1

I am asked to calculate the matrices ($\operatorname{ad} e_{1}$), ($\operatorname{ad} e_{2}$), and ($\operatorname{ad}e_{3}$), given $e_{1} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, $e_{2} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, and $e_{3} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ where these matrices are the basis for $\mathfrak{sl}(2,\mathbb{F})$.

From other examples I have seen, I understand that calculating $\operatorname{ad}_{\xi}$ where $\xi = ae_{1} + ae_{2} + ae_{3}$ would consist of calculating $[\xi,\mathfrak{g}]$. This much I can do, and am satisfied with the answer.

However, the present question that I am being asked throws me off just a bit. First off, there is no subscript on the (ad x). I understand that there is are differences between, say (ad $y$) and ($\operatorname{ad}_{y}$), but I don't fully understand them.

Furthermore, I understand that (ad $\mathfrak{g}$) is the isomorphism between $\mathfrak{g}$ and $\mathfrak{gl}(2, F)$, and I feel comfortable with that definition, but be darned if I can calculate (ad $\mathfrak{g}$) or in the present case ($\operatorname{ad} e_{1}$).

Any help would be appreciated.

Davide Giraudo
  • 172,925
the_photon
  • 209
  • I think that the notations $ad$ $y$ and $ad_y$, or $ad(y)$ if you want, mean the exact same thing (for $y$ an element of the Lie algebra). $ad$ is a homomorphism (not an iso for $n>1$, since for dimension reasons it's not surjective) from a (say, $n$-dimensional) Lie algebra over $F$ to $\mathfrak{gl}(n, F)$. Since your Lie algebra $\mathfrak{sl}(2,F)$ is 3-dimensional, you expect it to be a homomorphism to $3 \times 3$-matrices. Then proceed as in Lord Shark's answer. – Torsten Schoeneberg Jul 08 '17 at 19:25

1 Answers1

1

$\newcommand{\ad}{\text{ad}\,} \newcommand{\g}{\mathfrak{g}}$ Here $\ad z$ is the map from $\g=\mathfrak{sl}(2,F)$ to itself defined by $\ad z:w\mapsto[x,w]$. Here you want to find the matrix of each $\ad e_i$ with respect to the basis $e_1,e_2,e_3$.

First of all, $\ad e_1:e_1\mapsto 0$, $e_2\mapsto -2e_1$, $e_3\mapsto-e_2$ so $\ad e_1$ has matrix $$\pmatrix{0&0&0\\-2&0&0\\0&-1&0}$$ etc.

Angina Seng
  • 158,341