[proofplan] We remove from the dual projective space precisely the hyperplanes that are tangent to $X$ at some point. These tangent hyperplanes form the image of an algebraic incidence variety whose fibers over $X$ have dimension $N-n-1$, so the image has dimension at most $N-1$ inside the $N$-dimensional dual projective space. Therefore its complement is a nonempty Zariski open dense set. A hyperplane in this complement meets $X$ transversely at every intersection point, and the holomorphic [implicit function theorem](/page/Implicit%20Function%20Theorem) then gives smoothness and the expected dimension. [/proofplan] [step:Define the incidence variety of tangent hyperplanes] Let $V := \mathbb{C}^{N+1}$, so that $\mathbb{P}^N_{\mathbb{C}} = \mathbb{P}(V)$ and $(\mathbb{P}^N_{\mathbb{C}})^* = \mathbb{P}(V^*)$. For $x = [v] \in X$, let \begin{align*} \widehat{T}_xX \subset V \end{align*} denote the affine cone over the embedded projective tangent space $T_xX \subset T_x\mathbb{P}^N_{\mathbb{C}}$. Since $X$ is smooth of dimension $n$, the [vector space](/page/Vector%20Space) $\widehat{T}_xX$ has complex dimension $n+1$. Define the incidence subset \begin{align*} \mathcal{I} := \{(x,[\ell]) \in X \times \mathbb{P}(V^*) : \ell(w)=0 \text{ for every } w \in \widehat{T}_xX\}. \end{align*} Thus $[\ell]$ lies over $x$ exactly when the hyperplane $H_\ell$ contains the embedded tangent space to $X$ at $x$, equivalently when $H_\ell$ is tangent to $X$ at $x$. The assignment $x \mapsto \widehat{T}_xX$ is algebraic because $X \subset \mathbb{P}^N_{\mathbb{C}}$ is a smooth projective submanifold and the embedded tangent spaces are cut out locally by the Jacobian matrix of local defining equations for $X$. Hence $\mathcal{I}$ is a closed algebraic subvariety of $X \times \mathbb{P}(V^*)$. [guided] We want to isolate the bad hyperplanes, meaning the hyperplanes that fail to meet $X$ transversely. A hyperplane $H_\ell$ fails transversality at a point $x \in X \cap H_\ell$ exactly when its tangent space contains the tangent space of $X$ at $x$. In projective language this is the condition that the linear form $\ell$ vanish on the affine tangent cone $\widehat{T}_xX \subset V$. Let $V := \mathbb{C}^{N+1}$. Then points of $\mathbb{P}^N_{\mathbb{C}}$ are one-dimensional subspaces $[v]$ of $V$, and hyperplanes are parametrized by one-dimensional classes $[\ell]$ of nonzero linear forms $\ell \in V^*$. For a fixed point $x=[v] \in X$, the embedded projective tangent space $T_xX$ lifts to a vector subspace \begin{align*} \widehat{T}_xX \subset V. \end{align*} Because $X$ is smooth of complex dimension $n$, this lifted tangent space has dimension $n+1$: it contains the line corresponding to the point $x$ and the $n$ tangent directions of $X$ at $x$. We therefore define \begin{align*} \mathcal{I} := \{(x,[\ell]) \in X \times \mathbb{P}(V^*) : \ell(w)=0 \text{ for every } w \in \widehat{T}_xX\}. \end{align*} This is the incidence variety of tangent hyperplanes. The condition is algebraic in $x$ and $[\ell]$: locally, the tangent space of $X$ is described by the kernel of a Jacobian matrix of local defining equations, and requiring $\ell$ to vanish on that kernel is equivalent to polynomial rank conditions. Thus $\mathcal{I}$ is a closed algebraic subvariety of $X \times \mathbb{P}(V^*)$. [/guided] [/step] [step:Compute the dimension of the incidence variety] Let \begin{align*} p: \mathcal{I} &\to X \\ (x,[\ell]) &\mapsto x \end{align*} be the first projection. For a fixed $x \in X$, the fiber $p^{-1}(x)$ is the projective space of nonzero linear forms on $V$ that vanish on $\widehat{T}_xX$. Since $\dim_{\mathbb{C}} V = N+1$ and $\dim_{\mathbb{C}} \widehat{T}_xX = n+1$, the annihilator \begin{align*} (\widehat{T}_xX)^\perp := \{\ell \in V^* : \ell(w)=0 \text{ for every } w \in \widehat{T}_xX\} \end{align*} has dimension $N-n$. Hence \begin{align*} p^{-1}(x) \cong \mathbb{P}^{N-n-1}_{\mathbb{C}} \end{align*} whenever $n<N$. Therefore \begin{align*} \dim_{\mathbb{C}} \mathcal{I} = \dim_{\mathbb{C}} X + \dim_{\mathbb{C}} \mathbb{P}^{N-n-1}_{\mathbb{C}} = n + (N-n-1) = N-1. \end{align*} If $n=N$, then $X=\mathbb{P}^N_{\mathbb{C}}$ because $X$ is a closed complex submanifold of the connected manifold $\mathbb{P}^N_{\mathbb{C}}$ with the same dimension. In that case every hyperplane section is itself a hyperplane $\mathbb{P}^{N-1}_{\mathbb{C}}$, so the theorem holds with $U=(\mathbb{P}^N_{\mathbb{C}})^*$. [guided] We now count how many tangent hyperplanes there can be. Define the first projection \begin{align*} p: \mathcal{I} &\to X \\ (x,[\ell]) &\mapsto x. \end{align*} Fix $x \in X$. The fiber $p^{-1}(x)$ consists of all projective classes $[\ell]$ such that $\ell$ vanishes on the vector space $\widehat{T}_xX$. This is the projectivization of the annihilator \begin{align*} (\widehat{T}_xX)^\perp := \{\ell \in V^* : \ell(w)=0 \text{ for every } w \in \widehat{T}_xX\}. \end{align*} The dimension of this annihilator is computed by linear algebra. Since $\dim_{\mathbb{C}} V=N+1$ and $\dim_{\mathbb{C}}\widehat{T}_xX=n+1$, the quotient $V/\widehat{T}_xX$ has dimension $N-n$, and its dual identifies with $(\widehat{T}_xX)^\perp$. Therefore \begin{align*} \dim_{\mathbb{C}}(\widehat{T}_xX)^\perp = N-n. \end{align*} After projectivizing, the fiber has dimension $N-n-1$: \begin{align*} p^{-1}(x) \cong \mathbb{P}^{N-n-1}_{\mathbb{C}}. \end{align*} Since $X$ has dimension $n$, the total incidence variety has dimension \begin{align*} \dim_{\mathbb{C}} \mathcal{I} = n + (N-n-1) = N-1. \end{align*} This is the key dimension drop: the dual projective space has dimension $N$, while the family of tangent hyperplanes has dimension at most $N-1$. There is one boundary case. If $n=N$, then $X$ is a closed complex submanifold of $\mathbb{P}^N_{\mathbb{C}}$ of full dimension. Since $\mathbb{P}^N_{\mathbb{C}}$ is connected, this forces $X=\mathbb{P}^N_{\mathbb{C}}$. Every hyperplane section is then a copy of $\mathbb{P}^{N-1}_{\mathbb{C}}$, so the theorem is immediate with $U=(\mathbb{P}^N_{\mathbb{C}})^*$. [/guided] [/step] [step:Remove the proper closed set of tangent hyperplanes] Assume now that $n<N$. Let \begin{align*} q: \mathcal{I} &\to \mathbb{P}(V^*) \\ (x,[\ell]) &\mapsto [\ell] \end{align*} be the second projection, and define \begin{align*} X^* := q(\mathcal{I}) \subset \mathbb{P}(V^*). \end{align*} Because $X$ is projective, $\mathcal{I}$ is projective, and the image of a projective variety under a morphism is Zariski closed (citing a result not yet in the wiki: proper image theorem for projective morphisms). Moreover, \begin{align*} \dim_{\mathbb{C}} X^* \leq \dim_{\mathbb{C}} \mathcal{I}=N-1. \end{align*} Since $\dim_{\mathbb{C}}\mathbb{P}(V^*)=N$, the subset $X^*$ is a proper Zariski closed subset of $\mathbb{P}(V^*)$. Define \begin{align*} U := \mathbb{P}(V^*) \setminus X^*. \end{align*} Then $U$ is Zariski open and dense in $\mathbb{P}(V^*)$. [guided] The bad hyperplanes are exactly the hyperplanes tangent to $X$ somewhere. We collect them by projecting the incidence variety to the second factor. Define \begin{align*} q: \mathcal{I} &\to \mathbb{P}(V^*) \\ (x,[\ell]) &\mapsto [\ell], \end{align*} and set \begin{align*} X^* := q(\mathcal{I}). \end{align*} This subset $X^*$ is the dual variety of tangent hyperplanes to $X$. We need two facts about $X^*$. First, it is Zariski closed. This follows because $X$ is projective, hence $\mathcal{I}$ is projective, and the image of a projective variety under a morphism is Zariski closed (citing a result not yet in the wiki: proper image theorem for projective morphisms). Second, it is proper. The dimension estimate from the previous step gives \begin{align*} \dim_{\mathbb{C}} X^* \leq \dim_{\mathbb{C}} \mathcal{I} = N-1. \end{align*} But the whole dual projective space has dimension \begin{align*} \dim_{\mathbb{C}}\mathbb{P}(V^*)=N. \end{align*} Therefore $X^*$ cannot equal $\mathbb{P}(V^*)$. Now define \begin{align*} U := \mathbb{P}(V^*) \setminus X^*. \end{align*} Since $X^*$ is a proper Zariski closed subset, its complement $U$ is nonempty, Zariski open, and dense. [/guided] [/step] [step:Show that hyperplanes in the open set meet $X$ transversely] Let $[\ell] \in U$. Suppose $x \in X \cap H_\ell$. Since $[\ell] \notin X^*=q(\mathcal{I})$, the pair $(x,[\ell])$ is not in $\mathcal{I}$. Therefore $\ell$ does not vanish on all of $\widehat{T}_xX$. Equivalently, the differential of the restricted linear equation is nonzero on $T_xX$. More precisely, choose a local holomorphic chart \begin{align*} \varphi: W \to \varphi(W) \subset \mathbb{C}^n \end{align*} for $X$ near $x$, and choose a homogeneous coordinate chart of $\mathbb{P}^N_{\mathbb{C}}$ containing $x$ in which $H_\ell$ is represented by a [holomorphic function](/page/Holomorphic%20Function) \begin{align*} f_\ell: \Omega \to \mathbb{C}. \end{align*} Then the restricted function \begin{align*} g_\ell: \varphi(W) &\to \mathbb{C} \\ y &\mapsto f_\ell(\varphi^{-1}(y)) \end{align*} satisfies $g_\ell(\varphi(x))=0$ and has nonzero differential at $\varphi(x)$. Hence $0$ is a regular value of the local defining function for $X \cap H_\ell$. [guided] Take $[\ell] \in U$ and let $x \in X \cap H_\ell$. Since $[\ell]$ is outside the image $X^*$, it is not tangent to $X$ at any point. In particular, the pair $(x,[\ell])$ is not in the incidence variety $\mathcal{I}$. By the definition of $\mathcal{I}$, this means that $\ell$ does not vanish on the whole lifted tangent space $\widehat{T}_xX$. We now translate that projective statement into a local analytic statement. Choose a holomorphic coordinate chart \begin{align*} \varphi: W \to \varphi(W) \subset \mathbb{C}^n \end{align*} on $X$ with $x \in W$. Also choose a homogeneous coordinate chart of $\mathbb{P}^N_{\mathbb{C}}$ containing $x$, in which the hyperplane equation $\ell=0$ is represented by a holomorphic function \begin{align*} f_\ell: \Omega \to \mathbb{C}. \end{align*} Restricting this function to $X$ through the chart gives \begin{align*} g_\ell: \varphi(W) &\to \mathbb{C} \\ y &\mapsto f_\ell(\varphi^{-1}(y)). \end{align*} The condition $x \in H_\ell$ gives $g_\ell(\varphi(x))=0$. The condition that $H_\ell$ is not tangent to $X$ at $x$ says exactly that the differential of $g_\ell$ at $\varphi(x)$ is not the zero [linear map](/page/Linear%20Map). Thus $0$ is a regular value of this local defining function. [/guided] [/step] [step:Apply the holomorphic implicit function theorem to obtain smooth hyperplane sections] By the holomorphic [implicit function theorem](/theorems/52) (citing a result not yet in the wiki: holomorphic implicit function theorem), the zero set \begin{align*} g_\ell^{-1}(\{0\}) \subset \varphi(W) \end{align*} is, near $\varphi(x)$, a smooth complex submanifold of dimension $n-1$ whenever $n\geq 1$. Since this holds at every point $x \in X \cap H_\ell$, the section $X \cap H_\ell$ is a smooth complex manifold of dimension $n-1$ if it is nonempty and $n\geq 1$. The set $X \cap H_\ell$ is closed in the projective variety $X$, hence projective. Therefore, for every $[\ell] \in U$, the hyperplane section $X \cap H_\ell$ is either empty or a smooth complex projective manifold of dimension $n-1$ when $n\geq 1$. If $n=0$, then $X$ is finite. The set of hyperplanes meeting $X$ is a finite union of proper hyperplanes in $\mathbb{P}(V^*)$, so after replacing $U$ by the nonempty Zariski open [dense subset](/page/Dense%20Subset) of hyperplanes avoiding $X$, every section $X \cap H_\ell$ is empty. This completes the proof. [/step]