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I am using the textbook Guillemin and Pollack's Differential Topology but I am asked to solve a question need this fact that Euler characteristic of sphere is $1 + (-1)^n$.

So may I ask if this is introduced in the text, or this topic is discussed somewhere else?

Thanks~

Community
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WishingFish
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    WishingFish, are you familiar with cohomology? The Euler characteristic is equal to the alternating sum of the Betti numbers of the cohomology groups (with $\mathbb{Z}$-coefficients). Equivalently, the Euler characteristic is the alternating sum of the dimensions of the cohomology groups as $\mathbb{Q}$-vector spaces (with $\mathbb{Q}$-coefficients). Do you know the cohomology of the $n$-sphere? – Amitesh Datta Jul 27 '13 at 00:07
  • No, I know some basic thing about DeRham cohomology group from differential topology, but nothing more. Is what you are talking covered in algebraic topology? Thank you, @AmiteshDatta – WishingFish Jul 28 '13 at 03:38
  • The de Rham cohomology of a smooth manifold $M$ is a graded $\mathbb{R}$-vector space: $H^{\text{de Rham}}(M)=\bigoplus_{n=0}^{\infty} H^n(M)$; $H^n(M)$ is the $n$th de Rham cohomology group of $M$. The Poincare series of this graded $\mathbb{R}$-vector space is the power series $F(x)=\sum_{n=0}^{\infty} (\dim {\mathbb{R}} H^n(M))x^n$. The evaluation of the Poincare series at $x=-1$, i.e., $F(-1)$, is defined to be the Euler characteristic of $M$. In other words, the Euler characteristic of $M$ is equal to $\sum{n=0}^{\infty} (-1)^n \dim _{\mathbb{R}}(H^n(M))$. – Amitesh Datta Jul 28 '13 at 05:17
  • If $M=\mathbb{S}^n$, then $H^0(M)=\mathbb{R}$, $H^n(M)=\mathbb{R}$, and $H^i(M)=0$ for $i\neq 0,n$. Therefore, the associated Poincare series is $1+x^n$ (it's a polynomial; as is, in fact, always the case for (finite-dimensional) smooth manifolds). If we evaluate this polynomial at $x=-1$, then we obtain the Euler characteristic of $\mathbb{S}^n$: i.e., $1+(-1)^n$. So, if you can compute the de Rham cohomology of the sphere, then you can compute the Euler characteristic of the sphere. Can you compute the de Rham cohomology of the sphere? – Amitesh Datta Jul 28 '13 at 05:21

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I don't have a reference for you, but the computation isn't bad at all. As a CW-complex, $S^n$ is an $n$-cell whose boundary is glued to a $0$-cell. The Euler characteristic of an $n$-dimensional CW-complex is the alternating sum $d_0-d_1+d_2-d_3+\ldots+(-1)^nd_n$ where $d_i$ is the number of cells of dimension $i$. It follows that the Euler characteristic of $S^n$ is $1+(-1)^n$.

Jared
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    Alternatively, inductively, an $n+1$-sphere is gotten by gluing $2$ $n$-cells to an $n-sphere$, so you can make all the $d_i=2$. – Thomas Andrews Jul 27 '13 at 00:16
  • Hi Jared, I haven't learned $n$-cell and CW-complex. Will it be covered in algebraic topology? For some reason I am doing differential topology first. – WishingFish Jul 28 '13 at 04:04
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A good reference is the Hirsch's book "Differential Topology", page 134. If you don't have the book, I will write here the proof:

" Let $P$ the north pole and $Q=-P$ the south pole. Let $\sigma: S^n - P \rightarrow \mathbb{R}^n$ and $\tau: S^n - Q \rightarrow \mathbb{R}^n$ be the stereographic projections. The coordinate change $\tau \sigma^{-1}= \sigma \tau^{-1}: \mathbb{R}^n-0 \rightarrow \mathbb{R}^n-0$ is given by $x\mapsto \frac{x}{|x|^2}$.

Let $f$ be the vector field on $S^n-P$ whose representation via $\sigma$ is the identity vector field on $\mathbb{R}^n$. Then $f(x) \rightarrow 0$ as $x \rightarrow P$ and we define $f(P)=0$. Thus, $f:S^n \rightarrow TS^n$ has zeroes only at $P$ and $Q$.

In $\tau$ coordinates $f$ corresponds to the vector field $x \mapsto -x$ on $S^n-Q$. Thus $f$ is $C^{\infty}$.

The idendity map of $\mathbb{R}^n$ has degree $1$ or $0$, the antipodal map has degree $(-1)^n$. Therefore $Ind_P f=1$, $Ind_Q f=(-1)^n$.

Thus, we have proved $\chi(S^n)=1+(-1)^n$ "

Renato Moreira
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In Hatcher's algebraic topology, he has a formula $$\chi(M)=\sum_n(-1)^n dim_{\mathbb{Z}}(H_n(M))$$ with $H_n$ being the $n$-th singular homology group. For an $n$-sphere $S^n$, we have $H_0(S^n)=H_n(S^n)=\mathbb{Z}$ and is $0$ otherwise. So, $$\chi(S^n)=(-1)^0\cdot 1+(-1)^n\cdot 1=1+(-1)^n.$$

kid111
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