Euler Characteristic is not Dependent upon Triangulation

Theorem

Let $S$ be a surface.

Let $T_1$ and $T_2$ be triangulations of $S$.

Let:

$\map {\chi_1} S$ be the Euler characteristic of $S$ as calculated using $T_1$

$\map {\chi_2} S$ be the Euler characteristic of $S$ as calculated using $T_2$.

Then:

$\map {\chi_1} S = \map {\chi_2} S$

Proof

Recall:

\(\ds \map {\chi_1} S\)

\(=\)

\(\ds \map {\chi_2} S\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \map v {T_1} - \map e {T_1} + \map f {T_1}\)

\(=\)

\(\ds \map v {T_2} - \map e {T_2} + \map f {T_2}\)

Definition of Euler Characteristic of Surface

where:

$\map v {T_1}$ denotes the order of $T_1$ (that is, the number of its vertices)

$\map e {T_1}$ denotes the size of $T_1$ (that is, the number of its edges)

$\map f {T_1}$ denotes the number of faces of $T_1$.

This theorem requires a proof. In particular: Probably do it by induction on number of faces, once we have a rigorous definition of triangulation, and have the basics of the theory down. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Euler characteristic