Merci beaucoup à Eric et Karin pour l'invitation et à Happy Birthday Dork. Je suis très heureux d'en parler. Bonne occasion. Je vais vous partager un travail joint avec Sylvie Pecha, Kourouche, Johannes Zinger, Yan Cheng Cao et Jaime Castillo Medina. C'est disparu dans plusieurs papiers durant les dernières 10 ans. Ah, ok, je vais scroller comme ça, oui. Ok, donc je pense que la plupart de vous connaissent plusieurs valeurs de data, donc elles sont données par cette liste d'intérêts là-bas. Et les NJs sont des integers positifs et la première doit être plus grande que 2, pour que la série se convertisse. Et ça fait un peu de sens sur ça. Si vous avez un N1+, etc., pour un NJ, c'est strictement plus grand que un J pour un NJ, vous pouvez faire le sens de cette série. Donc, comme vous le savez, l'intagère K est equal au détail et le summe de l'argument est appelé le weight. Ok. Peut-être un peu moins bien connu, c'est la fonction multiple. Vous pouvez laisser l'argument aller au nombre complexe et ensuite vous faites le sens de la suivante série. Bien sûr, ça ne convertit pas à chaque fois, mais le domaine de convergence a été étendu par ces 3 valeurs 20 ans plus tard. La série convertisse a donné la vraie partie du summe du J, c'est strictement plus grand que un J. Et plus que ça, ces autorités ont prouvé que vous pouvez externer ça dans une fonction homomorphique. Vous pouvez externer ça dans une fonction homomorphique de valeurs K complexes dans la whole C to the K. Et c'est possible de décrire précisément où la singularité est. Et elles sont décriées ici. Donc, la singularité de l'homogénial est parmi les parties homogénieuses, en parlant d'un J, et ce sont ceux d'A to the K, dans la C to the K, avec E to Z1 equal to 1, or Z1 plus Z2 is in this set, or there's a J in 3 up to K, so that Z1 up to plus plus ZJ is the sum is smaller than J, smaller equal than J. Ok, here you recognize the fact that the simple zeta function is homomorphic everywhere except at Z equal to 1. Here you jump from 2 to 2, it's related to the fact that the Bernoulli numbers with odd rank all vanish except the first one. And from dimension 3, you don't have any gaps like this. Ok, so... Of course, you probably know quasi-shuffle relations. Here is an example. If you multiply 2 simple zeta values, you get a linear combination of double and simple zeta values.  Ok, this gives you these 3 terms. Ok. That's possible to write the most general quasi-shuffle relations, so to speak. So you multiply together 2 multiple zeta values of depth P and Q respectively. And you sum over all the quasi-shuffles which are surjective maps from 1 up to P plus Q on to 1 up to P plus Q minus the number of contractions. And these surjections are subject to shuffle-like, like, like, like, like, shuffle-like constraint. Sigma 1, strictly smaller than etc. Sigma P, the same for the second packet. But you can have contraction terms. In that case, you sum the arguments. Dominique, Frédéric Chappelton asked in the Q&A, he suggested if you full screened your slides, Ah, ok, ok, ok. Let me see, yeah, yeah, yeah. Ok. Let me try just a second. Play d'écran. Is it better? Ah, yeah. Ok. Merci Frédéric. Ok, so let's continue. So much for these quasi-shuffle relations. And of course, you may have to sum maybe two terms but not more, ok? Because it can happen that some sigma in the first row is equal to some sigma in the second row. Then you have to sum. Ok. So now there are shuffle relations because of the integral representation. So, most of you also know that you can have multiple zeta value by means of an iterative integral and you iterate the same the same number of integrals as the weight and you integrate dt1 over t1 and then up to n1 minus 1 and then dtn1 up to over 1 minus tn1 and then you start with n2, etc. Ok. And then, sorry, from this representation you have a second way to express the product of two multiple zetas as a linear combination of multiple zetas. These are the shuffle relations. So let me give the canonical example. You multiply zeta of 2 by itself and then you have to integrate fourfold integral of this form and you integrate over a product of simplices. And this product of simplices you can rearrange it into six different simplices and it amounts to shuffle two words with two letters and you get 4 times zeta of 3,1 plus 2 times zeta of 2,2. And then you have also regularization relations. The first of one was already known by Euler zeta of 2,1 is equal to zeta of 3 and you obtain it as follows. Of course, zeta of 1 doesn't exist. But ok, you say it's something. It's something determinate. Zeta of 1,2 it's also divergent. So it's marked in red. And you apply quasi shuffle relation formally and also shuffle relations formally. Ok, and then you can equate the two right hand side terms. And of course, these two infinite terms cancel and you are left with zeta of 3 is equal to zeta of 2,1. Ok, so of course there is a big conjecture related to the period conjecture of Konsevich and Nagy saying that no other relations occur among multiple zeta values. There has been some steps in that direction. They might look tiny but they are really crucial. It was done about 20 years ago by Rivoyle and Zudeline after the first pioneering work d'Apéry who proved that zeta of 3 is irrational. Ok, so it's a good point to make some historical remarks. Double zeta values were already known by Leonardo Euler as well as a lot of relations relating double and simple ones. I say almost all because you have some, for example at weight 12 you have some relations coming from modular forms by Gengel Kaneko and Zagier, for example. There was not known in the 18th century but apart from these very difficult relations a lot of them were already known. And then there's a gap which is not completely empty I could cite, for example works by Nielsen but then I directly jump to the 80s and I should mention Jean Écal because up to my knowledge it's the first place when you can see the multiple zetas in a whole in a full generality in les fonctions résurgentes in the second volume. And then of course came these works by Don Zagier, Michael Hoffman and the integral representation is attributed to Maxim Konsevich et Don Zagier in his 1994 paper in ICM and this was the starting point to the modern approach with periods for mixed state motifs. And of course I must mention the breakthrough by Francis Brown 8 years ago. Now we know that any multiple zeta value is a linear combination with rational coefficient of multiple zeta values with arguments equal to 2 or 3. This was the conjecture by Hoffman. Ok, so very quickly I mentioned multiple polylogarithms in one variable so you start with integral representation of multiple zetas but instead of integrating 0 to 1, you stop at some t between 0 and 1 and quite easy computation can show you that you have also a nested some representation of these multiple polylogarithms. Ok but I want to express these polylogarithms in another way so I introduce these two functions of t 1 over 1 minus t and I introduce the operators multiplication by these functions big x is multiplication by small x this map big y is multiplication by small y and the operator r is the integral the primitive operator this is well known it is a rotabaxter operator 0 et avec these three operators you can define you can find a formula for your polylogarithms you start with a constant function equal to 1 and you reply to it this row of operators ok and this gives you the polylogarithms and of course when your row of integers is a converging row when you set variable t to 1 you recover the multiple zeta ok there's the word description ok maybe you know all this I'll be quite quick you look at an alphabet with an infinite number of letters y star is the set of words with letters in y you look at the linear span and you define the quasi shuffle product which is just an abstract way of rewriting the quasi shuffle relations the sum is looking the same because on the span of y itself you have a kind of internal product given by y i multiplied by y j is equal to y i plus j and then you can define the quasi shuffle product like this and for later use you can also use the shuffle product this is the same sum but you stick to r equal to 0 I mean you stick to ordinary shuffles and you want to discard all the contraction terms so here is an example of quasi shuffle product and an example of shuffle product because here you have these two contraction terms when you multiply y2 by y3 or y2 by y1 you get these two contracted terms in the shuffle product you discard them oh I keep trying to write my talk ok things that I probably should try to understand but I don't want ok so shall I continue ok so I I call convergent words the words which doesn't which don't start with y1 and for me zeta quasi shuffle of words is just zeta of the corresponding arguments but then you can extend zeta quasi shuffle linearly and applying to linear combination of words and the quasi shuffle relations are just summed up by this relation ok so here is an example so this is the ordinary way of writing the quasi shuffle relation and this is the word way you can see the correspondence I go to to the work to give a meaning to this sum and we succeeded but just looking at the quasi shuffle relations so what is about exactly we can find the character of this algebra q of z equipped with the quasi shuffle product with values in the complex number and q on z, q of z is the the span of words with letters in z ok reflecting that I also want non positive arguments in my extending multiple zeta values ok so there exists a character so that phi of v is equal to zeta quasi shuffle of v for any convergent word converging word and also sorry if you take a word so that zeta of n1 up to nk can be defined by analytic continuation then I want that the value of the character coincide with this analytically continued value so in particular phi of minus 1 minus n is equal to the non value for zeta of minus n and for in depth 2 you have also some values which can be reached by analytic continuation and our character coincide with this so here is the table we obtained in depth 2 so you can see for example some diagonals some diagonals they have the same argument everywhere except at the upper upper right corner these diagonals are values obtained by analytic continuation this one also etc the other diagonals they are not obtained that way and for example there is a work by Liguon Bin Chang which was published a little bit before us and the table of values is different but of course the values obtained by analytic continuation they are the same fortunately but I'll go back to the fact that there's no uniqueness for this problem ok so maybe I'll skip the sketch of proof because time is running but just a few words we first go to regularization that means that we were able to obtain a character with value with values in the miromorphic functions this was not so trivial to prove that this intermediate character this regularized character takes values in the miromorphic functions it has to respect quasi shuffle and not shuffle that's why we have to twist with the so called non-exponential but all this works and then we used Birkhoff Conkramer decomposition of our character with values in the miromorphic functions we use minimal subtraction scheme and and then we evaluate the positive part 0 so this is a quick reminder of the Birkhoff Conkramer decomposition you can write your character like this as a convolution product of two characters 5 plus takes values in the functions which are holomorphic at 0 and this one basically takes values in the pole part and there are recursive formulas for these two characters 5 minus and 5 plus for 5 minus you project you will project this red expression on the pole part and this is the Birkhoff's preparation map and the same expression if you project on the on the other part on the positive part you get the positive character 5 plus and now here is the definition of our normalised multiple zeta values you evaluate 5 plus at z equal to 0 ok so now we have one solution we even have one parameter family of solution because you can you can decide what zeta of one is for example but now we want to describe all solution to the problem the problem was to find multiple zeta values at any argument of any sign so that the quasi shuffle relations are still verified ok so this is our work with Kurush and Johannes Zinger and there's a renormalisation group behind all this ok so you can start with any commutative connected filter of algebra and you look at commutative unital K algebra and you look at the group of characters of H with values in A the product is given by convolution and here is our first result you look at the right co-ideal with respect to the reduced ok so you want the delta delta tilde of N included in N tensor H and the co-unit applied to N is equal to 0 then this set is a subgroup subgroup of the group of character so this set is the set of the characters which vanish on this right co-ideal and this is a subgroup of the group of character so here is the proof it's not very complicated you you just take alpha and beta in TA you want that alpha multiplied by beta inverse is still in TA so you compute this on an element of the co-ideal and you want to find 0 so you you say that ok beta minus 1 is given by composing with antipode so now you display what is convolution ok this is swiddler's style notation and then you replace the antipode by its recursive expression here ok and then you can you can compare this term with that one here and you recover alpha of w minus beta of w plus a sum of alpha minus beta of w prime and because n is the right co-ideal these terms here vanish and the whole sum above vanish vanishes and you are done ok so we call it the renormalization group associated to the co-ideal n and now we look at zeta for us for the moment it's a partially defined character that means that it sends one and it's multiplicative whenever each term is defined ok and you look at xi zeta a is equal to those phi's in the group of characters so that phi restricted to the co-ideal is equal to zeta and we showed that it's a principal homogenous space on the renormalization group ta so we have a free transitive action and the proof is ok I'll skip it but it's quite the same than the proof that ta was a group ok so now we apply that to our special case the base field will be the field of regionals our hopf algebra is q of z star where z is the alphabet on relative integers and the target algebra will be just the complex numbers and the right co-ideal is the liner span of non singular words that means words y is equal to sorry w yn1 up to ynk is in big n if and only if n1 is different from 1 n1 plus m2 is not in this set and you have this third condition in depth bigger than 3 so there are the conditions to the definition of the singular locus so these are called non singular words and if you think a little bit about it big n is the right co-ideal for the deconcatenation co-product what matters is the left part of the decomposition if you want and moreover it is stable by contractions so sigma will be the complementary the set of singular words and it's related to the singular locus of the beginning ok and now our partially different character zeta will be zeta will be the multiple zeta value whenever it exists either by a direct convergent or by analytic continuation and now we can say that the set of all solution to our initial problem is given by x zeta c is equal to the renormalization group acting on a particular solution unfortunately we know that a particular solution exists ok and ok what else not so much a part that we are convinced that this group is very big at least it's infinite dimensional ok now so to speak we have renormalize the quasi shuffle relation what about the shuffle relations of course it's a more delicate question because shuffle relations if you go to the negative side that means that you have to shuffle 2 packets of cards with a negative number of cards which doesn't make so much sense so fortunately by going to q analogs we can have a reasonable glimpse on this so I start with a Jackson integral everybody knows ok here I have illustrated it with 2 values of q q equal to 1 half q equal to 3 quarters and you can see that if q goes to 1 your Jackson integral goes to the ordinary Riemann integral ok but the parameter q you can consider it as an indeterminate and it becomes the Jackson integral operator becomes a non-demorphism of this algebra tq of tq the I will use 2 indeterminates ok I will need a weight minus 1 rotabax operator given by pq so it is f of t minus plus f of qt plus f of q square t etc ok and here in our algebra a it's invertible and the inverse is given by dq f of t equal to f of t minus f of qt so we have modified Leibniz rule which it's just comes from the minus 1 weight minus 1 rotabax operator rotabax identity sorry and we have 3 identities which are equivalent because pq is the inverse of dq ok so now I define multiple polylogarithmes with parameter q so I recall the definition of the ordinary multiple polylogarithmes if you replace the Riemann integral operator by the Jackson integral operator you get your q multiple polylogarithmes ok so for later use I also need y bar of t which is equal to 1 minus t and big y bar will be the operator of multiplication by this function and here comes the Onno-Okuda-Zudelin model for q multiple z values so here is the definition you evaluate the q multiple polylogarithmes at q and explicitly this gives you this q of n1 up to nk this is also nested sum but you use q to the m1 on the numerator and you use quantum integers at the denominators ok of course when q goes to 1 in a reasonable way you recover your ordinary multiple z values when the world converges you can also express them in terms of the operator y bar sorry maybe I skipped something ah yes ok you see we have zqs and we have zq bars ok this is just a normalization of my multiple zetas by the 1-q up to minus the weight ok and this modified q multiple zetas can be expressed in terms of the operators pq and operator y bar ok other models for q multiple z values exist you evaluate your q multiple polylogarithmes at 1, not at q the Chao Bradley model like this and also the Barman-Kuhn model with multiple divisor functions and now Henrik Barman and Wolf-Kuhn have developed a model free approach of all these q multiple zeta values which is very nice but I have no time to develop this I'll stick to the Ono Okudazu-Dilin model ok because this model is interesting because you can extend to arguments of any sign it still makes sense that you have an estimate here I mean for us now q will be a complex number of modules strictly smaller than 1 and for any argument of any sign this formula for your modified q multiple zeta still makes sense provided you stick to the fact that the inverse of pq is equal to dq here ok so here are some examples with argument 0 or minus 1 for example and with corrosion we were able to describe double q shuffle relations and just before us Yoshihiro Takeyama did this in the Bradley model but only in the positive side and the expression is more complicated in this model ok so first of all the q shuffle relations we need an alphabet with 3 letters but with d equal to the inverse of p so a non-empty word with these letters writes uniquely like this with positive or negative powers of p and this is the definition of the shuffle zeta shuffle q zeta word description ok so now by abstracting the relation we had here I define a shuffle product on this word and we prove that it's commutative and associative and the q shuffle relations can be written like this the product of 2 q multiple zetas is the q multiple zeta of the shuffle in that sense of 2 words ok we have also q quasi shuffle relations so this quasi shuffle product here can be deduced from your ordinary quasi shuffle product by a kind of translation operator like this and at the end of the day you have q quasi shuffle relations like this and for example if you have 2 letters you have the 3 ordinary quasi shuffle terms plus 3 other terms with a weight drop ok, the weight is not conserved but if you look at the non modified q multiple zetas you have these 6 terms but the weight drop terms have 1 minus q in front so when q goes to 1 you end up with the ordinary quasi shuffle relations which is not a surprise of course ok there are no regularization relations because all this is defined for arguments of any sign so that's ok and you just have a change of coding from the quasi shuffle picture to the shuffle picture and repeat the summary of the situation you have quasi shuffle relations shuffle relations and that's it ok let me finish by an example of commutation I want to multiply zeta q of 1 with zeta q of 2 so with q quasi shuffle I have my 6 terms the 3 ordinary terms and the 3 weight drop terms with q shuffle I translated into words and then I use the shuffle relations in that sense so you have to play with this these rules here and at the end of the day you have this linear combination of words so you get this sum of the 4 terms ok so now I have to compare this result with the result above and you will have quite a lot of cancellations and you end up with this so it looks like very much the Euler's relation plus weight drop term this is more apparent in the non modified setting and interestingly enough of course you recover Euler's relation when q goes to 1 and this was already settled in the 30s by Bailey we thank Vadim Zoudilin for having pointed this to us and Bailey in this article attributes this identity to Bell and it was learned to him by Hardy it's quite interesting but now we have an algebraic framework for this ok so some open problems it would be interesting to find out other relations maybe there are some other relations among q multiple zetas than double shuffle we don't know and most importantly we would like to find a compatible co-product because ok this parameter q re yields a natural regularization of your multiple zeta values and you have a double shuffle picture which is very nice but when q goes to 1 in general all this explodes unless your world is convergent so we would like to have a renormalization and why not apply the Birkhoff concrime renormalization but for the moment we didn't succeed because we lack a reasonable co-product compatible with all this structure ok here are some references and thank you very much for your attention alright thank you Dominique are there questions a question in the chat actually in the chat it says what are the geometric or quantum or arithmetic applications of q MZT models ah this we don't know what are the geometric applications applications ok is it possible to precise a little bit the question here I will let ask or unmute and they can ask themselves potentially anyway I think in the chat it was more general geometric or quantum or arithmetic applications so any applications ok so our motivation was of course first of all to understand this model which was rather new at that point and once we notice that this model makes sense for arguments of any sign of course we wanted to understand that more closely but for other applications maybe it would be interesting to go to q analog of kz equations but we have to do that but we didn't do that yet alright I see David has a question yeah Dominique do you have anything to say about the limit q goes to a root of unity on the unit circle ah interesting no no but ah non that's a very good question yeah I think that was a question by Goulsh in the chat could you say a few more words on Henrik and Ulth model free project oh as far as you remember as far as you remember I would have to to dive into their work again but ah yes I mentioned this bracket notation related to this multiple derivative functions and they have a more general model with double brackets and they have a conjecture up to my knowledge it's not solved yet their conjecture is that double brackets coincide with simple brackets you don't create new objects by going to this more general setting and ok just a comment I saw Henrik Bachmann is among the viewers so anybody has questions to Henrik maybe they can catch him later in the break also I was just going to ask quickly about the stuff that not even from the main part of the talk but that nice table of rational numbers I can't remember the nice table of rational numbers what can you say about interpretations or patterns in those I mean clearly I see some factorials in the denominators and things like that but what can you say about that table ok it's it's mostly based on Bernoulli numbers so for example in the previous page you can see the values at these famous nice diagonals so you can see Bernoulli numbers in the other diagonals the formula is more complicated but you also have some of products of Bernoulli numbers so that's why you can see 691 691 here all these kinds of things Henrik can say a word if he wants I've set him so he can unmute yes hi there Dominic how are you thank you for your talk and the question of course was what about this generalized model so the idea is just that we take arbitrary polynomials in the numerator and then consider the space spent by these models but it turns out that I mean you mentioned for example this Schlesinger model so it turns out this space is also just spent by the Schlesinger model if you allow for indices if you allow them to be 0 and then this conjecture which Dominic mentioned this whole space of the Schlesinger model where you allow 0 as an index is the same as the space where all entries need to be greater equal to 1 so this is equivalent to the conjecture Dominic mentioned but yeah so this generalized idea is just we don't fix a model and just consider the space spent by these types of q-series and all these models which were mentioned and today are in there yeah but in fact among these models some do make sense at arguments of any sign and some do not so that's why we liked very much this model by Zoudeline Ono and Okuda I mean you consider and somehow in bigger space when you allow these negative indices we just consider this positive or positive part alright well in the interest of time why don't we all thank Dominic again


nto account existing benefits being retained for existing claimants in your analysis or what would the impact be of that has that been factor then our phase one of the group's report looked at this issue in terms of the transition and what we said was the primary focus in terms of delivery should be on the current claimants they should not have their benefits disrupted through this period of transition that was the most important thing that we said as number one in terms of the option appraisal we said over a period of time there should be a dual system the Scottish Government responded to that by saying that would be a very short period of time system operated between UK Government and the Scottish Government two year process maximum so I hope that answers your question there would be a period of time mainly because without that focus on the benefit we would disrupt people we would unlikely make people concerned about the security of their benefit and we all know how important that is if you're on a very low income to have certainty about your benefits whatever the change is taking place and my final question and the final question for this morning the issue of affordability I think you measured it in terms of GDP did you look at measuring affordability in terms of any other matrix? we looked at it in relation to GDP yes we looked at it in terms of not just that kind of relative affordability we looked at it what do we in Scotland currently pay for and what is the cost of that and we pay for what we get so in terms of a simple affordability measure there's no gap between what is raised in Scotland and then through the DWP payback so that was the simplest measure of that there was no gap if we had been looking at this in another jurisdiction of the United Kingdom we had found a gap we would have had to say to you yes there's a gap here which you're going to have to fill so not only was it affordable in terms of GDP that is more affordable in terms of percentage of GDP than the UK more affordable and a significant number of OECD countries and European countries importantly it was also currently affordable because it was paid for by Scottish taxpayers and therefore all those three men mean that we kind of moved on from that relatively quickly although I must say we did have something from the Institute of Fiscal Studies on our group and they went through all our figures so I'm quite certain we got that really clearly right I think the issue again going back to my point earlier on and I think one of the committee members made was I think the issue of affordabilities around not doing something and not changing what's there already is that it's far too costly both in terms of personal cost but also to the state that we have a system which is so complex and bureaucratic that actually disempowered people so I think the issue of affordability is much wider and that is that investing in a good system makes good economic sense on behalf of the committee and I thank you for first of all your evidence this morning but secondly for the effort and the work that you put into producing this report which I think is whether we agree with the idea of independence or not it's all thinking about just how we should look at this in the future because I think there is general agreement that the status quo is not an option and I think there is widespread acceptance that's the point and your contribution to that thinking is obviously in your work and it's been very valuable so thank you very much I'll suspend the meeting again for a few minutes to... how do you start to... Again, our fourth item of business today is consideration of the disabled persons badges of motor vehicles Scotland amendment regulations 2014 this instrument was considered by the Delegated Powers and Law Reform Committee at its meeting on the 10th of June the committee did not draw the attention of the parliament to the instrument on any grounds within its remit this morning to consider the instrument we are joined by Stuart Fwbister divisional solicitor of legal directorate Graham Thompson, transport accessibility and road safety team leader transport policy directorate and David Jamison, blue badge policy officer of the Scottish Government and right members to ask any questions they might have on the instrument Linda I think you'd indicated you might want Well I do, yeah it's a bit wider than that actually and I would appreciate your view on it it's a constituency query one which was also submitted to our your say session convener but the lady wasn't able to come along it's a very simple question about blue badges it's the idea of the potential of a duplicate blue badge and it's a very straightforward situation the lady has a blue badge disability car she's completely incapacitated and can't drive at the moment because of advanced problems and therefore has to take her blue badge of her car to put in her family or friends car who drives her to hospital therefore her original car is a badly parked and liable to end up with a fine or something because of this and I just wonder one whether you had looked at the potential of duplicate blue badges for specific circumstances to whether this was in fact something that could be at the discretion of the local authority would there be anything to stop the local authority making your exception cases like this to be helpful just as a general starting point the regulations and scheme just how they're worded they do limit a single badge to a person so it's not something that I think is being thought about in terms of duplicate badges per se and in some of the reasons I think probably why that's the case is just a desire generally just to keep the number of badges in circulation not the minimum but just to the need so a badge is given to the individual not to a particular vehicle so I guess that's the general reason why it is just the single badge to a person I think that the local authority because the regulations state that way the local authority doesn't have the discretion to necessarily apply it I think that the the overall blue badge scheme overall is designed to protect the give the individual a right to a blue badge and for a duplicate if you were to duplicate badges there would be a concern that say two badges could get used at once and that would be a concern overall so therefore it's not something that we have looked at certainly recently anyway I'm not saying it's never been looked at can I just say I get that the idea of trying to limit fraud that's perfectly acceptable but it seems to me that there are always things which come up it's all very well having the right to a blue badge but it seems to be you don't have the right to if you become so that you can't actually use your own car and you have to take your badge away because you happen to have a stucco on or something and can't drive that seems a bit harsh and and I wonder if you could perhaps have a think about that not asking you to change your legislation because I get that there is no discretion at all that can be applied by the local authority if it's not about giving a duplicate blue badge I wonder if you could have a think and perhaps get back to me about what could possibly be done in this case because it does seem to be penalising someone unduly certainly if there's a particular response to the year say that's been brought to your attention particular circumstances has been more than happy with those circumstances Ken and Annan Annabelle quickly please Thanks just to go over the numbers here can you tell me I would just look at the impact assessment to see how many people will be affected make sure I've got the numbers right roughly 100,000 people are entitled to the current DLA mobility component 60% of those is that right 60,000 people in other words qualify for a blue badge or sorry take up a blue badge and you're expecting 20% of those to lose that right under the new assessment 20% 12,000 people Department for Arts and Pensions estimates is roughly 20% reduction across all awards so roughly 12,000 people in Scotland will lose their blue badge that's roughly one 160% would be 60,000 so 20% or 60,000 would be 12,000 that's my arithmetic that's your ethnic text scene sound and of those how many will they'll be losing it but they'll be automatically they should have been entitled to a lifetime badge are they part of that 12,000 are they part of the 20% that they're losing or are they part of the 48,000 that are continuing I guess it's probably easier just to look at the 60,000 figure and we don't know how the provisions in these regulations that actually impact specifically against those who are perhaps expected to receive decreased awards when the reassessor PIP are if they receive no award under PIP but going from that 60,000 figure all of those who have an indefinite award would be able to continue to passport and then the second provision is all who have got a fixed term a higher rate mobility component daily award would if they were disputing their PIP decision once it comes through then they would have recourse to further blue badge for a period of time as they challenge that decision so the two new criteria they do in effect they should capture everybody within that 60,000 figure to some extent or the vast majority of which anyway the way I was thinking that those who should have qualified originally for a lifetime award will actually be a part of the 48,000 because the reason that they qualified first place will still apply I can't imagine they'll be downgraded in terms of the PIP assessment however I've thought that there's also covered those who are going to appeal their downgrade so that'll be the 12,000 is that right? this is the way I'm thinking of it basically 12,000 people are going to lose their blue badge if they appeal their downgrade under DLA to PIP while they're appealing they can keep their blue badge if their appeal is unsuccessful or after a year that will expire and they will lose their blue badge is that right? can I ask did the Government look I'm not sure if you can answer this did the Government look at just allowing those who currently had a blue badge to continue to keep the blue badge the people that currently have a blue badge are entitled under regulations we put in last year to keep that badge until expiry so that that's generally three years that they would or up to three years that they would have of the people that are losing their badge now we don't have exact figures which say this percentage but we reckon that the DWP estimates are approximately 70% of people are on lifetime or indefinite awards and that we're talking about 30% of people being on fixed fixed term awards so it's those 30% that we're talking about protecting from the giving the appeal some of those after the year will will lose their entitlement if the appeal and their appeal is unsuccessful we didn't we weren't able to consider keeping those people on indefinitely so those 30% we couldn't consider keeping those on indefinitely because of those 30% there'll be some that quite legitimately lose their award there's 30% our fixed term their fixed term for a reason because their condition may fluctuate it may change so therefore we couldn't see so you didn't look at it so basically if they're unsuccessful on appeal they will lose their badge double check one other thing what's happening with bus passes has that been dealt with the concessionary travel has that been dealt with a separate regulation altogether this isn't part of the scope of this but we're not aware of anything getting taken forward on concessionary travel necessarily the circumstances around concessionary travel are different and what was put in last year was more gave broader provision than the blue badge things so essentially with the blue badge we wanted to try and capture we we got the scheme the passporting and unpassporting benefits ensured parity and when ensured parity as far as we could between DLA and PIP last year so that was for new applicants to the blue badge scheme what we're doing now is trying to cover all the people that are existing bad shoulders as much as we can thanks very much please just back on what I was talking about earlier just struck me that the legislation which allows local authorities to issue blue badges in the first place to people I just wondered what level of discretion there was in there for local authorities in terms of who gets issued with the blue badge and whether in fact something could be done there in terms of discretion for duplicates I guess the eligibility criteria is quite clearly put in their regulations and to a large extent they need to follow that when they're making decisions as to entitlement it is of course always the responsibility of the local authorities to interpret what's the meaning of those eligibility criteria how they're read and then to apply them to their local circumstance really to a sense of I just wondered if they could have that discretion under that regulation you wouldn't be to bother running away worrying about it for this time. In addition to the regulations which are when it comes to eligibility fairly fairly clear what they should do we do publish an extensive code of practice that the local authorities follow and indicate that they want to follow in the vast majority of cases there are occasionally times where we have discussion groups with the local authorities in fact we had blue badge workshops with the local authorities earlier in the year where we discussed eligibility in the round and the while they essentially use the code of practice and use the regulations to say ok these are the standards that we've got and therefore they do have some flexibility as long as they do it within the regulations sorry I realise I'm not answering the very civil service speak that word thank you for that on that point I just want to go back briefly to the issue of discretion I would have thought but I may be wrong that the local authority must implement what is set forth in the various SSIs but if it wished to go further it would not be prevented by the SSI from doing so and that I've been trying to get to the number of that for a while because I have a particular constituent case and I don't know if you can really provide a definitive response today and I wonder looking then at the blue badge at the concessionary travel if the same would apply or would something different apply to the two questions that have if there's no clear cut definitive response today it would be helpful if some guidance could be given to the committee because these are issues that do come up quite a lot where somebody falls within a grey area what happens to them and it's certainly something I've received correspondence about Local authorities can't issue a badge or a travel permit to someone who isn't eligible in terms of the regulations but then obviously the categories of passported benefits are only part of the story the mobility assessments and things like that but local authority has no discretion to go beyond if someone doesn't fit within an eligible category then a badge can't be given There is always the other route that people can apply for so as Stuart indicated there that people can apply through a mobility assessment and ultimately that's where local authorities have discretion because they are the people that are conducting those independent mobility assessments and they are having to make a judgement yes there are we put in guidance as to how those should go but ultimately there's a judgement call as to whether someone gets a blue badge where they go through those criteria where they go through that non-passported route when they make the application direct to the local authority okay thank you okay finally do you have any under the current system or the system has been where a person lost their entitlement DLA or that entitlement was changed would they have and they lose the right to passport on to qualify for a blue badge do they lose that right immediately? The existing badge runs to its expiry date that's what was in the regulations that we put in last year if you lose that if you're not eligible in benefit terms your badge still runs on to its expiry date sorry maybe we're not making myself clear so let's say we've reached this stage where the badge has expired and your circumstances are now such under the scheme as was you lose your entitlement DLA or it's downgraded and you're no longer allowed to qualify for the blue badge under passporting arrangements do you lose that blue badge immediately? whether you're eligible at the point of application likewise if you're one of the categories we're putting in just now if you'd appealed the decision that takes you out of eligibility that if you like creates an interim eligibility for another badge while the appeal is running under the current system that's what this instrument that we're looking at today is putting in I'm aware of that I'm coming on to that in a minute if I'm asking what the system is do you have the interim passport entitlement if you seek to appeal a decision? no so essentially what mechanism you're putting in the Scottish Government is seeking to put in place now is an improvement to the current system because if someone's appealing they're now entitled to a year's stay of execution as it were a new development taking account of the new benefits system the personal independence payment and at the moment if you're appealing thank you as far as I've heard this is responding to a suggestion we made from this committee thanks very much I think we've exhausted the questions so I'm now inviting members to agree to note the regulations given that we suggested them I think that's a very good idea okay can I thank you gentlemen for coming this morning and answering the questions thanks a lot our fifth item this morning next item of business is a report back on a visit that Linda Fabiani, Alex Johnson and Ken Macintosh conducted to Glasgow Department of Work and Pension Service Centre on 3 June to examine the operation of practice of universal credit in Scotland I'll invite Alex to give a short summary of the visit I'll try and keep it very short convener the committee met in Glasgow with Mike Baker director of universal credit across the UK and his team and that included the manager of the Glasgow service centre Moira Wardson we took the opportunity to discuss with them the implementation process and to get some first hand experience and meet some of the people who were involved in that process the highlighted to us that the management team felt that universal credit was a better system than the legacy one and that claimants were finding finding work easier and faster of course this only applies to those who have access to universal credit that is single people in the specific areas in which it has been rolled out and it was explained to us that roll out had been cautious to allow the DWP to learn many lessons from the early implementation he mentioned that 800 issues had been flagged up through their feedback loops and that those were being dealt with we also had the opportunity to meet staff from the Inverness office who had been involved in implementing the pilot there they took the view that universal credit was easier to operate than the legacy systems the focus was more on employment the fact that claimants did not need to move between benefits anymore was seen as a major advantage by them as was the existence of a single phone number contact the majority of claimants 80% had completed their claims online without assistance although assistance is available at the job centre this was much higher rate than anticipated other issues that emerged in the final session included the fact that universal credit may be more expensive to administer than legacy systems because of the coaching costs might make savings through the administration of the new benefit the switch from paper to digital makes the system much faster the initial estimate available online of when claims will be paid and how much is likely to be paid is very popular staff who previously only handled phone calls now process applications online and vice versa operating with housing benefit and social landlords is a new area of activity for the DWP but we spoke to individuals who were engaged in that process there's also been a lower level of complaints than initially expected given that claimants are required to devote 35 hours a week to the job search process however there are not yet seem to be any public data on consumer satisfaction however the opportunity was welcomed to actually talk to people who were sitting at the vests dealing with claims in real time and the indication is that the individuals themselves who are working within the DWP are actually finding this a system that is flexible and allows them in their view to do their job effectively It's really helpful Alex as any of the other members who were on the visit of the point wonder I'd just like to add that I was very very impressed by the commitment shown by the staff in Glasgow in their view that they wanted to assist people and I think that was very very marked there's nothing I can disagree with from what Alex said the one caveat I would put is that we should bear in mind that this is a single person only pilot and I think it will be much more complicated when it starts to affect people beyond that Through an example where it had started with a single person and it got more and more complicated Yeah, there was a dog involved I think Yes Now just to echo that point I think that given the public concern that's been raised about the role of the universal credit I think that we're all struck by how welcoming this is the staff perspective this is not the recipients, this is the people running system I think we're all struck by how enthusiastic actually they were about it about the potential for improvement at the same time we're very conscious these are all young male mostly I think I've got the impression that there are more men than women but certainly all young single people are claimants Yes, claimants are the staff So I'm not sure it's difficult to jump to any conclusions from it but it was quite clear and it is a complex system it's got a live interaction with the tax of the HMRC website which is very good, very useful it allows them to make this initial estimate of how much they're going to receive which I think the claimants themselves find very useful and we're also struck by the low level of complaints about it again I think because people are told a figure quite clearly, quite early on an indicative figure, not the exact figure an indicative figure so some of the benefits were obvious from this the complications I think we'll see as it rolls out It's all helpful it keeps building up a picture of this system so thanks to the members for coming out and giving us a comprehensive report which we'll keep and high on as we move forward As we agreed at the start of the meeting that's all of our agenda items that have been taken in public and we now move into private session


The temperatures are getting cooler. The lengths of daylight are getting shorter, so that means our Bermuda grass is starting to slow down and get ready to go dormant for the winter. To make sure that we don't have brown grass throughout the winter, we oversee with perennial rye grass. That's a cool season grass that gives us a dark green color throughout the winter. It makes it pretty on TV and for our fans, but most importantly, it gives us stable consistent athletic fields for our student athletes. We started off the season with five consecutive home games. That was a challenge to say the least, but now we get to catch our breath, we get to love on the grass a little bit, give it some TLC, not only oversee, but top dress with sand and punch some solid-time holes to vent the field and really get it ready for our next home game. The nice thing about overseeing with rye grass in Alabama is we have a long period of great growing weather for rye grass in the fall. Temperatures are right around what, 80, 82 degrees during the daytime, down into the 50s at night, we're going to grow some rye grass. We use Dr. Hahn in our extension system not only for overseeing practices, but to get a feel for what's happening throughout the state. We can forecast through Dr. Hahn what insects are coming, particularly with our perennial rye grass. We have a lot of disease pressures late fall, early spring, so we can ask the extension system what diseases are you seeing on the golf courses and athletic fields. If we see something, we'll get Dr. Hahn to come over and help correctly diagnose the disease to make sure we're making the correct applications for the specific target. That connection with our extension system, it's critical because we can predict what's coming to Auburn. It's always nice to know what's coming before it hits you.  It's not a gigantic enterprise, so pretty much everybody knows everybody else.