Stability of Minkowski Space-time with a translation space-like Killing_vector field
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Transkript: Englisch(automatisch erzeugt)
00:16
first of all, I would like to thank the organizer for the invitation to talk here.
00:24
And so today I will talk to you about the stability of Minkowski with the translation symmetry. But first of all, I will introduce you a little the context.
00:43
So in general relativity the space-time is described by a manifold M equipped with a Lorentzian matrix G. So usually the manifold is of four dimensions. So Lorentzian matrix is a matrix with signature minus one, one, one, one.
01:06
And so Einstein equation links the geometry of the universe to the distribution of matter and energy which is present. And they can be written R mu nu
01:27
minus one half of R G mu nu equal T mu nu. So I will explain what it is. R mu nu is a rigid answer.
01:43
You can see it as a, so it describes the curvature of the metric and you can see it as a second-order non-linear operator on the metric G. And roughly speaking it is on the form G times two derivative of G plus some product of dG times dG.
02:06
It is a scalar curvature. It is a trace of the rigid answer. And so G mu nu, it is the metric, the Lorentzian metric.
02:21
And T mu nu, it is the energy-impulsion tensor. It is the object which describes the distribution of matter and energy. And its form depends on the matter model we study. It has a different form. Either we study
02:45
a perfect fluid or an electromagnetic field. But in our case, we will take T mu nu equal zero. So it is in vacuum. There is no matter. And if you take the trace of this equation, you obtain R minus one half,
03:04
R times one half times the trace of G mu nu equals zero. And so you obtain that R is equal to zero. And so Einstein equation in vacuum consists just in looking for Lorentzian metric G, which are each G flat, with rigid answer R mu nu is equal to zero.
03:27
So a particular solution is given by Minkowski metric. Minkowski space-time. This is
03:42
R4 equipped with the metric M, which is minus dt squared plus dx1 squared plus dx2 squared plus dx3 squared. So this metric is a flat metric, but it's not the only solution to
04:05
Einstein equation in vacuum. Actually, the set is broad and dynamical, and to see it, we can write the Einstein equation as a Cauchy problem. So the initial data for Einstein equation are triplets
04:28
sigma G bar k. You can see it as the data of what the space looks like at some instant t equals zero,
04:41
and what is its speed of propagation at t equals zero. Speed of deformation, sorry. So sigma is a 3D surface, or 3D manifold. G bar is a Riemannian metric, and k is a symmetric two-ton source.
05:14
And to solve Einstein equation with this initial data consists in finding
05:22
Mg, solution of Einstein vacuum equation, so with reach of G equals zero, and such that sigma embeds in M. G restricted to sigma is the Riemannian metric G bar, and k is the second fundamental form of
05:53
the embedding of sigma into M. But you can see k as the data at time t equals zero of the derivative with respect to time of
06:02
metric G. The initial data, they cannot be chosen arbitrarily because some of the the equation here depends only on the initial data. These are the equation R00
06:20
minus one-half of Rg0 equals zero, and R0i equals zero. So they can be written actually as R bar minus k square plus tau square equals zero, where R bar is a trace, is a Ricci, sorry, it's a scalar curvature
06:49
of G bar. So the trace of the Ricci tensor of G bar. k, so it's here, the norm is taken with respect to the metric G bar, and tau
07:03
is a trace of k. It's called the mean curvature. And so this is this equation. And the other equation can be written divergence of k minus gradient of tau is equal to zero. So these are equations which
07:24
depend only on the data of G bar and k. But once they are solved, Einstein equations are locally West-Poles. So this very important theorem is due to Ivan Shokke-Borya in the fifties.
07:48
And it says that, so for initial data, sigma G bar k with enough regularity, so with enough regularity, solution of the constraint, you have a local existence of solution to an Einstein equation.
08:27
But usually, this solution cannot be extended globally. So, we'll never see it again. Singularity may form, but if you consider perturbation of a particular
08:45
global solution, then you may hope that there is a global existence. And so this is the case for Minkowski space-time.
09:10
So the initial data for Minkowski space-time, this can be, for example,
09:20
S3 equipped with the Euclidean metric delta on the symmetric two tensor, which is zero. But it could be also other thing. You can also imagine Minkowski space-time that I draw here and draw an hyper-surface like this.
09:42
And then you will obtain another initial data set. But let's consider this. And if you take, so the stability of Minkowski, so proved by Christo Doulou and Feynman in the 90s,
10:09
says that if you take an initial data set R3, G bar k with,
10:20
so G bar and k have enough regularity, G bar is closed to delta k is small, in some sense. And also you assume that this initial data set is asymptotically flat. This means that G bar tends to delta at infinity with some decay rate, and k tends to zero at infinity.
10:47
Then you have a global existence of solution, and the convergence at infinity to Minkowski again.
11:05
So there is another proof of this theorem, which is due to Lindblad and Rodianski,
11:24
which uses wave coordinates. And I will explain to you later what wave coordinates are. So in this talk, we'll be also interested in the stability of Minkowski, but in a slightly different context, in the case, so in the presence of translation symmetry.
11:44
So we will study a manifold of the form R4 that I decompose in R2 times R, X3 times R3,
12:07
equipped with metric of the form G, the 3 plus 1 metric equal E minus 2 phi G plus E 2 phi G X3 squared,
12:26
where G is a Lorentzian metric on R2 plus 1, or R2 times RT.
12:40
And phi is a scalar function, and we assume that phi and G bar are independent of the coordinate X3.
13:07
Sorry, no, no, G, G. The Lorentzian metric in 2 plus 1. And then, Einstein vacuum equation, if you assume that G4 satisfies Einstein vacuum equation,
13:28
we obtain the system for phi and G, which is the Lambertian in the metric G of phi is equal to 0,
13:42
and the Ricci tensor of G is equal to D mu phi. So this is now a system of equation in R2 plus 1.
14:00
So there is a particular solution, trivial solution to this problem, which is actually Minkowski space term in 3 plus 1 dimension itself, because Minkowski space term has a lot of symmetry, in particular it has translation symmetry. And this solution, so Minkowski solution corresponds to the solution phi equals 0,
14:26
and G equals the Minkowski metric in 2 plus 1 dimension, that I call M2 minus DT squared plus DX1 squared plus DX2 squared.
14:43
And so the natural question one can ask is, is this solution stable for this system? Let me just do a small remark that the answer to this question
15:05
is not given by the stability of Minkowski space term by Christo Duluc-Lein-Horman or Leinblad-Ronensky, because perturbation we consider have translation symmetry, so in particular they are not asymptotically flat in 3 plus 1 dimension,
15:22
because then you cannot converge at infinity to the Euclidean metric if you have to remain the same in one direction. And so I will now give you the answer to this, so yes it is stable, but first of all, to be more precise,
15:51
I need to give you what are the initial data we will take.
16:02
So the initial data should be a solution of the constraint equation, so up there I wrote a constraint equation in vacuum, but if you are not in vacuum, if you have an energy input and tensor,
16:21
as it would be the case here, we have a right hand side in the equation, it just means that you have a right hand side on the constraint. So here this right hand side is given by the data at time t equals 0 of phi on dt phi, so this will be given, phi dt phi at time t equals 0 are given,
16:46
and then we can look for solution of the constraint equation of the form g-bar, so we look for solution R2 g-bar k,
17:00
and without loss of generality, since g-bar now is a metric on R2, we can assume that g-bar is in the conformal class of the Euclidean metric, so g-bar equals e 2 lambda times delta,
17:20
and we can also decompose k into its traceless part under its trace, we do nothing by this, we write k equals h plus one half of g times tau, tau is a trace of k, the mean curvature,
17:44
so h is traceless, and so the constraint equation consists now in an elliptic system
18:02
for lambda on h, for lambda on h with tau seen as a parameter in the equation,
18:27
so it's okay because lambda is one function on the traceless tensor, in symmetric traceless tensor in two dimensions, it has two components, two independent components,
18:41
and we have three constraint equations in two dimensions, so we have the right number of equations, or the right number of unknowns, and the theorem will be the following, so assume that phi, dt phi are in H2 times H1,
19:03
for simplicity, assume that they are compactly supported in symbol B0r, and assume that they are of size epsilon small,
19:25
and we also take some function theta in H1, compactly supported with integral one,
19:44
this theta will be actually the free data for tau, and then there exists a set of parameters, a0, a1, a2, c1, c2, g on big A in R,
20:10
and there exists a unique solution lambda H, solution of the constraint,
20:23
with lambda is equal to minus a0 times a cutoff function, which is one at infinity and zero in the middle, times logarithm of R,
20:40
plus, sorry, let's also take a delta, some number between zero and one, plus a big O of one over R, one plus delta, H, you can also write some development for H,
21:01
but I don't write it because it is matrices and it's complicated, and tau, tau is not quite free, tau is equal to e minus lambda, R times t over R times a1 cosineus of theta,
21:20
plus a2 sinus of theta, plus e minus lambda times big A times theta. So R and theta here, small theta,
21:42
are the polar coordinates centered in c1, c2. So let me do some small commands.
22:03
So here you see that tau is not so much a free parameter, the only freedom is in the choice of this function theta. The other part is imposed to have the right orthogonality condition
22:21
to inverse the analytic system for lambda on H. Also, the parameter a0, here you have lambda, so the first term in the development of lambda is log of R. This means that the metric G bar here
22:42
is asymptotically the metric in a cone with a deficit angle given by a0. And the parameter G, you don't see it appear actually, it appears in the development, in the asymptotic development of H and it corresponds to the angular momentum.
23:06
So just to give you precisely the initial data we consider.
23:30
So now let me state the stability theorem.
23:43
So now take phi dt phi, we will assume more regularities than H2 times H1. For example, let's assume H30 times H29.
24:03
Still now it is really, it's not a simplification, it's really we need it to assume that they are compactly supported, b0 R. And then take theta whatever,
24:23
for example 0 to simplify, but whatever. And then the initial data of the following theorem leads to a global solution. So there exists a global solution,
24:43
G phi, which is a development of the initial data given by the previous theorem. Such that, so to describe it, we actually can introduce two coordinate systems.
25:08
A first coordinate system in the C, the future of b0 R, the future of the initial data for phi.
25:22
So in this set we have a coordinate system, phi times t x1 x2. Such that we can have the estimate, phi is less than the constant times epsilon times 1 plus s 1 by 2,
25:44
1 plus q almost 1 by 2. So under G minus the Minkowski metric in 2 plus 1 is less than epsilon times 1 plus s almost 1 by 2.
26:06
Where, just to explain the notation, r theta are the polar coordinates,
26:24
s is r plus t and q is r minus t. And in the complement of C, so C complement,
26:40
we have another coordinate system, t prime x1 prime x2 prime. So there is no need in the complement of C to give you the estimate for phi because there phi is equal to 0 by definition. We are in the future of the set where phi is non-zero,
27:05
and in the complement of this set. So there phi is equal to 0. So we just need to know G and we will not compare G with the Minkowski metric
27:21
but with another metric G A. And we have that G minus G A is less than the constant times epsilon times 1 plus s almost one half times 1 plus q 1 plus delta.
27:44
Delta is a parameter between one half of 1 and the easy parameter that we have to put into this theorem to obtain the initial data. And so I will tell you what is this metric G A,
28:03
which is not Minkowski metric. Sorry, you should put here s prime and q prime because you take r prime theta prime
28:22
to be the polar coordinate associated to the prime coordinates and s prime q prime will be defined by r prime plus t prime, r prime minus t prime. So G A, I have to express it in prime coordinate. So I will put prime everywhere. So this is d prime squared plus dr prime squared
28:44
plus a term which is G times dq prime d theta prime plus r prime plus q prime times a zero plus a one cos theta prime plus a two sin theta prime
29:09
squared d theta prime squared. So this quantity G A zero, a one, a two are the quantity which are given by the theorem
29:22
for the initial data. This quantity are prescribed by the initial data. And this metric is not Minkowski metric because it has a deficit angle at infinity. As if you... Sorry, I'm sorry. So about this assumption. So you say that the phi is compactly supported
29:42
but the metric is a real perturbation. So for the metric you allow... Because you of course cannot prove Christo-Doulou-Clyman theorem for compactly supported. The metric itself you allow a general perturbation but phi is compactly supported. Is this what you mean?
30:00
Yes, the metric perturbation is given by the theorem of which gives you the initial data. The only freedom, once phi is given, the only freedom is some small freedom in the choice of the mean curvature. This is not really a freedom, it's just that what I explained here
30:24
is that even for Minkowski... Even for Minkowski metric you always have the freedom to choose, if you want, that t equals 0 would be the seglising, if you want.
30:41
The compactly supported perturbation, you cannot have the stability of the Minkowski. It's not the same theorem. If you have compactly supported perturbation. So how is this fitting with the stability of the Minkowski space, Christo-Doulou-Clyman? The thing is the whole perturbation
31:04
if you consider the perturbation now in 3 plus 1, it's not in the setting of Christo-Doulou-Clyman because with the translation symmetry you have to add another direction in which you are symmetric
31:21
and with this symmetry you are not asymptotically flat. So it's another setting. And in this setting you can assume phi is compactly supported on the fiber R2. But it's not compactly supported
31:41
if we go back to 3 dimensions and we add the extra direction because then, because we are symmetric we are not like in a, like if we were, maybe in a cylinder. Would you replace this with decay it would be enough as well?
32:00
If you were to take phi to be decaying instead of compactly supported? Then I don't, there are some technical difficulties that I don't know how to solve. So I don't know, I cannot give you an answer to this.
32:21
So now I will, in the remaining of the talk I will give you some elements of proof that will make you understand also why I need two coordinate systems to describe my solution
32:41
because it seems a little strange. Actually you can also introduce a global coordinate system but then if you take this global coordinate system you will have some, instead of GA which is quite nice you will have some ugly metric in this other coordinate system.
33:13
So some elements of proof.
33:21
So the strategy for proving this theorem is inspired from the proof of stability of Minkowski by Linblad Androniansky. So we will introduce wave coordinates. But before introducing them let's just write the Ricci tensor
33:41
in some coordinate system. I don't assume anything on this coordinate system and then you can just compute you know that the Ricci tensor is like the derivative of the Christopher symbol and the Christopher symbol are like the derivative of the metric. So just a very basic computation
34:01
leads that R mu nu is like one minus one half of the d'Alembertian in the metric G of the metric component G mu nu plus some quadratic term in dG plus some term
34:20
which are of the form G rho mu d mu H rho and you take the symmetric you also add the G rho nu d mu H rho where H rho is something which depends some quantity which depends
34:41
on the derivative of G which can be written like one over the square root of determinant of the metric G times d alpha of G alpha rho square root of the determinant of G when the indices are up
35:01
it just means that we take the inverse of the metric and this is just nothing but the d'Alembertian in the metric G coordinate function x rho and so if you take if you assume that
35:20
your coordinate function satisfy such that H rho equals zero you are in the so called wave coordinates and then our system
35:41
of equation which was there let's call it star in wave coordinate like a system of quasi linear wave equation so phi
36:01
is still d'Alembertian d'Alembertian is a metric G of phi equals to zero and d'Alembertian is a metric G of G mu nu is like minus 2 d mu phi d mu phi plus some quadratic term in dG
36:26
and so the question now is do we have a distance of global solution for this system for small data and to have an idea
36:40
we just give you what happened for a wave equation in 2 plus 1 the decay of a derivative
37:01
of phi phi satisfy wave equation in Minkowski with initial data let's say compactly supported and of size epsilon so like the assumption we have for phi then
37:21
a derivative of phi has the following decay decay like so epsilon because initial data are of size epsilon over 1 plus s one half 1 plus q
37:43
three half and if you take some not random derivative but what are called the good derivative you have more decay for phi
38:03
so the good derivative are the derivative with respect to s where s is r plus t or you can also take d theta of phi over r which is also a good derivative and then
38:23
you have more decay it's decay actually like epsilon over 1 plus s three half 1 plus q one half so this implies that if you
38:42
should now consider a non-linear wave equation of the form d'Alembertian of u equal du for example because it is like it seems to be
39:00
a good model for our problem then if you assume some bootstrap on u which is that u satisfies the same estimate as a solution to the free wave equation would satisfy so bootstrap the linear estimate
39:29
of du when you integrate and you obtain that d over dt of the integral of du squared will be less than the integral of du squared
39:40
times du and you estimate one of your derivative in L infinity with this estimate this is less than epsilon the integral of du squared
40:00
is the energy of u which we assume to be bounded because it's part of the linear estimate that the energy is bounded and so you obtain that this is less than epsilon square root of c times the integral of du squared and so the energy
40:29
of dt yes and so and so if you want
40:44
to bootstrap to recover the linear estimate you've done you need to assume that epsilon times square root of t is bounded and so you have existence in time t less
41:10
square root of epsilon times square root of t from that value or you assume that yes we assume that u satisfies
41:21
the same estimate that would satisfy the solution of d'Alembert's u equals 0 is that the conclusion or assumption that square root of integral of du squared is less than epsilon
41:47
times square root of it's a square root of I've done some integral of integral of
42:04
I think you're just integrating one over square root of t and you use square root of t yeah but you have y equal to y so you I assume that b is square on the left hand side which is that's human
42:20
you're not applying ground well I think that's a conclusion you're not applying ground well here assuming that on the right hand side you have everything you expect from the way that yeah and then you integrate you assume that energy is bounded on the right hand side and then you get the gross like square root of t on the left hand side right
42:53
let's assume that we have it's epsilon we integrate we obtain square root of t and if we wanted to get back to
43:00
this bounded we obtain that we have to assume that t is less than one over epsilon squared and so
43:24
but if we have some structure on the non-linearity we have if the long version u is like du time a good derivative of u that i naught d bar u then when we do the same thing if we for example if we integrate d bar u
43:40
we use the estimate we have for a good derivative of u then what we obtain is that the we have inside of epsilon time one over square root of t we have t to the three half and so this is integrable
44:00
and there is a global existence and so this result was due to for this particular case but earlier there was
44:20
some result by for a similar problem so unfortunately epsilon equation in wave coordinates don't have the good structure to see the structure we need to introduce a null frame so
44:41
l equals dt plus dr l bar equals dt minus dr and u equals d theta over r and in this frame the Einstein equation can be written so if we drop all the non-linearities
45:01
that involve a good derivative because then we have enough decay to have a global result then what remains is del inversion of phi is equal to zero minus g ll dq phi squared equal
45:20
to zero del inversion of g ll minus g ll dq squared g ll equals zero and del inversion of g l bar l bar minus g ll dq squared g l bar l bar equal dq phi squared so the only term
45:40
we kept as a quasi linear term which
46:00
doesn't have the null structure and then so this seems to be bad but we can as was shown for the stability of Minkowski in wave coordinate in three plus one the if we use the
46:21
wave coordinate condition we obtain that this gives you a relation between some derivative of g and in particular this gives you that a derivative of
46:41
g ll bad derivative so we can neglect them and we are left with the following system this
47:01
structure was this structure
47:20
was called the weak null structure and was introduced also by Lindblad and Ronians t and it allows them to prove global existence but with some logarithmic loss due to the fact that here there is no null structure
47:42
so in our case the
48:00
loss is more dramatic because of what I said here you have a growth in the energy like square root of t so actually this is just because we did not choose the right coordinates and
48:22
just a very quick word to see this we can assume that we have found an ideal coordinate system in this ideal coordinate system
48:45
every metric is ideal g minus we assume that in this ideal coordinate system we can write g minus m less than epsilon over
49:00
square root of t and then in this ideal coordinate system if we compute the R L bar bar which is only d q phi squared which is d q phi squared we can compute that it is like minus d q squared of
49:21
g u u plus some other term which are like epsilon over square root of t sorry t to the
49:51
3 this best one to deal with this bad term in the equation is not g L bar L bar as it would be
50:00
in wave coordinate but it is g u u and so what are called generalized wave coordinates which consists instead
50:20
of h rho equal to 0 we can choose this is equal to f rho another function as long
50:45
as it in this term is dq of g u and so we just impose that L alpha
51:00
h alpha is like the integral of dq phi squared r dq over r so impose this and
51:20
then if we impose this choice what we obtain is instead of this system this system is taken care of and what remains is a cubic non-linear term without null structure which is
51:40
g L L bar minus m L L bar times dq phi squared and for such now we such a structure which is really the equivalent
52:04
of just a very quick word to conclude this choice has some consequences it means that in the outside of the it leads to perturbation outside
52:21
the light cone which looks distinct coordinate system if we want to have a nice description of our solution so thank you very much for your attention
52:48
so I am a little bit confused about the way you treat the quadratic time so my understanding is that if you come from three dimensions to two dimensions and the null condition is satisfied you gain just one half
53:00
not a full order of decay so the same story with a cubic nonlinearity I mean the cubic nonlinearity will still have a local divergence normally unless you have structure so how do you but that is the same because this cubic term
53:21
without null structure it appears only for GL bar L bar which is a bad coefficient GL bar L bar only meets good derivative so that will grow logarithmically yeah GL bar L bar grow logarithmic it's like the point was decay for GL bar L bar is like log of T over square root of T
53:40
but also here when you say you have a good derivative I mean normally you gain just T to the one half right so you just quadratic nonlinearity so it's like the null condition should not be good enough you need more right no no it's good enough it's good enough because you gain one over T in particular because
54:00
you are using a derivative so the term which has only one derivative whenever you want to use a good derivative then you are restricted by what you can do in terms of the energy anyway we can talk about this so is this equivalent
54:20
to dimensions or no sorry is this equivalent to a matter which would be a scalar field I was wondering if you are aiming only for almost global so
54:45
without needing this assumption it's all a log issue
55:01
it's because it's equivalent to two plus one gravity with scalar field you would expect the same result without the restriction on emission data except as long as deficit angle is less than two what you mean the same result without the
55:20
compact support assumption no no without smallness assumption yes it's true that people expect that I presume there is no you did a polarised case here I presume there is no difficulty
55:48
to the yeah the nonlinearity you add have the null structure so it doesn't it's transparent yeah yeah yeah
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