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Can I get a spoiler on where this definition of the elements of a Lie group is headed?

In this lecture Alex Flournoy does a great job at introducing the elements of a Lie group. However, there is ultimately this unmotivated definition:

A general element of a Lie group can be written as

$$A = \exp \left ( g_A V^A \right)$$

where $g_A$ generates the transformation, and $V^A$ parameterizes it. So in the case of rotations in the Euclidean space, $SO(3)$ if $A$ is an $n \times n$ matrix, so is $g_A,$ while $V^A$ is an $n$ vector of parameters (angles of rotation in each plane: $xy,$ $xz$ and $yz$).

He later on explains that $g_A$ is better understood as a vector of matrices $g_A=\begin{bmatrix}R_{xy} & R_{yz} & R_{xz} \end{bmatrix},$ and the $V^A$ a vector $V^A=\begin{bmatrix}\phi&\theta&\alpha\end{bmatrix}^\top$ so that for the second entry, the Taylor series would be

$$\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\theta & - \sin \theta \\0 &\sin \theta & \cos\theta \end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1-\frac{\theta^2}2+\cdots & - \theta+\frac{\theta^3}{3!}+\cdots \\0 & \theta-\frac{\theta^3}{3!}+\cdots & 1-\frac{\theta^2}2+\cdots \end{bmatrix}$$

Is the inclusion of the exponential in the definition motivated to make the group operation (matrix multiplication) simply the addition of the exponents, as well as to show its algebraic nature by applying a Taylor expansion to the power series $\exp (x)=\displaystyle \sum_{k=0}^\infty \frac{x^k}{k!},$ with the ultimate intention of making it consistent with the definition:

A Lie group is the connected portion of a continuous group with analytic group composition function.

? Or is there a deeper reason perhaps connected to the charts of the group viewed as a manifold?

Antoni Parellada
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  • Not sure I fully understand your proposed answer to your question, but there are many "matrix Lie groups", subgroups of $M_{n\times n}$, in which the matrices in the tangent space (generator matrices) can be literally exponentiated to obtain an actual group element. Are you asking about something past this? – Keshav Oct 19 '20 at 03:22
  • @Keshav I am starting to try, and understand a bit the idea behind Lie groups using online courses, and there are approaches in which a more abstract approach with a continuous group, endowed with its operation, and a manifold atlas to Euclidean space seems to be all that's required, whereas others start off immediately with the exponential above. Why? – Antoni Parellada Oct 19 '20 at 03:30
  • That seems like a choice of teaching philosophy to me. Knowing about Lie groups means (eventually) knowing not just about the manifold + group structures, but also everything that follows from it, including the exponential map, Lie algebras, etc. For a physics course like the one you linked to, I imagine they are more interested in the very compute-able relationship between the group and the algebra via the exp map, so maybe it's natural to "start" there. – Keshav Oct 19 '20 at 03:50
  • @Keshav Both series are Physics, but yes, it makes sense that the exponential function ties directly with parallel transport along geodesics. – Antoni Parellada Oct 19 '20 at 04:46
  • (Note that geodesics relate to the Riem geo exp map, rather than the Lie group exp map.) – Keshav Oct 19 '20 at 04:50

1 Answers1

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Tentative answer based on the example on this presentation by XylyXylyX on $SL(2,\mathbb R)$ parameterized as

$$\begin{bmatrix} 1+a & b \\ c & \frac{1+bc}{1+a}\end{bmatrix}$$

The generators are found through the general formula

$$X_\mu(r\times r)=\lim_{\alpha \to 0}\frac{M(0,0,\dots,\alpha^\mu,0,0)-M(0,0,\dots,0)}{\alpha^\mu}$$

which applied to the example amounts to performing the partial derivative for each parameter for the total of $3$ parameters, which is the dimension of the manifold, and evaluating the result at $(0,0,0),$ corresponding to the first term of the Taylor series (linearization):

$$\begin{align} X_a=\left. \frac{\partial}{\partial a}M(a,0,0) \right|_{(0,0,0)}= \left. \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{(1+a)^2}\end{bmatrix}\right|_{(0,0,0)} &=\begin{bmatrix} 1 & 0\\ 0 & -1\end{bmatrix}\\[2ex] X_b=\frac{\partial}{\partial b}M(0,b,0) &= \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}\\[2ex] X_c=\frac{\partial}{\partial b}M(0,0,c) &= \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix} \end{align}$$

These three generators will form a basis for a vector space, and every linear combination $\varepsilon^\alpha X_\alpha$ will be an infinitesimal generator of group elements in a given direction away from the origin in the Euclidean parametric space, made possible by the manifold structures of the group $SL(2,\mathbb R)$.

Moving away from the origin can be viewed like a process of continuous compounding these displacements in any given direction. Given that the origin in the charted Euclidean space corresponds to the identity matrix, a point in the manifold (group) along a given direction (set by the linear combination of the basis vectors derived above) will be given by

$$\begin{align} \lim_{N\to \infty} \left (\mathbf I + \frac{\varepsilon^\mu X_\mu}{N} \right )^N &= \left( \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} + \frac 1 N \left( a \begin{bmatrix} 0 & 0 \\ 0 & -1\end{bmatrix} + b \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix} \right)\right)^N\\[2ex] &= \left( \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} + \frac 1 N \begin{bmatrix} a & b \\ c & -a\end{bmatrix} \right)^N\\[2ex] &= \exp\left(\begin{bmatrix} a & b \\ c & -a\end{bmatrix}\right)\\[2ex] &=\sum_{N=0}^\infty \frac 1 {N!}\begin{bmatrix} a & b \\ c & -a\end{bmatrix}^N \end{align}$$

Antoni Parellada
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  • Could you give me an $X\in sl(2,\mathbb{R})$ such that $e^{X} = \begin{bmatrix} -1 & 1 \ 0 & -1\end{bmatrix}$ (note that this matrix is clearly in $SL(2;\mathbb{R})$ and in the given parameteization we have $(a,b,c) = (-2, 1,0)$) – RyanK Oct 21 '20 at 18:12
  • @RyanK I am learning this topic on my own. At this point I am happy to have a vague idea of where the exponential map may possibly arise. I understand that the matrix as is has determinant $1.$ I am not sure how to handle the fact that it is the exponent of presumably an element of the set $X.$ I will keep in mind your question, as I make progress in understanding Lie groups, but please feel free to either edit my very tentative answer, or letting me know if it is wrong and why. If you decide to answer separately, and I understand the answer I will be happy to upvote and accept. – Antoni Parellada Oct 21 '20 at 23:15
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    I just wanted to point out that not every element of $SL(2;\mathbb{R})$ can be written as the exponential of a matrix in the associated Lie algebra, and the example matrix I gave cannot. Check out this Math Stack Link which addresses this problem. – RyanK Oct 22 '20 at 00:34