 the summer school, I think. You should be very proud of yourself to be in this intense course, courses, I mean. So let me, I said yesterday that it should be easy to prove that for, if the underlying state is product state, then the twist product, the correlation function that looks like, I'm sorry, the function that looks like a correlation function, but based on not the usual product, but based on the twist product, we'll vanish. It's not too trivial, so I thought I would explain with the proper reference. You can read a little bit general situation in the paper of mine. So let's just remind you, we ask the question whether the Torico state on a sphere, which is non-degenerate, I mean, there's no encoded qubit in this situation, can we generate this state using unitary circuit where the circuit is a small depth? The answer is no, it takes a linear depth, and the reason was not so trivial. All the previous techniques just failed, so we had to introduce a new one, and the one ingredient was to think about the twist product. We managed to show that the twist product is covariant under this small depth quantum circuit conjugation, and you really need this. So let me just explain the lemma. Suppose we are working with a product state, and there are two stabilizers, not PAUI, not potentially non-PAUI, stabilizer, stabilizer just means that if you multiply P, then it is just a eigenstate with value one. With value one, and they are supported on the annulus region I drew. So the P is supported on this annulus, and the Q is supported on that, the other annulus. And the second condition is somewhat nontrivial. I only want to think about the stabilizers, non-PAUI, possibly, that are themselves a shallow quantum circuit. The in the Torico case, the string operators, one particular stabilizer, PAUI stabilizer, happens to be a quantum circuit of depth one. But I want to consider everything, but under the condition that it's shallow circuit. Then you can show that this twist product, no matter what P and Q are, as long as the three assumptions are made, then this must be one. Individually, without this term, then P is a stabilizer, so the correlation function, the second term in the correlation function is one, so they will vanish. So why is that true? So here is the argument calculation. So yeah, this is the expression I want to evaluate. And observation is that Q itself is a quantum circuit of small depth. So it has a small light cone. And the product state is entirely specified by the local projections. If I tell you, oh, in this state, I don't know how entanglement it is, but in this state, this qubit is in the zero state and that qubit is in the zero state and so on, that fully specifies the underlying state. So if I look at the bigger circle here, and I know that if this part is product state, underlying is all zero, and then I apply Q, it may not stabilize the underlying state because I only know that the underlying state is product state on this small region. Outside, I don't know. But still, it is a small depth quantum circuit. The observable sitting at the smaller circle can only be affected by the larger patch. So in particular, if I apply the operator that tells me, oh, this middle qubit is in the zero state, that projector will still be telling me after the application of Q that the underlying state at that position is in the zero state. And this argument applies everywhere. So in other words, I can pull up one projector, a single qubit projector acting on a single site here and bring it closer to Q, even though P and Q are in a funny linked diagram product. So you pull up all those projectors, then from the viewpoint of Q, it only sees pi. It has all the information to stabilize the underlying state. So Q will evaluate, will be absorbed into this product state, but there, it's just identity. So it disappears and you are evaluating P. So, well, P was assumed to be a stabilizer, so it vanishes. Well, the value is one. Okay, so the lemma is proven that way. The second condition was crucial. Now, how do you use this for the sake, for the complete, completing the argument for the complexity? You remember the identity, this. So fixing the state, I changed the observable. And if P and Q satisfy all these three conditions, it will, the evolved, time-involved version by some small depth quantum circuit U, which is unknown to me, will still be satisfied these conditions on the evolved state. So with respect to that state, I have another set of those. And for those, I still know that it has to be valid to zero twist correlation function. And it should continue to hold as long as I have this shape. And this shape fails to be true when the circuit depth is so big that you no longer have a separation between these supports, and you no longer have a separation between these two crossing points. So that's the, so you, since you're considering all possible operators, and then you show that it's vanishing, finding one non-counter example to this, shows that under my state cannot be this. So that's why you prove the inequality between the potentially evolved state from the Tori coast. Okay, any questions? YQ, what? YQ pi is equal to pi here. Oh, so, so by, since Q is looking at only local patch to determine the local projector, I could pull up local projectors here as long as Q is bottom of the P. If Q is above P, then I can pull up the projector from this side by the same argument. So you eventually collect all the projectors that will ensure that that will suffice you to evaluate the Q value. The geometry is used. I am using the fact that if I look at this region, then I only see Q. If I look at this region, then Q is sitting on top of P so I can pull up the projector from the left side. If I look at this side, Q is below P so I can pull up the projector from below. So let's begin the fourth hour. The topic is tri orthogonal codes. So let me begin with the super basics remark. Why people care about Clifers plus T architecture? There are many universal gates set and the Solovec type theorem says that oh, any single cubic unitary can be synthesized by some pretty much arbitrary generating set. That's fine. And as long as we have a complete SEO2 unitary group and some entangling gate, we generate everything. And the overall gate complexity only differ by polylog factor. No big deal. But why do we care then about Clifers plus T? It's because of the fault tolerance. For other schemes, it's highly non-trivial how to make sure that every operation is accurate. But for Clifers plus T, we have developed a full theory of polystabilizer code to implement the Clifers part reliable. And then here's the non-trivial idea by Manik Neel and Sergei Bravy and Alexei Kitayev pioneered that we can distill one non-trivial non-Clifers state with which we can complement the Clifers operation to make a universal case set. That's the reason those T is important only because of the fault tolerance. So T state is nothing but this. So yeah, this benign looking single qubit state is acting behind all the quantum advantages and interesting phenomena that you can see behind the quantum computing. So let me remind you how to use this. You start with the T state and then you bring out some arbitrary data qubit and then you measure C and Z. And then you measure X and X. And depending on the measurement outcome, you can do some Clifers operation. And the result is that psi is as if the T gate was applied. The calculation is not that difficult, so let's do it. Then you don't have to do it again, ever again. So T state is here. And say arbitrary qubit state is here. And it suffices to consider a single qubit because by linearity, even if the single qubit is in-tang with something else, the result will hold. If I measure, say, ZZ outcome was plus one and I will be projecting onto a component that looks like this. So alpha 00 times beta. And then I measure out the second qubit, the ansula, in the X basis. And suppose I get plus one result. Then plus state has the same overlap with zero and the same overlap with one. Plus is. So the amplitude are exactly the same. So if you project onto the plus state, then they're just gone. And you're left with the correct face vector inserted at the correct position. And it's left to you to figure out what happens if I had a minus output or if I had a minus output there. One remark. These measurements are reversible measurements in the sense that you don't ever destroy your quantum underlying quantum state. And if you want to go back, then you can. This is another instance of the exercise we had around that board before. If there's a stabilizer group, if there are some non-logical operator, then exactly one generator is kicked out and a new measurement is coming in. The same thing is happening here. Everything is measurement. So now this is thrown away and this goes away. And there is some classically controlled Clifford unitary that you have to apply depending on the measurement. Now we want to talk about the coding. So we want to talk about how to distill a TKT states. And for that matter, let's think about the noise model first. The T state is a one state in a two-dimensional QB table space. What's the orthogonal state? It's going to be the Z times T because, yeah, they are orthogonal. So if I write down the most general state that is possibly noisy in this basis, then I would have a majority of weight on this true state. And by trace preserving a property, I must have epsilon there and something there. Potentially, the density matrix would look like this. And here comes the neat trick of Clifford twirling. So I'm building up the motivation why we consider a specific class of codes and this is the reason behind. So Clifford twirling maps this density operator down to diagonal density operator. How do you do that? Well, the channel is this. So this is a concrete channel that you actually want to implement. So with the problem that you have, so you are implementing your quantum computer and there is a completely random classical beat. So if you have a classical circuit, if it heads up, then you do nothing. If it's down, then you apply this unitary. But what is this unitary? This is a stabilizer, non-poly stabilizer of the T state. This was, so you can directly see why that is a T state. And this is Clifford. For the sake of completeness of this course, let me just briefly tell you about the Clifford hierarchy. So it's a hierarchy. So I'm building up a set of unitaries in that fashion. Level one is just nothing but the Pali group. That's level one. You may want to call it level zero, but there's a good reason that is to start with the one. You will see later. And the level K is inductively defined. Level K is a set of unitaries on whatever number of qubits you have, such that for all Pali operators, you're conjugating P results in the level K minus one set. So you pick up a unitary from here, you conjugate any arbitrary element there, then the level goes down by one step. So at level two, this is Clifford. And level three and so on, and it goes all the way up to infinity. One remark is that only level one and two are a group. Everything else is just a set. Why? It's a simple argument. So if it were a group, then it will in particular contain this particular element. This is Clifford, you can calculate. And since you have a Clifford and T, you can generate a dense subset of SU2. And it's a group, which means you can approximate arbitrary elements of the SU2. Right, but that's not going to happen because these groups are discrete. Yes? Oh, yeah, yeah, yeah. This is level one. So level one is distinguished. This is a starting point. Everything else just follows from it. So that T belongs to level three. Okay, so I left you an exercise to show that actually this parallel Clifford-Torling is automatically done by the injection circuit. That's a nice feature about this. So you don't ever have to, in practice, have to implement this. The randomness of measurement automatically does the job of the flicking coin for you. So then my effective reduced dense matrix for the possibly noisy T-state is either the perfect state or the Z-contaminated version of T. So all I have to do is to catch those errors. And for that purpose, a one infinite family of codes is known, which is called the triorthogonal codes. So before I define triorthogonal codes, I'm going to define triorthogonal matrix. It's a binary matrix. So there are K rows and S rows. For the convenience, I just separate it into two pieces. And the number of columns is going to be N, the number of qubits. And a binary matrix is called triorthogonal if the following is true. These rows have alt-weight. These rows have even weight. That's the first condition. The second condition is self-orthogonality. If you pick true rows, say VA and VB, the superscript denotes the row index. That's my convention for this hour. And you take everything, you take a dot product, then this should be 0 mod 2. Just a usual self-orthogonality over the binary field. The third condition is interesting. Some of our J, VJ, A, VJB, PC, it's a triplovelet. You take a bit-white end of the three rows, all distinct, then it should be 0 mod 2. So this is another instance where 2 is very different from 3. Well, like 3-set is empty and complete, but 2-set is easy. 2-local commuting Hamiltonian problem is easy. 3-local harder. And so on. And this is another instance. And, well, to be frank, calculation with a... Oh, yeah. This is definition of tri-orthogon matrix. Now, tri-orthogon code, on the other hand, is defined by the tri-orthogon matrix by declaring that, oh, my B string on this row is going to be my x-logical operator and z-logical operator. That makes sense because they are all dweights, so their overlap is always all dweights, so they anti-commute, but different rows will commute because of the second condition. And I declare that this row, each row here will generate x-stabilizer. And then the z-stabilizer will be just the orthogon complement of all the rows here. That's the maximum choice you can make. That's it. Usually, when you talk about poly-stabilizer codes, there's a freedom for you that you can choose the logical basis. It's up to you what operator you call it, logical x. But here I am specifying that particular logical operator. So, yeah. The corresponding CSS code is called a tri-orthogonal code. Calculation with the tri-orthogon code is not easy. In particular, how do you show a certain code that admits a transversal T? You are no more in the poly world. You have to deal with all the phase factors and signs and so on. It's difficult. So, there's a subclass of tri-orthogonal code that I am going to focus on this hour. And that is called the... This is my terminology. Let me know if you have better one. Level 3, be visible. So, this one has a cleaner definition, at least. So, I'm going to consider a subspace of some number of bits. It's a vector space of dimension M. I'm going to consider a subspace. And I'm going to pick a vector whose entries are plus minus 1 or plus minus 3. Okay, that sounds weird. Why would you care 1 and 3? But you will see why. These are the odd numbers, module 8. So, with respect to this T, V is called level 3, be visible if whatever factor you give me, I enumerate the coordinates 0 and 1. And I insist that these are integers. Mathematically, that's a weird thing to do. F2 is equivalent class of integers, module 2. So, formally, it's all, set of all odd numbers or set of all even numbers. Here, I insist that the set of all even numbers is represented by integer 0, and set of all odd numbers is represented by integer 1. So, I promoted the quotient, the equivalence class into the vanilla integers. And then I enumerate these T-vector T1 and tm. And I take a dot product. So, remember this component is an integer. I sum it over, and I insist for this definition that it is 0, module 8. The level 3 comes from the fact that 8 is 2 to the 3. Okay? So, and it's left to you to show that if you have any generating matrix for V, then that generating matrix is an instance of triarthogon matrix. So, let's build up. Well, now, any level 3 divisible code gives rise to the triarthogon matrix. So, I can think about the triarthogon code. But it's boring code, kind of, because any odd weight rose. Well, this equation tells me that, well, T is always odd. So, if I take the equation, module 2, then it just means that everything is even weight. So, there's no encoded qubit in the corresponding triarthogon code. But let's study that trivial space first. So, I'm going to show that the that zero, that code that encodes zero logical qubits will have some unique code state. And I'm claiming that, so there are m qubits. And according to T, I apply T to the power there. So, if T1 was plus 1, I apply T. If T2 was minus 3, then I apply T to the minus 3 and so on. So, every component, since T are all odd number, every component is some keyword matrix times T. Okay? So, this is my my thing. I want to show this. How can I do that? Well, if you are too confused, always you expand it. You know, a more familiar thing. Let's start with the all zero. Well, if I have all zero, then all the Z stabilizer will be satisfied. So, I don't have to worry about that. Then the code state projector onto the code space will have form like this G, where G is the stabilizer group consisting of all X operators properly normalized, but let's not worry about normalization. So, this one will make we will turn this zero into the bit stream specified by the bit vector element G. And that ranges over all the level 3D visible space V. Right? Now I apply T. Okay? Then what happens? T is right. T is inserting this face vector whenever it says 1. So, and T to some power will insert this face vector to that exact same power. So, here the face vector we pick up will be the i pi over 4 T i V i. That's the face vector I pick up. Now, here the weird promotion is in action. V was a binary bit, but when it comes to this equation this is a literal integer. Right? Now, okay. So, yeah, pi should have been 2 pi in all literature, but unfortunately history is not. So, 2 pi over 8 it is. Now, what's the condition for the level 3 divisibility? This sum is 0 mod 8. So, this gone. Yes, it's 1. So, this equality is now now let's do well, this is perhaps amusing to you. Yes. It is a non-clifford stabilizer. Not even a power, of course. Now, let's imagine a particular, well let's assume more towards that V. Suppose I have a generating matrix for the V, where I mean put differently, let the row span of a binary matrix be the V. Okay? With respect to some T here. Some T vector. And suppose I have some 1's. How many? Say K. And I think about the block decomposition of this matrix and suppose that I got 0 and some other bit strings in here and there. This is not too restrictive because whenever you receive a generating matrix you take a Gauss elimination to find the row echelon form then you will typically have this identity. Up to, well you will always have this identity. In this form up to row column permutations. Clear? Now I want to interpret this. So, let this collection of bits be say, I don't know A piece. And let this be B part. So set of the qubits according to the columns of this matrix. Then, what's my the code state there at the side there as a bipartite state between A and B. It has a stabilizer. It is uniquely specified a stabilizer state set of stabilizers that is specified this matrix. What does it do? This is an X operator so I'm going to call it X1, X1 bar. No reason, no particular reason I just insist that I just write it this way. So A and B and so in general for I that much and some identity acting here and some something in B B piece. And also there is a Z type stabilizer because my orthogonal I insisted that I take the orthogonal complement for the Z side. So, in the stabilizer group for this state I also have ZiA and ZiB bar. I wrote it bar but it's really simple. Whatever operator you have on that part. Is that clear? Now let's interpret again without calculating anything. If I just delete these number of columns and just look at this then all these conditions are satisfied. The identity matrix part if you take a bitwise overlap they will just cancel off. There's no repeated ones along one column. So self-orthogonality is okay trial orthogonality is okay. The only exception is the first part. I demanded the first K be odd weight. But if I just look at the right hand side these consist of odd weight rows. They consist of even weight rows. So this piece precisely defines some trial orthogonal code in coding K qubits. And XI bar is precisely the logical operator of that trial orthogonal code and Zi is also the logical operator. Everything else is just a stabilizer acting on this B part only. There's no support on A side. So this state is nothing but some K number of unprotected qubit and K logical qubit in coding in a trial orthogonal code forming K bell pairs. Clear? No, the other way around. The top part is odd weight. This part is either. Odd minus 1. Right, so yes. Oh yeah, thanks. So yes, so this thing is like we had a K qubit and some encoded qubit. Everything is now maximally entangled with a some well with a specific logical qubit in that trial orthogonal code. Now this calculation says that if I apply T to the T1 and T to the TK and some transversor operation T, we'll say tilde then I get invariant state. Now, forget about what happens here. When does bell pair remains invariant when you apply it by a unitary that is bipartite? When the right, so if this is the K, if this equality holds then V transpose must be U inverse. Did I get it right? Yes. So basically the same thing. So in other words the transversor T applied according to the prescription of my T vector, small T vector is enacting precisely the opposite of this T operation which is evidently applying T gate on a particular qubit. Therefore I conclude that if I take a level 3 divisible code making a generating matrix to arrange the rows and columns in such a way that there is an identity matrix on the top left corner read off the right block construct a triathlon code then the resulting code admits a transversor T. Yes, that's the next thing I'm going to say. So, remember you know, due to the Pauley, sorry the Cooley-Fortorily our error model is purely Z. All the Cooley-Fortorily operations are assumed to be perfect but the goal is to catch the Z errors upon the acting this T's. So, whatever error you may have here when you apply the T gate, well T piece will just pass through. It implements something useful for us. Z part remains and that's exactly the same as if you had a code and Z errors are happening. So if you just do the usual error correction procedure for the triathlon code, you will be able to catch those Z errors. All that matters in this business is the X stabilizer because they are going to catch the Z errors. So, if the triathlon code has a large Z distance, then you're in good shape. So, well I just spoke it in words but so yeah, KQ bits apply an encoding map into a triathlon code. Well, we only verified for the for the slightly small class where it is coming from the level 3 divisible apply transversal T and then inverse encode. I say inverse encode instead of decoding because decoding is somewhat differently in the quantum computer literature. Decoding usually means that you only correct errors, but I literally mean the inverse unitary that you, inverse isometry that you have here. You do it inverse then you're left with some K bits and some syndrome Q bits and if they are reporting no errors, then you use this then that's the distillation protocol. So let's now examine one important example that is of that comes in an infinite family and that is the color code in three dimensions. Yes. Oh, so let me emphasize again. All Clifford operations are assumed to be perfect. There's no error is going to occur in this encoding stab or there. The only possible position where error might occur is in this transverse of T insertion because you are going to use the faulty gate or faulty T state and there because of the Clifford twirling technique we may assume the error model is purely stochastic Z error so all that remains is to catch those Z errors and upon if the syndrome says okay then just you can use that and with the you could correct but there's little motivation to correct because since everything is perfect except for that Z error, rejecting gives you a higher quality and rejection probability is not that high well, you have to work in the region where rejection probability is not high except this is high. Yeah, I'm sorry. Rejection is low. No. Because we don't care. Right. Clifford operations are assumed to be perfect so we don't care about that. This is an example and here comes a sum of topological elements color code in 3D let's assume that we have a tessellation of Euclidean three space. We do have Z stabilizers Z stabilizer is assumed well Z stabilizer is whatever it can be it should commute with X logical operators and it should commute with Z stabilizer so that's just the kernel of this matrix as an operation from right to left so everything in that kernel is declared to be Z stabilizers. Yes. So all the data is in that matrix. So you pick your favorite four colorable tessellation of R3. Four colorable tessellation means that you are filling your three space using volumes that are touching each other and you're filling a number one out of four in such a way that no neighboring volumes have same color. That's all. Existence there exists and there are some general considerations we can talk about. If you are in the problem session like two times ago there's a working prescription you can take and then the definition of the color code is that I assign X stabilizers for every three cell the cubits are on vertices so the colorable tessellation gives you a cell structure, there's a volume, there's a phase, there's a line and a point every point you put a cubit and X stabilizer is going to be the product of X around all the vertices for a given three cell and Z stabilizer is assigned to every two cell. This is the definition. So now oh there's an interesting calculation that I thought I would do but I don't have time so let me just state it. I give you the definition of the visible code in terms of well for every elements in the vector space this dot product must be zero. How do you check that? Efficiently. That's in general a hard problem but because we are restricting the level it becomes an efficient algorithm. Okay, so let me state that proposition. So again on the generating matrix first condition T dot some row A is zero mode eight VA is a one row here number two T dot VA dot VB is zero mode four again I'm using the previous promotion any binary bit is promoted to the genuine integer number three T dot is mode two. So you see the clear pattern every time you add in more component to VB add in you reduce the modular thing by vector of two. So well there are only row worth of this conditions to check. Row squared, row cubed so this is efficient. Yeah in the in the lecture notes there is a calculation to show why this is the case. So let's apply this criteria for this 3D color code. Okay, so first thing we have to define we have to find the T. How do you define T? One observation well crucial observation is that the vertex set of this four colorable tessellation of three space is bipartite. When is graph bipartite? To general fact when is bipartite? If every cycle is even length then it's bipartite what is the cycle in this lattice? It will go through some edges and go here and there and come back. Right, so it actually depends on the fact that the first homology of my R3 is zero. Therefore any closed loop is coming from the boundary of two-chain and two-chain here is some two-cell so it suffices for us to check that every two-cell comes with an even number of edges but why is that? Every two-cell is a boundary of a three-cell so like some two-cell is a boundary of a three-cell now this phase is a unique joint boundary between two exactly two because of the colorability of two different colors now an edge is going to be the intersection of three three-cells you think about you're blowing the bubble and you make three spheres and there's a vertical line that goes on so one edge is going to be joint boundary of three exactly three-cells but how many how many different kinds of three-cells can I bring to see an edge? Here one color is color A is fixed here color B is fixed there are two remaining colors because we're working with the four color of the lattice and since no same color can be repeated as you go along this loop the color of the remaining two must alternate so therefore this is even length therefore the graph is bipartite so now the T vector is defined by that bipartition on the one party you assign plus one, the other party you assign minus one we got the T we can check this so T the third condition is that T dot VA now VA is the this one it's a three-cell and we have to show that the dot product between the T and V is zero I'm actually going to show that the stronger statement that for every vertex on a three-cell there is an equal number of plus one and minus one why is that? here's my three-cell there are some two-cells going on and this three-cell itself comes with one color therefore the two-cells on this three-cell is a three-colored tessellation of a sphere so you can think of it's generally the general situation this thing is repeated it's three-colored now pick one color say one and think about the collection of all a union of color one cells so color one is here and maybe here they cannot be adjacent because of the colorability and there could be here and so on and this union contains well gives a partition of my vertices there's no repetition and every vertex belongs to at least one a well obvious from the picture but for every plaquette it goes along the one-cycle there's an even cycle so there's equal number of plus and one and minus one therefore on the whole three-cell there is equal number of plus one and minus one okay great what about the second means that I should take an intersection of two three-cells which means I work with the two-cell two-dimensional cell but we just argued that there is equal number of plus one and minus one great what about a triple intersection when do you have a triple intersection of three three-cells if they are far apart they cannot intersect if they are very close together they are all along an edge hmm sorry they share a vertex but they also share an edge it's a three-color rule that is but the edge consists of two vertices one is plus one the other is minus one okay great so yeah by the choice of T according to the bipartition gives us the color code to be the level three divisible one example actually comes with the boundary so I'm going to draw all the edges in white if they are on the if they are on the face of this tetrahedron and red if they belong to the if they are inside like this so there are four three-cells it's bipartite there are exactly 15 points and you can assign T as according to the prescription and this is divisible at level three and this is the the famous code 1513 triothalbina there are nicer pictures than my drawing on the internet so go ahead and look it up it is useful for magic state distillation magic state is a example of magic state and distillation yes this is the distillation procedure usually it's used just for that it's largely unexplored for non-triothalbina general polyestabilizer codes well now we have good LDPC codes triothalbina codes we don't even know if there is a good triothalbina codes I was yeah let me use like a five minutes to do some theory piece we are actually back to the first hour I'm going to show oh yeah first we worked on the three dimensions you may wonder that's too high so you may wonder can we find something nice in the two dimensions let me understand why here we are working with the polyestabilizer code in two dimensions local the usual sense of my four hours and assume the most strong error correcting criteria that any disk like region is correctable in particular this entire disk on the torus is going to be correctable now since we are using the periodic boundary condition I can redraw the same figure it's like this and sliding a lot a little bit further it would look like this clear slide this thing over here using the periodic boundary condition and slide this thing over a little bit okay so the black thing is correctable now using the cleaning lemma remember if there's a correctable region on some subset of qubits then you can find a complete set of logical representatives on the complement we proved it by the dimension counting so in particular all the logical operator can be found here I'm not assuming it is the torus code I'm just deriving it under that error correction criteria okay now let's draw it a little bit bigger we found a union of two strips that supports all logical operators and well the argument is topological so I could have another region that is a union of two vertical and horizontal strips that supports another complete set of logical representatives now let's think about whether we can do the similar thing as here we apply some transversor gate it's unitary in particular it is a quantum circuit of some small depth and then hope that some interesting is induced at the logical space now after the application of that small depth quantum circuit transversor operation assumed to induce logical operation in the logical space my my red strip will fatten the post transversor gate I have a complete set of logical operators on that slightly thickened region that's no problem so all I have to do to figure out what the induced logical operation is I only have to look at these strips and the fattened operator and let the evolved operator be U and let any logical operator support it on on yellow Bp well I can use the chalk and now consider this group commutator well U is itself a quantum circuit the starting point was operator conjugated by some small depth U itself is a supported on the red region P is also a now look at this region or look at here the red thing appears here and there twice so by the light cone canceling argument anything that was supported here will cancel off what about here the same argument they will drop off from P so I end up with the operator if I take a group commutator to be an operator supported on the crossing points so sum operator A tensor with the sum operator and B but we've done this many times it's not this like region far separated they are both correctable simultaneously so and we know that everything here is a logical operator P is chosen to be logical U is assumed to be logical because it is a result of the small depth quantum circuit deformation that we want to study it's induced action so everything is logical supported on two disc like regions that is correctable so all it can be just a scalar what was P P was arbitrary poly logical operator now let's work exclusively on the logical space so it's a finite QB problem if U is a small I'm sorry if U is an operator it will have some poly operator expansion because poly operator is a complete basis now this equation says that if I conjugated U by P then it says it is a scalar multiple of U how can that be possible so P U well poly operators commute up to a sign that potentially depends on J Q are complete basis so the coefficient must match in particular sign cannot depend on J must be uniform but I could choose P to be some poly operator that only selects particular poly operator Q so I could I could choose a sign to depend on J if there were at least two terms in this sum so for this equation to make sense the J can only assume one value and P U P is equal to some yeah I'm sorry U consists of single poly operator some Q but what does that mean U is a result of the logical operators evolution my evolution did something it transformed the poly operator to a poly operator U is a unitary U is a Clifford that's the Bravais clinic bound so in two-dimensional poly stabilizer code if the code distance is sufficiently large then you can only have Clifford operation implemented using shallow depth physical operation you can generalize the higher dimension and the result is that as you go up the dimension this becomes the U becomes the induced operation becomes an element of the dimension level of elements so yeah and this is one of the reasons that dimension I am sorry the level must start with one so that it matches beautifully that's all I have to say is there any question yes yes yeah this is the proof ah ah ah well so the keyword that you want to remember is to take the group commutator because we are considering the poly operator and its conjugation by small depth quantum circuit all the representatives will themselves shallow depth quantum circuit so as you take the commutators except for the possible crossing point everything will cancel off it will reduce the dimensionality until you reach the zero dimension then you are left a scalar and if you just keep track of what has happened in that induction you end up with the Clifford hierarchy definition so yeah I have used some knowledge from the first class so it's all connected and even in this very very algebraic looking stuff if you start looking at the topological codes the dimensionality starts kicking well thanks a lot for your attention and we enjoyed it