# A Matrix Lie groups and its Lie Algebra

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1. Dec 22, 2016

### Xico Sim

Hi!

I'm studying Lie Algebras and Lie Groups. I'm using Brian Hall's book, which focuses on matrix lie groups for a start, and I'm loving it. However, I'm really having a hard time connecting what he does with what physicists do (which I never really understood)... Here goes one of my questions - very basic, but very important. It is about how one finds a Lie Algebra given its Lie Group. I believe I understood the method exposed in the book, but I am unable to relate it precisely to what I see done in physics textbooks.

Thanks!

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2. Dec 22, 2016

### Staff: Mentor

They simply cut the Taylor expansion of $\exp$ after the linear term and write the error as $O(\varepsilon^2)=O(t^2)$ for small $t=\varepsilon$ around $t=0$. In the case of $SL(2)$ this is especially convenient, for
$$\exp{\left(t\cdot \begin{bmatrix}0&0\\1&0\end{bmatrix}\right)} = \begin{bmatrix}1&0\\t&1\end{bmatrix}=1+t\cdot X$$
and similar for the other two basis vectors $\begin{bmatrix}0&1\\0&0\end{bmatrix}\, , \,\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ of $\mathfrak{sl}(2)$.

In general one has to consider paths within the group through the identity element and calculate the tangent vectors of them in $t=0$.

3. Dec 22, 2016

### Xico Sim

I understand that. What I want to know is why they do it. Also, why the characteristics one finds for $X$ in such an approximated formula for $U$ (e.g. one finds that $X$ is hermitian in the su(2) case) transfers to all elements of g. Finally, I think I've seen people refer to such an $X$ (an $X$ such that $U \approx I+i\epsilon X$) as a generator (i.e. a basis element) of g. Why is that true?

4. Dec 22, 2016

### Staff: Mentor

I'm not sure, whether I have the ultimate answer to this, or whether there is a specific one at all.
The easy answer would be, because of sloppiness and a linear approximation is good enough for calculations. One could also say, that locally there is no difference and differentials (or tangent vectors) are simply a linear approximation, so let's write them linearly. It also makes sense formally, as $U^{-1}$ becomes $-X$ and conjugation as well as transposition are linear and can be pulled into the sum.
I don't think, that this approximation is used to translate $\text{ unitary }$ to $\text{ skew-hermitian }$. If at all, then it is
$$-X=-\exp(U)=\exp(U^{-1})=\exp(\overline{U}^\tau)=\overline{\exp(U)}^\tau=\overline{X}^\tau$$
The usage of the word "generator" by physicists has to be taken with a big amount of generosity. Mathematically the word generator in the context of vector spaces makes no sense. It's reserved for group elements in a presentation of the group. It's highly confusing and by no means automatically the basis vectors. Locally they span the tangent space $\mathfrak{g}$ and can thus be called infinitesimal generators. But sometimes something different is meant: the basis transformations of a representation of $\mathfrak{g}$ or even of $G$, which isn't much better either. The representations are usually the goal and why tangent spaces are considered here, e.g. for the ladder operator. (O.k. it's only half the truth, but I think the linear approximation for calculations is the other half.) Latest at BCH even physicist take more care.

5. Dec 22, 2016

### Xico Sim

I don't see how... maybe I'm missing something trivial?: Suppose $U^{-1} = X$. Then,
$$U^{-1}U \approx -X(I + i \epsilon X) = -X - i \epsilon X^2 \ne I$$
Which is absurd.

1. You wrote $X=exp(U)$. But $X \in \mathbf{g}$ and $U \in \mathbf{G}$, so I would expect it to be $U=exp(X)$, and even this is questionable, since in general we only know that $U=e^{X_{1}}...e^{X_{m}}$ for some ${{X_i}}$ (and also assuming G connected).
2. I would say that in fact $X=X^{\dagger}$ because of the following: Let $X$ be an element of su(2). We have $X\in\mathfrak{su}(2)\ \iff\forall t\in\mathbb{R},\,e^{itX}\in\text{SU(2) }$. In particular, $(e^{itX})^{\dagger}e^{itX}=\mathbb{I}_{2}\,\Rightarrow e^{-itX^{\dagger}}e^{itX}=\mathbb{I}_{2}\,\Rightarrow X^{\dagger}=X$.

6. Dec 22, 2016

### Staff: Mentor

Right. I assume physicists think a lot more in terms of infinitesimals which they identify with tangents or better with small changes in the direction of the tangent. And the power $\{-1\}$ in a function is a minus sign in the differential, a infinitesimal generator of change. In the same sense the determinant ($\cdot$) becomes the trace ($+$) and $1$ becomes $0$. Analytically there is still the error margin $O(t^2)$. This only vanishes in special cases like the ones in my first post where $X^2=0$ or everything is on the diagonal where we get the ordinary exponential function. And although locally, i.e. in a small neighborhood of the identity matrix, unitary matrices behave like their tangent matrices, one cannot identify them - only in approximation which is often sufficient for it shows the main rate of change. You could as well ask, why we would bother Taylor series at all, if we simply could take $1+f\,'(x)$ as linear approximation. Sometimes it is good enough, but you won't try to prove $f(x)=1+f\,'(x)$
Yes, that was a mistake. It had to be $U^{-1}=(\exp(X))^{-1}=\exp(-X)=\overline{\exp(X)}^\tau=\overline{U}^\tau=U^\dagger$
Nope. $\mathfrak{su}(n)$ are skew-Hermitian matrices (see above). I don't see your $U:=e^{itX}$ in $SU(n)$ because $U^{-1}=e^{-itX}=\overline{e^{itX}}=\overline{U} \neq \overline{U}^\tau=U^\dagger$.

The Lie algebra of SU(n), denoted by su(n), can be identified with the set of traceless antihermitian n×n complex matrices, with the regular commutator as Lie bracket. Particle physicists often use a different, equivalent representation: the set of traceless hermitian n×n complex matrices with Lie bracket given by −i times the commutator.

7. Dec 23, 2016

### Xico Sim

Sorry, but I still have the same problem I had before:

Hm I think your second equal sign is unjustified. Let me try and compute that:
$U^{-1}=e^{-itX}=e^{-itX^{\dagger}}=e^{(itX)^{\dagger}}=(e^{itX})^{\dagger}=U^\dagger$
As expected. (I used the assumption that $X$ is hermitian on my second equal sign).

Exactly: this is in accordance with what I'm saying: "...the set of traceless hermitian n×n complex matrices..." (using the physicists convention).

8. Dec 23, 2016

### Staff: Mentor

Why? You don't need this representation as a product. Given any skew-Hermitian matrix, send it via the diffeomorphism $\exp$ into its group and get a unitary matrix as result. The other way around take a path through $\mathbb{1} \in G$ and differentiate. We may talk about the transportation of the structure (algebra $\rightarrow$ group), but then we have to talk about BCH.
Only if $X^\dagger =X.$
But in this case it should be stated, as it is mathematically completely unusual. In mathematics, $[A,B]=AB-BA$ in the case $A,B$ are linear transformations. This would result in $[A,B]^\dagger = (AB-BA)^\dagger = (AB)^\dagger- (BA)^\dagger=B^\dagger A^\dagger - A^\dagger B^\dagger = BA-AB = -[A,B]$ and an Abelian Lie algebra, which is (in your words) absurd unless $char\, \mathbb{F} =2$.
So if you change this convention, someone should tell.