The "textbook exercise" on Euler characteristic | Euler characteristic #1

if you draw a graph on a plane where all vertices are connected to each other and the edges don't cross then there is a relation between its vertices edges and faces the number of vertices V minus the number of edges e plus the number of faces F has to equal to two no matter what graph it is this is known as the oiler characteristic formula however the right hand side really depends on the surface you draw the graph on it's to on a plane and if you draw the graph on a Taurus it can still be two but it's not always two for example in this case which you can pause and check V minus E + f is zero it turns out that for the Taurus there are only three cases it could be zero like here or it could be one like this case which again you can pause and check or it could be two in general V minus E + f is between 2 - 2 G and 2 where G is known as the genus or very roughly the number of holes on the surface for example the genus of this double Taurus is two because there are two holes in it the Taurus has G equal 1 because there is just one hole so for the Taurus V minus E + f is between 0 and 2 and therefore it can only be 0 1 or two this theorem might be different from what you have seen because while this is true for all graphs without self intersections most literature deals with a specific kind of graph known as triangulation for those kinds of graphs V minus E + f takes the smallest part possible value 2 - 2 G in the rare occasion that the general case is mentioned the proof is basically left as exercise because it's supposed to be just an exercise I thought this was easy and attempted it this video is about my thought process When approaching this rather than riger I will focus more on the inition of where the result comes from I'll let you be the judge of whether this should just must be an exercise let's start with the plane because I didn't know how to prove V minus E + fals 2 in this case the usual way is to start with just one vertex the base case in this case there is just one vertex so V is one there are no edges so e is zero the whole plane is a connected region itself so f is one so in the base case V - e + f = 2 and if we subsequently add more edges to it that is the inductive step there are two possibilities the first possibility is we add an edge and a vertex in which case v and E both increase by one so V minus E + f remains unchanged the second possibility is that we add an edge to close off a region let's say we join the bottom blue and purple vertices then we form a cycle to close off a region in addition to the original phase we now have two phases so F increases by one but of course e also increases by one because we have added an edge but V remains unchanged so V minus E + f is still two so whether you connect to a new Vertex or form a cycle every time you add an edge V minus E plus F remains unchanged does this kind of argument extend to the Taurus well the base case is still exactly the same there is only one vertex so V equal 1 there are no edges yet so e is zero and the whole Taurus is not separated by edges yet so there is one continuous region and F is one so in the base case for the Taurus V minus E + f is still two if we subsequently add more edges to it one possibility is to connect to a new vertex in which case both V and E increase by one and so V minus E + f remains unchanged at two this possibility is similar to the case for the plane

the other possibility is that the new Edge doesn't connect to a new vertex and forms a cycle instead for example in this case when we connect the yellow and pink vertices this way e goes up by one but because we didn't add any new vertices V remains unchanged however this cycle doesn't help us create a phase there is still just one phase because you can go to The Other Side by going under so with f and V unchanged but e increasing by one the value of V minus e+ F actually decreases by one and is no longer two I've deliberately chosen this cycle because it's a non-c contractable loop that is you can't contract this Loop down to a point continuously on the Taurus no matter what you do you can't eliminate that hole this is a key feature that separates a Taurus from a plane but that's not the full reason why we decrease V minus E + f when we complete the loop let's say we have this graph already and want to close the lower loop but before that let's check the v e and f there are six vertices Al together so V is six for reges there are three on top that form the upper Loop two edges on the incomplete lower loop and one Edge that connects the two so altogether we have e equal 6 finally there is still just one continuous region even if you are apparently stuck here you can still go around the Taurus and get out so it is still one big continuous region altogether V - e + f is 1 if we now complete the lower loop then e will increase by one because we have added an edge but actually F will also increase by one this is because closing the lower loop genuinely creates a new region if you are stuck inside the strip you can't go outside without passing through the edges so with both E and F increasing by One V minus E + f remains unchanged so even if you have an extra non-c constructible Loop you can't decrease the value of vus e + f by more than 1 but maybe that's because the two Loops are kind of the same what if the two non-contract Loops are genuinely different so one of these Loops can't be deformed to another a more technical jargon for this is the two Loops are not homo topic let's count v e and f before closing the second Loop there are altogether five veres so V is five for the edges there are three on top that form a complete Loop and there are two that form an incomplete Loop so altogether e is also five the whole Taurus is still one big continuous region so f is one which means that V minus e + f is equal to 1 now let's complete the loop by completing the loop e has increased by One V hasn't changed because we haven't added any new vertices but f is actually still one so the value of V minus E + f has decreased by one the reason why f is still one is that if you want to go from the blue point to the orange point you can go under the Taurus to get to this purple Point first and then go around the Taurus to get to the Orange Point throughout your journey you haven't crossed any of the edges so this whole thing is still one big continuous region by considering different non-contract Loops we can make V minus E + f decrease by two from 2 to zero can it decrease even further well we know the answer should be no because in the case of a Taurus this theorem says it can't be below zero but what's stopping us from closing a non-contractible loop again the best way to think about this is that this phas after separated by edges is topologically equivalent to a plane these two sides separated by the loop can't connect with each other unless we go the long way around the Torus so

topologically it's the same as cutting along the loop and unwrapping the whole thing as a cylinder but then by the exact same logic these two sides are still separated by an edge so again topologically it is the same as cutting along this line and unwrapping it to become just a normal plane because we previously established that on the plane V minus E e+ F remains unchanged when we add edges when we add any further edges on the Taurus on the left V minus E + f also will not decrease any further this particular configuration has vus e plus F to be zero because this quantity doesn't decrease further when we add more edges it will never dip below zero of course we haven't covered all the possible cases but now at least we have the very rough intuition that V minus E + f decreases when we complete a different non-c constructible Loop using a similar intuition if we introduce more holes for the Taurus then roughly speaking each hole can contribute two of these non-contractible Loops so in general V - e + f is bounded below by 2 - 2 G Technically when we have more holes Loops like this in the middle are still non-c contractable but they won't decrease the value of V minus E + f because closing this Loop creates a new phase these two regions are now actually separated by this Loop so 2 - 2G is really the lower bound but is my idea for this supposed textbook exercise even remotely valid I Googled a bit harder and found this stack exchange answer basically it states that there are three cases when we add an edge we have already Illustrated each of them separately but this answer phrases them a bit more precisely in a different way so my thought process was actually not bad I put the link to the answer in the description while we did prove this more General inequality which will be useful in the next video there are plenty of reasons why most people only considered the special case this lower bound known as Oiler characteristic usually denoted as Kai is useful in distinguishing surfaces with different gener yes the plural of genus is gener not only is it useful in differentiating between different surfaces oil characteristic appears in places seemingly unrelated to graphs from ve fields to something called moris functions you can go to the description to see what those are but I want to highlight one other place where Oiler characteristic comes up these B notot B1 and B2 are called Betty numbers defined using homology an important Concept in algebraic topology these Betty numbers generalize to higher Dimensions so you can use this alternate ating sum of Betty numbers to Define oiler characteristic for higher dimensional spaces not just Services if you want to know more about homology which Betty numbers are defined with check out atus new video on his channel Alf zero as always thanks for the patrons like And subscribe I'll see you soon bye-bye