# Two Pet Peeves

By
**Not Even Wrong**
on April 21, 2017

I was reminded of two of my pet peeves while taking a look at the appendix A of this paper. As a public service to physicists I thought I’d go on about them here, and provide some advice to the possibly confused (and use some latex for a change).

**Don’t use the same notation for a Lie group and a Lie algebra**.

I noticed that Zee does this in is “Group Theory in a Nutshell for Physicists”, but thought it was unusual. It seems other physicists do this too (same problem with Ramond’s “Group Theory: a physicist’s survey”, the next book I checked). The argument seems to be that this won’t confuse people, but, personally, I remember being very confused about this when I first started studying the subject, in a course with Howard Georgi. Taking a look at Georgi’s book for that course (first edition) I see that what he does is basically only talk about Lie algebras. So, the fact that I was confused about Lie groups vs. Lie algebras wasn’t really his fault, since he was not talking about the groups.

The general theory of Lie groups and Lie algebras is rather complicated, but (besides the trivial cases of translation and U(1)=SO(2) groups) many physicists only need to know about two Lie groups and one Lie algebra, and to keep straight the following facts about them. The groups are

- SU(2): the group of two by two unitary matrices with determinant one. These can be written in the form

$$\begin{pmatrix}

\alpha & \beta\\

-\overline{\beta}& \overline{\alpha}

\end{pmatrix}$$

where \(\alpha\) and \(\beta\) are complex numbers satisfying \(\alpha^2+\beta^2=1\), and thus parametrizing the three-sphere: unit vectors in four real dimensional space. - SO(3): the group of three by three orthogonal matrices with determinant one. There’s no point in trying to remember some parametrization of these. Better to remember that a rotation by a counter-clockwise angle theta in the plane is given by

$$\begin{pmatrix}

\cos\theta & -\sin\theta\\

\sin\theta & \cos\theta

\end{pmatrix}$$

and then produce your rotations in three dimensions as a product of rotations about coordinate axes, which are easy to write down. For instance a rotation about the 1-axis will be given by

$$\begin{pmatrix}

1&0&0\\

0&\cos\theta & -\sin\theta\\

0&\sin\theta & \cos\theta

\end{pmatrix}$$

The relation between these two groups is subtle. Every element of SO(3) corresponds to two elements of SU(2). As a space, SO(3) is the three-sphere with opposite points identified. Given elements of SO(3), there is no continuous way to choose one of the corresponding elements of SU(2). Given an element of SU(2), there is an unenlightening impossible to remember formula for the corresponding element of SO(3) in terms of \(\alpha\) and \(\beta\), but to really understand what’s going on, you need to identify points in \(\mathbf R^3\) with traceless two by two self-adjoint matrices by for instance

$$(x_1,x_2,x_3)\leftrightarrow x_1\sigma_1 +x_2\sigma_2+x_3\sigma_3=\begin{pmatrix} x_3&x_1-ix_2\\x_1+ix_2&-x_3\end{pmatrix}$$

Then the SO(3) rotation corresponding to an element of SU(2) is given by

$$\begin{pmatrix} x_3&x_1-ix_2\\x_1+ix_2&-x_3\end{pmatrix}\rightarrow \begin{pmatrix}

\alpha & \beta\\

-\overline{\beta}& \overline{\alpha}

\end{pmatrix}\begin{pmatrix} x_3&x_1-ix_2\\x_1+ix_2&-x_3\end{pmatrix} \begin{pmatrix}

\alpha & \beta\\

-\overline{\beta}& \overline{\alpha}

\end{pmatrix}^{-1}$$

Since most of the time you only care about two Lie groups, you mostly only need to think about two possible Lie algebras, and luckily they are actually the same, both isomorphic to something you know well: \(\mathbf R^3\) with the cross product. In more detail you have

- su(2) or \(\mathfrak{su}(2)\): Please don’t use the same notation as for the Lie group SU(2). These are traceless self-adjoint two by two complex matrices, identified with \(\mathbf R^3\) as above except for a factor of two.

$$(x_1,x_2,x_3)\leftrightarrow \frac{1}{2}\begin{pmatrix} x_3&x_1-ix_2\\x_1+ix_2&-x_3\end{pmatrix}$$

Under this identification, the cross-product corresponds to the commutator of matrices.You get elements of the group SU(2) by exponentiating elements of its Lie algebra.

- so(3) or \(\mathfrak{so}(3)\): Please don’t use the same notation as for the Lie group SO(3). These are antisymmetric three by three real matrices, identified with \(\mathbf R^3\) by
$$(x_1,x_2,x_3)\leftrightarrow \begin{pmatrix}

0&-x_3&x_2\\

x_3&0 & -x_1\\

-x_2&x_1&0

\end{pmatrix}$$

Under this identification, the cross-product corresponds to the commutator of matrices.You get elements of the group SO(3) by exponentiating elements of its Lie algebra.

If you stick to non-relativistic velocities in your physics, this is all you’ll need most of the time. If you work with relativistic velocities, you’ll need two more groups (either of which you can call the Lorentz group) and one more Lie algebra, these are

- SL(2,C): This is the group of complex two by two matrices with determinant one, i.e. complex matrices

$$\begin{pmatrix}

\alpha & \beta\\

\gamma& \delta

\end{pmatrix}$$

satisfying \(\alpha\delta-\beta\gamma=1\). That’s one complex condition on four complex numbers, so this is a space of 6 real dimensions. Best to not try and visualize this; besides being six-dimensional, unlike SU(2) it goes off to infinity in many directions. - SO(3,1): This is the group of real four by four matrices M of determinant one such that

$$M^T\begin{pmatrix}-1&0&0&0\\

0&1&0&0\\

0&0&1&0\\

0&0&0&1\end{pmatrix}M=\begin{pmatrix}-1&0&0&0\\

0&1&0&0\\

0&0&1&0\\

0&0&0&1\end{pmatrix}$$

This just means they are linear transformations of \(\mathbf R^4\) preserving the Lorentz inner product.

The relation between SO(3,1) and SL(2,C) is much the same as the relation between SO(3) and SU(2). Each element of SO(3,1) corresponds to two elements of SL(2,C). To find the SO(3,1) group element corresponding to an SL(2,C) group element, proceed as above, removing the “traceless” condition, so identifying \(\mathbf R^4\) with self-adjoint two by two matrices as follows

$$(x_0,x_1,x_2,x_3)\leftrightarrow\begin{pmatrix} x_0+x_3&x_1+ix_2\\x_1-ix_2&x_0-x_3\end{pmatrix}$$

The SO(3,1) action on \(\mathbf R^4\) corresponding to an element of SL(2,C) is given by

$$\begin{pmatrix} x_0+x_3&x_1+ix_2\\x_1-ix_2&x_0-x_3\end{pmatrix}\rightarrow \begin{pmatrix}

\alpha & \beta\\

\gamma & \delta

\end{pmatrix}\begin{pmatrix} x_0+x_3&x_1+ix_2\\x_1-ix_2&x_0-x_3\end{pmatrix} \begin{pmatrix}

\alpha & \beta\\

\gamma& \delta

\end{pmatrix}^{-1}$$

As in the three-dimensional case, the Lie algebras of these two Lie groups are isomorphic. The Lie algebra of SL(2,C) is easiest to understand (please don’t use the same notation as for the Lie group, instead consider sl(2,C) or \(\mathfrak{sl}(2,C)\)), it is all complex traceless two by two matrices, i.e. matrices of the form

$$\begin{pmatrix}a&b\\

c&-a\end{pmatrix}$$

For the isomorphism with the Lie algebra of SO(3,1), go on to pet peeve number two and then consult a relativistic QFT book to find some form of the details.

**Keep track of the difference between a Lie algebra and its complexification**

This is a much subtler pet peeve than pet peeve number one. It really only comes up in one place, when physicists discuss the Lie algebra of the Lorentz group. They typically put basis elements \(J_j\) (infinitesimal rotations) and \(K_j\) (infinitesimal boosts) together by taking complex linear combinations

$$A_j=J_j+iK_j,\ \ B_j=J_j-iK_j$$

and then note that the commutation relations of the Lie algebra simplify into commutation relations for the \(A_j\) that look like the \(\mathfrak{su}(2)\) commutation relations and the same ones for the \(B_j\). They then announce that

$$SO(3,1)=SU(2) \times SU(2)$$

Besides my pet peeve number one, even if you interpret this as a statement about Lie algebras, it’s not true at all. The problem is that the Lie algebras under discussion are real Lie algebras, you’re just supposed to be taking real linear combinations of their elements. When you wrote down the equations for \(A_j\) and \(B_j\), you “complexified”, getting elements not of \(\mathfrak{so}(3,1)\), but what a mathematician would call the complexification \(\mathfrak{so}(3,1)\otimes C\). Really what has been shown is that

$$ \mathfrak{so}(3,1)\otimes C = \mathfrak{sl}(2,C) + \mathfrak{sl}(2,C)$$

It turns out that when you complexify the Lie algebra of an orthogonal group, you get the same thing no matter what signature you start with, i.e.

$$ \mathfrak{so}(3,1)\otimes C =\mathfrak{so}(4)\otimes C =\mathfrak{so}(2,2)\otimes C$$

all of which are two copies of \(\mathfrak{sl}(2,C)\). The Lie algebras you care about are what mathematicians call different “real forms” of this and they are different for different signature. What is really true is

$$\mathfrak{so}(3,1)=\mathfrak{sl}(2,C)$$

$$\mathfrak{so}(4)=\mathfrak {su}(2) + \mathfrak {su}(2)$$

$$\mathfrak{so}(2,2)=\mathfrak{sl}(2,R) +\mathfrak{sl}(2,R)$$

For details of all this, see my book.

Note: Posting this and heading home for the evening, haven’t checked some signs, and tomorrow morning will likely make some typographical improvements. If you want to check the signs, please do….