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In the framework of Quantum Physics, I have to explain to some of my colleagues what is a Lie group, a Lie algebra and their connections with the exponential map. This is mainly to make them understand the mathematical background behind Pauli matrices and the fact that they generates $SU(2)$.

I googled "Lie group" with "generators", along with "integral curves" and "tangent space" to understand how to present in a general form the link between Lie groups and algebras. But the mathematical formulation is mostly too far from my background and that of my colleagues.

So, I would like to start from a matrix Lie group $G$ parameterized by $\vec{a}=(a_1,\dots,a_p)^t\in\mathbb{R}^p$ such that: \begin{equation} M(\vec{a}+\vec{b})=M(\vec{a})M(\vec{b})\;,\quad M(\vec{0})=I\;, \quad M(-\vec{a}) = [M(\vec{a})]^{-1} \end{equation}

with $M(\vec{a})$ a $n\times n$ matrix belonging to $G$.

As far as I understood the many online available tutorials on Lie groups, one way to develop the link between Lie groups and algebras is to exhibit the tangent space at the Lie group neutral element ($I$) through integral curves.

So, first I define a one parameter curve within the Lie group: \begin{equation} \begin{split} \mathbb{R}&\mapsto G\\ t &\mapsto M(t)=M(\vec{a}(t)) \end{split} \end{equation} such that $M(0)=I$. Then the key point is that the Lie group parameters $\vec{a}$ are function of $t$.

This curve is an integral curve means it complies to: \begin{equation}\label{eq:0} M(t+h)=M(t)M(h) \end{equation} Then, by deriving this equation, one gets : \begin{equation} \frac{\,d M}{\,dt}=\frac{\,d M(t)}{\,dt}\biggl|_0 M(t) \end{equation} then: \begin{equation} M(t) = e^{m t} \end{equation} with \begin{equation} m=\frac{\,d M(t)}{\,dt}\bigg|_0 = \sum_{i=1}^p \frac{\partial M}{\partial a_i}\bigg|_0\frac{\,d a_i}{\,dt}\bigg|_0 = \vec{\sigma}\cdot\vec{n} \end{equation} where \begin{equation} \vec{\sigma}=\biggl(\frac{\partial M}{\partial a_1}\bigg|_0, \dots, \frac{\partial M}{\partial a_p}\bigg|_0 \biggr)^t \end{equation} is the a vector of matrices called the infinitesimal generators of $G$, through the exponential map. They form a basis for the tangent space of $G$ at $I$, which happens to be also a Lie algebra.

This is what I am able to understand, but there are several obscure points for me:

  • what is the meaning of the derivatives $\frac{\,d a_i}{\,dt}$?

  • integral curve: the equality $M(t+h)=M(t)M(h)$ implies that $\vec{a}(t+h)=\vec{a}(t)+\vec{a}(h)$, which has no reason to occur unless there are some specific assumptions. As far as I understand this assumption it that the curve be integral. Could you help me to demonstrate clearly how this holds?

  • in tutorials, this demonstration makes use of the "left invariant" of $G$, I am unable to give this notion its right place here.

  • is my demonstration correct ? I know that the Lie algebra only generates the connected to unity ($I$) part of $G$. But here, this seems apply to any element of $G$? or maybe the integral curve notion guarantees that $M(t)$ is connected to $I$?

  • also, I read several times that one-parameter subgroups of $G$ are always of this kind, what does it mean exactly?

I would be grateful to you if you could give me some more insights about this, so that I can make this clear to my colleagues too.

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  • $\begingroup$ WP. $\endgroup$ Mar 18, 2023 at 16:48
  • $\begingroup$ related/possible duplicate: physics.stackexchange.com/q/133758/50583 $\endgroup$
    – ACuriousMind
    Mar 18, 2023 at 16:51
  • $\begingroup$ @CosmasZachos I am not sure the wikipedia page on Pauli matrices really answers my questions, which are more general than Pauli matrices and SU(2) $\endgroup$
    – deb2014
    Mar 18, 2023 at 20:23
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    $\begingroup$ @CosmasZachos : your latest post is what I would like to better understand, could you develop ? $\endgroup$
    – deb2014
    Mar 18, 2023 at 21:44
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    $\begingroup$ Even writing $M(\vec a+\vec b)$ is problematic. Better write $M(\vec a\circ\vec b)$ since composition of elements is usually nowhere near a sum. $\endgroup$ Mar 19, 2023 at 19:18

2 Answers 2

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I don't know what you intend to study next with your colleagues, but you can avoid all this mathematico-mathematical considerations with tangent spaces and stuff, especially for an introduction to physicists.

If I were you, I would go roughly that way :

1. A Lie group $G$ is a (multiplicative) topological group. Its topology permits to define a notion of continuity (hence limits, derivatives, etc.). The group operation of $G$ is continuous and differentiable (as well as the inverse map $g \mapsto g^{-1}$).

2. It can be shown that every Lie group $G$ is associated to a Lie algebra $\mathfrak{g}$; it is to be recalled that an algebra is basically a vector space equipped with a bilinear product $-$ in other words, a vectorial multiplication $\mathfrak{g} \times \mathfrak{g} \rightarrow \mathfrak{g}$.

At this step, you can mention that $\mathfrak{g}$ is made of the left-invariant vector fields of the tangent space of the identity element of $G$, etc., and invite your colleagues to read the associated litterature if they are curious about it.

3. $G$ and $\mathfrak{g}$ are linked by the exponential map $-$ which is surjective, it is not an isomorphism $-$ and differentiation, i.e. $\exp : \mathfrak{g} \rightarrow G$ and $\frac{\mathrm{d}}{\mathrm{d}t} : G \rightarrow \mathfrak{g}$, and more precisely $A=e^{tX}\in G$ and $X = \left.\frac{\mathrm{d}A}{\mathrm{d}t}\right|_{t=0} \in \mathfrak{g}$.

4. Some manipulations are easier within the algebra, while others are simpler to deal with inside the group, that is why we are always going back and forth between the Lie group and its algebra with the help of the exponential map and differentiation. In consequence, every map $\phi : G \rightarrow G$ on the group can also be associated to a map $\Phi : \mathfrak{g} \rightarrow \mathfrak{g}$ on the algebra, and vice versa $-$ more precisely : $\Phi = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} \circ \phi \circ \exp$.

5. The basis vectors of the Lie algebra are called the generators and they are described by the commutation relations $[X_i,X_j] = f_{ij}{}^kX_k$, where $f_{ij}{}^k$ are the so-called "structure constants". You can also mention that the commutator over $\mathfrak{g}$ is associated to conjugation in $G$ (the two adjoint maps).

6. After parametrizing the elements of $SU(2)$, you can apply the differentiation rule to find the elements of $\mathfrak{su}(2)$ and deduce that the Pauli matrices form its basis.

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  • $\begingroup$ Thank you very much ! It is true my colleagues may not need this depth of understanding but, you know, any time you want to explain something to someone, it is better to have a deeper level of understanding than the one you aim to teach. $\endgroup$
    – deb2014
    Mar 20, 2023 at 14:30
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I am fleshing out my remark, as invited, possibly clarifying your talk. My gut says you could take n=2, p=3, for concreteness, sticking to SU(2), and leave generalizations to the rare genius most physicists have for "monkey-see, monkey-do".

So, parameterize the matrix Lie group by $\vec{a}=(a_1, a_2, a_3)^T=t\hat n$ near the identity where $\hat n$ is a unit 3-vector, and t=0 describes all curves going through the identity. So $\frac{d \vec a}{dt}=\hat n$ near the identity: tangent space coordinates. This is the only way to justify the wild $\vec a + \vec b$ simplification you started with.

I must now rectify your upended expressions, $$ M(t+h)=M(t)M(h) ~~~\leadsto \\ \frac{ d M(t+h)}{ dh}\biggl|_{h=0}=M(t) \frac{ d M(h)}{\,dh}\biggl|_{h=0} \leadsto \\ \frac{ d M(t )}{ d t} =M(t) \frac{ d M(h)}{\,dh}\biggl|_{h=0} \equiv M(t) ~m \implies \\ M(t) = e^{m t}, $$ where, in the last step, the matrix ODE is solved given the boundary conditions at the origin.

Note that multiplying M by a rigid matrix S on the left does not affect the Lie algebra element $$ m= M^{-1} \frac{ d M(t )}{ d t} . $$ It is said to be left-invariant. Moreover, the elaborate ("current") expression on the right is in the Lie algebra! Your matrix group is now represented in tangent space coordinates, composing linearly for a fixed $m=\hat n \cdot \vec \sigma$. Each $m=\hat n \cdot \vec \sigma$ defines an abelian subalgebra, and hence subgroup whose trivial composition law puzzled you. (But different ms compose in the nonabelian Gibbs formula I sent you to in WP in my original comment. That is what I would emphasize to an audience, not the turgid trivialities!)

Subsequently as you noted, \begin{equation} m=\frac{\,d M(t)}{\,dt}\bigg|_0 = \sum_{i=1}^3 \frac{\partial M}{\partial a_i}\bigg|_0 n_i = \vec{\sigma}\cdot \hat {n} \end{equation} where \begin{equation} \vec{\sigma}={\Large \nabla}_{\!\!\vec a} ~ M \bigg|_{\vec a =0}\equiv \biggl(\frac{\partial M}{\partial a_1}\bigg|_0, \frac{\partial M}{\partial a_2}\bigg|_0, \frac{\partial M}{\partial a_1}\bigg|_0\biggr )^T. \end{equation}

You may now go to the WP article, consider (physicists stick an i in front of the generators) $$ e^{i\vec \sigma \cdot \vec a }=e^{i\vec \sigma \cdot \hat n ~t}= M(t)=1\!\! 1 \cos t+ i\hat n \cdot \vec\sigma \sin t , $$ and illustrate the above statements and relations, especially that the above left-invariant "current" expression in the algebra is actually independent of t, as constructed!

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    $\begingroup$ Thank you very much ! Indeed, $\vec{a}=t\vec{n}$ with $\vec{n}$ fixed (an integral curve ?) explains for me why $\vec{a}(t+h) = \vec{a}(t) + \vec{a}(h)$. $\endgroup$
    – deb2014
    Mar 20, 2023 at 14:32
  • $\begingroup$ If you accept the answer, you may click on the checkmark or upvote. $\endgroup$ Mar 21, 2023 at 19:22

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