What is intuitively the Hodge dual of a $p$-form?
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Carroll in his textbook "Spacetime and geometry" defines the Hodge dual of a $p$-form $A$ on an $n$-dimensional manifold as $$(star A)_{mu_1...mu_{n-p}}=dfrac{1}{p!}epsilon^{nu_1...nu_p} _{ mu_1...mu_{n-p}}A_{nu_1...nu_p}.$$
Although this is pretty straightforward, I find that I lack any intuitive/geometrical understanding of this. What does the dual actually mean?
general-relativity differential-geometry duality
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add a comment |
$begingroup$
Carroll in his textbook "Spacetime and geometry" defines the Hodge dual of a $p$-form $A$ on an $n$-dimensional manifold as $$(star A)_{mu_1...mu_{n-p}}=dfrac{1}{p!}epsilon^{nu_1...nu_p} _{ mu_1...mu_{n-p}}A_{nu_1...nu_p}.$$
Although this is pretty straightforward, I find that I lack any intuitive/geometrical understanding of this. What does the dual actually mean?
general-relativity differential-geometry duality
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2
$begingroup$
Related/possible duplicate: physics.stackexchange.com/q/313091/50583
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– ACuriousMind♦
Feb 17 at 17:34
add a comment |
$begingroup$
Carroll in his textbook "Spacetime and geometry" defines the Hodge dual of a $p$-form $A$ on an $n$-dimensional manifold as $$(star A)_{mu_1...mu_{n-p}}=dfrac{1}{p!}epsilon^{nu_1...nu_p} _{ mu_1...mu_{n-p}}A_{nu_1...nu_p}.$$
Although this is pretty straightforward, I find that I lack any intuitive/geometrical understanding of this. What does the dual actually mean?
general-relativity differential-geometry duality
$endgroup$
Carroll in his textbook "Spacetime and geometry" defines the Hodge dual of a $p$-form $A$ on an $n$-dimensional manifold as $$(star A)_{mu_1...mu_{n-p}}=dfrac{1}{p!}epsilon^{nu_1...nu_p} _{ mu_1...mu_{n-p}}A_{nu_1...nu_p}.$$
Although this is pretty straightforward, I find that I lack any intuitive/geometrical understanding of this. What does the dual actually mean?
general-relativity differential-geometry duality
general-relativity differential-geometry duality
edited Feb 17 at 19:32
Qmechanic♦
105k121901202
105k121901202
asked Feb 17 at 17:03
TheQuantumManTheQuantumMan
2,75231851
2,75231851
2
$begingroup$
Related/possible duplicate: physics.stackexchange.com/q/313091/50583
$endgroup$
– ACuriousMind♦
Feb 17 at 17:34
add a comment |
2
$begingroup$
Related/possible duplicate: physics.stackexchange.com/q/313091/50583
$endgroup$
– ACuriousMind♦
Feb 17 at 17:34
2
2
$begingroup$
Related/possible duplicate: physics.stackexchange.com/q/313091/50583
$endgroup$
– ACuriousMind♦
Feb 17 at 17:34
$begingroup$
Related/possible duplicate: physics.stackexchange.com/q/313091/50583
$endgroup$
– ACuriousMind♦
Feb 17 at 17:34
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
There's a simple way to visualize differential forms that helps here.
One of the most important properties of differential $p$-forms is that they are tensors that can naturally be integrated over a $p$-dimensional submanifold. The simplest example is a one-form, which can be integrated over a one-dimensional curve $C$,
$$int_C A = int A_mu dx^mu.$$
Unlike the integration of a vector,
$$int mathbf{F} cdot dmathbf{x}$$
this integral requires no metric to define. Similarly, a two-form can be naturally integrated over a surface, for example
$$int_S F = int F_{munu} dx^mu dx^nu$$
could be interpreted as a magnetic flux if the surface is spatial.
This leads directly to the geometric interpretation of differential forms. The familiar visualization of a one-form is a set of hypersurfaces, i.e. submanifolds of codimension one. Similarly, two-forms may be interpreted as submanifolds of codimension two, and so on. In three dimensions, a one-form is a bunch of planes, a two-form is a bunch of lines, and a three-form is a bunch of points.
You can easily count the number of planes a curve goes through, the number of lines a surface intersects, and the number of points a volume contains. These are the geometric interpretations of integrals of one-forms, two-forms, and three-forms over curves, surfaces, and volumes. (The main thing missing from this picture is the sign; you should also think of the surfaces, etc. as coming with orientations, though that's harder to draw.)
The wedge product can be interpreted as taking the geometric intersection of the differential forms involved. For example, the wedge product of two one-forms in 3D is a set of intersections of surfaces, which is a set of curves, i.e. a two-form.
The interpretation of the Hodge dual, which is explicitly defined via the metric, is that it is the "orthogonal complement" of a differential form. The dual of a one-form in 3D is a set of curves perpendicular to the planes of the original one-form, so that the local densities of surfaces in both are proportional.
For completeness, here are a few more facts.
First, the exterior derivative $d$ corresponds to taking the boundaries of the surfaces associated with a differential $p$-form. That is why $d^2 = 0$ is just the usual "boundary of a boundary is zero" principle, and why a closed differential form is annihilated by $d$; its visualization is in terms of closed surfaces.
Second, one can also interpret the contraction of a $p$-form with a $(p, 0)$ antisymmetric tensor geometrically. A $(p, 0)$ antisymmetric tensor is called a $p$-blade and is constructed as the antisymmetric product of $p$ vectors. It can be interpreted as the $p$-dimensional volume element corresponding to the parallelepiped spanned by the vectors; for example the $dx^mu dx^nu$ above is a $2$-blade that is a small square in Cartesian coordinates.
Third, my interpretation is not the most common one. Most books treat $p$-forms in the exact same way as $p$-blades, i.e. as small volume elements. This is the point of view advanced in the three other answers to this question. It's a little simpler, but to use it you have to explicitly use the metric to convert forms to blades. My interpretation just generalizes the intuition we are usually given for $1$-forms as families of hypersurfaces, without needing to use the metric to modify the object.
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I'm not sure the expression "it's easy to see" is as appropriate to most people as to e.g. you, but upvoted. :-)
$endgroup$
– StephenG
Feb 17 at 19:30
$begingroup$
@StephenG Actually, in practice "it's easy to see" is just code for "I'm too lazy to check"! After coming back I realized that particular statement was just wrong, thanks for that.
$endgroup$
– knzhou
Feb 17 at 20:05
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@knzhou great answer, but I have a question: how can we see the "orthogonal complement" interpretation for the Hodge dual in a bit more generality so that the explanation of the example that you give for the dual of the 1-form comes out naturally?
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– TheQuantumMan
Feb 17 at 20:39
1
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@TheQuantumMan I was about to edit to address this, but I see it's nicely explained in tparker's answer!
$endgroup$
– knzhou
Feb 17 at 21:08
add a comment |
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The $epsilon$ symbol is (proportional to) the full-rank "volume form", which you can think of as a little full-dimensional box sitting at every point on your manifold. (Note that this is a different intuitive picture from the equally valid one used in knzhou's answer.) Each index corresponds to a different direction, or to one edge of the box coming out of one particular corner. Contracting the $epsilon$ symbol with a $p$-form defined on a $p$-dimensional submanifold "collapses" the $p$ edges of the box that run parallel to the submanifold, which leaves a lower-dimensional box spanned by the $n-p$ remaining edges, which are clearly all perpendicular to the $p$ edges that got collapsed. (Note that by using the notion of "perpendicular", you're implicitly using the metric, which is hidden inside the mixed signature of the $epsilon$ symbol in the equation you wrote above.) That motivates why the Hodge dual of a $p$-form is an $n-p$ form, and also why it's the local, point-by-point "orthogonal complement" of the original form, as knzhou explained.
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add a comment |
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I don't know if this is too basic for you, but think of freshman electromagnetism, where in order to define flux we associate an area vector with a small area on a surface. The area would more naturally be described by two vectors, $u$ and $v$, lying in its plane, such that the parallelogram they span has the area we're talking about. When we represent this area instead by a single vector $a$ perpendicular to the surface, we're taking the Hodge dual of the rank-2 tensor $u^av^b$. This works in three dimensions because 1+2=3, so the Hodge star converts a rank-2 tensor to a rank-1 tensor.
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add a comment |
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If you think of a blade as a subspace together with a real number, then the Hodge dual is the dual subspace together with the same real number.
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add a comment |
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4 Answers
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4 Answers
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$begingroup$
There's a simple way to visualize differential forms that helps here.
One of the most important properties of differential $p$-forms is that they are tensors that can naturally be integrated over a $p$-dimensional submanifold. The simplest example is a one-form, which can be integrated over a one-dimensional curve $C$,
$$int_C A = int A_mu dx^mu.$$
Unlike the integration of a vector,
$$int mathbf{F} cdot dmathbf{x}$$
this integral requires no metric to define. Similarly, a two-form can be naturally integrated over a surface, for example
$$int_S F = int F_{munu} dx^mu dx^nu$$
could be interpreted as a magnetic flux if the surface is spatial.
This leads directly to the geometric interpretation of differential forms. The familiar visualization of a one-form is a set of hypersurfaces, i.e. submanifolds of codimension one. Similarly, two-forms may be interpreted as submanifolds of codimension two, and so on. In three dimensions, a one-form is a bunch of planes, a two-form is a bunch of lines, and a three-form is a bunch of points.
You can easily count the number of planes a curve goes through, the number of lines a surface intersects, and the number of points a volume contains. These are the geometric interpretations of integrals of one-forms, two-forms, and three-forms over curves, surfaces, and volumes. (The main thing missing from this picture is the sign; you should also think of the surfaces, etc. as coming with orientations, though that's harder to draw.)
The wedge product can be interpreted as taking the geometric intersection of the differential forms involved. For example, the wedge product of two one-forms in 3D is a set of intersections of surfaces, which is a set of curves, i.e. a two-form.
The interpretation of the Hodge dual, which is explicitly defined via the metric, is that it is the "orthogonal complement" of a differential form. The dual of a one-form in 3D is a set of curves perpendicular to the planes of the original one-form, so that the local densities of surfaces in both are proportional.
For completeness, here are a few more facts.
First, the exterior derivative $d$ corresponds to taking the boundaries of the surfaces associated with a differential $p$-form. That is why $d^2 = 0$ is just the usual "boundary of a boundary is zero" principle, and why a closed differential form is annihilated by $d$; its visualization is in terms of closed surfaces.
Second, one can also interpret the contraction of a $p$-form with a $(p, 0)$ antisymmetric tensor geometrically. A $(p, 0)$ antisymmetric tensor is called a $p$-blade and is constructed as the antisymmetric product of $p$ vectors. It can be interpreted as the $p$-dimensional volume element corresponding to the parallelepiped spanned by the vectors; for example the $dx^mu dx^nu$ above is a $2$-blade that is a small square in Cartesian coordinates.
Third, my interpretation is not the most common one. Most books treat $p$-forms in the exact same way as $p$-blades, i.e. as small volume elements. This is the point of view advanced in the three other answers to this question. It's a little simpler, but to use it you have to explicitly use the metric to convert forms to blades. My interpretation just generalizes the intuition we are usually given for $1$-forms as families of hypersurfaces, without needing to use the metric to modify the object.
$endgroup$
$begingroup$
I'm not sure the expression "it's easy to see" is as appropriate to most people as to e.g. you, but upvoted. :-)
$endgroup$
– StephenG
Feb 17 at 19:30
$begingroup$
@StephenG Actually, in practice "it's easy to see" is just code for "I'm too lazy to check"! After coming back I realized that particular statement was just wrong, thanks for that.
$endgroup$
– knzhou
Feb 17 at 20:05
$begingroup$
@knzhou great answer, but I have a question: how can we see the "orthogonal complement" interpretation for the Hodge dual in a bit more generality so that the explanation of the example that you give for the dual of the 1-form comes out naturally?
$endgroup$
– TheQuantumMan
Feb 17 at 20:39
1
$begingroup$
@TheQuantumMan I was about to edit to address this, but I see it's nicely explained in tparker's answer!
$endgroup$
– knzhou
Feb 17 at 21:08
add a comment |
$begingroup$
There's a simple way to visualize differential forms that helps here.
One of the most important properties of differential $p$-forms is that they are tensors that can naturally be integrated over a $p$-dimensional submanifold. The simplest example is a one-form, which can be integrated over a one-dimensional curve $C$,
$$int_C A = int A_mu dx^mu.$$
Unlike the integration of a vector,
$$int mathbf{F} cdot dmathbf{x}$$
this integral requires no metric to define. Similarly, a two-form can be naturally integrated over a surface, for example
$$int_S F = int F_{munu} dx^mu dx^nu$$
could be interpreted as a magnetic flux if the surface is spatial.
This leads directly to the geometric interpretation of differential forms. The familiar visualization of a one-form is a set of hypersurfaces, i.e. submanifolds of codimension one. Similarly, two-forms may be interpreted as submanifolds of codimension two, and so on. In three dimensions, a one-form is a bunch of planes, a two-form is a bunch of lines, and a three-form is a bunch of points.
You can easily count the number of planes a curve goes through, the number of lines a surface intersects, and the number of points a volume contains. These are the geometric interpretations of integrals of one-forms, two-forms, and three-forms over curves, surfaces, and volumes. (The main thing missing from this picture is the sign; you should also think of the surfaces, etc. as coming with orientations, though that's harder to draw.)
The wedge product can be interpreted as taking the geometric intersection of the differential forms involved. For example, the wedge product of two one-forms in 3D is a set of intersections of surfaces, which is a set of curves, i.e. a two-form.
The interpretation of the Hodge dual, which is explicitly defined via the metric, is that it is the "orthogonal complement" of a differential form. The dual of a one-form in 3D is a set of curves perpendicular to the planes of the original one-form, so that the local densities of surfaces in both are proportional.
For completeness, here are a few more facts.
First, the exterior derivative $d$ corresponds to taking the boundaries of the surfaces associated with a differential $p$-form. That is why $d^2 = 0$ is just the usual "boundary of a boundary is zero" principle, and why a closed differential form is annihilated by $d$; its visualization is in terms of closed surfaces.
Second, one can also interpret the contraction of a $p$-form with a $(p, 0)$ antisymmetric tensor geometrically. A $(p, 0)$ antisymmetric tensor is called a $p$-blade and is constructed as the antisymmetric product of $p$ vectors. It can be interpreted as the $p$-dimensional volume element corresponding to the parallelepiped spanned by the vectors; for example the $dx^mu dx^nu$ above is a $2$-blade that is a small square in Cartesian coordinates.
Third, my interpretation is not the most common one. Most books treat $p$-forms in the exact same way as $p$-blades, i.e. as small volume elements. This is the point of view advanced in the three other answers to this question. It's a little simpler, but to use it you have to explicitly use the metric to convert forms to blades. My interpretation just generalizes the intuition we are usually given for $1$-forms as families of hypersurfaces, without needing to use the metric to modify the object.
$endgroup$
$begingroup$
I'm not sure the expression "it's easy to see" is as appropriate to most people as to e.g. you, but upvoted. :-)
$endgroup$
– StephenG
Feb 17 at 19:30
$begingroup$
@StephenG Actually, in practice "it's easy to see" is just code for "I'm too lazy to check"! After coming back I realized that particular statement was just wrong, thanks for that.
$endgroup$
– knzhou
Feb 17 at 20:05
$begingroup$
@knzhou great answer, but I have a question: how can we see the "orthogonal complement" interpretation for the Hodge dual in a bit more generality so that the explanation of the example that you give for the dual of the 1-form comes out naturally?
$endgroup$
– TheQuantumMan
Feb 17 at 20:39
1
$begingroup$
@TheQuantumMan I was about to edit to address this, but I see it's nicely explained in tparker's answer!
$endgroup$
– knzhou
Feb 17 at 21:08
add a comment |
$begingroup$
There's a simple way to visualize differential forms that helps here.
One of the most important properties of differential $p$-forms is that they are tensors that can naturally be integrated over a $p$-dimensional submanifold. The simplest example is a one-form, which can be integrated over a one-dimensional curve $C$,
$$int_C A = int A_mu dx^mu.$$
Unlike the integration of a vector,
$$int mathbf{F} cdot dmathbf{x}$$
this integral requires no metric to define. Similarly, a two-form can be naturally integrated over a surface, for example
$$int_S F = int F_{munu} dx^mu dx^nu$$
could be interpreted as a magnetic flux if the surface is spatial.
This leads directly to the geometric interpretation of differential forms. The familiar visualization of a one-form is a set of hypersurfaces, i.e. submanifolds of codimension one. Similarly, two-forms may be interpreted as submanifolds of codimension two, and so on. In three dimensions, a one-form is a bunch of planes, a two-form is a bunch of lines, and a three-form is a bunch of points.
You can easily count the number of planes a curve goes through, the number of lines a surface intersects, and the number of points a volume contains. These are the geometric interpretations of integrals of one-forms, two-forms, and three-forms over curves, surfaces, and volumes. (The main thing missing from this picture is the sign; you should also think of the surfaces, etc. as coming with orientations, though that's harder to draw.)
The wedge product can be interpreted as taking the geometric intersection of the differential forms involved. For example, the wedge product of two one-forms in 3D is a set of intersections of surfaces, which is a set of curves, i.e. a two-form.
The interpretation of the Hodge dual, which is explicitly defined via the metric, is that it is the "orthogonal complement" of a differential form. The dual of a one-form in 3D is a set of curves perpendicular to the planes of the original one-form, so that the local densities of surfaces in both are proportional.
For completeness, here are a few more facts.
First, the exterior derivative $d$ corresponds to taking the boundaries of the surfaces associated with a differential $p$-form. That is why $d^2 = 0$ is just the usual "boundary of a boundary is zero" principle, and why a closed differential form is annihilated by $d$; its visualization is in terms of closed surfaces.
Second, one can also interpret the contraction of a $p$-form with a $(p, 0)$ antisymmetric tensor geometrically. A $(p, 0)$ antisymmetric tensor is called a $p$-blade and is constructed as the antisymmetric product of $p$ vectors. It can be interpreted as the $p$-dimensional volume element corresponding to the parallelepiped spanned by the vectors; for example the $dx^mu dx^nu$ above is a $2$-blade that is a small square in Cartesian coordinates.
Third, my interpretation is not the most common one. Most books treat $p$-forms in the exact same way as $p$-blades, i.e. as small volume elements. This is the point of view advanced in the three other answers to this question. It's a little simpler, but to use it you have to explicitly use the metric to convert forms to blades. My interpretation just generalizes the intuition we are usually given for $1$-forms as families of hypersurfaces, without needing to use the metric to modify the object.
$endgroup$
There's a simple way to visualize differential forms that helps here.
One of the most important properties of differential $p$-forms is that they are tensors that can naturally be integrated over a $p$-dimensional submanifold. The simplest example is a one-form, which can be integrated over a one-dimensional curve $C$,
$$int_C A = int A_mu dx^mu.$$
Unlike the integration of a vector,
$$int mathbf{F} cdot dmathbf{x}$$
this integral requires no metric to define. Similarly, a two-form can be naturally integrated over a surface, for example
$$int_S F = int F_{munu} dx^mu dx^nu$$
could be interpreted as a magnetic flux if the surface is spatial.
This leads directly to the geometric interpretation of differential forms. The familiar visualization of a one-form is a set of hypersurfaces, i.e. submanifolds of codimension one. Similarly, two-forms may be interpreted as submanifolds of codimension two, and so on. In three dimensions, a one-form is a bunch of planes, a two-form is a bunch of lines, and a three-form is a bunch of points.
You can easily count the number of planes a curve goes through, the number of lines a surface intersects, and the number of points a volume contains. These are the geometric interpretations of integrals of one-forms, two-forms, and three-forms over curves, surfaces, and volumes. (The main thing missing from this picture is the sign; you should also think of the surfaces, etc. as coming with orientations, though that's harder to draw.)
The wedge product can be interpreted as taking the geometric intersection of the differential forms involved. For example, the wedge product of two one-forms in 3D is a set of intersections of surfaces, which is a set of curves, i.e. a two-form.
The interpretation of the Hodge dual, which is explicitly defined via the metric, is that it is the "orthogonal complement" of a differential form. The dual of a one-form in 3D is a set of curves perpendicular to the planes of the original one-form, so that the local densities of surfaces in both are proportional.
For completeness, here are a few more facts.
First, the exterior derivative $d$ corresponds to taking the boundaries of the surfaces associated with a differential $p$-form. That is why $d^2 = 0$ is just the usual "boundary of a boundary is zero" principle, and why a closed differential form is annihilated by $d$; its visualization is in terms of closed surfaces.
Second, one can also interpret the contraction of a $p$-form with a $(p, 0)$ antisymmetric tensor geometrically. A $(p, 0)$ antisymmetric tensor is called a $p$-blade and is constructed as the antisymmetric product of $p$ vectors. It can be interpreted as the $p$-dimensional volume element corresponding to the parallelepiped spanned by the vectors; for example the $dx^mu dx^nu$ above is a $2$-blade that is a small square in Cartesian coordinates.
Third, my interpretation is not the most common one. Most books treat $p$-forms in the exact same way as $p$-blades, i.e. as small volume elements. This is the point of view advanced in the three other answers to this question. It's a little simpler, but to use it you have to explicitly use the metric to convert forms to blades. My interpretation just generalizes the intuition we are usually given for $1$-forms as families of hypersurfaces, without needing to use the metric to modify the object.
edited 14 hours ago
answered Feb 17 at 17:33
knzhouknzhou
44.4k11121214
44.4k11121214
$begingroup$
I'm not sure the expression "it's easy to see" is as appropriate to most people as to e.g. you, but upvoted. :-)
$endgroup$
– StephenG
Feb 17 at 19:30
$begingroup$
@StephenG Actually, in practice "it's easy to see" is just code for "I'm too lazy to check"! After coming back I realized that particular statement was just wrong, thanks for that.
$endgroup$
– knzhou
Feb 17 at 20:05
$begingroup$
@knzhou great answer, but I have a question: how can we see the "orthogonal complement" interpretation for the Hodge dual in a bit more generality so that the explanation of the example that you give for the dual of the 1-form comes out naturally?
$endgroup$
– TheQuantumMan
Feb 17 at 20:39
1
$begingroup$
@TheQuantumMan I was about to edit to address this, but I see it's nicely explained in tparker's answer!
$endgroup$
– knzhou
Feb 17 at 21:08
add a comment |
$begingroup$
I'm not sure the expression "it's easy to see" is as appropriate to most people as to e.g. you, but upvoted. :-)
$endgroup$
– StephenG
Feb 17 at 19:30
$begingroup$
@StephenG Actually, in practice "it's easy to see" is just code for "I'm too lazy to check"! After coming back I realized that particular statement was just wrong, thanks for that.
$endgroup$
– knzhou
Feb 17 at 20:05
$begingroup$
@knzhou great answer, but I have a question: how can we see the "orthogonal complement" interpretation for the Hodge dual in a bit more generality so that the explanation of the example that you give for the dual of the 1-form comes out naturally?
$endgroup$
– TheQuantumMan
Feb 17 at 20:39
1
$begingroup$
@TheQuantumMan I was about to edit to address this, but I see it's nicely explained in tparker's answer!
$endgroup$
– knzhou
Feb 17 at 21:08
$begingroup$
I'm not sure the expression "it's easy to see" is as appropriate to most people as to e.g. you, but upvoted. :-)
$endgroup$
– StephenG
Feb 17 at 19:30
$begingroup$
I'm not sure the expression "it's easy to see" is as appropriate to most people as to e.g. you, but upvoted. :-)
$endgroup$
– StephenG
Feb 17 at 19:30
$begingroup$
@StephenG Actually, in practice "it's easy to see" is just code for "I'm too lazy to check"! After coming back I realized that particular statement was just wrong, thanks for that.
$endgroup$
– knzhou
Feb 17 at 20:05
$begingroup$
@StephenG Actually, in practice "it's easy to see" is just code for "I'm too lazy to check"! After coming back I realized that particular statement was just wrong, thanks for that.
$endgroup$
– knzhou
Feb 17 at 20:05
$begingroup$
@knzhou great answer, but I have a question: how can we see the "orthogonal complement" interpretation for the Hodge dual in a bit more generality so that the explanation of the example that you give for the dual of the 1-form comes out naturally?
$endgroup$
– TheQuantumMan
Feb 17 at 20:39
$begingroup$
@knzhou great answer, but I have a question: how can we see the "orthogonal complement" interpretation for the Hodge dual in a bit more generality so that the explanation of the example that you give for the dual of the 1-form comes out naturally?
$endgroup$
– TheQuantumMan
Feb 17 at 20:39
1
1
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@TheQuantumMan I was about to edit to address this, but I see it's nicely explained in tparker's answer!
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– knzhou
Feb 17 at 21:08
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@TheQuantumMan I was about to edit to address this, but I see it's nicely explained in tparker's answer!
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– knzhou
Feb 17 at 21:08
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The $epsilon$ symbol is (proportional to) the full-rank "volume form", which you can think of as a little full-dimensional box sitting at every point on your manifold. (Note that this is a different intuitive picture from the equally valid one used in knzhou's answer.) Each index corresponds to a different direction, or to one edge of the box coming out of one particular corner. Contracting the $epsilon$ symbol with a $p$-form defined on a $p$-dimensional submanifold "collapses" the $p$ edges of the box that run parallel to the submanifold, which leaves a lower-dimensional box spanned by the $n-p$ remaining edges, which are clearly all perpendicular to the $p$ edges that got collapsed. (Note that by using the notion of "perpendicular", you're implicitly using the metric, which is hidden inside the mixed signature of the $epsilon$ symbol in the equation you wrote above.) That motivates why the Hodge dual of a $p$-form is an $n-p$ form, and also why it's the local, point-by-point "orthogonal complement" of the original form, as knzhou explained.
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add a comment |
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The $epsilon$ symbol is (proportional to) the full-rank "volume form", which you can think of as a little full-dimensional box sitting at every point on your manifold. (Note that this is a different intuitive picture from the equally valid one used in knzhou's answer.) Each index corresponds to a different direction, or to one edge of the box coming out of one particular corner. Contracting the $epsilon$ symbol with a $p$-form defined on a $p$-dimensional submanifold "collapses" the $p$ edges of the box that run parallel to the submanifold, which leaves a lower-dimensional box spanned by the $n-p$ remaining edges, which are clearly all perpendicular to the $p$ edges that got collapsed. (Note that by using the notion of "perpendicular", you're implicitly using the metric, which is hidden inside the mixed signature of the $epsilon$ symbol in the equation you wrote above.) That motivates why the Hodge dual of a $p$-form is an $n-p$ form, and also why it's the local, point-by-point "orthogonal complement" of the original form, as knzhou explained.
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add a comment |
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The $epsilon$ symbol is (proportional to) the full-rank "volume form", which you can think of as a little full-dimensional box sitting at every point on your manifold. (Note that this is a different intuitive picture from the equally valid one used in knzhou's answer.) Each index corresponds to a different direction, or to one edge of the box coming out of one particular corner. Contracting the $epsilon$ symbol with a $p$-form defined on a $p$-dimensional submanifold "collapses" the $p$ edges of the box that run parallel to the submanifold, which leaves a lower-dimensional box spanned by the $n-p$ remaining edges, which are clearly all perpendicular to the $p$ edges that got collapsed. (Note that by using the notion of "perpendicular", you're implicitly using the metric, which is hidden inside the mixed signature of the $epsilon$ symbol in the equation you wrote above.) That motivates why the Hodge dual of a $p$-form is an $n-p$ form, and also why it's the local, point-by-point "orthogonal complement" of the original form, as knzhou explained.
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The $epsilon$ symbol is (proportional to) the full-rank "volume form", which you can think of as a little full-dimensional box sitting at every point on your manifold. (Note that this is a different intuitive picture from the equally valid one used in knzhou's answer.) Each index corresponds to a different direction, or to one edge of the box coming out of one particular corner. Contracting the $epsilon$ symbol with a $p$-form defined on a $p$-dimensional submanifold "collapses" the $p$ edges of the box that run parallel to the submanifold, which leaves a lower-dimensional box spanned by the $n-p$ remaining edges, which are clearly all perpendicular to the $p$ edges that got collapsed. (Note that by using the notion of "perpendicular", you're implicitly using the metric, which is hidden inside the mixed signature of the $epsilon$ symbol in the equation you wrote above.) That motivates why the Hodge dual of a $p$-form is an $n-p$ form, and also why it's the local, point-by-point "orthogonal complement" of the original form, as knzhou explained.
answered Feb 17 at 20:56
tparkertparker
23.4k148124
23.4k148124
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I don't know if this is too basic for you, but think of freshman electromagnetism, where in order to define flux we associate an area vector with a small area on a surface. The area would more naturally be described by two vectors, $u$ and $v$, lying in its plane, such that the parallelogram they span has the area we're talking about. When we represent this area instead by a single vector $a$ perpendicular to the surface, we're taking the Hodge dual of the rank-2 tensor $u^av^b$. This works in three dimensions because 1+2=3, so the Hodge star converts a rank-2 tensor to a rank-1 tensor.
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add a comment |
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I don't know if this is too basic for you, but think of freshman electromagnetism, where in order to define flux we associate an area vector with a small area on a surface. The area would more naturally be described by two vectors, $u$ and $v$, lying in its plane, such that the parallelogram they span has the area we're talking about. When we represent this area instead by a single vector $a$ perpendicular to the surface, we're taking the Hodge dual of the rank-2 tensor $u^av^b$. This works in three dimensions because 1+2=3, so the Hodge star converts a rank-2 tensor to a rank-1 tensor.
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add a comment |
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I don't know if this is too basic for you, but think of freshman electromagnetism, where in order to define flux we associate an area vector with a small area on a surface. The area would more naturally be described by two vectors, $u$ and $v$, lying in its plane, such that the parallelogram they span has the area we're talking about. When we represent this area instead by a single vector $a$ perpendicular to the surface, we're taking the Hodge dual of the rank-2 tensor $u^av^b$. This works in three dimensions because 1+2=3, so the Hodge star converts a rank-2 tensor to a rank-1 tensor.
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I don't know if this is too basic for you, but think of freshman electromagnetism, where in order to define flux we associate an area vector with a small area on a surface. The area would more naturally be described by two vectors, $u$ and $v$, lying in its plane, such that the parallelogram they span has the area we're talking about. When we represent this area instead by a single vector $a$ perpendicular to the surface, we're taking the Hodge dual of the rank-2 tensor $u^av^b$. This works in three dimensions because 1+2=3, so the Hodge star converts a rank-2 tensor to a rank-1 tensor.
answered Feb 17 at 22:40
Ben CrowellBen Crowell
51.3k6156302
51.3k6156302
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If you think of a blade as a subspace together with a real number, then the Hodge dual is the dual subspace together with the same real number.
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If you think of a blade as a subspace together with a real number, then the Hodge dual is the dual subspace together with the same real number.
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add a comment |
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If you think of a blade as a subspace together with a real number, then the Hodge dual is the dual subspace together with the same real number.
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If you think of a blade as a subspace together with a real number, then the Hodge dual is the dual subspace together with the same real number.
answered Feb 17 at 21:21
md2perpemd2perpe
93128
93128
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Related/possible duplicate: physics.stackexchange.com/q/313091/50583
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– ACuriousMind♦
Feb 17 at 17:34