Two Pet Peeves

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 his “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 \(\beta\). To really understand what’s going on, you need to do something like the following: identify points in \(\mathbf R^3\) with traceless two by two self-adjoint matrices by
$$(x_1,x_2,x_3)\leftrightarrow x_1\sigma_1 +x_2\sigma_2+x_3\sigma_3=\begin{pmatrix} x_3&x_1-ix_2\\x_1+ix_2&-x_3\end{pmatrix}$$
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:

  • su(2) or \(\mathfrak{su}(2)\): Please don’t use the same notation as for the Lie group SU(2). These are traceless skew-adjoint (\(M=-M^\dagger\)) two by two complex matrices, identified with \(\mathbf R^3\) as above except for a factor of \(-\frac{i}{2}\).
    $$(x_1,x_2,x_3)\leftrightarrow -\frac{i}{2}\begin{pmatrix} x_3&x_1-ix_2\\x_1+ix_2&-x_3\end{pmatrix}$$
    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:

  • \(SL(2,\mathbf C)\): 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 \(SL(2,\mathbf C)\) is much the same as the relation between SO(3) and SU(2). Each element of SO(3,1) corresponds to two elements of \(SL(2,\mathbf C)\). To find the SO(3,1) group element corresponding to an \(SL(2,\mathbf C)\) 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 \(SL(2,\mathbf C)\) 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 \(SL(2,\mathbf C)\) is easiest to understand (please don’t use the same notation as for the Lie group, instead consider \(sl(2,\mathbf C\)) or \(\mathfrak{sl}(2,\mathbf C)\)), 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 \(\mathfrak{so}(3,1)\otimes \mathbf C\). Really what has been shown is that
$$ \mathfrak{so}(3,1)\otimes \mathbf C = \mathfrak{sl}(2,\mathbf C) + \mathfrak{sl}(2,\mathbf C)$$

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.
$$ \mathfrak{so}(3,1)\otimes \mathbf C =\mathfrak{so}(4)\otimes \mathbf C =\mathfrak{so}(2,2)\otimes \mathbf C$$
all of which are two copies of \(\mathfrak{sl}(2,\mathbf C)\). 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
$$\mathfrak{so}(3,1)=\mathfrak{sl}(2,\mathbf C)$$
$$\mathfrak{so}(4)=\mathfrak {su}(2) + \mathfrak {su}(2)$$
$$\mathfrak{so}(2,2)=\mathfrak{sl}(2,R) +\mathfrak{sl}(2,R)$$

For details of all this, see my book.

This entry was posted in Uncategorized. Bookmark the permalink.

13 Responses to Two Pet Peeves

  1. Maxis says:

    Is it just me, or LaTeX isn’t rendering? You forgot to include the word “latex” after the dollar sign.

  2. --- says:

    Re the signs and typographic errors: in the definition of SU(2) you wat $\alpha\bar{\alpha} + \beta\bar{\beta} = 1$; so that the determinant is 1. Then you also actually get the 3-sphere.

  3. Peter Woit says:

    —,
    Thanks, fixed.

    Maxis,
    I’m doing this using MathJax, works for me. If people are having trouble with this let me know.

  4. Timothy P Keller says:

    I am trying to write up some notes for a course on Lie theory and applications I would like to teach when I retire ( but probably never will ). Your notes are some of my best references.
    It’s great that you posted this comment, and it’s a wonder it’s really necessary. Getting the notation right using Latex is so easy these days ….

  5. vmarko says:

    Peter,

    LaTeX isn’t rendering for me either.

    Note that MathJax generally does render in my browser (if I visit my own website, or arXiv, or various other places), and everything works fine — except for your website, which just displays LaTeX source instead.

    HTH,
    Marko

  6. Jeff M says:

    Peter

    First, the LaTeX is displaying fine for me, Safari 10.1. Second, physicists really can’t use different notation for the group and the algebra? In math, anything in Fraktur is the algebra. I assume this is still true, it certainly was when I took Lie Algebras in grad school in ’87.

  7. Fred P says:

    My broswers (Firefox, Edge, and Chrome) does not render the Latex because it views the Latex rendering as insecure. On Chrome, I could easily turn this off.

    Looking at your page source, I see at least part of the problem:

    http://www.math.columbia.edu/department/mathjax/MathJax.js?config=TeX-AMS-MML_HTMLorMML&ver=4.7.3

    is not https, which is how I’m seeing most of this page. Note that

    https://www.math.columbia.edu/department/mathjax/MathJax.js?config=TeX-AMS-MML_HTMLorMML&ver=4.7.3

    seems like it works, but my https everywhere extension does not automatically translate that link to https – which is why the rendering wasn’t working on my browsers.

  8. Peter Woit says:

    Fred P,
    I see the potential problem, changed the link to https.

    All, please let me know of any continuing problems with the Latex rendering.

  9. vmarko says:

    Yes, that fixed it, now it renders correctly for me too!

    🙂
    Marko

  10. vmarko says:

    Ok, now that I see the equations rendered correctly, shouldn’t the sums be “circled”, \(\oplus\), instead of an ordinary plus, \(+\), when writing sums of algebras at the end?

    I don’t want to sound like a nitpick, but an algebra is also a vector space, and its addition of vectors (usually denoted with an ordinary plus sign) should be distinguished from the notion of the direct sum of two algebras. Especially if the two algebras are actually two copies of the same algebra.

    HTH, 🙂
    Marko

  11. Peter Woit says:

    Marko,
    You’re probably right, but remember, here I’m trying to convince physicists that the distinctions mathematicians make are important ones….

  12. Narad says:

    Getting the notation right using Latex is so easy these days ….

    Aside from remembering to type {-}\sin instead of -\sin when it’s standing alone, among other things. 😉

  13. Al says:

    Peter- Maybe it’s ok to mention the March For Science tomorrow (Saturday):
    it’s supposed to be nonpartisan and a celebration of science…
    of course barely in the background are worries about climate change, budget cuts,
    the destruction of the EPA, NIH…
    Here’s TYT:
    https://www.youtube.com/watch?v=pCGFqbRC6do
    There are supposed to be marches in 500 cities worldwide.
    Information here:
    https://www.marchforscience.com/

Leave a Reply

Informed comments relevant to the posting are very welcome and strongly encouraged. Comments that just add noise and/or hostility are not. Off-topic comments better be interesting... In addition, remember that this is not a general physics discussion board, or a place for people to promote their favorite ideas about fundamental physics. Your email address will not be published. Required fields are marked *