Lie Algebra and Lie Group

[giulio_hep]

TL;DR Summary

Is the exponential an approximation to the first order?

Is it correct saying that the Exponential limit is an exact solution for passing from a Lie Algebra to a Lie group because a differential manifold with Lie group structure is such that for any point of the transformation the tangent space is by definition the Lie algebra: is that the underlying assumption to write a generalized transformation as that limit?
Or which is the intuition: how is the infinitesimal transformation in the neighborhood of the identity generalized to a finite distance?

 

[fresh_42]

Yes, this is generally true, i.e. the intuition. The exponentiation is the way back from straight to curved: $e^{x+y}\stackrel{\to}{=}e^x\cdot e^y$, and the reason why we usually try to solve a differential equation by an ansatz which involves the exponential function.

Lie group and Lie algebra are also related via $\operatorname{Ad}\circ \exp = \exp \circ \operatorname{ad}$, which connects the conjugation in the group with the left multiplication in the algebra.

 

[giulio_hep]

Oh, you are the author of the Insights , now I'm reading this one, and I see that in your reply your are mentioning a passage about which I have a question.
So it is written:

This formula can be visualized by the commutativity of the following diagram:

\begin{equation*} \begin{aligned} G &\stackrel{\operatorname{Ad}}{\longrightarrow} GL(\mathfrak{g}) \\ \exp \uparrow & \quad \quad \uparrow \exp \\ \mathfrak{g} &\stackrel{\operatorname{ad}}{\longrightarrow} \mathfrak{gl(g)} \end{aligned} \end{equation*}

between Lie groups (analytic manifolds in which group multiplication and inversion are analytical functions) in the top row and their tangent spaces at g=1 (Lie algebras) in the bottom row. It reflects an integration process, similar to the standard ansatz when solving differential equations by assuming an exponential function as solution. In this sense the adjoint representation of the Lie algebra is the differential of the adjoint representation of the Lie group, and the adjoint representation of the Lie group the integrated adjoint representation of the Lie algebra. It integrates 0∈g to 1∈G1, resp. the tangent space at g=1 to the connection component of the group identity. The differentiation process can be achieved by considering flows on the manifolds (cp. [6] or [12],[13]).

To be honest my question is only related to the left part of the diagram

\begin{equation*} \begin{aligned} G \\ \exp \uparrow \\ \mathfrak{g} \end{aligned} \end{equation*}

and my reasoning stems from following a video on youtube (I like it very much, so I believe that sharing is helpful...) where the author there is deriving (around minute 5) the exponential

from a basic relation that describes "how the generators generate the infinitesimal Lie group elements, that are infinitesimally close to the identity"

to an exponential relation that is supposed to relate "finite elements in a finite distance away, where we can no longer assume that we're dealing with infinitesimal distances".

So, my comment that I'm trying to write down in a confused way (I know that some math professionals can not stand me and so they can simply ignore my posts ) is that, for that exponential limit to hold at finite distance from the identity I guess we need to assume that:

a differential manifold with a Lie group structure is such that for any point (so, not only in the neighborhood of the identity) of the (finite, not infinitesimal) transformation the tangent space is by definition the Lie algebra... correct? ⁉

 

[fresh_42]

Varadarajan cheats a bit at this point, when we writes:
There is a bijection of vector fields $X\in \mathfrak{g}$ and flows $\xi$ in $G$ along $X$. It is customary to write $\xi_X(t) = \exp tX$.

So the notation is deliberate to some extent, but nevertheless justified, because we have: $\xi_X(s+t) \stackrel{(*)}{=} \xi_{sX}(t)$ and
$$
X_p(f)=\left( \dfrac{d}{dt} f( \xi_{X_p}(t) ) \right)_{t=0} = \left( \dfrac{d}{dt} f(p\exp(tX) \right)_{t=0}
$$
We can do all the calculations with flows along vector fields. But as these flows follow the rules $\xi_X(0)=1$ and $(*)$ which are the same for the exponential function, the notation makes sense. You can find a bit more about flows in https://www.physicsforums.com/insights/pantheon-derivatives-part-ii/ and the following parts (with examples for Lie derivatives and Lie algebras).

I would suggest to exponentiate $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ and see where you end up and what changes if you multiply it by a factor, but it is a bit nasty to exponentiate matrices. Furthermore, we try to get from a linear structure to an analytic structure, so a bit more is necessary for an exact treatment: namely the flows. They are the key, and especially their properties mentioned above. In the end it is the same answer as to the question: Why does an ansatz with the exponential function work, if we want to solve a differential equation?

 

[giulio_hep]

I would suggest to exponentiate $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ and see where you end up and what changes if you multiply it by a factor

Well,
so I can follow the Cayley-Hamilton Method and I have to find $e^{At}$ for
$A = \begin{bmatrix}0&-1\\1&0\end{bmatrix}$
The characteristic equation is $s^2 + 1 = 0$, and the eigenvalues are $λ_1 = +i$, $λ_2 = −i$
Hence $e^At = α_0 I + α_1 A$ and for $λ_1 = +i$, $λ_2 = −i$
$e^{it} = cos(t) + i sin(t) = α_0 + α_1i e^{−it} = cos(t) − i sin(t) = α_0 − α_1i$
so $α_0 = cos(t)$ and $α_1 = sin(t)$.
Then
$e^{At} = cos(t)I + sin(t)A = \begin{bmatrix}cos(t) &s in(t) \\− sin(t) & cos(t) \end{bmatrix}$

Not sure which is your idea behind the suggestion but I notice that this is the solution as symplectic integrator for Hamilton's equations of the harmonic oscillator. From the standpoint of physics we can define a map $ϕ$ that updates the state of the system over a length of time $Δt$
where $ϕ′$ be the Jacobian matrix of $ϕ$

$\phi' = \begin{pmatrix} \frac{\partial \phi}{\partial q} & \frac{\partial \phi}{\partial p} \end{pmatrix}$

and the map $ϕ$ is symplectic if its Jacobian $ϕ′$ satifies:

$\phi'^T J \phi' = J$

where

$J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$

So which is the conclusion? Again, I still see that

we denote the map as $\tilde {\phi}(q, p)$ since it is an approximation to the true flow $\phi(q, p)$

we showed that the Leapfrog integrator, which is an approximate map, is symplectic for the harmonic oscillator

 

[fresh_42]

My idea was to show, that ordinary exponentiation actually leads from a tangent vector to a group element, so that the convention to write flows (= integral curves!) by $\exp$ makes sense.

 

[giulio_hep]

Ok, so your talking about the Circle group

, which has

as its Lie algebra .
Indeed, in that case If θ is any local angle coordinate on an open set U ⊂

, then T = ∂/∂θ on U and

Because left translations are rotations, which preserve T , it follows that T is left-invariant, and therefore T spans the Lie algebra of

As far as I can see, it is not only a neighbourhood of the origin in

that is mapped homeomorphically by the exponential map to a neighbourhood of the identity in

: the exponential map holds for any point (am I right? is that specific of this simple example?)

 

[fresh_42]

I thought of $SU(2)$ and $\mathfrak{su}(2)$ and took just a matrix which could be exponentiated more or less easily. Your general statement has to be false (cp. the orthogonal group), but the question remains whether it is true at least for the component of $G$ which contains the $1$ or if an entire path in $G$ translates to one in $\mathfrak{g}$ via exponentiation. I don't know the answers, to be honest.

What if we take an example which is less closely related? Say we take the local Lie group defined by $x \cdot y = \dfrac{2xy-x-y}{xy-1}$ for real numbers $x,y$ around $0$. It has $0$ as neutral element and $x^{-1}= \dfrac{x}{2x-1}$ as inverse. It is a one dimensional local Lie group. How would you exponentiate this?

I think that exponentiation works fine with some matrix groups, because they are defined by determinants and multiplications, which translates to traces and additions of the corresponding Lie algebras, and both are fine with the local exponentiation at the identity. But already the condition $\det X = 1$ needs an evaluation point $I$ to become the trace, so entire paths would probably not work. Nevertheless the convention to write flows by the exponential function makes sense for the properties of integral curves, and it makes the formulas look nicer.

 

[giulio_hep]

I thought of $SU(2)$ and $\mathfrak{su}(2)$ and took just a matrix which could be exponentiated more or less easily. Your general statement has to be false (cp. the orthogonal group)

Yes, I'm focused on matrix group and a single counter-example would suffice for me.
Yet $SO(3)$ is still equal its the exponentiated Lie Algebra $so(3)$ as shown in this youtube lesson 29 (I already mentioned another video of the serie), which in turn is based on Robert Gilmore 's "Lie Groups, Lie Algebras, and Some of Their Applications".

Actually, googling around, I've also find "Notes on Differential Geometry and Lie Groups" of Jean Gallier, where in chapter 5, it is said that

One proof of the surjectivity of the exponential, $exp: so(n, 1) → SO_0(n, 1)$, is due to Nishikawa... (1983). ... However, such a proof was also given by Marcel Riesz as early as 1957

and

What we will do to close this section is to give a relatively simple proof that the exponential map, $exp: so(1, 3) → SO_0(1, 3)$, is surjective

So, yeah, the math to fully answer my question seems rather complex and maybe not within my reach .

But, finally, in the same pdf, the chapter 5 of Gallier, it is written

As we will see in the next theorem, the map exp: so(n) → SO(n) is well-defined and surjective. The map $exp: o(n) → O(n)$ is well-defined, but it is not surjective, since there are matrices in $O(n)$ with determinant −1.

which is I guess it is the counter-example I was looking for and that you mentioned in your quote above "(cp. the orthogonal group)"

So (at least for the moment) I will be satisfied with this answer!