I saw that a cell-decomposition of a genus g non-orientable surface is $D^0\cup D^1\cup ...\cup D^g$. Can anyone explain why this is true?

  • 1
    $\begingroup$ I think you are confusing two things. The simplest cell-decomposition for a genus 1 nonorientable surface has three cells, a $0$-cell, a $1$-cell and a $2$cell, which doesn't match your count. What you have written looks instead like a cell decomposition for $\mathbb{RP}^g$, or projective $g$-space. (The $g$-dimensional projective plane.) $\endgroup$ – Cheerful Parsnip Dec 7 '11 at 21:17
  • $\begingroup$ Oh yeah. That's why it didn't make sense to me. Thanks $\endgroup$ – 908979 Dec 7 '11 at 21:31

As Grumpy Parsnip pointed out, the decomposition fits $g$-dimensional real projective space (listed among examples here), not a surface. A surface can't have cells of dimension more than $2$.

A non-orientable closed surface of genus $g$ can be constructed out of one $0$-cell, $g$ $1$-cells, and one $2$-cell. (Reference)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.