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In the Question Which complete graphs can be embedded on a torus? the poster says, that the complete graph $K_5$ can be embedded on a torus.

But the Euler characteristic of the torus is 0 and the Euler characteristic of $K_5$ is 2. Shouldn't they be the same if $K_5$ can be embedded on the torus?

gt6989b
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xXx
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  • Have you read the comments on the linked question? This is explained in detail there. – saulspatz Dec 06 '18 at 13:20
  • But $K_4$ is planar, so clearly embeds in torus. Do you mean $K_5$ ? – coffeemath Dec 06 '18 at 13:23
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    @coffeemath sorry i met K5 – xXx Dec 06 '18 at 13:27
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    You can draw $K_5$ in the plane using only one crossing. But then that crossing can go through the torus' hole. – coffeemath Dec 06 '18 at 13:29
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    https://www.bing.com/images/search?view=detailV2&ccid=lUP1E%2bV1&id=5954BBA228D194D5C43728EFAB06560ABAA95BA2&thid=OIP.lUP1E-V1rXX0zCet-O5l4QAAAA&mediaurl=https%3a%2f%2fwww.learner.org%2fcourses%2fmathilluminated%2fimages%2funits%2f4%2fk5torus.png&exph=267&expw=366&q=embed+k5+on+a+torus&simid=608024482616640872&selectedIndex=0&ajaxhist=0 – user614287 Dec 06 '18 at 17:24
  • I believe $K_7$ can be embedded in a torus. I recall seeing a nice picture of that in a book. – bof Dec 06 '18 at 23:57
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    Nice image can be found here https://math.stackexchange.com/questions/2698746/is-the-following-embedding-of-k5-on-torus-not-a-2-cell-embedding-if-not-how – Michael T Mar 09 '25 at 15:55

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The Euler characteristic of $K_5$ is not $2$. A graph does not have an Euler characteristic.

A graph can have genus, which is the minimum genus of any orientable surface it can be embedded in; since $K_5$ can be embedded in a torus, which has genus $1$, we know that $K_5$ has genus at most $1$. (Since $K_5$ is not planar, we know it does not have genus $0$, therefore it has genus exactly $1$.)

For orientable surfaces, the genus $g$ and the Euler characteristic $\chi$ are related by $\chi = 2-2g$, so if you wanted to, you could define the Euler characteristic of a graph by this formula. This is not commonly done. By this definition, the Euler characteristic of $K_5$ would be $0$ - the same as that of the torus.

An equivalent way to phrase this definition would be that the Euler characteristic of a graph is the maximum Euler characteristic of any orientable surface it can be embedded into. So we'd conclude that whenever a graph is embedded in an orientable surface, its Euler characteristic must be less than or equal to that of the surface, and that's true here, so we're fine.

Misha Lavrov
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  • I seem to recall seeing a very nice picture in a book (in some bygone millennium) of a map on a torus with $7$ regions, each colored with a different color, and each bordering all the others. – bof Dec 07 '18 at 00:10
  • @bof I think even more impressive than that is the Szilassi polyhedron: topologically a torus, it has seven hexagonal sides, and any two sides have an edge in common. (So its dual graph gives an embedding of $K_7$ in the torus.) – Misha Lavrov Dec 07 '18 at 00:15