I want to understand the graph on the surface of genus 1. Let $G=(V,E,F)$ be a graph with $V,G,F$ denoting vertex, edge, and face respectively. Then the genus one condition give us that $V-E+F=0$, known as the Euler characteristics.
Also when we delete the graph we should get a disk that is boundary one surface.
I know one example of such a graph with 1, vertex, 2 edges and 1 face. That is drawing a square and glueing the opposite edge. What I don't understand why the degree of the face is three? I only see two edges.
What is the $Aut$ of such a graph? Is there is a picture to visualize?
I know that the degree of a face is the number of edges adjacent to the face, counted with multiplicity 2 if both sides of the edge belong to the face.
Can anyone give me an example of a graph in genus one with two faces with degree 3,1 with a picture? In particular of degree
First for you question,
No, this is impossible. If you have only two faces with degree 3 and 1, then by hand-shaking lemma, you must have 2 edges. But then by Euler's formula, if you have an emebedding of this graph with genus $1$, then you must have $\vert V\vert = 0$, your graph is empty. Contradicting the hypothesis on the faces. Such an embedding of a graph cannot exist.
It seems to me that you are mixing two notions : a graph and its embeddings. A graph is defined by its vertices and edge sets $G=(V,E)$. The faces depends on the actual embedding. The same graph can have different number of faces depending on its embedding (its drawing).
The genus $g$ of an embedding does verify : $V-E+F=2-2g$. The genus of a graph $G$ is the minimum $g$ such that $G$ can be embedded in a surface of genus $g$.
Your definition of "drawing a square and glueing the opposite edge" is not clear. If you do so, you end up with one single vertex, with two loops (kind of $\infty$ sign), hence a graph of genus $0$ not 1. Each "interior face of the loops" have degree $1$, while the outer face has degree 2.
If you force this graph into a surface of genus $1$, by placing the loops around the torus, then you end up with a graph with only 1 face (see below the poor picture). This face as degree 4, verifying the hand shaking lemma $\sum_{f\in F}d(f) = 2\vert E\vert$, counting twice any bridge as you mentionned.