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In the plane, Euler's Polyhedral formula tells us that $V - E + F = \chi$, where for graph embeddings we have that $\chi = 1$. Alternatively, we can think of a graph embedding as a simplicial $1$-complex embedded in the plane.

My question is about if there exists a straightforward generalization of the value of the Euler characteristic to higher dimensions. In particular, for a graph embedded in 3-space, or equivalently a simplicial 2-complex or simplicial 3-complex embedded in $\mathbb{E}^3$, is there a simple expression that defines the Euler characteristic as a constant value, as there is for the plane? (namely $\chi =1$).

Thank you!

EDIT: I am aware that there is a general formula for $\chi$ if you know the number of faces of a given dimension, I am looking to determine those faces by knowing what the Euler characteristic is from external information.

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According to wiki, the Euler's characteristic $\chi = v_0 - v_1 + v_2 - v_3 + \cdots + (-1)^d v_d$, where $d$ is the dimension, $v_i$ is the number of $i$-faces. When $d=2$, the formula reads $\chi = v_0 - v_1 + v_2 = V- E+F$. Is this what you need? – Shuhao Cao May 11 '12 at 0:43
How I want to use this, is to determine some of the $v_{i}$'s without knowing that information. I'm looking for the value of $\chi$ in more generality. For graph embeddings in the plane $\chi =1$, for graph embeddings (or simplicial 2-complexes, simplicial 3-complexes) in $\mathbb{E}^3$, is there also a simple answer? – Samuel Reid May 11 '12 at 1:17
The Euler characteristic isn't constant in higher dimensions. However, it has many equivalent characterizations and can be computed in lots of different ways; the Wikipedia article should have more details. – Qiaochu Yuan May 11 '12 at 1:25
@QiaochuYuan: I am interested in the "lots of different ways" you mentioned. I have already read the wikipedia article thoroughly. Do you happen to know if given the conditions that a simplicial $3$-complex $\mathcal{K}$ in $\mathbb{E}^3$ (for now, I want to generalize to simplicial $k$-complex) is homogeneous and connected, if there exists some characterization of the Euler characteristic of $\mathcal{K}$ in terms of the number of vertices of $\mathcal{K}$? – Samuel Reid May 11 '12 at 1:32
@Samuel: the number of vertices shouldn't determine the Euler characteristic in general. What do you mean by homogeneous? – Qiaochu Yuan May 11 '12 at 1:49

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