[Math] This example of (infinitesimal) generators of $U(2)$ and the adjoint representation.

lie-algebraslie-groupsrepresentation-theory

In trying to learn about $U(N)$ gauge Yang-Mills theory, I've found that what the adjoint representation actually is…is very a slippery concept for me. So I've tried to brute force generate an adjoint representation….here is what I have:

$U(2)$ is the set of all $2 \times 2$ unitary matrices. It is easy to see that the following matrices form a basis for the Lie Algebra of $U(2)$ (aka its infinitesimal generators):
$$
\left(\ T^{1}, \ T^{2}, \ T^{3}, \ T^{4} \ \right) = \left(\ \frac{1}{2} \sigma_{1}, \ \frac{1}{2} \sigma_{2}, \ \frac{1}{2} \sigma_{3}, \ \frac{1}{2} \mathbb{I} \ \right)
$$

…where $\sigma_{j}$ are the Pauli matrices. ${\ \bf NOTE:}$ I am using the ${\bf physicist's\ convention}$, such that $\left[ T^{a}, T^{b} \right] = i f_{abc} T^{c}$, where $f_{abc}$ are my structure constants.

I'm trying to play around with the adjoint representation ($R_{\mathrm{Adj}}$) of the above. In particular, I want to look at what the generators of the adjoint representation are (in reference to the $T^{a}$ above).

These are the matrices $T^{a}_{\mathrm{Adj}} = R_{\mathrm{Adj}} (T^{a})$. There will be $4$ of these, which are of size $4 \times 4$. It is a well-known result that these matrices have the elements determined by:
\begin{eqnarray*}
\left[ T^{a}_{\mathrm{Adj}} \right]_{bc} = – i f^{a}_{\ bc}
\end{eqnarray*}

The game I'm trying to play is to write out these matrices explicitly. Something is happening which is freaking me out.

Since $[\sigma_{j}, \sigma_{k}] = 2 i \epsilon_{jk\ell}\sigma_{\ell}$, this means that $f_{abc} = \frac{1}{2} \epsilon_{abc}$ for $(a,b,c) \in \{1, 2, 3\}^{3}$. Also, I ${\bf think}$ that any structure constants involving the index $4$ should be zero since we're just talking about the identity. By my calculations, this means that the only non-zero structure functions are:
$$
f_{123} = \tfrac{1}{2}, \
f_{132} = – \tfrac{1}{2}, \
f_{213} = – \tfrac{1}{2}, \
f_{231} = \tfrac{1}{2}, \
f_{312} = \tfrac{1}{2}, \
f_{321} = – \tfrac{1}{2}, \
$$

Which tells me that the four generators of the adjoint representation are given by:
$$
\left(\ T^{1}_{\mathrm{Adj}}, \ T^{2}_{\mathrm{Adj}}, \ T^{3}_{\mathrm{Adj}}, \ T^{4}_{\mathrm{Adj}} \ \right) = \left(\ \frac{i}{2} \left[ \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right], \ \frac{i}{2} \left[ \begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right], \ \frac{i}{2} \left[ \begin{matrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right], \ \mathbb{O} \ \right) \\
$$

I am super alarmed that the fourth generator in the adjoint representation is zero….did I do something wrong? Or is it okay that it is the zero matrix? Something is not sitting right here.

Best Answer

As @Mark van Leeuwen reassures you, for a non-simple & non-semi-simple Lie group such as $U(2)\sim SU(2)\times U(1)/Z_2$ there is no suprise there. The adjoint representation of SU(2) is real, so the U(1) rephasing is trivial and projected out.

To reassure yourself, consider that this triplet is the fundamental of O(3), so your group matrices are real orthogonal (freshman class rotations!) and considering unitary ones is overkill. If you insisted, you might spatchcock on a trivial phase to your real rotations, but it's kind of silly...

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