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Show that preimage is an embedded surface
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I have given a function
$$f:mathbbR^4 to mathbbR^3: (x,y,z,u) mapsto (xz-y^2, yu-z^2,xu-yz)$$
and I want to show that $f^-1(0)setminus0$ is an embedded surface.
If it would be an embedded curve I could verify that $f$ is an submersion to get a 1-dim submanifold. But I don't know how to show that this is an embedded surface.
manifolds
asked Jul 18 at 19:33
Andreas K
575
add a comment |Â
up vote
1
down vote
favorite
I have given a function
$$f:mathbbR^4 to mathbbR^3: (x,y,z,u) mapsto (xz-y^2, yu-z^2,xu-yz)$$
and I want to show that $f^-1(0)setminus0$ is an embedded surface.
If it would be an embedded curve I could verify that $f$ is an submersion to get a 1-dim submanifold. But I don't know how to show that this is an embedded surface.
manifolds
asked Jul 18 at 19:33
Andreas K
575
1st get $f$'s the derivative
– janmarqz
Jul 18 at 19:44
1
Are you sure this is right? It seems to consist of the $x$-axis, the $u$-axis, and the curve $(t^2,t,1,1/t)$. Moreover, $f$ has rank $3$ along the non-linear curve. I was expecting all the components of $f$ to be homogeneous.
– Ted Shifrin
Jul 18 at 19:45
sorry there was a mistake on the last component.
– Andreas K
Jul 18 at 19:51
The derivative is given by: $$Df(x,y,z,u) = beginpmatrix z & -2y & x & 0\ 0 & u & -2z & y\ u & -z & -y & xendpmatrix$$ which has rank 3 for all $(x,y,z,u) in mathbbR^4 setminus 0$ So, $f^-1(0)setminus0$ is not an embedded surface but an 1-dim submanifold of $mathbbR^4$. Is this argumentation right?
– Andreas K
Jul 18 at 20:14
@AndreasK... but at $(x,0,0,0)$ and $(0,0,0,u)$ has rank two
– janmarqz
Jul 18 at 21:03
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have given a function
$$f:mathbbR^4 to mathbbR^3: (x,y,z,u) mapsto (xz-y^2, yu-z^2,xu-yz)$$
and I want to show that $f^-1(0)setminus0$ is an embedded surface.
If it would be an embedded curve I could verify that $f$ is an submersion to get a 1-dim submanifold. But I don't know how to show that this is an embedded surface.
manifolds
asked Jul 18 at 19:33
Andreas K
575
I have given a function
$$f:mathbbR^4 to mathbbR^3: (x,y,z,u) mapsto (xz-y^2, yu-z^2,xu-yz)$$
and I want to show that $f^-1(0)setminus0$ is an embedded surface.
If it would be an embedded curve I could verify that $f$ is an submersion to get a 1-dim submanifold. But I don't know how to show that this is an embedded surface.
manifolds
asked Jul 18 at 19:33
Andreas K
575
edited Jul 18 at 19:50
asked Jul 18 at 19:33
Andreas K
575
asked Jul 18 at 19:33
Andreas K
575
asked Jul 18 at 19:33
Andreas K
575
575
1st get $f$'s the derivative
– janmarqz
Jul 18 at 19:44
1
Are you sure this is right? It seems to consist of the $x$-axis, the $u$-axis, and the curve $(t^2,t,1,1/t)$. Moreover, $f$ has rank $3$ along the non-linear curve. I was expecting all the components of $f$ to be homogeneous.
– Ted Shifrin
Jul 18 at 19:45
sorry there was a mistake on the last component.
– Andreas K
Jul 18 at 19:51
The derivative is given by: $$Df(x,y,z,u) = beginpmatrix z & -2y & x & 0\ 0 & u & -2z & y\ u & -z & -y & xendpmatrix$$ which has rank 3 for all $(x,y,z,u) in mathbbR^4 setminus 0$ So, $f^-1(0)setminus0$ is not an embedded surface but an 1-dim submanifold of $mathbbR^4$. Is this argumentation right?
– Andreas K
Jul 18 at 20:14
@AndreasK... but at $(x,0,0,0)$ and $(0,0,0,u)$ has rank two
– janmarqz
Jul 18 at 21:03
add a comment |Â
1st get $f$'s the derivative
– janmarqz
Jul 18 at 19:44
1
Are you sure this is right? It seems to consist of the $x$-axis, the $u$-axis, and the curve $(t^2,t,1,1/t)$. Moreover, $f$ has rank $3$ along the non-linear curve. I was expecting all the components of $f$ to be homogeneous.
– Ted Shifrin
Jul 18 at 19:45
sorry there was a mistake on the last component.
– Andreas K
Jul 18 at 19:51
The derivative is given by: $$Df(x,y,z,u) = beginpmatrix z & -2y & x & 0\ 0 & u & -2z & y\ u & -z & -y & xendpmatrix$$ which has rank 3 for all $(x,y,z,u) in mathbbR^4 setminus 0$ So, $f^-1(0)setminus0$ is not an embedded surface but an 1-dim submanifold of $mathbbR^4$. Is this argumentation right?
– Andreas K
Jul 18 at 20:14
@AndreasK... but at $(x,0,0,0)$ and $(0,0,0,u)$ has rank two
– janmarqz
Jul 18 at 21:03
1st get $f$'s the derivative
– janmarqz
Jul 18 at 19:44
1st get $f$'s the derivative
– janmarqz
Jul 18 at 19:44
1
1
Are you sure this is right? It seems to consist of the $x$-axis, the $u$-axis, and the curve $(t^2,t,1,1/t)$. Moreover, $f$ has rank $3$ along the non-linear curve. I was expecting all the components of $f$ to be homogeneous.
– Ted Shifrin
Jul 18 at 19:45
Are you sure this is right? It seems to consist of the $x$-axis, the $u$-axis, and the curve $(t^2,t,1,1/t)$. Moreover, $f$ has rank $3$ along the non-linear curve. I was expecting all the components of $f$ to be homogeneous.
– Ted Shifrin
Jul 18 at 19:45
sorry there was a mistake on the last component.
– Andreas K
Jul 18 at 19:51
sorry there was a mistake on the last component.
– Andreas K
Jul 18 at 19:51
The derivative is given by: $$Df(x,y,z,u) = beginpmatrix z & -2y & x & 0\ 0 & u & -2z & y\ u & -z & -y & xendpmatrix$$ which has rank 3 for all $(x,y,z,u) in mathbbR^4 setminus 0$ So, $f^-1(0)setminus0$ is not an embedded surface but an 1-dim submanifold of $mathbbR^4$. Is this argumentation right?
– Andreas K
Jul 18 at 20:14
The derivative is given by: $$Df(x,y,z,u) = beginpmatrix z & -2y & x & 0\ 0 & u & -2z & y\ u & -z & -y & xendpmatrix$$ which has rank 3 for all $(x,y,z,u) in mathbbR^4 setminus 0$ So, $f^-1(0)setminus0$ is not an embedded surface but an 1-dim submanifold of $mathbbR^4$. Is this argumentation right?
– Andreas K
Jul 18 at 20:14
@AndreasK... but at $(x,0,0,0)$ and $(0,0,0,u)$ has rank two
– janmarqz
Jul 18 at 21:03
@AndreasK... but at $(x,0,0,0)$ and $(0,0,0,u)$ has rank two
– janmarqz
Jul 18 at 21:03
add a comment |Â
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1st get $f$'s the derivative
– janmarqz
Jul 18 at 19:44
1
Are you sure this is right? It seems to consist of the $x$-axis, the $u$-axis, and the curve $(t^2,t,1,1/t)$. Moreover, $f$ has rank $3$ along the non-linear curve. I was expecting all the components of $f$ to be homogeneous.
– Ted Shifrin
Jul 18 at 19:45
sorry there was a mistake on the last component.
– Andreas K
Jul 18 at 19:51
The derivative is given by: $$Df(x,y,z,u) = beginpmatrix z & -2y & x & 0\ 0 & u & -2z & y\ u & -z & -y & xendpmatrix$$ which has rank 3 for all $(x,y,z,u) in mathbbR^4 setminus 0$ So, $f^-1(0)setminus0$ is not an embedded surface but an 1-dim submanifold of $mathbbR^4$. Is this argumentation right?
– Andreas K
Jul 18 at 20:14
@AndreasK... but at $(x,0,0,0)$ and $(0,0,0,u)$ has rank two
– janmarqz
Jul 18 at 21:03