It is known that a closed orientable surface of genus $g$ has Euler characteristic $2-2g$. According to this, the open disc being of genus $0$ should have Euler characteristic $2$, but this contradicts the fact that the disc is contractible so has a Euler characteristic $1$. Thank you for your clarification!!
Euler characteristic of a surface
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10Is the open disc a closed surface? – 2012-10-23
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0by closed we mean compact without boundary so yes it is a closed surface i think – 2012-10-23
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6The *open* disc is compact?! Come on... – 2012-10-23
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0it is not of course but i thaught the genus formula would apply!! since the open disc is an orientable surface and without boundary.. So compacteness is essential in the formula $\chi=2-2g$, otherwise it does not apply? – 2012-10-23
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2That's correct-you already demonstrated in your question that $\chi=2-2g$ need not apply for surfaces which aren't closed. That's why the theorem is stated for closed surfaces. – 2012-10-23
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3So, now that you know the answer, you can write it up and post it. Later, you can accept it. – 2012-10-23
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0I tried to write an answer! – 2012-10-23
2 Answers
First of all we need to clarify what is your definition of the Euler characteristic of a space $X$. For me, the Euler characteristic of a finite dimensional CW complex can be defined to be the alternating sum over the number of cells in each dimension. As it turns out, for a finite dimensional CW complex $X$ the Euler characteristic is also equal to
$$\sum_{i=0}^k (-1)^i \textrm{rank} H_i(X)$$
For the closed orientable surface $M_g$ of genus $g$, we have $H_0(X) = \Bbb{Z}$, $H_1(X) = \Bbb{Z}^{2g}$, $H_2(X) = \Bbb{Z}$ and so
$$\chi(M_g) = 1 - 2g + 1 = 2- 2g.$$
Now for the open/closed disk $X$ (it does not matter which one) we see that $H_0(X) = \Bbb{Z}$, and all higher homology groups vanish because it is contractible. Hence
$$\chi(X) = 1.$$
It is known that a closed orientable surface of genus $g$ has Euler characteristic $2-2g$. Being closed is essential here. Indeed, consider for example the open disc. This surface is orientable of genus zero but is not compact hence it is not a closed surface so it does not verify the formula $\chi=2-2g$ and in fact the actual $\chi$ of the open disc is $1$ since the disc is contractible. Moreover the closed disc is compact but it has a boundary so it is not a closed surface neither and hence the formula does not apply; in fact the closed disc is also contractible and has $\chi=1$.