# How to determine the matrix of adjoint representation of Lie algebra?

My questions will concern two pages:

and

http://mathworld.wolfram.com/KillingForm.html

In the first page, we know the basis of four matrix $\{e_1,e_2,e_3,e_4\}$, and my try to find their adjoint representations is (taking example of $e_2$): $$\hbox{ad}_{e_2}e_1=-e_2,\\\hbox{ad}_{e_2}e_2=0,\\\hbox{ad}_{e_2}e_3=e_1-e_4,\\\hbox{ad}_{e_2}e_4=-e_3.$$ Then in the basis $\{e_1,e_2,e_3,e_4\}$, we can write the matrix of adjoint representation of $e_2$ as: $$\hbox{ad}(e_2)=\left[\begin{array}{cccc}0 & 0 & 1 & 0\\-1 & 0 & 0 & 1\\0 & 0 & 0 & 0\\0 & 0 & -1 & 0\end{array}\right]$$ just like the result in the page. Now my questions:

Q1. If my try is right, now we read the second page ("killing form") and let's do the same calculations with the basis $[X,Y,H]$. I find the matrix of $\hbox{ad}(Y)$ as $$\hbox{ad}(Y)=\left[\begin{array}{ccc}0 & 0 & 2\\0 &0 & 0\\-2 & 0 & 0\end{array}\right]$$ but not the result in the page (just its transposition). If this page is right, my precedent result should be $$\hbox{ad}(e_2)=\left[\begin{array}{cccc}0 & -1 & 0 & 0\\0 & 0 & 0 & 0\\1 & 0 & 0 & -1\\0 & 1 & 0 & 0\end{array}\right].$$ What should it be?

Q2. We have the fomula of Lie algebra: $\hbox{ad}_XY=[X,Y]$. What are the relationships between $\hbox{ad}(X)$ and $\hbox{ad}_X(Y)$?

Q3. In the page of "killing form", how does he get $B=\left[\begin{array}{ccc}8 & 0 & 0\\0 & -8 & 0\\0 & 0 & 8\end{array}\right]$?

Thanks!

• I think $\hbox{ad}_{e_2}e_4=-e_3$ should be $\hbox{ad}_{e_2}e_4= e_2$ – jjgoings Jul 14 '16 at 16:31

$$\newcommand{\ad}{\operatorname{ad}}$$ Answer to Q1:

You shouldn't bother too much with this, it's just a matter of notation. Anyway, I think there's a mistake in their $$\ad(Y)$$ in the sense that, if they want to be coherent with the first page, they should have your $$\ad(Y)$$ and not the transpose of it.

The relatiion is simply that $$\ad_X(Y)$$ is the second column of $$\ad(X)$$. In your example $$\ad_X(Y)$$ is the $$2\times 2$$ matrix $$XY-XY = \begin{pmatrix} 2 & \phantom{-}0 \\ 0 & -2 \end{pmatrix}$$, which corresponds to the vector $$\begin{pmatrix}0 \\ 0 \\ 2\end{pmatrix}$$ in the basis $$X,Y,Z$$. This means that $$\ad_X(Y)$$ is expressed as the linear combination $$0\cdot X + 0\cdot Y + 2\cdot Z = \begin{pmatrix}0 \\ 0 \\ 2\end{pmatrix} \cdot \begin{pmatrix}X \\ Y \\ Z\end{pmatrix}$$
They just use the defining formula $$B(X,Y)=Tr(\ad(X)\cdot\ad(Y))$$. By the basic theory of bilinear forms we know that $$(i,j)$$-entry of the resulting matrix is given by $$Tr(\ad(e_i)\cdot\ad(e_j))$$ where in our case $$e_1=X$$, $$e_2=Y$$ and $$e_3=H$$. As an example, the entry $$(2,2)$$ is computed by $$Tr(\ad(Y)\cdot\ad(Y)) = Tr\; \begin{pmatrix} -4 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -4 \\ \end{pmatrix} = -8$$
• Thank you for your answer. Q3 is OK. For Q1, which sense should we take to have the matrix of $ad(Y)$(by column or row)? For Q2, I wanted to say that $ad_X(Y)=[X,Y]$ is of dimension $2*2$ in the example, but $ad(Y)$ is of dimension $3*3$ where $3$ is the number of element matrix. I don't know the relationship between the two matrix. Thanks again. – Martial Dec 13 '13 at 10:03
• Yes, your explanation is clear. Any way you give the best answers. But I still think that how to write the matrix of $adX$ is important. – Martial Dec 13 '13 at 12:52