 Okay, let me maybe start, so the plan for today's lecture is the following, so first I'll give a quick review of the definition of filling area function and then we'll basically say two theorems that, restate the two theorems that I'd like to prove and then give the aim basically of what today's lecture is, so I've already mentioned we will need some concepts from geometric measure theory, these are integral currents in metric spaces and then we need rescalings of metric spaces and pass to a limit, these are called asymptotic cones and then we'll show basically the main property that we'll need in the proof of our theorems, this is the persistence of a quadratic isometric inequality when you pass to a cone. So let me maybe just review quickly, so we're giving a metric space x, think of a remoney manifold if you feel more comfortable and a closed Lipschitz curve, for this we defined the filling area as the smallest parametrized house of area of a disc that extends our curve, so then using this filling area with discs we defined the so-called filling area function where you basically look at all curves of a given length at most r and you try to find the one that is hardest to fill, in Euclidean space this would just be plain circles of radius r, so basically in the rest of my lectures I would like to prove the following theorem, two theorems and additional stuff as well but this is going to be the main aim, first the following theorem that we already stated yesterday, so if you have a geodesic metric space and your filling area function for very large r, sufficiently large r, is a bit better than in Euclidean space, then actually it already has to grow only linearly, so we said that's a gromophobolicity condition or in another sense that if you have such a bound then from far away every geodesic triangle looks like a tripod. I would just like to make one comment on this theorem which actually I'll play a role later, so we said so that this gromophobolicity that we'll prove, this is a large scale concept right, so that only depends basically on distances at large scales and not on local geometry okay, however this guy here, this filling area function of course you know there's also taken into account you know local stuff right, if you know for example if you have I mean the worst case would be something like you know you have a little hole in your space and you're a two-dimensional space, well then you know your filling function will be infinity right, so or if you don't have a hole but you you make a lot of little hills here, then you can destroy your function right away just by changing your metric space locally okay, so this guy here is actually also depending on local data which of course we don't want okay, so we will actually prove a much more general theorem okay which will get rid of this local behavior, so we'll take a different function okay which in principle can be a lot smaller okay and if we can bound this new function which will only depend on the large scale geometry okay and we can bound it by such a thing then we get the same thing okay, so this is not the optimal theorem because you have something that depends on local data and but you won't only want it to to rely on global data, we'll see that a bit later okay, the second theorem is the following, so we'll prove that there exists, we'll give a specific example of an ill-puted league group for which we can prove that the filling area function is not exactly polynomial okay, so for no alpha and a real number it can grow like this as r goes to infinity okay, so in both proofs we'll use one the same basic idea okay, so well we want a bound or we want to you know know something about the growth of this function right as r goes to infinity, so now this is rather hard to control when things go to infinity, so what we'll do is in a given metric space x in order to make this you know something that is not just large scale but actually at all scales, we'll rescale the metric okay, we'll rescale the metric by factor so that means we shrink, we blow down the metric space okay, one can make sense then of a limit or actually there's going to be many limits okay, many possible limits of such a sequence they're called asymptotic codes, I'm going to explain what that means and then we'll analyse what we would like to show then okay, we would of course you know we still need to bound this guy here somehow or do something with this filling function so what we'd like to do is we would like to show if the original space has a quadratic at most a quadratic filling function on the large scale then actually in this limit we'll have an honest quadratic as a pragmatic inequality so that means for all scales okay, so intuitively you think this is quite clear right, I mean that's a thing like this should happen because you know you re-scale curves and you know in the end so if you take a curve in the limit actually this should come from bigger and bigger curves that you rescale okay, each one of those you know for large enough r has a good filling so you hope you can pass to a filling in here okay, so however the problem here is that if we define the filling area function via Lipschitz discs right, area of Lipschitz discs so you would need if you wanted to you know make this trivial you would need a bound on the Lipschitz constant right so you can use some Marcella as a quality type thing and this I think is in general you can't expect to have a bound on the Lipschitz constant like this okay only in very special cases so what we'll actually need to do we'll need to use a homological version so the question is is this true I don't this is in general will not be true okay, however it is true for a homological version okay so it means where instead of using Lipschitz discs okay where you would need a Lipschitz uniform bound on Lipschitz constant will use surfaces okay so and these surfaces will be integral currents okay, integral currents we already know you know you have good compactness and closure properties and so so this is much you know will be actually what we want okay, okay so my next aim now is to just give a very brief review of this theory of currents in metric spaces okay which is was developed by Ambrosio and Kirchheim around 2000 so let me just I'm not sure whether you're familiar with this theory so I'll go quite quickly but I'll try to cover at least the basics of course most things absolutely without proof so let me just recall you've already last week with with Camillo he introduced on integral currents and currents in Euclidean space so what are these so in Euclidean space the field that are flaming current right it's just a continuous linear functional on smooth compactly supported m-forms okay so now the first questions of course in this theory so if we want to do a thing like that in metric spaces you know what is this good substitute of an m-form okay so because we don't can't really make sense of that so idea of the Georgie in around 95 in a small paper he outlined this that maybe one can use m plus one tuples of Lipschitz functions on your metric space as a substitute for m-forms okay so of course you'll ask yourself right away well you know what at least in Euclidean space should such a tuple what kind of smooth n-form or m-form should this represent well as we will see if x is Euclidean space then you should think that such a tuple should represent f times now that's the exterior derivative right so that's just a one form for the wedge product of one form here okay so this will just use it as an idea first okay and just because remember how for example so if t is a current in Euclidean space you know how did we define the boundary or how did Federer Flaming define the boundary just by duality with the exterior derivative right so if you take an m to the boundary of t of an m minus one form now this is just t the exterior derivative of this okay so now if you look at this correspondence right so if you take the exterior derivative of this form here right what you get is df wedge d pi one and so on right okay so now this will actually you know give in some sense an indication on how on how one should define the boundary okay so our metric currents they will be just linear functionals on such tuples okay and in the Euclidean case we had that this should be continuous right so we'll have a continuity property okay so then the second thing is well how do you for a Lipschitz map or a smooth map how do you define in Euclidean space how do you find the push forward via a Lipschitz or say a smooth map right you just basically pull back again by duality you pull back your differential form okay so now if you pull back such a differential form by a smooth map well then you simply compute a compute that you get this formula so that gives us an indication will give us an indication again how to define the push forward of a metric current okay so everything I'm gonna say now will you know is was developed by Ambrosia and Kirchheim in in their breakthrough theory on metric currents and so whenever you see definition or theorem or proposition this will always be due to Ambrosia Kirchheim in their 2008 paper okay so okay we start with a complete metric space and now as I already said the the space of M forms okay should be Lipschitz tuples of Lipschitz functions okay so here I use the notation lip X is the space of all Lipschitz functions so here that should be the M-fold product of Lipschitz functions and the first F the first argument there should actually be bounded Lipschitz function okay so in Euclidean space remember there you take smooth compactly supported forms right okay well if you're in a for example in the infinite dimensional but not space of course it doesn't make sense you know to to to have the first argument F for example and that this does would be compactly supported right that doesn't make any sense because it would just have to be zero and so instead of taking compactly supported you take just bounded Lipschitz functions okay we'll also actually need two more spaces one more space this is the space of bounded Borel functions on X okay and in the following I will denote by lip F the Lipschitz constant of a map from one metric space to another please slow me down if you have questions or yeah okay so now I'm broken keyhand they find turns as following as follows so this is a function on these test forms on these tuples which is well okay they should be in Euclidean space it's a linear function right so they should be linear in each component okay so it should be linear so it is a function on such tuples and it should just be linear in each component okay then in Euclidean space you have a continuity so they assume a continuity property as well okay so which just says the following if you have a sequence of such forms Lipschitz forms okay such that the pi and the first one you you you for the moment you just fix okay and then you have a sequence of those with Lipschitz constants bounded and if they converge point-wise if each pi i and converges point-wise to pi i then you want to have convergence okay so maybe I should say right away so what in Euclidean theory right so you have basically continuity there you want sequences right where the derivatives also converge okay and here so this continuity property is or continuity axiom is much stronger okay so you want even if the derivatives right they're just bound the functions okay as long as you have point-wise conversions you want already that your that that that it converges okay so this is a stronger a stronger kind of continuity and this already gives an indication why you know if x is Euclidean space you know this is not going to be the same space as the fitter-flaming currents of finite pass okay so okay now so we said that the pi i's or the f pi i's they should represent in the Euclidean theory such a form here right so now of course if a pi i here is constant right then this m form here is just the zero form okay so zero okay so now a priori right now so if this is just a linear functional a continuous linear functional well you know why should it depend on in some sense a derivative the change of pi i rather than pi i themselves okay so that's the next that's the next axiom so if the pi i if one of the pi i is constant on the support of f then you would like your t to be zero okay so that in some sense just means that the t should in some sense depend on d pi i rather than pi i the values of pi i itself and lastly unlike in the Euclidean theory you know here we impose a finite mass condition right away so we require that there exists a finite Borel measure mu which is concentrated on a sigma compact set so that means on the countable union of compact sets such that we can bound t of f pi i's by the product of the Lipschitz constant times this integral for every Lipschitz Lipschitz form any questions okay for those who want to copy down I'll just leave it quickly so okay the the space of m currents of finite mass now it's just you know the space of all all these currents right okay so we denote it by as in Euclidean space but be aware if x is equal to our end this is not the feeder of lemming metric current space of finite mass okay okay one remark this is not difficult to see the first part that whenever you have a metric current then it extends uniquely to a functional which I will denote again by t where the first argument is just a bounded Borel function okay so this simply follows from the fact that the Lipschitz the bounded Lipschitz functions if you fix you know such a measure mu okay as you know as before that basically bounds you know t and this is the space of bounded Lipschitz function is dense in the space L1 x mu okay so this is dense so then one can show that this new functional now that this satisfies all the four properties when f is not Lipschitz bounded but actually only Borel and bounded okay so the first and so the the multilinearity is is clear this is a direct consequence basically of the approximation okay then the continuity this is also easy there you need this bounded mass property the bounded mass property for is also clear it's also not difficult to prove the only the only problematic thing is the locality so if pi i is constant so beforehand we had pi i constant on the support of f right when f is Lipschitz and bounded then that's this is zero right okay so now this is not completely clear right if I here have just a Borel function okay then t of f and I will usually abbreviate just pi to say pi 1 through pi n so this is just you know approximated by here you basically have to choose Lipschitz functions okay so then once you approximate this while the pi i might not be constant on so this is t f say n pi okay where this is Lipschitz and bounded so now you know this is not going to be constant anymore the pi i is not going to be constant on the support of this anymore so in order to prove that this locality property still holds and you need all the axioms you need continuity you need locality and you need to find that mass okay so I think just do we have yeah I think just to to give you an idea how one uses these things to play around a little bit with the with these axioms to get a bit of a grip on it and I'll present the argument huh okay so let's maybe go back to so that we have all the definitions ready okay so what I would like to prove is that this locality property here also holds when I approximate and f is just a band of bro function okay so I'll do actually an easier case so first of all you know I can always when I look at f and pi one until pi m I can always just you know fix all the other pies except that the I th one okay and then basically I can assume that actually have it just a current where I have one of the one pie here okay so which then means that it's just a one current okay so I will actually just do the argument in a bit the special case okay so I will assume that it's a one current by just fixing all the other pie eyes and I will assume you know that my f which is bound the barrel that this is just indicator function characteristic function on a borrow set okay so suppose this is a borrow set and that the pie okay we only have one pie now that this is constant on b okay so of course you know by just subtracting this constant I can of course assume that this is zero on b so now what I would like to show is that if I take the indicator function on b and plug in the pie that this is zero okay so from this case you get very easily then you know to just via the the finite mass property you get to the case when f when this indicator function is replaced by borrow functions okay so we'll prove this by contradiction so we suppose that this is not zero okay so suppose that this is say two is two epsilon which is bigger than zero in absolute value so now I can choose so fix maybe fix a mu just a measure that satisfies this okay fix mu and so now I can first what we're going to do we want to approximate this by lips its functions right okay but the problem is if we approximate the specific ellipses function this is not going to be constant okay so what we're going to do we're also going to approximate this by another ellipses functions that are basically constant on the approximation here okay so that's that's the main idea how do you do that well first we can find a closed set you'll see why we want the closed set so closed so simply because mu is a finite borrow measure we get the closed set with mu of so a subset of B that such that the complement is strictly smaller than epsilon so now it follows of course that so claim we have that T now if we replace B by C then this is strictly bigger than this okay so this basically just follows from applying the finite mass condition and using this okay or maybe I should also say you know we can assume without loss of generality we can of course assume that the ellipses constant of pi is smaller equal to one okay just by multiplying your function by by the inverse of the ellipses constant okay so now we define suppose so this I think is okay okay and now we start approximating and we'll see why closed is a why we want to closed so for Delta we define now first a G Delta which approximates this guy here basically or actually we just have to approximate this guy so this is the following ellipses function so this is on X so I'll just basically draw the graph so this is my closed set C okay I draw the Delta neighborhood so this is the Delta neighborhood which I denote by C Delta which is basically all the X in your space such that the distance to C is smaller equal to Delta okay so that's my C Delta and now I just take the indicator function on C and then by the distance function divided by two over Delta or something like that you let it go down very quickly say this is maybe smaller equal to Delta or this is exactly Delta half okay so this we can do just using the Delta function at sorry the distance function okay so that gives us ellipses function so we also so now this is going to approximate basically our indicator function here okay and now we also want to approximate the pies because we want them basically to be you know constant on the support of one of those guys okay so how do we do that well that's a simple trick so for Delta again we now define first a function from R to R which is basically just the following function so here we have zero here here we have Delta so we just make it around zero in the little neighborhood Delta neighborhood we make it zero and then otherwise it's basically like the we have slope one here and going slow down here so then so that's of course one ellipses so that's the slope one right okay so now if we set Pi say Delta to be row Delta composed with Pi then this is zero in a Delta neighborhood of so this is this is equal to zero on the Delta neighborhood okay so now you know so this guy here is zero on outside the Delta neighborhood so what we get is that what we get is that what we clearly get is that if I take t and I take g Delta and pi Delta then this is zero for every Delta so now we're almost done so now just note the following a few things so note firstly that t g Delta oops t g Delta and pi okay by basically the finite mass property okay this goes to t1c and pi okay so that just follows from four okay simply because well you know the integral of g Delta minus one c okay so this is d mu it's just smaller equal to mu so this guy has support in in c Delta so this is just c Delta over yes c Delta minus c okay and that goes to zero right because the the count of the the countable intersection of c Delta so Delta goes to zero it's just c okay and then so that means there exists a Delta say Delta zero bigger than zero such that so this guy here is bigger than bigger than epsilon so that we have t g Delta zero pi is bigger than epsilon and also we can of course just because that goes to zero we can also assume that at the same time this goes this is smaller than epsilon okay so now second thing we know so now if you this thing and now we just use the continuity okay so we have that g let me see what do I want yes so now I use the g Delta zero down here okay so that's a fixed function now and if I take now pi Delta and I let so that's Delta going to zero of course and I let Delta go to zero then this converges to t g Delta zero and pi okay so that's just the continuity axiom here okay and because this guy here sorry this guy here is one lip shits right so this is always the same constant okay so that means we find a Delta which is maybe smaller than which is smaller than Delta Delta zero okay so that this is again you know that this guy here is very is bigger than epsilon so that means we get that there exists a Delta one in zero Delta zero such that t of g Delta zero pi Delta one is bigger than epsilon simply because this guy here is bigger than epsilon in absolute value okay so now we're basically done because now we just have we can put all these things together so we have epsilon is smaller than the absolute value of of g Delta zero and pi Delta one that's this one okay so but this guy here because it's always the zero for if the the Deltas are the same so we get here that this is t g Delta zero minus g Delta one pi Delta one that's just because if you take just these two guys right then it's zero okay and so this now by the finite mass property this gives you smaller than g Delta zero minus g Delta one okay and so this guy here they have support in in the neighborhood of so this is smaller or equal to this is smaller or equal to mu of c Delta zero c c which is smaller than epsilon and that gives you contradiction so that shows that actually this also holds for the for the for F just in behind the portal okay okay so so this is a strengthening of the of the content of the of the locality property okay so now a proposition so now we just had fixed in you right but now of course you'd like in some sense to define the mass and the mass should basically be the smallest mu that you can you can find okay that still holds you know that still satisfies this property okay so there exists actually a smallest plural measure that satisfies this and will denote it by by by the measure of t okay or the mass of t and this you can show one can show that this is given by the following formula so basically the supremum of now the sum of tf n's and pi n so here I abbreviated pi 1n until pi mn just by pi n okay so I will do that very often from now on I just have pi will actually represent pi 1 through pi m so this should this should remind you of course these formula down here should remind you of the again of the Euclidean thing where basically the measure of t of u was something like the supremum okay of t omega where omega is bounded smooth form smaller equal to one okay and the support of omega is in you okay so this is basically an analog an analog of that so now you can define the mass of the total mass like the mass of the on the whole space of t just as the measure of t on the whole of x okay and an immediate corollary now of the proposition that we had is that this mass is lower semi-continuous okay simply because beforehand it was given by a supremum of you know basically plugging in test functions right so then this clearly passes to the limit and there you get lower semi-continuit so just a quick remark we'll use that a bit later you can check easily that the space of m currents becomes a complete metric space when you take the metric the mass metric so now we can make some constructions okay so we can define the the boundary of a t when m is bigger equal to 1 and as you see it's exactly you know using this intuition that a f pi 1 through pi m should correspond to f d pi 1 which d pi m okay so then the exterior derivative where that you want if you take it here right we said that this is df which d pi 1 so this should correspond by the george's idea to this is one times this so the one f pi 1 and so on okay and you define it exactly like this okay so now it follows immediately that the boundary satisfies the first three property so multilinearity is of course clear okay continuity is also clear right because you basically there you just take functions here so it's continuous t is continuous so if you take this and they converge right then you get exactly what you want okay so the only thing that you still have to think maybe a little bit is the locality axiom right so the locality axiom that would now suddenly mean that pi a certain pi i is constant on the support of f right but now the support of f if you plug this tuple here into or that should actually be now m minus one right okay if you plug this in right then this is constant on this guy here okay but now using just so this you can of course also write as t because we have this extension this unique extension right you can write this as we can just take right one as indicator function of support plus the support comp complement okay plus t and now this on the x minus the support f and pi one and so on so now because we had this extension right we said we proved that this extension satisfies again the locality axiom for for just boral bounded functions okay so now this guy here the pi one of the pi i say pi one okay is constant on the support of f so it's actually constant on the support of this guy here okay so this then becomes zero okay and f is constant of course on the support of this guy here so this becomes also zero and you're done okay so that gives you this and of course you know it's clear that if you take twice the boundary then this gives you zero simply because you have then t1 1 f and so on but this guy here is constant on the support of this and you know the locality axiom gives you right away that this is zero so now for ellipsis map from our complete metric space x into another complete metric space y you can define simply by this and if you remember the formula that we had before for the pullback for the pullback of by ellipsis map of this guy here so this we already said this is just composition here composition here and so one okay so and then this should correspond just you know clearly to the tuple f composition with phi and then you compose everything and so this is exactly no corresponds to this so in some sense this intuition what one should have you know has been confirmed so far right so and here it's trivial to check that push forward of a current of finite mass is again the current now in the in the target space and you have this inequality for the for the mass of the push forward case this is straightforward okay also just because we extended t to ban the Borel functions in the first argument okay now you get you can of course define the restriction just by adding here in the indicator function on a Borel set okay and then you know by what we showed this is clearly then a metric current okay and you can easily check that's the mass of the restriction is the restriction of the mass so I should say one more thing about this correspondence here and since I'll make it very brief because otherwise I will not have time to give you any of the applications to this filling area problem I won't note it down here but just here what do you have so in Euclidean theory all right of course f d pi one wedge d pi two so this is f d minus f d pi two d pi one so if you switch those then you know the M form becomes the negative right okay so in our in the definition of a current you know this was not an axiom right this was not a condition that you actually should have an alternating property however you can actually prove this okay so they on Brossier-Kirchheim they proved that if you switch any i's and j's pi i's and pi j's let's maybe do it for the first okay but this is negative t f and now pi two and pi one and then leaving the rest okay so that you have actually the same thing and of course you can also ask yourself so when I do when I multiply here by a function right so the exterior derivative that was just by the by the product rule you would get tau times the exterior derivative plus no pi m times the derivative of this okay so you should say okay well you should have a correspondence there again and again this is true actually you can prove this okay so that's actually quite surprising I think that this you know this is such a such a nice correspondence at this actually all is true okay so I will not note it down here but the correspondence goes further than what I state here so we haven't given a single example yet so let's give the first basic example so for any l1 function so the following integral so pi again you should think as pi one through pi m okay so then we know by Radimach a theorem the derivative you know the pi is is is differentiable almost everywhere and you know the differences of course are bound to borrow so we can take the determinant and we take this integral okay so that defines a current in the sense of unproved okay and so what is not you know the continuity axiom right this is the only thing that is is not is not evident right so multi-linearity is clear just by multi-linearity of the determinant and then then then locality is also clear if one of the pi i's is constant then you know that is the determined zero the finite mass property is also clear the only thing that is is is is not straightforward is the continuity but this actually follows from from from the weak continuity of determinants okay so that you can check that this is this is actually also true so now normal currents in the Euclidean theory they were defined as those currents whose boundary mass is finite okay we already said that if you take a current well m should be bigger or equal to one right we can make sense of this boundary we said it satisfies all the axioms except maybe the the finite mass axiom so if that finite axiom axiom axiom is also satisfied then you call it a normal current okay and actually normal currents in in a Ambrose Keigham theory in Rn is actually the same as feather-flemming normal currents so there you have a one-to-one you have an isomorphism between the two now the first important theorem that we will use we'll need to use is that if x is a compact metric space and you have a bounded sequence of normal currents so bounded in the sense that mass and boundary mass are bounded then you have a sub sequence that converges to some limit current which is again a normal current okay so remember so of course one has the same thing so in Euclidean case you just have you just have that they are continuously near functional so they're elements of the dual space and there you have you have weak star compactness right by Banachalooglu you get you get something so here you have to work a bit because you don't just have a a dual space okay now the object that we really want is actually not normal currents but integer rectifiable currents okay so for zero-dimensional so we will split the definition in two first for zero-dimensional these are just sums of point-deck evaluations so here for a point in our metric space X we'll use this to denote the zero current so a zero current remember so we have to the test forms I said I should maybe have emphasized that this is this cross lip to the M okay so when M is zero actually I will just this means that there's nothing here so this is just a functional on on on bounded lips its functions and this is just evaluation so really what you can think of integer rectifiable current and the integer means is that you have an integer integer multiplicity here so this is just a finite set of points with multiplicities okay now remember what is an integer rectifiable current of higher dimension in in Rn right so basically this is you integrate and now here you integrate on an countably rectifiable set okay so basically this is union countable union of by lip sheets pieces or lip sheets pieces or you know even C1 manifolds or parts of pieces of C1 manifolds okay and you know you have a multiplicity function and then you integrate this you also have of course there you have a you know at almost every point you have a tangent so you have an orientation so it's given like this okay so we're here this is the union of lip sheets pieces or by lip sheets pieces of certain sets so now the definition here at first sight looks different okay so a finite mass current of dimension bigger equal to one is called integer rectifiable first if its measure is concentrated on a countably Hm rectifiable set so this by this I mean you know it's a it's a borrow measurable and it's the countable union of lip sheets pieces of lip sheets images of subsets of RM okay and it vanishes on negligible world sets okay so so far so good right it lives on a rectifiable set on a countable Hm rectifiable set now what is integer so far you know there's no integer rectifiable right the integer so we have to make sense of this somehow so and this is done as follows whenever you have from so here is your rectifiable set right so whenever you take a lip sheets map down into your RM okay then it should look like one of the guys that we saw okay basically just integration and integration of that so plus a integer clear clearly an integer multiplicity okay so actually you want to prove that this is just the same definition as this okay namely that's the following thing so first let's maybe denote the space of integer rectifiable and currents like with a curly mx as usual okay so then one has the following represent representation there so if you have an integer rectifiable current okay then there exist by lip sheets pieces so basically so have by lip sheets pieces here so that makes out a countable Hm rectifiable set and on each one you basically just have integration of an integer multiplicity function okay so yeah and so you can represent it like this and you know the math you don't have any cancellation that means that the math is exactly this okay so then this is very close to this you know if this is a countable union you could just as well write this of course as this where you have to fight the Ki you push it forward with the f and with the phi right so you have maybe a guy like this and then you know you basically have to just take the push forward there okay so that's very close to this guy here okay now integral currents these are integer rectifiable currents whose boundaries have finite mass okay sometimes actually you know in the Euclidean theory people define it as those whose boundary is again a rectifiable, an integer rectifiable current right okay but actually one can prove that this is the same this is the celebrated boundary rectifiability theorem okay which also holds in this setting okay very very now not surprisingly anymore but I mean I think that was a big surprise that you would have all these the key properties of the Phaedon Fleming theory you also have here okay and now the closure theorem which I think you already saw in the Euclidean case is also true in the in arbitrary complete metric spaces okay so if you have the weak limit oh I forgot to say it last time so in then in this lower semi-continuity right so this is you take point-wise limits right if you take so we had lower semi-continuity so weak convergence here just means point-wise convergence that means whenever you fix Lipschitz forms okay then you want to have that this converges to the limit and the closure theorem says that if a sequence which is bound in the sense converges point-wise to a normal current okay then this limit is actually a again an integral current so this is this is yes okay so you see the all the key features of the Euclidean theory are actually hold in complete generality of of of complete metric spaces okay so that was the I think so yes that is the very brief very brief review introduction without any proofs of this theory which now we will use in order to to attack our problem namely you know to prove these two theorems and the first step was will be to say okay if we have a filling area function that is quadratic in the space then the more generalized filling area function where you use currents rather than Lipschitz discs you know is actually also quadratic in any limit of rescaling okay I think maybe it's a good time to to make a break here and in ten minutes we'll start defining these limits of rescaling of metric spaces okay I think I'll continue so yeah this is of course a very was only very very brief and as you see we'll use it basically as a black box almost because I didn't prove anything so now we will look at rescaling of metric spaces okay so we said that we take a metric space we take a sequence of rescaling factors we want to shrink the metric space so we take our ends going to zero and we look at this at this thing okay so before you know taking a thing like that what we will do we'll just look at sequences of metric spaces okay just take this dn and you know we'll just look at sequences right away okay so we would like to make we would like to make sense of limits of sequences of metric spaces okay so there is a notion of so-called Gromov house of convergence however you know so there you need maybe you know it maybe you not and there one needs you know that basically all the spaces and are uniformly on every scale they uniformly about the same okay they have that means that you know each ball can be covered by uniform number of epsilon balls okay and then you can expect convergence okay so but however you know this is in general of course you don't have for example if you just take no so you glue to a copy of our say you just glue hairs on at each integer point these are copies of the interval zero infinity right so now if I start to rescale the metric for example so you know by a factor of one half well then this guy he'll remove here so then you have suddenly you know at each point here then you do that you know smaller and smaller and you'll see that you have more and more hair hairs growing out here okay and you know clearly there will not be a Gromov house of a Gromov house of limit okay so so in general what we want and you know the spaces that we we care about right now no you can't expect in general this okay so now nevertheless one can define a limit a certain kind of limit of arbitrary sequences of metric spaces how is that done so we start with a sequence of metric spaces xn and dn and basically a little bit like when you know doing the construction for the completion of a metric space no so an element of the limit should basically just be a sequence of elements in xn okay so you would like that in some sense a point in the limit space should just be an equivalence class of xn's where each xn is in an xn okay so we will need base points for CY and we define now x hat as all the sequences where each xn is in the space xn and we would like that they're all a bounded distance from our base point okay so you have like a point of observation Pn in each space and then you just take points sequence of points so now how would be like to the how will we take the distance okay so of course if they're all the same copies right and all the points are the same well then you basically want that the the distance of the limit points should just be the distance of the points or the limit of the distance okay so since we have that since we have that the all xn's are a bounded distance from from from the base points right so actually this sequence here is a bounded sequence right okay so that means this bounded sequence has at least a subsequence bounded sequence of distances has a subsequence that converges so now the problem is of course you know how which subsequence should I take okay in general it's pretty difficult if your space is big into actually in a consistent way to always choose subsequences it's just not possible okay so if you have proper spaces then you know then the separable spaces that that that works but but but in general not okay and proper spaces are not separable okay but however there is a device which makes a consistent choice of conversion subsequence in some sense okay however for this you will actually need Zorn's Lemmel okay so this device is called a non-principle ultra filter okay which in some sense it extracts convergent subsequences of guys like this you know for any bounded sequence okay what is a non-principle ultra filter so this is a finitely additive measure on on n okay so on all the function from all subsets of n which has only values in zero and one okay basically says which subsets have full measure and which you know have zero measure okay so which ones you disregard and which ones you still consider okay furthermore you'd like that it's a probability measure in the sense you know that so you only have finitely additive not because otherwise you know and you would like that finite sets that they all have zero measures okay so of course you can't expect you know that it's a sigma additive measure right because otherwise you know the whole n would have to have measure zero right so that's so you have finitely only finitely additive you can expect at the most okay and one can prove that such a thing exists okay and now we said that we would like to extract for given sequences xn xn prime we would like to choose a sub-séquence of this this is a bounded guy okay and actually our non-principle ultra filter does exactly this okay so why is that there is an easy proposition that you can so you don't have that so it doesn't it doesn't it doesn't say that you know so for example if you take all the odds or all the the evens right so I mean one of them will have zero and the other one has has one okay that basically means the sub-sequences that you choose in some sense you know are going to be the ones in the even ones okay okay so now you know every sequence actually of real numbers of bounded you know every bounded sequence of real numbers you know you can associate and so-called ultra limit okay actually this is much this is true much more generally namely whenever you have a compact house-dorft topological space so not necessarily with a metric we will only use metrics and a sequence in there so for example here we would just have the bounded interval in R so then there exists a unique guy such that whenever you take a neighborhood of this point you know all the indices ends such that Xn is in here this has measure one so this guy here is of course unique you this is this is very easy to show because as we already said you can't have two sets right that are disjoint that yeah and so the so this is easy to show that that you have such a thing and so that means now that we can always so let me maybe say we will just note the limit the ultra limit okay the Z will call it ultra limit and will just denoted by Lim omega of Zn okay so that means here for any bounded so that because that's a bounded sequence now right we will get such an ultra limit okay so and it's not hard to show of course right because here all the ends in a neighborhood that satisfy this this has to be an infinite set right and you can make at least now if you're in a metric space or in R you know you make this neighborhoods you it's just excellent intervals make smaller and smaller and smaller and you easily see that there exists a subsequence of you know of of of of of of the ends now if Z is a is a is a compact metric space you have a subsequence that converges to the ultra limit okay so this you see very easily so let me maybe note this down because we'll use that later so easy if Z is compact metric space then given such a sequence there exists a subsequence Z and J that converges to the limit the ultra limit okay so this is so this is this is trivial okay okay so in some sense this can be thought of you know choosing in a consistent way and subsequences or limits for subsequences so now we can define the distance or a pseudo pseudo metric okay in our x hat which is just a bounded sequences right we can just take an ultra limit rather than a limit because limit in general doesn't exist okay so this is straightforward that this is a pseudo metric so it just might be that you know certain points of course if you take two points that converge right or sequences that converge in distance right then this will get in general get get they will have the same limit so that you know this should be now we can just identify all the sequences that have this distance or pseudo distance zero and this will be our metric space okay which is called the ultra limit so the ultra limit by definition of such a sequence of metric spaces with white base points with respect to such a non-principle ultra filter it's just a triple okay it's x hat when you mod out or you know equivalence relations of sequences where you know two sequences are the same when they have the same ultra distance limits okay and then with them that becomes a distance and you also have a base point which is just the the equivalence relation equivalence class of the pn so now let's maybe just give a few examples so when x is a proper metric space so that means every closed ball of finite radius is compact okay for example rn and if you take the constant sequence okay so then you just take you know bounded sequences constant sequence with a constant base point then it's not hard to show that the this ultra limit with respect to any non-principle ultra filter is isometric to the space itself so for those who know what a Gromov-Hausdorff limit is and I won't go into that so if you have proper metric spaces a sequence of proper metric spaces and the sequence of base points and if this sequence converges in the pointed Gromov Hausdorff sense to a limit space x infinity p infinity then any ultra limit is isometric to this limit space so this is in some sense a generalization of the first example okay again this is not it's not it's not very difficult however you have to be careful okay so if I take now non proper spaces so say for example your x is a infinite dimensional Banach space or you know any for example infinite set with a discrete metric okay then if you take the constant sequence then the ultra limit is in general not the space itself okay so this is simply because you know in an infinite dimensional Banach space you know you have many sequence or most sequences don't have a conversion subsequence right okay so then you can't you can't actually expect anything so so in general or all you know every every time you have an infinite dimensional Banach space the ultra limit of the or it's called the ultra products of you know the constant sequence this is much much bigger space okay so this is yeah so basically it has to do with the second tool right so there because there you have compact you have a sequential there you have you have that it's a weak start compact and then then you get okay so it's it's a much much bigger space it's comparable to the second tool okay so now let's go back to our problem of when we have special sequences of metric spaces so we said that we want to rescale and pass to a limit so now we just fix our space and we want to rescale the metric right with factors that go to zero and we'll also now consider base points okay so now here we know we can make sense of a ultra limit okay and when you have an ultra me like that it's called asymptotic care so the way you should really think about this is that so let's maybe look at the specific example let's for example just say Z in R okay so say our base point it's just always P and is equal to zero right okay so now you rescale your metric so that means you know these points become closer and closer to each other okay but you know for example if you go at distance say distance a hundred okay so you'll always have again points basically where that our distance to this guy here more or less a hundred right even after rescaling it will just be basically the guy that is around 1 over Rn distance okay will become this this guy here okay so then it's pretty straightforward to check that this guy here any ultra any asymptotic cone which usually I will just denote by by by by no Z or sorry that's the space and W and omega sorry and this will just be R okay and so maybe let me go one further that's the notation that we will usually use we will just not make mention of the Rn's and the Pn's and but you know they're always implicitly in the background there will be there there will be there so in some sense what you should think how you should think of asymptotic cones is that you basically go further and further away from from the space and you try you know you look at the space from further and further away right so if you do that for example in Z if you go further and further away you know in some sense an observer will see these points closer and closer and closer together okay and that's you go to a limit as you go in infinitely far away you know then basically that is your what you see is here asymptotic cone so in some sense that's how you should think of asymptotic cone should be the space as seen from infinitely far away okay and so what are the base points well and you know so basically of course certain parts of the of the space can get lost right so if I'm very far in this space over there but my observation point is is basically here and my rescaling factor you know is not fast enough to zero right basically there's only one portion of the space that that's that will be still be visible and points over here they just go to infinity okay so okay one crucial observation or a theorem that one can prove that Gromov proved is the following so remember a geodesic metric space is called Gromov hyperbolic if any triangle is thin so triangles they basically look like this you know basically here the distance is always smaller than a certain delta fixed okay so now if you go further and further away from the space and you look at bigger and bigger triangles well what will you see well basically you know this distance here from far away becomes very very small so if you you know if you take a scaling factor RN you go RN away right you basically this becomes delta RN okay so as you go further and further away these guys here they look more like the same guy so what you'll see infinitely from infinity will basically see a tripod okay so anything that is bounded okay it's the asymptotic cone of a bound metric space will just be one point okay so here you know that's yeah so if x is Gromov hyperbolic so if these are thin then every asymptotic cone so that means for any choice of RN going to zero and PNs and non-principle ultra filters and you'll actually get a metric tree which is a geodesic metric space where every geodesic triangle is zero thin or in other words it looks like a tripod okay however also the converse is true okay so if every asymptotic cone of x is a metric tree then all there exists a delta such that all these guys are delta thin okay actually you don't even need for every non-principle ultra filter you can fix one non-principle ultra filter and if this is true okay for all choices of PNs and RNs then you'll get that it's Gromov hyperbolic okay so this is not immediately evident right so because what could happen is that you could you have to avoid the following thing so you can have very big big bigger and bigger triangles right and somehow where where the distance here we said beforehand it should be bounded by a delta okay but in principle it could just be bounded by an amount so we have here a guy in here it could be bounded by something that is just sublinear in the diameter of this guy right so then you know when you rescale basically make this triangle you know of always of size one this will be sublinear okay so this distance will still go to zero okay however you know you don't have a uniform number delta you just have a delta times RN right so this is this is the problem so there in this direction you actually have to clearly prove something okay so that's that's not 100% clear so yeah because I don't have a lot of time I think I'll just skip this and we'll leave this as this okay so now we have the two things that we need we have this theory of currents where we have you know closure theorems and compactness theorems that we can use that where we have a chance to pass two limits okay and we have the notion of rescaling okay so now we'll formulate we'll formulate the basically this persistence of quadratic filling functions to the limit namely to asymptotic currents okay so we'll be a little bit more general because as I said at the beginning today that you know this theorem that I stated the filling area function in the metric space is really you know considers both local and global data and we only you want to make it independent of the local behavior of a metric space okay so so we will actually we will actually do the following thing okay so we'll have our metric space X in here okay so in order to make it in order to make it independent our filling area function from local data so that means for example if you have a hole here I mean you don't have fillings in general okay so what we'll do we'll actually take a bigger space so this is our X we'll work with bigger spaces why okay where that are nicer okay so we allow any metric space Y that is contains X okay which is at finite distance of X okay so that means this is smaller than some big constant L what do I mean by this just if I take all the every point in Y is at this it's at most L okay so this I'll sometimes call a thickening of the space so then you know here you suddenly have something in there okay or I call it at bounded distance oops sorry that's the wrong direction okay so now we have these two metric spaces will assume both are geodesic and Y is complete because we want to use currents in there right so we assume it's complete so now for a Lipschitz curve in X we'll define the filling area now without a zero down there that means you know it's not necessarily a disc so this will be the infimil mass of a integral two current in Y whose boundary is exactly the curve okay so here actually I identify S1 basically with zero one with the endpoints identified so the curve you know will just be a curve on zero one and it should be a closed curve so that's the current it's basically the push forward of the unit interval which is just a curve right okay so now for a given curve I have here a integral current that fills it okay and I take the infimil curve okay so now one remark which I should make so of course so if you take just as your currents you take now push forwards so you take remember beforehand for the filling area with a zero this was just basically Lipschitz Lipschitz maps from the disc right okay so now of course a Lipschitz map on the disc gives you a current just by pushing forward the indicator function on D via phi okay so we could always take this guy here okay so then you know the boundary is still this and so now you know you think okay so now we have a lot more choices right okay so we have a lot more choices so this guy here should actually be strictly smaller than the filling area or should be sorry not strictly small but it should should be bounded by the filling area okay so this however is not necessarily true it's only true up to a universal factor so why why is this why is this so basic the the basic reason is that so remember for the area of a map for a Lipschitz map we used a param parametrized house drift measure okay so it was basically the integral of you know of a Jacobian in some sense right oh yes sorry yes you're right here that's a mistake there should be a what no no there should be sorry that should be in X itself this should be next itself this is still in X yes but only using discs that's the one that we defined before for area of fees right filling yeah the supreme yeah the the in female area of of discs right okay so the reason why you don't have here one okay is basically that the area the definition of area is different okay so the mass of this guy here okay will not be in general the area of fee even if fee is injective okay if he is not injective right you could have cancellation in a current so if you do something like this you go back like this as a you know this is just a slice should be something like this you get cancellation right for for current so this will be just as a current this would be invisible this would be zero right so that's the first you know why you don't have necessarily this right but even if it's injective it's not going to be the same because if it's injective right we said that this is the two dimensional house to measure of the image okay on the other hand here how did we define the mass right we basically defined the mass as taking the supremum so here as basically taking the supremum all over all f pies where this is one lip sits and this is bound by one right so now if you if you do this for this s here so that gives you the integral of f fee the determinant of so this is the determinant of d now we have pi and fee right okay so that actually gives you if you take the supremum of that that gives you a different kind of area okay so this usually what this is called is a this this area is called the from of mass star area okay and this is just comparable to the house to measure you know up to a certain factor so just keep this in mind but this will not play a big role for us at all I can actually okay so that is this and now we define the generalized homological filling area function homological because we use we use currents okay and generalized because we use basically curves in x and fillings in maybe a bigger space okay and so this is just defined exactly the way we defined it in in in for for discs right just by replacing now and the filling area zero with the filling area in y using integral currents otherwise it's exactly okay so we will also abbreviate f a x to mean f a x x so if we stay in the same space we want the filling in the same space we will just use this abbreviation which is similar to the one we have before and clearly again this guy here is going to be comparable or small or equal to a constant times the filling area zero now maybe I should make a quick remark concerning this I wait yes we can of course also instead of just considering curves right we should we should could also consider integral one currents okay so that means you know not just one boundary curve but we could use for example so many curves or more generally just an integral one current without boundary okay so for this we can also define the filling area okay just exactly the same way except that t need not be a curve but it can be an integral one current okay so we can define then what is usually called you know a quadratic isoprometric inequality of any kind of isoprometric inequality for integral currents if now for every integral current we have integral one current we have an integral two current whose boundary is t and whose mass is bounded by a constant times the mass squared of this so this isn't and what one can show is actually that filling area function the way we defined it is being quadratic is equivalent to having a quadratic isoprometric inequality okay so has a quadratic isoprometric inequality for integral currents is the same as having having FA y so here we are both times in the same space why huh okay so so this is again the abbreviation FA y y okay so this is basically just uses the following fact that whenever I have an integral current one current without boundary then I can write it as the infinite sum of closed curves okay so then there exists curves see I they're all injective such that lips its curve such that t is basically just the push forward of these current of these curves okay and you know the mass basically is the same as the mass of the of the thing and then you use also that you know r squared is convex okay so that actually also works whenever you have that this is basically bound to a convex function then you get an isoprometric inequality with the same convex function okay so r squared is just is not you could take r cubed or anything like that you know simply you need something that is convex because once you know you take the fillings of each of these curves I do get s eyes whose boundaries are basically the push forwards of the sea eyes then you want to take the sum of all these s eyes right you take basically the mass of this is bounded by the sum of the masses for each one of those you have an isoprometric inequality like that so you have to basically make sure that the things persist okay so that's that's yeah so they're basically the same concept whether you take curves or integral currents so no worries about that okay so now we have the proposition that we we have been aiming for so now x a geodesic space okay and y is a bigger space which is at bounded distance which is just means that any point is at bounded distance from from x okay so then if we have that the generalized filling area function is at most quadratic on a large scale right then every asymptotic cone has an honest quadratic isoprometric inequality that means any curve can be filled can be filled okay so now we're gonna we're gonna prove this okay so yeah I hope I have time so the idea is not difficult I mean so basically as I already said what we're using is simply that whenever you have a curve in an asymptotic cone you can approximate it in some sense by curves in our space right bigger and bigger curves okay you fill each of those okay you can fill them by integral currents in there and you know then pass to you want to pass to a limit okay so the only problem is the following of course right so somehow you have bigger and bigger things you have maybe currents that look more and more like this so somehow you have to make sure that you can pass to a limit for these currents okay so so the problem is that they might become wilder and wilder and wilder and then how do you do convergence right so that's the big problem the only thing that we've seen is that if s and are all in a compact metric space okay if you have currents say in i2 or anything in a compact metric space that have bounded mass and bound boundary mass then you have a subsequence that converges okay so but if your fillings say that is your curve cn that approximates in some sense your final curve in x omega okay so and you take s n's here okay so we'll have to push these guys into a compact metric space in order to be able to you know take a limit right so but if they these guys become wilder and wilder I mean you I don't know I mean you can't you can't hope that you can actually put them into a common metric space which is compact right so that is what we have to deal to with but however if we have a quadratic guys metric inequality then we have much nicer fillings okay so basically if you take an area minimizer if there exists an area minimizing filling here right then you can't have anything like this okay because that is definitely not area minimizing you basically could cut it here and fill something in which is much better okay so that's yeah so let me see okay so yes the direct consequences is the thing that I state the first if you have such a thing then you get this for everything okay so that's that's just the direct consequence so now yes preparation for the proof now will state a proposition of ambrosia kirchheim which says that if you have isoprometric inequality a quadratic one then we actually get nicer fillings okay so suppose y is now a complete metric space and suppose it has a quadratic isoprometric inequality for integral currents okay so then for any integral cycle and for any epsilon there exists a filling of T which is as close to the optimal in area or in mass as you as you want plus which has the following growth lower growth of the mass so basically at every point if I take an R ball where the R you know is smaller such that you don't hit the boundary what you have in here is at least quadratic in R okay so for example this it definitely excludes a thing like this right because if you take an R ball here so this is a thin tube right whose area or mass is basically just R times the circumference which is very extremely small okay so this means you get actually things that have that and you know if there exists an area minimizer then this is a relatively easy computation okay so I'll maybe show that I'll maybe show that afterwards okay so this and this you can make sense offer this actually works in any dimension okay so two-dimensional currents that's nothing special you can do that for higher dimensional when you have instead of a quadratic you have the Euclidean equivalent of the quadratic as point you can call so Euclidean isoprometric okay so now I'd like to show you the proof using this lemma of this proposition of this proposition so so proof of proposition using theory of integral currents and using this lemma so it's in two steps so first of all we would like to use the lemma so we need a space where we can where we have a quadratic isoprometric inequality okay so our filling function right this just considers curves in x with fillings in y so maybe there's actually you know curves in y they cannot be filled at all right so our space y might be not good enough however you can always find a space which so there exists y prime shewdesic and complete with the following properties well so first of all y this y here is in y prime isometrically okay and y prime is at finite distance of y and y prime has an honest quadratic isoprometric inequality for currents as quadratic isoprometric inequality for integral one current sorry okay so any curve or more generally any sum of curves can be filled nicely okay so you have to thicken up your space correctly so this is a bit of a of a technical thing that you have to go through however this is not it's not so difficult okay for any curve basically you have to glue in the disk and then you have to make sure that everything works out but that's that's not not a big deal okay so now the second step so now we take any asymptotic cone so let X be just X we take a scaling sequence we take observation points and a non-principle ultra filter and we take this asymptotic cone and and so let C be now a curve a Lipschitz curve let's say from a closed Lipschitz curve into the asymptotic cone so this is following picture so now we would like to approximate this so I claim that the following it is sufficient to do the following okay so we choose a partition a finite partition of the interval say tm equals to one a partition okay and so I claim it's enough to show show that there exists se prime Lipschitz with the following properties first of all it coincides it coincides with the point with these partition points so that means C prime at TI is just C of TI and the length on the sub intervals C prime this is smaller or equal to the length of the C's on these sub intervals and there exists so now what do we have we have the following let's maybe make this a bit easier picture like this so we have a curve okay that's my C C prime that's my C and I can fill a I can fill this so there exists a two current in Y prime whose boundary is just the C prime the push forward and the mass of S is bounded by say the isopropymetic constants which I say is called C times the the the length of my original curve so that means we have here we have a current S and so why is this why is this sufficient well we can just go you know basically one step at a time we first for example we divide into say oh maybe I have one more point okay let's take one more point okay so now I do the same thing so I make sure that all the lengths here are the same length so the length divided by five for example okay so now I continue the same thing so this guy here what length does it has it has smaller equal to two times the length divided by five right so if I take this curve so now I can divide this again by five one two three four five and I can just take the same procedure if such a thing exists right I can take the same thing and I can divide I get an S and another S right so I can do that with everyone and then I go smaller and smaller and smaller okay so then if you add up everything everything works okay because basically in the in the case step right you have five to the K if you divided by five five to the K curves of you know two to the K over five to the K times times the length okay but you seems you can fill each one quadratically you get this squared which is two to the two K over five to the two K times five to the two K still gives you four to the K over five to the K and you know that is a geometric series right okay so so this is actually enough and so now we can do our approximate approximation the way we announced so now we just denote X I to be the C of T I okay so let me just take now we just have to get our C prime right so now these are our points they're in the asymptotic cone so these you can write as X n I's as equivalence classes of sequences right so now we can go to the original space and we can take there exists lips it's there exists curves now to X with this rescaled metric CN TI is X n I and such that CN on TI TI plus one this is a geodesic so now you have these points in your original space say X R and D and you just take geodesics so now I have only like two three minutes so now clearly because these are bounded sequences right so and we have only fine at the uniform number of points in our partition so we get actually that the C ends as curves in here are uniform elliptics so now since we have a quadratic isoprometric inequality in Y prime this good lemma here that tells us that we have good fillings for these curves okay so we can take fillings like this by the lemma there exists SN in an integral current in this space Y prime now let me rescale the metric because we have a quadratic isoprometric inequality that doesn't change you know it still has a quadratic the same quadratic isoprometric inequality such that the boundary is exactly this curve CN to push forward and such that that we have you know two and three in dilemma in so now I can define metric spaces or subsets of my metric space Y prime simply as my curve union the support of SN with this rescaled metric okay so a picture is the following okay so that could be my CN it might backtrack so it might actually you know cancel something out so and here we have our SN okay so now the claim or it follows directly from you know the the third condition there okay that now we have a sequence of metric spaces that this is uniformly compact okay in the following sense so for every epsilon there exists an N which only depends on epsilon and not on on the little N such that you can cover AN by a number at most N of epsilon balls and at the same time the supremum the diameters are bounded of the AN's bounded by a fixed constant lambda okay so let me just quickly say why this is and then actually I I guess I have to stop a bit short I'll just basically explain then what one can do so why is this the case well where any you take any point in the support of SN right so if I and that is at distance bigger equal to say epsilon fourth okay so now if I take the epsilon fourth ball the open one it has a certain amount of area certain amount of mass okay so now you just pick more and more of these points that are all at distance at least epsilon half right so how many points can you find like this well each one has this amount of mass so you know because our so you can get at most the mass of SN you know and divided by this number where r is equal to epsilon fourth okay but since this guy here is bounded by the length of by the length of the curve right by CN CN has uniformly bounded length you know this is a bounded a bounded number okay uniformly bounded number so this you get okay so now I think I have to stop sorry and that's stupid now can you give me five minutes sorry about that just give me five minutes and I'll explain quickly the ideas and then maybe I'll just put the things on the slides and put them on the on the web so you can actually yeah okay so we get this uniform compactness and now Gromov's compactness theorem which I'll just take as a black box box which says exactly what I'm stating now Gromov's compactness theorem says that there exists a compact metric space whenever you have a sequence of uniformly compact sequence of metric spaces there exists a compact metric space and there exists isometric embeddings of your spaces of all of your spaces in this compact C isometric embeddings okay so now you that means you can put them all into into one metric space one big metric space Z okay which is still compact okay so now of course you can push forward your SN these now are integral two currents and they all have you know they're basically it's a bounded sequence because the masses right they are uniformly bounded and the boundary curves are just CN which are uniformly yeah so you can by the closure and compactness theorem you can get a subsequence at least that goes to an integral two current in your Z so this is the closure and compactness which we stated okay so these converge up to a subsequence and the same thing by a sort of say let's call it you can assume that your curves the push forwards of your curves right the curves they are in all AN now that they converge uniformly to a Lipschitz curve so now the only thing that you have to do you have to push these forward to your asymptotic cone okay so principle you know these curves here that the CN's you know they approximate in some sense your original curve or at least you know at a finite number of points they approximate so you want to push that forward and you want to push the limit forward as well okay the only problem that you have now is of course that if you know you're choosing here a subsequence but if you're non-principle ultra filter chose a different subsequence in some sense right so this guy here this subsequence could actually be an negligible subsequence okay so your omega of these Nj's they could be yeah so there you have to be a little bit careful okay but this you can do I mean you there's a certain subset that you can so you the subset a in z which is basically all the ultra limits of say psi N AN's where the AN's are in AN okay so these this subset of ultra limits okay this you can push forward that's no problem okay and now you find a subsequence because you're in a compact matrix space you find a subsequence you know such that basically the psi NJ's ANJ's okay they converge in the howster sense to a subs subset of the A okay and the currents then you just take a subsequence again okay as soon as you lie in this guy in here then nothing can happen okay so I'm sorry this was a bit quick and a bit messy but yeah so maybe I'll put the put the the idea or the details in the in the in the notes are there any questions sorry for going over time so much