OK, so maybe I will stop. So I'm really glad to see that so many of you decided to come despite it's really late, really late in the day. I hope you will find it's worthwhile. So earlier today, I discussed the space. In particular, we obtained some understanding of this homogeneous space. We saw that this space can be viewed as the unit tangent bundle of a certain hyperbolic surface, except you have to be careful at the elliptic points maybe. But more or less, it's the unit tangent bundle of a certain hyperbolic surface of genus zero. Today, or now, I will discuss the space where we put the general d instead of 2 here. And we will have a different way to visualize this space. Namely, we will see that this space with d instead can be viewed as the space of all Euclidean lattices in dimension d, having covolume 1. And then I will discuss the geometry of this space a bit. OK, so I will have the following notation. We let G be equal to SLDR. So it's larger dimension now. This is a d square minus 1 dimensional, d group. And let gamma be the following discrete subgroup, SLDZ, the subgroup of all integer matrices in here. It is actually a lattice, but that is something which I won't exactly prove it. But we will see parts of the proof of the fact that gamma is a lattice in G. It's not obvious. But it is discrete, and that is obvious. And also, let X be the homogenous space here. So G mod gamma equals SLDZ slash SLDR. OK. So the first thing is I want to make precise this claim that this space can be identified with a family of all Euclidean lattices in dimension d. So what is a lattice in Rd? The following is you can take it as a definition, or you can take it as a fact. So a lattice in Rd. And here Rd is also a lead group with addition as the group operation. And we have a hard measure. That's Lebesgue measure. And because I have a lead group here, I have already defined what the lattice is. It's a discrete subgroup such that the quotient manifold has finite volume with respect to Lebesgue. So in that sense, the following is a fact. So a lattice, or Euclidean lattice, we call all these lattices Euclidean lattices. That is a set of the form. So it turns out that all lattices will have the following form. It's just the set of all integer linear combinations of some given basis of Rd. So I write it like this. All integer linear combinations, just to write out explicitly what this notation means, it's the set of all linear combinations like this, where V1 up to Vd is some basis, so R linear basis of Rd. And this is very easy to prove. I won't prove it at least. So any discrete subgroup of this lead group would be of this form except I would have some set of linearly independent vectors, not necessarily a spanning set. So it might be a shorter sum. But then it's easy to see that the co-volume is finite if and only if it's really a spanning set of vectors. So then it's a basis. And we can easily write out the fundamental domain now also for Rd modulo L. So I write it out explicitly. Then the following f, we can take the set consisting of all linear combinations of the following form, where now x1 up to xd are arbitrary numbers in the interval 0 to 1. This is a fundamental domain, or we can call it fundamental cell. It's really a parallelogram. It's a fundamental cell or domain or parallelogram of this quotient space, Rd modulo L. Or if I should use this type of notation mod out by the lattice from the left, I would write Rd mod L like this. These are the same because Rd is commutative, so there's no problem. OK. And the co-volume of this lattice. So in other words, the volume, no, the co-volume, well, is by definition the volume with respect to our measure. And I've fixed the far measure to be the big measure. It's the volume of this quotient. That's the volume of any fundamental domain. So it's the volume in particular of this fundamental domain. And it's easy to see or it's known that this volume is equal to the determinant of the, OK. So it's equal to the volume of f. And it's also equal to the determinant of the matrix I get by taking these as rho vectors, v1, v2, up to vd. This is a d by d matrix. So all the vectors. I will view them as rho vectors in this talk. I take this determinant and I take the absolute value. So that is the co-volume of L, just to set notation. OK, and maybe a picture. If d is equal to 2, then a lattice might look as follows. It could be a set like this, something. Maybe the drawing isn't very good. Then for example, here, v1, 1 might be this vector. v2 might be this vector. And then the fundamental cell would be this parallelogram. Sorry, maybe the picture is too small, but OK. And the first result I wanted to state is that we have a bijection from this homogenous space, x, onto the set of all Euclidean lattices of co-volume 1. So consider the following map, going from the homogenous space to the set of lattices in Rd. Let me call it j if I need it. It maps any point or any coset in x to the lattice zdg. And this should be understood as what I get by taking all integer vectors. And to each such integer vector, I multiply it by g. So by definition, this is this set. I take mg, where m runs through all the integer vectors in Rd. And remember, a vector is always a row vector. So this multiplication makes sense and gives a new vector in Rd. So in other words, this is the standard integer lattice, but then to which I have applied the map g, just moving it in some way, point wise. So the claim is that this map is a bijection onto the set of lattices of co-volume 1. So it's not onto this set, but onto the set of lattices of co-volume 1. Often we will identify x. Here is x. We will identify x with this set of lattices, at least in our mind. It's very useful to do this identification often. OK, so I won't write much of the proof, but it's a really good exercise to do this, and we will discuss it in the tutorial. But just maybe I will talk, say some words about the proof. So first, let's see what this map is. So if I have some g, and I'm looking at the coset gamma g, OK, what lattice do I get? Well, we can name the entries of g by naming the row vectors. So let v1 be the top row vector of g, let v2 be the next row vector, and so on. And if I do this, then clearly, the corresponding lattice, or the image under my map, sorry, is by definition zdg. And if you look at what this is, you carry this out, you get the set of linear combinations, of integer linear combinations of v1 and so on up to vvd. So now it is clear that we get the lattice of co-volume 1, because the co-volume, according to what I said there, is the determinant of these row vectors, and that's just the determinant of g. So that's one fact. And then you have to check that this map is well defined. So here, if I have one coset, just one point in x, that such a point can be represented by many different g elements. But to go from one to the other, you multiply by some element of gamma, that would mean that we here insert a matrix gamma in gamma. And the point is that multiplying zd by gamma from the right gives back zd. So that's the reason why this is a well-defined map. And then what more? The fact that it's subjective is really clear from the definition, if you look there. And then you have to check injectivity, and that's more or less checking that modding out by gamma is exactly what you have to do in order to keep track of the non-uniqueness in choosing a basis of a lattice. So that's a good exercise to do. OK, so let me stress again that if d is equal to 2, we now have two really strong and useful pictures of this homogenous space x. So if d is equal to 2, this is the unit tangent bundle of the modular hyperbolic surface. And it is also the space of covalent one lattices in the plane. And this will be really useful in later lectures. OK, so I want to discuss the main topic of the talk is to discuss something about the geometry of this space x. It's a non-computative, non-compact space. And what does it look like at infinity and so on? But first, OK, I want to do some more basic things. So first, I want to discuss an extension, a slight extension of this. So also for applications, it's really useful to have the space of affine Euclidean lattices. So affine lattices in RD. So I define an affine lattice to be a translate of a lattice. So an affine lattice in RD is a set L prime in RD of the form L prime equals W plus L, meaning point-wise addition, where W is some vector in RD and L is a lattice in RD. And we define the covolume. Maybe it's not really proper to call it covolume of L prime, but we define it to be the same as the covolume of L. And then the space of affine lattices of covolume one can be identified with another homogenous space. So I just wanted to state this, because we will need it in later lectures. So for that, I have to introduce a new Lie group. I will call it G prime here. It's the affine special linear group of order D. And this will be the group of all affine linear maps on RD, having determinant one. But I define it in a concrete way as the semi-direct product of SLDR and RD. And semi-direct product of groups means, as I said, it's just the Cartesian product. But then I have to tell you what the group law is. So the group law is that having G1, W1, and then having another element, G2, W2, in this Cartesian product, the product is defined to be G1, G2, W1 multiplied by G2 plus W2. And this group G prime acts on RD by affine linear maps, just like G also acts on RD by linear maps. And the action is from the right. I'm multiplying vectors, row vectors, from the right by my matrices. So the action is as follows, that if I take a vector V in RD, then I want to act on it by an element G comma W in G prime. This is defined as V multiplied by G plus W. So it's an affine linear map. OK. And so I just want to state the analog of this. Maybe I won't even state it. I will just say it in words. But let's also introduce gamma prime to be the group of integer elements in ASLDR. So gamma prime is ASLDZ. And yeah, to save some time, maybe I just state it in words. So theorem one prime, it says that this homogenous space, x prime equals G prime modulo gamma prime, this can be identified with a set of all affine lattices with covalent one. So I call it, let's say, equal the set of affine lattices covalent one of covalent one. And the map is the same as here, ZDG, except now G is an element in ASLDR. OK. So we have that out of the way. And OK. And then before I come to the geometry of x, I also have to introduce some good parametrization of G. And what I will introduce is the evasava decomposition of G. You have already seen this for SLDR, SL2R. And now we'll do the more general thing for SLDR. Maybe I can even, you will remember this, I hope. OK. So now I have to introduce some subgroups. So just as for SL2, I will let A be the subgroup of diagonal matrices with positive entries. So A1, A2, up to AD, and zeros of the diagonal. And I will let N be the group of upper triangular matrices with ones along the diagonal. So 1, 1, 1. And then here I have arbitrary real entries here. 1, 3, N, 2, 3, and so on. But here I have zeros. And all these arbitrary real entries. And I will let K be SOD. So the group of all rotations of RD. Just to remind you, it's a group of all elements in SLDR such that K times K transpose is equal to the identity matrix. OK. So it's all rotations of RD. Now here's the evasava decomposition. So what it says is this, that G is equal to NAK. And every element in G has a unique decomposition as a product of elements like this. So to state it in more precise terms, for any element in G that exists unique elements, some N in N and some A in A and some K in K, so that G is equal to NAK. And I guess we won't use it very explicitly, but it's also the fact that this map, going from the Cartesian product of these three groups onto that one, it's a diffeomorphism. The fact that it's C1 is clear at once, but also it's inverse is smooth. So both ways they are smooth. OK. I wanted to say a few words about the proof. This is actually just Gram-Schmidt orthogonalization, if you want. But I wanted to have it on the board to see what it works like. And we will need it later to look back at this proof. So it's maybe an outline. So let G be given. And again, we call the row vectors of G V1, V2, and so on. And the trick now is to apply Gram-Schmidt orthogonalization to this basis, V1, V2, to VD. So apply Gram-Schmidt orthogonalization or orthogonalization process to. And I apply it to the way I've set up things here. I should apply it to the basis in the opposite order. So I start with VD, then VD minus 1, and so on down to V1 in this order. And this gives me, so remember what does the Gram-Schmidt process do? Well, first we take VD, and this is a basis, since determinant of G is equal to 1, so it's non-zero. So first, we take VD, and we just multiply it by constant to get a unit vector. And then we take VD minus 1, and we add an appropriate constant times my unit vector in order to get the vector which is orthogonal to VD. And then we scale it by some real number to get the unit vector also. And then we take the next vector, we subtract the appropriate linear combination to be orthogonal to the previous vectors, and then we scale it by real number to get the unit vector. And in this way, we will get unit vectors, and they will be mutually orthogonal. OK, so we get an ON, or the normal basis, WD, WD minus 1, down to W1, and it has the following property. Well, so I can state it in many ways, but I want to state it as follows. The vector VJ belongs to the linear span of WD, WD minus 1, down to WJ. For every J, this is the case. But also, I have control on the sign here of the WJ vector. So actually, VJ belongs to the following type of linear combination. You take a positive real number times WJ. And then plus something in the span of the remaining vectors. So this is true for every J. And this is if in each scaling step, I make sure to scale by a positive real number, then I get this. OK, and now if I write out this fact in matrix form, I get what I want. So I had G equals this matrix. And now I can view this as the following matrix. Remember that this is a D by D matrix, and this is also D by D matrix because I've recorded the row vectors length D. And I get what I've written here can be written out as a matrix product. Here I have to just write out the appropriate D by D matrix. And the thing is that, first of all, VD is equal to some positive number times WD. So it's just AD, where this is a positive number. And I will have zeros here.  And then I have some real number there. And it continues like this. So I will get an upper triangular matrix with positive real numbers along the diagonal, and then some real numbers here. So the point is that these A1 to AD are positive numbers. And then if you look at this, it's easy to see that this is in the product A times N. You can easily work out what such products are and so on. And it's unique. And this, by construction, it's in K. So I guess immediately it's an orthogonal matrix, but then the determinant also has to be 1. Sorry, I defined it wrong there. But the determinant has to be 1 because this has positive determinant and this has determinant 1. No, no, no. Something is wrong, yeah. So I'm done. So this is an orthogonal basis, the orthogonal matrix. But it also has determinant 1 by looking at the signs of the determinants of these. And I have made a mistake here. Of course, what I defined was the orthogonal group. I was going, I should have written also determinant of K is equal to 1. It's OK because here, OK, OK, sorry. So I did not make a mistake. And really, everything is clear somehow because I was working inside this, so OK. I shouldn't, I just, yeah. OK, but that is it. In order for the proof to make good sense, it should be NA. But it's nice to check that they don't commute, certainly not, but NA is equal to AN as a set. So that ends the proof of Iwasova decomposition. It is easy to go in and check that we have uniqueness. Next, I wanted to just say a few words about harm measure on G. So there are many ways to write out an explicit harm measure on G, but they are all a little bit complicated on G. So for us, maybe one important format will be to write out the harm measure in terms of this parameterization. So I write it out here. And I will not prove it. It's not genuinely difficult, but you have to work a bit with the Jacobians or something. So mu, I will call my harm measure mu. And here is one harm measure. By the way, remember that harm measure is unique up to a positive constant. And I should really stress if it's a left or right harm measure, but it turns out for this group, any harm measure is both left and right invariant. So then I can say just harm measure. And this is fairly general feature in these lectures because whenever there exists a lattice inside G, the harm measure has to be left and right invariant. But OK, we haven't yet proved that G has a lattice. OK, so here's a formula. If I have this parameterization, then a harm measure is given by dn times a certain factor A. Oh, I should define A. A j divided by A i. I have to have this product. And here I should say that. So A, my diagonal matrix A, I always let A j be the j's diagonal entry of A. And then times a harm measure on the group A, which I write out explicitly, j from 1 to d minus 1. I will say a few words about this. And then dk. So here, dk is some harm measure on SOd. Harm measure on SOd. So on k, in other words. I will not write it out explicitly. And I guess we will not really need it very often. We can often work with just some harm measure. And this is some harm measure on the group n. And that is very easy to write out explicitly. I won't do it. But in words, you can just take the product of Lebesgue measures with respect to the entries. It's gone. So product of Lebesgue measures in all the entries will give a harm measure on n. And then in fact, this measure here, dAj divided by aj, product from j going from 1 to d minus 1, this is a harm measure on A. So note that when I write like this, these variables are dependent.  So ad is equal to 1 over a1 times ad minus 1, if you like. The product of all entries is, of course, 1. So in order to make sense here, I should take a product of d minus 1, Lebesgue measures. OK, so there is this factor arising. And well, you have to compute it simply. Actually, it turns out that going back to somebody's question, if you would instead use the parameterization a and k, then we would have the same formula without this factor. So it's a bit easier. But we often want this parameterization. It has a clear geometric sense somehow, for many reasons. OK, I think I hope we can give some good exercise in the tutorial. There are other ways to define harm measure. But as I said, it's a bit difficult. So let me just point out that harm measure on GL, if I have the full general linear group, then it's easy to give harm measure. Harm measure on GL dR. If I let the matrix in G be given, if I call the entries x11, x12, x21, and so on, then simply Lebesgue measure with respect to these variables times determinant raised to minus d gives a Lebesgue measure. So d new G, harm measure on that group, can be taken to be determinant of G raised to minus d. And then just d squared dimensional Lebesgue measure. So it's just a remark. And when you have this, somehow SLDR sits inside here as a co-dimension 1 sub-manifold. And it's not too hard. You would like to somehow take a delta function requiring that determinant is equal to 1. And you can make sense of that. But I don't want to discuss it more now. Say again, on K, no. So I leave that somehow. And it's often possible to just say that this is a harm measure on K. OK, so now I want to discuss the geometry of x, the geometry of G mod gamma. Or in other words, the geometry of the space of all Euclidean lattices of Covell in 1. So this is a non-compact space. And it's interesting to try to understand what does it look like at infinity. And well, one nice question is to ask, if you have a sequence of lattices of Covell in 1, then when is such a sequence not relatively compact? So when is it impossible? When does such a sequence tend to infinity? When is it possible to find a subsequence which converges to some lattice? And when is it not? And the answer is given by Mahler's criterion, which I will say a bit later. But first I will introduce Siegel sets. And somehow Siegel sets give a more precise description of the geometry at infinity than just having a criterion for when we tend to infinity. There are many kind of really different lattices that lie far out in the cusp of x. There are many different ways to tend to infinity. So this is captured by Siegel sets. OK, so what one would really like to do is, of course, to find a fundamental domain for, would like to find some fundamental domain for g mod gamma. And it can be done. Minkowski has given a precise description of really an exact fundamental domain. But it's more and more complicated to kind of decode it. It has more and more bounding sites and so forth. So we won't go into this. A Siegel set is a set which contains a fundamental domain. And it has a rather simple structure. So it can often be used if you want to integrate over x, and we just need an upper bound, for instance. Then we can replace it by an integral of the Siegel domain. OK, so I will give a definition. So remember that A is the group of diagonal matrices with positive entries. Now I define the following subset. For any positive number t, let A t be the following subset. It's a set of all diagonal matrices, always with positive entries and with determinant 1, such that the i plus first entry is less than or equal to t times A i for all i. So when I say for all i, of course, it means for i equals 1, 2, 3 up to d minus 1. So if t is equal to 1, this is simply saying that the entries should be monotonically decreasing. But I will actually have to take t slightly larger than 1. So it means that the entries are more or less decreasing. But they may increase a little bit, but they can't increase much when I go from one entry to the next. And OK, now I will define a Siegel domain. So for any t positive and any compact subset, so I will call it fn, this is a compact subset of n, I define s within the xfn and t to be simply the product set, fn times At times k. And this is a point-wise product, if you like. And this makes sense if you remember the Ivatova decomposition. Because this is a subset of n, this is a subset of A, and this is all of k. So it really sits element-wise, so to say, or factor-wise. It sits inside Na k, which we saw was equal to g. OK. So here's a Siegel set. And now, I don't know, maybe it's bad notation to call it fn. I want to remember that it's a subset of n simply. I could call it f here. And now I will make a specific choice. But any such set is called a Siegel set. And sometimes you actually work with other Siegel sets. So here's maybe today's main result. If I make the following choice, fn, or OK, sorry, if it's bad notation, I won't use any other fn. So let me fix fn to be the following. This is a subset of n, and I will define it by giving inequalities on the entries. And the requirement is simply that all these entries have absolute value less than or equal to a half. Clearly it is a compact subset of n. Sorry, ij. And this is for all ij, where it makes sense. And then also, let's take t to be 2 over square root of 3. So I make these specific choices. And then the corresponding Siegel set, I will call s, because I'm going to talk just about this now. So s is the Siegel set coming from this compact subset of n and with my specific choice of t. And the claim is that this s contains a fundamental domain for g mod gamma. So in other words, if I take gamma times s, point-wise, set multiplication, then I get all of g. So any element in g can be written as a product of some element in gamma and some element in my Siegel domain. OK, so I have several, before I hope to outline the proof of this, but at first I will speak about several consequences. Perhaps first, let me note that this 2 over square root of 3 is the same as the inverse of, if you remember, the fundamental domain we had when d is equal to 2. We had the following fundamental domain in the upper half plane for the modular group acting on the hyperbolic surface. And here is minus 1 half, here is 1 half, here is 1, here is minus 1. And so this is a fundamental domain. Now, if you do the translation from this subset of g to some subset of the unit tangent bundle, then it turns out that the Siegel domain, which I have here, is exactly the following rectangle. Sorry, it's really a horizontal line going from that point to that point. So I overshoot by a little bit. And it's the unit tangent bundle lying above that triangle. So it's a three-dimensional infinite rectangle. That is the Siegel domain in the case d is equal to 2. If you go from the Iwazawa decomposition notation to the unit tangent bundle picture. And the point is that the height of this point here is square root of 3 over 2. So it's the same as this t, if you do it. So that's a good exercise to do to get some picture of the Siegel domain in the simplest case. And maybe I should also say a word about lattices. If we think about lattices, what this says is that given an arbitrary lattice, I can represent it by some matrix in g. But if I choose the lattice basis correctly, then I will get some element in s. And I have more than one choices of such lattice basis. But the point is that any lattice has some lattice basis such that the corresponding matrix lies in s.  So it's not a really well-defined concept. A lattice can have many reduced lattice basis. But very roughly what it means, if I have a lattice and I speak about its reduced lattice basis, such a basis will have the vectors of such a basis will not be unnecessarily long. And the angles between the vectors will not be too small. And you will see more precisely this soon. OK. So a first corollary of this. And it's not a trivial corollary. You have to work a bit. But a corollary is that the harm measure of the quotient space, g mod gamma, this is finite. So gamma is a lattice. I promised I haven't proved anything yet. But it follows from this that gamma is a lattice. And the proof is by noticing or by showing, by working a bit harder and showing that the harm measure of this set s is actually finite. So I should write this. So gamma is a lattice in g. Remember that the harm measure of this quotient is by definition the harm measure of any fundamental domain for g. And it's a simple fact that any two fundamental domains will have the same measure. And now since I said that this shows that s contains a fundamental domain, then it follows that the harm measure of this must be less than or equal to the harm measure of s. And now you verify by some work that the harm measure of s is finite. Then we get this. So how do you verify that the harm measure is finite? Well, you go back to the formula we had for harm measure. And then you can integrate. That formula is well adapted for integrating over this because it gives the harm measure decomposed with respect to harm measures on this. So the only difficult thing is to integrate over at with respect to the measure I wrote out. So it's a nice exercise to show that this d minus 1 dimensional integral. And now it's a really concrete integral with respect to d minus 1 variables to check that it is a finite integral. And let me also point out a somewhat strong effect. And this is really difficult to prove or fairly complicated. I will definitely not try to prove it. That actually s contains only finitely many representatives of any point in x. So that is a stronger fact than this, at least a posteriori. So if I take the supremum over the number of point representatives for any coset, this is finite. So it's just a remark. So the single set doesn't overshoot by too much no matter where on x we are looking. And then also I will tell you the answer to the question I threw out earlier. If you have a sequence of lattices, when does this sequence diverge? And when can you find the subsequence which converges? Well, the answer is the following. Corollary 2. This is also a corollary of theorem 3, Mahler's criterion. So in other words, I want to give a condition for a given set. If I have some set c, a subset of x, or a subset of g mod gamma, when does this set c have compact closure? And the answer is that c has compact closure if and only if there is some ball in Euclidean space that all lattices in L avoid, except some ball around the origin that all lattices avoid, except all lattices, of course, contain the origin itself. Or in other words, if there is some lower bound on the shortest non-zero vector, so in order to tell if a lattice is far out in the cusp, you just look at the shortest non-zero vector of the lattice. If that vector is really short, then the lattice is far out in the cusp. So OK, a precise statement. So it has compact closure if and only if there exists some positive number r such that every lattice in the set c or corresponding to points in the set c has intersection with the unit ball. This is the unit ball of radius r around the origin. And this intersection should contain no other lattice points than the origin itself. And this should be true for all g or all gamma g in c. OK, so this is really a pleasant exercise. Again, I think to think about when you have given theorem 3. So let me just say two sentences or something. So the hard part of this is to check that if this condition is fulfilled, so if there is such a ball that this is a void, except for the origin, then I want to check that if I have a sequence of such lattices, I can find the convergent subsequence. And what you do is for each such lattice, take a representative in the Siegel set. So then we have a sequence of elements in this Siegel set, a sequence of elements here. And now, OK, originally I wanted to prove that I have a sequence of convergent elements here. But now I can actually prove that I have a sequence. I can find a subsequence that converges as elements of g, of the league of g. And that immediately implies that we converge also as elements of the quotient. And OK, so how do you find such a subsequence? Well, we have compacts. If you look at the definition of the Siegel domain, I have compact sets here and here. So it's easy to extract subsequences so that I converge in this component and in this component. So now what remains to show is that I can find the convergent subsequence here. And well, then you have to look at the picture. What does the lattice look like in terms of these a1 and a2 and so on? And well, basically, ad is the length of more or less the shortest vector of the lattice. So the point is that I have a lower bound on ad. The ad is bounded from below by positive number. And if you have such a bound from below, then these conditions and the condition that the product is equal to 1, they imply that I can extract a subsequence which converges, if you think about it. OK, and now I really don't want to go over time. So I've at least tried to sell theorem 3, but I've managed to have no time remaining to give the proof. Let me just say, I can begin a little bit. I was going to give a little bit of an outline of the proof. I can maybe say one or two sentences. The first thing to note is that this set fn is a fundamental domain for the following. Let me set gamma n equals the intersection of gamma and n. So this is the group of all upper triangular intergeomatrices. And then note that if I take point wise product of gamma n with my chosen fn, the one which I chose in theorem 3, this is equal to all of n. And in fact, this is easy to check by just writing out the product. And in fact, it's a fundamental domain even for the quotient. So I have not much over representation except that the boundary. OK, so if I note that, then I'm looking at the following set, gamma times s. I want to prove that this is all of s. And then I can use, so this is equal to gamma times fn times at times k. But here, from gamma, I can break out the fact that gamma n, this is no more than gamma. And then I use the fact that this is equal to all of n. So I get the following gamma times n times at of k. So now given an element g in the group g, I want to prove that the orbit gamma g meets this set. So I want to prove that for any given g, I can multiply from the left by gamma, and I can fall into this set. And if I interpret that in terms of lattices, if I let l be the corresponding lattice, then what I want to prove is the following. I want to prove that there exists a lattice basis, v1 up to vd of l such that, OK, I'm not doing much, but such that this particular matrix is in this domain, nat, k. And OK, what I wanted to give and to go back to maybe motivate some things I said earlier is to just point out what this means geometrically for the lattice basis. So we have a given lattice basis or a given element in SLDR, and I showed you how to do the Vasavvadi composition. So if you look back at that proof, the proof was applied Gram-Schmidt or Togonalization process to this basis, but backwards. If you go into that proof, then you can read off what it means for this matrix to lie in this set. And I want to just write out what it means. So the thing within the box over there is equivalent to the following. It's equivalent to saying that the distance of the vector v i plus 1 to a certain subspace, v i plus 2. I will define this subspace soon. That distance is less than or equal to t, the parameter t times the distance of the vector v i to the subspace v i plus 1. And this should hold for all i. So this is the geometrical interpretation of the diagonal entry, a i. And here I have to tell you what the subspace v i is.  And then also v d plus 1 is the zero space. So in particular, when i is equal to d minus 1, the largest value, then this distance is just the length of v d. So that length should be less than or equal to t times this distance and so forth. And now the task is to construct such a basis. And you do it basically by a greedy algorithm. You pick v d to be the shortest vector of the lattice, then take v i v d minus 1 to be the shortest possible vector lying outside of the subspace and so forth. And you can check that it works. OK, sorry for going over time. Thank you.


Hey guys real estate gaming here. I hope you all had a great weekend by the way And today we are back on the scary elevator because they have reworked on Michael Myers and they have added two new floors Which is epic and as we can see we got Michael back here. Oh So it says fine Laurie strode for a bad reward, okay So we have to find Laurie if we get the Michael Myers map and we're here with the whole squad man What's up everybody, but let's go straight into this elevator see what's up with it. Okay. We got Jason Boris Oh, there he is. Oh my god. He's already slashed away Typical Jason. Okay, so let's go over here. Let's try to see no I was gonna try to make it through the hatch to get some points man And also I forgot they have like a rebirth system in this game. I have a total of eight rebirths Wow, which a lot of people don't really rebirth because then you have to do everything over Which is not cool. Okay lockdown my garlic this elevator has oh my god is that blood that is blood gushing out on the side Okay, hopefully it doesn't get filled in there with blood Okay, who do we got? Oh my god.   He's over there chilling by the door Okay, where's my boy Chucky at? Oh look Spongebob. What's up Spongebob?  Oh God he got someone okay. Nice. He managed to survive that Hey, hey slender man. Oh Man, how tall is slender man? He's like six feet, right? He's pretty tall He's a tall man every 51. Okay. Let's go check out every 51. Yo See how this goes bros What's this one where they release like multiple killers look at an ID to escape the floor well nice We just located an ID Okay, let's go back here. Oh my god. The rake is chasing somebody down. No, don't break him to me Oh great. Now he's chasing us Somebody up on the door. Oh nice. Did we make it in here? I'll attempt them out. I'll help you out man Okay, nice. Wait, where did the rake go? He was oh, there you go. He's chasing down son Ah, leave the poor bacon hair alone.      What is wrong with that man? All right. Oh look somebody's dressed as Freddy Krueger. That's pretty neat. I like that. It's pretty spot-on yo I must be the morph. That's why I'm digging this bunch about one. It's so hilarious Okay, let's see three two one get ready to rumble you guys Oh, it's Sonic. He actually got to be quick because you know Sonic is a fast murderer. He's quick on his feet Look there. He is. He doesn't even run. He just like floats around Whatever man. Let's go over here. Just try to escape through the hatch Sonic don't come this way. You know what? Let's go up here Okay, there you go escape, but voila. We made it out of there safe and sound What is that? Starter crate advanced lantern speed coil. What's up? It's like a little crate. What's that's pretty crazy I don't know about that. Hey hatch floors. Okay, so all those floors got hatches I totally forgot about this lobby update. It's pretty neat Wait what you like skitties? I mean I like cats too, but I'm more of a dog person man Where's the dog music? I see p96. Oh my god. We got the shy guy here And there he is welcoming everybody with a nice warm screaming hug.  He's jumping everywhere Yo, you didn't chill out my boy He's screaming and running and slashing. Oh my god, somebody need to get this man some milk Okay, let's go this right now. Oh, did he was just right in our face smacking us? You know what?  Oh no. He just took somebody down. What recipe to that man? Okay, so we got Freddy Cougar to see if we got any other interesting skins got Freddy Cougar spongebob Then who else do we got here? Others on Leo's wings. Those are some nice wings Leo. Then we got jolly over here. What's up jolly? Oh man Hey, happy birthday. Wait, what?  It's the killer from happy death day to you or happy oops. They know happy death day There you go. Happy of day is a Roblox game. Oh my god. Yo, can you relax man? Do you guys hear this guy laughing like a maniac? Oh My goodness so much for a baby face, right? Look at the blood. It's gushing out of the ceiling over there Okay, so this is definitely a crime scene that we were on. Oh God, okay. Nice. We managed to survive that They run Okay, you guys get ready all Samara Okay, I doubt this is a big for tomorrow. I forgot what floor. I mean what game was a big for tomorrow Okay, no, this is the algae one Dude, I feel like this is one of the oldest floors in the scary elevator like he's never changed it in forever Oh, and there she is My god, Samara is just going crazy over there Dude, she got two people at once. How explain this to me tomorrow. Okay. She must have some evil powers Dude, she climbs up the bookshelf.   That was pretty mean.  Hello, I think this is a new floor. Oh, this is kind of like flee the facility Oh my god, is that the red dress girl? What is the red dress girl?  No Rest in peace to Spongebob man Oh my god, dude, she is fast Hey, yo, you need to relax man. Let's go in her house Okay, we need to hack this to get out of here everybody get on one side click the clack. Look, I'm flicking it I'll pretend you like I was hacking the computer. Oh Man All right, so so far we got in a couple. I believe we got both of the new floors already Cuz I never seen that red dress girl floor before. Oh Oh, I like this music. It was creepy. I think that music was from a dead by daylight Yeah, that was crazy, man. Oh No, I'm dancing again Okay, let's summon the noob So he can be a distraction Come over here Jason.  I believe it's still back here somewhere. Oh, there you are Okay, I'm like is this still back here or not? Oh my god, that Jason is creepy. He's just holding Laurie That's hostage right there. Oh No, gg man a Ronald.   Oh my god. This is the most evil McDonald's I've ever seen in my entire life Ronald, what are you doing? Oh my god. This man is out here serving Big Macs left and right Look at this food. It's already bitten They're gonna serve bitted food. That is nasty. Yeah, only blocks of cola. Can I have a block of cola? No, oh man. Oh, you know, you got to get that McDonald's sprite. Yeah, let me get a sprite, man All the badges there which I believe we have already Dude, that was the best floor so far.  Okay, oh Ronald again. Okay, whatever I guess we love McDonald's so much. We're going back for a second time This isn't your next happy meal get a free Ronald McDonald. Don't thank you, man Dude, he really does have a big Mac and he's smacking people with it Oh my god. We need a Maccham, Mac Ronald McDonald Okay, the Maccafe. Let's get the badge. Oh, we didn't have a badge. Nice.  It sounds like he squeaks every time he he smacks somebody with the with the burger on his hand Oh, no, right. So we managed to survive Ronald McDonald's wrath two times I mean, how terrifying could he be right? He's a clown. Now they're pretty scary, man Okay, three two one. What's up, man? He's punched by we got Leo Got Jolly right there So guys, okay three two one Oh, both selected hunters. Oh, this is where we can fight back We can finally fight the monsters back everybody attack Oh my god. No, I'm the only one out here Okay, you know what? I did not want to die You were so close to dying Look at the zombies. There's so many of them Come on guys. I know we can do this all together. We stand strong as one Take that zombie. No. No. Oh my god. He almost got me Okay, the table's almost turned on us, but thankfully we managed to pull through. Oh Ben Wait, Ben 10 or Ben drowns. I believe this is Ben drowns This should handle a Ben 10 like a Ben 10 exe that would be cool And then like every time he turns into a different monster Hey, that's a good idea, man. Oh god Where's Ben at Where is he at?  Even though it's full of blood All right, where's mr. Ben? Yo, oh my god, this dude is teleporting Need to chill out Ben. What that was pretty quick Okay, hopefully we get to the Michael Myers one next because That's like the only floor we haven't gotten besides all the other new ones So let's see if it's the next one. Please let it be Oh, man And camcrystal like done Jason again. All right, you guys well I'm tired of returning to camcrystal like thank you. What thank you guys for watching Thank you guys for the main support if you guys enjoyed the video Make sure to slap the like button if you're new to the channel Make sure to subscribe and turn that notification bell on and I will see you guys in the next one. Bye


Everybody for us tier ripping 2022 top stadium club baseball hobby eight box number two Who does number two work for? You can always find the results of late in sports cards calm. Thanks for joining everybody. Good luck. Oh, dude.  Where's he going Dodgers going to Dodgers?  I believe you did I recognize that name Better go to the baseball reference. I Want to say he did Yeah, like a couple seasons or something You know, he was actually a pretty good catcher. Yeah, he's actually pretty underrated red foil Tatis Ted Williams He was a right along Really five years So we pull it on license Gianna Navarro. It's gonna go to the Rays Red foil Alex Kiriloff, Adrian Beltrae, Jaren Duran Jose Lobatone, yeah Remember Sean Faggins Yeah, actually like for a right Mountcastle. I'll be your play for the Rays. I just wondered So they end up pasta my mind with those guys Sean but CHO and then he is only that right red foil Reese Hoskins Yeah, I know John Is very sad rookie auto Brian De La Cruz Marlins Team of the future Jazz Chisholm. I think we have one of those insta visions coming up here. He has an insta vision case hit Bobby Whit jr. Very nice.  You were saying big way. That's it. That's that's a fairly decent wit right there power zone Devours get say a Suzuki Red foil Mookie More than one like insta vision. I mean you get one insta vision a case one gold mints at a case So there are multiple case hits. I don't think you get more than like one gold mints Or one more than one insta vision right well Joe Carter. I don't know for sure. This is like the 10th box. I've opened so black foil Yordon Alvarez But I'll drop the checklist link if you want to check it out Totsies box topper should be a member's only parallel. Yeah, there's a few things that are in every case I don't like case hits like one a case, but I Don't know if you can get more than one insta vision or more than one gold minted black and white show a Ohtani angels Feel little field and fly ball there Warm ups rookie base auto cubbies Greg Deekman Team of the future John at the India red foil Hulu Orias. It's Kepler. I'm moving right along guys There's the Ortiz there hobby bias destiny we looking on spots Still the was it the Jays left on a still the Jays left on bowman chrome hobby case over 20 Red foil Carlos Rodin red foil Max Kepler Cal Raleigh black foil Miguel Cabrera Tigers Griffey and Hank Jayrod base Red foil, Sonny Gray Yankee legend Sonny Gray a Don trail Willis auto for the Marlins nice detrain That's a sequel with the World Series trophy Riley Red foil indoor a Chrome Clayton Kershaw Dodgers Triumvirates Derek Jeter Yankees are you see up? All right. All right a rookie blue foil Jeremy Pena Number 24 out of 50 That is a big one for Houston Wow Sick card Congrats Astros Jared P. Yeah, we will get that magged for sure Hey, thank you Tony baseball is all sold out tonight guys just football and Star Wars left Red foil Chris Taylor.  She's a big USA win today. It's another one's huge win Buster Posey box topper She's that big a Patriots win Thursday night Jake. Oh, yeah, huge job. It was awesome. It's so great That big winnings the bills It was weird to like Tom Brady was starting And like ready mosses. Yeah, like Teddy Bruce. He was also playing. It was really weird, but I was cool with a man It was a big win Yeah, it's very weird, I'm not gonna complain though. Yeah Matt Patricia was on the sideline, which was amazing red foil a cunea on the destiny insert braves Welcome Butler made a goal line That happened against the bills for sure a Refractor David Wright There you go Mets chrome refractor one every four boxes a red foil rookie O'Neill cruise pirates Yeah, a lot of opportunities a lot of mistakes too though the tactics were just were just bad I in my opinion. I thought it was pretty rough pretty pretty rough tactics for the US Never never really adjusted Black foil Andrew Jones Let's go bucks. Whoo looking good Black and white rookie Josh Lowe Tampa Bay That is true 510 that is that is true rookie auto Glen auto Rangers Base auto there. I'm not sure Not sure what you missed.  They're gonna have better record than the Bucks Oof You hate to see it Just combined teams for one year New England buccaneers with a Tampa Bay Patriots, then we'll just win it all Sounds better, right? Yeah, the Tampa Bay Patriots. Yeah red Ronald Cunea. That's a cool one. All right foil Jeff Bagwell black rookie Seth beer The backs that's right Patriots gonna go on a 20-year dynasty when Tom Brady comes back. That's right Can't wait Red foil rookie Reed Tadmer's Angels Yeah, our same club is good. It's definitely a good rip is a good time. What's some nice cards today, too? Colton Welker Rockies rookie base auto Jayrod team of the future insert Mariners Rookie red foil raise Josh Lowe Chrome a one-a-box rookie Cal Raleigh big dumper Well Mariners triumvirates Ken Griffey junior Mariners Ricky Henderson box-topper A's they had a parade of winter gardens there. I live in winter garden.  What was the parade for? Was it That's so much worse Oh Christmas season, it's for the Christmas season Florida. It was like 85 degrees today Is not there's no holiday season here rookie auto Marlins Edward Cabrera Ready for a rookie Gavin sheets White Sox. Yeah, Santa's cool. You know, it's cooler is getting a bagel sandwich on your Saturday morning Sean was denied a bagel sandwich. Yeah, be a lot cooler if Santa's it was on Christmas and you can actually see Santa but no a Chromorothraktor A's Mark McGuire There you go one every four boxes for the A's a red foil Kirby Puckett. I'll complain to the city for you Sean I'll go to the next city council meeting Black foil Patrick wisdom. That'll be like an episode of Parks and Rec. I think Actually, I think that's what every city council meeting is just like an episode of Parks and Rec Red foil for do go It's one of those on our garden is called NYC Bagel and pizza or whatever Jeremy Payne your rookie Stroves pre-solid a Chrome refractor auto for the Brewers Christian Yelich number 17 of 25 You go Brewer's nice card there Chrome refractor auto Yeah, Merry Christmas bagels. That's right.  He's saying go Dolphins, ma'am or sir.  I'm gonna watch World Cup during the day That's for sure. That's about all I'm gonna be watching during the day and then I'm have some have plans in the evening So probably not much red foil rookie Luke Williams for the Marlins Brian Taylor Cruz rookie Otto Marlins What a waddle the entire parade did the parade would have been 12 hours long Shawn just waddling for two miles Red foil Marcus Strowman You can have some stiff shoulders you have to go to a masseuse like our practice Rookie chrome wander Franco Won a box wander rookie chrome Nice card for the rays a triumvirates mookie bets Dodgers Mike trout box topper Or the angels so again everybody just footballing and a Star Wars tonight Fill the Star Wars break by the end of the night, then I'll use the force on Jake Red foil Tyler O'Neill Throw him into the end of the big pit we have at the end in the warehouse the big abyss You'll help pose though a rookie Otto for the Rangers Crazy how Star Wars is always some deep mysterious like pit you just throw someone into Kind of weird, huh? Red foil Andrew McCutcheon Say ah Suzuki Dude Sean should get in my house You guys might I know that I only said it like once on stream But so my wife went to that same bagel place we go we go probably at least once a week for breakfast and she went on Like Wednesday or something or no like like Thursday. I don't know what it was. It was a day this week and The guy in there knows us and she bought a dozen bagels and he's like so I hear it takes an extra And he gave her like an extra like eight bagels that were like he's like these little older So you probably want to eat these first. We got like 20 bagels what I've been eating like two bagels a day I've it's been awesome. I read Wilson Contreras. We haven't gone grocery shopping like a week because we just have bagels Cream cheese they get from Philly. They get like cheesecakes there from New York and stuff. It's great. It's great So many bagels red foil Bobby Dalbeck We should all have our team emblems Even though mine's not that Julia Rodriguez Ricky for the Mariners All my teams think actually except the except the Bruins the Bruins are freaking awesome. Actually every other team stinks Magic Pads Red Sox and US men's national team oof red foil Trevor Rogers Everything bagels are good.  The red foil Vladimir Guerrero auto expose going to the nationals That one's number 29 to 50 nice hit gnats We should tag this one That's a sweet one about to sign a ball right there. See you have key prediction I Who's supposed to who's supposed to who's in it? Who's supposed to be in it? I? Guess Ohio State now rookie red drew LSD backs even though they literally just got whooped by one of the teams It's in the playoffs. So why put them in anyway?  I think so Stefan two big ten and Two SEC teams. I don't know. I haven't really paid attention to college in quite a while black Babe Ruth Because of this stuff that keeps going on like this. I I get I get like less interested every year USC terrible defense terrible o-line Chipper Jones Braves box topper the Georgia and Bama play this season Georgia B. Bama. I'm pretty sure. Yeah, so we're gonna put Bama on Ohio State and When the two top teams there have already beaten them this season. What are we doing? What are we doing? It's a wander base black and white for the Dodgers Walker Bueller Rookie Otto Oliver Ortega angels Red foil Tyler O'Neill Which I'm only good for whatever's next Karthi Red foil Trevor story Red Sox Rookie red Matt brash Mariners More importantly, when is baseball coming back black foil you Darvish Padres? Red foil Bobby Dalback Rookie base on a Marlins Nick Fortes Red foil John Crook We go David Ortiz Jeremy Pena rookie a Chrome Nolan Ryan for the Rangers Triambritz Rafael Devers Red Sox We just pulled the Vladimir Guerrero auto to 50 we pulled a Christian yellow to 25 last night I pulled a Bobby a Bray you auto power zone. He'll your Rodriguez Mariners Matt pulled a Mike trout yesterday one of the breaks Red foil Jorge Solaire red foil Walker Bueller Black foil Machado all right a couple more Salvador Perez box topper Bruhan Rookie Red Destiny Bryce Harper Phillies Red foil do garrick and a red foil hunter green rookie for the Reds Black foil Tim Anderson white Sox Red foil Luis Castillo for the Reds Jeremy Payne your rookie for Stroze Spencer Strider braves rookie base auto Solid hit for the Braves there red foil Alfonso Reeve us rookie for the Cubs Red foil Sandberg for the Cubs Destiny Cabrera Ricky red foil Colton Welker Rockies Red foil Marcus Semi and Rangers That is a wander photo variation right there Yep wander from go photo variation nice card for the Rays First one I've noticed I'm sure there've been more but they're really tough to spot unless you know what it looks like You'd have to turn around every single card and look so Cmp code 105 last three digits the base is Zero six zero there you go raise nice card there black foil Trevor story Red foil Buxton Spencer Watkins Orioles rookie base auto Team of the future Julio Rodriguez Mariners red foil Delos Darrell strawberry Chrome Albert Pujols nice card for the Cardinals And Trey Umbridge Shohayotani angels last box box topper Miguel Cabrera Arizona Torkelson Red foil Logan Webb Red foil Cowrapean jr. Come on. Come on. Let's close out strong here Red foil Javi Baez Greg Deekman rookie auto Cubs Sensor Torkelson team of the future Bobby with jr. Rookie base card for the Royals Red foil Casey Meis Powerzone Wander Franco You more here Red foil Kershaw Jayrod base rookie black foil Max Scherzer Noob Bar Muller Red foil Delar Darrell strawberry Auto Lopez blue jays base rookie auto Red foil wants Soto Nationals Hey, what's up bray and a Cowrapean jr. Chrome Orioles here a triumvirates Paul Gulch made Cardinals That'll do it Give me a sequel recap, right? So we had that wanderer image variation And it's division Bobby with jr. Royals one o'case autos now Don trail Willis auto Lopez Spencer Watkins Nick Fortes Patrick Mazzica to Brian Dela Cruz autos Greg Deekman twice Glenn auto Oliver Ortega Colton Welker Edward Edward Cabrera you help ozo Spencer Strider Vladimir Guerrero, Jr. Red foil to 50. I'm only goes to the Nats via the Expos Number 29 to 50 and a Christian yellow chrome refractor auto number 25 Brewers and Rookie blue foil Jeremy Pena Astros number 24 50 That's gonna do it. Thanks again everybody. We got Bowman chrome coming up


Even if you're in Israel, you can still get a taste of France. My name is India, and today we're going to try macaroons so delicious that you might start speaking French. Wow, c'est bon! The macaroons is an iconic French pastry.  If you've ever tried making macaroons, you would probably know that they're considered one of the most difficult desserts to make. Making this is no easy task because macaroons are extremely sensitive to humidity and temperature. Even getting the timing right when baking is a challenge. But that didn't stop these sisters. I am Karim. And I'm Deborah. Yes, they're twins and the founders of... Dreamcake! And they just might be the macaroon masters of Israel. But this isn't just their family business. It's their dream turned reality. The sisters came to Israel seven years ago from France. We didn't even know how to speak Hebrew. But with trial and error and a lot of sugar, these sisters learned to speak Hebrew and perfect the macaroons. After just three years in business, they've already had clients like Google and Chanel. Twinscakes has 36 different flavors of macaroons, including special ones like Oreo, pina colada and lemon meringue. Salty caramel is my personal favorite. What is the key to the perfect macaroons? Well, it takes persistence to try and try again. Making macaroons is a lot like our own journeys. It's a delicate process from the beginning to end. And it's okay if you don't get it right the first time. From France to Israel! This is the magic of macaroons. Thanks for watching and I'll see you next time.


What's up everyone? It's your girl, Kenzie here from Leighton Sports Cards and I'm ripping a personal four. Joseph D. We've got a 2022 Panini Illusions football hobby. All right, best of luck here, Joseph. Thanks so much for your order. Let's see a big hit here. Let's see a Hollow Heroes case hit. Only pulled like one or two of those. What do we have here? How about a Tray Lance Auto? 12 of 25. All right, now we're starting with some football fire here. There you go. Probably going to get that starting job now that what's-a-name got hurt. Amazing Tom Brady. Mr. Irrelevant again. To 199, George Kittlebucks. And to 399, Funkadelic. Can you pick it? Here's to see where Jimmy G lands too. We have a Nate Burleson on the Suprile of the Auto. 47 of 75. That's how it was coming out already for you. Ooh, I like Apex. PJ, that's one of the games I play when I do play. Shining Stars, Joe Burrow. I like Apex and... Oh, I'm blanking on the other one. We've got to 399, Alec Pierce. Overwatch. Out of 299, King of Cards, Peyton Manning. Those are the two games that I play. Rock pretty good. Pick it. A Tremac Bride on the Instant Impact Patch. And my guy, Nashor. That nation. Terri Sox. The 499, Cooper Cup. And Mystique, Malik Willis. We have a Malik Willis Instant Impact Patch. And Jahan Dotson, Amazing Rookie. O'Neal Cruz Super was pulled. Wow, a lot of big cards from Gilda being pulled this morning. To 175, Saquon, someone said the Babe Ruth got pulled this morning. And the 199, Joe Burrow. King of Cards. You like that? That's pretty good. Now we've got a Greg Dulich on the Rookie Auto 299. Broncos. Broncos Country. That's right. And Desmond Ritter on the Detonation Record.  To 299, Alvin Camara. And Funkadelic, Dan the Man Marino. All right, Joseph, that'll do it for your illusions box. Congrats again. Nice box. There were some nice autos coming your way.


Hi everybody, Joe for jaspyscasewrakes.com coming at you with 2022 Panini National Treasures, FIFA, Road to World Cup Soccer, one box picker team number eight. Final box of the case that we were working on last night, mark that picker team eight. Thanks to the people who bought their teams straight up, appreciate it. Traps ended up with last bot mojo with L3. And here is the results of the random number block randomizer that we did for those teams right down there. That's a good set of teams too. Next one box break from fresh case is sold out. We sold it out straight up. So we'll be working on that. And then the next, the second box from that case is down to 17. We can definitely run that back tonight if we want to, ooh, sort of a fatter box here. Thick book or just extra hits, fat hits, who knows, but a little bit thicker than some of the other ones I've seen. All right. So we've got a book, some thick hits, and you can see that plate on the bottom. I don't know if that's an autograph spot, but there'll be a plate for, well it's a one of one. It could be for the number block team. Let's see which team it's going to be for. Good luck everybody. We're going to lead off with Marcel Sabitzer, 58 out of 99 for Austria. That'll go to Daniel. And there's former Liverpool man, Sadio Mane, 28 out of 99. Kelly picked up Senegal straight up. And we've got a fat book here, 17 out of 49 for Columbia, Jumbo triple material booklets. Nice. There's James Rodriguez, there's Juan Cuadrado, and there's Rafael Santos Borre. Nice. Columbia picked up straight up by Nicholas. And we've got, for Portugal, Rafael Guerrero, four, or Gutiero maybe, four out of 99. Well that'll be for Marc, 18 out of 75 for Brazil, that's Richard Alisson. And that is for Brazil, eight, Brazil's a number block team. So that'll go to Marc, Marc S and Brazil. Out of 49, we've got another Portugal player, Bruno Fernandez, 10 out of 49. So that goes to zero, Jeff, with this one. All right, here come the autographs we've got. We've got Callum Hudson-Odoi. Three out of 99 for the three Lions. England is a number block team. And that's Marc S with number three, last bot Mojo. Out of 25, we've got Juan Foyth, six out of 25, Prodigy, Dual Relic, and Auto. Argentina picked up straight up by Wenxun for Brazil. National Archives, Inc., 57 out of 99, that's Gilberto Silva. It's John Ryder with number seven. And we have 37 out of 99, that's Hernan Crespo for Argentina, that's for Wenxun. And the plate, just a plate. Not autographed or anything, that no relic. But it's a one of one for Chile, that's Arturo Vidal. Daniel Smith picking up Chile, straight up. And Daniel, all aboard the Big Hit Express, whoo-hoo. And there you go, gang. I'm Joe for Jaspy's Case Breaks dot com. Thanks for watching. Thanks for breaking with us. I'll see you next time for the next one. Bye-bye.


And obviously they didn't work out in your favor tonight, but just how tough do you think this Blue Jays team is going to be throughout the course of this season? They're a great bull club. They got a great offense that we saw today. Up and down, they got guys looking in for power, got good speed, up and down, and then a great pitching staff. They got a good starting rotation, good back end of the bullpen. So they're going to be tough all year, but we're looking forward to it. Much of the pitching staff you guys have seen Mano before, he's given you problems before. Why do you think that he is so tough to get to? He's got good stuff. He can work that fastball in there, mixes up between a sinker and a foreseamer, kind of good sweeping slider. Works both sides of the plate well. It's just tough to pinpoint one pitch and do some damage. So we had a couple of opportunities there with some guys on, and it was going to get the job done. How frustrated were you in that, I believe, sixth inning, the cold third strike? I was upset, but I can voice my opinion, but I can look back on that bat, and there's a couple of pitches that I missed that, if I put a better swing on them, then I'm on base, and we don't even get to that position. So nothing I can really do there.


ASMR Alexander on censored tube hello it used to be better when you read comics and whisper especially the word okay triggered I'd like to review your old videos and now there is not enough quite whistle yeah I'm sorry ASMR Alexander but these are interesting times and central centralized power attacked my family right they prevented my family from working they prevented my family from doing commerce they try to take away my bodily autonomy my freedom and as you can tell I'm not a type of person to bend the knee and take it up to shoot right if someone tries to bully me or some institution tries to bully me or my family I stand my ground right I do not tolerate bullies and I never have and if it means we have to change our lifestyles our work to deal with such monsters I behave accordingly right that's one thing I've never tried to be as a hypocrite so but we will at some point get back into doing lots of comic book readings and I've done livestream comic book readings right and at some point we'll get into editing comic book videos again so again it's it's I had no choice about it right me and my family were attacked I will defend myself this is my way of defending myself right


The Tony General of the Federation and Minister of Justice Abubakar Malami has said that the issue of pardoning IPUB leader Masi Namikano is premature. Malami stated this on a television program on Wednesday. According to him there is no need to talk about pardon as Masi Kano has not been convicted by any court of law. The AGF also said the government is yet to assess the security implication of a pardon that the IPUB leader is ever convicted. Hello, hope you enjoyed the news. Please do subscribe to our YouTube channel and don't forget to hit the notification button so you get notified about fresh news updates.


Ahoy, ahoy, and welcome to the video. Today, I'm going to do a reading of SCP-3691 by Fishish. You can find it on the SCP Wiki in the link in the description below, and it is under a Creative Commons share-alike attribution license. Item Number SCP-3691 Object Class Uncontained Special Containment Procedures All known footage of Incident 3691-A has been recovered by the Foundation. Access to the event have been anesthetized per Standard Foundation Protocol. A non-anomalous replica of SCP-3691 has been placed at the site of the discovery. Description SCP-3691 is a bronze casting of Augusta Rodin's La Pencière, which was situated in Albany, New York until June 13, 2022. The statue's creator is unknown. No records exist of the statue being commissioned or built since the city was founded, and no plaques of any kind are present on its base. SCP-3691 had been, since the time of its discovery, unable to be moved or damaged in any capacity, although the statue was found to be physically immovable. An early hypothesis, as to the nature of the anomaly, was that the effect was memetic. Attempts to move the statue induced frustration, and all individuals assigned to the task. This hypothesis was abandoned when no evidence of any secondary memetic properties was found. SCP-3691 displayed a significant deviation from its established anomalous effects during Incident 3691-A. Incident 3691-A.  The following interaction was recorded on camera. Begin log. SCP-3691 becomes animate, the statue stands up and looks at the camera. SCP-3691 size. SCP-3691, sorry, had to think about some stuff. SCP-3691 walks away from the frame. Following Incident 3691-A, SCP-3691 was reclassified as uncontained. Efforts to locate the entity are underway. Thank you very much for watching. If you enjoyed the video hit the subscribe button and then hit the notification bell next to that so you're notified when I upload new videos. And then head on over to patreon.com forward slash de-submarine and pledge at any level like everybody here on the screen already has, including Centuriki, who has pledged at $100. It is nice to know that I am not alone out here, and I will see you all again on Tuesday.


Hi, this is Uchoal Paradwaj. In this tutorial, we are going to understand how to write the YAML configuration file to generate a documentation website for your project using YADOC. Now, there are other tutorials in this channel explaining the functioning of YADOC. However, this tutorial is specifically designed to define a minimal YAML configuration file for YADOC. Let's get started. As you can see on the screen is my fork of the GitHub repository of Suzy Server. The original repository already used YADOC and has the YAML file defined in it. I've removed that file and will try to create one trying to explain its content. First is the author name, which will usually be the name of the owner of the project. Then there is the project name followed by the version name. The build section is first is the theme which can either be a Sphinx built in theme or a custom pi by theme. It is followed by the source where the documentation for the website is stored and then there is the logo, the location where the logo of the project is located. The next section is the publish section. The section is the publish section which defines the deployment for the documentation website. Here we are choosing GitHub pages and for that we define the doc URL for the GitHub pages. Now that the configuration file is created, let's test this using the documentation generator form at EADoc. After the documentation generation is over, we will see a preview link opening which we can check the documentation that will be generated. Thank you for watching this tutorial. In the following tutorials, we will extend the configuration file to handle some advanced configurations. Thank you.


Ukraine to use Abrams tanks for well-planned operation. Kharilov Budenov, head of Ukraine's main directorate of intelligence, said that Ukraine will use Abrams tanks, the first of which should be in Ukraine next week, for well-planned breakthrough operations. He said this in an interview with The Warzone. Budenov said that now on the battlefield the use of armored vehicles in almost all major fronts is minimized. The reason for this is anti-tank minefields, which are a problem for tracked vehicles and a large number of FPV kamikaze drones that can hit equipment. Now, according to Budenov, armored vehicles are used only for evacuation or for the rapid transformation of infantry teams to a certain place. However, they do not participate in hostilities. Answering the journalist's question about whether, in that case, the Abrams tanks, the first part of which would be in Ukraine next week, will be effective, Budenov replied that they are very much expected in Ukraine. They should be used in a very tailored way for very specific well-crafted operations because if they are used at the front line and just in the combined arms fight, they will not live very long on the battlefield. They need to be used in those breakthrough operations, but very well-prepared, he added. When asked by the journalist whether Budenov is sure of this, given the June situation near Mala Tokmachka, where a certain number of armored vehicles were destroyed, the head of Ukrainian intelligence denied this information. Actually, there wasn't that much material that was destroyed. There was a lot of damaged material, and by now it's repaired. The number of those that were destroyed was not that high, but it's the very example we've just talked about. So, if we just deploy some battalion tank groups into the battlefield somewhere, just as long as it gets under the range of artillery it will get hit, he said.


Hey everyone, you're about to watch a video that's going to get thousands of views. Very cool, but your ad needs to be right here. Give me a call. Kentucky Transportation Cabinet District 12 is preparing for a bridge replacement project at Long Fork in Pike County. The two-phase project will begin Monday, December 11th when motorists can expect some delays and one lane traffic for the time being. Construction is expected to be complete in Spring of 2024. The intersection of 1469 and 3414, they're on the right fork of Long Fork. So that bridge is going to be replaced in two separate sections. So we're not going to have a separate detour. People are actually going to be able to travel along the bridge. We're going to work on, we're essentially splitting the bridge in half. So we're going to work on replacing one section of the bridge while traffic flows on the alternating section. And then we're going to flip-flop that. And so once this section is complete, we'll change the traffic patterns. Traffic can flow on that section. And then we'll work on demo and repairing and replacing the opposite lane. Peak traffic times will likely cause some delays in the mornings and evenings. As always, be mindful of any additional signage and construction workers when in the area. There's peak travel times, and people know when those are. So those are in the mornings when we're the commute, when the kids are on the school bus, when you're trying to get to work, and in the evenings for those commutes. When you're coming home, when the kids are on the bus, those times. So during those peak travel times, traffic could get backed up. And so I like to remind people, be aware of those times.  This is going to be ongoing for quite some time. So know ahead that traffic is probably going to get backed up during those peak travel times. And just go ahead and plan to give yourself more time during those times, especially. Reporting from Out in Top News, I'm Nick Colum.


The President of the United States and Mrs. Reagan. We share with the President who quit disbelief. You and your countrymen have worked hard in realizing what you referred to as I've already been well-being. As we welcome you once again to our country, you note that many changes have occurred since you were last week. We can face the future with confidence. Our two people share a desire for liberty and progress that can and will triumph over adversity. We'll do so as friends. Mr. President, welcome. Security and individual freedom. The President has very often had access to this same element as the time of his representatives. Mr. President, I have received enthusiasm from the people of Ivoire, his economic vitality, the state of Croquet and other personalities. They have all been struck by the demand for what you have. You and your countrymen have worked hard. Thanks to this fair response for their work and recognizing the importance of this sector of your economy, Ivoire can be proud of their solid economic progress and independence. More than that, we welcome you once again in the United States. We note that many changes have occurred since your last visit to the White House. We can face the future with confidence. Our people share a desire for freedom and progress that can and will triumph over adversity. Mr. President, welcome. Mr. President, I am very honored. You have told me everything about myself, and I am proud to be able to tell you that I am interested in knowing your opinion. Like him, attached to the eternal values of peace and justice, the ties that have developed between the United States and I have been crossed over more than 20 years. I extend them to you entirely and look forward to being able to discuss with you the problems that concern me and which I cannot conceive of being resolved without effective enforcement. I thank you most warmly for having afforded me the opportunity once again to greet the American people through you The ambassador to the White House. I remember. Souvenir. Souvenir. The word is I can't remember what you learned. Souvi.. Souvi.. Souvi.. to find out where was the desk cafe and as I kept thinking about it words began to come back so I met a gendarme and I said pardon me she's a grand fan what am I a cafe and he told me and then my friend who was driving said what did he say and I said I haven't the smartest idea your decisions and start modifications