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Producing isothermal coordinates from a solution of the Beltrami Equation
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Suppose that $f colon U to V$ is a diffeomorphism of planar domains. Its differential $Df$ can be pointwise expressed the sum of a complex-linear and complex-anti-linear mapping: Given a tangent vector $xi$,
$$Dfxi = f_z xi + f_overlinezoverlinexi.$$
If $f$ is orientation preserving, then $|f_z|>|f_overlinez|$ and the Beltrami coefficient of $f$ is the defined to be the complex-valued function
$$mu_f:= fracf_zf_overlinez.$$
The diffeomorphism $f$ defines a Riemannian metric on $U$ via the pullback of the usual Euclidean inner product on $V$: given tangent vectors $xi$ and $zeta$,
$$langle xi, zeta rangle_f := langle Dfxi, Dfzeta rangle_textEuc = xi^T(Df^TDf)zeta.$$
Note that the entries of the positive symmetric matrix $Df^TDf$ determine the Beltrami coefficient of $mu$: if $Df^TDf=beginbmatrixE & F \ F & G\ endbmatrix$, then
$$mu_f = fracE - G + 2iFE+G+2sqrtEG-F^2.$$
Now, suppose that a Riemannian metric $g$ on $U$ is given, and its expression as a matrix is given by $beginbmatrixE_g & F_g \ F_g & G_g\ endbmatrix$. Moreover, suppose that we have an orientation preserving diffeomorphism $f colon U to V$ so that
$$mu_f = fracE_g- G_g + 2iF_gE_g+G_g+2sqrtE_gG_g-F_g^2 =:mu_g.$$
My understanding is that the pullback metric $langle cdot, cdot rangle_f$ is conformally equivalent to $g$, but I am unable to verify this. This is essentially stated as obvious on the wikipedia pages
https://en.wikipedia.org/wiki/Beltrami_equation
https://en.wikipedia.org/wiki/Isothermal_coordinates
https://en.wikipedia.org/wiki/Quasiconformal_mapping
but I am unable to follow the reasoning.
My attempt at a proof was as follows. Since $mu_f =mu_g$, we may write
$$Dfxi = f_z(xi + mu_goverlinexi).$$
Writing $Wxi := mu_goverlinexi$, we may then note that as $W$ is symmetric,
$$Df^TDf = |f_z|^2(textId + W)^2.$$
We may calculate
$$(textId + W)^2 = beginbmatrix (1+textRe mu_g)^2 + (textIm mu_g)^2 & 2textIm mu_g \2textIm mu_g &(1-textRe mu_g)^2 + (textIm mu_g)^2endbmatrix.$$
My hope was then to use the definition of $mu_g$, plugging in the real and imaginary parts, and then see the relationship between $Df^TDf$ and $g$. Unfortunately, my calculations just lead to a mess and don't simplify.
The key point I am missing is the following: how exactly do the complex number $mu_g$ and the real number $|f_z|^2$ determine the three real numbers $E_g, F_g, G_g$?
Thanks for reading.
riemannian-geometry conformal-geometry quasiconformal-maps
asked Jul 22 at 14:03
Kevin
111
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up vote
2
down vote
favorite
Suppose that $f colon U to V$ is a diffeomorphism of planar domains. Its differential $Df$ can be pointwise expressed the sum of a complex-linear and complex-anti-linear mapping: Given a tangent vector $xi$,
$$Dfxi = f_z xi + f_overlinezoverlinexi.$$
If $f$ is orientation preserving, then $|f_z|>|f_overlinez|$ and the Beltrami coefficient of $f$ is the defined to be the complex-valued function
$$mu_f:= fracf_zf_overlinez.$$
The diffeomorphism $f$ defines a Riemannian metric on $U$ via the pullback of the usual Euclidean inner product on $V$: given tangent vectors $xi$ and $zeta$,
$$langle xi, zeta rangle_f := langle Dfxi, Dfzeta rangle_textEuc = xi^T(Df^TDf)zeta.$$
Note that the entries of the positive symmetric matrix $Df^TDf$ determine the Beltrami coefficient of $mu$: if $Df^TDf=beginbmatrixE & F \ F & G\ endbmatrix$, then
$$mu_f = fracE - G + 2iFE+G+2sqrtEG-F^2.$$
Now, suppose that a Riemannian metric $g$ on $U$ is given, and its expression as a matrix is given by $beginbmatrixE_g & F_g \ F_g & G_g\ endbmatrix$. Moreover, suppose that we have an orientation preserving diffeomorphism $f colon U to V$ so that
$$mu_f = fracE_g- G_g + 2iF_gE_g+G_g+2sqrtE_gG_g-F_g^2 =:mu_g.$$
My understanding is that the pullback metric $langle cdot, cdot rangle_f$ is conformally equivalent to $g$, but I am unable to verify this. This is essentially stated as obvious on the wikipedia pages
https://en.wikipedia.org/wiki/Beltrami_equation
https://en.wikipedia.org/wiki/Isothermal_coordinates
https://en.wikipedia.org/wiki/Quasiconformal_mapping
but I am unable to follow the reasoning.
My attempt at a proof was as follows. Since $mu_f =mu_g$, we may write
$$Dfxi = f_z(xi + mu_goverlinexi).$$
Writing $Wxi := mu_goverlinexi$, we may then note that as $W$ is symmetric,
$$Df^TDf = |f_z|^2(textId + W)^2.$$
We may calculate
$$(textId + W)^2 = beginbmatrix (1+textRe mu_g)^2 + (textIm mu_g)^2 & 2textIm mu_g \2textIm mu_g &(1-textRe mu_g)^2 + (textIm mu_g)^2endbmatrix.$$
My hope was then to use the definition of $mu_g$, plugging in the real and imaginary parts, and then see the relationship between $Df^TDf$ and $g$. Unfortunately, my calculations just lead to a mess and don't simplify.
The key point I am missing is the following: how exactly do the complex number $mu_g$ and the real number $|f_z|^2$ determine the three real numbers $E_g, F_g, G_g$?
Thanks for reading.
riemannian-geometry conformal-geometry quasiconformal-maps
asked Jul 22 at 14:03
Kevin
111
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose that $f colon U to V$ is a diffeomorphism of planar domains. Its differential $Df$ can be pointwise expressed the sum of a complex-linear and complex-anti-linear mapping: Given a tangent vector $xi$,
$$Dfxi = f_z xi + f_overlinezoverlinexi.$$
If $f$ is orientation preserving, then $|f_z|>|f_overlinez|$ and the Beltrami coefficient of $f$ is the defined to be the complex-valued function
$$mu_f:= fracf_zf_overlinez.$$
The diffeomorphism $f$ defines a Riemannian metric on $U$ via the pullback of the usual Euclidean inner product on $V$: given tangent vectors $xi$ and $zeta$,
$$langle xi, zeta rangle_f := langle Dfxi, Dfzeta rangle_textEuc = xi^T(Df^TDf)zeta.$$
Note that the entries of the positive symmetric matrix $Df^TDf$ determine the Beltrami coefficient of $mu$: if $Df^TDf=beginbmatrixE & F \ F & G\ endbmatrix$, then
$$mu_f = fracE - G + 2iFE+G+2sqrtEG-F^2.$$
Now, suppose that a Riemannian metric $g$ on $U$ is given, and its expression as a matrix is given by $beginbmatrixE_g & F_g \ F_g & G_g\ endbmatrix$. Moreover, suppose that we have an orientation preserving diffeomorphism $f colon U to V$ so that
$$mu_f = fracE_g- G_g + 2iF_gE_g+G_g+2sqrtE_gG_g-F_g^2 =:mu_g.$$
My understanding is that the pullback metric $langle cdot, cdot rangle_f$ is conformally equivalent to $g$, but I am unable to verify this. This is essentially stated as obvious on the wikipedia pages
https://en.wikipedia.org/wiki/Beltrami_equation
https://en.wikipedia.org/wiki/Isothermal_coordinates
https://en.wikipedia.org/wiki/Quasiconformal_mapping
but I am unable to follow the reasoning.
My attempt at a proof was as follows. Since $mu_f =mu_g$, we may write
$$Dfxi = f_z(xi + mu_goverlinexi).$$
Writing $Wxi := mu_goverlinexi$, we may then note that as $W$ is symmetric,
$$Df^TDf = |f_z|^2(textId + W)^2.$$
We may calculate
$$(textId + W)^2 = beginbmatrix (1+textRe mu_g)^2 + (textIm mu_g)^2 & 2textIm mu_g \2textIm mu_g &(1-textRe mu_g)^2 + (textIm mu_g)^2endbmatrix.$$
My hope was then to use the definition of $mu_g$, plugging in the real and imaginary parts, and then see the relationship between $Df^TDf$ and $g$. Unfortunately, my calculations just lead to a mess and don't simplify.
The key point I am missing is the following: how exactly do the complex number $mu_g$ and the real number $|f_z|^2$ determine the three real numbers $E_g, F_g, G_g$?
Thanks for reading.
riemannian-geometry conformal-geometry quasiconformal-maps
asked Jul 22 at 14:03
Kevin
111
Suppose that $f colon U to V$ is a diffeomorphism of planar domains. Its differential $Df$ can be pointwise expressed the sum of a complex-linear and complex-anti-linear mapping: Given a tangent vector $xi$,
$$Dfxi = f_z xi + f_overlinezoverlinexi.$$
If $f$ is orientation preserving, then $|f_z|>|f_overlinez|$ and the Beltrami coefficient of $f$ is the defined to be the complex-valued function
$$mu_f:= fracf_zf_overlinez.$$
The diffeomorphism $f$ defines a Riemannian metric on $U$ via the pullback of the usual Euclidean inner product on $V$: given tangent vectors $xi$ and $zeta$,
$$langle xi, zeta rangle_f := langle Dfxi, Dfzeta rangle_textEuc = xi^T(Df^TDf)zeta.$$
Note that the entries of the positive symmetric matrix $Df^TDf$ determine the Beltrami coefficient of $mu$: if $Df^TDf=beginbmatrixE & F \ F & G\ endbmatrix$, then
$$mu_f = fracE - G + 2iFE+G+2sqrtEG-F^2.$$
Now, suppose that a Riemannian metric $g$ on $U$ is given, and its expression as a matrix is given by $beginbmatrixE_g & F_g \ F_g & G_g\ endbmatrix$. Moreover, suppose that we have an orientation preserving diffeomorphism $f colon U to V$ so that
$$mu_f = fracE_g- G_g + 2iF_gE_g+G_g+2sqrtE_gG_g-F_g^2 =:mu_g.$$
My understanding is that the pullback metric $langle cdot, cdot rangle_f$ is conformally equivalent to $g$, but I am unable to verify this. This is essentially stated as obvious on the wikipedia pages
https://en.wikipedia.org/wiki/Beltrami_equation
https://en.wikipedia.org/wiki/Isothermal_coordinates
https://en.wikipedia.org/wiki/Quasiconformal_mapping
but I am unable to follow the reasoning.
My attempt at a proof was as follows. Since $mu_f =mu_g$, we may write
$$Dfxi = f_z(xi + mu_goverlinexi).$$
Writing $Wxi := mu_goverlinexi$, we may then note that as $W$ is symmetric,
$$Df^TDf = |f_z|^2(textId + W)^2.$$
We may calculate
$$(textId + W)^2 = beginbmatrix (1+textRe mu_g)^2 + (textIm mu_g)^2 & 2textIm mu_g \2textIm mu_g &(1-textRe mu_g)^2 + (textIm mu_g)^2endbmatrix.$$
My hope was then to use the definition of $mu_g$, plugging in the real and imaginary parts, and then see the relationship between $Df^TDf$ and $g$. Unfortunately, my calculations just lead to a mess and don't simplify.
The key point I am missing is the following: how exactly do the complex number $mu_g$ and the real number $|f_z|^2$ determine the three real numbers $E_g, F_g, G_g$?
Thanks for reading.
riemannian-geometry conformal-geometry quasiconformal-maps
asked Jul 22 at 14:03
Kevin
111
asked Jul 22 at 14:03
Kevin
111
asked Jul 22 at 14:03
Kevin
111
asked Jul 22 at 14:03
Kevin
111
111
add a comment |Â
add a comment |Â
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