# Orthogonal basis of a Cartan of a Lie algebra with respect to Killing form.

I am trying to understand orthogonal basis of a Cartan of a Lie algebra with respect to Killing form.

For example, let $g=sl_2 = \text{Span}\{h, E, F\}$. Then a Cartan of $g$ is $\mathfrak{h} = \text{Span}\{h\}$. By definition, the Killing form is $K(x, y) = tr(ad(x)ad(y))$, $x, y \in g$. Therefore $K(h, h) = tr(ad(h)ad(h))$. We have $$[h,h]=0, \\ [h,E]=2E, \\ [h,F]=-2F.$$ Therefore the matrix of $ad(h)$ is \begin{align} ad(h) = \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -2 \end{matrix} \right). \end{align} Hence $K(h, h) = tr(ad(h)ad(h)) = 8$ and $tr(ad(\frac{1}{2\sqrt{2}}h)ad(\frac{1}{2\sqrt{2}}h))=1$. Therefore the orthonormal basis of the Cartan $\mathfrak{h}$ is $\{ \frac{1}{2\sqrt{2}}h \}$. Is this correct?

Let $g=sl_3 = \text{Span}\{h_1, h_2, E_1, E_2, E_3, F_1, F_2, F_3\}$. Then a Cartan of $g$ is $\mathfrak{h} = \text{Span}\{h_1, h_2\}$. We have $K(h_i, h_j) = tr(ad(h_i)ad(h_j))$. We have $$[h_1,h_1]=0, \\ [h_1,h_2]=0, \\ [h_1,E_1]=2E_1, \\ [h_1,E_2]=-E_2, \\ [h_1,E_3]=E_3, \\ [h_1,F_1]=-2F_1, \\ [h_1,F_2]=F_2, \\ [h_1,F_3]=-F_3,$$ $$[h_2,h_1]=0, \\ [h_2,h_2]=0, \\ [h_2,E_1]=-E_1, \\ [h_2,E_2]=2E_2, \\ [h_2,E_3]=E_3, \\ [h_2,F_1]=F_1, \\ [h_2,F_2]=-2F_2, \\ [h_2,F_3]=-F_3.$$ Hence $K(h_1, h_1) = tr(ad(h_1)ad(h_1)) = 12$ and $K(h_2, h_2) = tr(ad(h_2)ad(h_2)) = 12$. What is the orthogonal basis of $\mathfrak{h}$? Thank you very much.

Edit: we have \begin{align} ad(h_1) = diag(0, 0, 2, -1, 1, -2, 1, -1), \\ ad(h_2) = diag(0, 0, -1, 2, 1, 1, -2, -1). \end{align} Therefore $K(h_1, h_1) = K(h_2, h_2) = 12$, $K(h_1, h_2) = -6$. Therefore $h_1$ is orthogonal to $h_1 + 2 h_2$. Hence $\{h_1, h_1+2h_2\}$ is an orthogonal basis of $\mathfrak{h}$. Is this correct?

• Why would you want to normalise for the Killing form? The scalar factor involved in the definition of the Killing form is not important, only orthogonality really matters. – Marc van Leeuwen Jun 9 '16 at 8:53

## 1 Answer

Indeed $h_1$ and $h_2$ have equal lengths, and make an (obtuse) angle of $\frac{2\pi}3$; hence $2h_2+h_1\perp h_1$.