 We've determined that if a graph has genus g, then v plus f minus e is some constant value. Today we call the value the Euler characteristic. For genus zero graphs, v plus f minus e is two. For genus one graphs, v plus f minus e is zero. So what about genus g graphs? We find that if g is a connected graph with genus g, then v plus f equals e plus two minus two g. A full proof is somewhat technical. However, we can present a sketch of the proof. So suppose v plus f minus e is some function of the genus g. We know that f of zero is two, and so the question is how does f change as g changes? So let's begin with a graph that is embedded on a k-handled sphere. If the graph is embedded on a k-handled sphere, then there must be an n-cycle around the base of each handle. And that's the technical part. We can smush that cycle until it just fits over the handle. And just a note, we say that it's around the base of the handle in quotes because technically speaking there's nothing inside the cycle. Or if you really want to get into it, what's inside the cycle is actually the handle. So now we'll cut the handle, and in doing so we'll actually produce two copies of the cycle, one that's still on the sphere, and one that's on the handle. And now we'll retract the handle, kind of absorbing it into the rest of the sphere. And we get a new graph. Now note that three things have occurred. We've reduced the number of handles by one, we've increased the number of edges by n, the number of edges in the cycle, and we've also increased the number of vertices by n, the number of edges in the cycle, which equals the number of vertices. And the fourth of the three things we've done is we've increased the number of faces by 2. And what this means is decreasing g by 1 will increase the quantity by 2. And since f of 0 is 2, it follows that f of g is 2 minus 2g. Now that we have an extension of Euler's formula, we can use it in other contexts. For example, for a simple connected planar graph, we found that E is less than or equal to 3b minus 6. By essentially the same logic, we find that a connected genus g graph will satisfy, and this might help us to do things like find a lower bound on the genus of an arbitrary graph. For k10, for example, we have 10 vertices and 10 choose to 45 edges. And so we can substitute these into our formula and find. And since the genus has to be a whole number, we know that the lower bound of the genus of k10 is 4. One of the other immediate consequences of Euler's formula is that for planar graphs, we proved that the minimum degree had to be less than 6. And we can do the same analysis for graphs with genus g. So I suppose g is a connected genus g graph where the minimum degree is k. Again, by the handshake here, the sum of the degrees is twice the number of edges. And if our minimum degree is k, we have. If we substitute this into our formula relating the edges and the vertices, we find. So let's prove that if g is a connected graph with genus 1, the minimum degree is less than or equal to 6. So if g equals 1, we know. So any graph that can be embedded on a torus must have a vertex of degrees 6 or less. Now, because this minimum degree depends on the number of vertices, it can get a little tricky. So for example, let's see if we could find a genus 2 graph where the minimum degree is 8. Our formula relates the minimum degree to the number of vertices. So if we substitute the minimum degree, we can find a corresponding value of v. So we'll substitute our minimum degree into our formula and find. And so we find that if a genus 2 graph has 6 or fewer vertices, the minimum degree is actually less than or equal to 5. Remember, even though you can do the algebra, the variables actually mean something. And in this case, we're talking about graphs with 6 or fewer vertices. So it's impossible for any vertex to have a degree greater than 5. And so the minimum has to be less. What if we had more than 6 vertices? If v is greater than 6, then since v has to be a whole number, v is greater than or equal to 7. And so we have. And since our minimum degree has to be a whole number, the degree is less than or equal to 7. So if g has more vertices, the minimum degree is less than or equal to 7. And so no genus 2 graph can have a minimum degree of 8. This leads to some interesting possibilities. Suppose we want to characterize the genus 2 graphs with minimum degree 7. Now, if we want our minimum degree to be 7, we have. And so in order to have a minimum degree of 7, the graph can have at most 12 vertices. And also in order to have the degree 7, the graph must have at least 8 vertices. So a necessary condition for a genus 2 graph to have minimum degree 7 is for the number of vertices to be somewhere between 8 and 12. And because the number of edges in a simple graph is determined by the number of vertices, this means that it's sometimes possible to prove theorems that are true for most graphs. For example, most genus 2 graphs have minimum degree less than or equal to 6. In fact, while it's necessary for the genus 2 graph with minimum degree 7 to have between 8 and 12 vertices, we don't know if it's sufficient. And in fact, it's possible that no genus 2 graph has minimum degree 7. And since there are only a finite number of genus 2 graphs with between 8 and 12 vertices, we can in principle check them out. We won't, that's actually a research project, but we could. So for example, we want to prove that for most graphs of genus 4, the minimum degree is less than or equal to 7. So if our genus is 4, then our minimum degree satisfies. Now in order for the minimum degree to be greater than 7, we have to have that minimum degree to be 8 or more. And so we find. So if the number of vertices is greater than 18, the minimum degree has to be less than or equal to 7. And with 8 or fewer vertices, the minimum degree also has to be less than or equal to 7. So the only genus 4 graphs where the minimum degree is greater than 7 are those where the number of vertices is between 9 and 18.