Non-linear path between symplectic forms in $mathbbR^4$

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Give an example of a pair of symplectic forms $omega_0,omega_1$ in $mathbbR^4$, which:

$1)$ induce the same orientation (i.e., the volume forms $omega_0wedgeomega_0$ and $omega_1wedgeomega_1$ provide the same orientation)

$2)$ have some degenerate convex combination (i.e., $omega_t:=(1-t)omega_0+tomega_1$ is degenerate for some $tin[0,1]$)

$3)$ admit a smooth $1$-parameter family of symplectic forms joining them (i.e., symplectic forms $eta_t$ varying smoothly on $t$ with $eta_0=omega_0$ and $eta_1=omega_1$)

Consider the forms:

$$omega_0:=dxwedge dy+dzwedge dw+dxwedge dz$$
$$omega_1:=dxwedge dy+dzwedge dw+4 dywedge dw$$

They define the same orientation and are both symplectic because:

$$omega_0wedgeomega_0=omega_1wedgeomega_1=2dxwedge dywedge dzwedge dw$$

Furthermore, we can check that $omega_twedgeomega_t=2(1-2t)^2dxwedge dywedge dzwedge dw$, so $omega_t$ is degenerate $Leftrightarrow t=1/2$.

Geometrically, I have a strong feeling that we can find $eta_t_t$ by taking the segment between $omega_0$ and $omega_1$ and making a slight deviation around the point $omega_1/2$.

How could I do this formally?

edited Jul 27 at 23:56

asked Jul 27 at 23:28

rmdmc89

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up vote
2
down vote

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Give an example of a pair of symplectic forms $omega_0,omega_1$ in $mathbbR^4$, which:

$1)$ induce the same orientation (i.e., the volume forms $omega_0wedgeomega_0$ and $omega_1wedgeomega_1$ provide the same orientation)

$2)$ have some degenerate convex combination (i.e., $omega_t:=(1-t)omega_0+tomega_1$ is degenerate for some $tin[0,1]$)

$3)$ admit a smooth $1$-parameter family of symplectic forms joining them (i.e., symplectic forms $eta_t$ varying smoothly on $t$ with $eta_0=omega_0$ and $eta_1=omega_1$)

Consider the forms:

$$omega_0:=dxwedge dy+dzwedge dw+dxwedge dz$$
$$omega_1:=dxwedge dy+dzwedge dw+4 dywedge dw$$

They define the same orientation and are both symplectic because:

$$omega_0wedgeomega_0=omega_1wedgeomega_1=2dxwedge dywedge dzwedge dw$$

Furthermore, we can check that $omega_twedgeomega_t=2(1-2t)^2dxwedge dywedge dzwedge dw$, so $omega_t$ is degenerate $Leftrightarrow t=1/2$.

Geometrically, I have a strong feeling that we can find $eta_t_t$ by taking the segment between $omega_0$ and $omega_1$ and making a slight deviation around the point $omega_1/2$.

How could I do this formally?

edited Jul 27 at 23:56

asked Jul 27 at 23:28

rmdmc89

1,7311520

add a commentÂ |Â

up vote
2
down vote

favorite

Give an example of a pair of symplectic forms $omega_0,omega_1$ in $mathbbR^4$, which:

$1)$ induce the same orientation (i.e., the volume forms $omega_0wedgeomega_0$ and $omega_1wedgeomega_1$ provide the same orientation)

$2)$ have some degenerate convex combination (i.e., $omega_t:=(1-t)omega_0+tomega_1$ is degenerate for some $tin[0,1]$)

$3)$ admit a smooth $1$-parameter family of symplectic forms joining them (i.e., symplectic forms $eta_t$ varying smoothly on $t$ with $eta_0=omega_0$ and $eta_1=omega_1$)

Consider the forms:

$$omega_0:=dxwedge dy+dzwedge dw+dxwedge dz$$
$$omega_1:=dxwedge dy+dzwedge dw+4 dywedge dw$$

They define the same orientation and are both symplectic because:

$$omega_0wedgeomega_0=omega_1wedgeomega_1=2dxwedge dywedge dzwedge dw$$

Furthermore, we can check that $omega_twedgeomega_t=2(1-2t)^2dxwedge dywedge dzwedge dw$, so $omega_t$ is degenerate $Leftrightarrow t=1/2$.

Geometrically, I have a strong feeling that we can find $eta_t_t$ by taking the segment between $omega_0$ and $omega_1$ and making a slight deviation around the point $omega_1/2$.

How could I do this formally?

edited Jul 27 at 23:56

asked Jul 27 at 23:28

rmdmc89

1,7311520

Give an example of a pair of symplectic forms $omega_0,omega_1$ in $mathbbR^4$, which:

$1)$ induce the same orientation (i.e., the volume forms $omega_0wedgeomega_0$ and $omega_1wedgeomega_1$ provide the same orientation)

$2)$ have some degenerate convex combination (i.e., $omega_t:=(1-t)omega_0+tomega_1$ is degenerate for some $tin[0,1]$)

$3)$ admit a smooth $1$-parameter family of symplectic forms joining them (i.e., symplectic forms $eta_t$ varying smoothly on $t$ with $eta_0=omega_0$ and $eta_1=omega_1$)

Consider the forms:

$$omega_0:=dxwedge dy+dzwedge dw+dxwedge dz$$
$$omega_1:=dxwedge dy+dzwedge dw+4 dywedge dw$$

They define the same orientation and are both symplectic because:

$$omega_0wedgeomega_0=omega_1wedgeomega_1=2dxwedge dywedge dzwedge dw$$

Furthermore, we can check that $omega_twedgeomega_t=2(1-2t)^2dxwedge dywedge dzwedge dw$, so $omega_t$ is degenerate $Leftrightarrow t=1/2$.

Geometrically, I have a strong feeling that we can find $eta_t_t$ by taking the segment between $omega_0$ and $omega_1$ and making a slight deviation around the point $omega_1/2$.

How could I do this formally?

edited Jul 27 at 23:56

asked Jul 27 at 23:28

rmdmc89

1,7311520

edited Jul 27 at 23:56

asked Jul 27 at 23:28

rmdmc89

1,7311520

asked Jul 27 at 23:28

rmdmc89

1,7311520

asked Jul 27 at 23:28

rmdmc89

1,7311520

add a commentÂ |Â

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The specific factors $1$ in $1. dx wedge dz$ and $4$ in $4. dy wedge dw$ were not important in order to make $omega_0$ and $omega_1$ symplectic forms with the same orientation. Change these two factors for $t$-dependent functions themselves, for instance

$$ omega_0(t) = dx wedge dy + dz wedge dw - 2(t-1/2) dx wedge dz , , \ omega_1(t) = dx wedge dy + dz wedge dw + 8(t-1/2) dy wedge dw , .$$

The forms $omega'_t = (1-t) omega_0(t) + t omega_1(t)$ are such that $omega'_0 = omega_0(0) = omega_0$ and $omega'_1 = omega_1(1) = omega_1$, and they are all nondegenerate for $t in [0,1]$ since
$$ omega'_t wedge omega'_t = 2(1 + undersetge , 0 mbox for t in [0,1]underbrace16t(1-t)(t-1/2)^2) , dx wedge dy wedge dz wedge dw ; . $$

(Dividing $omega'_t$ by $sqrt1 + 16t(1-t)(t-1/2)^2$ then yields a path $omega''_t$ with constant associated volume form, if that matters to you.)

edited Jul 28 at 15:28

answered Jul 28 at 14:34

Jordan Payette

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$$ omega_0(t) = dx wedge dy + dz wedge dw - 2(t-1/2) dx wedge dz , , \ omega_1(t) = dx wedge dy + dz wedge dw + 8(t-1/2) dy wedge dw , .$$

(Dividing $omega'_t$ by $sqrt1 + 16t(1-t)(t-1/2)^2$ then yields a path $omega''_t$ with constant associated volume form, if that matters to you.)

edited Jul 28 at 15:28

answered Jul 28 at 14:34

Jordan Payette

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up vote
1
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accepted

$$ omega_0(t) = dx wedge dy + dz wedge dw - 2(t-1/2) dx wedge dz , , \ omega_1(t) = dx wedge dy + dz wedge dw + 8(t-1/2) dy wedge dw , .$$

(Dividing $omega'_t$ by $sqrt1 + 16t(1-t)(t-1/2)^2$ then yields a path $omega''_t$ with constant associated volume form, if that matters to you.)

edited Jul 28 at 15:28

answered Jul 28 at 14:34

Jordan Payette

2,681147

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up vote
1
down vote

accepted

$$ omega_0(t) = dx wedge dy + dz wedge dw - 2(t-1/2) dx wedge dz , , \ omega_1(t) = dx wedge dy + dz wedge dw + 8(t-1/2) dy wedge dw , .$$

(Dividing $omega'_t$ by $sqrt1 + 16t(1-t)(t-1/2)^2$ then yields a path $omega''_t$ with constant associated volume form, if that matters to you.)

edited Jul 28 at 15:28

answered Jul 28 at 14:34

Jordan Payette

2,681147

$$ omega_0(t) = dx wedge dy + dz wedge dw - 2(t-1/2) dx wedge dz , , \ omega_1(t) = dx wedge dy + dz wedge dw + 8(t-1/2) dy wedge dw , .$$

(Dividing $omega'_t$ by $sqrt1 + 16t(1-t)(t-1/2)^2$ then yields a path $omega''_t$ with constant associated volume form, if that matters to you.)

edited Jul 28 at 15:28

answered Jul 28 at 14:34

Jordan Payette

2,681147

edited Jul 28 at 15:28

answered Jul 28 at 14:34

Jordan Payette

2,681147

answered Jul 28 at 14:34

Jordan Payette

2,681147

answered Jul 28 at 14:34

Jordan Payette

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