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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 $\text{Tr}\ T^a T^b T^c T^d$:

We start with the following:

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

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?

Pxx
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  • You forgot the δ term in the anticommutator of two generators. – Cosmas Zachos Oct 02 '19 at 21:39
  • Related. – Cosmas Zachos Oct 02 '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 $SU(N)$. – Pxx Oct 02 '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 02 '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. – Pxx Oct 02 '19 at 22:04
  • Well, (2.8-2.10) of my MacFarlane et al link for SU(N) surely gives you enough Fierz-rearrangement identities to confirm that in this case... – Cosmas Zachos Oct 02 '19 at 22:08
  • @CosmasZachos That's probably true, but the article costs 42€. – Pxx Oct 02 '19 at 22:13
  • This should be free, if the springer one is paywalled.... – Cosmas Zachos Oct 02 '19 at 22:30
  • @CosmasZachos Yes it is, thanks! I already know the identities that you mentioned, and was not able to progress using them, but still this looks like a great paper to have! – Pxx Oct 02 '19 at 22:43

1 Answers1

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Let's address the problem for the fundamental of su(N). You may then trivially embed the solution into u(1) by inserting the identity tagged by an extraneous index 0 into the quadrilinear once;twice;thrice; four times, converting it to a trilinear; bilinear; zero; and scalar number, respectively in the trace-- all of them trivially compliant with cyclicity. A solution for su(N) solves u(N). There might be an elegant or diagrammatic solution depicting the troubling symmetries of u(N) introduced by the identity, but if you want to reach the truth, you start by grinding, opting for elegance only in the end.

So you shamelessly utilize the "canned" identities for the fundamental of su(N) in the canonical source, Macfalane, Sudbery, and Weisz 1968, or, unpaywalled, namely (2.9), (2.10), (2.22), etc...

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 $$\operatorname{Tr}\ T^a T^b T^c T^d = \frac{1}{4N} \delta^{ab}\delta^{cd}+\frac{1}{8} (-f^{abe}f^{cde} + d^{abe}d^{cde})\\ +\frac{i}{8}(f^{abe}d^{cde} +f^{cde} d^{abe}) .$$

You may proceed to process the real part by use of (2.10) and (2.22) to check cyclicity. Here, let me check cyclicity for the imaginary part, the last term, and let you do the rest.

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.
Cosmas Zachos
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