[proofplan] We prove the result locally at an arbitrary point of $U$. In a holomorphic coordinate on the curve and normal holomorphic coordinates on the target, the pullback metric has coefficient equal to the squared norm of the velocity vector of the holomorphic curve. Differentiating the logarithm of this coefficient gives the Gaussian curvature of the pullback metric; the Kähler normal-coordinate computation separates this curvature into the ambient holomorphic sectional curvature minus a nonnegative second fundamental term. The assumed bound on the ambient holomorphic sectional curvature then gives the required upper bound. [/proofplan]

[step:Write the pullback metric in a holomorphic coordinate on the curve]Fix $p \in U$. Since $df_p \neq 0$ and nonvanishing of $df$ is assumed on all of $U$, the pullback form $f^*\omega_Y$ is positive on $T_p\Sigma$ and hence defines a smooth conformal metric near $p$. Choose a holomorphic coordinate chart $(V,z)$ on $\Sigma$ with $p \in V \subset U$ and $z(p)=0$. Choose holomorphic coordinates $(W,w_1,\dots,w_m)$ on $Y$ around $f(p)$ such that $f(V) \subset W$ after shrinking $V$ if necessary. Write the coordinate representation of $f$ as the holomorphic map \begin{align*} F: z(V) &\to \mathbb{C}^m, \\ \zeta &\mapsto (F_1(\zeta),\dots,F_m(\zeta)). \end{align*} Let the Kähler metric coefficients of $\omega_Y$ in the coordinates $w=(w_1,\dots,w_m)$ be the smooth functions \begin{align*} g_{\alpha\bar\beta}: W &\to \mathbb{C}, \qquad 1 \leq \alpha,\beta \leq m, \end{align*} so that \begin{align*} \omega_Y = i\sum_{\alpha,\beta=1}^m g_{\alpha\bar\beta}(w)\,dw_\alpha\wedge d\bar w_\beta. \end{align*} Define the smooth positive function \begin{align*} \lambda: z(V) &\to (0,\infty), \\ \zeta &\mapsto \sum_{\alpha,\beta=1}^m g_{\alpha\bar\beta}(F(\zeta))\, \frac{\partial F_\alpha}{\partial z}(\zeta)\, \overline{\frac{\partial F_\beta}{\partial z}(\zeta)}. \end{align*} Then on $V$, \begin{align*} f^*\omega_Y = i\,\lambda(z)\,dz\wedge d\bar z. \end{align*}[/step]

[guided]We first reduce the statement to a calculation in one complex variable. The assumption $df_q \neq 0$ for every $q \in U$ means that, at each point of $U$, the complex-[linear map](/page/Linear%20Map) $df_q:T_q^{1,0}\Sigma \to T_{f(q)}^{1,0}Y$ is nonzero. Since $T_q^{1,0}\Sigma$ is one-dimensional, this implies that $f^*\omega_Y$ is positive on nonzero tangent vectors at $q$. Therefore $f^*\omega_Y$ is a smooth Kähler metric on $U$. Fix $p \in U$, choose a holomorphic coordinate $z$ on $\Sigma$ with $z(p)=0$, and choose holomorphic coordinates $w_1,\dots,w_m$ on $Y$ near $f(p)$. After shrinking the coordinate neighbourhood of $p$, the map $f$ is represented by a holomorphic map \begin{align*} F: z(V) &\to \mathbb{C}^m, \\ \zeta &\mapsto (F_1(\zeta),\dots,F_m(\zeta)). \end{align*} In these coordinates the target metric is written as \begin{align*} \omega_Y = i\sum_{\alpha,\beta=1}^m g_{\alpha\bar\beta}(w)\,dw_\alpha\wedge d\bar w_\beta, \end{align*} where each $g_{\alpha\bar\beta}:W\to\mathbb{C}$ is smooth and the Hermitian matrix $(g_{\alpha\bar\beta})$ is positive definite. Pulling back $dw_\alpha$ gives \begin{align*} f^*(dw_\alpha)=\frac{\partial F_\alpha}{\partial z}\,dz, \end{align*} because $F_\alpha$ is holomorphic. Hence \begin{align*} f^*\omega_Y = i\sum_{\alpha,\beta=1}^m g_{\alpha\bar\beta}(F(z)) \frac{\partial F_\alpha}{\partial z} \overline{\frac{\partial F_\beta}{\partial z}} \,dz\wedge d\bar z. \end{align*} Thus the pullback metric has the conformal coefficient \begin{align*} \lambda(\zeta) = \sum_{\alpha,\beta=1}^m g_{\alpha\bar\beta}(F(\zeta))\, \frac{\partial F_\alpha}{\partial z}(\zeta)\, \overline{\frac{\partial F_\beta}{\partial z}(\zeta)}. \end{align*} The positivity of $\lambda$ follows from positive definiteness of $(g_{\alpha\bar\beta})$ and from $df \neq 0$ on $U$.[/guided]

[step:Choose normal target coordinates adapted to the tangent direction]It remains to prove the curvature bound at $p$. Choose Kähler normal holomorphic coordinates at $f(p)$ such that \begin{align*} g_{\alpha\bar\beta}(f(p)) &= \delta_{\alpha\beta}, \\ \frac{\partial g_{\alpha\bar\beta}}{\partial w_\gamma}(f(p)) &= 0, \\ \frac{\partial g_{\alpha\bar\beta}}{\partial \bar w_\gamma}(f(p)) &= 0 \end{align*} for all indices $\alpha,\beta,\gamma$, and such that \begin{align*} \frac{\partial F_1}{\partial z}(0) = a>0, \qquad \frac{\partial F_\alpha}{\partial z}(0)=0 \quad\text{for }2\leq \alpha\leq m. \end{align*} Here the last condition is achieved by a unitary change of the [normal coordinates](/theorems/2713), and $a = |df_p(\partial/\partial z)|_{\omega_Y}$. Define the complex numbers \begin{align*} b_\alpha := \frac{\partial^2 F_\alpha}{\partial z^2}(0), \qquad 1\leq \alpha\leq m. \end{align*} With these coordinates, \begin{align*} \lambda(0)=a^2. \end{align*}[/step]

[guided]We now choose coordinates that make the curvature calculation at $p$ as simple as possible. Kähler normal holomorphic coordinates at $f(p)$ mean that, at the point $f(p)$, the Hermitian matrix of metric coefficients is the identity and all first derivatives of those coefficients vanish: \begin{align*} g_{\alpha\bar\beta}(f(p)) &= \delta_{\alpha\beta}, \\ \frac{\partial g_{\alpha\bar\beta}}{\partial w_\gamma}(f(p)) &= 0, \\ \frac{\partial g_{\alpha\bar\beta}}{\partial \bar w_\gamma}(f(p)) &= 0. \end{align*} These identities are exactly what will remove the first-order metric terms when differentiating the pullback coefficient $\lambda$. Since $df_p\neq 0$, the vector $df_p(\partial/\partial z)\in T^{1,0}_{f(p)}Y$ is nonzero. After the normal coordinates are chosen, a unitary change of the coordinate frame preserves the identities above and sends this nonzero vector to the first coordinate direction. Thus we may arrange \begin{align*} \frac{\partial F_1}{\partial z}(0) = a>0, \qquad \frac{\partial F_\alpha}{\partial z}(0)=0 \quad\text{for }2\leq \alpha\leq m, \end{align*} where $a$ is the norm of the velocity vector: \begin{align*} a = \left|df_p\left(\frac{\partial}{\partial z}\right)\right|_{\omega_Y}. \end{align*} We also introduce the complex numbers \begin{align*} b_\alpha := \frac{\partial^2 F_\alpha}{\partial z^2}(0), \qquad 1\leq \alpha\leq m. \end{align*} These are the coordinate components of the second derivative of the holomorphic curve at $p$. Substituting the normalized first derivative into the definition of $\lambda$ gives \begin{align*} \lambda(0) &= \sum_{\alpha,\beta=1}^m \delta_{\alpha\beta} \frac{\partial F_\alpha}{\partial z}(0) \overline{\frac{\partial F_\beta}{\partial z}(0)} = |a|^2 = a^2. \end{align*}[/guided]

[step:Compute the logarithmic curvature of the pullback metric]For the conformal Kähler metric \begin{align*} i\,\lambda(z)\,dz\wedge d\bar z \end{align*} on a Riemann surface, the associated Riemannian metric is \begin{align*} 2\lambda(z)\,|dz|^2. \end{align*} The constant factor $2$ disappears after applying $\partial_z\partial_{\bar z}\log$, and the standard local formula for Gaussian curvature of a conformal metric gives \begin{align*} K_{f^*\omega_Y}(p) = -\frac{1}{\lambda(0)} \frac{\partial^2}{\partial z\,\partial\bar z} \log \lambda(0). \end{align*} Using the normal-coordinate identities from the previous step and holomorphicity of $F$, we compute \begin{align*} \frac{\partial \lambda}{\partial z}(0) &= \sum_{\alpha=1}^m b_\alpha\,\overline{\frac{\partial F_\alpha}{\partial z}(0)} = b_1\bar a, \\ \frac{\partial \lambda}{\partial \bar z}(0) &= a\bar b_1. \end{align*} Next, \begin{align*} \frac{\partial^2 \lambda}{\partial z\,\partial\bar z}(0) = \sum_{\alpha=1}^m |b_\alpha|^2 + a^4 \frac{\partial^2 g_{1\bar 1}}{\partial w_1\,\partial \bar w_1}(f(p)). \end{align*} Therefore \begin{align*} \frac{\partial^2}{\partial z\,\partial\bar z}\log \lambda(0) &= \frac{1}{\lambda(0)} \frac{\partial^2\lambda}{\partial z\,\partial\bar z}(0) - \frac{1}{\lambda(0)^2} \frac{\partial\lambda}{\partial z}(0) \frac{\partial\lambda}{\partial\bar z}(0) \\ &= \frac{1}{a^2} \left( \sum_{\alpha=1}^m |b_\alpha|^2 + a^4 \frac{\partial^2 g_{1\bar 1}}{\partial w_1\,\partial \bar w_1}(f(p)) \right) - \frac{1}{a^4}|b_1|^2a^2 \\ &= \frac{1}{a^2}\sum_{\alpha=2}^m |b_\alpha|^2 + a^2 \frac{\partial^2 g_{1\bar 1}}{\partial w_1\,\partial \bar w_1}(f(p)). \end{align*} Thus \begin{align*} K_{f^*\omega_Y}(p) = -\frac{1}{a^4}\sum_{\alpha=2}^m |b_\alpha|^2 - \frac{\partial^2 g_{1\bar 1}}{\partial w_1\,\partial \bar w_1}(f(p)). \end{align*}[/step]

[guided]The curvature of a conformal Kähler metric on a Riemann surface is obtained by differentiating the logarithm of its conformal coefficient. The Kähler form \begin{align*} i\,\lambda(z)\,dz\wedge d\bar z \end{align*} corresponds to the Riemannian metric \begin{align*} 2\lambda(z)\,|dz|^2. \end{align*} The usual conformal curvature formula for $e^{2u}|dz|^2$ is $K=-e^{-2u}\Delta u$, and with $e^{2u}=2\lambda$ and $\Delta=4\partial_z\partial_{\bar z}$ this becomes \begin{align*} K_{f^*\omega_Y}(p) = -\frac{1}{\lambda(0)} \frac{\partial^2}{\partial z\,\partial\bar z} \log \lambda(0), \end{align*} because the constant factor $2$ has zero logarithmic derivative. We now compute the two derivatives of $\lambda$. The normal-coordinate identities eliminate all first derivatives of the target metric coefficients at $f(p)$. Since $F$ is holomorphic, the functions $\partial F_\alpha/\partial z$ depend holomorphically on $z$, and their complex conjugates depend anti-holomorphically on $\bar z$. Therefore, at $z=0$, \begin{align*} \frac{\partial \lambda}{\partial z}(0) &= \sum_{\alpha=1}^m \frac{\partial^2F_\alpha}{\partial z^2}(0) \overline{\frac{\partial F_\alpha}{\partial z}(0)} = b_1\bar a, \\ \frac{\partial \lambda}{\partial \bar z}(0) &= \sum_{\alpha=1}^m \frac{\partial F_\alpha}{\partial z}(0) \overline{\frac{\partial^2F_\alpha}{\partial z^2}(0)} = a\bar b_1. \end{align*} For the mixed derivative, two types of terms remain. One differentiates the velocity vector of the curve and gives the squared norm of the second derivative; the other differentiates the target metric coefficient twice and records the ambient curvature contribution: \begin{align*} \frac{\partial^2 \lambda}{\partial z\,\partial\bar z}(0) = \sum_{\alpha=1}^m |b_\alpha|^2 + a^4 \frac{\partial^2 g_{1\bar 1}}{\partial w_1\,\partial \bar w_1}(f(p)). \end{align*} The logarithmic derivative formula gives \begin{align*} \frac{\partial^2}{\partial z\,\partial\bar z}\log \lambda(0) &= \frac{1}{\lambda(0)} \frac{\partial^2\lambda}{\partial z\,\partial\bar z}(0) - \frac{1}{\lambda(0)^2} \frac{\partial\lambda}{\partial z}(0) \frac{\partial\lambda}{\partial\bar z}(0). \end{align*} Since $\lambda(0)=a^2$, substituting the previous expressions yields \begin{align*} \frac{\partial^2}{\partial z\,\partial\bar z}\log \lambda(0) &= \frac{1}{a^2} \left( \sum_{\alpha=1}^m |b_\alpha|^2 + a^4 \frac{\partial^2 g_{1\bar 1}}{\partial w_1\,\partial \bar w_1}(f(p)) \right) - \frac{1}{a^4}|b_1|^2a^2 \\ &= \frac{1}{a^2}\sum_{\alpha=2}^m |b_\alpha|^2 + a^2 \frac{\partial^2 g_{1\bar 1}}{\partial w_1\,\partial \bar w_1}(f(p)). \end{align*} Multiplying by $-1/\lambda(0)=-1/a^2$ gives \begin{align*} K_{f^*\omega_Y}(p) = -\frac{1}{a^4}\sum_{\alpha=2}^m |b_\alpha|^2 - \frac{\partial^2 g_{1\bar 1}}{\partial w_1\,\partial \bar w_1}(f(p)). \end{align*} The first term is nonpositive. It is the extrinsic bending term of the holomorphic curve inside $Y$.[/guided]

[step:Identify the ambient term with holomorphic sectional curvature] In Kähler coordinates, the curvature tensor components are \begin{align*} R_{\alpha\bar\beta\gamma\bar\delta} = -\frac{\partial^2 g_{\alpha\bar\beta}}{\partial w_\gamma\,\partial \bar w_\delta} + \sum_{\mu,\nu=1}^m g^{\mu\bar\nu} \frac{\partial g_{\alpha\bar\nu}}{\partial w_\gamma} \frac{\partial g_{\mu\bar\beta}}{\partial \bar w_\delta}, \end{align*} where $(g^{\mu\bar\nu})$ denotes the inverse Hermitian matrix to $(g_{\mu\bar\nu})$. In Kähler normal coordinates at $f(p)$, all first derivatives of $g_{\alpha\bar\beta}$ vanish at $f(p)$, so the quadratic first-derivative term is zero there. Hence the curvature tensor component in the complex line spanned by $\partial/\partial w_1$ is \begin{align*} R_{1\bar 1 1\bar 1}(f(p)) = - \frac{\partial^2 g_{1\bar 1}}{\partial w_1\,\partial \bar w_1}(f(p)). \end{align*} Because $g_{1\bar 1}(f(p))=1$ and $g_{\alpha\bar\beta}(f(p))=\delta_{\alpha\beta}$, the holomorphic sectional curvature in the nonzero direction \begin{align*} \xi := df_p\left(\frac{\partial}{\partial z}\right) = a\frac{\partial}{\partial w_1}\bigg|_{f(p)} \end{align*} is \begin{align*} H_{\omega_Y}(\xi) = R_{1\bar 1 1\bar 1}(f(p)) = - \frac{\partial^2 g_{1\bar 1}}{\partial w_1\,\partial \bar w_1}(f(p)). \end{align*} Thus the previous step gives \begin{align*} K_{f^*\omega_Y}(p) = H_{\omega_Y}(\xi) - \frac{1}{a^4}\sum_{\alpha=2}^m |b_\alpha|^2 \leq H_{\omega_Y}(\xi). \end{align*} By hypothesis, $H_{\omega_Y}(\xi)\leq -\kappa$ because $\xi\neq 0$. Hence \begin{align*} K_{f^*\omega_Y}(p)\leq -\kappa. \end{align*} Since $p\in U$ was arbitrary, the inequality holds everywhere on $U$. [/step]