Mixing
How to find a global flat chart?
$begingroup$
Let $Vsubsetmathbb{R}^3$ with positive coordinates and let $T$ be an associated distribution with
$T=<yfrac{partial}{partial z}-zfrac{partial}{partial y}, zfrac{partial}{partial x}-xfrac{partial}{partial z}>$. How can I find a global flat chart for $T$ on $V$?
differential-geometry differential-topology vector-fields
asked Jan 20 at 20:03
KatherineKatherine
438310
$endgroup$
add a comment |
$begingroup$
Let $Vsubsetmathbb{R}^3$ with positive coordinates and let $T$ be an associated distribution with
$T=<yfrac{partial}{partial z}-zfrac{partial}{partial y}, zfrac{partial}{partial x}-xfrac{partial}{partial z}>$. How can I find a global flat chart for $T$ on $V$?
differential-geometry differential-topology vector-fields
asked Jan 20 at 20:03
KatherineKatherine
438310
$endgroup$
add a comment |
$begingroup$
Let $Vsubsetmathbb{R}^3$ with positive coordinates and let $T$ be an associated distribution with
$T=<yfrac{partial}{partial z}-zfrac{partial}{partial y}, zfrac{partial}{partial x}-xfrac{partial}{partial z}>$. How can I find a global flat chart for $T$ on $V$?
differential-geometry differential-topology vector-fields
asked Jan 20 at 20:03
KatherineKatherine
438310
$endgroup$
Let $Vsubsetmathbb{R}^3$ with positive coordinates and let $T$ be an associated distribution with
$T=<yfrac{partial}{partial z}-zfrac{partial}{partial y}, zfrac{partial}{partial x}-xfrac{partial}{partial z}>$. How can I find a global flat chart for $T$ on $V$?
differential-geometry differential-topology vector-fields
differential-geometry differential-topology vector-fields
asked Jan 20 at 20:03
KatherineKatherine
438310
asked Jan 20 at 20:03
KatherineKatherine
438310
edited Jan 21 at 14:17
Katherine
asked Jan 20 at 20:03
KatherineKatherine
438310
asked Jan 20 at 20:03
KatherineKatherine
438310
asked Jan 20 at 20:03
KatherineKatherine
438310
438310
add a comment |
add a comment |
1 Answer
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$begingroup$
Given: $T = langle xpartial_z - zpartial_y, zpartial_x - xpartial_zrangle$.
First we should check whether your given fields are indeed linearly independent. This is true if and only if $z neq 0$, because this is precisely when $$begin{bmatrix} 0 & z \ -z & 0 \ x & -xend{bmatrix}$$ has full rank. Meaning that $T$ is a distribution on $V$. Then you should check whether $T$ is indeed involutive. We have $$begin{align}[xpartial_z-zpartial_y, zpartial_x - xpartial_z] &= [xpartial_z, zpartial_x] - [xpartial_z,xpartial_z] - [zpartial_y, zpartial_x]+[zpartial_y, xpartial_z] \ &= (xpartial_x - zpartial_z)- 0 - 0- xpartial_y \ &= xpartial_x - xpartial_y - zpartial_z.end{align}$$Writing $$(x,-x,-z) = (0,-zf,xf) + (zg,0,-xg)$$leads to the system $$begin{cases}x = zg \ -x = -zf \ -z = xf-xg, end{cases}$$which has no solution ($f = g= x/z$ is not compatible with the last equation).
So unless I have made any computation mistake, your distribution $T$ is not involutive, hence not integrable, and you will not be able to find flat charts.
For the edited version, we were given $T= langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$.
We start following the same recipe. The matrix $$begin{bmatrix}0 & z\ -z & 0 \ y & -x end{bmatrix}$$again has full rank if $z neq 0$, which happens since all the points in $V$ have positive coordinates, so we indeed have a distribution on $V$. Next, we compute the Lie bracket to check whether $T$ is involutive or not: $$begin{align} [ypartial_z - zpartial_y, zpartial_x - xpartial_z] &=[ypartial_z, zpartial_x] - [ypartial_z, xpartial_z] - [zpartial_y, zpartial_x]+[zpartial_y, xpartial_z] \ &= ypartial_x-0-0-xpartial_y \ &=ypartial_x - xpartial_y.end{align}$$Writing $$(y,-x,0) = (0,-zf,yf)+(zg,0,-xg)$$leads to the system $$begin{cases} y = zg \ -x=-zf \ 0 = yf-xg,end{cases}$$which has the solution $f=x/z$ and $g=y/z$. So $T$ is involutive and hence integrable, so we will be able to find flat charts for $T$. The idea here is that if $X$ and $Y$ are commuting vector fields spanning $T$, the combined flow $$(t,s)mapsto Phi_{t,X}circ Phi_{s,Y}(x,y,z)$$gives a parametrization for the integral surface of $T$ passing through the point $(x,y,z)$. This understood, the idea here is to first find such fields. Since the first $2times 2$ submatrix of the coefficient matrix above has full rank, one can just do $$begin{bmatrix} X \ Y end{bmatrix}
= left(begin{bmatrix} 0 & -z \ z & 0end{bmatrix}^{-1}right)^top begin{bmatrix} ypartial_z-zpartial_y \ zpartial_x -xpartial_zend{bmatrix}.$$Once you have the expressions for $X$ and $Y$ you can solve two systems of ODEs to find the flows, and from this get the flat chart. Can you go on?
More comments: I kept thinking about this in the last couple of days, and indeed the algorithm above is not something natural to think of. So here's an easier way to think of it: using the formulas from classic vector calculus, you can prove that if ${bf N}$ is a non-vanishing vector field along some open set $Usubseteq Bbb R^3$, then ${bf N}^perp hookrightarrow TU$ is involutive if and only if $langle {bf N}, {rm curl}({bf N})rangle =0$. Taking the cross product, we can describe $T$ as ${bf N}^perp$ for $mathbf{N} = xzpartial_y+yzpartial_y+z^2partial_z$. Then ${rm curl}({bf N}) = -ypartial_x+xpartial_y$ is orthogonal to ${bf N}$ and so $T$ is involutive. Since $z>0$ on $V$, we may describe $T$ by the relation $${rm d}z = -frac{x}{z},{rm d}x - frac{y}{z},{rm d}y.$$Now, we have that $$left({rm d}x, {rm d}y, {rm d}z +frac{x}{z},{rm d}x+frac{y}{z},{rm d}z right)$$is a coframe for $T^*V$, and dualizing this back to a frame on $TV$ we get $$left(partial_x - frac{x}{z}partial_z, partial_y - frac{y}{z}partial_z, partial_zright).$$By linear algebra, the first two fields give a trivialization for $T$, and they automatically commute because we have already checked that $T$ is involutive (in fact, they commute if and only if $T$ is involutive -- see here). Thinking of it like this we avoid paranoia with submatrices and keeping track of whether coordinates should be permuted or not.
answered Jan 21 at 6:39
Ivo TerekIvo Terek
46.3k954142
$endgroup$
$begingroup$
I am very sorry! I had a typo, that should be $yfrac{partial}{partial z}$ instead of $xfrac{partial}{partial z}$.
$endgroup$
– Katherine
Jan 21 at 14:18
$begingroup$
So $T$ should be $ = langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$. I have fixed
$endgroup$
– Katherine
Jan 21 at 16:50
1
$begingroup$
Am I computing wrong? I got $X = z^2 partial_x - zx partial_z$ and $Y = zypartial_z -z^2 partial_y$ with $[X,Y] = 2z^2x partial_y - 2z^2 y partial_xneq 0$.
$endgroup$
– Jason DeVito
Jan 21 at 23:25
1
$begingroup$
I'll try to give a rough sketch: if $X_1,ldots, X_k$ is an involutive (local) frame for an involutive distribution $T hookrightarrow TM$ around some point $p$, you start writing $X_j = sum_i rho_{ij}partial_i$, where $rho$ takes values in ${rm Monomorphism}(Bbb R^k, Bbb R^n$. We may assume that the $rho e_1(p),ldots, rho e_k(p), e_{k+1}(p),ldots, e_n(p)$ is a basis for $T_pM$. Put these vectors in the column of a matrix $G(x)$; in particular know that $G(p)$ is non-singular. Reducing the neighboorhood if necessary [...]
$endgroup$
– Ivo Terek
Jan 23 at 5:04
1
$begingroup$
[...] , we may assume that $G(x)$ is also non-singular for $x$ near $p$, and write it in blocks as $$G(x)^{-1}begin{pmatrix} A(x) & 0 \ ast & {rm Id}end{pmatrix}$$Then $G(x)G(x)^{-1} = {rm Id}$ implies that $$rho(x)A(x) = begin{pmatrix} {rm Id} \ B(x)end{pmatrix}$$for some $B(x)$. So put $$V_r = sum_{j=1}^k a_{jr}X_j = partial_r + sum_{s=k+1}^nb_{sr}partial_s.$$Then the fact that $V_1,ldots, V_k$ is a commuting frame follows from $X_1,ldots, X_k$ being an involutive frame.
$endgroup$
– Ivo Terek
Jan 23 at 5:04
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$begingroup$
Given: $T = langle xpartial_z - zpartial_y, zpartial_x - xpartial_zrangle$.
First we should check whether your given fields are indeed linearly independent. This is true if and only if $z neq 0$, because this is precisely when $$begin{bmatrix} 0 & z \ -z & 0 \ x & -xend{bmatrix}$$ has full rank. Meaning that $T$ is a distribution on $V$. Then you should check whether $T$ is indeed involutive. We have $$begin{align}[xpartial_z-zpartial_y, zpartial_x - xpartial_z] &= [xpartial_z, zpartial_x] - [xpartial_z,xpartial_z] - [zpartial_y, zpartial_x]+[zpartial_y, xpartial_z] \ &= (xpartial_x - zpartial_z)- 0 - 0- xpartial_y \ &= xpartial_x - xpartial_y - zpartial_z.end{align}$$Writing $$(x,-x,-z) = (0,-zf,xf) + (zg,0,-xg)$$leads to the system $$begin{cases}x = zg \ -x = -zf \ -z = xf-xg, end{cases}$$which has no solution ($f = g= x/z$ is not compatible with the last equation).
So unless I have made any computation mistake, your distribution $T$ is not involutive, hence not integrable, and you will not be able to find flat charts.
For the edited version, we were given $T= langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$.
We start following the same recipe. The matrix $$begin{bmatrix}0 & z\ -z & 0 \ y & -x end{bmatrix}$$again has full rank if $z neq 0$, which happens since all the points in $V$ have positive coordinates, so we indeed have a distribution on $V$. Next, we compute the Lie bracket to check whether $T$ is involutive or not: $$begin{align} [ypartial_z - zpartial_y, zpartial_x - xpartial_z] &=[ypartial_z, zpartial_x] - [ypartial_z, xpartial_z] - [zpartial_y, zpartial_x]+[zpartial_y, xpartial_z] \ &= ypartial_x-0-0-xpartial_y \ &=ypartial_x - xpartial_y.end{align}$$Writing $$(y,-x,0) = (0,-zf,yf)+(zg,0,-xg)$$leads to the system $$begin{cases} y = zg \ -x=-zf \ 0 = yf-xg,end{cases}$$which has the solution $f=x/z$ and $g=y/z$. So $T$ is involutive and hence integrable, so we will be able to find flat charts for $T$. The idea here is that if $X$ and $Y$ are commuting vector fields spanning $T$, the combined flow $$(t,s)mapsto Phi_{t,X}circ Phi_{s,Y}(x,y,z)$$gives a parametrization for the integral surface of $T$ passing through the point $(x,y,z)$. This understood, the idea here is to first find such fields. Since the first $2times 2$ submatrix of the coefficient matrix above has full rank, one can just do $$begin{bmatrix} X \ Y end{bmatrix}
= left(begin{bmatrix} 0 & -z \ z & 0end{bmatrix}^{-1}right)^top begin{bmatrix} ypartial_z-zpartial_y \ zpartial_x -xpartial_zend{bmatrix}.$$Once you have the expressions for $X$ and $Y$ you can solve two systems of ODEs to find the flows, and from this get the flat chart. Can you go on?
More comments: I kept thinking about this in the last couple of days, and indeed the algorithm above is not something natural to think of. So here's an easier way to think of it: using the formulas from classic vector calculus, you can prove that if ${bf N}$ is a non-vanishing vector field along some open set $Usubseteq Bbb R^3$, then ${bf N}^perp hookrightarrow TU$ is involutive if and only if $langle {bf N}, {rm curl}({bf N})rangle =0$. Taking the cross product, we can describe $T$ as ${bf N}^perp$ for $mathbf{N} = xzpartial_y+yzpartial_y+z^2partial_z$. Then ${rm curl}({bf N}) = -ypartial_x+xpartial_y$ is orthogonal to ${bf N}$ and so $T$ is involutive. Since $z>0$ on $V$, we may describe $T$ by the relation $${rm d}z = -frac{x}{z},{rm d}x - frac{y}{z},{rm d}y.$$Now, we have that $$left({rm d}x, {rm d}y, {rm d}z +frac{x}{z},{rm d}x+frac{y}{z},{rm d}z right)$$is a coframe for $T^*V$, and dualizing this back to a frame on $TV$ we get $$left(partial_x - frac{x}{z}partial_z, partial_y - frac{y}{z}partial_z, partial_zright).$$By linear algebra, the first two fields give a trivialization for $T$, and they automatically commute because we have already checked that $T$ is involutive (in fact, they commute if and only if $T$ is involutive -- see here). Thinking of it like this we avoid paranoia with submatrices and keeping track of whether coordinates should be permuted or not.
answered Jan 21 at 6:39
Ivo TerekIvo Terek
46.3k954142
$endgroup$
$begingroup$
I am very sorry! I had a typo, that should be $yfrac{partial}{partial z}$ instead of $xfrac{partial}{partial z}$.
$endgroup$
– Katherine
Jan 21 at 14:18
$begingroup$
So $T$ should be $ = langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$. I have fixed
$endgroup$
– Katherine
Jan 21 at 16:50
1
$begingroup$
Am I computing wrong? I got $X = z^2 partial_x - zx partial_z$ and $Y = zypartial_z -z^2 partial_y$ with $[X,Y] = 2z^2x partial_y - 2z^2 y partial_xneq 0$.
$endgroup$
– Jason DeVito
Jan 21 at 23:25
1
$begingroup$
I'll try to give a rough sketch: if $X_1,ldots, X_k$ is an involutive (local) frame for an involutive distribution $T hookrightarrow TM$ around some point $p$, you start writing $X_j = sum_i rho_{ij}partial_i$, where $rho$ takes values in ${rm Monomorphism}(Bbb R^k, Bbb R^n$. We may assume that the $rho e_1(p),ldots, rho e_k(p), e_{k+1}(p),ldots, e_n(p)$ is a basis for $T_pM$. Put these vectors in the column of a matrix $G(x)$; in particular know that $G(p)$ is non-singular. Reducing the neighboorhood if necessary [...]
$endgroup$
– Ivo Terek
Jan 23 at 5:04
1
$begingroup$
[...] , we may assume that $G(x)$ is also non-singular for $x$ near $p$, and write it in blocks as $$G(x)^{-1}begin{pmatrix} A(x) & 0 \ ast & {rm Id}end{pmatrix}$$Then $G(x)G(x)^{-1} = {rm Id}$ implies that $$rho(x)A(x) = begin{pmatrix} {rm Id} \ B(x)end{pmatrix}$$for some $B(x)$. So put $$V_r = sum_{j=1}^k a_{jr}X_j = partial_r + sum_{s=k+1}^nb_{sr}partial_s.$$Then the fact that $V_1,ldots, V_k$ is a commuting frame follows from $X_1,ldots, X_k$ being an involutive frame.
$endgroup$
– Ivo Terek
Jan 23 at 5:04
|
show 7 more comments
$begingroup$
Given: $T = langle xpartial_z - zpartial_y, zpartial_x - xpartial_zrangle$.
First we should check whether your given fields are indeed linearly independent. This is true if and only if $z neq 0$, because this is precisely when $$begin{bmatrix} 0 & z \ -z & 0 \ x & -xend{bmatrix}$$ has full rank. Meaning that $T$ is a distribution on $V$. Then you should check whether $T$ is indeed involutive. We have $$begin{align}[xpartial_z-zpartial_y, zpartial_x - xpartial_z] &= [xpartial_z, zpartial_x] - [xpartial_z,xpartial_z] - [zpartial_y, zpartial_x]+[zpartial_y, xpartial_z] \ &= (xpartial_x - zpartial_z)- 0 - 0- xpartial_y \ &= xpartial_x - xpartial_y - zpartial_z.end{align}$$Writing $$(x,-x,-z) = (0,-zf,xf) + (zg,0,-xg)$$leads to the system $$begin{cases}x = zg \ -x = -zf \ -z = xf-xg, end{cases}$$which has no solution ($f = g= x/z$ is not compatible with the last equation).
So unless I have made any computation mistake, your distribution $T$ is not involutive, hence not integrable, and you will not be able to find flat charts.
For the edited version, we were given $T= langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$.
We start following the same recipe. The matrix $$begin{bmatrix}0 & z\ -z & 0 \ y & -x end{bmatrix}$$again has full rank if $z neq 0$, which happens since all the points in $V$ have positive coordinates, so we indeed have a distribution on $V$. Next, we compute the Lie bracket to check whether $T$ is involutive or not: $$begin{align} [ypartial_z - zpartial_y, zpartial_x - xpartial_z] &=[ypartial_z, zpartial_x] - [ypartial_z, xpartial_z] - [zpartial_y, zpartial_x]+[zpartial_y, xpartial_z] \ &= ypartial_x-0-0-xpartial_y \ &=ypartial_x - xpartial_y.end{align}$$Writing $$(y,-x,0) = (0,-zf,yf)+(zg,0,-xg)$$leads to the system $$begin{cases} y = zg \ -x=-zf \ 0 = yf-xg,end{cases}$$which has the solution $f=x/z$ and $g=y/z$. So $T$ is involutive and hence integrable, so we will be able to find flat charts for $T$. The idea here is that if $X$ and $Y$ are commuting vector fields spanning $T$, the combined flow $$(t,s)mapsto Phi_{t,X}circ Phi_{s,Y}(x,y,z)$$gives a parametrization for the integral surface of $T$ passing through the point $(x,y,z)$. This understood, the idea here is to first find such fields. Since the first $2times 2$ submatrix of the coefficient matrix above has full rank, one can just do $$begin{bmatrix} X \ Y end{bmatrix}
= left(begin{bmatrix} 0 & -z \ z & 0end{bmatrix}^{-1}right)^top begin{bmatrix} ypartial_z-zpartial_y \ zpartial_x -xpartial_zend{bmatrix}.$$Once you have the expressions for $X$ and $Y$ you can solve two systems of ODEs to find the flows, and from this get the flat chart. Can you go on?
More comments: I kept thinking about this in the last couple of days, and indeed the algorithm above is not something natural to think of. So here's an easier way to think of it: using the formulas from classic vector calculus, you can prove that if ${bf N}$ is a non-vanishing vector field along some open set $Usubseteq Bbb R^3$, then ${bf N}^perp hookrightarrow TU$ is involutive if and only if $langle {bf N}, {rm curl}({bf N})rangle =0$. Taking the cross product, we can describe $T$ as ${bf N}^perp$ for $mathbf{N} = xzpartial_y+yzpartial_y+z^2partial_z$. Then ${rm curl}({bf N}) = -ypartial_x+xpartial_y$ is orthogonal to ${bf N}$ and so $T$ is involutive. Since $z>0$ on $V$, we may describe $T$ by the relation $${rm d}z = -frac{x}{z},{rm d}x - frac{y}{z},{rm d}y.$$Now, we have that $$left({rm d}x, {rm d}y, {rm d}z +frac{x}{z},{rm d}x+frac{y}{z},{rm d}z right)$$is a coframe for $T^*V$, and dualizing this back to a frame on $TV$ we get $$left(partial_x - frac{x}{z}partial_z, partial_y - frac{y}{z}partial_z, partial_zright).$$By linear algebra, the first two fields give a trivialization for $T$, and they automatically commute because we have already checked that $T$ is involutive (in fact, they commute if and only if $T$ is involutive -- see here). Thinking of it like this we avoid paranoia with submatrices and keeping track of whether coordinates should be permuted or not.
answered Jan 21 at 6:39
Ivo TerekIvo Terek
46.3k954142
$endgroup$
$begingroup$
I am very sorry! I had a typo, that should be $yfrac{partial}{partial z}$ instead of $xfrac{partial}{partial z}$.
$endgroup$
– Katherine
Jan 21 at 14:18
$begingroup$
So $T$ should be $ = langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$. I have fixed
$endgroup$
– Katherine
Jan 21 at 16:50
1
$begingroup$
Am I computing wrong? I got $X = z^2 partial_x - zx partial_z$ and $Y = zypartial_z -z^2 partial_y$ with $[X,Y] = 2z^2x partial_y - 2z^2 y partial_xneq 0$.
$endgroup$
– Jason DeVito
Jan 21 at 23:25
1
$begingroup$
I'll try to give a rough sketch: if $X_1,ldots, X_k$ is an involutive (local) frame for an involutive distribution $T hookrightarrow TM$ around some point $p$, you start writing $X_j = sum_i rho_{ij}partial_i$, where $rho$ takes values in ${rm Monomorphism}(Bbb R^k, Bbb R^n$. We may assume that the $rho e_1(p),ldots, rho e_k(p), e_{k+1}(p),ldots, e_n(p)$ is a basis for $T_pM$. Put these vectors in the column of a matrix $G(x)$; in particular know that $G(p)$ is non-singular. Reducing the neighboorhood if necessary [...]
$endgroup$
– Ivo Terek
Jan 23 at 5:04
1
$begingroup$
[...] , we may assume that $G(x)$ is also non-singular for $x$ near $p$, and write it in blocks as $$G(x)^{-1}begin{pmatrix} A(x) & 0 \ ast & {rm Id}end{pmatrix}$$Then $G(x)G(x)^{-1} = {rm Id}$ implies that $$rho(x)A(x) = begin{pmatrix} {rm Id} \ B(x)end{pmatrix}$$for some $B(x)$. So put $$V_r = sum_{j=1}^k a_{jr}X_j = partial_r + sum_{s=k+1}^nb_{sr}partial_s.$$Then the fact that $V_1,ldots, V_k$ is a commuting frame follows from $X_1,ldots, X_k$ being an involutive frame.
$endgroup$
– Ivo Terek
Jan 23 at 5:04
|
show 7 more comments
$begingroup$
Given: $T = langle xpartial_z - zpartial_y, zpartial_x - xpartial_zrangle$.
First we should check whether your given fields are indeed linearly independent. This is true if and only if $z neq 0$, because this is precisely when $$begin{bmatrix} 0 & z \ -z & 0 \ x & -xend{bmatrix}$$ has full rank. Meaning that $T$ is a distribution on $V$. Then you should check whether $T$ is indeed involutive. We have $$begin{align}[xpartial_z-zpartial_y, zpartial_x - xpartial_z] &= [xpartial_z, zpartial_x] - [xpartial_z,xpartial_z] - [zpartial_y, zpartial_x]+[zpartial_y, xpartial_z] \ &= (xpartial_x - zpartial_z)- 0 - 0- xpartial_y \ &= xpartial_x - xpartial_y - zpartial_z.end{align}$$Writing $$(x,-x,-z) = (0,-zf,xf) + (zg,0,-xg)$$leads to the system $$begin{cases}x = zg \ -x = -zf \ -z = xf-xg, end{cases}$$which has no solution ($f = g= x/z$ is not compatible with the last equation).
So unless I have made any computation mistake, your distribution $T$ is not involutive, hence not integrable, and you will not be able to find flat charts.
For the edited version, we were given $T= langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$.
We start following the same recipe. The matrix $$begin{bmatrix}0 & z\ -z & 0 \ y & -x end{bmatrix}$$again has full rank if $z neq 0$, which happens since all the points in $V$ have positive coordinates, so we indeed have a distribution on $V$. Next, we compute the Lie bracket to check whether $T$ is involutive or not: $$begin{align} [ypartial_z - zpartial_y, zpartial_x - xpartial_z] &=[ypartial_z, zpartial_x] - [ypartial_z, xpartial_z] - [zpartial_y, zpartial_x]+[zpartial_y, xpartial_z] \ &= ypartial_x-0-0-xpartial_y \ &=ypartial_x - xpartial_y.end{align}$$Writing $$(y,-x,0) = (0,-zf,yf)+(zg,0,-xg)$$leads to the system $$begin{cases} y = zg \ -x=-zf \ 0 = yf-xg,end{cases}$$which has the solution $f=x/z$ and $g=y/z$. So $T$ is involutive and hence integrable, so we will be able to find flat charts for $T$. The idea here is that if $X$ and $Y$ are commuting vector fields spanning $T$, the combined flow $$(t,s)mapsto Phi_{t,X}circ Phi_{s,Y}(x,y,z)$$gives a parametrization for the integral surface of $T$ passing through the point $(x,y,z)$. This understood, the idea here is to first find such fields. Since the first $2times 2$ submatrix of the coefficient matrix above has full rank, one can just do $$begin{bmatrix} X \ Y end{bmatrix}
= left(begin{bmatrix} 0 & -z \ z & 0end{bmatrix}^{-1}right)^top begin{bmatrix} ypartial_z-zpartial_y \ zpartial_x -xpartial_zend{bmatrix}.$$Once you have the expressions for $X$ and $Y$ you can solve two systems of ODEs to find the flows, and from this get the flat chart. Can you go on?
More comments: I kept thinking about this in the last couple of days, and indeed the algorithm above is not something natural to think of. So here's an easier way to think of it: using the formulas from classic vector calculus, you can prove that if ${bf N}$ is a non-vanishing vector field along some open set $Usubseteq Bbb R^3$, then ${bf N}^perp hookrightarrow TU$ is involutive if and only if $langle {bf N}, {rm curl}({bf N})rangle =0$. Taking the cross product, we can describe $T$ as ${bf N}^perp$ for $mathbf{N} = xzpartial_y+yzpartial_y+z^2partial_z$. Then ${rm curl}({bf N}) = -ypartial_x+xpartial_y$ is orthogonal to ${bf N}$ and so $T$ is involutive. Since $z>0$ on $V$, we may describe $T$ by the relation $${rm d}z = -frac{x}{z},{rm d}x - frac{y}{z},{rm d}y.$$Now, we have that $$left({rm d}x, {rm d}y, {rm d}z +frac{x}{z},{rm d}x+frac{y}{z},{rm d}z right)$$is a coframe for $T^*V$, and dualizing this back to a frame on $TV$ we get $$left(partial_x - frac{x}{z}partial_z, partial_y - frac{y}{z}partial_z, partial_zright).$$By linear algebra, the first two fields give a trivialization for $T$, and they automatically commute because we have already checked that $T$ is involutive (in fact, they commute if and only if $T$ is involutive -- see here). Thinking of it like this we avoid paranoia with submatrices and keeping track of whether coordinates should be permuted or not.
answered Jan 21 at 6:39
Ivo TerekIvo Terek
46.3k954142
$endgroup$
Given: $T = langle xpartial_z - zpartial_y, zpartial_x - xpartial_zrangle$.
First we should check whether your given fields are indeed linearly independent. This is true if and only if $z neq 0$, because this is precisely when $$begin{bmatrix} 0 & z \ -z & 0 \ x & -xend{bmatrix}$$ has full rank. Meaning that $T$ is a distribution on $V$. Then you should check whether $T$ is indeed involutive. We have $$begin{align}[xpartial_z-zpartial_y, zpartial_x - xpartial_z] &= [xpartial_z, zpartial_x] - [xpartial_z,xpartial_z] - [zpartial_y, zpartial_x]+[zpartial_y, xpartial_z] \ &= (xpartial_x - zpartial_z)- 0 - 0- xpartial_y \ &= xpartial_x - xpartial_y - zpartial_z.end{align}$$Writing $$(x,-x,-z) = (0,-zf,xf) + (zg,0,-xg)$$leads to the system $$begin{cases}x = zg \ -x = -zf \ -z = xf-xg, end{cases}$$which has no solution ($f = g= x/z$ is not compatible with the last equation).
So unless I have made any computation mistake, your distribution $T$ is not involutive, hence not integrable, and you will not be able to find flat charts.
For the edited version, we were given $T= langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$.
We start following the same recipe. The matrix $$begin{bmatrix}0 & z\ -z & 0 \ y & -x end{bmatrix}$$again has full rank if $z neq 0$, which happens since all the points in $V$ have positive coordinates, so we indeed have a distribution on $V$. Next, we compute the Lie bracket to check whether $T$ is involutive or not: $$begin{align} [ypartial_z - zpartial_y, zpartial_x - xpartial_z] &=[ypartial_z, zpartial_x] - [ypartial_z, xpartial_z] - [zpartial_y, zpartial_x]+[zpartial_y, xpartial_z] \ &= ypartial_x-0-0-xpartial_y \ &=ypartial_x - xpartial_y.end{align}$$Writing $$(y,-x,0) = (0,-zf,yf)+(zg,0,-xg)$$leads to the system $$begin{cases} y = zg \ -x=-zf \ 0 = yf-xg,end{cases}$$which has the solution $f=x/z$ and $g=y/z$. So $T$ is involutive and hence integrable, so we will be able to find flat charts for $T$. The idea here is that if $X$ and $Y$ are commuting vector fields spanning $T$, the combined flow $$(t,s)mapsto Phi_{t,X}circ Phi_{s,Y}(x,y,z)$$gives a parametrization for the integral surface of $T$ passing through the point $(x,y,z)$. This understood, the idea here is to first find such fields. Since the first $2times 2$ submatrix of the coefficient matrix above has full rank, one can just do $$begin{bmatrix} X \ Y end{bmatrix}
= left(begin{bmatrix} 0 & -z \ z & 0end{bmatrix}^{-1}right)^top begin{bmatrix} ypartial_z-zpartial_y \ zpartial_x -xpartial_zend{bmatrix}.$$Once you have the expressions for $X$ and $Y$ you can solve two systems of ODEs to find the flows, and from this get the flat chart. Can you go on?
More comments: I kept thinking about this in the last couple of days, and indeed the algorithm above is not something natural to think of. So here's an easier way to think of it: using the formulas from classic vector calculus, you can prove that if ${bf N}$ is a non-vanishing vector field along some open set $Usubseteq Bbb R^3$, then ${bf N}^perp hookrightarrow TU$ is involutive if and only if $langle {bf N}, {rm curl}({bf N})rangle =0$. Taking the cross product, we can describe $T$ as ${bf N}^perp$ for $mathbf{N} = xzpartial_y+yzpartial_y+z^2partial_z$. Then ${rm curl}({bf N}) = -ypartial_x+xpartial_y$ is orthogonal to ${bf N}$ and so $T$ is involutive. Since $z>0$ on $V$, we may describe $T$ by the relation $${rm d}z = -frac{x}{z},{rm d}x - frac{y}{z},{rm d}y.$$Now, we have that $$left({rm d}x, {rm d}y, {rm d}z +frac{x}{z},{rm d}x+frac{y}{z},{rm d}z right)$$is a coframe for $T^*V$, and dualizing this back to a frame on $TV$ we get $$left(partial_x - frac{x}{z}partial_z, partial_y - frac{y}{z}partial_z, partial_zright).$$By linear algebra, the first two fields give a trivialization for $T$, and they automatically commute because we have already checked that $T$ is involutive (in fact, they commute if and only if $T$ is involutive -- see here). Thinking of it like this we avoid paranoia with submatrices and keeping track of whether coordinates should be permuted or not.
answered Jan 21 at 6:39
Ivo TerekIvo Terek
46.3k954142
edited Jan 29 at 6:18
answered Jan 21 at 6:39
Ivo TerekIvo Terek
46.3k954142
answered Jan 21 at 6:39
Ivo TerekIvo Terek
46.3k954142
answered Jan 21 at 6:39
Ivo TerekIvo Terek
46.3k954142
46.3k954142
$begingroup$
I am very sorry! I had a typo, that should be $yfrac{partial}{partial z}$ instead of $xfrac{partial}{partial z}$.
$endgroup$
– Katherine
Jan 21 at 14:18
$begingroup$
So $T$ should be $ = langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$. I have fixed
$endgroup$
– Katherine
Jan 21 at 16:50
1
$begingroup$
Am I computing wrong? I got $X = z^2 partial_x - zx partial_z$ and $Y = zypartial_z -z^2 partial_y$ with $[X,Y] = 2z^2x partial_y - 2z^2 y partial_xneq 0$.
$endgroup$
– Jason DeVito
Jan 21 at 23:25
1
$begingroup$
I'll try to give a rough sketch: if $X_1,ldots, X_k$ is an involutive (local) frame for an involutive distribution $T hookrightarrow TM$ around some point $p$, you start writing $X_j = sum_i rho_{ij}partial_i$, where $rho$ takes values in ${rm Monomorphism}(Bbb R^k, Bbb R^n$. We may assume that the $rho e_1(p),ldots, rho e_k(p), e_{k+1}(p),ldots, e_n(p)$ is a basis for $T_pM$. Put these vectors in the column of a matrix $G(x)$; in particular know that $G(p)$ is non-singular. Reducing the neighboorhood if necessary [...]
$endgroup$
– Ivo Terek
Jan 23 at 5:04
1
$begingroup$
[...] , we may assume that $G(x)$ is also non-singular for $x$ near $p$, and write it in blocks as $$G(x)^{-1}begin{pmatrix} A(x) & 0 \ ast & {rm Id}end{pmatrix}$$Then $G(x)G(x)^{-1} = {rm Id}$ implies that $$rho(x)A(x) = begin{pmatrix} {rm Id} \ B(x)end{pmatrix}$$for some $B(x)$. So put $$V_r = sum_{j=1}^k a_{jr}X_j = partial_r + sum_{s=k+1}^nb_{sr}partial_s.$$Then the fact that $V_1,ldots, V_k$ is a commuting frame follows from $X_1,ldots, X_k$ being an involutive frame.
$endgroup$
– Ivo Terek
Jan 23 at 5:04
|
show 7 more comments
$begingroup$
I am very sorry! I had a typo, that should be $yfrac{partial}{partial z}$ instead of $xfrac{partial}{partial z}$.
$endgroup$
– Katherine
Jan 21 at 14:18
$begingroup$
So $T$ should be $ = langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$. I have fixed
$endgroup$
– Katherine
Jan 21 at 16:50
1
$begingroup$
Am I computing wrong? I got $X = z^2 partial_x - zx partial_z$ and $Y = zypartial_z -z^2 partial_y$ with $[X,Y] = 2z^2x partial_y - 2z^2 y partial_xneq 0$.
$endgroup$
– Jason DeVito
Jan 21 at 23:25
1
$begingroup$
I'll try to give a rough sketch: if $X_1,ldots, X_k$ is an involutive (local) frame for an involutive distribution $T hookrightarrow TM$ around some point $p$, you start writing $X_j = sum_i rho_{ij}partial_i$, where $rho$ takes values in ${rm Monomorphism}(Bbb R^k, Bbb R^n$. We may assume that the $rho e_1(p),ldots, rho e_k(p), e_{k+1}(p),ldots, e_n(p)$ is a basis for $T_pM$. Put these vectors in the column of a matrix $G(x)$; in particular know that $G(p)$ is non-singular. Reducing the neighboorhood if necessary [...]
$endgroup$
– Ivo Terek
Jan 23 at 5:04
1
$begingroup$
[...] , we may assume that $G(x)$ is also non-singular for $x$ near $p$, and write it in blocks as $$G(x)^{-1}begin{pmatrix} A(x) & 0 \ ast & {rm Id}end{pmatrix}$$Then $G(x)G(x)^{-1} = {rm Id}$ implies that $$rho(x)A(x) = begin{pmatrix} {rm Id} \ B(x)end{pmatrix}$$for some $B(x)$. So put $$V_r = sum_{j=1}^k a_{jr}X_j = partial_r + sum_{s=k+1}^nb_{sr}partial_s.$$Then the fact that $V_1,ldots, V_k$ is a commuting frame follows from $X_1,ldots, X_k$ being an involutive frame.
$endgroup$
– Ivo Terek
Jan 23 at 5:04
$begingroup$
I am very sorry! I had a typo, that should be $yfrac{partial}{partial z}$ instead of $xfrac{partial}{partial z}$.
$endgroup$
– Katherine
Jan 21 at 14:18
$begingroup$
I am very sorry! I had a typo, that should be $yfrac{partial}{partial z}$ instead of $xfrac{partial}{partial z}$.
$endgroup$
– Katherine
Jan 21 at 14:18
$begingroup$
So $T$ should be $ = langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$. I have fixed
$endgroup$
– Katherine
Jan 21 at 16:50
$begingroup$
So $T$ should be $ = langle ypartial_z - zpartial_y, zpartial_x - xpartial_zrangle$. I have fixed
$endgroup$
– Katherine
Jan 21 at 16:50
1
1
$begingroup$
Am I computing wrong? I got $X = z^2 partial_x - zx partial_z$ and $Y = zypartial_z -z^2 partial_y$ with $[X,Y] = 2z^2x partial_y - 2z^2 y partial_xneq 0$.
$endgroup$
– Jason DeVito
Jan 21 at 23:25
$begingroup$
Am I computing wrong? I got $X = z^2 partial_x - zx partial_z$ and $Y = zypartial_z -z^2 partial_y$ with $[X,Y] = 2z^2x partial_y - 2z^2 y partial_xneq 0$.
$endgroup$
– Jason DeVito
Jan 21 at 23:25
1
1
$begingroup$
I'll try to give a rough sketch: if $X_1,ldots, X_k$ is an involutive (local) frame for an involutive distribution $T hookrightarrow TM$ around some point $p$, you start writing $X_j = sum_i rho_{ij}partial_i$, where $rho$ takes values in ${rm Monomorphism}(Bbb R^k, Bbb R^n$. We may assume that the $rho e_1(p),ldots, rho e_k(p), e_{k+1}(p),ldots, e_n(p)$ is a basis for $T_pM$. Put these vectors in the column of a matrix $G(x)$; in particular know that $G(p)$ is non-singular. Reducing the neighboorhood if necessary [...]
$endgroup$
– Ivo Terek
Jan 23 at 5:04
$begingroup$
I'll try to give a rough sketch: if $X_1,ldots, X_k$ is an involutive (local) frame for an involutive distribution $T hookrightarrow TM$ around some point $p$, you start writing $X_j = sum_i rho_{ij}partial_i$, where $rho$ takes values in ${rm Monomorphism}(Bbb R^k, Bbb R^n$. We may assume that the $rho e_1(p),ldots, rho e_k(p), e_{k+1}(p),ldots, e_n(p)$ is a basis for $T_pM$. Put these vectors in the column of a matrix $G(x)$; in particular know that $G(p)$ is non-singular. Reducing the neighboorhood if necessary [...]
$endgroup$
– Ivo Terek
Jan 23 at 5:04
1
1
$begingroup$
[...] , we may assume that $G(x)$ is also non-singular for $x$ near $p$, and write it in blocks as $$G(x)^{-1}begin{pmatrix} A(x) & 0 \ ast & {rm Id}end{pmatrix}$$Then $G(x)G(x)^{-1} = {rm Id}$ implies that $$rho(x)A(x) = begin{pmatrix} {rm Id} \ B(x)end{pmatrix}$$for some $B(x)$. So put $$V_r = sum_{j=1}^k a_{jr}X_j = partial_r + sum_{s=k+1}^nb_{sr}partial_s.$$Then the fact that $V_1,ldots, V_k$ is a commuting frame follows from $X_1,ldots, X_k$ being an involutive frame.
$endgroup$
– Ivo Terek
Jan 23 at 5:04
$begingroup$
[...] , we may assume that $G(x)$ is also non-singular for $x$ near $p$, and write it in blocks as $$G(x)^{-1}begin{pmatrix} A(x) & 0 \ ast & {rm Id}end{pmatrix}$$Then $G(x)G(x)^{-1} = {rm Id}$ implies that $$rho(x)A(x) = begin{pmatrix} {rm Id} \ B(x)end{pmatrix}$$for some $B(x)$. So put $$V_r = sum_{j=1}^k a_{jr}X_j = partial_r + sum_{s=k+1}^nb_{sr}partial_s.$$Then the fact that $V_1,ldots, V_k$ is a commuting frame follows from $X_1,ldots, X_k$ being an involutive frame.
$endgroup$
– Ivo Terek
Jan 23 at 5:04
|
show 7 more comments
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