su(N)

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Completeness

Suppose we write the inner product as

\left\langle t^a | t^b \right\rangle \equiv \operatorname{tr}\left(\frac{t^{\dagger a}}{\sqrt{\lambda}} \frac{t^b}{\sqrt{\lambda}}\right) = \delta^{ab}\,.

Viewed as N^2\,-dimensional vectors in complex Euclidean space, this is just the Euclidean norm. We therefore have a subspace spanned by n\, orthonormal vectors. If these were a complete set, then we would have the completeness relation

\left| t^a\right\rangle \left\langle t^a \right| = \mathbf{1}_{N^2\times N^2}\,,

(summation implied). However, since the generators are traceless, they are all orthogonal to \left|1\right\rangle\,, the N^2\,-dimensional vector corresponding to \frac{1}{\sqrt{N}}\delta_{rs}\,. For \mathfrak{su}(N)\,, n = N^2 - 1\,, so we simply project out the subspace \left|1\right\rangle\,:

\left| t^a\right\rangle \left\langle t^a \right| = \mathbf{1}_{N^2\times N^2} - \left| 1\right\rangle \left\langle 1\right|\,.

In matrix form, this becomes (keeping in mind that the generators are also hermitian)

\frac{1}{\lambda} t^a_{pq} t^a_{rs} = \delta_{pr}\delta_{qs} - \frac{1}{N} \delta_{pq}\delta_{rs}\,.

Adjoint representation

Consider the tensor product of the antifundamental representation and fundamental representations of SU(N)\,, i.e., \mathbf{N} \otimes\bar{\mathbf{N}}\,. It transforms as

X_j^{\mbox{ }i}\, = \psi_j \psi^{* i}\,,
\to U_j^{\mbox{ }l}\psi_l \psi^{* m} {U^{*}}_m^i \,,
= U_j^{\mbox{ }l} X_l^{\mbox{ }m} {U^{*}}_m^i \,,
= U_j^{\mbox{ }l} X_l^{\mbox{ }m} (U^\dagger)_m^{\mbox{ }i} \,,
= U_j^{\mbox{ }l} X_l^{\mbox{ }m} (U^{-1})_m^{\mbox{ }i} \,.

Now the trace (singlet) part is invariant, so transforms as the \mathbf{1}\,. The trace-free part stays trace-free under a transformation owing to the cyclic property of the trace. The trace-free part of X\, can be written as a linear combination of some basis of N^2-1\, traceless matrices. There happen to be N^2-1\, generators of SU(N)\,, which are also traceless. For convenience, assume X\, is traceless (since the trace-part transforms trivially), so that X \in \mathfrak{su}(N)\,:

X = X^a t^a\,.

Therefore X \to U X U^{-1}\,, which is just the adjoint representation of SU(N)\,, so

\mathbf{N} \otimes\bar{\mathbf{N}} = \mathbf{1} \oplus \mathbf{Adj}\,.


Quadratic Casimir

SU(2)\, is a subgroup of SU(N)\,, since we can consider the subgroup of N\times N\, matrices leaving all but the two indices fixed. Any complete representation of SU(N)\, is therefore a (possibly reducible) representation of SU(2)\,. Three generators of SU(N)\, generate a (possibly reducible) representation of SU(2)\,. Therefore \operatorname{tr}(t^a t^b) = d_{r,SU(N)} \delta^{ab} = d_{r, SU(2)} \delta^{ab}\, where the indices a\, and b\, range over values which are defined for both groups. This assists in obtaining the Casimirs.

The \mathbf{N}\, of SU(N)\, decomposes into N-2\, singlets (\mathbf{1}\,) and one doublet (\mathbf{2}\,) SU(2)\,. From the point of view of SU(2)\,,

d_{\mathbf{N},SU(N)} = (N-2)d_{\mathbf{1}, SU(2)} + d_{\mathbf{2}, SU(2)} = \frac{1}{2}\,,

and

c_\mathbf{N} = d_\mathbf{N} \frac{\dim G}{\dim \mathbf{N}} = \frac{1}{2} \frac{N^2-1}{N}\,.

For \mathbf{N}\otimes\bar{\mathbf{N}}\,,

\left[ \mathbf{N}\otimes\bar{\mathbf{N}} \right]_{SU(N)}\, = \left[\mathbf{2} \oplus (N-2)\mathbf{1}\right]_{SU(2)} \otimes \left[\mathbf{2} \oplus(N-2)\mathbf{1}\right]_{SU(2)} \,,
= \left[\mathbf{2}\otimes\mathbf{2} + 2(N-2)\mathbf{2}\otimes\mathbf{1} +(N-2)^2\mathbf{1}\otimes\mathbf{1}\right]_{SU(2)}\,,
= \left[\mathbf{3} + \mathbf{1} + 2(N-2)\mathbf{2} +(N-2)^2 \mathbf{1}\right]_{SU(2)}\,,
\mathbf{Adj}_{SU(N)}\, = \left[\mathbf{3} + 2(N-2)\mathbf{2} +(N-2)^2 \mathbf{1}\right]_{SU(2)}\,.

It follows that

d_{\mathbf{Adj},SU(N)}\, = \left[R_\mathbf{3} + 2(N-2)R_\mathbf{2} +(N-2)^2 R_\mathbf{1}\right]_{SU(2)}\,,
= 2 + 2(N-2)\frac{1}{2} +(N-2)^2 0\,,
= N\,.

Consequently, since \dim \mathbf{Adj} = \dim G\,,

c_A = N\,.

Summary

c_\mathbf{N} = \frac{1}{2} \frac{N^2-1}{N}\,,
c_A = N\,.

Fierz identities

A = A^a t^a\, and B = B^a t^a\,, then

\lambda\,\operatorname{tr}(AB) = \lambda^2 A^a B^a = \operatorname{tr}(A t^a) \operatorname{tr}(t^a B)\,.

See adjoint representation).

\operatorname{tr}(t^aAt^aB) \propto \operatorname{tr}(A) \operatorname{tr}(B)\, for \mathfrak{u}(N)\,.

References

[1] [2]

  1. ↑ H. Georgi (1999). Lie Algebras in Particle Physics. Westview Press. ISBN 978-0738202334. 
  2. ↑ Walter Pfeifer (2003). The Lie Algebras su(N): An Introduction. Birkhäuser Basel. ISBN 978-3764324186. 
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