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My questions will concern two pages:

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

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!

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

Answer to Q2:

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 $$ 0\cdot X + 0\cdot Y + 2\cdot Z = \begin{pmatrix}0 \\ 0 \\ 2\end{pmatrix} \cdot \begin{pmatrix}X \\ Y \\ Z\end{pmatrix} $$

Answer to Q3:

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 $$ Tr(\ad(Y)\cdot\ad(Y)) = Tr\; \begin{pmatrix} -4 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -4 \\ \end{pmatrix} = -8 $$

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  • $\begingroup$ 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. $\endgroup$ – Martial Dec 13 '13 at 10:03
  • $\begingroup$ See my edited answer. I hope everything is clear now ;-) $\endgroup$ – Abramo Dec 13 '13 at 10:31
  • $\begingroup$ 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. $\endgroup$ – Martial Dec 13 '13 at 12:52

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