# Cyclicity of $\text{Tr}\ T^a T^b T^c T^d$ for the unitary group $U(N)$?

I am trying to calculate the trace of $$4$$ generators of $$U(N)$$, i.e. $$\text{Tr}\ T^a T^b T^c T^d$$? I found a plausible result, but I would also like to show that the result is cyclic with respect to the indices, as the trace is. Here is first my calculation of $$\text{Tr}\ T^a T^b T^c T^d$$:

$$T^a T^b = \frac{1}{2} \left( [T^a,T^b]+\lbrace T^a,T^b\rbrace \right) \tag{1}$$

with

$$[T^a,T^b] := if^{abc} T^c \tag{2}$$

and

$$\lbrace T^a, T^b \rbrace = d^{abc} T^c \tag{3}$$

which can be shown using the completeness relation for $$U(N)$$:

$$T^a_{ij}T^a_{lk} = \frac{1}{2}\delta_{ik}\delta_{jl} \tag{4}$$

and the definition of $$d^{abc}$$:

$$d^{abc}:= 2\ \text{Tr} (\lbrace T^a,T^b\rbrace T^c) \tag{5}$$

Putting everything together, I find:

$$T^a T^b T^c T^d = \frac{1}{4} (if^{abe}T^e + d^{abe}T^e)(if^{cdf}T^f + d^{cdf}T^f) \tag{6}$$

and thus

$$\text{Tr}\ T^a T^b T^c T^d = \frac{1}{8} (-f^{abe}f^{cde} + if^{abe}d^{cde} + id^{abe}f^{cde} + d^{abe}d^{cde}) \tag{7}$$

Now the issue is the following: if I change the indices cyclically, I get:

$$\text{Tr}\ T^b T^c T^d T^a = \frac{1}{8} (-f^{bce}f^{dae} + if^{bce}d^{dae} + id^{bce}f^{dae} + d^{bce}d^{dae}) \tag{8}$$

which should be equal to $$(7)$$ because of the cyclicity of the trace. How can I show that?

• You forgot the δ term in the anticommutator of two generators. – Cosmas Zachos Oct 2 '19 at 21:39
• – Cosmas Zachos Oct 2 '19 at 21:43
• @CosmasZachos But this is $U(N)$, not $SU(N)$. Where would the $\delta$ come from? I think it should not be there because the completeness relation is different than in $SU(N)$. – Jxx Oct 2 '19 at 21:50
• I see. You have incorporated the identity of the U(1) into the algebra and count the δ term as a d term. You may have to look into the embedding of SU(N) in U(N) to use the well known identities for the latter. – Cosmas Zachos Oct 2 '19 at 22:00
• @CosmasZachos I think that working in $U(N)$ or $SU(N)$ does not change the final question: how is the cyclicity of the trace fulfilled by the right-hand side of $(8)$, no matter if there is an additional $\delta$ term or not? I guess that it means that $f^{abe}f^{cde}=f^{bce}f^{dae}$ but I can't manage to prove it. Maybe this is not even true. – Jxx Oct 2 '19 at 22:04

For the fundamental of su(N), $$\operatorname{Tr}\ T^a T^b =\delta^{ab}/2$$, $$[T^a,T^b]=if^{abe}T^e,\\ \{ T^a,T^b\}=\frac{1}{N}\delta^{ab} +d^{abe}T^e, \Longrightarrow \\ T^aT^b= \frac{1}{2N}\delta^{ab} +\frac{1}{2}d^{abe}T^e +\frac{i}{2} f^{abe}T^e,$$ so that $$\operatorname{Tr}\ T^a T^b T^c T^d = \frac{1}{4N} \delta^{ab}\delta^{cd}+\frac{1}{8} (-f^{abe}f^{cde} + d^{abe}d^{cde})\\ +\frac{i}{8}(f^{abe}d^{cde} +f^{cde} d^{abe}) .$$
The last term is proportional to the the tensor with the real and imaginary parts displayed explicitly, $$f^{abe}d^{cde} +f^{cde} d^{abe} \equiv M^{[ab],(cd)}+ M^{[cd],(ab)}\equiv N^{abcd} .$$ By virtue of (2.9), $$f^{abe}d^{cde} + f^{cbe}d^{dae}+ f^{dbe}d^{ace}=0,$$ you see that $$N^{abcd}= M^{[ab],(cd)}+ M^{[cd],(ab)}\\ =- M^{[cb],(da)} - M^{[db],(ac)}- M^{[ad],(bc)} - M^{[bd],(ca)}\\= M^{[bc],(da)}+ M^{[da],(bc)} = N^{bcda}.$$ You may repeat this step twice more if the full cyclicity were not self-evident.
• OK, let me stick in the real part as well. Eliminating the ff bilinears through (2.10) and utilizing (2.22) suitably generalizing 3 to N, I believe you get $$\frac{1}{8N} ( \delta^{ab}\delta^{cd}+ \delta^{da}\delta^{bc}-3 \delta^{ac}\delta^{bd} )+\frac{1}{4} ( d^{dae}d^{bce} + d^{abe}d^{cde}),$$ with the requisite cyclic symmetry.