# Nonorientable surfaces: genus or demigenus?

The genus $g$ of a closed, orientable surface is the maximum number of disjoint simple closed curves that can be drawn on the surface without disconnecting it. In terms of the Euler characteristic, $\chi = 2 - 2g$.

For closed nonorientable surfaces, Wikipedia defines its nonorientable genus, demigenus or Euler genus $k$ as the number of cross-caps attached to a sphere. In terms of the Euler characteristic, $\chi = 2 - k$.

Here we have four different "numbers" for any given surface:

1. The maximum number of disjoint simple closed curves that can be drawn on the surface without disconnecting it
2. $g = (2 - \chi)/2$
3. The number of cross-caps attached to a sphere
4. $k = 2 - \chi$

My question: What is the genus of a nonorientable surface?

The first paragraph of page 3 of this article defines the genus of a nonorientable surface using $(1)$. This is compatible with Wikipedia which avoids defining the genus of a nonorientable surface.

Wolfram Mathworld defines the genus of a nonorientable surface using $(4)$. The Manifold Atlas Project defines the genus of a nonorientable surfaces using $(3)$.

Assuming that all of the above are consistent, the only way I can resolve this conflict is:

Possible answer: Could it be that for orientable surfaces $(1) = (2)$, but for nonorientable surfaces $(1) = (3) = (4)$, so that definition $(1)$ works for everything?

Is there even a consensus among mathematicians? If there is, could I have a reference for this consensus?

• Your "possible answer" is correct. Do you want to see a proof? Definition 3 is indeed the most common one and its source (the "atlas") is the most professional one of all you link. I do not believe there is a reference for a consensus on anything in mathematics, including genus. – Moishe Kohan Apr 21 '14 at 18:15

Here is a proof of equivalence of 1, 3 and 4. Below, $$S$$ is a compact connected surface without boundary.
Definition. A maximal cut in $$S$$ is a 1-dimensional submanifold $$L\subset S$$ such that $$S\setminus L$$ is connected and contains no nonseparating simple loops. The latter condition just means that $$S\setminus L$$ is the 2-dimensional sphere with $$q=q(S,L)$$ points removed. (To see the latter, you can use the classification of compact surfaces.) In particular, $$\chi(S_L)=2-q$$.
Let $$L$$ be a maximal cut in $$S$$ and let $$S_L$$ denote the surface with boundary such that $$S_L\setminus \partial S_L$$ is homeomorphic to $$S\setminus L$$. Then $$S$$ can be reconstructed from $$S_L$$ as follows: represent $$\partial S_L$$ as a disjoint union of circles: $$C_1\cup C_1' \sqcup ... \sqcup C_{n}\sqcup C_n' \cup A_{1} \sqcup ... A_m,$$ $$q=2n+m$$. Identify each $$C_i$$ to $$C_i'$$, $$i\le n$$ via a homeomorhism and identify each $$A_i$$ to itself via an antipodal map of the circle. The space $$S_L/\sim$$ equipped with the quotient topology is homeomorphic to $$S$$.
Using this, we can compute the Euler characteristics: $$\chi(S)= \chi(S_L)= 2-q.$$ To see the first identity, triangulate $$S_L$$ so that the identification maps of the boundary are simplicial and just count (with the usual sign) the simplices in $$S$$ using simplices in $$S_L$$.
In particular, we see that $$q$$ is independent of the maximal cut. Moreover, this shows that 1 is equivalent to 4 (orientation is irrelevant). To prove equivalence of 1 and 3 in the non-orientable case, note that we can obtain $$S$$ from the surface $$S_L$$ (which is the sphere with $$q$$ holes) by adding a cross-cap to each hole ($$q$$ cross-cups in total). In the more modern terminology, we represent $$S$$ as the connected sum of $$q$$ projective planes. Equivalently, $$S=S_L/\sim$$, where we set $$n=0, m=q$$ holes with my notation above.