Optimal way to prove smooth homotopy between polynomials
$begingroup$
I am trying to prove that given $p: hat{mathbb{C}} to hat{mathbb{C}}$ a polynomial given by $p(z)=a_nz^n+...+a_0$ then, $p$ is smoothly homotopic to the polynomial $q(z)=a_nz^n$. I am using the homotopy $F: hat{mathbb{C}} times [0,1] to hat{mathbb{C}}$, given by $F(z,t)=a_nz^n+t(a_{n-1}z^{n-1}+...+a_0)$. I can prove that this is an homotopy (its continous), the smooth part is what troubles me, since it looks like a bunch of calculations, and a lot of ''dirty'' work.
Lets say I want to keep my hands clean...How can I go about proving that $F$ is smooth? I have thought about approximating the homotopy by smooth maps but really do not know if I actually can, and if I can, how do I assure the smooth map I took, ''connects'' $p$ to $q$?
If it helps: I want to use this to prove that the Brouwer degree of a polynomial at $hat{mathbb{C}}$ is its actual degree.
Thanks in advance
differential-topology homotopy-theory
asked Nov 28 '18 at 21:32
Bajo FondoBajo Fondo
410315
$endgroup$
add a comment |
$begingroup$
I am trying to prove that given $p: hat{mathbb{C}} to hat{mathbb{C}}$ a polynomial given by $p(z)=a_nz^n+...+a_0$ then, $p$ is smoothly homotopic to the polynomial $q(z)=a_nz^n$. I am using the homotopy $F: hat{mathbb{C}} times [0,1] to hat{mathbb{C}}$, given by $F(z,t)=a_nz^n+t(a_{n-1}z^{n-1}+...+a_0)$. I can prove that this is an homotopy (its continous), the smooth part is what troubles me, since it looks like a bunch of calculations, and a lot of ''dirty'' work.
Lets say I want to keep my hands clean...How can I go about proving that $F$ is smooth? I have thought about approximating the homotopy by smooth maps but really do not know if I actually can, and if I can, how do I assure the smooth map I took, ''connects'' $p$ to $q$?
If it helps: I want to use this to prove that the Brouwer degree of a polynomial at $hat{mathbb{C}}$ is its actual degree.
Thanks in advance
differential-topology homotopy-theory
asked Nov 28 '18 at 21:32
Bajo FondoBajo Fondo
410315
$endgroup$
$begingroup$
$F:mathbb{C}timesmathbb{R}tomathbb{C}$ is a polynomial in $(z,t)$. Is it not given, in your class, that such functions between euclidean spaces are smooth?
$endgroup$
– Prototank
Nov 29 '18 at 14:06
$begingroup$
@Prototank : Yes... but my function is defined in $hat{mathbb{C}}$ (i.e. $S^2$), and there is a subtlety there with the north pole, in fact differentiability elsewhere is given. My problem is pretty much at $N=(0,0,1)$ or $infty$ if you will.
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:19
$begingroup$
I think it is fine, since $F$ fixes $infty$ for all $t$.
$endgroup$
– Prototank
Nov 30 '18 at 0:24
add a comment |
$begingroup$
I am trying to prove that given $p: hat{mathbb{C}} to hat{mathbb{C}}$ a polynomial given by $p(z)=a_nz^n+...+a_0$ then, $p$ is smoothly homotopic to the polynomial $q(z)=a_nz^n$. I am using the homotopy $F: hat{mathbb{C}} times [0,1] to hat{mathbb{C}}$, given by $F(z,t)=a_nz^n+t(a_{n-1}z^{n-1}+...+a_0)$. I can prove that this is an homotopy (its continous), the smooth part is what troubles me, since it looks like a bunch of calculations, and a lot of ''dirty'' work.
Lets say I want to keep my hands clean...How can I go about proving that $F$ is smooth? I have thought about approximating the homotopy by smooth maps but really do not know if I actually can, and if I can, how do I assure the smooth map I took, ''connects'' $p$ to $q$?
If it helps: I want to use this to prove that the Brouwer degree of a polynomial at $hat{mathbb{C}}$ is its actual degree.
Thanks in advance
differential-topology homotopy-theory
asked Nov 28 '18 at 21:32
Bajo FondoBajo Fondo
410315
$endgroup$
I am trying to prove that given $p: hat{mathbb{C}} to hat{mathbb{C}}$ a polynomial given by $p(z)=a_nz^n+...+a_0$ then, $p$ is smoothly homotopic to the polynomial $q(z)=a_nz^n$. I am using the homotopy $F: hat{mathbb{C}} times [0,1] to hat{mathbb{C}}$, given by $F(z,t)=a_nz^n+t(a_{n-1}z^{n-1}+...+a_0)$. I can prove that this is an homotopy (its continous), the smooth part is what troubles me, since it looks like a bunch of calculations, and a lot of ''dirty'' work.
Lets say I want to keep my hands clean...How can I go about proving that $F$ is smooth? I have thought about approximating the homotopy by smooth maps but really do not know if I actually can, and if I can, how do I assure the smooth map I took, ''connects'' $p$ to $q$?
If it helps: I want to use this to prove that the Brouwer degree of a polynomial at $hat{mathbb{C}}$ is its actual degree.
Thanks in advance
differential-topology homotopy-theory
differential-topology homotopy-theory
asked Nov 28 '18 at 21:32
Bajo FondoBajo Fondo
410315
asked Nov 28 '18 at 21:32
Bajo FondoBajo Fondo
410315
asked Nov 28 '18 at 21:32
Bajo FondoBajo Fondo
410315
asked Nov 28 '18 at 21:32
Bajo FondoBajo Fondo
410315
asked Nov 28 '18 at 21:32
Bajo FondoBajo Fondo
410315
410315
$begingroup$
$F:mathbb{C}timesmathbb{R}tomathbb{C}$ is a polynomial in $(z,t)$. Is it not given, in your class, that such functions between euclidean spaces are smooth?
$endgroup$
– Prototank
Nov 29 '18 at 14:06
$begingroup$
@Prototank : Yes... but my function is defined in $hat{mathbb{C}}$ (i.e. $S^2$), and there is a subtlety there with the north pole, in fact differentiability elsewhere is given. My problem is pretty much at $N=(0,0,1)$ or $infty$ if you will.
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:19
$begingroup$
I think it is fine, since $F$ fixes $infty$ for all $t$.
$endgroup$
– Prototank
Nov 30 '18 at 0:24
add a comment |
$begingroup$
$F:mathbb{C}timesmathbb{R}tomathbb{C}$ is a polynomial in $(z,t)$. Is it not given, in your class, that such functions between euclidean spaces are smooth?
$endgroup$
– Prototank
Nov 29 '18 at 14:06
$begingroup$
@Prototank : Yes... but my function is defined in $hat{mathbb{C}}$ (i.e. $S^2$), and there is a subtlety there with the north pole, in fact differentiability elsewhere is given. My problem is pretty much at $N=(0,0,1)$ or $infty$ if you will.
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:19
$begingroup$
I think it is fine, since $F$ fixes $infty$ for all $t$.
$endgroup$
– Prototank
Nov 30 '18 at 0:24
$begingroup$
$F:mathbb{C}timesmathbb{R}tomathbb{C}$ is a polynomial in $(z,t)$. Is it not given, in your class, that such functions between euclidean spaces are smooth?
$endgroup$
– Prototank
Nov 29 '18 at 14:06
$begingroup$
$F:mathbb{C}timesmathbb{R}tomathbb{C}$ is a polynomial in $(z,t)$. Is it not given, in your class, that such functions between euclidean spaces are smooth?
$endgroup$
– Prototank
Nov 29 '18 at 14:06
$begingroup$
@Prototank : Yes... but my function is defined in $hat{mathbb{C}}$ (i.e. $S^2$), and there is a subtlety there with the north pole, in fact differentiability elsewhere is given. My problem is pretty much at $N=(0,0,1)$ or $infty$ if you will.
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:19
$begingroup$
@Prototank : Yes... but my function is defined in $hat{mathbb{C}}$ (i.e. $S^2$), and there is a subtlety there with the north pole, in fact differentiability elsewhere is given. My problem is pretty much at $N=(0,0,1)$ or $infty$ if you will.
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:19
$begingroup$
I think it is fine, since $F$ fixes $infty$ for all $t$.
$endgroup$
– Prototank
Nov 30 '18 at 0:24
$begingroup$
I think it is fine, since $F$ fixes $infty$ for all $t$.
$endgroup$
– Prototank
Nov 30 '18 at 0:24
add a comment |
1 Answer
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First write $z=x+iy$ and $a_k=b_k+ic_k$ for all $k$. Then we can rearrange $F(x,y,t)=a_n(x+iy)^n+t(a_{n-1}(x+iy)^{n-1}+...+a_0)$ so that $F(x,y,t)=u(x,y,t)+iv(x,y,t)$, where the component functions $u$ and $v$ are elements of $mathbb{R}[x,y,t]$. Then, since $u$ and $v$ are just polynomials in several variables, they are infinitely differentiable in each variable. It follows that $F$ is also inifinitely differentiable in each variable. So $F:mathbb{R}^3tomathbb{R}^2$ is smooth.
answered Nov 29 '18 at 16:39
PrototankPrototank
1,105820
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$begingroup$
Does $F$ being smooth from $mathbb{R}^3$ to $mathbb{R}^2$ assures smoothness in $S^2 times [0,1]$ to $S^2$?
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:22
$begingroup$
Think about it this way, if $F$ was defined as in $F(z,t)=t(a_nz^n+..+a_0)$ then in $mathbb{R}^3$ it would be contiunous and smooth as in your proof above. But you can actually check, this function is not even continous on $S^2$
$endgroup$
– Bajo Fondo
Nov 30 '18 at 19:36
add a comment |
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1 Answer
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$begingroup$
First write $z=x+iy$ and $a_k=b_k+ic_k$ for all $k$. Then we can rearrange $F(x,y,t)=a_n(x+iy)^n+t(a_{n-1}(x+iy)^{n-1}+...+a_0)$ so that $F(x,y,t)=u(x,y,t)+iv(x,y,t)$, where the component functions $u$ and $v$ are elements of $mathbb{R}[x,y,t]$. Then, since $u$ and $v$ are just polynomials in several variables, they are infinitely differentiable in each variable. It follows that $F$ is also inifinitely differentiable in each variable. So $F:mathbb{R}^3tomathbb{R}^2$ is smooth.
answered Nov 29 '18 at 16:39
PrototankPrototank
1,105820
$endgroup$
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Does $F$ being smooth from $mathbb{R}^3$ to $mathbb{R}^2$ assures smoothness in $S^2 times [0,1]$ to $S^2$?
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:22
$begingroup$
Think about it this way, if $F$ was defined as in $F(z,t)=t(a_nz^n+..+a_0)$ then in $mathbb{R}^3$ it would be contiunous and smooth as in your proof above. But you can actually check, this function is not even continous on $S^2$
$endgroup$
– Bajo Fondo
Nov 30 '18 at 19:36
add a comment |
$begingroup$
First write $z=x+iy$ and $a_k=b_k+ic_k$ for all $k$. Then we can rearrange $F(x,y,t)=a_n(x+iy)^n+t(a_{n-1}(x+iy)^{n-1}+...+a_0)$ so that $F(x,y,t)=u(x,y,t)+iv(x,y,t)$, where the component functions $u$ and $v$ are elements of $mathbb{R}[x,y,t]$. Then, since $u$ and $v$ are just polynomials in several variables, they are infinitely differentiable in each variable. It follows that $F$ is also inifinitely differentiable in each variable. So $F:mathbb{R}^3tomathbb{R}^2$ is smooth.
answered Nov 29 '18 at 16:39
PrototankPrototank
1,105820
$endgroup$
$begingroup$
Does $F$ being smooth from $mathbb{R}^3$ to $mathbb{R}^2$ assures smoothness in $S^2 times [0,1]$ to $S^2$?
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:22
$begingroup$
Think about it this way, if $F$ was defined as in $F(z,t)=t(a_nz^n+..+a_0)$ then in $mathbb{R}^3$ it would be contiunous and smooth as in your proof above. But you can actually check, this function is not even continous on $S^2$
$endgroup$
– Bajo Fondo
Nov 30 '18 at 19:36
add a comment |
$begingroup$
First write $z=x+iy$ and $a_k=b_k+ic_k$ for all $k$. Then we can rearrange $F(x,y,t)=a_n(x+iy)^n+t(a_{n-1}(x+iy)^{n-1}+...+a_0)$ so that $F(x,y,t)=u(x,y,t)+iv(x,y,t)$, where the component functions $u$ and $v$ are elements of $mathbb{R}[x,y,t]$. Then, since $u$ and $v$ are just polynomials in several variables, they are infinitely differentiable in each variable. It follows that $F$ is also inifinitely differentiable in each variable. So $F:mathbb{R}^3tomathbb{R}^2$ is smooth.
answered Nov 29 '18 at 16:39
PrototankPrototank
1,105820
$endgroup$
First write $z=x+iy$ and $a_k=b_k+ic_k$ for all $k$. Then we can rearrange $F(x,y,t)=a_n(x+iy)^n+t(a_{n-1}(x+iy)^{n-1}+...+a_0)$ so that $F(x,y,t)=u(x,y,t)+iv(x,y,t)$, where the component functions $u$ and $v$ are elements of $mathbb{R}[x,y,t]$. Then, since $u$ and $v$ are just polynomials in several variables, they are infinitely differentiable in each variable. It follows that $F$ is also inifinitely differentiable in each variable. So $F:mathbb{R}^3tomathbb{R}^2$ is smooth.
answered Nov 29 '18 at 16:39
PrototankPrototank
1,105820
answered Nov 29 '18 at 16:39
PrototankPrototank
1,105820
answered Nov 29 '18 at 16:39
PrototankPrototank
1,105820
answered Nov 29 '18 at 16:39
PrototankPrototank
1,105820
1,105820
$begingroup$
Does $F$ being smooth from $mathbb{R}^3$ to $mathbb{R}^2$ assures smoothness in $S^2 times [0,1]$ to $S^2$?
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:22
$begingroup$
Think about it this way, if $F$ was defined as in $F(z,t)=t(a_nz^n+..+a_0)$ then in $mathbb{R}^3$ it would be contiunous and smooth as in your proof above. But you can actually check, this function is not even continous on $S^2$
$endgroup$
– Bajo Fondo
Nov 30 '18 at 19:36
add a comment |
$begingroup$
Does $F$ being smooth from $mathbb{R}^3$ to $mathbb{R}^2$ assures smoothness in $S^2 times [0,1]$ to $S^2$?
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:22
$begingroup$
Think about it this way, if $F$ was defined as in $F(z,t)=t(a_nz^n+..+a_0)$ then in $mathbb{R}^3$ it would be contiunous and smooth as in your proof above. But you can actually check, this function is not even continous on $S^2$
$endgroup$
– Bajo Fondo
Nov 30 '18 at 19:36
$begingroup$
Does $F$ being smooth from $mathbb{R}^3$ to $mathbb{R}^2$ assures smoothness in $S^2 times [0,1]$ to $S^2$?
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:22
$begingroup$
Does $F$ being smooth from $mathbb{R}^3$ to $mathbb{R}^2$ assures smoothness in $S^2 times [0,1]$ to $S^2$?
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:22
$begingroup$
Think about it this way, if $F$ was defined as in $F(z,t)=t(a_nz^n+..+a_0)$ then in $mathbb{R}^3$ it would be contiunous and smooth as in your proof above. But you can actually check, this function is not even continous on $S^2$
$endgroup$
– Bajo Fondo
Nov 30 '18 at 19:36
$begingroup$
Think about it this way, if $F$ was defined as in $F(z,t)=t(a_nz^n+..+a_0)$ then in $mathbb{R}^3$ it would be contiunous and smooth as in your proof above. But you can actually check, this function is not even continous on $S^2$
$endgroup$
– Bajo Fondo
Nov 30 '18 at 19:36
add a comment |
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$begingroup$
$F:mathbb{C}timesmathbb{R}tomathbb{C}$ is a polynomial in $(z,t)$. Is it not given, in your class, that such functions between euclidean spaces are smooth?
$endgroup$
– Prototank
Nov 29 '18 at 14:06
$begingroup$
@Prototank : Yes... but my function is defined in $hat{mathbb{C}}$ (i.e. $S^2$), and there is a subtlety there with the north pole, in fact differentiability elsewhere is given. My problem is pretty much at $N=(0,0,1)$ or $infty$ if you will.
$endgroup$
– Bajo Fondo
Nov 29 '18 at 23:19
$begingroup$
I think it is fine, since $F$ fixes $infty$ for all $t$.
$endgroup$
– Prototank
Nov 30 '18 at 0:24