[proofplan] We encode the critical equations by the smooth map $F=(\partial_{\theta_1}\phi,\dots,\partial_{\theta_N}\phi)$ and use the nondegeneracy condition to identify $C_\phi=F^{-1}(0)$ as a regular level set of codimension $N$. Homogeneity gives conicity of $C_\phi$ and equivariance of $j_\phi$ under positive scaling in the cotangent fibers. The main point is to prove that $j_\phi$ is an immersion; this follows by combining the tangent equations defining $T C_\phi$ with the injectivity of the transpose of $dF$ at critical points. Finally, the pullback of the tautological one-form on $T^*X$ is $d(\phi|_{C_\phi})$, so the pullback of the symplectic form vanishes; since the immersed submanifold has dimension $n$ inside the $2n$-dimensional symplectic manifold $T^*X$, it is Lagrangian. [/proofplan]

[step:Realize the critical set as a regular level set] Define the smooth map \begin{align*} F:\Omega \to \mathbb{R}^N,\quad (x,\theta)\mapsto (\partial_{\theta_1}\phi(x,\theta),\dots,\partial_{\theta_N}\phi(x,\theta)). \end{align*} Then $C_\phi=F^{-1}(\{0\})$. At every $(x,\theta)\in C_\phi$, the differential \begin{align*} dF_{(x,\theta)}:T_{(x,\theta)}\Omega \to \mathbb{R}^N \end{align*} has component covectors $d_{x,\theta}(\partial_{\theta_a}\phi)(x,\theta)$ for $1\leq a\leq N$. By the nondegeneracy hypothesis, these $N$ covectors are linearly independent, so $dF_{(x,\theta)}$ is surjective. Hence $0\in\mathbb{R}^N$ is a regular value of $F$. By the regular level set theorem (citing a result not yet in the wiki: Regular Level Set Theorem), $C_\phi=F^{-1}(\{0\})$ is a smooth submanifold of $\Omega$ of codimension $N$. Since $\Omega$ is an open subset of the smooth manifold $X\times\mathbb{R}^N_0$, it has dimension $n+N$, and therefore \begin{align*} \dim C_\phi=(n+N)-N=n. \end{align*} [/step]

[step:Use homogeneity to prove conicity of the critical set and of the parametrization] Fix $r>0$. Since $\phi$ is positively homogeneous of degree $1$ in $\theta$, differentiating the identity $\phi(x,r\theta)=r\phi(x,\theta)$ with respect to $\theta_a$ gives \begin{align*} \partial_{\theta_a}\phi(x,r\theta)=\partial_{\theta_a}\phi(x,\theta) \end{align*} for each $1\leq a\leq N$. Therefore, if $(x,\theta)\in C_\phi$, then $F(x,r\theta)=F(x,\theta)=0$, so $(x,r\theta)\in C_\phi$. Thus $C_\phi$ is conic. Differentiating the same homogeneity identity with respect to the base variable $x$ gives \begin{align*} d_x\phi(x,r\theta)=r\,d_x\phi(x,\theta). \end{align*} Hence, for the positive scaling action $r\cdot(x,\theta):=(x,r\theta)$ on $C_\phi$ and $r\cdot(x,\xi):=(x,r\xi)$ on $T^*X\setminus 0$, we have \begin{align*} j_\phi(r\cdot(x,\theta))=r\cdot j_\phi(x,\theta). \end{align*} The hypothesis $d_x\phi(x,\theta)\neq 0$ on $C_\phi$ ensures that $j_\phi(x,\theta)\in T^*X\setminus 0$ for every $(x,\theta)\in C_\phi$. [/step]

[step:Prove that the phase parametrization is an immersion]Let $(x,\theta)\in C_\phi$ and let \begin{align*} v\in T_{(x,\theta)}C_\phi \end{align*} satisfy \begin{align*} d(j_\phi)_{(x,\theta)}(v)=0. \end{align*} Choose a coordinate chart $(U,\kappa)$ on $X$ near $x$, with coordinate functions $x_1,\dots,x_n$, and use the standard coordinates $\theta_1,\dots,\theta_N$ on $\mathbb{R}^N_0$. In these local coordinates, write \begin{align*} v=(\delta x,\delta\theta)\in\mathbb{R}^n\times\mathbb{R}^N. \end{align*} Since $j_\phi$ has base component $x$, the equality $d(j_\phi)_{(x,\theta)}(v)=0$ implies $\delta x=0$. Since the fiber component of $j_\phi$ is $d_x\phi$, the same equality also gives \begin{align*} d_{x,\theta}(\partial_{x_i}\phi)(x,\theta)(v)=0 \end{align*} for every $1\leq i\leq n$. Because $v\in T_{(x,\theta)}C_\phi=T_{(x,\theta)}F^{-1}(\{0\})$, we also have \begin{align*} d_{x,\theta}(\partial_{\theta_a}\phi)(x,\theta)(v)=0 \end{align*} for every $1\leq a\leq N$. Since $\delta x=0$, the vector $v$ is vertical, so write $v=(0,\delta\theta)$. The preceding equations say exactly that the covector in $T_{(x,\theta)}^*\Omega$ obtained by pairing $\delta\theta$ with the transpose of $dF_{(x,\theta)}$ is zero: \begin{align*} (dF_{(x,\theta)})^\top(\delta\theta)=0. \end{align*} The map $dF_{(x,\theta)}$ is surjective by nondegeneracy, hence its transpose is injective. Therefore $\delta\theta=0$, and consequently $v=0$. Thus $d(j_\phi)_{(x,\theta)}$ is injective at every point of $C_\phi$, so $j_\phi$ is an immersion.[/step]

[guided]We prove the immersion statement directly from the linear algebra of the phase equations. Fix a point $(x,\theta)\in C_\phi$ and take a tangent vector \begin{align*} v\in T_{(x,\theta)}C_\phi \end{align*} such that \begin{align*} d(j_\phi)_{(x,\theta)}(v)=0. \end{align*} The goal is to prove $v=0$. Work in a coordinate chart $(U,\kappa)$ on $X$ near $x$, with coordinates $x_1,\dots,x_n$, together with the standard coordinates $\theta_1,\dots,\theta_N$ on $\mathbb{R}^N_0$. In these coordinates, write \begin{align*} v=(\delta x,\delta\theta)\in\mathbb{R}^n\times\mathbb{R}^N. \end{align*} The map $j_\phi$ has two pieces: its base component is $x$, and its fiber component is the covector $d_x\phi$. Therefore the equality $d(j_\phi)_{(x,\theta)}(v)=0$ first forces the base variation to vanish: \begin{align*} \delta x=0. \end{align*} The same equality also says that the first variation of each fiber coordinate $\partial_{x_i}\phi$ vanishes: \begin{align*} d_{x,\theta}(\partial_{x_i}\phi)(x,\theta)(v)=0 \end{align*} for every $1\leq i\leq n$. Now we use that $v$ is tangent to the critical set. Since \begin{align*} C_\phi=F^{-1}(\{0\}),\quad F=(\partial_{\theta_1}\phi,\dots,\partial_{\theta_N}\phi), \end{align*} a tangent vector to $C_\phi$ lies in the kernel of $dF_{(x,\theta)}$. Thus \begin{align*} d_{x,\theta}(\partial_{\theta_a}\phi)(x,\theta)(v)=0 \end{align*} for every $1\leq a\leq N$. At this point the vanishing of the base variation matters. Since $\delta x=0$, the vector $v$ is purely vertical, so $v=(0,\delta\theta)$ for some $\delta\theta\in\mathbb{R}^N$. The equations coming from $d(j_\phi)(v)=0$ say that $\delta\theta$ kills all mixed second derivatives $\partial_{\theta_b}\partial_{x_i}\phi$, while the tangent equations say that $\delta\theta$ kills all second derivatives $\partial_{\theta_b}\partial_{\theta_a}\phi$. Together, these are exactly the statement \begin{align*} (dF_{(x,\theta)})^\top(\delta\theta)=0. \end{align*} Here $(dF_{(x,\theta)})^\top:\mathbb{R}^N\to T_{(x,\theta)}^*\Omega$ is the transpose of the surjective [linear map](/page/Linear%20Map) $dF_{(x,\theta)}:T_{(x,\theta)}\Omega\to\mathbb{R}^N$. The nondegeneracy hypothesis says that the component covectors of $dF_{(x,\theta)}$ are linearly independent. Equivalently, $dF_{(x,\theta)}$ is surjective. A surjective linear map has injective transpose, so the equation \begin{align*} (dF_{(x,\theta)})^\top(\delta\theta)=0 \end{align*} implies $\delta\theta=0$. Since also $\delta x=0$, we get $v=0$. Therefore the kernel of $d(j_\phi)_{(x,\theta)}$ is zero at every point, which proves that $j_\phi$ is an immersion.[/guided]

[step:Pull back the tautological one-form to an exact one-form] Let \begin{align*} \pi:T^*X\setminus 0\to X,\quad (x,\xi)\mapsto x \end{align*} be the cotangent projection. Let $\alpha\in\Omega^1(T^*X\setminus 0)$ be the tautological one-form, defined by \begin{align*} \alpha_{(x,\xi)}(W)=\xi(d\pi_{(x,\xi)}W) \end{align*} for every $(x,\xi)\in T^*X\setminus 0$ and every $W\in T_{(x,\xi)}(T^*X\setminus 0)$. Let \begin{align*} \iota:C_\phi\hookrightarrow\Omega \end{align*} denote the inclusion map. Let \begin{align*} \operatorname{pr}_X:\Omega\to X,\quad (x,\theta)\mapsto x \end{align*} denote the restriction to $\Omega\subset X\times\mathbb{R}^N_0$ of the projection onto the $X$ factor. For $(x,\theta)\in C_\phi$ and $v\in T_{(x,\theta)}C_\phi$, we compute \begin{align*} (j_\phi^*\alpha)_{(x,\theta)}(v)=\alpha_{j_\phi(x,\theta)}(d(j_\phi)_{(x,\theta)}v)=d_x\phi(x,\theta)(d\pi_{j_\phi(x,\theta)}d(j_\phi)_{(x,\theta)}v). \end{align*} Since $\pi\circ j_\phi(x,\theta)=x$, the last expression is \begin{align*} d_x\phi(x,\theta)(d(\pi\circ j_\phi)_{(x,\theta)}v)=d_x\phi(x,\theta)(d\operatorname{pr}_X{}_{(x,\theta)}v). \end{align*} Because $v\in T_{(x,\theta)}C_\phi$, the critical equation $d_\theta\phi(x,\theta)=0$ gives \begin{align*} d\phi(x,\theta)(v)=d_x\phi(x,\theta)(d\operatorname{pr}_X{}_{(x,\theta)}v). \end{align*} Thus \begin{align*} j_\phi^*\alpha=\iota^*(d\phi)=d(\phi|_{C_\phi}). \end{align*} [/step]

[step:Differentiate the pullback identity to obtain the Lagrangian condition] Let \begin{align*} \omega:=-d\alpha \end{align*} be the standard symplectic form on $T^*X\setminus 0$ associated to the tautological one-form $\alpha$. From the identity $j_\phi^*\alpha=d(\phi|_{C_\phi})$, functoriality of exterior differentiation gives \begin{align*} j_\phi^*\omega=-d(j_\phi^*\alpha)=-d(d(\phi|_{C_\phi}))=0. \end{align*} Therefore the immersed submanifold parametrized by $j_\phi:C_\phi\to T^*X\setminus 0$ is isotropic. The ambient symplectic manifold $T^*X\setminus 0$ has dimension $2n$, and we proved $\dim C_\phi=n$. An isotropic immersed submanifold of dimension half the ambient symplectic dimension is Lagrangian by the maximal isotropic criterion (citing a result not yet in the wiki: Maximal Isotropic Submanifolds are Lagrangian). Hence \begin{align*} \Lambda_\phi=j_\phi(C_\phi) \end{align*} is an immersed Lagrangian submanifold of $T^*X\setminus 0$, with the immersed structure induced by $j_\phi$. Since $C_\phi$ is conic and $j_\phi$ is equivariant under positive scaling, $\Lambda_\phi$ is conic. [/step]

[step:Conclude embeddedness under the embedding hypothesis] If $j_\phi:C_\phi\to T^*X\setminus 0$ is an embedding, then its image $\Lambda_\phi=j_\phi(C_\phi)$ is an embedded submanifold, and the preceding pullback computation shows that the restriction of the symplectic form to $\Lambda_\phi$ vanishes. Since $\dim \Lambda_\phi=\dim C_\phi=n$, the embedded submanifold $\Lambda_\phi$ is Lagrangian. Since smooth manifolds are taken to be Hausdorff and second countable, if $j_\phi$ is a proper injective immersion, then the standard proper injective immersion theorem (citing a result not yet in the wiki: Proper Injective Immersion is an Embedding) implies that $j_\phi$ is an embedding. Applying the previous paragraph gives that $\Lambda_\phi$ is an embedded conic Lagrangian submanifold of $T^*X\setminus 0$. [/step]