$2-$cell embedding of a graph in a surface
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I want to prove that every graph $G$ embeddable in the projective plane has a vertex of degree $leq 5$. If I suppose the graph has a $2-$cell embedding (an embedding in a way every face is homeomorphic to a disk) we can extend it to a triangulation and in a triangulation we can easily prove using Euler's formula that there are exactly $3n-3$ edges, from which we get that $delta(G)leq 2frac{3n-3}{n}<6$ so we get the result.
In the general case, when the embedding is arbitrary, I wanted to approach it proving that every graph which can be embedded in the projective plane, can be seen as a subgraph of a graph with a $2-$cell embedding. I don't know if this is true, but at the least, if the graph is planar, we can add edges to make it into a maximal planar graph, so that the outer face (in a plane embedding) is a triangle, and drawing it inside the cycle $4564$ in the following $2-$cell embedding of $K_{6}$ (edges of the same color are identified) would make it a subgraph of a $2-$cell embeddable graph.
I think the same approach would work to prove that planar graphs can be drawn as subgraphs of $2-$cell embeddable graphs in any compact and closed surface.
But what happens if the graph is not planar? Is there a better approach for this?
graph-theory
asked Jan 10 at 19:15
David MolanoDavid Molano
1,368720
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add a comment |
$begingroup$
I want to prove that every graph $G$ embeddable in the projective plane has a vertex of degree $leq 5$. If I suppose the graph has a $2-$cell embedding (an embedding in a way every face is homeomorphic to a disk) we can extend it to a triangulation and in a triangulation we can easily prove using Euler's formula that there are exactly $3n-3$ edges, from which we get that $delta(G)leq 2frac{3n-3}{n}<6$ so we get the result.
In the general case, when the embedding is arbitrary, I wanted to approach it proving that every graph which can be embedded in the projective plane, can be seen as a subgraph of a graph with a $2-$cell embedding. I don't know if this is true, but at the least, if the graph is planar, we can add edges to make it into a maximal planar graph, so that the outer face (in a plane embedding) is a triangle, and drawing it inside the cycle $4564$ in the following $2-$cell embedding of $K_{6}$ (edges of the same color are identified) would make it a subgraph of a $2-$cell embeddable graph.
I think the same approach would work to prove that planar graphs can be drawn as subgraphs of $2-$cell embeddable graphs in any compact and closed surface.
But what happens if the graph is not planar? Is there a better approach for this?
graph-theory
asked Jan 10 at 19:15
David MolanoDavid Molano
1,368720
$endgroup$
add a comment |
$begingroup$
I want to prove that every graph $G$ embeddable in the projective plane has a vertex of degree $leq 5$. If I suppose the graph has a $2-$cell embedding (an embedding in a way every face is homeomorphic to a disk) we can extend it to a triangulation and in a triangulation we can easily prove using Euler's formula that there are exactly $3n-3$ edges, from which we get that $delta(G)leq 2frac{3n-3}{n}<6$ so we get the result.
In the general case, when the embedding is arbitrary, I wanted to approach it proving that every graph which can be embedded in the projective plane, can be seen as a subgraph of a graph with a $2-$cell embedding. I don't know if this is true, but at the least, if the graph is planar, we can add edges to make it into a maximal planar graph, so that the outer face (in a plane embedding) is a triangle, and drawing it inside the cycle $4564$ in the following $2-$cell embedding of $K_{6}$ (edges of the same color are identified) would make it a subgraph of a $2-$cell embeddable graph.
I think the same approach would work to prove that planar graphs can be drawn as subgraphs of $2-$cell embeddable graphs in any compact and closed surface.
But what happens if the graph is not planar? Is there a better approach for this?
graph-theory
asked Jan 10 at 19:15
David MolanoDavid Molano
1,368720
$endgroup$
I want to prove that every graph $G$ embeddable in the projective plane has a vertex of degree $leq 5$. If I suppose the graph has a $2-$cell embedding (an embedding in a way every face is homeomorphic to a disk) we can extend it to a triangulation and in a triangulation we can easily prove using Euler's formula that there are exactly $3n-3$ edges, from which we get that $delta(G)leq 2frac{3n-3}{n}<6$ so we get the result.
In the general case, when the embedding is arbitrary, I wanted to approach it proving that every graph which can be embedded in the projective plane, can be seen as a subgraph of a graph with a $2-$cell embedding. I don't know if this is true, but at the least, if the graph is planar, we can add edges to make it into a maximal planar graph, so that the outer face (in a plane embedding) is a triangle, and drawing it inside the cycle $4564$ in the following $2-$cell embedding of $K_{6}$ (edges of the same color are identified) would make it a subgraph of a $2-$cell embeddable graph.
I think the same approach would work to prove that planar graphs can be drawn as subgraphs of $2-$cell embeddable graphs in any compact and closed surface.
But what happens if the graph is not planar? Is there a better approach for this?
graph-theory
graph-theory
asked Jan 10 at 19:15
David MolanoDavid Molano
1,368720
asked Jan 10 at 19:15
David MolanoDavid Molano
1,368720
edited Jan 10 at 19:23
David Molano
asked Jan 10 at 19:15
David MolanoDavid Molano
1,368720
asked Jan 10 at 19:15
David MolanoDavid Molano
1,368720
asked Jan 10 at 19:15
David MolanoDavid Molano
1,368720
1,368720
add a comment |
add a comment |
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