What is up beautiful people of YouTube? Welcome back to the Johnny Q channel. I'm Johnny Q Today's video is all about the perfect vlogging set up the perfect YouTube camera for Vlogging YouTube and just everyday life this video will be for those who are just getting into YouTube Getting into filmmaking or photography and you need you know a simple setup that won't break the bank But it's not gonna be super cheap that you're not satisfied with it So this video is for those people who just need that kind of basic vlogging setup And for all the people that know what they're doing all the veterans I would still recommend as like a side or second camera the 60 mark 2 is all you need Okay, guys, let's just get right into it. Let's not waste any time Let's talk about the perfect vlogging setup and the perfect YouTube camera. I'm just gonna come out and say enough said it before The Canon 60 mark 2 is the perfect the absolute most perfect vlogging and YouTube and everyday camera set a brief interruption It's all good little buddy just wants to play Xbox Where was I the Canon 60 mark 2 is in my opinion the best camera to have for video for starter for people who are starting to Vlog to film anything and just photography in general this flip-out screen is Everything I don't have to explain it. You know what it does and see what it does Yeah, it's a flip screen. What else do you want from a camera? I mean, I don't I don't understand why all these cameras are coming out and they're so phenomenal But they just don't have a flip screen every camera in 2019 should have like it should be required By the FDA by President Trump It should be required that every camera that comes to market should have a flip screen It is so beneficial for youtubers videographers photographers who might not need a flip screen But let me tell you the day will come when you need to get that one Angle from above for that wedding the day will come when you're you know on the ground and you don't want to get dirty Because the day will come trust me. Why do I love this camera? Let's see the next best thing Oh, I upgraded from the ADD now the ADD after I want to say after 15 minutes or 20 minutes I have to Google this It would actually video file would split into two files So if I went over a certain amount of time on my SD card It would say it would have like a display saying you've reached the limits Your video file has been split up into two files the 60 mark to gives you one file So you could be filming for like half an hour. It'll be one file for half an hour Another big reason why I love this Canon 60 mark to look at that That is a full frame sensor It is a full frame sensor. I just I love it. I had a Canon ADD It was a crop sensor at 1.6. I think this is full frame. You don't need a crop sensor Why else do I love this ISO? It shoots well in low light both video and photography You can get too dark, but it does shoot very well It's got 26.4 megapixels a phenomenal camera for both photos and Video it's this is very minute, but on the ADD the dials weren't as user-friendly These dials are it was a lot better feels a lot sturdier if he was a lot just When you spin it, you know, you're spinning it You know that you're getting quality spins one of the best things about this camera is that it's super light now The Canon ADD was light, but this camera is very light almost as light as a Canon ADD Now I don't have my tripod with me But I have a tripod adapter here and I could hold this out with one hand and not be tired Like I can I can do this all all day. I don't have my microphone on here But I'm using my microphone and it's the road road mic just a basic road mic But I could be doing this vlogging off all day switch it over and be fine Like it's that light paired with a red lens to the perfect vlogging lens And I'll do a separate video on that but quickly the Canon 16 to 35 2.8 version 3 the best lens for vlogging There is both great and phenomenal imagery and both great and phenomenal video It's it's a workhorse like this thing will literally you I've run it through its paces in low light It has autofocusing manual focus no image stabilization in both The body and lens but this is a great lens and camera a combo for vlogging for YouTube creation For any kind of film really like this camera shoots very well on 24 frames a second at 1080p 60 frames at 1080p 30 at 1080p and that's about it if you are a beginner just Getting into the creative of photography or video the Canon 60 Mart 2 is probably the best you can find and it's priced at 1300 bucks just for the body. I mean that's pretty good deal used. I'm sure you could find Lowest I've seen like 700 bucks. If not 800 bucks is the lowest So if you're looking for a vlogging camera if you're looking for a YouTube camera If you're looking for an overall just a really good vlogging setup I mean the Canon 60 Mart 2 is your best bet Also, if you can spend $2,000 get the 16 to 35 2.8 third version. So yeah guys, that's pretty much it I just wanted to make you know a video encapsulating the you know the best beginner or best overall You know kind of my vlog set up and what I use because I'm getting a lot of comments and emails asking dude What do you use for camera? What kind of lens do you use? So if you wanted to know this is what I'm using for everything right now for both photo and film the 60 Mart 2 with the Canon 16 to 35 2.8 version 3 and Thank you so much for watching. I'll catch you next time on the Johnny Q channel. Peace


So, let me just recall in five minutes what I said in the two previous lectures. So, I want to be considering discrete subgroups in P, S, O, N plus one, C, and discrete. Then, we define the notion of limit set, which we call the Kulkarni limit set, which when N equals one coincides with the usual definition, is just the set of accumulation points of the orbits. In the general case, you have to add to that set something else so that to ensure that the action on the complement is properly continuous. So, in that case, you take the set of accumulation points of all orbits, then you take the you look at the complement of that set and you look at compact sets in that complement and you look at the points where compact sets accumulate. And the union of all of that is the Kulkarni limit set that we've been looking at. And so, the first day, we spoke basically about the classical case, N equals one. Yesterday, I gave families of examples mostly in dimension three, and yet they extend to higher dimensions, which in many cases, this was a very, very good concept, but I mentioned at the end, I gave examples of this, what we call a type of short key groups in CPN, in CPN for N, OTH, for which this is not always a minimal, I mean, there are examples in which this contains a proper closed invariance of set where such that the action in the complement is properly discontinuous. Okay, so in higher dimensions, it's clear that to have a good understanding of this situation, we need to do something else. We have to look at more refined things, and the Adolfo Guillot and Maya Ramendes, they have good things in that direction. I think I'm sure for next meeting, we'll be able to say something. Okay, I mean, there are already good results in that sense. So far, in higher dimensions, it's kind of an experimental science. Okay, so we're coming in. So now I want to focus on the case N equals two. Yes, so I want to look at automorphisms, holomorphic automorphisms of CP two. So that would be my focus today. And my claim is that in that dimension, this is the good concept of limit set. Let me start with some examples. The first one we already saw it, consider a metric lambda one, lambda two, lambda three, zero, zero, with lambda one greater than lambda two, greater than lambda three, greater than zero. We have three one-dimensional eigenspaces in C3 where we pass to the quotient. We have three fixed points that we denote E1, E2, E3. These are the fixed points. And this is an exercise to show what one is attractive. The other is a subtle point. The other one is repelling. And it's an exercise to show that this is the set of fixed points, of accumulation points. These are the fixed points and these are the accumulation points of orbits of points in CP two. And then in that case, we have three invariant lines, E1, E2, E3. And the Gulkarni limit set is the union of these two, which one of these is attracting and the other one is repelling. And the one which is attractive is repelling for the inverse. So we call this the repelling invariant lines of this, which is a loxodromic element. I say that in dimension three we call an element loxodromic if it has a lifting, which is diagonalizable or not, with at least one eigenvalue of non-equal to one. So this is a loxodromic element and this loxodromic element has two invariant lines in its limit set. One is repelling, the other is attracting. Another example. Consider now lambda, lambda, lambda to the minus two, zero, zero, lambda. Okay, so now we have two distinct eigenvalues, different to one. Okay, now we have two distinct eigenvalues and we have one invariant line corresponding to this block and one fixed point corresponding to this block. In this case the Gulkarni limit set will consist of one line, e2, e1 and one point. Okay, so I say that for loxodromic elements the limit set was going to be the union of two projective sub-spaces of Cp2. This is an example here. We have two lines, here we have one line and one point. Another example. Consider one, one, zero, zero, one, one, zero, zero, one. Okay, in this case one can check that one is non-diagonalizable and one is the only eigenvalue. We have one fixed point. E1 is a fixed point and we have one invariant line but it's not an eigenspace. Okay, so in this case the limit set is just the line e1, e2. Okay, so here we have examples where the limit set is one line, one line and one point, two lines. Now, for example, now take one of these groups, take this one, I mean the group generated by this matrix and now take the group, for example, lambda one, lambda two, lambda three as before and add one generator, zero, zero, one, zero, one, zero, zero, sorry, one, zero, zero. This matrix permutes the invariant lines so when you take the two generators, now you get a group with exactly three lines as limit set in a position. Each two of them meet at one point and no three of them meet at the same point. Okay? So in this case we have a limit set with three lines. What? Exactly, exactly. Yeah, but now e1 is here and the other one is here. I mean the bigger the value, the bigger the values. I mean, this is, you need carefully, you get the three lines as limit set. Now, okay. Maybe with this example. Now, okay, now let me give you two more families of examples that have come in previous lectures. So now consider gamma, Fuxian group, sorry, in isometries, orientation preserving isometries of H2R and such that its limit set is the whole circle. So a lattice in the group of isometries, either co-compact or co-finet. And think of it as a subgroup u1, 1. So we are actually thinking of this as a complex hyperbolic one-dimensional space. Okay? Then you have a natural embedding into pu21, as John explained us yesterday. And this is naturally a subgroup of p, s, l, 3, c. Okay? So you have this group acting on Cp2 in a natural way. So you have Cp2 inside, you have a round ball, which serves as model for hyperbolic two-dimensional complex space. And then the group, since it is as a group here, acts on Cp2 preserving this ball. Okay? But because of how it is constructed, it is also preserving a complex line. So we have a copy of Cp1, whose intersection with this ball is a copy of the complex one-dimensional hyperbolic space. Okay? Now, in this case, we have the, say, chain Greenberg limit set, which is going to be exactly. Because of this, the limit set is going to be this circle. The intersection of this line with the boundary sphere. Okay? So I told you that when you have a subgroup of p, u2, 1, the Kulkarni limit set is the union of all projective lines tangent to the three sphere at points in the chain Greenberg limit set. So to get this guy, to get lambda koulm, we take, we look at each point in this circle, and we look at the unique projective line tangent to S3 at the given point. Okay? So we have one projective line for each point in this circle. And in this case, it's very easy to describe, which are those projective lines. L is a line in Cp1. We have a projective duality. We have some point outside, the polar point of the line. And then for each point in this circle, we have the unique projective line joining these two. Okay? So we have a family of projective lines, parametrized by this circle, and all passing through the given point. Okay? So in this case, we have infinitely many lines in the limit set, but all of them meet at one point. So we can say that there are exactly two in general position. A set of lines in Cp2 is in general position, and not three of them meet at a point. Okay? So we have infinitely many two in general position. Let me say that in that case, it's also, you have a very nice clean description of the complement, the region of this, the Gulkarni region of this continuity, which is, which one? Take one point in the complement. Okay? At that point, this point together with this, determine a unique projective line. Okay? So you look at the projective line, given by these two. The drawing is not very good, because the line is going out of this space, but this is not. Okay? Now, that projective line meets this projective line in exactly one point. Two lines in Cp2 meet at one point. Okay? And that point has to be away from this circle. So you get a projection from the complement of the limit set into this projective line minus the circle. You can easily check that that's actually a holomorphic bundle with fiber C and base S2 minus a circle. So topologically, you have two copies of R4. Okay? So in this case, the Gulkarni region of this continuity consists of two copies of R4, or two open four balls. Now let me consider the same setting up to here, but now gamma... Now consider gamma in S021. Okay? And this naturally lives in Su21, which can be projected to Pu21. Okay? So we can look at this group here, but also John explained it to us just a bit. And actually what I will say now is part of a joint work with John Parker and Angel Cano, which should appear soon in the Asian Journal of Mathematics. Okay? So just look at this setting. We have now this Foxian group, co-finate or co-compact Foxian group, regarded as a group of automorphisms in CP2. Again, it's preserving... Well, this is in PSL3C. So again, it's acting on CP2, preserving the complex hyperbolic space, which is a ball. So we have more or less the same picture. Okay? So this hyperbolic two-space, complex hyperbolic two-space is CP2. But now, because of how the group is lying here, it is preserving the points in CP2, which can be represented by real coordinates. Okay? That's a Lagrangian two-dimensional plane in CP2. Okay? So we have a similar drawing, but now this is a copy... which is a copy of R3, sorry, of Rp2. It's a real projective space, which is being preserved by the action. So the action is preserving a ball and a real projective space. The intersection now is, again, is a two-dimensional disc, which serves as model for real hyperbolic two-space. Now it is a different model. Now it is the Beltrani model. It's a copy of R2. Okay? Now again, the limit set, the Schrodinger limit set is going to be, again, a circle. Actually, now it's a circle, but to be correct, it's a copy of Rp1, which is homeomorphic to this one. Okay? We have this disc. If we look at this projective space, R minus the disc, this is a Mobius band. Okay? So the complement in this projective-reprojective space of this disc is a Mobius band. Now, again, the Kulkarni limit set is going to be the union of all lines, projective lines tangent to S3 at points in this circle. So far it's just exactly as before. But now this set looks very different, very, very different. Now, one can show that if you take, if you consider two such lines, any two projective lines in CP2 meet at exactly one point. Okay? Well, that point is a point in this Mobius band. Okay? So any two lines here meet in exactly one point in this Mobius band. And through each point in this Mobius band, there pass exactly two such lines. Okay? So what you have now is a, this set consists of infinitely many lines parametrized by the circle. But now all of them are in general position. Two of them meet at one point and not three of them meet at any point. Okay? The meeting points of every pair of lines is parametrized by this Mobius band. Okay? And then in this case it's not totally trivial, but one can show that the complement now consists of three copies of the four-dimensional disc. One can find explicit projections and one can control the topology of the fibers and everything. It's very beautiful, I think. Okay? So in this case the complement has three components. In that case it has two components. Now we have infinitely many lines in general position. Here we have infinitely many lines, but only two in general position. Okay? So I remind you that in the classical dimension, complex dimension one, the limit set consists of one point, two points or infinitely many points. The equivalent statement now in dimension two is the following theorem, which is essentially due to the limit set. I mean, not everything is in one paper and some statements are due to one, others are due to the other, others are due to the three of them, so I just put everything together. The set, lambda cool, always contains at least projective. This statement is actually true in all dimensions, a great theoretical tool. That's something that Karn and myself proved. Two can consist one, two, or it can have or contain one, two, three, infinitely many lines. So if it has more than three lines, it has infinitely many lines. Three, it can have one, two, three, four, or infinitely many lines in general position. So I gave you an example where you have infinitely many lines, but only two in general position. You can construct examples in which you can have exactly three lines in general position, infinitely many lines, but only three in general position, examples in which you have exactly four lines in general position, infinitely many lines in general. Four, if it has isolated points, then there is exactly one such point. So there's an isolated point, there's at most one, and if that happens, the limit set consists of one point and one line, and the group is virtually cyclic. So we have this theorem, and here I'd like to make a remark what about the discontinuity region, the number of components in this discontinuity region? In dimension, in complex dimension one, the region can have zero, one, two, four, infinitely many connected components. So it can be empty, or it can be connected to components or infinitely many components. In dimension two, zero, one, three, four, let me put it this way. We have examples, there exist examples with zero, one, two, three, four, or infinitely many components. But we don't know if that's a theorem or we just, there are more possibilities, okay? So are you already getting excited? It's easy to construct examples with zero, one, two, and three components. This is not so easy, but this also is known as something that these guys did. And yes, please. You are seeing if that can happen. Here, in this example, yeah, in this example, the region of this continuity is a bundle over C, let me say, over a disk. So topologically, that is just a product. And then when you take the quotient, what you get is a, it's a disk bundle over the Riemann surface that you have a quotient of the disk divided by the action of the Fuchsian group. Yes, the same in the other case. In that case, you have three components. Each has this type. Of course, the bundles are not the same. No, no, but the dual action is also on CP2, okay? So it's not the classical limit set. It's again, you're asking, are they on CP2? Yeah. Oh, yeah, the dualist point. They are very interesting relations. And for many of these results, they have done, you have to look at the dual space to get information. They are not exactly the same. It's different things, but they give information in each of the other one. In fact, in the last year I'm going to state, I will say just something about that. So may I now recall that, well, in dimension two, in dimension one, we have that the limit set is many things. It has many nice properties. It's the largest set where the action is properly discontinuous. It is also the, sorry, the complement. CP1 minus the limit set, sorry. It is a region of continuity. And lambda is the closure of set of fixed, of loxodromic elements. And one more property. Let me assume, sorry. G non-elementary. So this has cardinality greater than two. And one more point. And lambda is minimal. All orbits are dense. And it has many properties, but let me state this. We have a theorem, again, Navarrete and myself. Lambda Q has at least three lines in general position. Let me put this in memory one minute. Let me just recall this example I gave. Lambda one, lambda two, lambda three, zero, zero, with the distinct norms. You know, it has, lambda Q has two lines in general position. This lambda one, E1, E2, E3. So the limit set consists of these two lines. And in this case, one has that there is, there is not a largest set where action is properly discontinuous. There are rather two maximal such sets. Because if you take this line, union, this point, the action on the complement is properly discontinuous. And if you take this line, union, this point, also, those two are maximal regions, but none of them is the largest. So there is not one largest region where the action is properly discontinuous. And in that case, the region of the continuity coincides with the Kulkarni region of discontinuity. And it's Cp2 minus the two lines. So there are two maximal regions. No, there is not one largest. And then coincides with the continuity region. And the theorem is going to say that this is somehow the only possibility. I mean, you can have infinitely many lines, but the only possibility is that you have only two lines in general position. It's the only possibility to have troubles. So the theorem says that if you have at least three lines in your position, then everything is fine. Then one, there exists a largest, the action is continuous. And this is the complement coincides with the continuity region. This was two, three. Lambda Q is the closure repulsive invariant lines, loxodromic. And to mimic this, I would like to state this, but this is... We need one more condition. And even so, this statement is quicker, but I think it's nice, let me say it. If G, the group, is called strongly irreducible, which means that there exists no invariant... No projective sub-spaces, finite orbit. Irreducible, yeah. So there are no periodic points and there are no lines that after a finite number of times return to be line, okay? Now, a comment which is not obvious is that this is not an additional hypothesis. This is a stronger hypothesis. If you have this, you automatically have this condition, okay? If you have that, then the group, minimally, the dual of the limit set, okay? The action on the lines can be non-minimal. But if you consider the lines, then the orbit of each line is dense in the limit set, okay? Now, this is the proof who has... Oh, yes, let me go by stages. This and this was already proved by these three guys. That's already published. And the idea is first to prove this. And this is, in fact, more or less trivial using... Well, this is not more. This is an obvious consequence of the Cartan-Montel theorem in this dimension, which, you know, Montel's theorem tells you that if you have a family in CP1 minus three points, the family is normal. In higher dimensions, you have a theorem in CP2 that if you have a limit set, if you have a region, a family on a region which omits at least five lines in general position, then it is normal, okay? And that was improved up to having three lines. If you have three lines in general position, if you avoid three lines in general position, you have a normal family, okay? Then you get this for free, okay? Now, then you prove that the action here is properly discontinuous. And you also prove that the region is maximal. You cannot get a continuity going into the limit set, okay? So putting all of that together, you get the first two statements. Well, it depends on the hypothesis you need. The nice thing, I mean, the big difference is that here we are removing projective sub-spaces. That's a very nice picture. He has a more complicated life in this sense. In this sense, I mean, it's much harder to control, okay? Okay, now, the hardest part or the new part is this. To express it as the closure of invariant lines, repulsive invariant lines of loxodromic elements. For that, the proof is very long and somehow technical, and it has several stages. First step is obvious is just to notice that all these groups have finite index of groups with no elliptic elements. And elliptic elements make no difference for the limit set, so you can forget about elliptic elements. Now, second step, look at parabolic. Look at purely parabolic. So elements, whose elements are all parabolic elements. And then after some work, you can show that limit set consists either one line, or let's say a cone of projective lines parametrized by S1, one common. So it is the limit set is as in one of the examples I gave you before, in which you have one circle, one point away, and for each point in the circle, you have one projective line. So you have infinitely many lines, but only two of them in general position, okay?  Then you must have lots of dramatic elements. Then there are three cases where the number of lines, let me denote it by this in general position, is three, four, or greater than five. Okay, I told you that the number of lines in the limit set in general position can be three, or can be one, two, three, four, or infinitely many. Okay? So in this case, these two cases are special in the sense that they have to be done independently by hand. And in both cases one has to do the classification of such groups. So the classification of four groups with exactly four lines in general position was done by these three guys, and it's already published. The case with three lines is new. And then using those classifications, you go case by case, and you see that the statements are true. And then the last case, the most interesting somehow, under the idea, well, this is the obvious one, to show that there are so many loxodromic lines that if you have a point in the limit set and a neighborhood, there's always an loxodromic line passing through that neighborhood. So you can approach every point in the limit set by a possible invariant line of an loxodromic element. But this you have to do quite a lot of work. And I think I should stop here. I have a few minutes, but...