 Okay, thanks Zach. Good morning everyone, or afternoon, wherever you are, it's morning here. I'm going to talk about a meeting of hypercube graph on orientable surfaces. I worked on this with Dr. Robert Fransilovstov from the University of Maine. So started off with the idea of drawing certain graphs on surfaces with restrictions like no intersection except vertices or in such a way that you end up with exactly one component. So it was a homework problem which was a motivation and it turns out there are more than one ways in which you can embed a graph into a surface. So the natural question to ask was can we find all of them? So that turned out to be a more complicated question. So instead of focusing on all embeddings, I decided to focus on the one that's on minimal genius surface and one on a maximal genius surface. And again instead of working on K and N, decided to work on hypercube graphs. So here's an example of three hypercube graphs Q1, Q2, Q3. It's exactly what you would think it is going to be. We are going to label the vertices in a certain manner where connected vertices differ by exactly one bit or there which is to say the hamming distance is one. And then we will be using the binary notation for the vertices throughout the stop, the switching between binary and decimal. So that's the convention that we're using for labeling our vertices. What's coming up is I'm going to define what a rotation system is, a boundary walk algorithm, ABC rotation system that we came up with something called an adjustment switch theorem, minimal embedding and the phase to the maximum. So those are the main results that I'll be building up to and some conclusions and unanswered questions. For the sake of the idea, we'll be working with polygons a lot in this talk. So essentially, gluing polygons gives you back surfaces. In our case, we will be gluing polygons with directed edges so that you not only recover the surface, but you also recover the graph that's embedded on that surface. And then one of the results that we will be using is the v minus e plus f is two minus two times the genus where v minus e plus f is called the Euler characteristic. Now this comes in super handy for finding the limits lower and upper limits for what the genus of the embedding of certain graphs are. All right. So the one way of representing the embeddings is to draw it, which is really complicated and it's not always possible. So we did another one in which we can talk about these embeddings called the rotation system. So right here I have is q2 embedded on here showing the front and the back side of the sphere. Let's see. So that's one representation. Another way to represent it is by looking at the edges around each vertex. So let's say I look at vertex zero. Going in a counterclockwise direction, I can make a list of all the other vertices it's connected to. So it starts with two, one, followed by four, which means I can encode this by saying that the rotation system around zero is two, one, four. Similarly, if I look at one, it's always counterclockwise. So it's zero, three, five, five, zero, five, zero, three. Doesn't matter as long as it's counterclockwise. So essentially this is a rotation system, which is it tells you the ordering, the counterclockwise ordering of the edges around each vertex. So once you have the drawing or the embedding, you can find the rotation system by just looking at each vertex. So obviously the next thing is if I'm given the rotation system, am I able to recover the embedding? So we can do that and that's called finding the boundary walk. So essentially the rotation system is going to give us the polygons, which when glued together, they give us back the surface in which the graph is embedded. Now here's an example of a rotation system right here. A slight change in notation here. The first entry zero through seven is just a vertex and the remaining three entries in each array is the rotation system. So for each vertex, we have an approach edge and we have a departure edge and the enclosed area we're talking about. So you can start to walk at any edge. Say I start at the edge zero one. Now to find the next entry or to find where I should go next, I'm going to look at the rotation system entry for one and now I'm looking for zero because that's my approach edge. Now because everything is counterclockwise, the next edge or the departure has to be the next entry in the rotation system, which is three, which gives me the next entry. And then I repeat this process. I look at the rotation system at three and then I look at the entry before three, which is one. That's my approach edge and my departure edge is going to be the next the number. Repeat for seven, six, four. So we repeat this until the first two entries repeat because that means we've come back to where we started giving us a polygon. So in this case, it's length six where it starts repeating. Now to find the remaining walk, we know we haven't, for example, the edge zero two is not appearing on the boundary walk. So we can start with zero two and then it ends and we keep continuing this until we have covered all the possible edges. Now the total boundary walk length is always going to be two times the number of edges because four zero would be different from zero four because in one case, the approach edge and the departure edge get interchanged. Okay, so that's how you make the boundary walk. So in this particular case, the rotation system gives us four hexagons. So that can be glued together to get the surface that the graph lives on. Let's see how using the calculation for v minus c plus f gives me zero, which gives me that the Euler characteristic is zero. Hence it should be living in a genus one surface. So let's look at these polygons. Here's one top and bottom of the torus. That's the first polygon with sort of a neck attack and the interior corresponds to that part. Same with the green region, the purple and the gray. So what we have is a bunch of polygons with directed edges and if you follow the instruction for gluing these polygons, preserving the direction, we're going to recover a one-hole torus with q2 on it. And the q2 will be such that it has a rotation system that okay. Now we defined a specific rotation system for our purposes. We call it the ABC rotation system, which stands for alternate bit change rotation system. The process is very simple. So let's say I'm in q3, no, that's right, q4. And I'm on vertex number 10. So I write 10 in binary form, which is 1010. The first entry in the rotation system is obtained by complimenting the rightmost bit. So again, 1011. The next entry is obtained by complimenting the bit to the left of the rightmost and so on. Now because this is q4, the each edge has a degree of, or each vertex has a degree of four, is make sure I have the right rotation system for each vertex. So that's just the alternate bit change rotation system. We can also define something called the reverse alternate bit change rotation system. When it's complimenting from the right, we start complimenting from the leftmost bit. So the ABC rotation system gives a nice embedding all components of size 2n. So that's why the previous example, we got a bunch of hexagons for q3, right? So 2n, six sites. One more thing that we need to know before the main result is something that so the question is let's say I start with this particular rotation system and I pick one of three vertex and I make one switch. How does that change the genius of the embedding or if it does? So basically this theorem that we have here, which I won't be going to the proof, gives us a control over one of three things that can happen. Either there is no change in the number of face, meaning no change in the genius or the three, you have three faces that combine to form one. So the genus, the number of face decreases by two or you have three faces that you start with and it breaks down into, oh sorry, one face and it breaks down into three faces. So basically all we need to know is we have control over how we can make the switches to make polygons. So one example is here, you know, before switching there are three different faces and after switching they combine to form one single face. So similarly we have control over breaking one face into three or having no change at all. So a change in rotation system, one single change increases the genius by one, decreases by one, is it unchanged? It all tells us that if one embedding has an odd number of face then all embedding has odd and if one is even then all has, all of them have even number of faces. That's just a result of the adjustment. The minimal embedding is a graph on the surface of minimal genus. It's also called the genus of graph. So what is known as serum due to lower than franckary is that the genus of an n cube is given by expression. So what we do is we divide an explicit way of finding the embedding itself. So the bound can be found by using the relation v minus e plus f equals 2 minus 2g and substituting values for v e and f for q n. Now one important thing here is for minimal embedding we want there to be as many faces as possible. So theoretical maximum face corresponds to having each face being as small as possible which in the case of a hyper cube would be something of site four. I'm going to skip this. So what we did was we define a four cycle rotation system. You start with any vertex, say vertex zero. You give it the abc rotation system and then vertices which are at an even hamming distance from zero also have the abc rotation system but the rest of the vertices that is odd hamming distance have the reverse abc rotation system and it turns out that that gives us the minimal embedding. Sketch of the proof right here. So we started in vertex with the vertex in binary notation. No matter what walk we start because they differ by one bit the a ith bit is going to be complimented. Now because this is an odd distance away it's going to have the reverse abc which means we start complimenting from the left towards the right. So the next entry is going to have the right bit ai plus one complimented but this is now an even hamming distance from the original vertex so it's going to have the abc which means that we start complimenting towards the left. So ai switches back to its original value and then this has a reverse abc so we switched the right one and we get back to where we started. So this basically shows you that all walks in this particular system are size four which is what we said is required for the embedding to be minimal. So all you have to do is just have abc reverse abc at different vertex and you get minimal embedding. Some examples the first one that we saw was the dimensional in the four-dimensional case it forms a nice lattice which we will be coming back to later. So you got abc reverse abc abc reverse abc the minimal on genus one surface. To keep this picture in mind we'll come back to this when we talk about maximal embedding. Okay maximal embedding is a meeting of a graph on a surface of maximal genus. Every component is only morphic to an open disk meaning you can't just add handles without there being anything there. First of all is to explore different kinds of maximal embedding. For maximal embedding you have to have the least number of faces which is theoretically is two for q and it's two because of just in theorem which said that genus the number of faces changes or changes by two. So if you have even number the least possible x2. If I look at some embedding for q3 this is one where you start with abc and then you switch one vertex and you get two components of size 18 and 6. Then you start with abc on all of them you switch from 0 to 1 to get another vertex with 14 and 10 sizes. Then you make a switch at 0, 1 and 3 and you end up with something which has equal so it's two components of size 12 each. And then you switch at 0 and 5 is this nice little embedding where one face is as small as possible and the other is as large as possible. Now this is the one that we're focused on and we call this a big face maximal embedding. One face is as small as possible the other is the largest possible. So it seemed like a natural maximal embedding to go after. In fact if we start with a minimal you can make switches and using the just in change theorem which says if you make the appropriate changes you can combine these three faces into one and you keep doing that and you can end up with the maximal embedding where you have one face of size 4 and the remaining and this can be obtained after seven switches. And this picture that I told you to think of before we start with a minimal embedding and it just amounts to add seven handles to create the maximal embedding which is neat I think. So we have what's known is Mark Youngerman's work that any four connected graph can always be maximally embedded thus Q1 can be maximally embedded. It uses these graphs and does not demonstrate a maximal embedding. So what we have is a recursive for the kind of maximal embedding particularly the big face embedding. For that we need a lemma so I'm going to skip through this so basically what the lemma says is you start with an embedding and you add a pair of edges you can do it in such a way everything can be controlled so that you don't change the genus off the surface. By making sure the rotation system is changing a particular way you can make sure the genus doesn't change. I am going to skip over the sketch of proof in time. Okay and the theorem says that given Qn I can have two boundary walks one size four and the other remaining made up of the remaining edges. It's true for two proof is by induction so essentially we start with it's true for Q minus one we take two copies of Q minus one and then we have the small face which we connect with a tube. Outside this tube we can use one of the results from before about switching. Now I take this vertex and I can switch in such a manner that's one bottom face and the back face combined to form one single face then I can switch along this other vertex to make the result of the previous operation g front face and s not combined to form g not and what I'm left with is this green the four sided thing which is actually the small face for Qn. Now this is not Qn yet because we haven't connected remember these are copies of Qn minus one so we haven't connected all the edges but we have the lemma. The lemma says that you can add missing edges in pairs without affecting the genus which means that you can recursively construct a maximal big face embedding for Qn. Okay that's the end so we define the rotation system basically construction of minimal embedding maximal embeddings. I wanted to make so I could find the sequence of change for Q4 to go from minimal to maximal. Open question is like how do you do that for every Qn? Is there a certain third of doing that? One thing that I didn't talk about is there is a particular embedding that I found which consisted of two islands which combined gave you the Q4. I could define it for Q4 and yeah what's the minimum number of changes to go from minimal to maximal? What are the possible combination of face sizes that can be obtained for a particular graph and an algorithm to see what the actual embedding looks like because drawing it is a real pain and that's it.