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Homology groups and Euler characteristic of tetrahedron

by Octavian on Monday, June 29, 2009

Suppose we have a tetrahedron with its triangulation

K = \{ p_0, p_1, p_2, p_3, (p_0p_1), (p_0p_2), (p_0p_3), (p_1p_2), (p_1p_3), (p_2p_3),

(p_0p_1p_2), (p_0p_1p_3), (p_0p_2p_3), (p_1p_2p_3) \}.

Analog with my last two posts, I will find homology groups associated with K and its Euler characteristic.

Since K doesn’t have 3-simplex, then B_2 (K) = \{ 0 \}. We have

C_2 (K) = \{ i(p_0p_1p_2) + j(p_0p_1p_3) + k(p_0p_2p_3) + l(p_1p_2p_3) | i,j,k,l \in \mathbb{Z} \}

For z \in Z_2 (K) \subset C_2 (K), \partial_2 z = 0. Therefore,

i((p_1p_2) - (p_0p_2) + (p_0p_1)) + j((p_1p_3) - (p_0p_3) + (p_0p_1)) + k((p_2p_3) - (p_0p_3) + (p_0p_2)) + l((p_2p_3) - (p_1p_3) + (p_1p_2)) = 0

(i+j)(p_0p_1) + (k-i)(p_0p_2) + (-j-k)(p_0p_3) + (i+l)(p_1p_2) + (j-l)(p_1p_3) + (k+l)(p_2p_3) = 0

And we conclude that j = -i, k = i, l = -i.

Hence,

Z_2 (K) = \{ i((p_0p_1p_2) - (p_0p_1p_3) + (p_0p_2p_3) - (p_1p_2p_3)) | i \in \mathbb{Z} \}

And H_2 (K) = Z_2 (K) / B_2 (K) \cong \mathbb{Z}.

We know that B_1 (K) = im (\partial_2), then

B_1 (K) = \{ (i+j)(p_0p_1) + (k-i)(p_0p_2) + (-j-k)(p_0p_3) + (i+l)(p_1p_2) + (j-l)(p_1p_3) + (k+l)(p_2p_3) | i,j,k,l \in \mathbb{Z} \}

Since Z_1 (K) = \{ z | \partial_1 z = 0, z \in C_1 (K) \}, and

C_1 (K) = \{ i(p_0p_1) + j(p_0p_2) + k(p_0p_3) + l(p_1p_2) + m(p_1p_3) + n(p_2p_3) | i,j,k,l,m,n \in \mathbb{Z} \}

we will have

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

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

such that we can conclude

k = -i-j

m = i-l

n = j+l

Then,

Z_1 (K) = \{ i(p_0p_1) + j(p_0p_2) + (-i-j)(p_0p_3) + l(p_1p_2) + (i-l)(p_1p_3) + (j+l)(p_2p_3) | i,j,l \in \mathbb{Z} \}

Define a surjective homomorphism f : Z_1 (K) \to \mathbb{Z} such that

f(i(p_0p_1) + j(p_0p_2) + k(p_0p_3) + l(p_1p_2) + m(p_1p_3) + n(p_2p_3)) = i+j+k+l+m+n

Since im(f) = \mathbb{Z} = f(B_1(K)), then

Z_1 (K) / ker(f) \cong f(B_1 (K)) = \mathbb{Z}, while Z_1 (K) \cong \mathbb{Z}

Hence ker(f) = \{ 0 \}. And it means

H_1 (K) = Z_1 (K) / B_1 (K) \cong \{ 0 \}

We also know that B_0 (K) = im(\partial_1), then

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

Since

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

is just the set of 0-simplexes, we have

Z_0 (K) = C_0 (K)

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

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

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

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

Euler characteristic of this tetrahedron can be computed using

\chi (K) = \sum_{r=0}^2 (-1)^r b_r (K)

where

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

such that we have the result

\chi (K) = 2

This result also can be computed directly using Euler theorem,

\chi (K) = \text{number of vertices} - \text{number of edges} + \text{number of faces}

\chi (K) = 4 - 6 + 4 = 2

From → Topology

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