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6.1 Finding the equations on u and v [03LE]

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6.1 Finding the equations on uu and vv

We now calculate the conditions on the functions u(x,y),v(x,y):ℝ2→ℝu(x,y),v(x,y):\mathbin{\mathbb{R}}^{2}\rightarrow\mathbin{\mathbb{R}} for the 3-fold NN of (30) to be special Lagrangian.

Proposition 6.1

Let u,v:ℝ2→ℝu,v:\mathbin{\mathbb{R}}^{2}\rightarrow\mathbin{\mathbb{R}} be continuous, and let a∈ℝa\in\mathbin{\mathbb{R}}. Define

N={(z1,z2,z3)∈ℂ3:Re(z1z2)=u(Re(z3),Im(z1z2)),Im(z3)=v(Re(z3),Im(z1z2)),|z1|2−|z2|2=a}.\begin{split}N=\Bigl\{(z_{1}&,z_{2},z_{3})\in\mathbin{\mathbb{C}}^{3}:\mathop{\rm Re}(z_{1}z_{2})=u\bigl(\mathop{\rm Re}(z_{3}),\mathop{\rm Im}(z_{1}z_{2})\bigr),\\ &\mathop{\rm Im}(z_{3})=v\bigl(\mathop{\rm Re}(z_{3}),\mathop{\rm Im}(z_{1}z_{2})\bigr),\quad|z_{1}|^{2}-|z_{2}|^{2}=a\Bigr\}.\end{split} (31)

Then

  • (a)

    If a=0a=0, then NN is a singular special Lagrangian 33-fold in ℂ3\mathbin{\mathbb{C}}^{3} if u,vu,v are differentiable and satisfy

    ∂u∂x=−2​(u2+y2)1/2​∂v∂yand∂u∂y=∂v∂x,\frac{\partial u}{\partial x}=-2\bigl(u^{2}+y^{2}\bigr)^{1/2}\frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}, (32)

    except at points (x,0)(x,0) in ℝ2\mathbin{\mathbb{R}}^{2} with u⁡(x,0)=0u(x,0)=0, where u,vu,v need not be differentiable. The singular points of NN are those of the form (0,0,z3)(0,0,z_{3}), where z3=x+i​v​(x,0)z_{3}=x+iv(x,0) for x∈ℝx\in\mathbin{\mathbb{R}} with u⁡(x,0)=0u(x,0)=0.

  • (b)

    If a≠0a\neq 0, then NN is a nonsingular special Lagrangian 33-fold in ℂ3\mathbin{\mathbb{C}}^{3} if and only if u,vu,v are differentiable on all of ℝ2\mathbin{\mathbb{R}}^{2} and satisfy

    ∂u∂x=−(4​u2+4​y2+a2)1/2​∂v∂yand∂u∂y=∂v∂x.\frac{\partial u}{\partial x}=-\bigl(4u^{2}+4y^{2}+a^{2}\bigr)^{1/2}\frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}. (33)

Proof. We shall give the proof for part (a). Part (b) is similar but more complicated, and will be left to the reader. Let a=0a=0, let NN be defined by (31), and let 𝐳=(z1,z2,z3)∈N{\bf z}=(z_{1},z_{2},z_{3})\in N. For 𝐳\bf z to be a nonsingular point of NN, we need uu and vv to be differentiable at (x,y)=(Re(z3),Im(z1​z2))(x,y)=\bigl(\mathop{\rm Re}(z_{3}),\mathop{\rm Im}(z_{1}z_{2})\bigr) in ℝ2\mathbin{\mathbb{R}}^{2}, and for the derivatives of the three functions

Re(z1​z2)−u⁡(Re(z3),Im(z1​z2)),Im(z3)−v⁡(Re(z3),Im(z1​z2)),|z1|2−|z2|2\mathop{\rm Re}(z_{1}z_{2})-u\bigl(\mathop{\rm Re}(z_{3}),\mathop{\rm Im}(z_{1}z_{2})\bigr),\quad\mathop{\rm Im}(z_{3})-v\bigl(\mathop{\rm Re}(z_{3}),\mathop{\rm Im}(z_{1}z_{2})\bigr),\quad|z_{1}|^{2}-|z_{2}|^{2}

on ℂ3\mathbin{\mathbb{C}}^{3} to be linearly independent at 𝐳\bf z.

Now if z1=z2=0z_{1}=z_{2}=0 then |z1|2−|z2|2|z_{1}|^{2}-|z_{2}|^{2} has zero derivative at 𝐳\bf z. Thus points of the form (0,0,z3)(0,0,z_{3}) in NN will be singular. Clearly, these occur exactly when z3=x+i​v​(x,0)z_{3}=x+iv(x,0) for x∈ℝx\in\mathbin{\mathbb{R}} with u⁡(x,0)=0u(x,0)=0. Also, as |z1|2−|z2|2=0|z_{1}|^{2}-|z_{2}|^{2}=0, such points occur in NN only when a=0a=0. We shall see that these are the only singular points in NN, provided uu and vv are differentiable.

To prove part (a) we need to show that each 𝐳∈N{\bf z}\in N not of the form (0,0,z3)(0,0,z_{3}) is a nonsingular point of NN, and the tangent space T𝐳​NT_{\bf z}N is a special Lagrangian 3-plane ℝ3\mathbin{\mathbb{R}}^{3} in ℂ3\mathbin{\mathbb{C}}^{3}. As NN is U(1)\mathbin{\rm U}(1)-invariant, it is enough to prove this for one point in each orbit of the U(1)\mathbin{\rm U}(1)-action (29). Since |z1|=|z2||z_{1}|=|z_{2}| on NN, each U(1)\mathbin{\rm U}(1)-orbit in NN contains one or two points (z1,z2,z3)(z_{1},z_{2},z_{3}) with z1=z2z_{1}=z_{2}.

Thus it is enough to show that T𝐳​NT_{\bf z}N exists and is special Lagrangian for points 𝐳=(z1,z1,z3){\bf z}=(z_{1},z_{1},z_{3}) in NN with z1≠0z_{1}\neq 0. In our next lemma we identify T𝐳​NT_{\bf z}N at such a point. The proof is elementary, and is left as an exercise.

Lemma 6.2

Let 𝐳=(z1,z1,z3)∈N{\bf z}=(z_{1},z_{1},z_{3})\in N, with z1≠0z_{1}\neq 0. Set x=Re(z3)x=\mathop{\rm Re}(z_{3}) and y=Im(z12)y=\mathop{\rm Im}(z_{1}^{2}). Then NN is nonsingular at 𝐳\bf z, and T𝐳​N=⟨𝐩1,𝐩2,𝐩3⟩ℝ,T_{\bf z}N=\langle{\bf p}_{1},{\bf p}_{2},{\bf p}_{3}\rangle_{\scriptscriptstyle\mathbb{R}}, where

𝐩1\displaystyle{\bf p}_{1} =(i​z1,−i​z1,0),\displaystyle=(iz_{1},-iz_{1},0), (34)
𝐩2\displaystyle{\bf p}_{2} =((2z1)−1∂u∂x(x,y),(2z1)−1∂u∂x(x,y),1+i∂v∂x(x,y))and\displaystyle=\bigl((2z_{1})^{-1}{\textstyle\frac{\partial u}{\partial x}}(x,y),(2z_{1})^{-1}{\textstyle\frac{\partial u}{\partial x}}(x,y),1+i{\textstyle\frac{\partial v}{\partial x}}(x,y)\bigr)\quad\text{and} (35)
𝐩3\displaystyle{\bf p}_{3} =((2​z1)−1​(∂u∂y​(x,y)+i),(2​z1)−1​(∂u∂y​(x,y)+i),i​∂v∂y​(x,y)).\displaystyle=\bigl((2z_{1})^{-1}({\textstyle\frac{\partial u}{\partial y}}(x,y)+i),(2z_{1})^{-1}({\textstyle\frac{\partial u}{\partial y}}(x,y)+i),i{\textstyle\frac{\partial v}{\partial y}}(x,y)\bigr). (36)

Now define ×:ℂ3×ℂ3→ℂ3\times:\mathbin{\mathbb{C}}^{3}\times\mathbin{\mathbb{C}}^{3}\rightarrow\mathbin{\mathbb{C}}^{3} as in (2), and apply Proposition 2.4 with 𝐫=𝐩1{\bf r}={\bf p}_{1} and 𝐬=𝐩2{\bf s}={\bf p}_{2}. Clearly 𝐩1{\bf p}_{1} and 𝐩2{\bf p}_{2} are linearly independent, and ω⁡(𝐩1,𝐩2)=0\omega({\bf p}_{1},{\bf p}_{2})=0. So Proposition 2.4 shows that ⟨𝐩1,𝐩2,𝐩1×𝐩2⟩ℝ\langle{\bf p}_{1},{\bf p}_{2},{\bf p}_{1}\times{\bf p}_{2}\rangle_{\scriptscriptstyle\mathbb{R}} is the unique SL 3-plane in ℂ3\mathbin{\mathbb{C}}^{3} containing ⟨𝐩1,𝐩2⟩ℝ\langle{\bf p}_{1},{\bf p}_{2}\rangle_{\scriptscriptstyle\mathbb{R}}.

Therefore ⟨𝐩1,𝐩2,𝐩3⟩ℝ\langle{\bf p}_{1},{\bf p}_{2},{\bf p}_{3}\rangle_{\scriptscriptstyle\mathbb{R}} is an SL 3-plane if and only if 𝐩3∈⟨𝐩1,𝐩2,𝐩1×𝐩2⟩ℝ{\bf p}_{3}\in\langle{\bf p}_{1},{\bf p}_{2},{\bf p}_{1}\times{\bf p}_{2}\rangle_{\scriptscriptstyle\mathbb{R}}. Combining equations (2), (34) and (35) gives

𝐩1×𝐩2=(z¯1​(∂v∂x+i),z¯1​(∂v∂x+i),−i​∂u∂x).{\bf p}_{1}\times{\bf p}_{2}=\bigl(\bar{z}_{1}({\textstyle\frac{\partial v}{\partial x}}+i),\bar{z}_{1}({\textstyle\frac{\partial v}{\partial x}}+i),-i{\textstyle\frac{\partial u}{\partial x}}\bigr). (37)

So suppose 𝐩3=α​𝐩1+β​𝐩2+γ​𝐩1×𝐩2{\bf p}_{3}=\alpha{\bf p}_{1}+\beta{\bf p}_{2}+\gamma{\bf p}_{1}\times{\bf p}_{2}. As the first two coordinates are equal in 𝐩2,𝐩3{\bf p}_{2},{\bf p}_{3} and 𝐩1×𝐩2{\bf p}_{1}\times{\bf p}_{2} but not in 𝐩1{\bf p}_{1}, we see that α=0\alpha=0. Taking real parts in the third coordinate gives β=0\beta=0. And comparing real multiples of i​z¯1i\bar{z}_{1} in the first coordinate shows that γ=12​|z1|−2\gamma={\textstyle\frac{1}{2}}|z_{1}|^{-2}.

Thus T𝐳​NT_{\bf z}N is special Lagrangian if and only if 𝐩1×𝐩2=2​|z1|2​𝐩3{\bf p}_{1}\times{\bf p}_{2}=2|z_{1}|^{2}{\bf p}_{3}. By (36) and (37), this reduces to

∂u∂x=−2​|z1|2​∂v∂yand∂u∂y=∂v∂xat (x,y).\frac{\partial u}{\partial x}=-2|z_{1}|^{2}\frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}\quad\text{at $(x,y)$.} (38)

But u=Re(z12)u=\mathop{\rm Re}(z_{1}^{2}) and y=Im(z12)y=\mathop{\rm Im}(z_{1}^{2}) by (31), so that |z1|4=u2+y2|z_{1}|^{4}=u^{2}+y^{2}, and |z1|2=(u2+y2)1/2|z_{1}|^{2}=(u^{2}+y^{2})^{1/2}. Substituting this into (38) gives equation (32), which proves part (a) of Proposition 6.1. Part (b) is left to the reader. □\square

Equations (32) and (33) are nonlinear versions of the Cauchy–Riemann equations. For if we replace the factors 2​(u2+y2)1/22(u^{2}+y^{2})^{1/2} and (4​u2+4​y2+a2)1/2(4u^{2}+4y^{2}+a^{2})^{1/2} in (32) and (33) by 1, the equations become

∂u∂x=−∂v∂yand∂u∂y=∂v∂x,\frac{\partial u}{\partial x}=-\,\frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x},

which are the conditions for u−i​vu-iv to be a holomorphic function of x+i​yx+iy. We may therefore expect the solutions of (32) and (33) to have qualitative features in common with solutions of the Cauchy–Riemann equations.

Note that (32) is the case a=0a=0 of equation (33), so we will often use (33) to refer to both, without assuming a≠0a\neq 0. Following Harvey and Lawson [9, Th. III.2.7], who use results of Morrey, we may prove:

Proposition 6.3

Any solutions u,vu,v of (33) are real analytic, except in the case a=0a=0 at points (x,0)(x,0) with u⁡(x,0)=0u(x,0)=0.

Now holomorphic functions on ℂ\mathbin{\mathbb{C}} are determined uniquely by their values on ℝ\mathbin{\mathbb{R}}. In the same way, solutions of (33) on ℝ2\mathbin{\mathbb{R}}^{2} are determined by their values on the xx-axis. We state this in the following proposition, which may be proved using the Cauchy–Kowalevksy Theorem [17, p. 234].

Proposition 6.4

Let UU be an open neighbourhood of ww in ℝ\mathbin{\mathbb{R}} and u′,v′:U→ℝu^{\prime},v^{\prime}:U\rightarrow\mathbin{\mathbb{R}} be real analytic functions. If either a≠0a\neq 0, or a=0a=0 and u′​(w)≠0u^{\prime}(w)\neq 0, then in an open neighbourhood VV of (w,0)(w,0) in ℝ2\mathbin{\mathbb{R}}^{2} there exist unique real analytic solutions u,v:V→ℝu,v:V\rightarrow\mathbin{\mathbb{R}} of (33) such that u⁡(x,0)=u′​(x)u(x,0)=u^{\prime}(x) and v⁡(x,0)=v′​(x)v(x,0)=v^{\prime}(x) for all x∈Ux\in U with (x,0)∈V(x,0)\in V.

Next we show that solutions u,v:ℝ2→ℝu,v:\mathbin{\mathbb{R}}^{2}\rightarrow\mathbin{\mathbb{R}} of (33) may be written in terms of a single potential f:ℝ2→ℝf:\mathbin{\mathbb{R}}^{2}\rightarrow\mathbin{\mathbb{R}}.

Proposition 6.5

Let u,v:ℝ2→ℝu,v:\mathbin{\mathbb{R}}^{2}\rightarrow\mathbin{\mathbb{R}} be solutions of (33). Then there exists a unique function f:ℝ2→ℝf:\mathbin{\mathbb{R}}^{2}\rightarrow\mathbin{\mathbb{R}} with

∂f∂x=u,∂f∂y=vandf(0,0)=0,satisfying∂2f∂x2+(4​(∂f∂x)2+4​y2+a2)1/2​∂2f∂y2=0.\begin{gathered}\frac{\partial f}{\partial x}=u,\quad\frac{\partial f}{\partial y}=v\quad\text{and}\quad f(0,0)=0,\\ \text{satisfying}\quad\frac{\partial^{2}f}{\partial x^{2}}+\Bigl(4\Bigl(\frac{\partial f}{\partial x}\Bigr)^{2}+4y^{2}+a^{2}\Bigr)^{1/2}\frac{\partial^{2}f}{\partial y^{2}}=0.\end{gathered} (39)

Conversely, all solutions of (39) yield solutions of (33).

Proof. Let u,v:ℝ2→ℝu,v:\mathbin{\mathbb{R}}^{2}\rightarrow\mathbin{\mathbb{R}} be solutions of (33), and define ff by

f⁡(x,y)=∫0xu⁡(s,0)​𝑑s+∫0yv⁡(x,t)​𝑑t.f(x,y)=\int_{0}^{x}u(s,0){\rm d}s+\int_{0}^{y}v(x,t){\rm d}t.

Then ∂f∂y​(x,y)=v​(x,y)\frac{\partial f}{\partial y}(x,y)=v(x,y) and f⁡(0,0)=0f(0,0)=0 are immediate, and

∂f∂x​(x,y)\displaystyle\frac{\partial f}{\partial x}(x,y) =u⁡(x,0)+∫0y∂v∂x​(x,t)​𝑑t\displaystyle=u(x,0)+\int_{0}^{y}\frac{\partial v}{\partial x}(x,t){\rm d}t
=u⁡(x,0)+∫0y∂u∂y​(x,t)​𝑑t\displaystyle=u(x,0)+\int_{0}^{y}\frac{\partial u}{\partial y}(x,t){\rm d}t
=u⁡(x,0)+[u⁡(x,t)]0y=u⁡(x,y),\displaystyle=u(x,0)+\bigl[u(x,t)\bigr]^{y}_{0}=u(x,y),

as ∂u∂y=∂v∂x\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}. This proves the first line of (39), and the second follows by substituting ∂f∂x=u\frac{\partial f}{\partial x}=u and ∂f∂y=v\frac{\partial f}{\partial y}=v into the first equation of (33). The converse is easy. □\square

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