 So yeah, let's start by reviewing yesterday First there's a correction in the in the in the next for those you you can access on the web page There's a typo on the last digit of this archive number that lets you Wrong paper Not that paper is wrong It's me if it's not a paper that I intended you can also look at my notes on the archive That that basically contains all the idea that I described yesterday with the with a little bit more technical detail So the today's topic is the homology and the elitist relation to error correcting codes and Later half I will focus on the Entanglement the rng transformation which is which can be think of it as a some encoding map into your code Yeah, we'll get to that and in the meantime We will be basically delivering the main idea why homology is a topical invariant So you may find it amusing that it's a completely a pure math question is now proved along the lines of coding theory So but let's before we begin. Let's just briefly review what we have done yesterday. We do it we derived a quantum singleton bound for the case of the Howly stabilizer code the code length number of number of physical cubits number of encoded cubits and the code distance are Related by this inequality. The main idea was for this proof was that The major step was to show that the number of encoded cubits is less than the entropy of the region That is a complement of the two individually correctable regions in the key Concept or equation that you I encourage you to remember is the code projector is a is a The projector onto the The space where G assumes identity where there's some ranges over all elements of as so it is indeed a projector And the correctable region means that whatever operator you give me if I sandwiched by the code space projector Then it becomes a scale of multiplication operationally whatever you want to do on the code space by acting on a correctable region all you can do is just a scale of multiplication So operationally it does nothing And I omitted for the time sake One important bound that I wanted to discuss That proof is very simple. So I just wrote it here and let me go through the argument The goal is to prove that in a geometrical local codes on the Euclidean space of dimension capital D The coats coats code distance can be only as large as the co-dimension one object volume So in case of two-dimension it can only be linear dimension in three dimensions It can be a surface area that always code mention one The argument is entirely one dimension It applies to arbitrary dimension because we can always slice the Euclidean capital D dimensional cube into slices of L to the D minus one worth of slice and that each slice is laid down like this in an interval fashion The universe thickness is order the locality of the stabilizer generator So it's a constant number. So I'm some fixed number and If this D is greater than that number Well, in particular if each interval individually was correctable on its own then Because of the locality that their union must be correctable as well Whatever logical operator you give me that are supported on the union I can test the commutativity with this stabilizer group individually and that should also be true So individually that must be trivial the combination must be trivial as well So the union becomes Correctable the other union must be also correctable, but there's nothing else beyond that So applying this inequality that number of encoded cubits must be at most the entropy of the complements of the two Individually correctable region we got nothing. So we can conclusion must be the cake is zero Including no no logical cubit. So whenever you have a non logical non trivial error correcting code You're very assumption the very beginning assumption must be violated, which is that the code distance at One of the at least one of the interval is not Correctable. No, it does not assume that it. Well, so Okay, forget about the capital D dimension just focus in the one dimension where there are only from some firethly many cubits per practice interval if I assume that the Each reach well the only assumption I am I'm using is that any interval that's whose length is larger than o of r Then it's correctable conclusion is that there's no encoded cubit Okay, let's let's Look at the homology man So I assume no knowledge about the homology And I while preparing this I realized the historical amazing coincidence The beginning of the 20th century was the era of invention of quantum mechanics homology Has a similar feature in the following sense. So let me start with the chain complex It's called complex so it might look complicated, but not It's just a series of linear spaces indexed by the integer there may be a Some infinitely many linear spaces you want to consider. Oh, yeah, some some ground field is fixed It doesn't have to be filled, but for the sake of simplicity. Let's assume field And there's a this is the linear space and that by error. I mean some linear map historically it is denoted by this boundary symbol because it's called boundary map and and there are chain of maps and it's called chain complex if The composition of the two consecutive maps Is equal to zero so useful picture is That you think of the linear space as a blob and then So each a C comes with a blob and the linear map will map this into some subspace like this So the image is a subspace of the next linear space The composition being zero means that if I do it twice then it should map to zero The actual map from here to here will be mapped along like this So you have this like a like a telescope Looking figure and that's the that's the pictorial understanding of this algebraic homology of the chain complex is defined by I Should write it here because it should be a lot. No So I define a quotient group So this is called a homology group Define for each index J So at this position what you look at is you look at the kernel of this map So it will be some some subspace that looks like this and By assumption the image of the previous map will we land on inside it somewhere here Because you know it I if I do go twice then I should go to it here So my pre-image must be my image must be here and I and I consider it the quotient Which means I consider only this piece That's called the homology. This is a purely algebraic definition There's no topology or anything Yeah, so it's called complex because it's got with a lot of data That's I think that's why it's called complex, but this is this is the homology And it's got invented in a topological context How and there there comes the correspondence with the quantum mechanics is Let's imagine a circle I can draw a circle with the two points with the two arcs and the one arc is like a interval and I late label it by V1 and V2 and Let this edge be V1 and V2 No problem with that. I Just named it and then I consider this is a this is amazing part to me. I consider the linear span of V1 and V2 and and call it to be C0 and Another another span E1 and E2 so E1 and E2 are just labels. Yes Jane minus one. Oh Yeah, Jake plus one. Sorry. Yeah the previous man. I'm sorry. Yeah, the one one one that comes before Yeah, so Out of nowhere, I just constructed a two-dimensional linear space whose basis is given by The vertices. Why would I consider linear combination vertices? Well in a quantum against you every every day you do that you consider superposition of You know mass being here and versus mass being there and you take a superposition you add them What does it mean even mean to add them? You just declared to take a linear combination here. I do the same thing There's no operation naturally defined to add two vertices together or Multiplied by a scalar, but I just declare that it is done And that's the amusing connection I find between quantum mechanics and the moloch And then I construct a linear map between the two going from the higher dimension to lower dimension how by taking the boundary of this edge So in matrix so my my My space comes with the preferred basis because they are labeled by the cells themselves so e1 Here have a boundary points Boundary points v1 and v2 so To not be distracted all let all the coefficients be z2. We don't worry about the sign so e1 E1 will be mapped on to this one plus one I you have v1 plus v2 And each two has the same boundary points my matrix would be like that So you know this chain complex two spaces one map Is a chain complex because I can always add zero maps At the end and at the front So one linear map is trivially a chain complex and what's the homology at the zero dimension? Colonel of this map, which is everything Modulo the image from the the previous map The rank of this matrix over the binary field is one. So I will have a one-dimensional model here What about this the kernel of this map will be the domain rank? I'm sorry that the division of the domain minus the rank of this map, which is also one modulo zero because there's nothing comes from the higher dimension. So Homology at zero dimension is also one dimensional and Homology at this at the dimension one is also one dimensional. That's it We can do this exercise once again with the triangle Say this f V1 V2 V3 V1 V2 V3 and How can I construct the map so the zero-dimensional map will be spanned by V1 V2 V3 One-dimensional cells are there are three of them e1 e2 e3 At dimension two, how many cells do I have just one f? So I construct a linear map according to our Quote-and-quote intuition tip by taking the boundary. What's the boundary of f? It consists of three edges so I Take the linear combination of the three cells together How about here? It's got three by three matrix E1 is mapped to v1 v2 so 1 1 like that e2 v1 v3 e3 goes to v2 and v3 and Right so Map is going along that direction and it's you it's you know Well, you can do calculation over your head that if I multiply this matrix to that that the result is zero So if I add join zero maps here and there Then the overall thing is a chain complex and you can compute the homology and so on That's the homology. Okay Now you may wonder oh Here's another like a little bit abstract point So I drew figures using points lines and faces and the formerly they call zero cell one cell and two cell depending on the Dimensionality, so there is a collection of those cells which is called state of the complex and Then we constructed a chain complex consists of linear space the span each spans by the cells themselves and Then we take the homology groups one at each dimension now Given a some topological space, whatever you you want to study The the the very prescription for decomposing into cells is up to you It's not a purely given I could have introduced the third self a third point for example Into the increase in number of vertices to set to three instead of two and the number of edges is now become three instead of two so This has no a priori canonical thing it's but Associated topological spaces are something meaningful. I want to study This map is canonical in the sense that if you have a cell structure, then it's there's a well-defined prescription There's only one thing you can do Here to here is completely algebraic. You can let the computer do amazing thing is that if you compose all these Transformations or viewpoint difference here to here is well-defined If highly nontrivial statement There are many arbitraryness going on here, but the initial to final is Well-defined and let's understand I'm not going to prove in a rigorous fashion. Let's understand how it is that Using some of the coding theory. Yes. Oh, this is just a picture I mean, so it is a it is a quotient group So, you know literal elements here does not you know, it's well The elements in this is obvious equivalence class module of this So there may be a two points in this space that are identified because of this The subspace image so in that sense it's not appropriate to draw a diagram for this quotient, but Yeah, it's just a picture Oh Well, so okay, if you if this equation is confusing don't think about the equation just think about this map chain of maps The rule is if you go twice, it's zero That's it. So let's let's not now now let the this Pure math stuff aside and then let's come back to error correction. You have may have worth syndrome errors and stabilizers abstractly you consider well, okay for the sake of concreteness and easy Visualization, let me come to all always fancy CSS where the stabilizer group is decomposed into a purely z part and purely x part So you only have to consider either purely x logical operator I'm sorry both purely x logical operator and purely z logical operator, but again for the sake of simplicity, I only consider Z errors and z logical operators only so in diagram you you have a poly group and if you are designing some code and this poly group now In the you know in my simplification consist of some some some types of products of Zs and Is that's it and If your error correcting code is going to detect some of the errors, then what you In a piece of paper what you do is given an error you find a The the set of all x stabilizers that are anti-committed with the given error So that becomes you know some yeah, this is as a set. This is a space of x stabilizers Well, maybe it's easier to think of x checks Because you're a check whether there is an error or not by measuring the x stabilizer and then on the other hand There's a some z stabilizers They are Polly operators So if you give me a label of z stabilizer, I realize as a Polly operator here If you give me two labels of a stabilizer generator I multiply them together and write down a Polly operator corresponding to the product and Then by definition of stabilizer they commute with the x stabilizers x checks so Anything that comes from picking a z stabilizer label realize as a Polly if I check against the x the x checks It will result in zero So I have a two linear maps that compose into zero So that fits into the The algebraic definition of homolog there. I constructed a part of a chain complex Three linear spaces two linear maps It is left you as an exercise to confirm that these are linear maps over z2 But that should be easy Z-Polly group Okay, let's instantiate this on a torus and and there comes the torque code I draw square lattice because that's the easiest to draw it doesn't have to be a square lattice and then I first consider the checks well by definition checks are associated with each vertex and Given the vertex I write down a tensor product of four Polly x. Oh, yeah, the cubits are laid out on on the edges What I'm defining? I'm defining a code. Yes Okay, why don't we do that with example So, yeah, the definition of the of the Tori code is that you give me a Graph on that can be drawn on a plane and I assign a One X type stabilizer that is a tensor product of X around a vertex On this form one mnemonic to keep in mind is that the vertex, you know There's a crossing and the letter X has a crossing so And if you have a Plot ket then there is a product of these around that block everywhere Okay, that's my definition of the my stabilizer group. So let's examine that map So I kept saying labels What I mean my label in this context is that I consider the plockets Yeah, you you tell me oh, I chose this plocket and that plocket Then my job the job of this this map is to write down the corresponding stabilizer associated with those plockets see Again, you give me some set of plockets. So in the in the formal language, you give me at a linear combination of plockets Which doesn't make sense intuitively that by the way, it's very innovative idea Then I write down the the job of this the first map here is to write down the product of policies around that plocket if there's a if you give me a near nearby plockets then the overlapping Z's will cancel off and This association is Z to Z to linear and the second map Okay, let's forget about the stabilizer for a moment. The second map is that if you give me a poly operator anywhere Oh, yeah, right another way of saying it is if you give me a linear combination of a firethly many edges Then I can write down a corresponding Z poly operator and I test the commutation relation with the vertex Stabilizers that are X type. So for this given Z There are two violations Here and here that are anti-commuting with the given Polly air so Z from Z to these two blue dots is my is the action of my second map Super clear now, let's compose the two maps You give me a plocket and my first map turns that into a product of Z's and my second map Tells me where to put the blue dots when it's where it is anti-commuting with the Z. There's nowhere and That's by design. I Declare that there are stabilizer group, which means they are commuting So whatever plocket you give me if I realize that the poly operator then I check then I test commutation relation against the X type laser it must vanish Composition of two maps is zero now. Let's compare that with the with the homology here so Your drawing is a is a decomposition of your two plane two-dimensional plane into two-dimensional cells One-dimensional cell and vertices the zero-dimensional cell My Z stabilizer is precisely the boundary of a given two cell that matches the conventional definition of Homology from two mentioned to two-dimensional one and the commutation relation turns out that Given edge the commutation anti-commutation occurs exactly at the end points of that So that's my boundary map at dimension one to down to dimension zero so my prescription from Z stabilizer to Z poly to X checks is nothing but the chain complex that arises from the Cellulation of my two plane and I interpret it as a two-cell one seven zero cell That's exactly the same. Yes Yes Even in the usual homology The boundary of a plocket is a loop the boundary points of any segments also can't cancel each other It's the same same thing Word by word setting Okay, so let's now that we have a some understanding of of the air patterns and the check operators in terms of a geometrical Setting let's show that this has large code distance. So you put the periodic boundary condition To make it concrete although I don't really need that let's suppose. Let's show that Say this large region is Correctable, how do you show that I? Didn't I didn't tell you how many qubits there are I just showed one geometric Covering of that region. I want to show that it's correctable. How do you do that? Well, let's apply this condition Or equivalently that Whatever operator you give me there's no it's not going to be a non non-trivial logical operator It is a multiple of a stabilizer Let's focus on Z again. So whatever Z operator you give me here In order to in order for it to be a chance to be a logical operator. It should come up with the average X check in terms of homology that means The Z operator interpreted as a linear combination of my one cells that should not have any boundary So in other words, it is a loop some some collection of loops Like here Now you have a loop on a two-dimensional plane. It will enclose some number of plaquettes But the product of Z stabilizer of that plaquette is precisely the boundary So I have just shown that the Z of how the operator supported on this ball-like region Is nothing if it is committing with this X checks is nothing but the product of Z stabilizers So it's the this entire ball is correctable Okay, now if you are designing a code well, we have checked the we got the stabilizer definition We we know that ball-like region is always correctable and how large can you? Choose a ball-like region on a under the period of boundary condition Yes, I'm I'm only doing for the Z I'm only doing against the X stabilizer and Z errors So just to look for the Z a logical operator, but you can do the same thing For the X by taking the dual lattice, but I'm not going to that So as long as this lob has a linear dimension smaller than the the period of boundary condition Then you're good So code distance must be at least the code the linear system size and that matches In terms of scaling the brahvi-tahal bound So by tessellating torus we have matched in dimension 2 But the question remains when you're analyzing an error correcting code you you need to figure out how many logical qubits there are Okay, let's figure out how many Z logical operators that are non-trivial We we know the formula it is going to be the the commitment of the X stabilizer modulo Z stabilizer But the committance is nothing but the zero the kernel of this second map Z stabilizer is nothing but the image of the first map So we are essentially computing the homology algebraic definition of homology Okay, so here we have a precise interpretation of our code in terms of The homology the topological homology all we have to do is to compute the homology of torus two-dimensional torus It is a well-known easy problem But let me not quote that result and let's get to how we can show that using Whatever tool we have that at now so the goal is Calculate this let me not answer this because that's going to be the Yeah, let's do it in five minutes, but let's let me just talk about the higher-dimensional generalization first So what would you do in the higher-dimensional setting? The lesson from looking at that diagram is that if I have a chain complex Then I maybe interpret the subspaces of the chain complex in terms of these elements Then I'm done, but I'm defining one air correcting code and indeed in high dimensions for example for concreteness consider the simple cubic lattice Like that and I throw in and I declare that my cubits are associated with the One cells the edges Then I have to assign According to that I am motivated to define that that my axis stabilizers are associated with the vertex catching any endpoints of this edge against the Z errors and that's so that leads me to Define X type stabilizer of weight six for each vertex on this hyper cubic lattice and I Want to define the Z stabilizer to be associated with each placket Wait for a stabilizer for each placket. So these are at the two Two cells cubits are the one cell and the X stabilizer at the zero cells So this prescription generalized to arbitrary higher dimensions Starting at dimension four something interesting happened so in in if you're if you are salulating forest space We get five different kinds of cells zero one two three four It's always D plus one and we got The boundary maps that forms a that gives a chain complex when we do the boundary assignment properly And I want to interpret this chain complex as a code. I Could do here. That's the old prescription putting all the cubits at the one cell Zero cells are for X two cells are the Z But I could also do at dimension two Declaring that my two cells are my cubits My boundary of any three cell is going to be the Z stabilizer For each one cell I design X stabilizer such that It captures the boundary of a given two cell. That's going to be my X stabilizer So in literature this choice is called the regulatory code To be more specific some people put it to come a tutorial code But yeah, and this would be like one comma two Tori code Or simply just a three-dimensional Tori code Well in dimension three you don't you only don't have this but you get up to see three and you may wonder Oh, I could choose I could construct my code here or here But it turns out they it turns out that you can actually take a dual lattice and they are kind of the same thing Not exactly the same, but they are kind of the same That's the higher-dimensional Tori code. Yes. Yes More general recipes the following so Well forget about the poly operators and stuff just to focus on the topology here So I have a cell cell structure the axes are at the co well Yeah co boundary of a point that the collection of one cells whose boundary includes that given point In terms of linear maps, that's precisely as transpose of the boundary map you define from dimension one to down Dimensional zero so if you take a transpose of that of that map and Interpret that is that is defined my X stabilizer Then that description taking transpose and insisting on the X stabilizer applies here So if I tell you just one then you wouldn't know how many dimensions that I am in But if I if I tell you one comma two then you can add them out to figure out the ambient dimension three and one means that My qubits are at the one cells in between the accent Yes, yeah, so yeah, right So the first coordinate just specifies the dimensionality of the cell where the qubits lie No There are the edges so yeah, right So I only gave you definition and I gave you a some recipe to construct any code on a Tessellated manifold from the topological they call the Yeah, the triangulation data and the choice of your of your dimension Now I want to compute the homology And this is a Right so but well if I if I give you a like a five by five grade then you could Write some some code to calculate the linear dimension no big deal, but mathematics always involves some infinity So in this case, I want to consider the all possible families of tessellation over my two tourists And I want to answer for the K value for for that entire family. How can I do that? So the my logic here is that I want to reduce I'm going to reduce the calculation to a finite calculation The infinitely many possibility comes down to finitely many situate, you know some possibility and you do just do it So the the the piece I'm going to explain is the reduction from that infinite family becomes a finite calculation And for that purpose, I'm gonna discuss the entanglement RG and to do that Let me briefly remind you with the Measurement dynamics versus unitary. It's a bit of this discourse, but I think it's interesting approach. So let me explain that So given a stabilizer group generated by the rows of My well, I just well imagine that I have written my generators in a poly string form in a one in each row No problem. So in this in this case My star the operator associated with the one vertex will occupy one row and the next vertex will occupy the next row as and so on So I just imagine that I have written down that and then Let's consider I measure a poly operator P on top of this stabilizer code state and Let's suppose further to make this story more interesting P is not logical operator So not logical means that it does not commute with the sum stabilizer So if I write down my P In some poly string form then you should be able to figure out. Oh This row is anti-committing with that that row is anti-committing with that and so on and those Those rows are not commuting with my P So, yeah, there's some completely general situation. No locality here Then I apply the Gauss elimination to reduce the number of generators that are anti-committing with P How far can I go? one Because I can multiply this row to that to cancel off the any anti-committing instant sign So it becomes a committing row Whatever there are more anti-committing row. I can always multiply this to that There's no locality just the algebra So I am end up in a situation where all Generators are commuting but one named to Q is anti-committing with P. I Measure P what happens to understand that situation just Imagine the simplest possible Q was say single QBZ and I'm measuring X So My zero state is stabilized by that Z if I measure X what happens It just forgets the the underlying states and initializes a state into either plus or minus state Depending on the measurement outcome. That's all the same thing happens in this scenario So the measuring P will kick out This Q and P itself will creep in but now the the the sign in front of the P as a stabilizer will depend on the measurement outcome, but the measurement outcome is always 50-50 random Okay, so the measurement dynamics and also has the summarized Given a stabilizer code if I give you an operator that is not logical And if I measure that on the on top of code space Then we can find we can reorder or rearrange the my generators in such a way that Everything else everything but one is preserved and that one is kicked out P comes in with a sign random clear Now that's the measurement dynamics part, but we can realize the measurement dynamics Post-selected measurement dynamics using unitary Consider this P plus Q not P times Q What what is this? This is well. Yeah, this is an operator. This is a self Hermitian operator and you square is going to be easy They are one Because there are QP poly operators. They always square to square to one What is this? Zero so you is unitary What about the post-measurement state? Let's consider this expression If you do calculation right, I didn't give an exercise I would do that, but writing slows me down So let me not do it you can Show that this is equal to u times psi and My calculation is the independent of the logical state of my code because whatever logical operator you you you well The logical operators are preserved under this dynamics because I can always multiply my anti-communist stabilizer to that logical operator and let them Preserved so you don't have to worry about whether the s is full rank or not so my unitary Somehow magically realized and transformed my underlying state to the state as if it was P was measured But the difference is that now I realized that in using unitary Q is the unique element of s that had and yeah unique generator of s that is anti-committing with P That's exercise so if if if measurement outcome of P was minus one then you would have a minus Q here and I'll let you figure out what happens afterwards Sorry Yes, the stabilizer code is changing Now let's mean let me you know just bootstrap that simple purely algebraic calculation into this local setting How can I do what does so the goal is to? Make the cells bigger You know when I when I say Infinite families of the tessellation of my two surface I was imagining that the t2 the topological space is fixed But my my cell relation is refined and they ever refined so it's very fine That's that's that that makes the infinity. I want to get rid of all the fine grained triangles or pockets or whatever and become very big Some constant and then in the end I want I wish to be remained with some constant number of cells That's what I mean by reducing the calculation into a finite thing. So abstractly there is a code space on an n qubit You know hubo space right and then I want to transform this code space into something else but Some remaining piece that I very well understand By by very well, I mean the completely disentangled qubits that I can safely forget about Let's do it on the on the circle using repetition code So I'm I'm gonna follow the same prescripts right same prescription as we had the Had here, but without this this piece only x checks with the z errors So the vertices are here qubits are here on the one cells And my and my x stabilizer is also say with the vertex Acting on two nearby edges Okay, that's the same prescription the co-boundary of a vertex is two edges So I assign x stabilizer there. It's a repetition code Now imagine I Imagine I measure out z here. Yes qubits are on the edges Well, the ticks are remarked because I want to signify the where the vertices are and The stabilizer generator is one per vertex given by the x times x Okay, and imagine I just measure z a single qubit poly just there. So my My stabilizer generator was had the least that looked like this and if I measure If I bring a z here then two are anti-commuting So according to the previous prescription I'm gonna choose this one to be my unique one and cancel off any anti-anti-commutation in the remainder So I multiply this to that that turns this Into this and everything else is now commuting so I'm fine. So I singled out one of my stabilizer generator And I if I measure z then the post measurement state will have this qubit On the eigenstate of z which is completely disentangled. It's a zero state So I managed to understand the code better Well better because I know at least one qubit is in the product state that is a completely disentangled But I could Realize this transformation that singles out the one disentangle qubit using your unitary So what I have done is I apply a unitary forms by these two poly matrix Operators using that formula. I apply it then one qubit is singles out So without worrying about what Clifford operator I I should apply to this entanglement one qubit You just measure keep measure now Okay, now I can measure this and that the remainder is again a repetition code on a Slightly coarser one one one less qubit So I measure here and there until I am left with the some Say three qubits and then you calculate Yes yes, no The probability you will have a zero a plus one outcome is 50% exactly But I don't really care I mean the purpose of talking about measurement is is easier to think about Because my my qubit is going to be certainly disentangled from the rest. That's all that's all my purpose and the argument Behind why I'm doing that is I want a unitary transformation from the old code space into new code space because I don't want to lose anything and I point it out that whenever my measurement is anti-committing with something my Measurement is can be replaced by some unitary Although it may depend on the measurement outcome But I'm I'm fine because the the measurement outcome only affects this is entangled qubit, which I'm going to throw away So using the similar strategy you can do You can map The original torque code on the on the white lattice into another torque code on the red lattice By measuring out See here see here see here and see there Oh Well, the fourth is he is unnecessary because because there was a stabilizer that goes around here So after you measure the three the fourth one is all anyway fixed and It is a little bit exercise that you had to make sure that If I measure this first then there is a some stabilizer that ends commits with that measurement And after those two measurements, there's some stabilizer still anti-committing with with that measurement So a little bit of care is needed, but that's possible and the end result is that And the result is that I managed to transform Though though the torque code on the white lattice into the torque code on the red lattice Tensored with some completely decent angle qubits and that transformation was enacted by some Unitary acting on this region only so Sad yet differently, I managed to transform a torque code on the white lattice Into the one course or red lattice using a some small depth quantum circuit who skates our geometry local So if I do it, you know indefinitely The end the end result is that I'm left with a torque code on say for qubits And then you can do the calculation Yeah So measurement was a conceptual vehicle that you that that allows you to think about the transformation more visually evident Unitary is the backup argument that my transformation that my post measurement state is a unitary result of my previous one So only I only need the existence of such a unitary. This is a one concrete realization. Yes Yes, yeah, there's a Some carries needed the measurement outcome must be post-selected here for this particular unitary that makes the Yeah, but this is just the existence of unitary is just a tool Right, right. Well, I could have equally said that. Oh, oh, let me show you that there exists a unitary circuit that this Entangles the torque code on the white lattice into into the red lattice. I could do that And you can actually calculate what it is, but how would you find that I? Find that's somewhat cumbersome So I introduced another trick to think about this problem Where you just think about the measurement and try to disentangle bunch And you realize that oh my measurement. I yeah, I thought about measurement But all I did was already achievable by unitary. Ah, that's non-trivial That some carries needed actually now as you measure. Well, yeah Making one measurement you could choose a one anti-commuting stabilizer that is nearby It might grow. Yes. So you wait that that's a non-trivial part. The trick is that you do it on a patch here and a very far distant patch there and Any manipulation any growth of any oil any violation of locality Violation will be will be happening in this Gauss elimination procedure So that's that's the step where a little bit of carries needed, but that's possible which I've not have not explained Yes No Right measurement was acting on the one qubit But the overall effect was as if you had applied unitary on two qubits It may grow up. So you need to be careful But in this particular instance, it's possible that you maintain the locality What what what? Why do I want that? This leaves on a smaller number of qubits So if I could iterate this process inductively then I can say that oh, here's my infinite family of Tori Kose But the calculation of number of encode because if qubits for entire family Reduces to my finite calculation and I'm I'm showing you the how the reduct the reduction is done Okay. Now now let me conclude How it it shows the the the homology is topological invariant the suggestive picture is that Given the topological space you make it you you consider two different allusions Say you have a two identical tourists here. I have Triangulated in a in some fashion here. I trying it in a different way And I want to show that the homology calculation from those two cellulations result in the same group How can I do that the usual argument is that you think about the the refinement of these such that the refinement would agree and at the same time you show that as you refine the homology doesn't change I Did the inverse thing? I did the starting with the very fine grained lattice and I went to the coarser But it's a unitary transformation. So homology in particular the code speak dimension should not change So I have exactly implemented this argument all right, so You bootstrapping this transformation you end up in a two logical qubits the answer is yeah this is Two-dimensional subspace you end up in a two logical qubits If you since this is a unit transformation You can think about this as an encoding encoding maps starting with a two logical qubit You build up your code space and you become a large Toricoate a state very highly entangled complicated state and you can think about that that and If you think a little bit more then you can show that the entire process has a linear depth in the linear dimension of the torus Yes, no No, the I'm completely ignoring the fault-holerance stuff here This transformation will actually Amplify your errors so in practice don't do it Yeah, so yeah the right is you right from the two logical pure You know not naive logical qubits now Inversing in this procedure you end up in a big torus which means interpret differently You have just wrote down an explicit quantum circuit that generates the Toricoate state out of a product state and that that the depth is growing with the with the with the system size and Then next well day after tomorrow. We're going to show that that's actually the best you can do So some some complexity argument. Sorry Maps of the chain complex the coding theory interpretation It's just that definition of a code one one Crucial difference when you encounter a chain complex in a topological context or any other If you are purely interested in a topological data of underlying space Then you know having a map and the having the dimension often suffices But when it comes to coding theory, you need to count the weight of each row and column Excellent question. There may be I don't know. I don't know a good one. Yeah. Thanks