 I will be talking about constructing faithful maps and this is joint work with RP, Ramtasad and we are both from TIFR, right, TIFR Mumbai, okay. So let us start the first question of course is what are faithful maps, right. So before we go into that definition, we need to look at the notion of algebraic independent. So again before algebraic independence, let us look at linear spaces. So we are, suppose we are in the vector space some R3 over some field R, okay. So suppose you are given these three vectors, they are linearly dependent, right. Because there is a linear combination of these vectors which gives the zero vector, this should be the zero vector anyway, okay. So they are linearly dependent. Now for algebraic dependence, what you need to look at is the space of polynomial. So let us focus on some, on the space of bivariate polynomials over some field say complex numbers. So these are clearly not linearly dependent over the field C because there cannot be any complex linear combination of these which is zero. However, if you allow a polynomial combination, then you can make this into, so because x square into y square is equal to xy whole square, there is a polynomial combination of these three polynomials which is zero. And so they are said to be algebraically dependent. So formally suppose you are given a bunch of polynomials over n variables, then they are said to be algebraically dependent if there is an annihilating polynomial, right. A polynomial which is non-zero as a polynomial, but if you plug in these FIs into it, then it becomes zero. Note that it is a k-variate polynomial, so same as the number of polynomials given and it is over the same field that the original polynomials are over, okay. So in such a case, these input polynomials are said to be algebraically dependent, otherwise they are algebraically independent, okay. So one thing to note is that the underlying field is very important, right. So for any prime p, suppose we have these two polynomials x to the p, plus y to the p and x plus y, over the complex numbers these are algebraically independent, right. But if the underlying field was fp, then they are dependent because x to the p plus y to the p is x plus y to the p in fp, right. So now that we have defined, seen what algebraic independence is, we move into the notion of algebraic rank. As before let us look at what linear rank is, right. So if you are given a bunch of vectors, then the size of the maximum linearly independent vectors in this set is the linear rank, right. So linear rank is well defined because any maximum linearly independent set of vectors has a unique size. So for example, the vectors we saw, the linear rank of this bunch of vectors is 2. So similarly, the algebraic rank is also a well defined concept because a set of, if you are given a set of polynomials, then the maximal set of algebraically independent polynomials in this set also has a unique size, okay. So that is why the notion of algebraic rank is well defined and again the example we saw, if the field was fp, then the algebraic rank of this is 1, whereas if the field was complex then the algebraic rank is 2, okay. So this is what algebraic rank is, it is the size of the largest or maximal algebraically independent subset of a set of polynomials. So now that we have the notion of algebraic rank, we want to ask whether we can come up with rank preserving maps. So again in the linear algebra, if you are given a set of vectors and you know that the linear rank of this set of vectors is k, then you can always extract out of it a bunch of k linearly independent subset and which will be the basis of the underlying space. So similarly, if you are given a bunch of polynomials which have algebraic rank k, intuitively it should depend on only k variables, right. So a faithful map is a notion which formalizes this. So if you are given a bunch of polynomials with algebraic rank k, a map phi which takes the, so these are over x1 through xn, n-variate polynomials. This is a map which maps each xi to a polynomial over just k variables and this map is said to be faithful. If after you substitute each xi by phi of xi's, then this set, so f1 of phi, f2 of phi up to fm of phi, these are now a bunch of polynomials over k variables and the algebraic rank of this set of polynomials is also k. So is that fine? So you are given a set of polynomials over n variables, you are mapping it to k-variate polynomials such that the algebraic rank of the resulting polynomials are also, algebraic rank is also k, ok. So the question of course is, can we construct faithful maps efficiently and of course it is of independent interest to come up with faithful maps, but a bonus is that if you come up with faithful maps you can also solve the polynomial identity testing questions for certain cases in which is a important question in algebraic circuit complexity. So what is the polynomial identity testing question, you are given a circuit an algebraic circuit. So what is an algebraic circuit, the leaves of it is a dag where the leaves are some either variables or field constants and every vertex will compute a polynomial in the most natural way and the polynomial computed by the circuit is the polynomial computed by the root, ok. And the polynomial identity testing question asks whether the polynomial computed by c is the 0 polynomial. A trivial upper bound on for solving this question is d plus 1 to the n if d is the degree of the polynomial being computed and n is the number of variables. So an approach to give a better upper bound would be to reduce the number of variables because the number of variables is in the exponent and you keep the degree under control keep it to poly d and ensure that after whatever changes you have made the circuit continues to stay non-zero if you started off with a non-zero circuit. So let us look at a special case where the input circuit actually looks like this, there is a smaller circuit c prime on top and into it are feeding in some polynomials f 1 through f m and you are guaranteed that the algebraic rank of these polynomials is k, ok and think of k to be much much smaller than n, right. So the question is if you are given some faithful map, yeah if you are given a faithful map then using the fact that the structure is like this and the algebraic rank is much smaller than n can you make the upper bounds closer to d to the k rather than d to the n, ok and the question the answer is that yeah I mean it is helpful, right. So suppose you are given a faithful map what is the most natural thing you will do instead of x i you will be plugging in phi of x i's, right. So phi i is just phi of x i. Now the property you want is that c after you plugged in c's instead of x's the resulting circuit should also be non-zero if you started off with a non-zero circuit, right. Now just look at the simple case when yeah simple case when k and m are same are equal that is the rank is basically all of f 1 through f m the algebraic rank is k that is all of them are algebraically independent. In this case if c prime is a non-zero circuit then this can never be giving you a it can never become a zero circuit because after you plug in f i's, right. So if phi is faithful then you know that even after plugging in phi's the resulting circuit which is c of phi will not be zero. So you have the required property but it is a fact and it can be proved when not very difficult proof to is that even in the general case when you are just given that even in the general case when the algebraic rank is smaller than the set of polynomials there also if you plug in a faithful map then the resulting circuit c of phi continues to be non-zero if you started off with a non-zero circuit, ok. And yeah I mean this was proved in B. M. So this is B. M. B. K. Mitman Saxena and ASSS later but yeah cool. So now that we know what faithful maps are and how to how it can be used we will go into how we can construct faithful maps, right. So what is the question you are given a set of polynomials you want to come up with a map phi such that a map phi which maps each of the x i's into a k variate into k variate polynomials such that the algebraic rank after you plugged in phi's should be equal to the algebraic rank of the original polynomial. So it turns out that if you if phi was a random affine transformation of this sort so s i j's are new variables, a i's are new variables and you just plug in a random values for s i j's and a i's. It turns out that this is a faithful map, right. So the question then becomes can you construct them deterministically, right. That is yeah reduce the number of yeah that is cool. So what is it turns out that this question was already answered over characteristic 0 field, ok. So before I mean the first question is so now you want to give a map such that the algebraic rank stays the same, right. So how are you going to test whether the algebraic rank stays the same or not, right. So for that the question becomes how do you test the algebraic rank of a set of polynomials. So one approach could be that find the annihilating polynomial but it turns out that that is a hard problem. It is like sharp p hard to compute even the constant degree, right constant term. So you need a different criterion, ok. So it turns out that over characteristic 0 fields there is a criterion called the Jacobian criterion, ok. So what you want to do is you want to capture the algebraic rank via the linear rank of some matrix. And so what is the rank going to be suppose you are given sorry what is the matrix going to be? Suppose you are given a set of polynomials over n variables. The Jacobian of this set of polynomials is defined in the following way. So the i jth entry is going to be the partial derivative of f i with respect to x j, ok. And the Jacobian criterion says that if f has is characteristic 0 then the algebraic rank of the input polynomials is the same as the linear rank of this matrix, ok. So now we know that at least over characteristic 0 fields you can capture the algebraic rank via the linear rank of a matrix which we will call the Jacobian, ok. So now that we have captured we know how to test for algebraic test the algebraic rank or rather find the algebraic rank. What you want to do is come up with a faithful map. So we start off with a generic linear transformation because we know that that is a faithful map and look at the Jacobian of f of phi, right. It turns out that you get a nice matrix decomposition of the following form. The Jacobian of f of phi is the old Jacobian with each entry in the matrix substituted with each entry in the matrix being substituted with phi times some matrix M phi, ok. So what do we need? We need this phi to be such that the rank of the original Jacobian should be equal to the rank of the Jacobian after phi has been substituted. It turns out that this can be done if f i's are if f i's have some structure, ok and the a i's generally handle that part. And the second more important thing is that you want M phi to be such that whatever the rank of phi of the Jacobian is should stay the same after you multiplied it with M phi, right. So the rank of this matrix should be the same as the rank of phi of the Jacobian. And these are what are called linear rank extractors and it turns out that Gabison and Raj had shown that if M phi has the following structure. So if the ijth entry of M phi is S to the ijth thing S is a new variable. Then it turns out that the rank of phi of the Jacobian is equal to the rank of the new Jacobian over yeah rank of the new Jacobian over the field fs, right. Because now this this matrix will have polynomials coming from the field f together with S. So now once we know this, ok yeah. So it turns out that the ijth entry of M phi is just S ij. So now if you know that this is this is a linear rank extractor then the following map becomes a faithful map. So it is really that simple for characteristic zero fields. And so BMS showed that this was a rank extractor and used it to give PITs for certain set of in certain cases. And Agrawal, Sahas, Saptarishi, Saxena used the same maps to give PITs for more classes of circuits, ok. So now the question is what happens over final characteristics. The main problem is that the Jacobian criterion is false, ok. So then what do you do? You need sorry. I am confused because then in the testing the algebraic rank that is randomize, right. Yeah, yeah. So, but will you start up by saying that you want to construct these math written in a statement? Yeah, so how do you want? The yeah it is a randomized test but I mean you. So it is a randomized test because you need to find the rank of this when it is the randomization comes only in the fact that you need to find the rank of this matrix. But this the testing the rank does not come into the algorithm it is just an in the analysis. And the I mean this rank is you can check the rank over FS when it turns out yeah it does not answer your question. But the testing the testing algebra. So that for random s i is the the random in a transformation would give you a face yeah, yeah. So then like what is the. So now this is over just one variable and this this has an extra variable s, right. So to get it what happens is that there is no one one faithful map what we will be what this gives is a family of faithful maps for which is indexed by the variable s and it turns out that for some good cases a very few you just need to run over a few s's does that make sense I mean. So this is a family of faithful maps which is indexed by the variable s. So for different different s's this is of faithful map if you put s at random it turns out to be faithful map what you want to give is a small family of s's for which any if you give any set f 1 through f m one of these f i's is going to be a faithful map and it turns out that for some structured f i's and when when everything is good. So for finite characteristic field the Jacobian criterion is false right. So what we need to do is we need to look at look further into the Taylor expansion in some sense. So it turns so if you look at the Taylor expansion of any polynomial around a point z then the first few terms is what is captured by the Jacobian right. So what Pandesina Babu and Saxena said is that for finite characteristic fields we need to look further into the expansion and till when do we have to look you have to look till the degree equal to the inseparable degree of f i's many terms. So this is a linear part because each x i is has degree one it turns out that for characteristic finite characteristic fields you need to look at terms up up till the degree of x is equal to the inseparable degree ok. And what is the inseparable degree given a set of polynomial f i's you I mean I am not going to be defining it I am not going to be defining it. Given a set of polynomial f i's you can think of this as a parameter which somewhat captures how many repeated roots it has in the extension field where it completely factorizes ok. But for you do not need to go into that just think of it as a parameter which is fixed given a set of polynomial ok. So PSS what they showed is that for finite characteristic fields you need to look at this new operator which is h t which looks at the Taylor expansion up till the degree t at component of x. And what they essentially show is that the algebraic rank of f i's are captured by a matrix of by this matrix ok, but it that is not exactly what they say. What they say is that suppose you are given a set of polynomials they are algebraically independent if and only if this matrix has full rank over f z. So z is a new variable which has been introduced by h's and it is not just that the linear rank of h t f 1 to h t f k has to be full. You have to look at look at this matrix where the v i's come from for every v i coming from some space i t. Think of i t to be some i t actually looks like this, but forget it. The main point is i t is a pretty ugly looking space of polynomials and f 1 through f k are algebraically independent only if and only if for every polynomial every k polynomials coming from this space this matrix is linearly independent ok. So, finally, we can come to our result suppose you are given a set of matrix polynomials whose algebraic rank is k and inseparable degree is given to be t. Then we can construct a faithful map or a family of faithful maps with a new variable s such that the algebraic rank of this is equal to k yeah is equal to k. The difference is that the earlier thing was giving the algebraic rank over f s. So, these polynomials are now over f s the underlying field is f s. It was giving the characteristic zero setting they were being able to give a faithful map with the algebraic rank being equal even over f s we are not able to do that because of the ugliness of the i t. But yeah we can give it over f and it turns out that for p i t applications this is enough. So, now for yeah for some structured f these are just examples. For some structured f we can give such faithful maps and we can use it for p i t ok. So, just a very high level overview of the proof it will follow very similar to the characteristic zero setting only that I mean yeah the calculations are much more tedious. So, the first step is the algebraic rank is now being captured by the linear rank of the p s s Jacobian we will call that matrix the p s s Jacobian. And for a generic linear map you again write the p s s Jacobian after you the new p s s Jacobian in terms of the old p s s Jacobian which looks something like this which is similar to what it was earlier. There are I mean there are issues here of course, because there are these v i s which are there, but it turns out what we essentially saw is that for every v i there exist some u i s in the original thing such that this happen. And what do we need? We need the p s s after you substituted phi to be equal to the p s s rank originally this is again similar to what it was earlier. Again this can be done if f are some structured polynomials, but the most important thing is that can we come up with the m phi such that the rank is preserved the rank of this times this should be equal to the rank of this. So, earlier this m phi has a had a very nice structure now it turns out that it is much more complicated. So, the original m phi if you can think of it is just in this block 1. So, what is this matrix? This is labeled by monomials in x of degree up till t and monomials in y of degree up till t everything other than these blocks is 0. So, it is a block diagonals it has a block diagonal structure and each block is like I mean the rows and column should have the same degree then for those are the blocks basically this is when both of them have degree 1 this is when both of them have degree 2 and for degree t ok. So, it has this structure and the bigger problem is that earlier we had that m phi of i j was just s i j. So, if you knew what the a linear rank extractor you could just put that directly in here these blocks are correlated right. So, you only have a handle on what you can substitute this first block to be once the first block is fixed all the other blocks are also fixed right. So, that is why yeah that is where the hurdle comes in, but so again taking inspiration from the previous case what we do is that we define this first block in this way ok. So, earlier it was just s to the i j now that now it is going to be replaced by a weight of i earlier weight of i was just i, but now it has to change slightly and again weight of i is not very good to look at. So, it, but it turns out that if you for some correct definition everything works out nicely and you get a faithful map which has this form ok. So, this is essentially a faithful map with a lot of things hidden, but yeah ok. So, that almost the end what I want to look at are some open threads as I said we were only able to construct f faithful maps I do not think we whatever this the maps that we gave are also f s faithful. So, maybe we will need to come up with a different type of maps which are f s faithful and they improve dependence on inseparable degree as I mean what we are giving is a family of maps and we can give, we can say that deterministically give you a small family only if the inseparable degree is small. So, if only if it is bounded can we give a poly-sized family of faithful maps. So, it would be nice if we can handle higher inseparable degree the yeah I mean the main problem is that the PSS Jacobian is very big if the inseparable degree is not bounded that is where the main problem comes in, but yeah still can we can we do that and more recently Gouh Saxena and Sina Babu have come up with a different characterization of in algebraic independence, but this is not I mean yeah it is not algorithmic right, but still can yeah it is not algorithmic, but it does not depend on inseparable degree. So, can we make an algorithmic version of it somehow and so that we can use it for the idea. Thank you.