Mixing
What does $du^2$ notation mean in differential geometry?
I'm trying to read Lectures on quasiconformal mappings by Ahlfors, and I have trouble understanding some notation. I have a very basic understanding of differential geometry, so bear with me here.
On page 6 he takes a function $f(z) = u + iv$, $z = x + iy$, and writes
$du = u_x dx + u_y dy$
$dv = v_x dx + v_y dy$
and then says "in classical notation one writes"
$du^2 + dv^2 = (u_x^2 + v_x^2) dx^2 + (2u_x u_y + 2 v_x v_y) dx dy + (u_y^2 + v_y^2) dy^2$
I would like to thoroughly go through this as an exercise in differential geometry.
My understanding of the first equations is this: we have the tangent spaces at a specific point $z$ and $f(z)$, both of which are $mathbb{R}^2$. We have the differential forms $dx, dy$ on the tangent space of the domain of $f$, which act by projecting a vector onto first/second coordinate, and these linear transformations between $mathbb{R}^2$ and $mathbb{R}$ obviously also form a linear space. The function $f$ induces a linear transformation between these spaces, and it can be shown that it's given by these first two equations.
Now I'm a bit lost what he means here by $du^2$, because I don't think it's literally taking the square of $u$? And what object is $dx dy $, is that a wedge product? Basically the entire meaning of the last equation escapes me, thanks in advance.
complex-analysis differential-geometry
asked Nov 20 '18 at 10:30
TheMountainThatCodes
1486
|
show 2 more comments
I'm trying to read Lectures on quasiconformal mappings by Ahlfors, and I have trouble understanding some notation. I have a very basic understanding of differential geometry, so bear with me here.
On page 6 he takes a function $f(z) = u + iv$, $z = x + iy$, and writes
$du = u_x dx + u_y dy$
$dv = v_x dx + v_y dy$
and then says "in classical notation one writes"
$du^2 + dv^2 = (u_x^2 + v_x^2) dx^2 + (2u_x u_y + 2 v_x v_y) dx dy + (u_y^2 + v_y^2) dy^2$
I would like to thoroughly go through this as an exercise in differential geometry.
My understanding of the first equations is this: we have the tangent spaces at a specific point $z$ and $f(z)$, both of which are $mathbb{R}^2$. We have the differential forms $dx, dy$ on the tangent space of the domain of $f$, which act by projecting a vector onto first/second coordinate, and these linear transformations between $mathbb{R}^2$ and $mathbb{R}$ obviously also form a linear space. The function $f$ induces a linear transformation between these spaces, and it can be shown that it's given by these first two equations.
Now I'm a bit lost what he means here by $du^2$, because I don't think it's literally taking the square of $u$? And what object is $dx dy $, is that a wedge product? Basically the entire meaning of the last equation escapes me, thanks in advance.
complex-analysis differential-geometry
asked Nov 20 '18 at 10:30
TheMountainThatCodes
1486
2
Yes, it is literally taking the square, as the expansion shows. $dx,dy$ is the element of area in the $(x,y)$ coordinates.
– Yves Daoust
Nov 20 '18 at 10:34
2
I could be the symmetrization of the tensor product from the two 1-forms, i.e. $mathrm d x mathrm d y (v,w) := mathrm{Sym} ( mathrm d x otimes mathrm d y ) (v,w) = frac{1}{2}left( mathrm d x(v) mathrm d y(w) + mathrm d y(v) mathrm dx(w)right)$? Then the equations reads as a equation for 2-froms. (I think something like this is defined in John M Lee's 'intro to smooth manifolds' book )
– Steffen Plunder
Nov 20 '18 at 10:38
@YvesDaoust I'm confused, isn't the left hand side a one-form? And shouldn't the right hand side be one as well then?
– TheMountainThatCodes
Nov 20 '18 at 10:41
1
No, the differentials are squared.
– Yves Daoust
Nov 20 '18 at 10:45
1
Indeed. (And the differential $d(u^2)=2u,du$ is also not a one-form.)
– Yves Daoust
Nov 20 '18 at 10:59
|
show 2 more comments
I'm trying to read Lectures on quasiconformal mappings by Ahlfors, and I have trouble understanding some notation. I have a very basic understanding of differential geometry, so bear with me here.
On page 6 he takes a function $f(z) = u + iv$, $z = x + iy$, and writes
$du = u_x dx + u_y dy$
$dv = v_x dx + v_y dy$
and then says "in classical notation one writes"
$du^2 + dv^2 = (u_x^2 + v_x^2) dx^2 + (2u_x u_y + 2 v_x v_y) dx dy + (u_y^2 + v_y^2) dy^2$
I would like to thoroughly go through this as an exercise in differential geometry.
My understanding of the first equations is this: we have the tangent spaces at a specific point $z$ and $f(z)$, both of which are $mathbb{R}^2$. We have the differential forms $dx, dy$ on the tangent space of the domain of $f$, which act by projecting a vector onto first/second coordinate, and these linear transformations between $mathbb{R}^2$ and $mathbb{R}$ obviously also form a linear space. The function $f$ induces a linear transformation between these spaces, and it can be shown that it's given by these first two equations.
Now I'm a bit lost what he means here by $du^2$, because I don't think it's literally taking the square of $u$? And what object is $dx dy $, is that a wedge product? Basically the entire meaning of the last equation escapes me, thanks in advance.
complex-analysis differential-geometry
asked Nov 20 '18 at 10:30
TheMountainThatCodes
1486
I'm trying to read Lectures on quasiconformal mappings by Ahlfors, and I have trouble understanding some notation. I have a very basic understanding of differential geometry, so bear with me here.
On page 6 he takes a function $f(z) = u + iv$, $z = x + iy$, and writes
$du = u_x dx + u_y dy$
$dv = v_x dx + v_y dy$
and then says "in classical notation one writes"
$du^2 + dv^2 = (u_x^2 + v_x^2) dx^2 + (2u_x u_y + 2 v_x v_y) dx dy + (u_y^2 + v_y^2) dy^2$
I would like to thoroughly go through this as an exercise in differential geometry.
My understanding of the first equations is this: we have the tangent spaces at a specific point $z$ and $f(z)$, both of which are $mathbb{R}^2$. We have the differential forms $dx, dy$ on the tangent space of the domain of $f$, which act by projecting a vector onto first/second coordinate, and these linear transformations between $mathbb{R}^2$ and $mathbb{R}$ obviously also form a linear space. The function $f$ induces a linear transformation between these spaces, and it can be shown that it's given by these first two equations.
Now I'm a bit lost what he means here by $du^2$, because I don't think it's literally taking the square of $u$? And what object is $dx dy $, is that a wedge product? Basically the entire meaning of the last equation escapes me, thanks in advance.
complex-analysis differential-geometry
complex-analysis differential-geometry
asked Nov 20 '18 at 10:30
TheMountainThatCodes
1486
asked Nov 20 '18 at 10:30
TheMountainThatCodes
1486
asked Nov 20 '18 at 10:30
TheMountainThatCodes
1486
asked Nov 20 '18 at 10:30
TheMountainThatCodes
1486
asked Nov 20 '18 at 10:30
TheMountainThatCodes
1486
1486
2
Yes, it is literally taking the square, as the expansion shows. $dx,dy$ is the element of area in the $(x,y)$ coordinates.
– Yves Daoust
Nov 20 '18 at 10:34
2
I could be the symmetrization of the tensor product from the two 1-forms, i.e. $mathrm d x mathrm d y (v,w) := mathrm{Sym} ( mathrm d x otimes mathrm d y ) (v,w) = frac{1}{2}left( mathrm d x(v) mathrm d y(w) + mathrm d y(v) mathrm dx(w)right)$? Then the equations reads as a equation for 2-froms. (I think something like this is defined in John M Lee's 'intro to smooth manifolds' book )
– Steffen Plunder
Nov 20 '18 at 10:38
@YvesDaoust I'm confused, isn't the left hand side a one-form? And shouldn't the right hand side be one as well then?
– TheMountainThatCodes
Nov 20 '18 at 10:41
1
No, the differentials are squared.
– Yves Daoust
Nov 20 '18 at 10:45
1
Indeed. (And the differential $d(u^2)=2u,du$ is also not a one-form.)
– Yves Daoust
Nov 20 '18 at 10:59
|
show 2 more comments
2
Yes, it is literally taking the square, as the expansion shows. $dx,dy$ is the element of area in the $(x,y)$ coordinates.
– Yves Daoust
Nov 20 '18 at 10:34
2
I could be the symmetrization of the tensor product from the two 1-forms, i.e. $mathrm d x mathrm d y (v,w) := mathrm{Sym} ( mathrm d x otimes mathrm d y ) (v,w) = frac{1}{2}left( mathrm d x(v) mathrm d y(w) + mathrm d y(v) mathrm dx(w)right)$? Then the equations reads as a equation for 2-froms. (I think something like this is defined in John M Lee's 'intro to smooth manifolds' book )
– Steffen Plunder
Nov 20 '18 at 10:38
@YvesDaoust I'm confused, isn't the left hand side a one-form? And shouldn't the right hand side be one as well then?
– TheMountainThatCodes
Nov 20 '18 at 10:41
1
No, the differentials are squared.
– Yves Daoust
Nov 20 '18 at 10:45
1
Indeed. (And the differential $d(u^2)=2u,du$ is also not a one-form.)
– Yves Daoust
Nov 20 '18 at 10:59
2
2
Yes, it is literally taking the square, as the expansion shows. $dx,dy$ is the element of area in the $(x,y)$ coordinates.
– Yves Daoust
Nov 20 '18 at 10:34
Yes, it is literally taking the square, as the expansion shows. $dx,dy$ is the element of area in the $(x,y)$ coordinates.
– Yves Daoust
Nov 20 '18 at 10:34
2
2
I could be the symmetrization of the tensor product from the two 1-forms, i.e. $mathrm d x mathrm d y (v,w) := mathrm{Sym} ( mathrm d x otimes mathrm d y ) (v,w) = frac{1}{2}left( mathrm d x(v) mathrm d y(w) + mathrm d y(v) mathrm dx(w)right)$? Then the equations reads as a equation for 2-froms. (I think something like this is defined in John M Lee's 'intro to smooth manifolds' book )
– Steffen Plunder
Nov 20 '18 at 10:38
I could be the symmetrization of the tensor product from the two 1-forms, i.e. $mathrm d x mathrm d y (v,w) := mathrm{Sym} ( mathrm d x otimes mathrm d y ) (v,w) = frac{1}{2}left( mathrm d x(v) mathrm d y(w) + mathrm d y(v) mathrm dx(w)right)$? Then the equations reads as a equation for 2-froms. (I think something like this is defined in John M Lee's 'intro to smooth manifolds' book )
– Steffen Plunder
Nov 20 '18 at 10:38
@YvesDaoust I'm confused, isn't the left hand side a one-form? And shouldn't the right hand side be one as well then?
– TheMountainThatCodes
Nov 20 '18 at 10:41
@YvesDaoust I'm confused, isn't the left hand side a one-form? And shouldn't the right hand side be one as well then?
– TheMountainThatCodes
Nov 20 '18 at 10:41
1
1
No, the differentials are squared.
– Yves Daoust
Nov 20 '18 at 10:45
No, the differentials are squared.
– Yves Daoust
Nov 20 '18 at 10:45
1
1
Indeed. (And the differential $d(u^2)=2u,du$ is also not a one-form.)
– Yves Daoust
Nov 20 '18 at 10:59
Indeed. (And the differential $d(u^2)=2u,du$ is also not a one-form.)
– Yves Daoust
Nov 20 '18 at 10:59
|
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Yes, it is literally taking the square, as the expansion shows. $dx,dy$ is the element of area in the $(x,y)$ coordinates.
– Yves Daoust
Nov 20 '18 at 10:34
2
I could be the symmetrization of the tensor product from the two 1-forms, i.e. $mathrm d x mathrm d y (v,w) := mathrm{Sym} ( mathrm d x otimes mathrm d y ) (v,w) = frac{1}{2}left( mathrm d x(v) mathrm d y(w) + mathrm d y(v) mathrm dx(w)right)$? Then the equations reads as a equation for 2-froms. (I think something like this is defined in John M Lee's 'intro to smooth manifolds' book )
– Steffen Plunder
Nov 20 '18 at 10:38
@YvesDaoust I'm confused, isn't the left hand side a one-form? And shouldn't the right hand side be one as well then?
– TheMountainThatCodes
Nov 20 '18 at 10:41
1
No, the differentials are squared.
– Yves Daoust
Nov 20 '18 at 10:45
1
Indeed. (And the differential $d(u^2)=2u,du$ is also not a one-form.)
– Yves Daoust
Nov 20 '18 at 10:59