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Generalisation of elementary result for embedded surfaces
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If we have a triangle $T$ in $mathbbR^3$ and we consider the 3 canonical projections
$$pi_xy(x,y,z)=(x,y)$$
$$pi_xz(x,y,z)=(x,z)$$
$$pi_yz(x,y,z)=(y,z)$$
Then we have a pythagoras-type relation for the area of the triangle with respect to its projections $T_xy=pi_xy(T)$,
$T_xz=pi_xz(T)$ and $T_yz=pi_yz(T)$, as follows:
$$A(T)=A(T_xy)^2+A(T_xz)^2+A(T_yz)^2$$
In fact, this is actually the pythagorean theorem when considering bivectors, but this is irrelevant for the question. The fact is that this result still holds true when considering other parallelepipeds apart from triangles. My question is: is this result true for embedded surfaces $ShookrightarrowmathbbR^3$? or, if it is not true in this sense, is there any kind of generalization?. I think the framework of differential forms should have something to say. I am really interested in the general case, specially when considering seifert surfaces of knots. Thanks in advance.
differential-geometry surfaces geometric-topology knot-theory
asked 2 days ago
galois1989
157
add a comment |Â
up vote
0
down vote
favorite
If we have a triangle $T$ in $mathbbR^3$ and we consider the 3 canonical projections
$$pi_xy(x,y,z)=(x,y)$$
$$pi_xz(x,y,z)=(x,z)$$
$$pi_yz(x,y,z)=(y,z)$$
Then we have a pythagoras-type relation for the area of the triangle with respect to its projections $T_xy=pi_xy(T)$,
$T_xz=pi_xz(T)$ and $T_yz=pi_yz(T)$, as follows:
$$A(T)=A(T_xy)^2+A(T_xz)^2+A(T_yz)^2$$
In fact, this is actually the pythagorean theorem when considering bivectors, but this is irrelevant for the question. The fact is that this result still holds true when considering other parallelepipeds apart from triangles. My question is: is this result true for embedded surfaces $ShookrightarrowmathbbR^3$? or, if it is not true in this sense, is there any kind of generalization?. I think the framework of differential forms should have something to say. I am really interested in the general case, specially when considering seifert surfaces of knots. Thanks in advance.
differential-geometry surfaces geometric-topology knot-theory
asked 2 days ago
galois1989
157
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If we have a triangle $T$ in $mathbbR^3$ and we consider the 3 canonical projections
$$pi_xy(x,y,z)=(x,y)$$
$$pi_xz(x,y,z)=(x,z)$$
$$pi_yz(x,y,z)=(y,z)$$
Then we have a pythagoras-type relation for the area of the triangle with respect to its projections $T_xy=pi_xy(T)$,
$T_xz=pi_xz(T)$ and $T_yz=pi_yz(T)$, as follows:
$$A(T)=A(T_xy)^2+A(T_xz)^2+A(T_yz)^2$$
In fact, this is actually the pythagorean theorem when considering bivectors, but this is irrelevant for the question. The fact is that this result still holds true when considering other parallelepipeds apart from triangles. My question is: is this result true for embedded surfaces $ShookrightarrowmathbbR^3$? or, if it is not true in this sense, is there any kind of generalization?. I think the framework of differential forms should have something to say. I am really interested in the general case, specially when considering seifert surfaces of knots. Thanks in advance.
differential-geometry surfaces geometric-topology knot-theory
asked 2 days ago
galois1989
157
If we have a triangle $T$ in $mathbbR^3$ and we consider the 3 canonical projections
$$pi_xy(x,y,z)=(x,y)$$
$$pi_xz(x,y,z)=(x,z)$$
$$pi_yz(x,y,z)=(y,z)$$
Then we have a pythagoras-type relation for the area of the triangle with respect to its projections $T_xy=pi_xy(T)$,
$T_xz=pi_xz(T)$ and $T_yz=pi_yz(T)$, as follows:
$$A(T)=A(T_xy)^2+A(T_xz)^2+A(T_yz)^2$$
In fact, this is actually the pythagorean theorem when considering bivectors, but this is irrelevant for the question. The fact is that this result still holds true when considering other parallelepipeds apart from triangles. My question is: is this result true for embedded surfaces $ShookrightarrowmathbbR^3$? or, if it is not true in this sense, is there any kind of generalization?. I think the framework of differential forms should have something to say. I am really interested in the general case, specially when considering seifert surfaces of knots. Thanks in advance.
differential-geometry surfaces geometric-topology knot-theory
asked 2 days ago
galois1989
157
asked 2 days ago
galois1989
157
asked 2 days ago
galois1989
157
asked 2 days ago
galois1989
157
157
add a comment |Â
add a comment |Â
1 Answer
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If you have a surface like a portion of the "drop wave function", then by increasing the number of waves you can make the surface area as large as you want while keeping the areas of the three projections constant. So, under my interpretation of your question of whether it holds for embedded surfaces, "no."
But, if you approximate the surface with a triangulation, then you can project each triangle individually, use the Pythagorean-type equation, and add up all the results. If you take the limit of finer and finer triangulations, it converges to the true surface area.
This is similar to using differential forms to compute the area of a surface in $mathbbR^3$. If you take the standard volume form $dxwedge dywedge dz$, you can specialize it to the surface by using the surface's normal vector field to partially evaluate the $3$-form to get a $2$-form. This $2$-form is in some way tangent to the surface, and one can imagine that it covers the surface in infinitesimal parallelograms, whose total area is that of the surface. $2$-forms and bivectors are related. The $2$-form can be written in terms of the basis $dxwedge dy$, $dxwedge dz$, $dywedge dx$, which correspond to the three projections you give.
Anyway, the point of this impressionistic account of differential forms is to suggest that they do have something to say about surface area of embedded surfaces.
answered 2 days ago
Kyle Miller
6,704723
Yes, good observation that wave-type foldings increase total area but projected areas remain constant. Thanks!!
– galois1989
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If you have a surface like a portion of the "drop wave function", then by increasing the number of waves you can make the surface area as large as you want while keeping the areas of the three projections constant. So, under my interpretation of your question of whether it holds for embedded surfaces, "no."
But, if you approximate the surface with a triangulation, then you can project each triangle individually, use the Pythagorean-type equation, and add up all the results. If you take the limit of finer and finer triangulations, it converges to the true surface area.
This is similar to using differential forms to compute the area of a surface in $mathbbR^3$. If you take the standard volume form $dxwedge dywedge dz$, you can specialize it to the surface by using the surface's normal vector field to partially evaluate the $3$-form to get a $2$-form. This $2$-form is in some way tangent to the surface, and one can imagine that it covers the surface in infinitesimal parallelograms, whose total area is that of the surface. $2$-forms and bivectors are related. The $2$-form can be written in terms of the basis $dxwedge dy$, $dxwedge dz$, $dywedge dx$, which correspond to the three projections you give.
Anyway, the point of this impressionistic account of differential forms is to suggest that they do have something to say about surface area of embedded surfaces.
answered 2 days ago
Kyle Miller
6,704723
Yes, good observation that wave-type foldings increase total area but projected areas remain constant. Thanks!!
– galois1989
2 days ago
add a comment |Â
up vote
1
down vote
accepted
If you have a surface like a portion of the "drop wave function", then by increasing the number of waves you can make the surface area as large as you want while keeping the areas of the three projections constant. So, under my interpretation of your question of whether it holds for embedded surfaces, "no."
But, if you approximate the surface with a triangulation, then you can project each triangle individually, use the Pythagorean-type equation, and add up all the results. If you take the limit of finer and finer triangulations, it converges to the true surface area.
This is similar to using differential forms to compute the area of a surface in $mathbbR^3$. If you take the standard volume form $dxwedge dywedge dz$, you can specialize it to the surface by using the surface's normal vector field to partially evaluate the $3$-form to get a $2$-form. This $2$-form is in some way tangent to the surface, and one can imagine that it covers the surface in infinitesimal parallelograms, whose total area is that of the surface. $2$-forms and bivectors are related. The $2$-form can be written in terms of the basis $dxwedge dy$, $dxwedge dz$, $dywedge dx$, which correspond to the three projections you give.
Anyway, the point of this impressionistic account of differential forms is to suggest that they do have something to say about surface area of embedded surfaces.
answered 2 days ago
Kyle Miller
6,704723
Yes, good observation that wave-type foldings increase total area but projected areas remain constant. Thanks!!
– galois1989
2 days ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you have a surface like a portion of the "drop wave function", then by increasing the number of waves you can make the surface area as large as you want while keeping the areas of the three projections constant. So, under my interpretation of your question of whether it holds for embedded surfaces, "no."
But, if you approximate the surface with a triangulation, then you can project each triangle individually, use the Pythagorean-type equation, and add up all the results. If you take the limit of finer and finer triangulations, it converges to the true surface area.
This is similar to using differential forms to compute the area of a surface in $mathbbR^3$. If you take the standard volume form $dxwedge dywedge dz$, you can specialize it to the surface by using the surface's normal vector field to partially evaluate the $3$-form to get a $2$-form. This $2$-form is in some way tangent to the surface, and one can imagine that it covers the surface in infinitesimal parallelograms, whose total area is that of the surface. $2$-forms and bivectors are related. The $2$-form can be written in terms of the basis $dxwedge dy$, $dxwedge dz$, $dywedge dx$, which correspond to the three projections you give.
Anyway, the point of this impressionistic account of differential forms is to suggest that they do have something to say about surface area of embedded surfaces.
answered 2 days ago
Kyle Miller
6,704723
If you have a surface like a portion of the "drop wave function", then by increasing the number of waves you can make the surface area as large as you want while keeping the areas of the three projections constant. So, under my interpretation of your question of whether it holds for embedded surfaces, "no."
But, if you approximate the surface with a triangulation, then you can project each triangle individually, use the Pythagorean-type equation, and add up all the results. If you take the limit of finer and finer triangulations, it converges to the true surface area.
This is similar to using differential forms to compute the area of a surface in $mathbbR^3$. If you take the standard volume form $dxwedge dywedge dz$, you can specialize it to the surface by using the surface's normal vector field to partially evaluate the $3$-form to get a $2$-form. This $2$-form is in some way tangent to the surface, and one can imagine that it covers the surface in infinitesimal parallelograms, whose total area is that of the surface. $2$-forms and bivectors are related. The $2$-form can be written in terms of the basis $dxwedge dy$, $dxwedge dz$, $dywedge dx$, which correspond to the three projections you give.
Anyway, the point of this impressionistic account of differential forms is to suggest that they do have something to say about surface area of embedded surfaces.
answered 2 days ago
Kyle Miller
6,704723
answered 2 days ago
Kyle Miller
6,704723
answered 2 days ago
Kyle Miller
6,704723
answered 2 days ago
Kyle Miller
6,704723
6,704723
Yes, good observation that wave-type foldings increase total area but projected areas remain constant. Thanks!!
– galois1989
2 days ago
add a comment |Â
Yes, good observation that wave-type foldings increase total area but projected areas remain constant. Thanks!!
– galois1989
2 days ago
Yes, good observation that wave-type foldings increase total area but projected areas remain constant. Thanks!!
– galois1989
2 days ago
Yes, good observation that wave-type foldings increase total area but projected areas remain constant. Thanks!!
– galois1989
2 days ago
add a comment |Â
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