Thank you, Ahmed, for the introduction. It's a pleasure and honor to be speaking at this seminar. Today I want to discuss about the so-called equivalent amount of one number conjecture with coefficients in HECA algebra. So these form a complex web of interlocking conjectures, so I would like to start with concrete questions in one specific example. So fix for the moment p equal to 3 and I consider the following three Eigencass forms. So F1 is of weight 2 and level gamma 014 and it is the Eigencass form attached to the elliptic curve of equation y squared equal x cubed minus 7x minus 6. F2 is also of weight 2 and level 716 and this one is attached also to an elliptic curve. The elliptic curve y squared equals x cubed minus 67x plus 926 and F3 is of weight 4 and level 1640 and coefficients in a rather large extension of q. And this three Eigencass forms are all congruent modulo p. So let me say a word about why I chose this one. I chose this one because they are congruent but they are not obviously related in any other way, meaning they are not the same weight, at least this one is under the same weight, so they do not occur in the homology of the same geometric object. Also, neither of them is free ordinary, so they do not belong to a common HEDA family and in fact this one has a3 equal to zero, whereas these two has a3 different from zero, so they also do not belong to a Coleman family. So they are congruent but apart from that there are not many there are not many relations between them. Let me introduce a few further notation. So for f4 it means for the one f4 it means congruence relative just to the usual key expansion. Yes. Yes, yes. So in fact if you want to be more precise, because this one has not q expansion in z, there is a number field and a single prime over 3 such that this congruence holds. The coefficients are in an extension of degree 15 and there is a single prime in the ring of integer of the extension over 3 such that this congruence holds more precisely. So let me introduce a couple of few more notation. So q infinity will be zp extension of q, gamma is Galois group and as usual lambda will denote the complete group algebra zp double bracket gamma. And we will be interested in the variation of the special values of this L function. So more precisely we consider the L function of fi twisted by some character chi, evaluated at 1 and because I'm interested in chaotic variation I will remove the order factor at p and I will quotient by some period. And here chi is a finite order character of gamma. And I want to ask the following questions about these values. So first can we make sense of them in terms of algebraic objects like in the Bursch and Stundler conjecture for instance. So for f1 and f2 this is the Bursch and Stundler conjecture but for f4 as many people know this is the Block-Cato conjecture. The order factor at p is removed. This is important because it is not defined or because the conjecture should be adjusted. So I am going to state a precise conjecture but the Block-Cato conjecture applies to any, well, the version of the Block-Cato conjecture that I will present applies to any partial function. Second can we predict something about these values? So for instance can we predict the piezic valuation? And I want to distinguish these two questions because for instance the Bursch and Stundler conjecture says that the piezic valuation of this special value is related to the order of char but of course the order of char is hard to compute. So we express something hard in terms of something hard and here I want to know if we can say something definite and easy about this piezic valuation. Three because these forms are congruent we could ask are they congruent? Module p. And finally four are they related in any way? So can we exploit these congruence to say using these values can we say something for one eye can we say something for the other? So if we know the values for f1 can we say something about f2 and f3. So at least about question one can we make sense of them there's a very definite answer in terms of the Block-Cato conjecture for the motive vf attached to a diagonal form f. So this is the motive constructed by Scholl in 1990. And in terms of its three realizations the petty realization the RAM and for MEP the etal-cladic realization. So the only thing that I will need is that inside the petty realization there is a class that I will call that's what I will call delta f. And what is delta f? It's the path from the cusp 0 to the cusp infinity on the modular curve and thanks to Euler-Ponkerizuality you can see this as a class in the petty realization of vf. And so let me state the so-called equivalent amagawa number conjectures with coefficients in lambda for the forms f. This is a conjecture due to Cato in 93. And it says the following if you take an old lattice inside the etal realization and if you put h i iwasawa to be the etal homologing a spec z with p inverted with coefficients in t tensor lambda iwasawa then first of all h2 iwasawa should be torsion as lambda module h1 iwasawa should be ranked one and all the other h i should vanish and second there exists a class zf inside h1 iwasawa satisfying two properties. The characteristic ideal of h2 iwasawa which is defined because h2 iwasawa is torsion should be equal to the characteristic ideal of h1 iwasawa over zf so these are all lambda modules so that's the first property and the second property relates zf to special values of alpha function in the following way you fix an integer n and an integer s between 1 and k minus 1 where k i recall is the weight of f and then first of all you can project h1 iwasawa in the etal homology of z with p inverted but with roots of unity added with coefficients in vf then you can localize this group at p so this gives the galois homology of the extension qp zps with coefficient in the same sheet then there is a dual exponential map from this group to the didgerard zero of vf everywhere it's vfp and that is isomorphic to a certain space sf which is a space of cusp form on which the Hecker algebra acts by dragging values of f tensor over f with fpgn and here what are my notation so f has coefficients in a number field big f and p divides p and gn is the galois group of this extension so so you have these maps and zf is in this group so you can send it into that group and the first property and the first fact is that no no it's isomorphic to the description of an explicit description of this space and so zf is sent to a rational subspace inside this periodic space and furthermore now that we know it's in this rational subspace we can transfer it with c and that's in fact if chi is a finite order character of gamma there is a pegged map depending on chi which sends sf tensor with cgn to the materialization of f and let me recall that there is a class delta f in here and this sends zf tensor one to the special value of l of the dual motive of f twisted by chi and evaluated at s so this class zf and the element h2 iuasawa compute all special values of the motive vf dual twisted by chi and at the integers s so that's the statement of the conjecture and it answers it settles question one above can we make sense of these special values well we can in terms of h2 iuasawa and this class zf let me recall the following theorem which is due to kato in 2004 that in this conjecture one in this conjecture one the statement one and two b are true and under my hypothesis so so remains statement two a so this statement and kato proved that the characteristic ideal of h2 iuasawa divides the characteristic ideal of h1 iuasawa over zf under my hypothesis and uh in fact knows that this the maps are such that it defines you more or less so so so what oh so okay so what what you mean the meaning the precise meaning of this theorem so two b means that the image under certain maps is equal to something and this means just that the something is in the image so so kato constructed a class zf yes he showed that the image under these maps is indeed um delta f times the special value of the function okay and the map is also injective or not so many of these maps are isomorphisms but not all of them of course because this one's a lambda module and then you but the injective i asked you about the injective because uh yes it is injective record rank one okay yeah yeah it is torsion so under under this my hypothesis actually h1 iuasawa is free over and quiet it is always torsion free of course so with that conjecture and theorem in hand i want to comment a bit uh again on our forms f1 f2 and f3 so for them it's actually not hard to see using kato's theorem that conjecture one is true for f1 and f2 and that the special values so s equal one for them are all pid units for one and two so using the theorem it's not hard to establish this statement but the special value of f3 at one is not a pedicule and so we see that first of all special values are not congruent even though never this is never this is always a pedicule it independently of car and this is never a pedicule and so we see that these values are not congruent and we see also that in so far as these values shed light on these values this cannot be an obvious process because of its discrepancy and so the question that's a question from tokyo so are you claiming for all kai so these values are that's unit for all kai yes all kai but you are now considering the super singular case yes you say a3 equal 0 yes is that okay isn't it no that's wrong this is not these are not the values interpreted by a pedicule function right so pedicule functions for in a super singular case for example and so plus minus pedicule functions should be unit that's okay i think yes it's not trivial okay anyway anyway so takashi said this one two is not not so good no that's okay that's okay oh you cannot read not for a bit yeah go ahead okay okay so in order to relate these values we have to introduce fecule algebra so that's that's what i'm going to do now question from vijin you mentioned that some of them is ordinary but some are not ordinary right no no no they are all not ordinary no no no not ordinary okay all all not ordinary yes yes okay on the other hand one of them has a3 equals 0 and the other two have a3 different from 0 so one of them is infinite slope and two of them are finite slope okay okay so so now from from now on p is not three anymore it's just any odd prime and and i consider roba an irreducible modular gq representation which is unremifed outside the finite sets of primes which i will denote sigma roba and i will also fix a finite set of prime sigma which contains sigma roba but which might be strictly larger and i'm going to introduce the following fecule algebra t sigma so this is the inverse limit on the weight of fecule algebra of weight k and here t sigma k uh is the reduced fecule algebra generated by heco operators tl for l not in sigma uh generated over some discrete variation ring o and this is inside the endomorphism ring of all modular forms of a certain level and weight less than k so to roba corresponds a maximal ideal m of this fecule algebra and i mentioned that the localization of t sigma at m is believed to be equidimensional of crew dimension four and this is often known to be true uh one more notation if a is a minimal prime ideal of the reduced ring the tm sigma then i will write r a for the quotient the the integral domain quotient t sigma tm sigma over a and uh i would like to briefly uh draw these rings so starting with uh with uh r a so we have an irreducible component like this and and what i've drawn here is more precisely the spectrum of r a with p inverting and if i want to draw the full spectrum of tm sigma then maybe there's another there's another irreducible component crossing the first maybe something like this and so the full picture is the spectrum of tm sigma with p inverted and because i've inverted p these rings are dimension three but i've drawn um surfaces so the meaning being that at any point on this surface there's an extra dimension which corresponds to the cyclotomic deformation on the cyclotomic variable so so if this point corresponds to a modular form f then conjecture one describes the special values the r function on that line and now if i assume that i'm looking at my f one and f two and f three but it's easy to see that f one and f three are on different irreducible components so conjecture one describes the special values on that line and on that line but i want to relate them so i i need to move first on this surface and then from that surface to that surface and that is uh that is the point of the equivalent tm i get one number conjecture with coefficients in hyperalgebra so now i uh oh one last uh thing on on the whole space so what rings are you the second ring yes sigma yes and this is a localization of the hyperalgebra okay reducible components yes okay and what is the line so so the reducible component so where is it okay so in fact r a is dimension four okay so r a one of a piece i mentioned three so so in fact this r a is not the surface it's a space okay but one dimension of this space is due to the presence of possible psychotomic twist so very all these rings are lambda algebra while lambda is this zp double bracket gamma so i from my picture i've removed this dimension but you have to imagine that on each point there is this line corresponding to a twist so if f1 is this point this point would be f1 tensor by some character but it is it still lies in the reducible component it still lies in the irreducible component you do as if it is okay oh no yeah yeah okay as if i were moving outside of it no oh irreducible component our space is dimension three okay okay and there is no preferred way to cut down the dimension two here is a conjecture that's conjecture two and this is an equivalent amagawa number conjecture but with coefficients in this vector ring tm sigma and one possible formulation is that there exists a z sigma and a delta sigma and for all minimal prime ideal z a and delta a delta a such that first of all the natural projection from from the recurring onto r a induces an isomorphism from delta sigma tensor r a with delta a which sends z sigma tensor one to z a and second if lambda is a modular point of r a so if this is a system of eigenvalue of a eigen cospons then delta a tensor with lambda was our should be canonically isomorphic to the following objects you take the determinant of the complex the italic homology complex of z with p inverted with coefficients in some lattice so it was our lattice in the homology of the module of four and then there is a supplementary term which is this lattice and you take the plus part so the invariant part under complex conjugation so we should have this isomorphism and such a canonical isomorphism and this canonical isomorphism should send z a tensor one to the class z f of the first conjecture and tensor with the dual of the delta class of the first conjecture so meaning meaning that if you take the period map of the image for lambda of z a tensor one for any modular point lambda then this should be equal to the special value at p of v f star one chi s so that so if you remember the first conjecture there was an extra class delta here and now I have incorporated it on that side so that it disappears on this side so so this conjecture roughly states that this class z sigma can compute all special values of all modular forms appearing as points on the ring t m sigma which kind of a check is supposed to be that's exactly the right question so yeah what is delta sigma so of course yes so this conjecture is quite meaningless if we don't at least propose a candidate for delta sigma and for the canonical isomorphism appearing so here is the answer you take up a compact open subgroup into the other in the idyllic point of gl2 outside of infinity and p and define the completed homology of tame level up and coefficients o as the inverse limits on all u u p u lower p of the homology of the modular curve of level up u p u lower p u upper p with coefficients in o so here u p is compact open in gl2 of q p now I take the direct limit on all this level and I point out that inside this completed homology there is a free module of rank one z sigma a module over the action of the hacker algebra which is generated by a certain class delta sigma which is more or less again the path from zero to infinity but seen in this completed homology and and with this z sigma we are going to build the delta sigma so delta sigma at least a candidate for delta sigma is the determinant of the hacker algebra of the et al homology with compact support of z with sigma inverted with coefficients in the galois representation t sigma with with coefficients in the hacker algebra tensor the determinant of z sigma this z sigma minus five and this is how you should think of this object so this this is of algebraic origin or homological origin so you should think of this as homological as algebraic special value you should think of this part as the part predicting the special value and you should think of this part with the delta in it as the actual analytic special value and so the conjecture one way to rephrase the conjecture is to know that it amounts to this part the determinant z determinant part is canonically isomorphic to some module x sigma inside a total fraction ring of tm sigma the second part likewise it's there is a canonical isomorphism to some module y sigma inside this total portion ring and so inside this total portion ring x sigma tensor y sigma is equal to the hacker algebra and so this statement so so we have statement like this and all these isomorphisms should be compatible with the maps first from tm sigma to an irreducible component and from an irreducible component to a modular port but that's a way so a way to understand the conjecture is that this y sigma is the periodic variation of the actual special values this x sigma the periodic variation of the formula predicting the values this statement is that they are equal over the full hacker ring and this compatibility says that if they are equal over the full hacker ring they are equal at each regular point that's how to understand roughly the conjecture so before stating a theorem on this conjecture let me illustrate it by going back once again to our example yeah so t i i actually i i meant to write it down but then forget so t subscript sigma is the galois representation with coefficients in the hacker algebra so it's the big galois representation to our forms f1 f2 and f3 so which belongs to different irreducible components as i said before uh so for them sigma will be the set of prime 2 3 5 19 and 41 and i will point out that the conjecture gives us two ways of computing special values now because if you want to compute special value you can take the class z sigma inside delta sigma and then you map it to uh zf tensor delta f star you raise the top board so that you can compare these oh sure the first okay so so the composition of these two isomorphisms one to the irreducible component then to the modular point tell you that z sigma has to go to zf tensor delta f star which for the period map maps to the special value with p removed in my case s will be one but you can also take delta sigma inside z sigma and map it to some delta sigma f star that will be inside uh the better realization of the motive vf and then compute the special value with respect to this class so so maybe you can answer it with zf and that one is mapped through the period map not to the same special value but with the the special value with all oiler factors at sigma removed so this l sigma is by definition the product for all prime in sigma over all prime sigma of the oiler factor at this prime l times the full L function so so we see a difference we see a difference in these two process and uh exploiting this difference we can now explain what we noticed earlier because if you take l equal to 41 then you will see that the oiler factor at l of the first of our modular font is not a free addict unit and uh if you translate so so this means that this l sigma uh this special value with sigma uh oiler factors removed so this l sigma will have a non-free addict unit part here but if you translate this in terms of um the module y sigma above you will see that this means uh that the module y sigma is not isomorphic to tm sigma which is perfectly fine in terms of the conjecture because it's just x sigma tensor y sigma which should be isomorphic to tm sigma so this means that x sigma is also not isomorphic to tm sigma but that's fine uh however the product of all oiler factors for the modular form f3 that is a piedicule and so again if we translate the fact that y sigma is not tm sigma plus this statement implies that the the piedic valuation of this special value is strictly positive and in fact it implies that this piedic valuation has to be exactly equal to the piedic valuation of this oiler factor so this is very briefly speaking how the conjecture can relate the special values and the oiler factors of one modular form to the special value and oiler factors of another modular form in uh um under the hypothesis that they are congruent so now let me state a theorem so I remind you that p is odd and I will state the theorem and simplifying assumptions chosen for brevity so assume that the Galois representation rho bar is subjective so its image is all of gl2 of some finite extension of fp and also that the local Galois representation rho bar restricted to qp is irreducible and then there's a mild condition uh at primes which are in sigma but not in sigma rho bar and I will state it if needed um in the sketch of proof no no you you choose sigma arbitrarily arbitrarily yeah but in the conjecture I mean and uh but for this theorem well I can tell you the condition do you want to well I won't understand yeah so um the theorem will apply to to um my example of course so the the condition is that if l is congruent to plus or minus 1 modulo p so of course when p is equal to 3 this is going to happen every time then you have to know something about the shape of the local Galois representation at l and in fact if l is equal to minus 1 modulo p which is the case here I guess the local Galois representation has to be reducible which is also the case because uh 41 divides only 1 16 40 so you are Steinberg at uh at error okay but if you had the modulo form so of course not with these conjectures but with uh a super singular a super case fiddle I mean um local Galois representation at 41 then the theorem would not apply um you will I mean at least the proof would not apply and and four uh there exists a modulo point of this Hecker algebra such that conjecture one is true so let me remind you that conjecture one was the original equivalent omega one number conjecture of kato then conjecture two is true so the full equivalent omega one number conjecture with coefficients in Hecker algebra is true and in particular conjecture one is true for all modulo points of tmc1 so if you apply this theorem to our example as I mentioned it's easy to check that the conjecture one is true for f1 and so it implies in particular that it is true for f2 and f3 but I should point out that conjecture two is stronger than the sum of conjecture one for all modulo points because conjecture one for all modulo points describes the variation just in this cyclotomic line whereas conjecture two describes the variation on a full Hecker Hecker ring so it's conjecture one for all modulo points plus congruences between these values so in the remaining 10 minutes or so I will briefly sketch the proof that's correct eight minutes or so so the proof proceeds in three steps three main steps and the first step is descent and this means so this can be sum up in a following proposition the modules x sigma and y sigma of the conjecture compatible with the map tm sigma onto ra and modulo maps ra onto lambda y wasawa in the following sense um x sigma tensor y sigma is sum module m inside the total portion ring of tm sigma so the conjecture says that it is tm sigma but some module m and you can specialize on the left and specialize here and this is this is a community diagram and same with this map so so once you know the relative position of x sigma and y sigma you know the relative position of our images at all module points that all irreducible components I should point out that neither these maps this uh neither I mean x sigma does not go to x a and y sigma does not go to y a but the tensor product goes to the tensor product and the idea of proof of this statement so it's the tensor product of what uh this is about tm sigma and you know it's over i but this is inside r r a the fraction ring which is a finite extension of our a is an irreducible component of the oh yeah okay it's a minimal parameter yeah it's a minimal parameter and you have the same community diagram for lambda y wasawa in which case this is over lambda so this relies on the purity of the galvan representation attached to module of forms some variants of hiara's lemma for competitive logic because if you recall x sigma is something like a determinant of ethyl homology y sigma is something leaving incomplete homology and these are the tools used for the first step if we can establish this first step then notice that uh then it is enough to show that x sigma tensor y sigma contains tm sigma and that there exists a point a module of point such that uh very much from this module of point x is lambda y wasawa because if the containment was strict then in this community diagram we would have a strict containment here so so once you know the compatibility of x sigma and y sigma uh you know it is enough to prove the containment and inequality on one point so we are going to bootstrap this equality to inequality at every point but first we need to prove this containment so that's the second step so the second step is the containment x sigma tensor y sigma here contains tm sigma and this is achieved by a tele-wide system method meaning to construct a projective system of artinian quotient of tm sigma whose limit is a regular ring so for the benefits of experts uh in the original tele-wide formulation this works only if in the minimally modified case and in that case tele-wide constructs such a system and shows that the limit is a regular ring but we all want so that would correspond to sigma is equal to sigma rubar and so that explains the mind hypothesis on l in sigma minus sigma rubar so in the general case one needs to analyze the singularity of the local deformation ring at l in sigma not in sigma rubar to to get this result so so this is a tele-wide system in the in Kissing's formulation and in Kissing's formulation in general you don't get the regular ring and I really need a regular ring so I need this analysis and so I need this hypothesis so the local ring is talking about yeah different from p so exactly I'm looking at the frame deformation ring of rubar restricted to gql and different from tm and that's one of the situation yes yes yes yes so I mean yeah and in some so so these uh these frame deformation this frame deformation ring are never regular rings but the irreducible components might be and uh generally very good this is but still i think the singularity is a lot understood even yeah so you're in dimension two sorry I mean dimension two and in my case as soon as there is one singularity in an irreducible component dimension so that's step two and so what is the use of this regular ring it's because now that I'm over a regular ring I can apply the Euler system method regular ring I will call b and then this way uh this produces uh containment for many points so you can find many points s and and make sense of this xs and ys such that you have such a containment and then you can deduce all that containment and then you just appeal to hypothesis four and hypothesis four tells you that there is one point at which they coincide so they coincide and they start by Tokyo then Beijing and then yes oh there any question so at the last point why why you need to do gravity ah um in in it was our theory so so Carl Rubin observed that if you have um so you want to prove a main conjecture so it's an equality of characteristic ideals and Carl would be absurd but if the characteristic ideals are or maybe a divisibility of characteristics they absurd that if this does not hold then you can find a discrete variation ring for it for which it is very false and that will yield a little prediction but if you are over a non regular ring then possibly something that is there is no divisibility for the ring but there is a divisibility at any image through uh to this iteration ring because to just pattern normalization so maybe the non divisibility is destroyed by normalization so so i mean um phrase in these terms maybe this does not hold at any point because at the point of singularity of this ring this sees to hold and your system method will not see this over t and sigma but if i go to be infinity then i resolve the singularities so i can really detect that so that's what's happening is there one question yeah thank you that's it from Tokyo question from Beijing probably stupid question so in your step wait can you hear me i think okay so in your step one you said you can you need only to show one point to deduce you need to show conjecture one at one point to deduce it for all the points yes um do you uh i would believe that if you need to show one point on each electrical component is it true that you just need to show one point on all all space how do you come and pass through the intersection and go ah jump to the other yeah so um so i can repeat the question maybe so so the question was um i claim in the theorem that if there is uh equality at one point then there is equality everywhere and the question was uh i can believe that there is equality at one point there is equality on the on the irreducible component containing this point but how to pass from uh one irreducible component to the other and and that's so that's that's an excellent question so that's precisely the point so um it it's so i'm sorry i need to to write down something to answer this question so so in terms of fundamental lines this is a statement saying okay so so we have some x a and so y a and this contains our a and then there is a point x such that there is an equality and so we deduce equality on the irreducible component so now how do we pass from this irreducible component to the full space and the point is that x sigma and so y sigma maps isomorphically to this um to this uh module and and i really have a containment on that space as well but this is a highly this is a highly non-ompius uh statement because as i mentioned but probably too briefly it's not true that either of these of these parts map to these parts but um the way the the error term can so so and a more conceptual a more conceptual way to save is to answer the question how do you pass to the other irreducible components is to think of pediatric interpolation so where should i look happy i want here because okay sorry is to think about pediatric interpolation of the local langvance correspondence so it's very easy to periodically interpolate the local langvance correspondence on an irreducible component but thanks to the result of emberton and helman helman his co-author uh gilmos we know that we can actually interpolate the local langvance correspondence on the full space and that's exactly what i do so that's exactly what i make use of so at at the point of intersection there is nevertheless another factor but which is not defined uh as a determinant or and that's how i pass from one component to another component and this is hidden in uh in this so you need each component to intersect with another or i mean it's so the center was connected right oh it's connected no so you need to go through intersections to one well i mean um if i were really moving in uh bottom up way i guess so but in fact you can uh reverse the direction and the result i mean you construct you construct your automotive representation over the full space and then you specialize so but but i mean part of it is that you can specialize i put into our intersection and at point of intersection it does interpolate in a non-obvious way but it does this is the position by the way is only known for gf2 and g3 yes yes yes but you are in g2 yeah so yeah other question from begin my questions okay that'd be all for the you question is okay so if not let's thank the speaker again


Good evening, my name is Sarah and I work in the community ravenment service. So we are a team of occupational therapists and physiotherapists now and we work as part of a regaining independent service, so it's a short-term fast-paced service and we provide support and coordinate sort of long-term care plans for adults in West Sussex across the county. Historically we're split into different teams but to my knowledge we're now in a sort of situation where we're sort of covering cross county which has a lot of benefits to people because obviously we're spread out across the county living in different locations and now have a mix of home and a bit of office working so it's been a really interesting team to join. I joined West Sussex last year, I qualified in 2019 having previously worked in the acute setting and I've been pleasantly surprised actually how well suited I feel to the social care role and I think any experience you gain whether it be through practice placements or in your first role as qualified I think there's never any team you work in where the knowledge isn't then taken forward to your new role. I think my experience working in acute has really informed my understanding of bridging the gap between health and social care so I found that really to have been good stead for for my role as a social services OT. So as I said I work in RIS which is Regaining Independence. We're a team of OTs and we have one physiotherapist now and we work in partnership with Essex Care's Limited the care provider that cover care for adults within the whole of the county and we have very much goals focused short-term fast-paced so we're trying to really establish from from day one when we receive a referral to establish what's important to customers what is meaningful to them and really focus on reablement and determining what needs are longer term if they have needs identified. We work very closely with the social workers and we work towards assessments that are connected to the social care act so social care act is our bread and butter go to for information in terms of how we assess and well-being and putting customers needs and wishes first. My experience to date since joining the team eight months ago nine months ago now is it's been incredibly supportive I can't fault the team really in terms of their flexibility the professional approach to supervision absolutely oodles to support with in terms of learning if there's anything you want to learn the sky's the limit really I think there's plenty of scope to if you have a particular interest within community OT and whether it be moving and handling whether it be a condition-specific training yeah the sky's the limit I think if you express an interest in a particular area I think the management and supervision has been very open to supporting you in your interests and I think just mentioning before about flexibility of working I mean life happens people have commitments outside of work and health needs and personal needs and I can't say that I haven't ever been supported with that so yeah really open to managing work and life and being realistic about trying to be the best practitioner you can sort of really understanding well-being for ourselves as practitioners in terms of early practice support so I'm due to start the early practice framework next week and I've had time to to actually take on and learn about my role first so I've taken on a case load and really get myself stuck in and really understand more about the role before starting that I didn't have that opportunity in the NHS it was very much a preceptorship from month one and I think it's nice to really find your feet first and then take the opportunity to embark on further training so that's been really really good application I think really swatting up on again the CareRats as I mentioned earlier thinking about social care resources that are available online and also just making reference to some of the adaptations without delay documentation from the Royal College of OTs found that guidance really helpful to to support me in the interview questions I had when I applied for my role and have a good take measure I use my take measure almost every week so yeah definitely have a good take measure all sorts can come into my emails or from a telephone call I think we have a couple of different pathways so we have our community reablement customers that come in we also have a discharge to assess pathway the customers are discharged from hospital settings into the community and require ongoing assessment social care assessment and also customers that come our way that have got long-term conditions that sometimes need packages of the care starting and need some initial assessment to support before we refer on to the independent living service a very very case load every week's different learning every day and thoroughly enjoy the challenge and I quite like the faster pace of it so having that sort of variety but also being quick on your feet to to make some of those decisions I think for me a big a big challenge initially was stepping away from the stage of the acute acute setting so having professionals help professionals at your fingertips and having that supervision to hand informally and I found that quite quite a challenge initially in terms of stepping back and being more autonomous in my role and obviously with new ways of working we've relied heavily on on being home-based and having to book meetings in and times to speak to people in the team and but again that will change through the year as things begin to open up but I think it does give you the opportunity to really work on your time management skills being more autonomous practitioner and also your resilience as well managing your case load and I think there's always support there so it's never never feel like I'm too overwhelmed if the numbers are creeping up in terms of wait times and I feel like as long as you communicate regularly and manage your time well I think I think there's always support available


Hello. So, we continue our discussion on Fourier transforms of tempered distributions and we studied the one variable case in detail. Now, we go on to the several variables. Now, the Dirac delta, the Dirac delta is a as a distribution in R n. Now, the first question is show that the Fourier transform of the Dirac delta is a constant function 1. It will immediately follow from the definition exactly as the one variable case. So, exercise 16 that I showed you in the first slide we will skip over. This next question that we want to discuss is the Fourier transforms of the derivatives of the Dirac delta. When I said derivatives, what kind of derivatives are you talking about? Suppose we are talking about R 2, am I talking about del 2 by del x del y or am I talking about del 2 by del x squared or am I talking about del 2 by del y squared. So, take a multi index alpha alpha 1 alpha 2 alpha n and you take del del x 1 alpha 1 times del del x 2 alpha 2 times delta del del x n alpha n times. Compute the Fourier transform of this just apply the definition put the hat on the G and then transfer the derivatives out there and you can see that the Fourier transforms of the derivatives of the Dirac delta are basically monomers. So, if I take del del x 1 to the power alpha 1 del del x 2 to the power alpha 2 etcetera it will be chi 1 to the power alpha 1 chi 2 to the power alpha 2 to the power alpha n to the power alpha n. There will be a constant coming in because the Fourier transform exchanges differentiation and multiplication by the coordinate function, but there is an I factor floating around. So, there will be some I to the power mod alpha or something like that I leave it to you to figure out the details. So, that is easy find the Fourier transform of the derivative of the Dirac delta which what is the answer the expected answer is going to be a monomial. What about the Fourier transform of monomial suppose I take a monomial x to the power alpha what will be the Fourier transform no prices for guessing it will be the corresponding derivative of the Dirac delta, but there will be some constant because of the 1 upon I factor. The Fourier transform of the monomial x to the power alpha is going to be a derivative of the Dirac delta the corresponding derivative of the Dirac delta and the Fourier transform of the derivatives of the Dirac delta are the monomials. So, 16, 17 and 18 are through let us take the next question question number 19 I take the unit disk in R 2 x squared plus y squared less than or equal to 1 and I am looking at the characteristic function of D that is a function which is identically 1 on x squared plus y squared less than or equal to 1 and 0 outside the disk and that is a nice L 1 function if you like and so you can compute the Fourier transform using the directly the formula, but I do not want to compute the Fourier transform I want to compute the derivatives of this temper distribution because in L 1 function it is a temper distribution. The characteristic function is not differentiable in the classical sense, but in the classical sense you can still compute the derivative on x squared plus y squared less than 1 in the interior of the unit disk you can compute the derivative it is 0 in the exterior of the derivative x squared plus y squared greater than 1 it is 0. So, the derivative of the characteristic function of the disk D that is going to be 0 inside the disk and it is going to be 0 outside the disk. So, the support of this is going to be the circle x squared plus y squared equal to 1. So, it is going to be a distribution the derivatives of this distribution is going to be a distribution which has support along the unit circle x squared plus y squared equal to 1. What is this distribution? Let us find out. Let us compute the x partial derivative del u by del x.  So, the pairing is basically an integral over d del g by del x dx dy that is what you see in the slide here. Now, use a Gauss divergence theorem to simplify it. So, you got del g by del x and you can apply the Gauss divergence theorem and you can get minus integral over the boundary n 1 g ds. This equality how you apply the Gauss divergence theorem I am going to leave it to you as a vector calculus exercise. Gauss divergence theorem and you will ask me there should be a vector field in the background. What is that vector field? I can give you that vector field. g i plus 0 g that i j are the standard unit vectors in r 2 or an r 3 i j k are the standard unit vectors if you want to use the undergraduate vector calculus notations then i g plus 0 j plus 0 k what are the divergence of this del g by del x and so the double integral that you see is the divergence and so it is going to be what it is going to be n dot the vector field what is the vector field again g i what is the outer normal n n 1 i plus n 2 j. So, n 1 i plus n 2 j dot product with g i plus 0 j what is going to happen it is going to be n 1 g and that you are going to integrate over the boundary of the domain del d. So, that is how we get this. So, we get that the distribution del u by del x is obtained by restricting this function g on the circle x square plus y square d equal to 1 and integrating n 1 g d s. Similarly, you can calculate del u by del y again it is going to be minus integral over the boundary n 2 g d s. Now, I am going to ask you what is x del u by del x plus y del u by del y what is y del u by del x minus x del u by del y please compute these two and one case you are going to get 0 in the other case you are going to get something interesting. So, I leave it to you to complete the problem. So, now let us prove the theorem of Liouville that I have promised last capsule the last capsule we proved that if you take the Laplace's operator Laplacian of u if u is a tempered distribution Laplacian of u is also a tempered distribution what is its Fourier transform use those formulas or the Fourier transform exchange in the derivatives with multiplication by the coordinate function remember there is going to be an I floating around in one of the formulas there is going to be a minus sign floating around, but you are applying the formula twice because it is Laplacian. So, Fourier transform of Laplacian of u is minus mod chi squared u hat that is the function you need to keep in mind ok. So, now we are going to prove Liouville's theorem what is Liouville's theorem in complex analysis it says a bounded entire function is constant but an entire function satisfies the Laplace's equation right if f is entire then Laplacian of u is 0 Laplacian of v is 0 f is u plus i v Laplacian of f is 0. So, an entire function is a harmonic function it satisfies the Laplace's equation and I am saying that the entire function is bounded which means it is a tempered distribution it is a tempered distribution, but why talk about R2 why talk about complex analysis we will work in Rn we will generalize Liouville's theorem and say that if you have a tempered distribution in Rn which satisfies the Laplace's equation then the distribution is actually the polynomial further if you have a tempered distribution which satisfies the Laplace's equation and if that distribution happens to be in LP then u must be 0 if p is between 1 and infinity infinity excluded and if p is infinity that is if the function is bounded harmonic function then it must be constant. So, theorem 118 is a profound generalization of the classical Liouville's theorem from one complex analysis that you teach in undergraduate complex analysis courses and the proof is going to be very simple because we are developed all the machinery that we need to establish such results. So, let us take the Fourier transform of Laplacian of u equal to 0 as I said you are going to get minus mod chi squared u hat is 0 since 0 is in the right hand side I do not have to put the minus sign which means that the tempered distribution u hat vanishes away from the 0 because if chi is not equal to 0 I can divide by mod chi squared and u hat will be 0. So, u hat will be 0 away from the origin which means the support of u hat u hat is a distribution which has point support namely the origin and last time I stated the theorem without proof that a distribution with point support is a finite linear combination of Dirac delta and its derivatives. So, it is going to be summation mod alpha less than or equal to n c alpha delta not to the power alpha that is you have to take alpha derivative the Dirac delta and so what is u? u is going to be obtained by applying hat again and doing the reflection business. So, when I apply hat again I am going to get u of minus x when I take the Fourier transform of the derivative the Dirac delta remember the very first thing we did in the today's capsule that is going to be a monomial x to the power alpha and there is going to be some minus signs and powers of I floating around. So, in anyway those constants are going to be clubbed with this c alpha. So, u is going to be a polynomial in the variables x 1, x 2, x n. So, u is a polynomial and the only polynomials which are in LP are 0 if p is not infinity and the only polynomial that is in L infinity is the constant polynomial and that completes the proof of Liouville's theorem an interesting theorem in partial differential equation that we picked up since we are done a lot of work ok. Now, we talk about a concept called restriction or localization this concept is very familiar from elementary calculus you take a function say sin x on the real line and you want to talk about restricting the function to an open subset. So, you want to restrict sin x to the open interval minus pi by 3 to pi by 3 for example. So, there is a restriction operation and you will denote the restriction by f vertical stroke g we want to discuss the corresponding notion for distributions how do you localize a distribution restricting a function to an open set means localizing the function to the open set just taking the restriction the way to do that would be to take those functions which are compactly supported in g. So, we take a tempered distribution on the real line again we shall discuss it only for the one variable case the multivariable generalization is routine. So, take a tempered distribution in one variable then the restriction of this tempered distribution to an open set g simply means you pair you with those g's which have compactly supported in capital G that is equation 10.17 that you see in the slide. So, this is the business that we are talking about. So, we use the same notation you restricted to g to denote this map 10.17 the support of you is the complement to the largest open set. So, take the largest open set g such that the localization of u to this g is 0 and this g is an open set and the complement of this open set g is the support of the distribution. So, the support of the distribution can be reformulated in this language of localization. Again you know from elementary analysis supposing a two functions f and g and you know that f restricted to g equal to g restricted to g and this happens for all the open sets in f and g are equal. So, localizing is a way to compare two functions in small neighborhoods and see. For example, if you want to discuss whether the function is continuous you want to discuss whether the function is holomorphic you localize the function and you check continuity holomorphy these are local properties to. So, for example, if you if a function f is defined on an open set in a complex plane and to understand whether this function is holomorphic you take a small open ball or an open disc restrict the function to the open disc and check whether it is holomorphic in that in that open disc. So, restricting is a very important idea localizing is very important today. Now suppose if you have two distributions u and v such that u localize to g is the same as v localize to g that means that u minus v when localize to g is 0 that means that the support of the distribution u minus v is outside of this g that is its intersection with g is empty support is contained in g complement.  Now this raises the following question let us turn this around the question that can turn around is suppose I give you a smooth function on r minus 0 1 upon x is an example another example could be 1 upon sinh x hyperbolic sin x 1 upon hyperbolic sin x is a perfectly nice function away from the origin right what is hyperbolic sin x 1 half of e to the power x minus e to the power minus x the derivative is always positive is a strictly increasing function and it vanishes at the origin and it does not vanish anywhere else.  So, given a smooth function f definition 120 given a smooth function f on an open set g in r can you extend this smooth function to a distribution on the real line such that the distribution localized to g is the given smooth function. In other words it is an extension problem you are given a smooth function f on an open set can you extend it as a distribution 1 upon x is smooth on r minus 0 of course 1 upon x is not going to extend smoothly on the real line it won't even extend continuously on the real line forget about smoothness but I am not asking for a continuous extension I am not asking for a smooth extension I am asking for an extension which is a distribution which is a tempered distribution tempered distribution can be ugly no problem it can be like Dirac delta and its derivatives pv1 over x and also all sorts of things. So, the question is there a distributional extension. So, for example that is the next question is does the smooth function 1 upon sinh x the hyperbolic sin does it extend as a tempered distribution on the real line and is this extension unique. Obviously the extension cannot be unique right because if you have one extension add to it a Dirac delta or its derivatives I am going to get another extension. So, the extension is certainly not unique. Let us write 1 upon sinh x is 1 upon x into x upon sinh x minus 1 plus 1 upon x as a simple routine algebra look at the first piece here 1 upon x into x upon sinh x minus 1 remember x and sinh checks both have a simple 0 at the origin they are both holomorphic functions having a simple 0 at the origin. So, x upon sinh checks does not have a 0 at the origin the origin is a removable singularity and x upon sinh checks is a perfectly nice function near the origin and its value at the origin is 1 and a subtracted of 1 look at the Taylor series for x upon sinh checks what the Taylor series for x upon it is an even function right and it vanishes when I subtract of 1. So, x upon sinh checks is 1 so I subtract of 1 and it vanishes and so what are the next term in the Taylor series x squared and the x squared will cancel with the 1 upon x. So, this first piece out here is a perfectly nice function it's a smooth function and it's a tempered distribution because it it doesn't grow it grows in fact it's bounded and the second piece is 1 upon x and you know 1 upon x has an extension to the whole real line what is the extension it is p v 1 over x it extends as a tempered distribution. So, the answer is yes that it does extend and the extension is not unique what about 1 upon x squared remember that distributions can be differentiated as many times as you want take p v 1 over x and differentiate and see what you get you will get some other distribution try to restrict it or localize it to r minus 0 and see whether you get 1 upon x squared the next problem gamma i x modulus the whole squared on r minus 0 remember when x is not 0 i of x is purely imaginary x is not 0 and real number i of x is purely imaginary and non-zero where are the poles of the gamma function the poles of the gamma function are 0 minus 1 minus 2 minus 3 etcetera right. So, mod gamma i x the whole squared gamma function is a is a holomorphic function of the complex plane except for poles at 0 minus 1 minus 2 minus 3 etcetera and so what is this mod gamma i x the whole squared remember that what is gamma z bar it will be gamma of z bar gamma is real when x is real remember gamma restricted to the real axis is real of course it has got poles remove the poles of course so it means that gamma z bar is gamma of z bar. So, what is mod gamma i x the whole squared is gamma i x gamma i x bar in other words is gamma i x into gamma minus i x but gamma minus i x means multiply and divide by minus i x. So, gamma minus i x is going to be gamma 1 minus i x upon minus i x but what is gamma z into gamma 1 minus z the Euler's reflection formula pi by sin pi z. So, this object mod gamma i x the whole squared is basically pi upon minus i x sin i x sin i x is i sinh x and so the answer is this particular function simplifies to pi by x sinh pi x and now the question I am asking you is 1 upon x sinh pi x that is a smooth function on r minus 0 does it extend as a tempered distribution on the real line I think after doing question 20 and 21 it should become easy now the next question is a difficult question now you can say 1 upon x has been extended as a tempered distribution 1 upon x squared has been extended as a tempered distribution of the real line 1 upon x cube 1 upon x to the power 4 you will believe they will extend as tempered distribution. What about e to the power 1 upon x squared that is also smooth on r minus the origin will e to the power 1 upon x squared extend as a tempered distribution on the real line No, answer is a big no. How do you prove that there is no tempered distribution on the real line such that its localization to r minus 0 is exactly e to the power of 1 upon x square. That's a difficult question and that's why it's an optional exercise. So the problems of this kind of extending distributions from an open set to a whole space. For example, I could give you a smooth function like 1 upon 1 minus x square minus y square. Let us look at 1 upon 1 minus x square minus y square. This function is smooth everywhere except along the unit circle x square plus y square equal to 1. So, from the plane remove the circle x square plus y square equal to 1. On the complement 1 upon 1 minus x square minus y square is a perfectly nice smooth function. Is there a tempered distribution which when localized to the plane minus the circle gives you exactly 1 upon 1 minus x square minus y square. This kind of problems in other words, problems of an analogous nature in several variables become very interesting and they are highly non-trivial and call for sophisticated techniques from algebraic geometry. And these matters are closely related to another problem called the problem of division in distribution theory. This problem of division has been studied by a number of mathematicians, Malgrange, Aranpris, Hormander, Lausaceux and a whole lot of people have studied the problem of division and the literature on this problem on division is very vast. So, in the next capsule, we will take up an interesting discussion on distributional solutions of ODE's with polynomial coefficients. Now, we know that a distribution can be differentiated and a distribution can be multiplied by polynomials. We are going to talk only about tempered distributions in this course. And so, when I say distribution, I tacitly mean it is tempered and these differential equations that we are looking at have polynomial coefficients and it is an interesting question whether they have solutions which are other than the classical solutions. The next capsule will take up this very interesting question and we will see some new features arise when you cross a singularity and then this will be a very good place to stop this capsule. Thank you very much.


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