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My concrete questions are (for context see below):

  1. Is is true that $i$ as below embeds $M$ in $T^*\mathbb{R}^n$?
  2. Is it true that $M$ is Lagrangian in $\mathbb{C}^n$ if and only if $i(M)$ is Lagrangian in $T^*\mathbb{R}^n$?
  3. Is it true that Theorem 4.3 implies that $$dh=\text{proj}_{d\psi}y\iff dh=(\sum_ky_k\dot\psi_k)d\psi$$ as I derived below? How can I obtain the identity stated by Harvey (quoted below)?

Here is the context: In his book "Spinors and Calibrations", F.R. Harvey provides a definition of Lawlor necks (which I restate below as Definition 1) as $n$-dimensional submanifolds of $\mathbb{C}^n\cong\mathbb{R}^{2n}$, and then introduces an equivalent characterization by looking at intersections of Lawlor necks with $n$-planes $P_{\theta}$, which I define below.

Definition. Let $\theta=(\theta_1,\dots,\theta_n)\in\mathbb{R}^n$. Then we denote by $$P_\theta=\{(t_1e^{i\theta_1},\dots,t_1e^{i\theta_1})\in\mathbb{C}:t_k\in\mathbb{R}\}$$ the $n$-plane in $\mathbb{C}^n$ obtained by rotating $\mathbb{R}^n$ by $e^{i\theta}$.$\diamond$

Harvey remarks that the intersection of a Lawlor neck $M$ with $P_\theta$, for any $\theta$, is either empty or a compact hypersurface of $P_\theta$ (indeed, it can be easily shown that, if not empty, $M\cap P_\theta$ is an ellipsoid). In Theorem 7.78, Harvey proves that Lawlor necks are the only possible special Lagrangian submanifolds of $\mathbb{C}^n$ that admit this property.

In his proof, Harvey claims the following (which I cannot quite follow): ~verbatim quote~ "If $M$ is to consist of the union of hypersurfaces in a family of the $n$-planes $P_\theta$, and $M$ is to be Lagrangian, then $M$ must be of the form (set $w_j=R_je^{i\theta_j}$) $$M(\Gamma,h)=\left\{ w\in\mathbb{C}^n:\sum_{j=1}^nR^2_j\frac{\text{d}\theta_j}{\text{d}s}=\frac{\text{d}h}{\text{d}s}\right\}$$ for some curve $\Gamma$ in the $\theta$-space and some function $h(s)$ defined on $\Gamma$ (here, $\theta(s)$ parametrizes $\Gamma$). This can be reforumlated as a general fact about Lagrangian submanifolds with degenerate (one-dimensional) projection onto one of the Lagrangian axis planes, using the alternate symplectic coordinates $p_j=\frac{1}{2}R^2_j$ and $q_j=\theta_j$, $j=1,\dots,n$. (Note that $\text{d}p_j\wedge\text{d}q_j=\text{d}x_j\wedge\text{d}y_j$. See the article 'Calibrated Geometries' by Harvey-Lawson for a proof of $M_a=M(\Gamma,h)$.)" ~end of quote~

I am struggling with rigorously understanding how the identity $$\sum_{j=1}^nR^2_j\frac{\text{d}\theta_j}{\text{d}s}=\frac{\text{d}h}{\text{d}s}$$ is derived (what result from'Calibrated Geometries' by Harvey-Lawson is being used?).

Here is what I understand so far: The alternate symplectic coordinates let us embed a Lawlor neck $M$ into $T^*\mathbb{R}^n\cong\mathbb{R}^n\times\mathbb{R}^n$, via $$i:(r_ke^{i\theta_k})_{k=1}^n\mapsto((\theta_1,\dots,\theta_n),(r_1^2/2,\dots,r_n^2/2))$$ Since $\text{d}p_j\wedge\text{d}q_j=\text{d}x_j\wedge\text{d}y_j$, $M$ is Lagrangian in $\mathbb{C}^n$ if and only if $i(M)$ is Lagrnagian in $T^*\mathbb{R}^n$. Now $T^*\mathbb{R}^n$ is equipped with the natural projection $\pi:T^*\mathbb{R}^n\to\mathbb{R}^n$. Clearly, we have $\pi(P_\theta)=\{\theta\}$ (where we view $P_\theta$ as a subset of $T^*\mathbb{R}^n$ via $i$). Since the intersection of a Lawlor neck $M$ with each plane $P_\theta$ is either empty or a ($n-1$ dimensional) hypersurface, the projection $\pi:M\to\mathbb{R}^n$ is degenerate with constant rank $1$. This, I beleive, lets us apply the following result from 'Calibrated Geometries' by Harvey-Lawson (Theorem 4.3):

Theorem. Suppose $i(M)=X$ is an $n$-dimensional Lagrangian submanifold of $T^{\\*}\mathbb{R}^n$ whose projection $\pi:X\to\mathbb{R}^n$ is degenerate with constant rank $p$. Then there exists a unique pair of an $p$-dimensional submanifold $M$ of $\mathbb{R}^n$ and a real-valued function $h$ on $M$ so that $X$ is the affine subbundle $A$ of $T^{\\*}\mathbb{R}^n$ obtained by translating the normal bundle $N(M)$ of $M$ in the cotagent bundle $T^{\\*}M$ by the exterior derivative $\text{d}h$, i.e., $$A_x=N_x(M)+(dH)_x$$ for all $x\in M$, where $H$ is some smooth extension of $h$ to the ambient space $\mathbb{R}^n$.$\diamond$

In our case, the projection $\pi:i(M)\to\mathbb{R}^n$ has constant rank $1$, so if $M$ is Lagrangian, there exists a $1$-dimensional submanifold $\Gamma$ of $\mathbb{R}^n$ (say $\Gamma$ is parametrized by $\psi=(\psi_1,\dots,\psi_n):\mathbb{R}\to\mathbb{R}^n$ with $\vert{\psi}\vert\equiv1$) and a function $h:\Gamma\to\mathbb{R}$ (with $d_ph\in T^*\mathbb{R}^n$) so that, for $$(\psi_1(t),\dots,\psi_n(t),R_1,\dots,R_n)\in i(M)$$ we have that $(R_1,\dots,R_n)-d_{\psi(t)}h$ is an element of $N_{\psi(t)}(M)$. Denoting by $d\psi$ the covector associated with $\dot\psi(t)$, this is equivalent to $(R_1,\dots,R_n)-d_{\psi(t)}h\perp d\psi$. Since $d\psi$ is paralell to $dh:=d_{\psi(t)}h$, and $\vert{d\psi}\vert=1$, this is equivalent to $$dh=\text{proj}_{d\psi}y\iff dh=(\sum_ky_k\dot\psi_k)d\psi$$ where $y=\sum_ky_kd\theta_k$, $d\psi=\sum_k\dot\psi_k\cdot d\theta_k$. I feel like this gets me somewhere near the identity $$\sum_{j=1}^nR^2_j\frac{\text{d}\theta_j}{\text{d}s}=\frac{\text{d}h}{\text{d}s}$$ But I am uncertain about whether my steps so far are correct, and what the full derivation looks like. I am especially concerned that $X=i(M)$ does not appear to be an affine subbundle of $T^*\mathbb{R}$ (its fibres $X_p$ are not linear subspaces of $T_p^*\mathbb{R}$), which seemingly contradicts Theorem 4.3.

Since Harvey references the article by Harvey-Lawson as a whole, instead of pointing to a specific section, I am also not sure if Theorem 4.3 is the only relevant result. The article by Harvey-Lawson can be downloaded for free here.

Definition 1. (Lawlor Necks) Let $n\geq2$ and $a=(a_1,\dots,a_n)$ be a vector in $\mathbb{R}^n$ with $a_k\geq0$. We define $n$ functions $z^a_k\mathbb{R}\to\mathbb{C}$ as follows. First, set $$P^a:\mathbb{R}\longrightarrow\mathbb{R},\qquad y\mapsto\frac{1}{y^2}\left(\left(1+a_1y^2\right)\cdots\left(1+a_py^2\right)-1\right),$$ and, for $k\in\{1,\dots,n\}$, $$r^a_k:\mathbb{R}\longrightarrow\mathbb{R},\qquad y\longmapsto\sqrt{a_k^{-1}+y^2},$$ as well as $$\theta^a_k:\mathbb{R}\longrightarrow\mathbb{R},\qquad y\longmapsto a_k\int_0^y\frac{\text{d}y}{(1+a_ky^2)\sqrt{P^a(y)}}$$ Then we define $z^a_k:\mathbb{R}\to\mathbb{C}$ as follows, interpreting $(r^a_k,\theta^a_k)$ as polar coordinates; $$z^a_k:\mathbb{R}\longrightarrow\mathbb{C},\qquad y\longmapsto r_k(y)\cdot e^{i\cdot\theta^a_k(y)}.$$ We now use the functions $z^a_k$ to define the following submanifold of $\mathbb{C}^n$; $$M_a=\left\{(t_k\cdot z^a_k(y))\in\mathbb{C}^n:y,t_k\in\mathbb{R},\sum_kt_k^2=1\right\}.$$ We call $M_a$ a Lawlor neck. $\diamond$

querryman
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1 Answers1

1

The answers to questions 1. and 2. are yes. Here is a derivation for the identity stated by Harvey:

The alternate symplectic coordinates let us embed a Lawlor neck $M$ into $T^*\mathbb{R}^n\cong\mathbb{R}^n\times\mathbb{R}^n$, via $$i:(r_ke^{i\theta_k})_{k=1}^n\mapsto((\theta_1,\dots,\theta_n),(r_1^2/2,\dots,r_n^2/2))$$ Since $\text{d}p_j\wedge\text{d}q_j=\text{d}x_j\wedge\text{d}y_j$, $M$ is Lagrangian in $\mathbb{C}^n$ if and only if $i(M)$ is Lagrnagian in $T^*\mathbb{R}^n$. Now $T^*\mathbb{R}^n$ is equipped with the natural projection $\pi:T^*\mathbb{R}^n\to\mathbb{R}^n$. Clearly, we have $\pi(P_\theta)=\{\theta\}$ (where we view $P_\theta$ as a subset of $T^*\mathbb{R}^n$ via $i$). Since the intersection of a Lawlor neck $M$ with each plane $P_\theta$ is either empty or a ($n-1$ dimensional) hypersurface, the projection $\pi:M\to\mathbb{R}^n$ is degenerate with constant rank $1$. We can therefore apply the quoted Theorem (Theorem 4.3) from 'Calibrated Geometries' by Harvey-Lawson. Consequently, there exists a $1$-dimensional submanifold $\Gamma$ of $\mathbb{R}^n$ (say $\Gamma$ is parametrized by $\psi=(\psi_1,\dots,\psi_n):\mathbb{R}\to\mathbb{R}^n$ with $\vert{\psi}\vert\equiv1$) and a function $h:\Gamma\to\mathbb{R}$ (with $d_ph\in T^*\mathbb{R}^n$) so that a vector $v\in T_{\psi(t)}^*\mathbb{R}^n$ is an element of $A_{\theta(t)}$ if and only if $v\in \text{d}_{\psi(t)}H+N_{\psi(t)}(\Gamma)$, i.e., if and only if $\left(v-\text{d}_{\theta(t)}H\right)\perp\psi'(t)$. Since $\vert\psi'(t)\vert=1$, this is equivalent to $\operatorname{proj}_{\psi'(t)}v=\text{d}_{\(t)}H$. Noting that $$\operatorname{proj}_{\psi'(t)}v=(\psi'(t)\cdot v)\cdot \psi'(t)$$ where $\psi'(t)\cdot v$ denotes the dot product between $\psi'(t)$ and $v$, and expressing this condition in coordinates (i.e., choosing a basis for $\mathbb{R}^n$ and expressing $\psi'(t)$ and $v$ in terms of its dual basis) - and that $\text{d}_{\psi(t)}H=\dot h(t)\cdot\psi'(t)$, where we denote $h(s):=h(\psi(s))$ and $\dot h(t):=(\partial h/\partial s)(t)$ - directly gives rise to the expression stated by Harvey (if translated back into the polar coordinates of $\mathbb{C}^n$ via $i^{-1}$).

querryman
  • 155