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Let $G$ be a connected, directed graph with $v$ vertices and $e$ edges. According to Massey (Ch. VIII, Section 3), the euler characteristic satisfies \begin{align} v - e = \chi(G) = \text{rank} \, H_0(G) - \text{rank} \, H_1(G) = 1 - \text{rank} \, H_1(G). \end{align} Based on basic definitions, $H_{1}(G) = \ker \partial_1 / \text{im} \, \partial_2$, where $\partial_*$ is the (chain) boundary map. I'd like to compute $\text{dim} \ker \partial_1$ and $\text{dim} \, \text{im} \, \partial_1$.

From what I understand, if $\rho(\partial_1)$ is a matrix representation of $\partial_1$, then it has size $e \times v$. By Rank-Nullity, $\dim \ker \partial_1 + \dim \text{im} \, \partial_1 = v$. Since $G$ has no $2$-chains, then $H_1 \cong \ker \partial_1$. Thus, $\dim \ker \, \partial_1 = e- v + 1$ (also the number of circles in the homotopy type of $G$) and $\dim \, \text{im} \, \partial_1 = 2v - e - 1$.

I have a feeling that I've misunderstood something fundamental. Is my reasoning correct?

Edit: It appears that I switched rank with size, so $\dim \, \text{im} \, \partial_1 = v - 1$ consistent with the rank of $\rho(\partial_1) = v - 1$, as wckronholm points out. Now the question boils down to the following :Is $\dim \ker \partial_1 = 1$ or $e - v + 1$? It seems now it is $1$. However, $\text{rank} \, H_1(G) = 1 - \chi(G) = e - v + 1$, but $\text{rank} \, H_1(G) = \dim \ker \partial_1 - \dim \, \text{im} \, \partial_2$ and $\dim \, \text{im} \, \partial_2 = 0$, no?

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1 Answer 1

up vote 2 down vote accepted

The graph is connected (and finite, evidently) so the rank of $H_0(G)$ is $1$. Hence $\mathrm{dim}\; H_0(G) = \mathrm{dim} \; \mathrm{ker}\; \partial_0 - \mathrm{dim} \; \mathrm{im}\; \partial_1$. But $\mathrm{dim}\;\mathrm{ker} \;\partial_0 = v$ so this gives $\mathrm{dim}\;\mathrm{im}\;\partial_1 = v-1$.

It seems to me that, using your notation, the rank of $\rho(\partial_1)$ is $v-1$.

Edited to address question edits:

As you computed above, $\mathrm{dim}\;\mathrm{ker}\; \partial_1 = e-v+1$. This is consistent with rank-nullity since $\mathrm{dim}\;\mathrm{im}\; \partial_1 + \mathrm{dim}\;\mathrm{ker}\;\partial_1 = (v-1) + (e-v+1) = e = \mathrm{dim}\; C_1(G)$.

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