Mixing
What properties does an adjacency matrix have for triangular mesh of a sphere?
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Given an adjacency matrix $M$. Are there any simple properties that $M$ must have such that it represents the triangular mesh of a sphere. (examples are icosohedron, or geodesic spheres). What about for a torus?
One thing I can think of is that every line must belong to two and only two triangles. I think you would write this as $M_{ij}=1 implies (M^2)_{ij}=2$. Are there any other formulae?
Can we express this some way in terms of perhaps the cube $M^3$ of the adjacency matrix?
graph-theory adjacency-matrix
asked Jan 27 at 17:14
zoobyzooby
1,032716
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add a comment |
$begingroup$
Given an adjacency matrix $M$. Are there any simple properties that $M$ must have such that it represents the triangular mesh of a sphere. (examples are icosohedron, or geodesic spheres). What about for a torus?
One thing I can think of is that every line must belong to two and only two triangles. I think you would write this as $M_{ij}=1 implies (M^2)_{ij}=2$. Are there any other formulae?
Can we express this some way in terms of perhaps the cube $M^3$ of the adjacency matrix?
graph-theory adjacency-matrix
asked Jan 27 at 17:14
zoobyzooby
1,032716
$endgroup$
add a comment |
$begingroup$
Given an adjacency matrix $M$. Are there any simple properties that $M$ must have such that it represents the triangular mesh of a sphere. (examples are icosohedron, or geodesic spheres). What about for a torus?
One thing I can think of is that every line must belong to two and only two triangles. I think you would write this as $M_{ij}=1 implies (M^2)_{ij}=2$. Are there any other formulae?
Can we express this some way in terms of perhaps the cube $M^3$ of the adjacency matrix?
graph-theory adjacency-matrix
asked Jan 27 at 17:14
zoobyzooby
1,032716
$endgroup$
Given an adjacency matrix $M$. Are there any simple properties that $M$ must have such that it represents the triangular mesh of a sphere. (examples are icosohedron, or geodesic spheres). What about for a torus?
One thing I can think of is that every line must belong to two and only two triangles. I think you would write this as $M_{ij}=1 implies (M^2)_{ij}=2$. Are there any other formulae?
Can we express this some way in terms of perhaps the cube $M^3$ of the adjacency matrix?
graph-theory adjacency-matrix
graph-theory adjacency-matrix
asked Jan 27 at 17:14
zoobyzooby
1,032716
asked Jan 27 at 17:14
zoobyzooby
1,032716
edited Jan 27 at 18:03
zooby
asked Jan 27 at 17:14
zoobyzooby
1,032716
asked Jan 27 at 17:14
zoobyzooby
1,032716
asked Jan 27 at 17:14
zoobyzooby
1,032716
1,032716
add a comment |
add a comment |
1 Answer
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$begingroup$
We need $(M^3)_{ii}=2(M^2)_{ii}$ for a "locally planar" triangular mesh (for every edge incident with $i$, thee are two triangles with vertex $i$), but by itself this is weaker than your condition.
To distinguish between sphere and torus, we should look for the Euler characteristic $v+f-e$. We have $v$ vertices from the dimension of the matrix, $e=frac 12operatorname {Tr}M^2$ edges, and $f=frac 16operatorname {Tr}M^3$ triangular faces. By the condition in the first paragraph, $operatorname {Tr}M^3=2operatorname {Tr}M^2$, or (unsurprisingly) $3f=2e$., so that (given that we have a triangular mesh in the first place), the Euler characteristic is
$$dim V-frac 16operatorname{Tr}M^2. $$
answered Jan 27 at 17:35
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Did you mean the indices to be $ij$ instead of $ii$ ?
$endgroup$
– zooby
Jan 27 at 17:55
$begingroup$
Is the condition sufficient or do I need my condition too?
$endgroup$
– zooby
Jan 27 at 19:23
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We need $(M^3)_{ii}=2(M^2)_{ii}$ for a "locally planar" triangular mesh (for every edge incident with $i$, thee are two triangles with vertex $i$), but by itself this is weaker than your condition.
To distinguish between sphere and torus, we should look for the Euler characteristic $v+f-e$. We have $v$ vertices from the dimension of the matrix, $e=frac 12operatorname {Tr}M^2$ edges, and $f=frac 16operatorname {Tr}M^3$ triangular faces. By the condition in the first paragraph, $operatorname {Tr}M^3=2operatorname {Tr}M^2$, or (unsurprisingly) $3f=2e$., so that (given that we have a triangular mesh in the first place), the Euler characteristic is
$$dim V-frac 16operatorname{Tr}M^2. $$
answered Jan 27 at 17:35
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Did you mean the indices to be $ij$ instead of $ii$ ?
$endgroup$
– zooby
Jan 27 at 17:55
$begingroup$
Is the condition sufficient or do I need my condition too?
$endgroup$
– zooby
Jan 27 at 19:23
add a comment |
$begingroup$
We need $(M^3)_{ii}=2(M^2)_{ii}$ for a "locally planar" triangular mesh (for every edge incident with $i$, thee are two triangles with vertex $i$), but by itself this is weaker than your condition.
To distinguish between sphere and torus, we should look for the Euler characteristic $v+f-e$. We have $v$ vertices from the dimension of the matrix, $e=frac 12operatorname {Tr}M^2$ edges, and $f=frac 16operatorname {Tr}M^3$ triangular faces. By the condition in the first paragraph, $operatorname {Tr}M^3=2operatorname {Tr}M^2$, or (unsurprisingly) $3f=2e$., so that (given that we have a triangular mesh in the first place), the Euler characteristic is
$$dim V-frac 16operatorname{Tr}M^2. $$
answered Jan 27 at 17:35
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Did you mean the indices to be $ij$ instead of $ii$ ?
$endgroup$
– zooby
Jan 27 at 17:55
$begingroup$
Is the condition sufficient or do I need my condition too?
$endgroup$
– zooby
Jan 27 at 19:23
add a comment |
$begingroup$
We need $(M^3)_{ii}=2(M^2)_{ii}$ for a "locally planar" triangular mesh (for every edge incident with $i$, thee are two triangles with vertex $i$), but by itself this is weaker than your condition.
To distinguish between sphere and torus, we should look for the Euler characteristic $v+f-e$. We have $v$ vertices from the dimension of the matrix, $e=frac 12operatorname {Tr}M^2$ edges, and $f=frac 16operatorname {Tr}M^3$ triangular faces. By the condition in the first paragraph, $operatorname {Tr}M^3=2operatorname {Tr}M^2$, or (unsurprisingly) $3f=2e$., so that (given that we have a triangular mesh in the first place), the Euler characteristic is
$$dim V-frac 16operatorname{Tr}M^2. $$
answered Jan 27 at 17:35
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
We need $(M^3)_{ii}=2(M^2)_{ii}$ for a "locally planar" triangular mesh (for every edge incident with $i$, thee are two triangles with vertex $i$), but by itself this is weaker than your condition.
To distinguish between sphere and torus, we should look for the Euler characteristic $v+f-e$. We have $v$ vertices from the dimension of the matrix, $e=frac 12operatorname {Tr}M^2$ edges, and $f=frac 16operatorname {Tr}M^3$ triangular faces. By the condition in the first paragraph, $operatorname {Tr}M^3=2operatorname {Tr}M^2$, or (unsurprisingly) $3f=2e$., so that (given that we have a triangular mesh in the first place), the Euler characteristic is
$$dim V-frac 16operatorname{Tr}M^2. $$
answered Jan 27 at 17:35
Hagen von EitzenHagen von Eitzen
283k23272507
answered Jan 27 at 17:35
Hagen von EitzenHagen von Eitzen
283k23272507
answered Jan 27 at 17:35
Hagen von EitzenHagen von Eitzen
283k23272507
answered Jan 27 at 17:35
Hagen von EitzenHagen von Eitzen
283k23272507
283k23272507
$begingroup$
Did you mean the indices to be $ij$ instead of $ii$ ?
$endgroup$
– zooby
Jan 27 at 17:55
$begingroup$
Is the condition sufficient or do I need my condition too?
$endgroup$
– zooby
Jan 27 at 19:23
add a comment |
$begingroup$
Did you mean the indices to be $ij$ instead of $ii$ ?
$endgroup$
– zooby
Jan 27 at 17:55
$begingroup$
Is the condition sufficient or do I need my condition too?
$endgroup$
– zooby
Jan 27 at 19:23
$begingroup$
Did you mean the indices to be $ij$ instead of $ii$ ?
$endgroup$
– zooby
Jan 27 at 17:55
$begingroup$
Did you mean the indices to be $ij$ instead of $ii$ ?
$endgroup$
– zooby
Jan 27 at 17:55
$begingroup$
Is the condition sufficient or do I need my condition too?
$endgroup$
– zooby
Jan 27 at 19:23
$begingroup$
Is the condition sufficient or do I need my condition too?
$endgroup$
– zooby
Jan 27 at 19:23
add a comment |
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