Generators of a semi simple lie algebra must be traceless

Question

Consider a semi simple lie algebra. Show that if $T_a$ are the generators of a semi simple Lie algebra then $\text{Tr}T_a=0$.

Attempt: $[T_a, T_b] = ic^c_{ab}T_c \Rightarrow \text{Tr}[T_a, T_b] = ic^c_{ab}\text{Tr}(T_c) = 0$.

I also know that $g_{ab} = -c^c_{ad}c^{d}_{bc}\,\,(1)$. So since the l.h.s of the equation above is zero, I could multiply by another structure constant, thereby introducing the metric and then since the lie algebra is semi simple, multiply by the inverse of the metric and thus just have $\text{Tr}(T_a)$ on the l.h.s meaning it has to vanish.

That is my argument but I can't write it down because i don't see a way to introduce another structure constant because in the definition $(1)$, $d$ is contracted but in my expression $d$ is also contracted with the generator. So the placements of the indices mean I can't just multiply by another structure constant with the required indices without violating the summation convention.

Any hints on how to make progress? Thanks!

Answer 1

Every semisimple Lie algebra $L$ has a faithful representation to $\mathfrak{gl}_n(K)$, e.g., given by the adjoint representation $x\mapsto {\rm ad}(x)$. Because $L=[L,L]$ and because ${\rm ad}([x,y])=[{\rm ad}(x),{\rm ad}(y)]$ for all $x,y\in L$ it follows that ${\rm ad}(x)$ has trace zero for all $x\in L$. This holds for any faithful representation of $L$. Hence $L$, i.e., an isomorphic copy of it represented by matrices, consists of trace zero matrices. We have $L\hookrightarrow \mathfrak{sl}_n(K)$. Thus all elements have trace zero.

Edit: Suppose that the generators $T_a$ are already matrices. Then every $T_a$ is a sum of commutators $[T_b,T_c]$ because of $L=[L,L]$. Hence every $Tr(T_a)$ has trace zero, because all $[T_b,T_c]$ have trace zero, as you have shown.

Answer 2

Although Dietrich's answer is perfect; here's the completion of how you started (where only one simple step was missing)

$Tr[T_a, T_b] =i f_{ab}{}^c Tr[T_c] = 0 \quad (1)$

Then, because the Lie algebra is semi-simple, we indeed have a metric $g_{ab} = -f_{ac}{}^d f_{bd}{}^c$ which we can use to raise and lower Lie algebra indices. Now (1) has as free indices $a$ and $b$ so you can multiply it with $ g^{be} f_{de}{}^a $. Then do some raising and lowering of indices:

$g^{be} f_{de}{}^a f_{ab}{}^{c} Tr [T_c] = f_{de}{}^a f_{ca}{}^{e} Tr [T^c]= -g_{dc} Tr[T^c] = - Tr[T_d] = 0$

Answer 3

Just to add to the answers above, there is also a statement you can make for any Lie algebra:

For any Lie algebra $\mathfrak g$ one has an ideal $$ [\mathfrak g,\mathfrak g] = D\mathfrak g = \mathrm{span}\{[x,y]: x,y \in \mathfrak g\} $$ called its derived sub-algebra (even though it is always ideal).

If $\mathfrak g$ is semisimple, then it is a direct sum of nonabelian simple Lie algebras, say $\mathfrak g = \mathfrak g_1 \oplus \ldots \oplus \mathfrak g_k$ where for each $\mathfrak g_i$, the derived subalgebra $D\mathfrak g_i = [\mathfrak g_i,\mathfrak g_i]$ is nonzero because $\mathfrak g_i$ is nonabelian, and then since $\mathfrak g_i$ is simple, the ideal $D\mathfrak g_i$ must be all of $\mathfrak g_i$. Since $\mathfrak g$ is the direct sum of the $\mathfrak g_i$s (as a Lie algebra) it follows that $D\mathfrak g = D\mathfrak g_1 \oplus \ldots \oplus D\mathfrak g_k = \mathfrak g$.

The property $\mathfrak g = D\mathfrak g$, i.e, that $\mathfrak g$ is equal to its derived subalgebra $D\mathfrak g$, is all that is needed for $\rho(\mathfrak g) \subseteq \mathfrak{sl}(V)$ for any representation $\rho\colon \mathfrak g \to \mathfrak{gl}(V)$. Indeed it is straightforward to check that if $\theta\colon \mathfrak g_1 \to \mathfrak g_2$ is a homomorphism of Lie algebras then $\theta(D(\mathfrak g_1))\subseteq D(\mathfrak g_2)$: indeed

$$ \begin{split} \theta(D\mathfrak g_1) &= \mathrm{span}\{\theta([x,y]): x,y \in \mathfrak g_1\}\\ &= \mathrm{span}\{[\rho(x),\rho(y)]:x,y \in \mathfrak g_1\} \\&= \mathrm{span}\{[a,b]: a,b \in \theta(\mathfrak g)\} \\ &\subseteq \mathrm{span}\{[a,b]: a,b \in \mathfrak g_2\}\\ &= D\mathfrak g_2. \end{split} $$ But now since the trace of a commutator is always equal to $0$, clearly we have $D(\mathfrak{gl}(V)) \subseteq \mathfrak{sl}(V)$. Thus if $\rho\colon \mathfrak g \to \mathfrak{gl}(V)$ is any representation of a Lie algebra $\mathfrak g$ it is always the case that $\rho(D\mathfrak g) \subseteq \mathfrak{sl}(V)$. If $\mathfrak g$ satisfies $\mathfrak g = D\mathfrak g$ then this clearly implies $\rho(\mathfrak g) \subseteq \mathfrak{sl}(V)$, as is the case when $\mathfrak g$ is semisimple.

Another way to think about this is to notice that $\mathrm{tr}\colon \mathfrak{gl}(V) \to \mathfrak{gl}_1(\mathbb C)$ is a homomorphism of Lie algebras: since $\mathfrak{gl}_1 = \mathfrak{gl}_1(\mathbb C)$ is abelian, this is exactly the statement that $$ \mathrm{tr}([x,y])= [\mathrm{tr}(x),\mathrm{tr}(y)]=0, \quad \forall x,y \in \mathfrak{gl}(V). $$ But then if $\rho\colon \mathfrak g \to \mathfrak{gl}(V)$ is any homomorphism of Lie algberas, $\mathrm{tr}\circ \rho\colon \mathfrak g \to \mathfrak{gl}_1(\mathbb C)$ is also homomorphism of Lie algebras, hence $\mathrm{tr}(\rho([x,y]))=0$ for all $x,y \in \mathfrak g$, hence $\mathrm{tr}(\rho(D\mathfrak g))=0$.