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Morse lemma for holomorphic functions
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If $f:C^nto C$ is holomorphic in a neighbourhood of $0$ and $0$ is a nondegenerate critical point, then there is a neighbourhood $U$ of $0$ with a holomorphic local chart, namely a holomorphic invertible map
$varphi =(w_1,…,w_n):U→Vsubset C$,
such that $varphi(0)=0$ and $=f∘Æ^−1$ takes the form $f(w)=f(0)+w^2_1+…+w^2_n$.
Is there a good reference (understandable with the elementary knowledge of differential topology) to this?
It would be really helpful if somebody could write the proof?
reference-request differential-topology morse-theory
asked 2 days ago
user345777
22018
add a comment |Â
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If $f:C^nto C$ is holomorphic in a neighbourhood of $0$ and $0$ is a nondegenerate critical point, then there is a neighbourhood $U$ of $0$ with a holomorphic local chart, namely a holomorphic invertible map
$varphi =(w_1,…,w_n):U→Vsubset C$,
such that $varphi(0)=0$ and $=f∘Æ^−1$ takes the form $f(w)=f(0)+w^2_1+…+w^2_n$.
Is there a good reference (understandable with the elementary knowledge of differential topology) to this?
It would be really helpful if somebody could write the proof?
reference-request differential-topology morse-theory
asked 2 days ago
user345777
22018
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $f:C^nto C$ is holomorphic in a neighbourhood of $0$ and $0$ is a nondegenerate critical point, then there is a neighbourhood $U$ of $0$ with a holomorphic local chart, namely a holomorphic invertible map
$varphi =(w_1,…,w_n):U→Vsubset C$,
such that $varphi(0)=0$ and $=f∘Æ^−1$ takes the form $f(w)=f(0)+w^2_1+…+w^2_n$.
Is there a good reference (understandable with the elementary knowledge of differential topology) to this?
It would be really helpful if somebody could write the proof?
reference-request differential-topology morse-theory
asked 2 days ago
user345777
22018
If $f:C^nto C$ is holomorphic in a neighbourhood of $0$ and $0$ is a nondegenerate critical point, then there is a neighbourhood $U$ of $0$ with a holomorphic local chart, namely a holomorphic invertible map
$varphi =(w_1,…,w_n):U→Vsubset C$,
such that $varphi(0)=0$ and $=f∘Æ^−1$ takes the form $f(w)=f(0)+w^2_1+…+w^2_n$.
Is there a good reference (understandable with the elementary knowledge of differential topology) to this?
It would be really helpful if somebody could write the proof?
reference-request differential-topology morse-theory
asked 2 days ago
user345777
22018
asked 2 days ago
user345777
22018
asked 2 days ago
user345777
22018
asked 2 days ago
user345777
22018
22018
add a comment |Â
add a comment |Â
1 Answer
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The following proof comes from the book "The monodromy group" by Henryk Żołądek (the author says that the proof was suggested by T. Maszczyk).
First you need the lemma of Hadamard (see wikipedia for the proof which can be easily adapted to the complex case) :
Hadamard's lemma : Let $f : Bbb C^n to Bbb C$ be holomorphic with $f(0) = 0$. Then, there exists holomorphics maps $g_1, dots, g_n : Bbb C^n to Bbb C$ so that $f = sum x_ig_i(x)$.
Now we can prove the complex Morse lemma.
Proof of the Morse lemma :
Let $f : Bbb C^n to Bbb C$ be holomorphic with $f(0) = 0$, $d_0f = 0$ and a non-degenerate hessian at $0$. We can write $f(x) = sum x_ig_i(x)$ by Hadamard lemma. By hypothesis $g_i(0) = 0$ so we can apply Hadamard lemma again, and we can write $$ f(x) = sum_i,j x_ix_j g_ij(x) $$
Writing $h_ij = g_ij + g_ji$, we obtain that the matrix $H(x)$ with coefficients $h_ij(x)$ is symetric. Moreover, $H$ is not singular for small $x$, because $H(0)$ is the Hessian of $f$ at $0$.
Now we look at the quadratic form $q_x(xi) = sum h_ij(x)xi_i xi_j $. The form $q_x$ is diagonalisable at the origin, so by implicit function theorem there are local holomorphic coordinates $eta_i = sum_j alpha_ij(x)xi_i$ so that $q(eta) = sum_i eta_i^2$. Now we can take $z_i = sum_j alpha_ij(x)x_j$.
answered 3 hours ago
Nicolas Hemelsoet
4,735215
Could you please state the form implicit function theorem you are using here?
– user345777
2 hours ago
Sure, let $G(x,A) : Bbb C^n times Bbb C^m to Bbb C^r$ be holomorphic with $G(0,0) = 0$. Assume the jacobian matrix with respect to $A$ has maximal rank. Then, there is an open in $Bbb C^n$ containing $0$, and a change of variables $A = A(x)$ so that $G(x,A(x)) = 0$.
– Nicolas Hemelsoet
54 mins ago
Application : we take $x in Bbb C^n$ and $A in M_n times n(Bbb C)$. Let us look the application $G : Bbb C^n times Bbb C^n times n to Bbb C^n times n$ defined by $G_i,j(x,A) = (Aq_xA^-1)_i,j$. We would like $G_i,j = delta_ij$ in a neighborhood of $0$ and this is exactly what said this version of the implicit function theorem.
– Nicolas Hemelsoet
51 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The following proof comes from the book "The monodromy group" by Henryk Żołądek (the author says that the proof was suggested by T. Maszczyk).
First you need the lemma of Hadamard (see wikipedia for the proof which can be easily adapted to the complex case) :
Hadamard's lemma : Let $f : Bbb C^n to Bbb C$ be holomorphic with $f(0) = 0$. Then, there exists holomorphics maps $g_1, dots, g_n : Bbb C^n to Bbb C$ so that $f = sum x_ig_i(x)$.
Now we can prove the complex Morse lemma.
Proof of the Morse lemma :
Let $f : Bbb C^n to Bbb C$ be holomorphic with $f(0) = 0$, $d_0f = 0$ and a non-degenerate hessian at $0$. We can write $f(x) = sum x_ig_i(x)$ by Hadamard lemma. By hypothesis $g_i(0) = 0$ so we can apply Hadamard lemma again, and we can write $$ f(x) = sum_i,j x_ix_j g_ij(x) $$
Writing $h_ij = g_ij + g_ji$, we obtain that the matrix $H(x)$ with coefficients $h_ij(x)$ is symetric. Moreover, $H$ is not singular for small $x$, because $H(0)$ is the Hessian of $f$ at $0$.
Now we look at the quadratic form $q_x(xi) = sum h_ij(x)xi_i xi_j $. The form $q_x$ is diagonalisable at the origin, so by implicit function theorem there are local holomorphic coordinates $eta_i = sum_j alpha_ij(x)xi_i$ so that $q(eta) = sum_i eta_i^2$. Now we can take $z_i = sum_j alpha_ij(x)x_j$.
answered 3 hours ago
Nicolas Hemelsoet
4,735215
Could you please state the form implicit function theorem you are using here?
– user345777
2 hours ago
Sure, let $G(x,A) : Bbb C^n times Bbb C^m to Bbb C^r$ be holomorphic with $G(0,0) = 0$. Assume the jacobian matrix with respect to $A$ has maximal rank. Then, there is an open in $Bbb C^n$ containing $0$, and a change of variables $A = A(x)$ so that $G(x,A(x)) = 0$.
– Nicolas Hemelsoet
54 mins ago
Application : we take $x in Bbb C^n$ and $A in M_n times n(Bbb C)$. Let us look the application $G : Bbb C^n times Bbb C^n times n to Bbb C^n times n$ defined by $G_i,j(x,A) = (Aq_xA^-1)_i,j$. We would like $G_i,j = delta_ij$ in a neighborhood of $0$ and this is exactly what said this version of the implicit function theorem.
– Nicolas Hemelsoet
51 mins ago
add a comment |Â
up vote
2
down vote
The following proof comes from the book "The monodromy group" by Henryk Żołądek (the author says that the proof was suggested by T. Maszczyk).
First you need the lemma of Hadamard (see wikipedia for the proof which can be easily adapted to the complex case) :
Hadamard's lemma : Let $f : Bbb C^n to Bbb C$ be holomorphic with $f(0) = 0$. Then, there exists holomorphics maps $g_1, dots, g_n : Bbb C^n to Bbb C$ so that $f = sum x_ig_i(x)$.
Now we can prove the complex Morse lemma.
Proof of the Morse lemma :
Let $f : Bbb C^n to Bbb C$ be holomorphic with $f(0) = 0$, $d_0f = 0$ and a non-degenerate hessian at $0$. We can write $f(x) = sum x_ig_i(x)$ by Hadamard lemma. By hypothesis $g_i(0) = 0$ so we can apply Hadamard lemma again, and we can write $$ f(x) = sum_i,j x_ix_j g_ij(x) $$
Writing $h_ij = g_ij + g_ji$, we obtain that the matrix $H(x)$ with coefficients $h_ij(x)$ is symetric. Moreover, $H$ is not singular for small $x$, because $H(0)$ is the Hessian of $f$ at $0$.
Now we look at the quadratic form $q_x(xi) = sum h_ij(x)xi_i xi_j $. The form $q_x$ is diagonalisable at the origin, so by implicit function theorem there are local holomorphic coordinates $eta_i = sum_j alpha_ij(x)xi_i$ so that $q(eta) = sum_i eta_i^2$. Now we can take $z_i = sum_j alpha_ij(x)x_j$.
answered 3 hours ago
Nicolas Hemelsoet
4,735215
Could you please state the form implicit function theorem you are using here?
– user345777
2 hours ago
Sure, let $G(x,A) : Bbb C^n times Bbb C^m to Bbb C^r$ be holomorphic with $G(0,0) = 0$. Assume the jacobian matrix with respect to $A$ has maximal rank. Then, there is an open in $Bbb C^n$ containing $0$, and a change of variables $A = A(x)$ so that $G(x,A(x)) = 0$.
– Nicolas Hemelsoet
54 mins ago
Application : we take $x in Bbb C^n$ and $A in M_n times n(Bbb C)$. Let us look the application $G : Bbb C^n times Bbb C^n times n to Bbb C^n times n$ defined by $G_i,j(x,A) = (Aq_xA^-1)_i,j$. We would like $G_i,j = delta_ij$ in a neighborhood of $0$ and this is exactly what said this version of the implicit function theorem.
– Nicolas Hemelsoet
51 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The following proof comes from the book "The monodromy group" by Henryk Żołądek (the author says that the proof was suggested by T. Maszczyk).
First you need the lemma of Hadamard (see wikipedia for the proof which can be easily adapted to the complex case) :
Hadamard's lemma : Let $f : Bbb C^n to Bbb C$ be holomorphic with $f(0) = 0$. Then, there exists holomorphics maps $g_1, dots, g_n : Bbb C^n to Bbb C$ so that $f = sum x_ig_i(x)$.
Now we can prove the complex Morse lemma.
Proof of the Morse lemma :
Let $f : Bbb C^n to Bbb C$ be holomorphic with $f(0) = 0$, $d_0f = 0$ and a non-degenerate hessian at $0$. We can write $f(x) = sum x_ig_i(x)$ by Hadamard lemma. By hypothesis $g_i(0) = 0$ so we can apply Hadamard lemma again, and we can write $$ f(x) = sum_i,j x_ix_j g_ij(x) $$
Writing $h_ij = g_ij + g_ji$, we obtain that the matrix $H(x)$ with coefficients $h_ij(x)$ is symetric. Moreover, $H$ is not singular for small $x$, because $H(0)$ is the Hessian of $f$ at $0$.
Now we look at the quadratic form $q_x(xi) = sum h_ij(x)xi_i xi_j $. The form $q_x$ is diagonalisable at the origin, so by implicit function theorem there are local holomorphic coordinates $eta_i = sum_j alpha_ij(x)xi_i$ so that $q(eta) = sum_i eta_i^2$. Now we can take $z_i = sum_j alpha_ij(x)x_j$.
answered 3 hours ago
Nicolas Hemelsoet
4,735215
The following proof comes from the book "The monodromy group" by Henryk Żołądek (the author says that the proof was suggested by T. Maszczyk).
First you need the lemma of Hadamard (see wikipedia for the proof which can be easily adapted to the complex case) :
Hadamard's lemma : Let $f : Bbb C^n to Bbb C$ be holomorphic with $f(0) = 0$. Then, there exists holomorphics maps $g_1, dots, g_n : Bbb C^n to Bbb C$ so that $f = sum x_ig_i(x)$.
Now we can prove the complex Morse lemma.
Proof of the Morse lemma :
Let $f : Bbb C^n to Bbb C$ be holomorphic with $f(0) = 0$, $d_0f = 0$ and a non-degenerate hessian at $0$. We can write $f(x) = sum x_ig_i(x)$ by Hadamard lemma. By hypothesis $g_i(0) = 0$ so we can apply Hadamard lemma again, and we can write $$ f(x) = sum_i,j x_ix_j g_ij(x) $$
Writing $h_ij = g_ij + g_ji$, we obtain that the matrix $H(x)$ with coefficients $h_ij(x)$ is symetric. Moreover, $H$ is not singular for small $x$, because $H(0)$ is the Hessian of $f$ at $0$.
Now we look at the quadratic form $q_x(xi) = sum h_ij(x)xi_i xi_j $. The form $q_x$ is diagonalisable at the origin, so by implicit function theorem there are local holomorphic coordinates $eta_i = sum_j alpha_ij(x)xi_i$ so that $q(eta) = sum_i eta_i^2$. Now we can take $z_i = sum_j alpha_ij(x)x_j$.
answered 3 hours ago
Nicolas Hemelsoet
4,735215
answered 3 hours ago
Nicolas Hemelsoet
4,735215
answered 3 hours ago
Nicolas Hemelsoet
4,735215
answered 3 hours ago
Nicolas Hemelsoet
4,735215
4,735215
Could you please state the form implicit function theorem you are using here?
– user345777
2 hours ago
Sure, let $G(x,A) : Bbb C^n times Bbb C^m to Bbb C^r$ be holomorphic with $G(0,0) = 0$. Assume the jacobian matrix with respect to $A$ has maximal rank. Then, there is an open in $Bbb C^n$ containing $0$, and a change of variables $A = A(x)$ so that $G(x,A(x)) = 0$.
– Nicolas Hemelsoet
54 mins ago
Application : we take $x in Bbb C^n$ and $A in M_n times n(Bbb C)$. Let us look the application $G : Bbb C^n times Bbb C^n times n to Bbb C^n times n$ defined by $G_i,j(x,A) = (Aq_xA^-1)_i,j$. We would like $G_i,j = delta_ij$ in a neighborhood of $0$ and this is exactly what said this version of the implicit function theorem.
– Nicolas Hemelsoet
51 mins ago
add a comment |Â
Could you please state the form implicit function theorem you are using here?
– user345777
2 hours ago
Sure, let $G(x,A) : Bbb C^n times Bbb C^m to Bbb C^r$ be holomorphic with $G(0,0) = 0$. Assume the jacobian matrix with respect to $A$ has maximal rank. Then, there is an open in $Bbb C^n$ containing $0$, and a change of variables $A = A(x)$ so that $G(x,A(x)) = 0$.
– Nicolas Hemelsoet
54 mins ago
Application : we take $x in Bbb C^n$ and $A in M_n times n(Bbb C)$. Let us look the application $G : Bbb C^n times Bbb C^n times n to Bbb C^n times n$ defined by $G_i,j(x,A) = (Aq_xA^-1)_i,j$. We would like $G_i,j = delta_ij$ in a neighborhood of $0$ and this is exactly what said this version of the implicit function theorem.
– Nicolas Hemelsoet
51 mins ago
Could you please state the form implicit function theorem you are using here?
– user345777
2 hours ago
Could you please state the form implicit function theorem you are using here?
– user345777
2 hours ago
Sure, let $G(x,A) : Bbb C^n times Bbb C^m to Bbb C^r$ be holomorphic with $G(0,0) = 0$. Assume the jacobian matrix with respect to $A$ has maximal rank. Then, there is an open in $Bbb C^n$ containing $0$, and a change of variables $A = A(x)$ so that $G(x,A(x)) = 0$.
– Nicolas Hemelsoet
54 mins ago
Sure, let $G(x,A) : Bbb C^n times Bbb C^m to Bbb C^r$ be holomorphic with $G(0,0) = 0$. Assume the jacobian matrix with respect to $A$ has maximal rank. Then, there is an open in $Bbb C^n$ containing $0$, and a change of variables $A = A(x)$ so that $G(x,A(x)) = 0$.
– Nicolas Hemelsoet
54 mins ago
Application : we take $x in Bbb C^n$ and $A in M_n times n(Bbb C)$. Let us look the application $G : Bbb C^n times Bbb C^n times n to Bbb C^n times n$ defined by $G_i,j(x,A) = (Aq_xA^-1)_i,j$. We would like $G_i,j = delta_ij$ in a neighborhood of $0$ and this is exactly what said this version of the implicit function theorem.
– Nicolas Hemelsoet
51 mins ago
Application : we take $x in Bbb C^n$ and $A in M_n times n(Bbb C)$. Let us look the application $G : Bbb C^n times Bbb C^n times n to Bbb C^n times n$ defined by $G_i,j(x,A) = (Aq_xA^-1)_i,j$. We would like $G_i,j = delta_ij$ in a neighborhood of $0$ and this is exactly what said this version of the implicit function theorem.
– Nicolas Hemelsoet
51 mins ago
add a comment |Â
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