# Generators of ${\rm SU}(n)$ are traceless. Why?

A general element of the Lie group $${\rm SU}(n)$$ is written as $$g({\vec{\theta}})=e^{-i\sum_a\theta_a T_a}$$ where $$\theta_a$$ for $$(a=1,2,\ldots,n^2-1)$$ denotes $$n^2-1$$ real parameters. The unitarity $$g^\dagger g=I$$ demands that $$T_a^\dagger=T_a$$ i.e. the generators be hermitian. The unimodularity condition, $$\det g=+1$$, further tells that $$\det g=e^{-i\sum_a\theta_a{\rm Tr}(T_a)}=+1.$$ This only requires that $$\sum_a\theta_a{\rm Tr}(T_a)=2\pi n$$ where $$n$$ is any integer. What restricts the value of $$n$$ to be zero or in other words, what makes the generators traceless?

• A correct statement is that $\det(e^{-i \sum_a \theta_a T^a }) = \exp\{ \mathrm{Tr}\log e^{-i \sum_a \theta_a T^a } \}$. In general $\exp\{ \mathrm{Tr} \log e^{-i \sum_a \theta_a T^a } \} \neq \exp\{-i \sum_a \theta_a \mathrm{Tr}(T^a) \}$. The standard language around Lie Algebras is to consider 'infinitesimal' shifts from the identity as group elements. So $e^{i \epsilon T}$ as $\epsilon \to 0$. In particular, imposing $e^{i \epsilon T} \in SU(n)$ for all $\epsilon$ and taking the derivative at $\epsilon=0$ gives (exercise!) that $T$ is traceless and Hermitian.
– Joe
Oct 20, 2023 at 14:59
• Consider to accept one of the answers below. If you think none of them addressed your question properly, edit your question accordingly or write a comment under the respective answer. This also holds for many (!) of your other questions, where you have answers (which sometimes also have several upvotes) but you did not respond by any means. Oct 25, 2023 at 22:09

The zero trace condition for the Lie algebra is the counterpart of the unit determinant condition for the Lie group. Indeed, if $$M\in {\rm SU}(N)$$, then $$\det M=1$$. Now consider the exponential map $$\exp:\mathfrak{su}(N)\to \mathrm{SU}(N)$$, which for a matrix group and algebra is really the matrix exponential. Now recall that $$\ln\det \exp X=\operatorname{tr} X\tag{1}.$$

Since $$\exp X\in \mathrm{SU}(N)$$ we must have $$\det \exp X=1$$. As a result $$\ln\det \exp X=0$$ for any $$X\in \mathfrak{su}(N)$$ and hence $$\operatorname{tr}X=0$$ for any $$X\in \mathfrak{su}(N)$$.

The reason $$\operatorname{tr}X=2\pi n$$ with $$n\in \mathbb{Z}\setminus \{0\}$$ is not an option is exactly because of the fact that we can form any linear combination with complex coefficients of a given basis of generators as mentioned in comments to this post and in Connor's answer.

Nevertheless, we give another approach, which is more direct. Let $$M(t)=\exp tX$$ for $$t\in (-\epsilon,\epsilon)$$ be a short path with $$M(0)=1$$ and $$M'(0)=X$$. The derivative of $$\det M(t)$$ is easily calculated from (1) along the path. Indeed taking the derivative with respect to $$t$$ we find $$\dfrac{1}{\det (\exp tX)}\dfrac{d}{dt}(\det(\exp tX))= \operatorname{tr} X\tag{2}.$$

Since $$\exp tX\in \mathrm{SU}(N)$$ we have $$\det \exp tX=1$$ for all $$t\in (-\epsilon,\epsilon)$$ and the LHS of (2) is trivially zero. As such $$\operatorname{tr}X=0$$ follows.

Note this is not special to $${\rm SU}(N)$$: the argument applies to any matrix group where $$\det M =1$$. As I mentioned in the beginning: zero trace is the Lie algebra counterpart of unit determinant.

• Physicists stick the imaginary $i$ in the exponential and in that case, one can also entertain the possibility that ${\rm tr} X=2\pi n$ with $n\neq 0$. There lies my confusion. Oct 20, 2023 at 15:09
• @Solidification That doesn't change that you can multiply $T\mapsto \mathrm{i}T$ (this is a complex representation, after all) and get Gold's condition from this answer back. The $\mathrm{i}$ is just a convention for how we pick the basis of the complexified Lie algebra. Oct 20, 2023 at 15:55
• Gold, I edited it to read $\ln \det = tr \ln$ instead of $\det = e^{tr \ln}$. I think this may be more clear to @Solidification. Feel free to rollback if you do not like it.
– hft
Oct 20, 2023 at 20:24
• Although, now that I look at it for a third time, I'm not sure this is the best answer. I feel like it needs further explanation about why $tr X$ can't be $i2\pi n$... since $\ln 1 = \ln e^{i2\pi n}$... But I defer to others.
– hft
Oct 21, 2023 at 2:16
• @hft I have added a comment on why $\operatorname{tr} X=2\pi n$ doesn't work and also a quick proof by direct computation that $\operatorname{tr}X=0$.
– Gold
Oct 21, 2023 at 4:17

The definition of the generators is not by the exponential, but from differentials. This is why your reasoning is necessary but not sufficient.

Abstractly, the generators of a Lie group are its Lie algebra, i.e. the tangent space of the group at the origin. In particular for matrix groups, the definition can be made more concrete. Say you have a Lie group $$G\subset GL_n(\mathbb K)$$, then it's Lie algebra (generators) is: $$\mathfrak g = \{x\in M_n(\mathbb K)|\exists g\in\mathcal C^1((-\epsilon,\epsilon),G),g(0) = I_n,g'(0)=x\}$$ with $$\epsilon>0$$. Basically, it's all the possible derivatives of the Lie group at the identity, which for matrix groups can be seen as a subset of the matrices, but for Lie groups in general, you need to construct this space leading to the notion of tangent space.

In particular, for $$SU(n)$$, say you have a curve $$g$$ passing by the origin, since it's in the group, it satisfies: $$g^\dagger g = I_n\\ \det g = 1$$ Taking the derivative, using $$g(0)=I_n$$ and $$x = g'(0)$$, you get: $$x^\dagger +x = 0\\ \text{Tr}\, x = 0$$ so you do get the more stringent constraint of zero trace, not just an integer multiple of $$2\pi$$.

Btw, for spectral reasons, physicists add an extra $$i$$ to the Lie algebra, I used the mathematicians' convention.

Hope this helps.

• +1 but I think you mean det(g) =1, not Tr(g) =1. Oct 21, 2023 at 8:03
• Yes you're right, corrected.
– LPZ
Oct 21, 2023 at 8:11

The fact that \begin{align} \sum_a \theta_a \text{Tr}(T_a) = 2\pi n \end{align} must hold for all $$\theta_a$$, where $$\theta_a$$ is continuous, implies that $$n$$ must be zero, since $$n$$ is discrete.

• How does that imply that without any loss of generality, we can take ${\rm Tr}(T_a)=0$? Oct 20, 2023 at 15:11
• This answer is fine, this argument does prove that $n=0$. The r.h.s. is discrete and while the l.h.s. is continuous, hence the only possibility is $\text{tr}(T)=n=0$. Oct 20, 2023 at 17:37
• Connor's answer is correct. What's with the downvotes? Maybe he can add a little more detail along the lines of AccidentalFourierTransform's comment. Oct 20, 2023 at 18:54
• I edited it to make the answer a complete sentence and add the content suggested in @AccidentalFourierTransform 's comment. Feel free to roll back if you don't like the edit.
– hft
Oct 20, 2023 at 20:25

Consider the one-parameter group $$\mathbb{R} \ni s \mapsto e^{sA} \in G$$ generated by the element $$A$$ of the Lie algebra $$\mathfrak{g}$$ of a matrix Lie group $$G$$ like $$SU(N)$$.

For instance $$A= i\sigma_k$$ when $$N=2$$ or $$iT_k$$ in the general $$N$$ case. The factor $$i$$ is embodied in $$A$$, but if writing explicitly it, this proof does not change.

As is well known (*) $$\det(e^{sA})= e^{tr(sA)}= e^{s\: tr A}\tag{1}$$ is valid for all $$s$$.

If the group $$G$$ is special as $$SU(N)$$, the function above takes the value $$1= \det(e^{sA})$$ constantly.

Computing the $$s$$-derivative for $$s=0$$ of the rightmost term in (1), we find $$tr \:A=0.$$

(*) As a general elementary result of matrix theory, if $$B$$ is any nxn real or complex matrix, $$\det e^B = e^{tr\:B}$$, where the exponential is defined in terms of Taylor series.