Let $\mathfrak g$ be a simple Lie algebra. Then what is an orthonormal basis of $\mathfrak g\ $? It has been written in one of the notes that if $(I_{\nu})$ is an orthonormal basis of $\mathfrak g$ then the Casimir tensor $\Omega$ corresponding to the Killing form looks like $\Omega = \sum\limits_{\nu} I_{\nu} \otimes I_{\nu}.$ I think by orthonormality the author means that $K (I_{\nu}, I_{\mu}) = 0$ for $\nu \neq \mu$ and $K(I_{\nu}, I_{\nu}) = 1.$ If one can manage to find orthogonal basis then we can normalize it so as to make it orthonormal and that the Casimir tensor will be of the above form follows directly from the definition. But I don't know whether such a basis exists or not. Do anyone have any idea regarding this? Any help would be appreciated.
Thanks!
This is true for any arbitrary vector space $V$ over a quadratically closed field $k$ of characteristic zero on which a symmetric non-degenerate bilinear form $B$ is defined. For such a bilinear form first note that there always exists $0 \neq v \in V$ such that $B(v,v) \neq 0.$ Here's how the argument goes $:$
Let $0 \neq v \in V.$ If $B(v,v) \neq 0$ then we are through. So let us assume that $B(v,v) = 0.$ Since $B$ is non-degenerate we can find out a $0 \neq w \in V$ such that $B (v,w) \neq 0.$ If $B(w,w) \neq 0$ then we are again through. So assume that $B (w,w) = 0.$ Then we have by the symmetricity of $B$ $$\begin{align*} B(v + w, v + w) & = B(v,v) + B(w,w) + 2 B (v,w) \\ & = 2 B(v,w) \neq 0\ \ (\because \text {char} (k) = 0) \end{align*}$$
This proves our claim. Now the rest of the part can be proved using induction on the dimension of $V.$ If $\dim V = 1$ choose any $0 \neq v \in V.$ Then $V = \text {span}\ \{v\}.$ So we must have $B(v,v) \neq 0$ for otherwise $B$ would be identically zero and hence not non-degenerate. Now consider the vector $v_1 = \frac {v} {\sqrt {B(v,v)}}.$ Such a $v_1$ always exists as the underlying field $k$ is quadratically closed. Then $v_1$ spans $V$ and $B(v_1, v_1) = 1.$ So the singleton set $\{v_1\}$ serves the purpose of an orthonormal basis for $V$ if $\dim V = 1.$ This proves the base case for the induction.
Now assume that our result is true for any vector space $V$ with $\dim V \leq n - 1.$ Now consider the vector space $V$ with $\dim V = n.$ Take a $0 \neq v \in V$ with $B (v,v) \neq 0.$ WLOG we may assume that $B(v,v) = 1$; otherwise we can normalize $v$ accordingly (this is again possible since $k$ is quadratically closed). Let $W = \text {span}\ \{v\}.$ Now consider the vector subspace $W'$ of $V$ defined by $$W' : = \left \{x \in V\ |\ B(x,v) = 0 \right \}.$$ First of all observe that $W \cap W' = (0)$ because if $\alpha v \in W'$ for some $\alpha \in k$ then owing to the fact that $\alpha v \perp v$ and $\text {char} (k) = 0$ we must have $\alpha = 0$ as $B (v,v) \neq 0.$ Secondly, we have $V = W + W'$ as for any vector $x \in V$ the vector $x - \frac {B(x,v)} {B(v,v)} v \in W'$ so that $$x = \frac {B(x,v)} {B(v,v)} v + \left (x - \frac {B (x,v)} {B(v,v)} v \right ) \in W + W'.$$ Hence we find that $V = W \oplus W'.$ Also $B \big\rvert_{W' \times W'}$ is non-degenerate because if there exists $0 \neq z \in W'$ such that $B(z, W') = \{0\}$ then $B(z, V) = \{0\}$ since $z \perp W$ with respect to $B$ and $V = W \oplus W',$ contradicting the non-degeneracy of $B.$ Since $\dim W = 1$ it follows that $\dim W' = n - 1$ as $\dim V = n.$ So by induction hypothesis it follows that $W'$ has an orthonormal basis say $\{y_1, y_2, \cdots, y_{n - 1} \}.$ But then $\{v, y_1, \cdots, y_n \}$ would be the desired orthonormal basis for $V.$
QED