Differential of Hopf's map
Let $$h : mathbb{C^2} rightarrow mathbb{C times R} $$
$$h(z_1, z_2) = (2z_1z_2^*, |z_1|^2-|z_2|^2)$$
How do you find the differential of $h$ and show it is onto/surjective?
I know that I can express $h$ as $mathbb{R^4}$ instead of $mathbb{C^2}$, but then how do I proceed? Do I just differentiate with respect to each $x_1, x_2, x_3, x_4$ instead of $z_1, z_2$ given that $x_2, x_4$ are the imaginary part?
So, let's say $h(z_1, z_2)$ becomes $$h(x_1,x_2,x_3,x_4) = (2(x_1+x_2i)(x_3-x_4i), x_1^2+x_2^2-x_3^2-x_4^2) \ rightarrow (2(x_1x_3 + x_2x_4), 2(x_2x_3-x_1x_4), x_1^2+x_2^2-x_3^2-x_4^2)$$
But if I start differentiating $h$ with respect to each variable, I am getting a Jacobian matrix $mathbb{R^{3 times 4}}$. I think I am doing something wrong here.
complex-analysis differential-geometry differential-topology hopf-fibration
edited Dec 14 '18 at 10:57
Paul Frost
9,3862631
asked Dec 11 '18 at 21:05
J.Tegav
133
add a comment |
Let $$h : mathbb{C^2} rightarrow mathbb{C times R} $$
$$h(z_1, z_2) = (2z_1z_2^*, |z_1|^2-|z_2|^2)$$
How do you find the differential of $h$ and show it is onto/surjective?
I know that I can express $h$ as $mathbb{R^4}$ instead of $mathbb{C^2}$, but then how do I proceed? Do I just differentiate with respect to each $x_1, x_2, x_3, x_4$ instead of $z_1, z_2$ given that $x_2, x_4$ are the imaginary part?
So, let's say $h(z_1, z_2)$ becomes $$h(x_1,x_2,x_3,x_4) = (2(x_1+x_2i)(x_3-x_4i), x_1^2+x_2^2-x_3^2-x_4^2) \ rightarrow (2(x_1x_3 + x_2x_4), 2(x_2x_3-x_1x_4), x_1^2+x_2^2-x_3^2-x_4^2)$$
But if I start differentiating $h$ with respect to each variable, I am getting a Jacobian matrix $mathbb{R^{3 times 4}}$. I think I am doing something wrong here.
complex-analysis differential-geometry differential-topology hopf-fibration
edited Dec 14 '18 at 10:57
Paul Frost
9,3862631
asked Dec 11 '18 at 21:05
J.Tegav
133
Welcome to Math.SE ! I recommend that you share more of your work when asking a question. This will help to clarifiy the question and understand what is the difficult you encounter. For instance you could to show what you obtain if you differentiate $h$ both ways and make the question more precise.
– Tom-Tom
Dec 11 '18 at 21:14
add a comment |
Let $$h : mathbb{C^2} rightarrow mathbb{C times R} $$
$$h(z_1, z_2) = (2z_1z_2^*, |z_1|^2-|z_2|^2)$$
How do you find the differential of $h$ and show it is onto/surjective?
I know that I can express $h$ as $mathbb{R^4}$ instead of $mathbb{C^2}$, but then how do I proceed? Do I just differentiate with respect to each $x_1, x_2, x_3, x_4$ instead of $z_1, z_2$ given that $x_2, x_4$ are the imaginary part?
So, let's say $h(z_1, z_2)$ becomes $$h(x_1,x_2,x_3,x_4) = (2(x_1+x_2i)(x_3-x_4i), x_1^2+x_2^2-x_3^2-x_4^2) \ rightarrow (2(x_1x_3 + x_2x_4), 2(x_2x_3-x_1x_4), x_1^2+x_2^2-x_3^2-x_4^2)$$
But if I start differentiating $h$ with respect to each variable, I am getting a Jacobian matrix $mathbb{R^{3 times 4}}$. I think I am doing something wrong here.
complex-analysis differential-geometry differential-topology hopf-fibration
edited Dec 14 '18 at 10:57
Paul Frost
9,3862631
asked Dec 11 '18 at 21:05
J.Tegav
133
Let $$h : mathbb{C^2} rightarrow mathbb{C times R} $$
$$h(z_1, z_2) = (2z_1z_2^*, |z_1|^2-|z_2|^2)$$
How do you find the differential of $h$ and show it is onto/surjective?
I know that I can express $h$ as $mathbb{R^4}$ instead of $mathbb{C^2}$, but then how do I proceed? Do I just differentiate with respect to each $x_1, x_2, x_3, x_4$ instead of $z_1, z_2$ given that $x_2, x_4$ are the imaginary part?
So, let's say $h(z_1, z_2)$ becomes $$h(x_1,x_2,x_3,x_4) = (2(x_1+x_2i)(x_3-x_4i), x_1^2+x_2^2-x_3^2-x_4^2) \ rightarrow (2(x_1x_3 + x_2x_4), 2(x_2x_3-x_1x_4), x_1^2+x_2^2-x_3^2-x_4^2)$$
But if I start differentiating $h$ with respect to each variable, I am getting a Jacobian matrix $mathbb{R^{3 times 4}}$. I think I am doing something wrong here.
complex-analysis differential-geometry differential-topology hopf-fibration
complex-analysis differential-geometry differential-topology hopf-fibration
edited Dec 14 '18 at 10:57
Paul Frost
9,3862631
asked Dec 11 '18 at 21:05
J.Tegav
133
edited Dec 14 '18 at 10:57
Paul Frost
9,3862631
asked Dec 11 '18 at 21:05
J.Tegav
133
edited Dec 14 '18 at 10:57
Paul Frost
9,3862631
edited Dec 14 '18 at 10:57
Paul Frost
9,3862631
edited Dec 14 '18 at 10:57
Paul Frost
9,3862631
9,3862631
asked Dec 11 '18 at 21:05
J.Tegav
133
asked Dec 11 '18 at 21:05
J.Tegav
133
asked Dec 11 '18 at 21:05
J.Tegav
133
133
Welcome to Math.SE ! I recommend that you share more of your work when asking a question. This will help to clarifiy the question and understand what is the difficult you encounter. For instance you could to show what you obtain if you differentiate $h$ both ways and make the question more precise.
– Tom-Tom
Dec 11 '18 at 21:14
add a comment |
Welcome to Math.SE ! I recommend that you share more of your work when asking a question. This will help to clarifiy the question and understand what is the difficult you encounter. For instance you could to show what you obtain if you differentiate $h$ both ways and make the question more precise.
– Tom-Tom
Dec 11 '18 at 21:14
Welcome to Math.SE ! I recommend that you share more of your work when asking a question. This will help to clarifiy the question and understand what is the difficult you encounter. For instance you could to show what you obtain if you differentiate $h$ both ways and make the question more precise.
– Tom-Tom
Dec 11 '18 at 21:14
Welcome to Math.SE ! I recommend that you share more of your work when asking a question. This will help to clarifiy the question and understand what is the difficult you encounter. For instance you could to show what you obtain if you differentiate $h$ both ways and make the question more precise.
– Tom-Tom
Dec 11 '18 at 21:14
add a comment |
1 Answer
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In fact you must understand $h$ as a map from $mathbb{R}^4$ to $mathbb{R}^3$ and the derivative of $h$ as a the derivative in the sense of real multivariable calculus. You have done this almost correctly (I corrected a typo), and you are right that the Jacobian $Jh(x)$ of $h$ at $x$ is a $3 times 4$-matrix. With respect to the standard bases of $mathbb{R}^4, mathbb{R}^3$ it is the matrix representation of the differential $Dh(x)$ of $h$ at $x$ which is linear map $Dh(x) : mathbb{R}^4 to mathbb{R}^3$.
You have to determine for which $x$ the map $Dh(x)$ is a surjection. This is equivalent to determining when $Jh(x)$ has maximal rank, i.e. rank $3$. You have
$$Jh(x) = left( begin{array}{rrrr}
2x_3 & 2x_4 & 2x_1 & 2x_2 \
-2x_4 & 2x_3 & 2x_2 & -2x_1 \
2x_1 & 2x_2 & -2x_3 & -2x_4 \
end{array}right) = 2 left( begin{array}{rrrr}
x_3 & x_4 & x_1 & x_2 \
-x_4 & x_3 & x_2 & -x_1 \
x_1 & x_2 & -x_3 & -x_4 \
end{array}right) = 2 M(x) .$$
You see that $Jh(0) = 0$. i.e. $Jh(0)$ has rank $0$. Let us show that the rank is $3$ if $x ne 0$. Denote by $M_i(x)$ the matrix obtained from $M(x)$ be deleting the $i$-th column. Easy computations show that
$$det M_1(x) = -x_2(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_2(x) = -x_1(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_3(x) = -x_4(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_4(x) = -x_3(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
At least one of these four expressions is $ne 0$ which proves our claim.
answered Dec 14 '18 at 11:45
Paul Frost
9,3862631
add a comment |
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In fact you must understand $h$ as a map from $mathbb{R}^4$ to $mathbb{R}^3$ and the derivative of $h$ as a the derivative in the sense of real multivariable calculus. You have done this almost correctly (I corrected a typo), and you are right that the Jacobian $Jh(x)$ of $h$ at $x$ is a $3 times 4$-matrix. With respect to the standard bases of $mathbb{R}^4, mathbb{R}^3$ it is the matrix representation of the differential $Dh(x)$ of $h$ at $x$ which is linear map $Dh(x) : mathbb{R}^4 to mathbb{R}^3$.
You have to determine for which $x$ the map $Dh(x)$ is a surjection. This is equivalent to determining when $Jh(x)$ has maximal rank, i.e. rank $3$. You have
$$Jh(x) = left( begin{array}{rrrr}
2x_3 & 2x_4 & 2x_1 & 2x_2 \
-2x_4 & 2x_3 & 2x_2 & -2x_1 \
2x_1 & 2x_2 & -2x_3 & -2x_4 \
end{array}right) = 2 left( begin{array}{rrrr}
x_3 & x_4 & x_1 & x_2 \
-x_4 & x_3 & x_2 & -x_1 \
x_1 & x_2 & -x_3 & -x_4 \
end{array}right) = 2 M(x) .$$
You see that $Jh(0) = 0$. i.e. $Jh(0)$ has rank $0$. Let us show that the rank is $3$ if $x ne 0$. Denote by $M_i(x)$ the matrix obtained from $M(x)$ be deleting the $i$-th column. Easy computations show that
$$det M_1(x) = -x_2(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_2(x) = -x_1(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_3(x) = -x_4(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_4(x) = -x_3(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
At least one of these four expressions is $ne 0$ which proves our claim.
answered Dec 14 '18 at 11:45
Paul Frost
9,3862631
add a comment |
In fact you must understand $h$ as a map from $mathbb{R}^4$ to $mathbb{R}^3$ and the derivative of $h$ as a the derivative in the sense of real multivariable calculus. You have done this almost correctly (I corrected a typo), and you are right that the Jacobian $Jh(x)$ of $h$ at $x$ is a $3 times 4$-matrix. With respect to the standard bases of $mathbb{R}^4, mathbb{R}^3$ it is the matrix representation of the differential $Dh(x)$ of $h$ at $x$ which is linear map $Dh(x) : mathbb{R}^4 to mathbb{R}^3$.
You have to determine for which $x$ the map $Dh(x)$ is a surjection. This is equivalent to determining when $Jh(x)$ has maximal rank, i.e. rank $3$. You have
$$Jh(x) = left( begin{array}{rrrr}
2x_3 & 2x_4 & 2x_1 & 2x_2 \
-2x_4 & 2x_3 & 2x_2 & -2x_1 \
2x_1 & 2x_2 & -2x_3 & -2x_4 \
end{array}right) = 2 left( begin{array}{rrrr}
x_3 & x_4 & x_1 & x_2 \
-x_4 & x_3 & x_2 & -x_1 \
x_1 & x_2 & -x_3 & -x_4 \
end{array}right) = 2 M(x) .$$
You see that $Jh(0) = 0$. i.e. $Jh(0)$ has rank $0$. Let us show that the rank is $3$ if $x ne 0$. Denote by $M_i(x)$ the matrix obtained from $M(x)$ be deleting the $i$-th column. Easy computations show that
$$det M_1(x) = -x_2(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_2(x) = -x_1(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_3(x) = -x_4(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_4(x) = -x_3(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
At least one of these four expressions is $ne 0$ which proves our claim.
answered Dec 14 '18 at 11:45
Paul Frost
9,3862631
add a comment |
In fact you must understand $h$ as a map from $mathbb{R}^4$ to $mathbb{R}^3$ and the derivative of $h$ as a the derivative in the sense of real multivariable calculus. You have done this almost correctly (I corrected a typo), and you are right that the Jacobian $Jh(x)$ of $h$ at $x$ is a $3 times 4$-matrix. With respect to the standard bases of $mathbb{R}^4, mathbb{R}^3$ it is the matrix representation of the differential $Dh(x)$ of $h$ at $x$ which is linear map $Dh(x) : mathbb{R}^4 to mathbb{R}^3$.
You have to determine for which $x$ the map $Dh(x)$ is a surjection. This is equivalent to determining when $Jh(x)$ has maximal rank, i.e. rank $3$. You have
$$Jh(x) = left( begin{array}{rrrr}
2x_3 & 2x_4 & 2x_1 & 2x_2 \
-2x_4 & 2x_3 & 2x_2 & -2x_1 \
2x_1 & 2x_2 & -2x_3 & -2x_4 \
end{array}right) = 2 left( begin{array}{rrrr}
x_3 & x_4 & x_1 & x_2 \
-x_4 & x_3 & x_2 & -x_1 \
x_1 & x_2 & -x_3 & -x_4 \
end{array}right) = 2 M(x) .$$
You see that $Jh(0) = 0$. i.e. $Jh(0)$ has rank $0$. Let us show that the rank is $3$ if $x ne 0$. Denote by $M_i(x)$ the matrix obtained from $M(x)$ be deleting the $i$-th column. Easy computations show that
$$det M_1(x) = -x_2(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_2(x) = -x_1(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_3(x) = -x_4(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_4(x) = -x_3(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
At least one of these four expressions is $ne 0$ which proves our claim.
answered Dec 14 '18 at 11:45
Paul Frost
9,3862631
In fact you must understand $h$ as a map from $mathbb{R}^4$ to $mathbb{R}^3$ and the derivative of $h$ as a the derivative in the sense of real multivariable calculus. You have done this almost correctly (I corrected a typo), and you are right that the Jacobian $Jh(x)$ of $h$ at $x$ is a $3 times 4$-matrix. With respect to the standard bases of $mathbb{R}^4, mathbb{R}^3$ it is the matrix representation of the differential $Dh(x)$ of $h$ at $x$ which is linear map $Dh(x) : mathbb{R}^4 to mathbb{R}^3$.
You have to determine for which $x$ the map $Dh(x)$ is a surjection. This is equivalent to determining when $Jh(x)$ has maximal rank, i.e. rank $3$. You have
$$Jh(x) = left( begin{array}{rrrr}
2x_3 & 2x_4 & 2x_1 & 2x_2 \
-2x_4 & 2x_3 & 2x_2 & -2x_1 \
2x_1 & 2x_2 & -2x_3 & -2x_4 \
end{array}right) = 2 left( begin{array}{rrrr}
x_3 & x_4 & x_1 & x_2 \
-x_4 & x_3 & x_2 & -x_1 \
x_1 & x_2 & -x_3 & -x_4 \
end{array}right) = 2 M(x) .$$
You see that $Jh(0) = 0$. i.e. $Jh(0)$ has rank $0$. Let us show that the rank is $3$ if $x ne 0$. Denote by $M_i(x)$ the matrix obtained from $M(x)$ be deleting the $i$-th column. Easy computations show that
$$det M_1(x) = -x_2(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_2(x) = -x_1(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_3(x) = -x_4(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
$$det M_4(x) = -x_3(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$
At least one of these four expressions is $ne 0$ which proves our claim.
answered Dec 14 '18 at 11:45
Paul Frost
9,3862631
answered Dec 14 '18 at 11:45
Paul Frost
9,3862631
answered Dec 14 '18 at 11:45
Paul Frost
9,3862631
answered Dec 14 '18 at 11:45
Paul Frost
9,3862631
9,3862631
add a comment |
add a comment |
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Welcome to Math.SE ! I recommend that you share more of your work when asking a question. This will help to clarifiy the question and understand what is the difficult you encounter. For instance you could to show what you obtain if you differentiate $h$ both ways and make the question more precise.
– Tom-Tom
Dec 11 '18 at 21:14