4. Lagrangian subspaces of complex spaces

Some of the results have obtained in Ref. [8], but because the Lagrangian subspaces of complex spaces are essential to establish the generalized volume formula in complex integral geometry, let us give an expository on the Kahler strut rue of generalized complex spaces.

Theorem 4.1. The set of Lagrangian subspaces of C<sup>2</sup> with L<sup>1</sup> norm is T<sup>2</sup>∪T<sup>1</sup>, where

$$\mathbb{T}^2 \coloneqq \{ \text{span}((z, 0), (0, w)) : z, w \in \mathcal{U}(1) \} \mathbb{H}(1) \times \mathcal{U}(1) \tag{8}$$

and

$$\mathbb{T}^1 := \left\{ P : P = \{ \lambda(z, w) : \lambda \in \mathbb{R}, z, w \in \mathcal{U}(1), \ zw \text{ is a constant in } \mathcal{U}(1) \} \right\} \mathbb{\#}\mathcal{U}(1). \tag{9}$$

Proof. First, we can show that

$$P = \{\lambda(z, w) : \lambda \in \mathbb{R}, z, w \in \mathcal{U}(1), \ zw \text{ is a constant in } \mathcal{U}(1)\}\tag{10}$$

is identical to some

$$\boldsymbol{P}' := \text{span}((z\_1, z\_1e^{i\theta}), (z\_2, \frac{z\_1^2 \overline{z\_2}}{\left| z\_1 \right|^2} e^{i\theta})) \tag{11}$$

where <sup>z</sup>1;z<sup>2</sup> <sup>∈</sup> <sup>C</sup>\f0g. For any <sup>λ</sup>ðei<sup>ϕ</sup>;e<sup>i</sup><sup>ψ</sup>Þ∈P, let <sup>z</sup><sup>1</sup> <sup>¼</sup> <sup>λ</sup>e<sup>i</sup><sup>ϕ</sup>, <sup>θ</sup> <sup>¼</sup> <sup>ψ</sup>−ϕ, we have <sup>P</sup> <sup>¼</sup> spanððz1;z1ei<sup>θ</sup>Þ, ðz2; z2 1z2 jz1j <sup>2</sup> <sup>e</sup><sup>i</sup><sup>θ</sup>ÞÞ ¼ <sup>P</sup>′ where <sup>z</sup><sup>2</sup> <sup>∈</sup> <sup>C</sup>\f0g.

We can get κ1ðz1; 0Þ,ð0;z2ÞÞ ¼ 0. On the other hand, for any

$$\lambda(z, w) = \lambda\_1(z\_1, z\_1) + \lambda\_2(z\_2, \frac{z\_1^2 \overline{z\_2}}{\left| z\_1 \right|^2}) \in \text{span}((z\_1, z\_1), (z\_2, \frac{z\_1^2 \overline{z\_2}}{\left| z\_1 \right|^2})),\tag{12}$$

where λ1;λ<sup>2</sup> ∈ R,

$$\begin{array}{rcl} \left|w\right|^2 &=& (\lambda\_1 z\_1 + \lambda\_2 \frac{z\_1^2 \overline{z\_2}}{\left|z\_1\right|^2})(\lambda\_1 \overline{z\_1} + \lambda\_2 \frac{\overline{z\_1}^2 z\_2}{\left|z\_1\right|^2})\\ &=& \lambda\_1^2 z\_1 \overline{z\_1} + \lambda\_1 \lambda\_2 \overline{z\_1} z\_2 + \lambda\_2 \lambda\_1 z\_1 \overline{z\_2} + \lambda\_2^2 \overline{z\_2} z\_2\\ &=& (\lambda\_1 z\_1 + \lambda\_2 z\_2)(\lambda\_1 \overline{z\_1} + \lambda\_2 \overline{z\_2})\\ &=& \left|z\right|^2,\end{array} \tag{13}$$

that implies j w <sup>z</sup> j ¼ 1. Therefore, we have

κðz;<sup>w</sup>Þððz1;z1Þ,ðz2; z2 1z2 jz1j <sup>2</sup> ÞÞ ¼ <sup>3</sup> <sup>2</sup> ðImðz2z<sup>1</sup> Þ þ <sup>3</sup> <sup>2</sup> Imð z2 1z2 jz1j <sup>2</sup> z<sup>1</sup> ÞÞ −1 <sup>2</sup> Imð <sup>z</sup> w j w <sup>z</sup> jð<sup>z</sup><sup>2</sup> 1z2 jz1j <sup>2</sup> z<sup>1</sup> −z1z<sup>2</sup> ÞÞ ¼ <sup>3</sup> <sup>2</sup> ðImðz2z<sup>1</sup> Þ þ Imðz1z<sup>2</sup> ÞÞ ¼ 0: (14)

So κ vanishes on spanððz1;z1Þ,ðz2; z2 1z2 jz1j <sup>2</sup> ÞÞ for any z1;z<sup>2</sup> ∈ C\f0g, Imð<sup>z</sup><sup>1</sup> z1 Þ≠0. Conversely, suppose that κ vanishes on a plane P spanned by ðz1;w1Þ and ðz2;w2Þ. We know that

4. Lagrangian subspaces of complex spaces

and

32 Lagrangian Mechanics

ðz2; z2 1z2 jz1j

<sup>T</sup><sup>1</sup> :<sup>¼</sup>

is identical to some

where λ1;λ<sup>2</sup> ∈ R,

that implies j

w

So κ vanishes on spanððz1;z1Þ,ðz2;

Proof. First, we can show that

<sup>2</sup> <sup>e</sup><sup>i</sup><sup>θ</sup>ÞÞ ¼ <sup>P</sup>′ where <sup>z</sup><sup>2</sup> <sup>∈</sup> <sup>C</sup>\f0g.

We can get κ1ðz1; 0Þ,ð0;z2ÞÞ ¼ 0. On the other hand, for any

jwj

<sup>z</sup> j ¼ 1. Therefore, we have

κðz;<sup>w</sup>Þððz1;z1Þ,ðz2;

ðz;wÞ ¼ λ1ðz1;z1Þ þ λ2ðz2;

<sup>2</sup> ¼ ðλ1z<sup>1</sup> <sup>þ</sup> <sup>λ</sup><sup>2</sup>

z2 1z2 jz1j

z2 1z2 jz1j

<sup>¼</sup> <sup>λ</sup><sup>2</sup>

¼ jzj 2 ;

Some of the results have obtained in Ref. [8], but because the Lagrangian subspaces of complex spaces are essential to establish the generalized volume formula in complex integral geometry,

<sup>P</sup> : <sup>P</sup> ¼ fλðz;w<sup>Þ</sup> : <sup>λ</sup> <sup>∈</sup> <sup>R</sup>;z;<sup>w</sup> <sup>∈</sup> <sup>U</sup>ð1Þ, zw is a constant in <sup>U</sup>ð1Þg

<sup>T</sup><sup>2</sup> :¼ fspanððz; <sup>0</sup>Þ,ð0;wÞÞ : <sup>z</sup>;<sup>w</sup> <sup>∈</sup> <sup>U</sup>ð1Þg≅Uð1<sup>Þ</sup> · <sup>U</sup>ð1<sup>Þ</sup> (8)

P ¼ fλðz;wÞ : λ∈ R;z;w ∈ Uð1Þ, zw is a constant in Uð1Þg (10)

z2 1z2 jz1j 2 e

<sup>2</sup>Þ∈spanððz1;z1Þ,ðz2;

z1 2z2 jz1j 2 Þ

> <sup>2</sup> Imð z2 1z2 jz1j <sup>2</sup> z<sup>1</sup> ÞÞ

> > z1 Þ≠0.

<sup>2</sup> ðImðz2z<sup>1</sup> Þ þ Imðz1z<sup>2</sup> ÞÞ

<sup>2</sup> z<sup>1</sup> −z1z<sup>2</sup> ÞÞ

<sup>2</sup>Þðλ1z<sup>1</sup> þ λ<sup>2</sup>

<sup>2</sup> ðImðz2z<sup>1</sup> Þ þ <sup>3</sup>

<sup>1</sup>z1z<sup>1</sup> <sup>þ</sup> <sup>λ</sup>1λ2z<sup>1</sup> <sup>z</sup><sup>2</sup> <sup>þ</sup> <sup>λ</sup>2λ1z1z<sup>2</sup> <sup>þ</sup> <sup>λ</sup><sup>2</sup>

−1 <sup>2</sup> Imð <sup>z</sup> w j w <sup>z</sup> jð<sup>z</sup><sup>2</sup> 1z2 jz1j

<sup>2</sup> ÞÞ for any z1;z<sup>2</sup> ∈ C\f0g, Imð<sup>z</sup><sup>1</sup>

z2 1z2 jz1j

<sup>2</sup>z<sup>2</sup> z<sup>2</sup>

<sup>i</sup><sup>θ</sup>Þ,ðz2;

where <sup>z</sup>1;z<sup>2</sup> <sup>∈</sup> <sup>C</sup>\f0g. For any <sup>λ</sup>ðei<sup>ϕ</sup>;e<sup>i</sup><sup>ψ</sup>Þ∈P, let <sup>z</sup><sup>1</sup> <sup>¼</sup> <sup>λ</sup>e<sup>i</sup><sup>ϕ</sup>, <sup>θ</sup> <sup>¼</sup> <sup>ψ</sup>−ϕ, we have <sup>P</sup> <sup>¼</sup> spanððz1;z1ei<sup>θ</sup>Þ,

z2 1z2 jz1j

z2 1z2 jz1j

¼ ðλ1z<sup>1</sup> þ λ2z2Þðλ1z<sup>1</sup> þ λ2z<sup>2</sup> Þ

<sup>2</sup> ÞÞ ¼ <sup>3</sup>

¼ <sup>3</sup>

¼ 0:

≅Uð1Þ: (9)

<sup>i</sup><sup>θ</sup>ÞÞ (11)

<sup>2</sup>ÞÞ, (12)

(13)

(14)

let us give an expository on the Kahler strut rue of generalized complex spaces.

Theorem 4.1. The set of Lagrangian subspaces of C<sup>2</sup> with L<sup>1</sup> norm is T<sup>2</sup>∪T<sup>1</sup>, where

<sup>P</sup>′ :<sup>¼</sup> spanððz1;z1<sup>e</sup>

$$(1+\frac{1}{2}|\frac{w}{z}|)Im(z\overline{z\_1}) + (1+\frac{1}{2}|\frac{z}{w}|)Im(w\overline{z\_2}\overline{w\_1}) + \frac{1}{2}Im(\frac{z}{w}|\frac{w}{z}|(w\_2\overline{z\_1} - w\_1\overline{z\_2})) = 0\tag{15}$$

holds for any ðz;wÞ∈ spanððz1;w1Þ,ðz2;w2ÞÞ. In the following argument, we divide it into three cases to discuss in terms of j w <sup>z</sup> j and <sup>w</sup> z j w z j.

The first case is that j w <sup>z</sup> j ¼ λ for some fixed λ > 0. Let ðz;wÞ ¼ λ1ðz1;w1Þ þ λ2ðz2;w2Þ for any λ1;λ<sup>2</sup> ∈ R, then jλ1w<sup>1</sup> þ λ2w2j ¼ λjλ1z<sup>1</sup> þ λ2z2j, that implies jw1j ¼ λjz1j, jw2j ¼ λjz2j and Reðw1w<sup>2</sup> Þ ¼ <sup>λ</sup><sup>2</sup> Reðz1z<sup>2</sup> <sup>Þ</sup>. It follows that <sup>w</sup><sup>1</sup> <sup>¼</sup> <sup>λ</sup>e<sup>i</sup><sup>θ</sup>z1, <sup>w</sup><sup>2</sup> <sup>¼</sup> <sup>λ</sup>e<sup>i</sup><sup>θ</sup>z2, or <sup>w</sup><sup>1</sup> <sup>¼</sup> <sup>λ</sup>ei<sup>θ</sup>z1, <sup>w</sup><sup>2</sup> <sup>¼</sup> <sup>λ</sup>e<sup>i</sup><sup>θ</sup> <sup>z</sup><sup>2</sup> 1z2 jz1j 2 for some θ∈½0; 2πÞ.

In the sub-case of <sup>w</sup><sup>1</sup> <sup>¼</sup> <sup>λ</sup>e<sup>i</sup><sup>θ</sup>z1, <sup>w</sup><sup>2</sup> <sup>¼</sup> <sup>λ</sup>e<sup>i</sup><sup>θ</sup>z<sup>2</sup> for some <sup>θ</sup>∈½0; <sup>2</sup>πÞ, by Eq. (15) we have

$$(1+\frac{\lambda}{2})Im(z\_2\overline{z\_1}) + (1+\frac{1}{2\lambda})\lambda^2Im(z\_2\overline{z\_1}) + \lambda Im(z\_2\overline{z\_1}) = (1+\lambda)^2Im(z\_2\overline{z\_1}) = 0,\tag{16}$$

which implies Imðz2z<sup>1</sup> Þ ¼ 0 and furthermore Imðw2w<sup>1</sup> Þ ¼ 0. That means ðz1;w1Þ and ðz2;w2Þ are colinear. So this case cannot occur.

However, for the other sub-case of <sup>w</sup><sup>1</sup> <sup>¼</sup> <sup>λ</sup>e<sup>i</sup><sup>θ</sup>z1, <sup>w</sup><sup>2</sup> <sup>¼</sup> <sup>λ</sup>e<sup>i</sup>θz<sup>2</sup> 1z2 jz1j <sup>2</sup> for some θ∈½0; 2πÞ, by Eq. (15) we have

$$(1+\frac{\lambda}{2})Im(\overline{z\_2}\overline{z\_1}) + (1+\frac{1}{2\lambda})\lambda^2 Im(\overline{z\_1}\overline{z\_2}) = (1-\lambda^2)Im(\overline{z\_2}\overline{z\_1}) = 0. \tag{17}$$

Then λ ¼ 1 or Imðz2z<sup>1</sup> Þ ¼ 0, but ðz1;w1Þ and ðz2;w2Þ cannot be colinear. So, we have λ ¼ 1 which gives

$$P = \text{span}((z\_1, z\_1e^{i\theta}), (z\_2, \frac{z\_1^2 \overline{z\_2}}{|z\_1|^2}e^{i\theta})),\tag{18}$$

where z1;z<sup>2</sup> ∈ C\f0g and Imðz1z<sup>2</sup> Þ≠0 for some θ∈½0; 2πÞ. This finishes the first case.

The second case is <sup>w</sup> z j w <sup>z</sup> j ¼ <sup>e</sup><sup>i</sup><sup>θ</sup> for some fixed <sup>θ</sup><sup>∈</sup> <sup>½</sup>0; <sup>2</sup>πÞ. Let <sup>w</sup><sup>1</sup> <sup>¼</sup> <sup>λ</sup>1e<sup>i</sup><sup>θ</sup>z1;w<sup>2</sup> <sup>¼</sup> <sup>λ</sup>2e<sup>i</sup><sup>θ</sup>z<sup>2</sup> for some λ1;λ<sup>2</sup> > 0. Then it follows from (15) that

$$\begin{array}{lcl} & \left(1+\frac{\lambda\_1}{2}\right)Im(z\_2\overline{z\_1}) + \left(1+\frac{1}{2\lambda\_1}\right)\lambda\_1\lambda\_2Im(z\_2\overline{z\_1}) + \frac{1}{2}(\lambda\_1+\lambda\_2)Im(z\_2\overline{z\_1})\\ & = & \left(1+\frac{\lambda\_1}{2}\right)Im(z\_2\overline{z\_1}) + \left(1+\frac{1}{2\lambda\_2}\right)\lambda\_1\lambda\_2Im(z\_2\overline{z\_1}) + \frac{1}{2}(\lambda\_1+\lambda\_2)Im(z\_2\overline{z\_1})\\ & = & (1+\lambda\_1)(1+\lambda\_2)Im(z\_2\overline{z\_1})\\ & = & 0 \end{array} \tag{19}$$

at the points ðz1;w1Þ and ðz2;w2Þ, which implies Imðz2z<sup>1</sup> Þ ¼ 0 and furthermore Imðw2w<sup>1</sup> Þ ¼ 0. Thus, z<sup>1</sup> and z2, w1, and w<sup>2</sup> are colinear, which implies that P equals a plane spanned by one vector from fðz1; <sup>0</sup>Þ,ðz2; <sup>0</sup>Þg and the other from fð0;w1Þ,ð0;w2Þg. Thus <sup>P</sup>∈T<sup>2</sup> .

The last case is the negative to the first one and the second one. It gives Imðz2z<sup>1</sup> Þ ¼ Imðw2w<sup>1</sup> Þ ¼ 0 and w2z<sup>1</sup> −w1z<sup>2</sup> ¼ 0 because of the linear independence, but the former implies the latter by linear transformation, so it is brought down to Imðz2z<sup>1</sup> Þ ¼ Imðw2w<sup>1</sup> Þ ¼ 0. Thus, we have <sup>P</sup>∈T<sup>2</sup> by the second case, and that concludes the proof.
