Mixing
How to show that this manifold has dimension 2?
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Q. Given that $M={mathbf{x} in mathbb{R}^4: x_1^2 + x_2^2 + x_3^2 + x_4^2 =1, x_1x_2 = x_3x_4}$. Show that $M$ is a smooth manifold of dimension 2.
I write $M = mathbf{F}^{-1}({mathbf{0}})$, where $mathbf{F}: mathbb{R}^4 rightarrow mathbb{R}^2$ is given by $mathbf{F} = begin{pmatrix} x_1^2 + x_2^2 + x_3^2 + x_4^2 -1 \ x_1x_2 - x_3x_4 end{pmatrix}$.
I calculate the derivative $Dmathbf{F} = begin{pmatrix} 2x_1 & 2x_2 & 2x_3 & 2x_4 \ x_2 & x_1 & -x_4 & -x_3end{pmatrix}$.
If I can show that $Dmathbf{F}$ has rank 2, then that will imply that $M$ has dimension 4-2 =2. But I am unable to show the rank is 2.
I could manipulate the constraints to get $(x_1 pm x_2)^2 + (x_3 mp x_4)^2 =1$, but am stuck after that.
linear-algebra multivariable-calculus smooth-manifolds
asked Jan 8 at 16:33
me10240me10240
450212
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add a comment |
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Q. Given that $M={mathbf{x} in mathbb{R}^4: x_1^2 + x_2^2 + x_3^2 + x_4^2 =1, x_1x_2 = x_3x_4}$. Show that $M$ is a smooth manifold of dimension 2.
I write $M = mathbf{F}^{-1}({mathbf{0}})$, where $mathbf{F}: mathbb{R}^4 rightarrow mathbb{R}^2$ is given by $mathbf{F} = begin{pmatrix} x_1^2 + x_2^2 + x_3^2 + x_4^2 -1 \ x_1x_2 - x_3x_4 end{pmatrix}$.
I calculate the derivative $Dmathbf{F} = begin{pmatrix} 2x_1 & 2x_2 & 2x_3 & 2x_4 \ x_2 & x_1 & -x_4 & -x_3end{pmatrix}$.
If I can show that $Dmathbf{F}$ has rank 2, then that will imply that $M$ has dimension 4-2 =2. But I am unable to show the rank is 2.
I could manipulate the constraints to get $(x_1 pm x_2)^2 + (x_3 mp x_4)^2 =1$, but am stuck after that.
linear-algebra multivariable-calculus smooth-manifolds
asked Jan 8 at 16:33
me10240me10240
450212
$endgroup$
$begingroup$
Hint: consider the determinants formed by the first two and the last two columns. Can they both be zero?
$endgroup$
– Wojowu
Jan 8 at 16:36
$begingroup$
I do not understand. To show that rank is 2, you are suggesting that every 2x2 determinant should be non-zero ?
$endgroup$
– me10240
Jan 8 at 16:41
1
$begingroup$
just one of them is enough
$endgroup$
– Carlos Campos
Jan 8 at 16:42
$begingroup$
A matrix has rank at least $r$ iff you can find some $rtimes r$ submatrix which has nonzero determinant.
$endgroup$
– Wojowu
Jan 8 at 16:45
$begingroup$
So if I consider the first determinant, it will be 0 if $x_1 + x_2 =0 $ or $x_1 - x_2 =0$. If $x_1 =x_2$, then I get $ (x_3 + x_4)^2 =1$. Otherwise I get $(x_3 -x_4)^2 =1$. I am still unable to see what it means.
$endgroup$
– me10240
Jan 8 at 16:54
add a comment |
$begingroup$
Q. Given that $M={mathbf{x} in mathbb{R}^4: x_1^2 + x_2^2 + x_3^2 + x_4^2 =1, x_1x_2 = x_3x_4}$. Show that $M$ is a smooth manifold of dimension 2.
I write $M = mathbf{F}^{-1}({mathbf{0}})$, where $mathbf{F}: mathbb{R}^4 rightarrow mathbb{R}^2$ is given by $mathbf{F} = begin{pmatrix} x_1^2 + x_2^2 + x_3^2 + x_4^2 -1 \ x_1x_2 - x_3x_4 end{pmatrix}$.
I calculate the derivative $Dmathbf{F} = begin{pmatrix} 2x_1 & 2x_2 & 2x_3 & 2x_4 \ x_2 & x_1 & -x_4 & -x_3end{pmatrix}$.
If I can show that $Dmathbf{F}$ has rank 2, then that will imply that $M$ has dimension 4-2 =2. But I am unable to show the rank is 2.
I could manipulate the constraints to get $(x_1 pm x_2)^2 + (x_3 mp x_4)^2 =1$, but am stuck after that.
linear-algebra multivariable-calculus smooth-manifolds
asked Jan 8 at 16:33
me10240me10240
450212
$endgroup$
Q. Given that $M={mathbf{x} in mathbb{R}^4: x_1^2 + x_2^2 + x_3^2 + x_4^2 =1, x_1x_2 = x_3x_4}$. Show that $M$ is a smooth manifold of dimension 2.
I write $M = mathbf{F}^{-1}({mathbf{0}})$, where $mathbf{F}: mathbb{R}^4 rightarrow mathbb{R}^2$ is given by $mathbf{F} = begin{pmatrix} x_1^2 + x_2^2 + x_3^2 + x_4^2 -1 \ x_1x_2 - x_3x_4 end{pmatrix}$.
I calculate the derivative $Dmathbf{F} = begin{pmatrix} 2x_1 & 2x_2 & 2x_3 & 2x_4 \ x_2 & x_1 & -x_4 & -x_3end{pmatrix}$.
If I can show that $Dmathbf{F}$ has rank 2, then that will imply that $M$ has dimension 4-2 =2. But I am unable to show the rank is 2.
I could manipulate the constraints to get $(x_1 pm x_2)^2 + (x_3 mp x_4)^2 =1$, but am stuck after that.
linear-algebra multivariable-calculus smooth-manifolds
linear-algebra multivariable-calculus smooth-manifolds
asked Jan 8 at 16:33
me10240me10240
450212
asked Jan 8 at 16:33
me10240me10240
450212
edited Jan 8 at 16:37
me10240
asked Jan 8 at 16:33
me10240me10240
450212
asked Jan 8 at 16:33
me10240me10240
450212
asked Jan 8 at 16:33
me10240me10240
450212
450212
$begingroup$
Hint: consider the determinants formed by the first two and the last two columns. Can they both be zero?
$endgroup$
– Wojowu
Jan 8 at 16:36
$begingroup$
I do not understand. To show that rank is 2, you are suggesting that every 2x2 determinant should be non-zero ?
$endgroup$
– me10240
Jan 8 at 16:41
1
$begingroup$
just one of them is enough
$endgroup$
– Carlos Campos
Jan 8 at 16:42
$begingroup$
A matrix has rank at least $r$ iff you can find some $rtimes r$ submatrix which has nonzero determinant.
$endgroup$
– Wojowu
Jan 8 at 16:45
$begingroup$
So if I consider the first determinant, it will be 0 if $x_1 + x_2 =0 $ or $x_1 - x_2 =0$. If $x_1 =x_2$, then I get $ (x_3 + x_4)^2 =1$. Otherwise I get $(x_3 -x_4)^2 =1$. I am still unable to see what it means.
$endgroup$
– me10240
Jan 8 at 16:54
add a comment |
$begingroup$
Hint: consider the determinants formed by the first two and the last two columns. Can they both be zero?
$endgroup$
– Wojowu
Jan 8 at 16:36
$begingroup$
I do not understand. To show that rank is 2, you are suggesting that every 2x2 determinant should be non-zero ?
$endgroup$
– me10240
Jan 8 at 16:41
1
$begingroup$
just one of them is enough
$endgroup$
– Carlos Campos
Jan 8 at 16:42
$begingroup$
A matrix has rank at least $r$ iff you can find some $rtimes r$ submatrix which has nonzero determinant.
$endgroup$
– Wojowu
Jan 8 at 16:45
$begingroup$
So if I consider the first determinant, it will be 0 if $x_1 + x_2 =0 $ or $x_1 - x_2 =0$. If $x_1 =x_2$, then I get $ (x_3 + x_4)^2 =1$. Otherwise I get $(x_3 -x_4)^2 =1$. I am still unable to see what it means.
$endgroup$
– me10240
Jan 8 at 16:54
$begingroup$
Hint: consider the determinants formed by the first two and the last two columns. Can they both be zero?
$endgroup$
– Wojowu
Jan 8 at 16:36
$begingroup$
Hint: consider the determinants formed by the first two and the last two columns. Can they both be zero?
$endgroup$
– Wojowu
Jan 8 at 16:36
$begingroup$
I do not understand. To show that rank is 2, you are suggesting that every 2x2 determinant should be non-zero ?
$endgroup$
– me10240
Jan 8 at 16:41
$begingroup$
I do not understand. To show that rank is 2, you are suggesting that every 2x2 determinant should be non-zero ?
$endgroup$
– me10240
Jan 8 at 16:41
1
1
$begingroup$
just one of them is enough
$endgroup$
– Carlos Campos
Jan 8 at 16:42
$begingroup$
just one of them is enough
$endgroup$
– Carlos Campos
Jan 8 at 16:42
$begingroup$
A matrix has rank at least $r$ iff you can find some $rtimes r$ submatrix which has nonzero determinant.
$endgroup$
– Wojowu
Jan 8 at 16:45
$begingroup$
A matrix has rank at least $r$ iff you can find some $rtimes r$ submatrix which has nonzero determinant.
$endgroup$
– Wojowu
Jan 8 at 16:45
$begingroup$
So if I consider the first determinant, it will be 0 if $x_1 + x_2 =0 $ or $x_1 - x_2 =0$. If $x_1 =x_2$, then I get $ (x_3 + x_4)^2 =1$. Otherwise I get $(x_3 -x_4)^2 =1$. I am still unable to see what it means.
$endgroup$
– me10240
Jan 8 at 16:54
$begingroup$
So if I consider the first determinant, it will be 0 if $x_1 + x_2 =0 $ or $x_1 - x_2 =0$. If $x_1 =x_2$, then I get $ (x_3 + x_4)^2 =1$. Otherwise I get $(x_3 -x_4)^2 =1$. I am still unable to see what it means.
$endgroup$
– me10240
Jan 8 at 16:54
add a comment |
1 Answer
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For every $x$, $rank(DF_x)leq 2$. Now $rank(dF_x)leq 1$ iff there is $lambda$ s.t.
$(*)$ $x_2=2lambda x_1,x_1=2lambda x_2,x_4=-2lambda x_3,x_3=-2lambda x_4$ (since $xin M$, $xnot=0$).
Consider the cases when $lambda=pm 1/2$ and $lambdanot=pm 1/2$ and deduce that there are no solutions of $(*)$ in $M$.
answered Jan 9 at 17:13
loup blancloup blanc
23k21850
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add a comment |
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$begingroup$
For every $x$, $rank(DF_x)leq 2$. Now $rank(dF_x)leq 1$ iff there is $lambda$ s.t.
$(*)$ $x_2=2lambda x_1,x_1=2lambda x_2,x_4=-2lambda x_3,x_3=-2lambda x_4$ (since $xin M$, $xnot=0$).
Consider the cases when $lambda=pm 1/2$ and $lambdanot=pm 1/2$ and deduce that there are no solutions of $(*)$ in $M$.
answered Jan 9 at 17:13
loup blancloup blanc
23k21850
$endgroup$
add a comment |
$begingroup$
For every $x$, $rank(DF_x)leq 2$. Now $rank(dF_x)leq 1$ iff there is $lambda$ s.t.
$(*)$ $x_2=2lambda x_1,x_1=2lambda x_2,x_4=-2lambda x_3,x_3=-2lambda x_4$ (since $xin M$, $xnot=0$).
Consider the cases when $lambda=pm 1/2$ and $lambdanot=pm 1/2$ and deduce that there are no solutions of $(*)$ in $M$.
answered Jan 9 at 17:13
loup blancloup blanc
23k21850
$endgroup$
add a comment |
$begingroup$
For every $x$, $rank(DF_x)leq 2$. Now $rank(dF_x)leq 1$ iff there is $lambda$ s.t.
$(*)$ $x_2=2lambda x_1,x_1=2lambda x_2,x_4=-2lambda x_3,x_3=-2lambda x_4$ (since $xin M$, $xnot=0$).
Consider the cases when $lambda=pm 1/2$ and $lambdanot=pm 1/2$ and deduce that there are no solutions of $(*)$ in $M$.
answered Jan 9 at 17:13
loup blancloup blanc
23k21850
$endgroup$
For every $x$, $rank(DF_x)leq 2$. Now $rank(dF_x)leq 1$ iff there is $lambda$ s.t.
$(*)$ $x_2=2lambda x_1,x_1=2lambda x_2,x_4=-2lambda x_3,x_3=-2lambda x_4$ (since $xin M$, $xnot=0$).
Consider the cases when $lambda=pm 1/2$ and $lambdanot=pm 1/2$ and deduce that there are no solutions of $(*)$ in $M$.
answered Jan 9 at 17:13
loup blancloup blanc
23k21850
answered Jan 9 at 17:13
loup blancloup blanc
23k21850
answered Jan 9 at 17:13
loup blancloup blanc
23k21850
answered Jan 9 at 17:13
loup blancloup blanc
23k21850
23k21850
add a comment |
add a comment |
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$begingroup$
Hint: consider the determinants formed by the first two and the last two columns. Can they both be zero?
$endgroup$
– Wojowu
Jan 8 at 16:36
$begingroup$
I do not understand. To show that rank is 2, you are suggesting that every 2x2 determinant should be non-zero ?
$endgroup$
– me10240
Jan 8 at 16:41
1
$begingroup$
just one of them is enough
$endgroup$
– Carlos Campos
Jan 8 at 16:42
$begingroup$
A matrix has rank at least $r$ iff you can find some $rtimes r$ submatrix which has nonzero determinant.
$endgroup$
– Wojowu
Jan 8 at 16:45
$begingroup$
So if I consider the first determinant, it will be 0 if $x_1 + x_2 =0 $ or $x_1 - x_2 =0$. If $x_1 =x_2$, then I get $ (x_3 + x_4)^2 =1$. Otherwise I get $(x_3 -x_4)^2 =1$. I am still unable to see what it means.
$endgroup$
– me10240
Jan 8 at 16:54