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I have some matrix lie algebra $\mathfrak{g}$ and I want to compute the matrix of the adjoint action of $x\in\mathfrak{g}$, $\mathrm{ad}_x:\mathfrak{g}\rightarrow\mathfrak{g}$ given by $y\mapsto[x,y]$. How can I proceed?

I have been told that it should be $\mathrm{ad}_x = x\otimes\mathrm{id} - \mathrm{id}\otimes x$, so I tried to compute it by fixing a basis of $\mathfrak{g}$, computing the tensor of the matrices $x$ and $\mathrm{id}$ and multiplying by the vector representation of some $y\in\mathfrak{g}$. This works fine until I use the canonical basis

$$e_1 = \left(\begin{array}{cc}1 & 0\\0 & 0\end{array}\right),\ \ e_2 = \left(\begin{array}{cc}0 & 1\\0 & 0\end{array}\right),\ \ e_3 = \left(\begin{array}{cc}0 & 0\\1 & 0\end{array}\right),\ \ e_4 = \left(\begin{array}{cc}0 & 0\\0 & 1\end{array}\right)$$

but when I try with a different basis it doesn't work anymore.

Daniel Robert-Nicoud
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1 Answers1

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You find the matrix of the adjoint action the same way you find the matrix of any linear operator. Choose a basis $\{v_1,\dots,v_n\}$ of $\mathfrak{g}$ (I'm assuming your Lie algebra is finite-dimensional here). Now apply your operator to each basis element. That is, compute $[x,v_j]$ for $j=1,\dots,n$, and express it as a linear combination of your basis elements: $[x,v_j] = \sum_i a_{ij} v_i$. Then the matrix of $\mathrm{ad}_x$ in the basis $\{v_1,\dots,v_n\}$ is the matrix $(a_{ij})$.

Let's compute a specific example. Take the Lie algebra $\mathfrak{sl}_2(\mathbb{C})$, with basis $$ e = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad f = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \quad h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. $$ Let's say we want to compute the matrix of $\operatorname{ad}_e$ in this basis. We have $$ \operatorname{ad}_e(e) = [e,e] = 0,\quad \operatorname{ad}_e(f) = [e,f] = h,\quad \operatorname{ad}_e(h) = [e,h] = -2e. $$ Thus, the matrix of $\operatorname{ad}_e$ in the basis $e,f,h$ is $$ \begin{pmatrix} 0 & 0 & -2 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}. $$

Alistair Savage
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  • Is this true for any matrix, $A$ such that $\text{ad}A := \bf{a} $, where $[A, v_j] = \sum_i \text{a}{i j} v_j$. Could you give a reference for this representation ? I am struggling to find any explicit working examples. – user91411 May 05 '23 at 15:26
  • @user91411 I've added a specific example. I hope that helps. – Alistair Savage May 07 '23 at 00:07
  • thanks, that helped. So by basis we mean just the elements of the algebra; we do not include the identity element. right ? – user91411 May 07 '23 at 13:33
  • @user91411 By basis, I mean a basis as a vector space. Lie algebras don't have identity elements. – Alistair Savage May 08 '23 at 21:16