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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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 $$tr \:A=0.$$

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(*) 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.