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Let $\underline G$ be an arbitrary Lie algebra. For arbitrary $X,Y\in\underline G$, we can define the adjoint of $X$ by $\text{ad}_X(Y)=[X,Y]$. Thus, $\text{ad}_X$ is a linear map on $\underline G$.

Given the adjoint, one defines the Killing Form $g(X,Y)=\text{Tr}(\text{ad}_X\circ \text{ad}_Y)$.

I don't understand what $\text{Tr}$ means in this context. I understand that $\underline G$ is a vector space, so $\text{ad}_X\circ \text{ad}_Y$ is a linear map on the vector space, thus something that we should be able to take a trace of. But as far as I can tell, there is no canonical basis of $\underline G$, so I don't know what basis to take the trace in. There's no inner product defined on $\underline G$, so I can't just pick an orthonomal basis.

What does this trace mean? It would be helpful to see a simple example, the specific context I'm interested in is when $\underline G$ is the Lie algebra of $SU(n)$.

glS
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Jahan Claes
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    The trace is independent of the choice of a basis. https://math.stackexchange.com/questions/72303/proving-the-trace-of-a-transformation-is-independent-of-the-basis-chosen – Torsten Schoeneberg Jun 27 '18 at 04:41
  • @TorstenSchoeneberg Right, of course! I was thinking of the trace as the operation $\text{tr}(M)=\sum_i \langle v_i,Mv_i\rangle$, which needs an inner product $\langle\cdot,\cdot\rangle$ and requires the basis $v_i$ to be orthonormal. This definition is is NOT invariant under change of basis (only orthogonal change of basis). But of course if you just consider the trace as the sum of diagonal elements, it should be obvious. Thank you! – Jahan Claes Jun 27 '18 at 17:33
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    @TorstenSchoeneberg I was wondering why literally no textbook explained this, but I guess it's because they all reasonably thought it was trivial. – Jahan Claes Jun 27 '18 at 17:34
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    It is not that textbooks consider this trivial. It is that this is something that should be covered in textbooks about linear algebra (and indeed, it is covered in many of them). Authors writing about Lie algebras will generally assume that the reader has a pretty deep knowledge of linear algebra. – Tobias Kildetoft Jun 28 '18 at 06:10

2 Answers2

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The trace can be done in any basis. The trace is invariant under an arbitrary change of basis. If $M$ is an arbitrary matrix, and $A\in \mathbb{GL}(n,\mathbb{C})$, then $$ \text{Tr}(A^{-1}MA)=\text{Tr}(M) $$ by the cyclic property of the trace.

So for $SU(N)$, you just pick any basis of the Lie algebra of skew-symmetric, traceless matrices, and you figure out the diagonal elements.

Jahan Claes
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It really means the trace of the resultant matrix. Let's take $\mathfrak{sl}\left(2\right)$ and consider the following base $$e=\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right),$$$$h=\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right),$$$$f=\left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right)$$ You have the following relations $$\left[h,\,e\right]=2e,\,\,\,\,\left[h,\,f\right]=-2f,\,\,\,\left[e,f\right]=h.$$ Then considering as order of the base $\left\{ e,\,h,\,f\right\}$ you get $$\text{ad}\left(e\right) =\left(\begin{array}{ccc} 0 & -2 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right),$$ $$\text{ad}\left(h\right) =\left(\begin{array}{ccc} 2 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -2 \end{array}\right),$$ $$\text{ad}\left(f\right) =\left(\begin{array}{ccc} 0 & 0 & 0\\ -1 & 0 & 0\\ 0 & 2 & 0 \end{array}\right).$$ Then do the multiplication of the matrices (ex $\text{ad}\left(e\right)\text{ad}\left(e\right)$) take the Trace and you will get the following coefficient for the killing form: $$\left(B\right)_{ij}=\left(\begin{array}{ccc} 0 & 0 & 4\\ 0 & 8 & 0\\ 4 & 0 & 0 \end{array}\right).$$ It's really just a straightforward calculation. You might find a lot of examples of this kind in Carter

Dac0
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