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Homology groups and Euler characteristic of circle S^1

by Octavian on Sunday, June 28, 2009

Take a circle S^1, with its triangulation to be a polyhedron K = \{ p_0, p_1, p_2, p_3, (p_0p_1), (p_1p_2), (p_2p_3), (p_3p_0) \}. We want to find homology groups associated with K, and then compute the Euler characteristics of circle S^1, using triangulation K.

Since K doesn’t have 2-simplex, Z_2 (K) = \{ 0 \}, B_2 (K) = \{ 0 \}, and it means H_2 (K) = \{0 \}. Also, B_1 (K) = \{ 0 \}. We have

C_1 (K) = \{ i(p_0p_1)+j(p_1p_2)+k(p_2p_3)+l(p_3p_0) | i,j,k,l \in \mathbb{Z} \}

If z \in Z_1 (K) \subset C_1 (K), then

\partial_1 z = i(p_1-p_0)+j(p_2-p_1)+k(p_3-p_2)+l(p_0-p_3)

0 = (l-i)p_0 + (i-j)p_1 + (j-k)p_2 + (k-l)p_3

Hence, i=j=k=l. And we have Z_1 (K) = \{ i ( (p_0p_1)+(p_1p_2)+(p_2p_3)+(p_3p_0) ) | i \in \mathbb{Z} \}, or Z_1 (K) \cong \mathbb{Z}.

Then it’s easy to see that H_1 (K) = Z_1 (K) / B_1 (K) \cong \mathbb{Z}.

Since every elements of C_0 (K) is boundaryless, then

Z_0 (K) = C_0 (K) = \{ ip_0 + jp_1 + kp_2 + lp_3 | i,j,k,l \in \mathbb{Z} \}

And also we have

B_0 (K) = \{ \partial_1 c | c \in C_0 (K) \}

B_0 (K) = \{ m(p_1-p_0)+n(p_2-p_1)+l(p_3-p_2)+s(p_0-p_3) | m,n,l,s \in \mathbb{Z} \}

B_0 (K) = \{ (s-m)p_0 + (m-n)p_1 + (n-l)p_2 + (l-s)p_3 | m,n,l,s \in \mathbb{Z} \}

Define a surjective homomorphism f: Z_0 (K) \to \mathbb{Z} by

f(ip_0 + jp_1 + kp_2 + lp_3) = i+j+k+l

Then ker(f) = f^{-1} (0) = B_0 (K), since (s-m)+(m-n)+(n-l)+(l-s) = 0.

Hence H_0 (K) = Z_0 (K) / B_0 (K) \cong im(f) = \mathbb{Z}.

The Euler characteristic of S^1 can be computed easily at this point. Since K is a triangulation of S^1, then

\chi(K) = \sum_{r=0}^1 {(-1)^r b_r (K)}

where

b_r (K) = dim H_r (K; \mathbb{R})

And it is easy to get this result

\chi(K) = 0

Hence the Euler characteristic of a circle is 0.

From → Topology

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