[proofplan] The strategy is to reduce the implicit equation $\Psi(x) = 0$ to a graph equation $\tilde{x}_n = \gamma(\tilde{x}')$ by two steps. First, we perform a rigid rotation that aligns $\nabla \Psi(z)$ with the $n$-th coordinate axis, transforming the problem into one where the partial derivative with respect to the last variable is nonzero. Second, we apply the [Implicit Function Theorem](/theorems/52) to solve for the last coordinate as a $C^k$ function of the remaining $n-1$ coordinates. The local identification of $\partial U \cap B(z,r)$ with the graph of this function then follows directly. [/proofplan] [step:Rotate coordinates to align $\nabla \Psi(z)$ with the $n$-th axis] Since $z \in \partial U$ and $\nabla \Psi(x) \neq 0$ for every $x \in \partial U$, the gradient $\nabla \Psi(z) \in \mathbb{R}^n$ is a nonzero vector. We choose an orthogonal matrix $R \in O(n)$ whose $n$-th row is the unit vector $\nabla \Psi(z) / |\nabla \Psi(z)|$. Such a matrix exists because any unit vector can be extended to an orthonormal basis of $\mathbb{R}^n$ (by Gram-Schmidt), and the matrix whose rows are this basis is orthogonal. Define the rotated and translated coordinates \begin{align*} \tilde{x} &:= R(x - z), \end{align*} and set \begin{align*} F: \mathbb{R}^n &\to \mathbb{R}, \\ \tilde{x} &\mapsto \Psi(R^{-1}\tilde{x} + z). \end{align*} Since $R$ is orthogonal and $\Psi$ is $C^k$, the composition $F$ is $C^k$. The point $z$ maps to the origin: $F(0) = \Psi(z) = 0$. By the chain rule, \begin{align*} \nabla F(0) &= R \, \nabla \Psi(z). \end{align*} The $n$-th component of $\nabla F(0)$ is the inner product of the $n$-th row of $R$ with $\nabla \Psi(z)$: \begin{align*} \frac{\partial F}{\partial \tilde{x}_n}(0) &= \frac{\nabla \Psi(z)}{|\nabla \Psi(z)|} \cdot \nabla \Psi(z) = |\nabla \Psi(z)| \neq 0. \end{align*} [guided] **Why do we rotate?** The [Implicit Function Theorem](/theorems/52) requires a nonzero partial derivative with respect to the variable we wish to solve for. In the original coordinates, there is no reason to expect $\partial \Psi / \partial x_n$ to be nonzero at $z$ — the gradient $\nabla \Psi(z)$ could point in any direction. The rotation remedies this: by aligning the $n$-th axis with the gradient direction, we guarantee that the partial derivative of $F$ with respect to the last variable is exactly $|\nabla \Psi(z)| > 0$. Concretely, define $\tilde{x} := R(x - z)$ where $R \in O(n)$ has $n$-th row equal to $\nabla \Psi(z) / |\nabla \Psi(z)|$. Such an $R$ exists: start with the unit vector $e := \nabla \Psi(z)/|\nabla \Psi(z)|$, extend it to an orthonormal basis $\{v_1, \dots, v_{n-1}, e\}$ of $\mathbb{R}^n$ via the Gram-Schmidt process, and take $R$ to be the matrix with rows $v_1, \dots, v_{n-1}, e$. Define the transformed defining function \begin{align*} F: \mathbb{R}^n &\to \mathbb{R}, \\ \tilde{x} &\mapsto \Psi(R^{-1}\tilde{x} + z). \end{align*} This is a $C^k$ function because $\Psi$ is $C^k$ and the map $\tilde{x} \mapsto R^{-1}\tilde{x} + z$ is a $C^\infty$ affine isomorphism. We verify the key properties: - **Zero condition.** $F(0) = \Psi(R^{-1} \cdot 0 + z) = \Psi(z) = 0$, since $z \in \partial U = \Psi^{-1}(0)$. - **Gradient computation.** By the chain rule applied to the composition $F = \Psi \circ L$ where $L(\tilde{x}) = R^{-1}\tilde{x} + z$ is affine with $DL = R^{-1} = R^\top$, \begin{align*} \nabla F(0) &= R \, \nabla \Psi(z). \end{align*} The $n$-th component is \begin{align*} \frac{\partial F}{\partial \tilde{x}_n}(0) &= (R \, \nabla \Psi(z))_n = \frac{\nabla \Psi(z)}{|\nabla \Psi(z)|} \cdot \nabla \Psi(z) = |\nabla \Psi(z)| > 0. \end{align*} This is the crucial nonvanishing condition that makes the Implicit Function Theorem applicable in the next step. [/guided] [/step] [step:Apply the Implicit Function Theorem to express $\tilde{x}_n$ as a function of $\tilde{x}'$] Write $\tilde{x} = (\tilde{x}', \tilde{x}_n)$ where $\tilde{x}' = (\tilde{x}_1, \dots, \tilde{x}_{n-1}) \in \mathbb{R}^{n-1}$. The function $F: \mathbb{R}^n \to \mathbb{R}$ satisfies: 1. $F(0) = 0$. 2. $F$ is $C^k$ with $k \ge 1$. 3. $\frac{\partial F}{\partial \tilde{x}_n}(0) \neq 0$. These are the hypotheses of the [Implicit Function Theorem](/theorems/52) (with $n-1$ playing the role of $n$ and $1$ playing the role of $m$ in the standard formulation). We conclude: there exist open neighborhoods $V \subset \mathbb{R}^{n-1}$ of $0$ and $W \subset \mathbb{R}$ of $0$, with $V \times W \subset \mathbb{R}^n$, and a unique $C^k$ function \begin{align*} \gamma: V &\to W, \\ \tilde{x}' &\mapsto \gamma(\tilde{x}') \end{align*} such that $\gamma(0) = 0$ and, for all $(\tilde{x}', \tilde{x}_n) \in V \times W$, \begin{align*} F(\tilde{x}', \tilde{x}_n) = 0 \quad &\Longleftrightarrow \quad \tilde{x}_n = \gamma(\tilde{x}'). \end{align*} [guided] **Verifying the hypotheses of the IFT.** The [Implicit Function Theorem](/theorems/52) requires three conditions for a $C^1$ map $F: U \to \mathbb{R}^m$ at a point $(x_0, y_0)$: (i) $F(x_0, y_0) = 0$, (ii) $F$ is at least $C^1$, and (iii) the partial derivative $D_y F_{(x_0, y_0)}: \mathbb{R}^m \to \mathbb{R}^m$ is an isomorphism. In our setting, $m = 1$, $x_0 = 0 \in \mathbb{R}^{n-1}$, $y_0 = 0 \in \mathbb{R}$, and we identify $y$ with $\tilde{x}_n$. We check each condition: 1. **Zero condition.** $F(0', 0) = F(0) = 0$, established in the previous step. 2. **Regularity.** $F$ is $C^k$ with $k \ge 1$, so in particular $F$ is $C^1$. 3. **Invertibility.** The partial derivative $D_{\tilde{x}_n} F_{(0', 0)}: \mathbb{R} \to \mathbb{R}$ is multiplication by $\frac{\partial F}{\partial \tilde{x}_n}(0) = |\nabla \Psi(z)| \neq 0$. A nonzero linear map $\mathbb{R} \to \mathbb{R}$ is an isomorphism. All hypotheses are satisfied. The IFT produces neighborhoods $V \subset \mathbb{R}^{n-1}$ of $0'$ and $W \subset \mathbb{R}$ of $0$, together with a unique $C^k$ function $\gamma: V \to W$ satisfying $\gamma(0) = 0$ and the equivalence $F(\tilde{x}', \tilde{x}_n) = 0 \Leftrightarrow \tilde{x}_n = \gamma(\tilde{x}')$ on $V \times W$. **Why $\gamma$ inherits $C^k$ regularity.** The standard IFT guarantees $\gamma$ is $C^1$ when $F$ is $C^1$. When $F$ has higher regularity $C^k$, repeated application of the implicit differentiation formula (differentiating the identity $F(\tilde{x}', \gamma(\tilde{x}')) = 0$) shows that $\nabla \gamma$ satisfies a $C^{k-1}$ equation, so $\gamma$ is $C^k$. This is why the $C^k$ regularity of $\Psi$ transfers to $\gamma$. [/guided] [/step] [step:Identify the local boundary with the graph of $\gamma$] Choose $r > 0$ small enough that $B(0, r) \subset V \times W$ in the rotated coordinates. Since $\tilde{x} = R(x - z)$ is an isometry, $B(0, r)$ in $\tilde{x}$-coordinates corresponds to $B(z, r)$ in $x$-coordinates. For any $x \in B(z, r)$, \begin{align*} x \in \partial U \cap B(z, r) \quad &\Longleftrightarrow \quad \Psi(x) = 0 \text{ and } x \in B(z, r) \\ &\Longleftrightarrow \quad F(\tilde{x}) = 0 \text{ and } \tilde{x} \in B(0, r) \\ &\Longleftrightarrow \quad \tilde{x}_n = \gamma(\tilde{x}') \text{ and } \tilde{x} \in B(0, r). \end{align*} The first equivalence uses $F(\tilde{x}) = \Psi(R^{-1}\tilde{x} + z) = \Psi(x)$ and the isometric correspondence of balls. The second equivalence applies the result of the previous step, valid because $B(0, r) \subset V \times W$. This gives the desired local graph representation: \begin{align*} \partial U \cap B(z, r) &= \bigl\{ x \in B(z, r) \mid \tilde{x}_n = \gamma(\tilde{x}') \bigr\}. \end{align*} For the moreover clause, the [boundary](/page/Boundary) $\partial U$ separates $B(z, r)$ into two connected components (for $r$ sufficiently small), characterized by $\tilde{x}_n > \gamma(\tilde{x}')$ and $\tilde{x}_n < \gamma(\tilde{x}')$. Since $\nabla \Psi(z)$ points in the direction of increasing $\Psi$, the sign of $\Psi$ determines which component is $U$. After possibly replacing $R$ by $-R$ (which replaces $\tilde{x}_n$ by $-\tilde{x}_n$ and $\gamma$ by $-\gamma$), we arrange that \begin{align*} U \cap B(z, r) &= \bigl\{ x \in B(z, r) \mid \tilde{x}_n > \gamma(\tilde{x}') \bigr\}. \end{align*} [/step]