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Consider an element $X\in\mathfrak g$ of some Lie algebra $\mathfrak g$. I understand that $\mathfrak g$ can be represented via its action on other elements of the same algebra, as $\operatorname{ad}(X)Y\equiv[X,Y]$, so that $\operatorname{ad}(X)\in\operatorname{GL}(\mathfrak g)$.

One often considers things such as the trace of these objects (e.g. when considering the Killing form), or more generally the "matrix elements" of operators such as $\operatorname{ad}(X)$.

However, I usually don't see any direct mention of the inner product with respect to which these things are defined. To properly define what something such as $\operatorname{ad}(X)_{ij}$ is, don't I need to be able to define uniquely the coefficient of the generator $X_i$ in the decomposition of $[X,X_j]$ (denoting with $\{X_i\}$ the elements of some basis for the algebra)?

Is there a canonical choice of such inner product? And on a similar note, do we just usually assume that the basis of the Lie algebra is orthonormal (or at least orthogonal) with respect to this inner product?

glS
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    Trace of a matrix is independent of the basis. – Moishe Kohan Mar 29 '19 at 17:15
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    One needs no inner product to define the trace of a linear map. In some common definitions it looks like one needs to choose a basis to define it, but then it should be immediately noted that the trace is actually independent from that choice. See e.g. https://math.stackexchange.com/q/72303/96384. – Torsten Schoeneberg Mar 29 '19 at 17:16
  • @MoisheKohan sure, but my question is how is it defined in the first place. Once I have an inner product and thus can talk of an orthogonal basis, then I understand that changing the basis will not change the trace – glS Mar 29 '19 at 17:17
  • @TorstenSchoeneberg doesn't that just say that the trace is independent of the basis? I might be misguided here, but without an underlying inner product how do I even talk of the matrix elements of the operator? Like even in practice, how do I find the component $X_i$ of a vector $X$ without using an inner product? – glS Mar 29 '19 at 17:20
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    Surely $\mathfrak{g}$ is assumed to be a finite-dimensional vector space over some field $k$ (otherwise, traces indeed make little sense without further effort). "Finite-dimensional" literally means you can choose a finite basis, and then you write linear maps as matrices with respect to that basis. This is kind of the main content of an elementary linear algebra course, isn't it? – Torsten Schoeneberg Mar 29 '19 at 17:27
  • @TorstenSchoeneberg mh, fair point. I guess this is what got me confused: I have some $X\in\mathfrak g$ and I want to find the matrix representation of $\operatorname{ad}(X)$. I start computing $[X,X_i]$ for some basis element $X_i\in\mathfrak g$ and get some $[X,X_i]\in\mathfrak g$, which is in itself a matrix if I'm considering some standard matrix Lie algebra. Now I'm looking for the component $X_k$ in $[X,X_i]$. How do I do this? I thought some kind of inner product ought to be used here, but it's true that this isn't the case, as one can uniquely define the coefficients even without one – glS Mar 29 '19 at 17:47
  • Yes, I can understand the confusion if $\mathfrak{g}$ is already given as a matrix Lie algebra. But matrices are vectors too, and one has to "forget" for this that they are also matrices. Although later in the theory, there is often interesting interplay between, say, properties of an element $X$ given as matrix, and properties of the corresponding adjoint map (written as "bigger" matrix) $ad(X)$. E.g. under some conditions one is nilpotent iff the other is. – Torsten Schoeneberg Mar 29 '19 at 17:56
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    I really do not understand the source of confusion: To define trace on endomorphisms of a finite-dimensional vector space you do not need an inner product, all you need is a basis. This would be discussed in any linear algebra class. Then you learn that $tr(ABA^{-1})=tr(B)$, hence, trace is independent of the choice of a basis. Are you asking for a definition of the trace without having to choose a basis? For this, see https://math.stackexchange.com/questions/1369839/coordinate-free-definition-of-trace?rq=1 – Moishe Kohan Mar 29 '19 at 18:18

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Consider $\mathfrak{sl}_2$ with basis $\{e,f,h\}$. You have $$ad(e)(e)=0,\;\;\;ad(e)(f)=h,\;\;\;ad(e)(h)=-2e$$ so the matrix of $ad(e)$ in this basis is $$ \begin{pmatrix}0&0&-2\\0&0&0\\0&1&0\end{pmatrix} $$ which has trace 0.

You can check that the matrix for $ad(h)$ is $$ \begin{pmatrix}2&0&0\\0&-2&0\\0&0&0\end{pmatrix} $$ which again has trace 0. Can you do $ad(f)$?

David Hill
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