x_C$. The two transported values therefore differ by the induced holonomy action on $K_{X,x_C}$. Step 1 shows that this induced holonomy action is trivial because every tangent holonomy transformation has determinant $1$. Performing this construction on every connected component, the fibrewise sections assemble to a globally defined smooth parallel section $\Omega$ of $K_X$; smoothness follows because parallel transport depends smoothly on the endpoint along smoothly varying local paths. Since $\nabla\Omega=0$, the section $\Omega$ has constant pointwise norm with respect to $\omega$ on each connected component of $X$. Also, the Kähler hypothesis gives $\nabla J=0$, so the Levi-Civita connection preserves the holomorphic tangent bundle $T_X$ and agrees there with the Hermitian Chern connection determined by the holomorphic structure and the metric induced by $\omega$. Passing to determinants, the induced connection on $K_X=\det(T_X^*)$ is the Hermitian Chern connection of the holomorphic line bundle $K_X$, so its $(0,1)$ part is the Dolbeault operator $\bar\partial_{K_X}$. Hence \begin{align*} \bar\partial_{K_X}\Omega=0 \end{align*} and $\Omega$ is holomorphic. Since parallel transport is an isomorphism on each fibre and $\Omega_{x_C}\neq 0$, the section $\Omega$ is nowhere vanishing. Therefore $K_X$ is holomorphically trivial, and condition 1 holds. [/step] [step:Show that a constant-norm holomorphic volume form is parallel] Assume condition 1. Let $H^0(X,K_X)$ denote the vector space of holomorphic global sections of $K_X$. Choose a nowhere-vanishing holomorphic section \begin{align*} \Omega\in H^0(X,K_X) \end{align*} whose pointwise norm $|\Omega|_\omega$ is constant. Let $\nabla^{K}$ denote the Hermitian Chern connection on $K_X$ induced by the Kähler metric. We prove that $\nabla^{K}\Omega=0$. Let $U\subset X$ be a coordinate neighbourhood with holomorphic coordinates $(z_1,\dots,z_n)$, and define the local holomorphic frame \begin{align*} e:U \to K_X|_U,\quad p \mapsto dz_1\wedge\dots\wedge dz_n|_p \end{align*} of $K_X|_U$. Since $\Omega$ is nowhere vanishing and holomorphic, there is a nowhere-vanishing [holomorphic function](/page/Holomorphic%20Function) \begin{align*} f:U\to\mathbb C \end{align*} such that \begin{align*} \Omega=f e. \end{align*} Define the local Hermitian metric coefficient by \begin{align*} h:U\to (0,\infty),\quad p\mapsto |e(p)|_\omega^2. \end{align*} The $(1,0)$ part of the Chern connection in this holomorphic frame satisfies \begin{align*} (\nabla^{K})^{1,0}(f e)=\partial f\otimes e+f\,\partial\log h\otimes e. \end{align*} The $(0,1)$ part is $\bar\partial_{K_X}$, and it vanishes on $\Omega$ because $\Omega$ is holomorphic. Because $f$ is nowhere zero, \begin{align*} \nabla^{K}\Omega=f\,\partial\log(h|f|^2)\otimes e. \end{align*} But \begin{align*} h|f|^2=|\Omega|_\omega^2 \end{align*} is constant on $U$ by hypothesis, so \begin{align*} \partial\log(h|f|^2)=0. \end{align*} Thus $\nabla^{K}\Omega=0$ on $U$. Since the coordinate neighbourhood $U$ was arbitrary, $\Omega$ is parallel on all of $X$. [guided] The goal is to turn “holomorphic and constant norm” into “parallel.” For a Hermitian holomorphic line bundle, this is a local computation with the Chern connection. Choose holomorphic coordinates $(z_1,\dots,z_n)$ on an [open set](/page/Open%20Set) $U\subset X$ contained in one connected component of $X$. They define the holomorphic frame $e:U\to K_X|_U$ by \begin{align*} e(p)=dz_1\wedge\dots\wedge dz_n|_p \end{align*} for $p\in U$. Because $\Omega$ is a nowhere-vanishing holomorphic section of $K_X$, there is a nowhere-vanishing holomorphic function \begin{align*} f:U&\to\mathbb C \end{align*} with \begin{align*} \Omega=f e. \end{align*} Let the Hermitian metric coefficient of $e$ be the function $h:U\to (0,\infty)$ defined by \begin{align*} h(p)=|e(p)|_\omega^2 \end{align*} for $p\in U$. Then the pointwise norm of $\Omega$ is \begin{align*} |\Omega|_\omega^2=h|f|^2. \end{align*} In a holomorphic frame of a Hermitian line bundle, the $(1,0)$ part of the Chern connection has connection form $\partial\log h$, while its $(0,1)$ part is the Dolbeault operator. Since both $e$ and $f$ are holomorphic, $\bar\partial_{K_X}(f e)=0$. For the $(1,0)$ part, \begin{align*} (\nabla^{K})^{1,0}(f e)=\partial f\otimes e+f\,\partial\log h\otimes e. \end{align*} Since $f$ is holomorphic and nowhere zero, $\partial f=f\,\partial\log f$. Also $\partial\log |f|^2=\partial\log f$, because $\partial\overline{f}=0$ for holomorphic $f$. Hence \begin{align*} \nabla^{K}\Omega=f\,\partial\log(h|f|^2)\otimes e. \end{align*} The constant-norm hypothesis says that $h|f|^2$ is constant on $U$. Therefore \begin{align*} \partial\log(h|f|^2)=0, \end{align*} and consequently \begin{align*} \nabla^{K}\Omega=0 \end{align*} on $U$. Since these coordinate neighbourhoods cover $X$, the section $\Omega$ is parallel globally. [/guided] [/step] [step:Use the parallel volume form to force determinant one holonomy] Let $x\in X$ and let $A\in\operatorname{Hol}_x(\omega)$. By Step 3, $\Omega$ is parallel, so parallel transport around any loop based at $x$ fixes the fibre element \begin{align*} \Omega_x\in K_{X,x}. \end{align*} Since $\Omega$ is nowhere vanishing, $\Omega_x\neq 0$. Step 1 identifies the induced action of $A$ on $K_{X,x}$ as multiplication by $\det(A)^{-1}$. Therefore \begin{align*} \det(A)^{-1}\Omega_x=\Omega_x. \end{align*} Since $\Omega_x\neq 0$, this gives $\det(A)=1$. As $A$ was arbitrary and Kähler holonomy is unitary, every element of $\operatorname{Hol}_x(\omega)$ lies in $SU((T_X)_x,\omega_x)$. Hence condition 2 holds. [/step] [step:Conclude the equivalence] Step 2 proves that condition 2 implies condition 1. Steps 3 and 4 prove that condition 1 implies condition 2. Therefore the two stated conditions are equivalent. [/step]