 Yeah, Matthew, please go ahead with the second part. Right. I titled part two positive formulas for the mole of Sagan case. That is what we're going to be talking about. So I'm going to be very specific now about something I mentioned in passing earlier, which is factorial elementary symmetric polynomials. We're going to define a specific factorial elementary symmetric polynomial e k k of X and Z. As the product as I goes from one to K of Xi minus C one. So you'll notice there are two indices there. The second K is the number of variables, and the first K is the degree. There are other factorial elementary symmetric polynomials that I won't go to the trouble of defining here, but e p K for zero less than equal to P less than equal to K, that do have more Z variables but in a sense, at least in this context, e K K is the most important factorial elementary symmetric polynomial. So, almost the basis of almost everything after this slide is a formula I found for multiplying an arbitrary double Schubert polynomial in X and Y by a factorial elementary symmetric polynomial in X and Z. So, I'll state the theorem, let you be a permutation p and K integers is K greater than equal to one, then there is a relation arrow K on S infinity. So this is the same relation as in satelius 1996 paper for multiplying an ordinary Schubert polynomial by an elementary symmetric polynomial. And there's also a set of positive integers p K u w for each W and S infinity, such that the product SU of X, Y times e p K of X and Z has the coefficient of SW X and Y, as you can see is is itself a factorial elementary symmetric polynomial in some Y variables and Z. So, as the degree of W increases both the degree and the number of variables of the factorial elementary symmetric polynomial in the coefficient decrease. So, as you can see variables are as usual, but the Y variables are written as Y sub p K of X. So, we're substituting into the factorial elementary symmetric polynomial, the indices of why that occur in the set p K u w. The explicit explicit definition of that set is the set of all values of the permutation u I such that I is less than or equal to K, and you I equals w I so so these are the values of the permutation that don't change before K as we move from you to W. I just give a quick example here. I've picked you 43512 p equals three K equals four and I've expanded the product using the Perry formula. And I've circled the elements of p K u w. So, as you can see, those are the indices of the Y variables in the factorial elementary symmetric polynomial that is in the coefficient. So, there, there's something having a formula for multiplying by a factorial elementary symmetric polynomial and X and Z. That allows you to do things that only having a formula for multiplying by a factorial elementary symmetric polynomial in X and Y that doesn't allow you to do. So we can actually pick apart other double Schubert polynomials and express them in terms of the factorial elementary symmetric polynomials in order to be able to compute the coefficients. In particular, there are these permutations known as dominant permutations such that if mu and S infinity is dominant, then the double Schubert polynomial S mu of XZ is itself a product of factorial elementary symmetric polynomials. So the factorial elementary symmetric polynomials that occur in this product are of the type where the degree is equal to the number of variables. And it's product as I goes from one to infinity of E lambda I mu lambda I mu, where lambda is a permutation, sorry, a partition that we associate to me. So the degree and the number of variables decreases as I increases, and it's a factorial elementary symmetric polynomial in X and Z I. So with my theory formula, we can multiply by Schubert polynomials corresponding to dominant permutations by substituting the index the variables. So, I'll, I'll mention one use of my theory formula which is not obtaining a positive formula, it's actually a method for efficiently computing all the coefficients. So, that is actually in most cases, even faster than existing methods for computing the ordinary structure constant C of BW. So what we do is given be an S infinity. We start with a dominant permutation and apply a divided difference to the Z variables to express SV of X and Z as a sum of products of factorial elementary symmetric polynomials in X with Z variables having their indices in varying sets. So there of course will be signs this is not a positive formula, but having an expansion of this form allows us allows us to apply the Perry formula to compute the product. Now, the way the program is implemented it's not quite as simple as that the divided difference is not computed explicitly. So the sum is done over a graph of permutations that ends up giving the same result but but results in fewer multiplications. And in the ordinary Schubert polynomial case we set Y and Z to zero and this is a lot faster, of course, then the case where we're trying to compute polynomials. So in this measurement, you can download my program, if you have Python 3.9 plus, you can run pip install shoot molt, and that will give you scripts for computing these coefficients which I'll get into a little more detail later. So the strategy in this paper which has the same name as the talk a mole of Sagan type formula for double Schubert polynomials is to compute SU of XY times SV of XZ by reducing to problem to be being a dominant permutation which, as I mentioned earlier, can be computed in the formula so the most general easily stated case where this is possible is when you and we have separated the sense, which I mentioned earlier, that it's an interesting considered in 2019 for ordinary Schubert polynomials. Now, when a permutation is dominant. So for a permutation u and s infinity, we define its code code of you to be the sequence such that code I have you is equal to the number of J greater than I such that you I it's greater than you J. So we're taking the inversion separating them into the indices where they occur and counting them. The permutation is dominant, precisely when code of you is not increasing so code of you is a partition. So we can associate to any permutation be a dominant permutation that the referee, who was looking at the paper suggests that I call the dominant approximation. We write this permutation as mu be so we define it recursively if the is dominant set mu be equal to be if the is not dominant. Let I be the maximal index such that code IV less than code I plus one of the, which is an index where v fails to be dominant and then we define mu be equals mu VSI. So the inversion terminates in a dominant permutation mu V such that L V inverse mu V is equal to L mu V minus L V. Now, you, you may notice that the coefficient C u v w y z are directional. Now, if, if you flip you and V, you don't get the same coefficient. So, we're going to be directional with our separated the sense condition here. So if you and we are an S infinity, you and we are said to have separated the sense, if there exists a piece such that you has no sense less than P, and V has no sense greater than P so we're distinguishing you and me. The first permutation is the one that has no sense less than P. Now I have found a way to remove this restriction, but the formula is entirely different. So in these polynomial terms if you in the satisfy this condition, then SU of x y is symmetric in x one through XP and SV of XZ has at most PX variables. So, if you and we have separated the sense, we have this quality of coefficients so for all w such that C u v w y z is not zero. So C w y z is equal to C u mu v w v inverse mu v, and why is he now this in the second coefficient, we're multiplying by a double super polynomial corresponding to a dominant permutation. So we can apply the Perry formula to compute this. So this is the main theorem of paper. So, suppose you and we have separated the sense, then C u v w y z is zero, unless the length of the w v inverse mu v is equal to the length of w, plus the length of the inverse mu v. If that condition is satisfied, then there exists a partition lambda v, such that C u v w y z is the sum over all lambda v Perry path. If you recall in the Perry formula there's a relation arrow K. So, the K in the arrow K in these paths are the parts of the partition lambda v. So it's u zero arrow lambda one v u one, etc, and the paths are from you to w v inverse mu v. So, what we're summing is a product of what I call weights and these weights correspond to edges in the path, and they're just the P area products. So, it's the product for. So, at index J. So, this contribution is the product over all I in P lambda J v u J minus one u J of why I minus CJ. So, there's actually bonus is actually, I didn't do this on purpose it just worked out that way in the formula in the main theorem, if we substitute equals why each term in the sum is a polynomial in the negative roots. So, this is one minus why I with non negative integer coefficients. So, this formula gives a separated descent formula in a covariant comology, where the directionality of the separated descent definition doesn't matter. So, just for the purpose of giving examples of when the supplies, I'll mention that this gives a positive formula for multiplying factorial sure polynomials. So, in particular, this applies when M equals K, which is the original mulliv sagan case. And it actually gives a mulliv sagan formula that is positive after substitution. For me, it generalizes the corner case, we can multiply a factorial sure polynomial in X and Y by a double Schubert polynomial in X and Z that has the same number, or fewer X variables. And again, this is grand positive after substitution. So, I'm going to draw some diagrams when I give examples of the formula. So, a lambda theory path from my formula can be visualized as an array of numbers. The numbers are the permutations in the path written vertically from left to right with a bar in column I under position lambda I. The perm in the product is contributed if a number above the bar in the column is the same as the number in the same row in the column to the left, in which case it will be circle. So, if the circle number is the number a, and B is the column it is in with column numbers starting at zero. And then we have a factor of YA minus CB, and they take the product over all circles in the diagram. So I have an example here there's a circle four above the bar in column three, because it's the same as the four in column two in the same row. And that gives us why four minus three. So I tried to find, you know, examples I could match in other papers. The, the logical place to look would be more live in Sagan's paper, however, there aren't any examples in their paper, but Knudsen and towel in their 2003 paper give an example for their more live Sagan formula in terms of puzzles. So I take that example and computed it my way. So in that case, UV and W are equal to 132. The partition lambda V is one one. So that means their bars under row one in columns one and two and there are two columns. UV is 312 and WV inverse mu V is also 312. So we're taking 11 peary paths from you which is 132 to WV inverse mu V which is 312 and I've drawn the diagrams here. So in the first one, there's a circle three in column two, which gives you why three minus E two. And then the second one there's a circle one in column one which gives you why one minus C one. Now of course this gives the same answer as a Knudsen and towels formula, but they computed as why three minus C one plus why one minus C two instead. When you substitute the why this becomes why three minus why one plus why one minus why two and why one minus why two is not positive. But you'll notice in my formula what you get is why three minus why two, which is positive, plus why one minus why one which is zero. So, the first term is positive and the second term vanishes so. So this is a positive formula for that coefficient. I'm going to show an example for multiplying two factorial sure polynomials with different numbers of X variables. Here I picked one in three variables in X and Y, and one in two variables and X and Z. Two is 13524 lambda V is 111 and WV inverse mu V is 41523. So it's 111 peary paths from 13524 to 41523. And this is just one coefficient. So my formula lets you compute all coefficients w for a given you and me. But of course, on a slide I can really only show one time. So, in this case as I mentioned when why and see a different polynomial with why coefficients must have the larger number of variables. I also wanted to show an example of the corner case, which is Schubert by sure. So I picked a short factorial sure polynomial in X and Y with three X variables, and the double Schubert polynomial s 1432 and X and Z, and that has exactly three X variables so the formula applies. So this is 13524 lambda V is 221 and WV inverse mu V is 461235. And you can see the diagrams representing the paths with the circle numbers and the corresponding weights. And as I mentioned in this case, the polynomial with why coefficients must be factorial sure polynomial button y equals the coefficients are completely. And they're always commutative but if you flip you and me, you get the same results so this fully generalizes to equivariant comodity, not cornered formula, but the case where it corners formula applies. And I also just wanted to give one example that doesn't involve factorial sure polynomials. My slides are intended to be used as a reference. This is probably more useful to look at after they're posted about this for this example has the benefit that it has two circles. So one is a circle for in column one, and the second is a circle for in column two so. So the contribution is why four minus C one times y four minus C two. And this is the only diagram. So the coefficient that we are computing is equal to this polynomial. So, my formula actually gives you a positive formula in terms of xi minus yj for double Schubert polynomials. This is because if you and w are both equal to the identity. This is a separated the sense case. So, and in that case the coefficient C one v one of x and y is equal to the double Schubert polynomial SV of x y. I'll give you an example here for the double Schubert polynomial s 1432 of x and y, which is a degree three. So, if you recall there's vanishing formula, if we set x equal to y, then the double Schubert polynomial vanishes. So, in order for that to be true, and for the formula to still be positive, each individual term must vanish. And that does in fact occur if you'll notice the, the Northwest for terms. I'll have a factor of x one minus y one which becomes zero when we substitute, and the southeast term has a factor of x two minus y two, which also vanishes when we substitute. So I wanted to explain why my formula is positive. So, if we multiply a double Schubert polynomial by a double Schubert polynomial corresponding to dominant permutation, which is a product of factorial elementary symmetric polynomials. We can multiply by the factorial elementary symmetric polynomials in any order, the multiplication is commutative, and we can apply the Perry formula in any order we want. Now, while this does give technically the same result. It can be expressed very very differently and the expression can have different properties. So my formula starts at index one and continues an increasing order and this is exactly what makes the specialized formula positive. So, the way this works out if there is a negative factor after substitution yj minus yi with j less than I, there also is before substitution, a factor of yp minus Cp for some P, which vanishes after substitution. Or a term either has only positive factors, or if it has a negative factor it also has zero factor and vanishes so every term is non negative. Other orders may sometimes be advantageous. I can lecture that the opposite order can be used to obtain the pipe dream formula for double Schubert polynomials. And that's this is just based on empirical data. It would be interesting to prove if this indeed holds. So that's the content of the paper a molybden type formula for double Schubert polynomials. Now, I mentioned that I have found a way to flip you and V and the separate descent condition. This also involves the period formula, however, it is rather an indirect use of it. So, the way we do this to find CU BW YZ, where you is dominant is we express the double Schubert polynomial SV of XZ. This is the way that it's, it's a sum of products of double Schubert polynomials it X, omitting I, and Z. And in each of these terms we have a product, which may be one of Xi minus various indices, ZJ. And being able to do this, because of the special form of the divided difference operator partial mu w for mu dominant dominant. This allows us to compute partial mu w applied to SV of X and Z for any V as a polynomial in Xi minus CJ with non negative integer coefficients, which is what allows us to flip you and be in the separated descent result. I'll mention that this formula after substitution is not positive. But you could just flip you and be to get the positive formula. I'm actually going to prove that this decomposition exists using my period formula. So, we can express any double Schubert polynomial SV of X and Z as partial the inverse mu V applied to the double Schubert polynomial SV of X and Z. Now it's very easy. I'm going to do this on the Z variables but it's equivalent. If you flip X and Z it's still a double Schubert polynomial. So we're out of a dominant permutation factorial elementary symmetric polynomial in X and Z I, and we're left with another double Schubert polynomial as a factor for that corresponds to a dominant permutation in X and Z, omitting I. Now we're applying a divided difference of to this so we can expand it in terms of the lateness formula. And what we get is a straight divided difference applied to the double Schubert polynomial, which gives us another double Schubert polynomial, and a skew divided difference applied to the factorial elementary symmetric polynomial. So the PR formula is what allows us to positively compute a skew divided difference applied to a factorial elementary symmetric polynomial. Hence, we have the positive formula of this form as I promised. So, given this decomposition. There's an algebraic result that the skew divided difference operator partial mu w alternatively applies a straight divided difference, and then pulls out an X variable and moves it to the front it permutes the indices in such a way that it moves X I to X one. So, working this out, you can get a formula for the result of C mu v C mu v w y z where mu is dominant, and that won't be in this paper, but will be in an upcoming paper. And as I promised, in abstract, I also have a formula for multiplying an arbitrary double Schubert polynomial in X and Y by a factorial complete symmetric polynomial. The formula is entirely similar, although the relation here which I write as fat arrow K for between you and W is different. Though it's still in subtly is 1996 paper. And here the coefficient is a factorial complete symmetric polynomial. And as the degree of W increases the degree of the factorial complete symmetric polynomial decreases but the number of variables increases, and it's a polynomial in Y sub q K of UW and Z, where q K of UW is the set of all W I such that I is less than or equal to K, all of them. Union all W I such that I is greater than K and UI is not equal to W I. I also have a formula for multiplying by a factorial short polynomial in X and Z corresponding to a partition of hook shape which I won't mention here in the interest time. I'm going to end with an advertisement for my program. If you download Python, at least version 3.9 what you'll probably get now is version 3.10 that's the current version, and you run pip install shoe malt that'll give you scripts shoe malt underscore p y, which I would put that way in order not to conflict with existing software named shoe malt that computes the ordinary structure constants shoe malt underscore double for double super polynomials in the same set of variables express Graham positively and shoe malt underscore YZ for them all I've seen in case and I added a recent feature to that that actually allows you to display the result positively it does, it does not do that by default. So I'd like to thank the organizers again and thank you all for attending.