Question

@Michele_Laino What is a Lie group? :D

Answer 1

A group of liers. :P XD

Answer 2

:D

Answer 3

Lie group is in general a group, which carries the structure of a smooth manifolds, such that the groups elements can be continuously varied.
In the case of a specific PDE , the groups of elements are the symmetry operators of that PDE, and those symmetry operators depend on a little quantity namely \epsilon, or \lambda

Answer 4

I am not sure I understand, for instance is the Hamiltonian operator or any of the operators of QM like momentum operator any of these group symmetry operators?

Or is it more like when you have a spherically symmetric PDE, how does this form a group?

Answer 5

A Lie group of a PDE is a more abstract concept, that doesn't refer to a single PDE or a single class of PDEs. Furthermore, the Hamiltonian operator is a particular operator which belongs to amore large class of operators, and those operators are considered when we study the analysis of Hamiltonian systems, using Lie algebras

Answer 6

Here is an example of Lie group G:
G = SO(2), namely, G is the subsequent set:
\[G = \left\{ {\left( {\begin{array}{*{20}{c}}
{\cos \theta }&{ - \sin \theta } \\
{\sin \theta }&{\cos \theta }
\end{array}} \right)\;:\quad 0 \leqslant \theta \leqslant 2\pi } \right\}\]
where \theta is the angle of rotation

Answer 7

namely a vector space now is enriched of the structure of smooth manifold, or differetiable manifold, since \theta can be varied contnuously, and we can make the differentiation of such of matrices in SO(2)

Answer 8

At the beginning it is not simple, nevertheless, if it can help, you can think about a manifold of dimension n, for example, like the euclidean space R^n

Answer 9

Ok, so this is essentially just saying rotation matrices form a group, and I see that at least in 2 dimensions they are commutative and associative. I am guessing that SO(3) is all the rotations in 3D and that would be associative but not commutative I believe just from my experience of playing with quaternions.

Answer 10

that's right!

Answer 11

so a Lie group is a group, which carries a distance, a differentiation operators with respect all of coordinates of our manifolds, furthermore it carries a structure of vector space,since R^n is a vector space

Answer 12

a simpler Lie group is:
G= R^r, and the operation of group is:
\[\left( {x,y} \right) \to x + y\]
namely the vector sum

Answer 13

Please keep in mind that our discussion is general, and it will be more simple, when we make the connection between Lie groups and PDEs

Answer 14

Is this related to parallel transport or the Levi-Civitta connection? It sounds like parallel transport and the Christoffel symbols help us move coordinates from one local coordinate system to another place but I haven't really done much of that yet but I think I knida get it. I'm sort of new to these concepts so I am not entirely sure if my intuition about these things is right.

Answer 15

I'm not sure, I don't think that there is some type of link between the parallel transport and Lie groups

Answer 16

Ok that's fine I'm just trying to juggle this all around in my head and understand what a Lie group is as genuinely as possible!

So what's this connection between Lie groups and PDEs that you're talking about?

Answer 17

more precisely, given a PDE, we have to find some operators, namely differential operators, such that the subsequent system of differential equation holds:
\[\left\{ \begin{gathered}
{\text{p}}{{\text{r}}^{\left( n \right)}}{\mathbf{v}}\left( \Delta \right) = 0 \hfill \\
\Delta = 0 \hfill \\
\end{gathered} \right.\]
where:
\[\Delta = 0\], is our PDE, and
the operator:
\[{\text{p}}{{\text{r}}^{\left( n \right)}}{\mathbf{v}}\]
is the so called "prolongation" of the unknown operator \[{\mathbf{v}}\]

Answer 18

two things are necessary:
1) we have to go from Lie groups to Lie algebras
2) we hvae to introduce the prolongation of a vector field, so first we have to define what a vector field is into a smooth manifolds

Answer 19

That differential system defines, the so called a classical or exact Lie symmetries of a PDE. It is necessary to keepo in mind that the set of all symmetry operators of a PDE is a Lie algebra associated to that PDE

Answer 20

If you want, we can start with the definition of a Lie algebra

Answer 21

Ok, that sounds like a good place to start. I like your examples given so far, I can see how they form groups so I'm interested how they turn into a Lie algebra. :)

Answer 22

ok!

Answer 23

we can say that a Lie algebra, is a vector space on the field R (real line), which is provided by an operator, called Lie parentheses:
\[\left[ { \cdot \;,\; \cdot } \right]\;:\;H \to H\]
such that the subsequent properties holds:

bilinearity:
\[\left[ {c{\mathbf{v}} + c'{\mathbf{v'}}\;,\;{\mathbf{w}}} \right] = c\left[ {{\mathbf{v}}\;,\;{\mathbf{w}}} \right] + c'\left[ {{\mathbf{v'}}\;,\;{\mathbf{w}}} \right]\]

Answer 24

\[\left[ {{\mathbf{v}},c{\mathbf{w}} + c'{\mathbf{w'}}\;} \right] = c\left[ {{\mathbf{v}}\;,\;{\mathbf{w}}} \right] + c'\left[ {{\mathbf{v}}\;,\;{\mathbf{w'}}} \right]\]

Answer 25

I believe the tensor product has this same property, is that right or is this slightly different?

Answer 26

I think that you are right, if you refer to the product or vector, or cross product between vectors, or if we refer to algebra of diadi, namely the algebra of the tensor product between two vector of a vector space:
\[{\mathbf{v}} \otimes {\mathbf{w}}\]
skew-symmetry:
\[\left[ {{\mathbf{v}},{\mathbf{w}}\;} \right] = - \left[ {{\mathbf{w}}\;,\;{\mathbf{v}}} \right]\]

Answer 27

and finally:
Jacoby identity:
\[\left[ {{\mathbf{u}},\left[ {{\mathbf{v}},{\mathbf{w}}\;} \right]} \right] + \left[ {{\mathbf{v}},\left[ {{\mathbf{w}},{\mathbf{u}}\;} \right]} \right] + \left[ {{\mathbf{w}},\left[ {{\mathbf{u}},{\mathbf{v}}\;} \right]} \right] = {\mathbf{0}}\]

Answer 28

Ahhh ok I don't mean to keep interrupting, but I have seen this sort of thing with the Riemann-Christoffel tensor, this looks like the Bianchi identity, I feel pretty comfortable with what you've said so far. =D

Answer 29

that's right! The Bianchi identities are a good example

Answer 30

more precisely, those identities are a good example of cyclicity

Answer 31

please keep in mind there are many other concepts which are important to understand, nevertheless we can introduce them when progressively, namely we need of them, at the right moment. An useful concept, is the vector field, and the prolongation of a vector field

Answer 32

I'm interested in the prolongation of an operator. Would the D'Alembertian be considered one of these operators perhaps or the Laplacian?

Answer 33

no, they are simply operators, the prolongation of an operator v, is a different operator which can be derived from that operator v, and never is equal to that operator v

Answer 34

It is necessary to start with a vector field on a manifold M

Answer 35

for example a vector field v, on a manifold M, whose dimension is n (please keep in mind M = R^n) is defined as below:
\[{\left. {\mathbf{v}} \right|_x} = {\dot \phi ^1}\left( \varepsilon \right)\frac{\partial }{{\partial {x^1}}} + ... + {\dot \phi ^n}\left( \varepsilon \right)\frac{\partial }{{\partial {x^n}}}\]

Answer 36

where:
\[\phi \left( \varepsilon \right) = \left( {{\phi ^1}\left( \varepsilon \right),...,{\phi ^n}\left( \varepsilon \right)} \right)\]
is a smooth curve on M, and whose tangent vector at point x, is:
\[\frac{d}{{d\varepsilon }}\phi \left( \varepsilon \right) = \left( {{{\dot \phi }^1}\left( \varepsilon \right),...,{{\dot \phi }^n}\left( \varepsilon \right)} \right)\]

Answer 37

I have seen a vector represented as an operator like this before, you can use the tensor notation of \[v = \dot \phi^i \partial_i \] I'm comfortable with this notation. This can operate on a position vector, which essentially gives you the components with respect to the covariant basis, as far as I know, just so you know where I stand so far. :)

Answer 38

that's right! it is the right compact form to write my expression above

Answer 39

next, in the subsequent example I will show a simple connection between a Lie groups and its generator. A generator is a special vector filed, which, as its name indicates, generates our Lie group of transformation

Answer 40

we start from this Lie group:
\[\Psi \left( {\varepsilon ;\left( {x,y} \right)} \right) = \left( {x\cos \varepsilon - y\sin \varepsilon ,\;\;x\sin \varepsilon + y\cos \varepsilon } \right)\]

Answer 41

Ok, this is SO(2) from earlier, I like what I see so far!

Answer 42

that group, as you can recognize, is a rotation of the x,y plane, and \epsilon, is a rotation. Furthermore, we can see that every transformation, depends on the point which has to be transformed, and by a parameter, namely \epsilon, well that parameter is called Lie parameter

Answer 43

exactly!
Now we can compute the component of our "infinitesimal generator", of such rotation, as follows:
\[\begin{gathered}
\xi \left( {x,y} \right) = {\left. {\frac{d}{{d\varepsilon }}} \right|_{\varepsilon = 0}}\left( {x\cos \varepsilon - y\sin \varepsilon } \right) = - y \hfill \\
\hfill \\
\eta \left( {x,y} \right) = {\left. {\frac{d}{{d\varepsilon }}} \right|_{\varepsilon = 0}}\left( {x\sin \varepsilon + y\cos \varepsilon } \right) = x \hfill \\
\end{gathered} \]

Answer 44

so the infinitesimal generator of our group above is represented by the subsequent vector field:
\[{\mathbf{v}} = - y{\partial _x} + x{\partial _y}\]

Answer 45

I feel like I have somehow seen this particular thing before but it looked a lot different than this. They were using the taylor series of e^x to compute a matrix exponential for rotations... That was about 2 years ago so I don't really remember much other than this vague idea, and it certainly wasn't in tensors.

At any rate, I suppose now that we have these infinitesimal generators, is the point that we can now integrate this to get any rotation (or really more generally integrate with respect to the Lie parameter to do something...?)

Answer 46

Also, is it possible to have multiple lie parameters at the same time, in a "Two dimensional sense"? I guess that's what you meant earlier when you said they tend to use \(\epsilon\) and \(\lambda\) earlier.

Answer 47

no, I meant \epsilon or \lambda, since some authors prefer \lambda, whereas some others authors prefer \epsilon
In my thesis I worked with only one-parameter Lie groups, sincerely I don't know if there exist multiparameter Lie groups.

Answer 48

of course, the Taylor series can be applied in order to find the so called "flow" associated to a vector field.

Answer 49

another example is:
\[\Psi \left( {\varepsilon ;\left( {x,y} \right)} \right) = \left( {\frac{x}{{1 - \varepsilon x}},\;\;\frac{y}{{1 - \varepsilon x}}} \right)\]

Answer 50

as, now you can easily check, the infinitesimal generator, is the subseqiuent vector field:
\[\Large {\mathbf{v}} = {x^2}{\partial _x} + xy{\partial _y}\]

Answer 51

subsequent*

Answer 52

So in general we can say that a general form of an infinitesimal generator of a given transformation, which is a group transformation, is given by the subsequent formula:
\[\Large {\mathbf{v}} = \sum\limits_{i = 1}^n {{\xi ^i}\left( {\mathbf{x}} \right)\frac{\partial }{{\partial {x^i}}}} = {\xi ^i}{\partial _i},\quad {\mathbf{x}} = \left( {{x^1},{x^2},...,{x^n}} \right)\]
and we operate in a n-manifolds, namely a manifolds whose dimension is n

Answer 53

Interesting, I can see perfectly how this works so far, so what can we do with these generators now that we have them for their respective groups?

Answer 54

when we know a generator of some transformation groups, associated with a given PDE, we can use that generator to find new solutions of that PDE

Answer 55

Woah.

Answer 56

Nevertheless the concept of a generator need to be extended. A way to understand it, is given by the subsequent simple example

Answer 57

Let's consider the heat equation, namely:
\[\Delta = {u_t} - {u_{xx}} = 0\]

Answer 58

As we can see, we have the second derivative (with respect to space of the unknow function, and the first derivative (with respect to time, we are in one dimension) of the unknow function

Answer 59

Ok good example, I have experience with this. Actually, is this related to finding the solution to the heat equation as a dirac delta and then integrating it with a function to make an arbitrary solution based on these? (This is something I partly learned at one point)

Answer 60

Not exactly, we haven't need to do that integration, since we use the infinitesimal generators. Our task is to search or compute those infinitesimal generators.
So our infinitesimal generator has to be able to operate in a space, or prolonged manifold, whose coordinates are:
\[\Large x,\;t,\;u,\;{u_x},\;{u_t},\;{u_{xx}}\]

Answer 61

from that, there is the necessity to enlarge our manifolds, in order to introduce new coordinates, namely all the first derivative of the function u, and the second derivative (with respect to the space) of the same function u

Answer 62

in other words, these variables:
\[\Large x,\;t,\;u,\;{u_x},\;{u_t},\;{u_{xx}}\]
are viewed like all independent variable from our infinitesimal generators

Answer 63

variables*

Answer 64

This is strange, but I'm with you!

Answer 65

ok! All that will be more simple, when we work on a PDE.
So, instead of considering the infinitesimal generator v, we have to consider its prolongation v*, which is defined as below:

Answer 66

\[\Large \begin{gathered}
{{\mathbf{v}}^*} = \xi \left( {x,t,u} \right)\frac{\partial }{{\partial x}} + \tau \left( {x,t,u} \right)\frac{\partial }{{\partial t}} + \phi \left( {x,t,u} \right)\frac{\partial }{{\partial u}} + \hfill \\
\hfill \\
+ {\phi ^x}\left( {x,t,u} \right)\frac{\partial }{{\partial {u_x}}} + {\phi ^t}\left( {x,t,u} \right)\frac{\partial }{{\partial {u_t}}} + {\phi ^{xx}}\left( {x,t,u} \right)\frac{\partial }{{\partial {u_{xx}}}} \hfill \\
\end{gathered} \]

Answer 67

where the functions:
\[\Large {\phi ^x},\;{\phi ^t},\;{\phi ^{xx}}\]
are the first-extensions (the first two) and the second extension (the third) of the function \[\phi \]

Answer 68

Our task is to determine, the unknow functions:
\[\Large \xi \left( {x,t,u} \right),\quad \tau \left( {x,t,u} \right),\quad \phi \left( {x,t,u} \right)\]

Answer 69

I assume the next step is to infinitely extend them? Then we have a polynomial that we can do convolutions on maybe? Is this invertible in a matrix sense that we can look at maybe? I'm excited haha.

Answer 70

theoretically we can speak about an infinitely prolonged vector field, practically we deal with finite prolongations, since an infinite prolongation means a PDE with partial derivatives of infinity order

Answer 71

Now, we have some formulas which we can use in order to compute the fu nctions
\[\Large {\phi ^x},\;{\phi ^t},\;{\phi ^{xx}}\]
starting from the unknow function \[\Large \phi \left( {x,t,u} \right)\]

Answer 72

All those formulas are recursive, nevertheless they are very simple formulas

Answer 73

In order to understand those formulas, we have to introduce the concept of total derivative of a function

Answer 74

Alright, is this just the regular total derivative that I would think of from multivariable calculus?\[d f = \frac{\partial f}{\partial t} d t + \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial u} du\] Or maybe I should write \[df = \partial_i f dx^i\]

Answer 75

yes! keep in mind that now our variables are the same unknow function u, together all its partial derivatives.

Answer 76

example, if we have the subsequent function:
\[\Large P = xu{u_x}\]

Answer 77

sorry, if we have the subsequent function:
\[\Large P = xu{u_{xy}}\]
then its total derivatives are:
\[\Large \begin{gathered}
{P_x} = u{u_{xy}} + x{u_x}{u_{xy}} + xu{u_{xxy}} \hfill \\
{P_y} = x{u_y}{u_{xy}} + xu{u_{xyy}} \hfill \\
\end{gathered} \]

Answer 78

Ok that clears it up I think, can we also calculate:

\[\large P_u = xu_x\]

Answer 79

ok! that's right!

Answer 80

in other words we have to apply the chain rule

Answer 81

and the Leibniz rule

Answer 82

sorry, no the chain rule, only the Leibniz rule

Answer 83

Ok, sounds good to me. :D

Answer 84

ok! Now we are ready for the formulas about the extension of the function \[\phi \]
Please keep in mind that our infinitesimal generator is:
\[{\mathbf{v}} = \xi \left( {x,t,u} \right)\frac{\partial }{{\partial x}} + \tau \left( {x,t,u} \right)\frac{\partial }{{\partial t}} + \phi \left( {x,t,u} \right)\frac{\partial }{{\partial u}}\]

Answer 85

I will rewrite the infinitesimal generator v as below:
\[{\mathbf{v}} = {\xi _1}\left( {{x_1},...,{x_m},u} \right)\frac{\partial }{{\partial {x^1}}} + {\xi _2}\left( {{x_1},...,{x_m},u} \right)\frac{\partial }{{\partial {x^2}}} + ... + \phi \left( {{x_1},...,{x_m},u} \right)\frac{\partial }{{\partial u}}\]

Answer 86

here are those formulas:
\[\Large \phi _i^{\left( 1 \right)} = {D_i}\phi - \left( {{D_i}{\xi _j}} \right){u_j},\quad i = 1,2,...,m\]
for the first-order extension

Answer 87

here is the formula for the k-th order extension of the function \[\Large \phi \]