# How is the trace of the adjoint of Lie algebra elements defined? [duplicate]

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?

• Trace of a matrix is independent of the basis. Mar 29 '19 at 17:15
• 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. math.stackexchange.com/q/72303/96384. 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
• 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? Mar 29 '19 at 17:27
• 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 math.stackexchange.com/questions/1369839/… Mar 29 '19 at 18:18

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 $$ad(f)$$?