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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?

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    Oh yeah. That's why it didn't make sense to me. Thanks2011-12-07

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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)