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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!

Anacardium
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    Yes, it means orthonormal with respect to the Killing form, except that the $1$ needs to be a $\pm 1$ because in general the Killing form may be indefinite. – Qiaochu Yuan Feb 16 '23 at 05:45
  • @QiaochuYuan$:$ How to prove existence of such orthogonal basis? Will the Gram-Schmidt orthogonolization work here? BTW, by definiteness of an inner product $(\cdot, \cdot)$ do you mean $(x,x) = 0 \implies x = 0\ $? But how does it affect the orthonormality? I have to revisit the proof of Gram-Schmidt process of orthogonalization once again. Also can I say that $K(x,x) \geq 0\ $? Any suggestion on your end? – Anacardium Feb 16 '23 at 05:52
  • @QiaochuYuan$:$ What I find that if $K(x,x) \neq 0$ then we are still through. But how to guarantee that? – Anacardium Feb 16 '23 at 06:17
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    What ground field do we assume here? The Killing form of a simple Lie algebra is a non-degenerate bilinear form. This does not exclude isotropic vectors (especially over $\mathbb C$, where also $K(x,x) \ge 0$ does not make sense in general), but Gram-Schmidt still works. Over $\mathbb R$, there is Sylvester's Inertia. – Torsten Schoeneberg Feb 16 '23 at 06:29
  • @TorstenSchoeneberg$:$ I am working over the field of complex numbers. In this case how to apply Gram-Schmidt unless $K(x,x) \neq 0\ $? It is clear that $K(x,x) \geq 0$ is not relevant and in fact, not needed to apply Gram-Schmidt. – Anacardium Feb 16 '23 at 06:37
  • You just need to find some $x$ with $K(x,x)\neq 0$. If $K(x,x)=0$, by non-degeneracy, to $x \neq 0$ there is $y$ with $K(x,y)\neq 0$ and thus either $K(y,y)\neq 0$ or $K(x+y, x+y) = 2K(x,y)\neq 0$. In any case, you should just find a good linear algebra textbook containing the fact: If $V$ is an $n$-dimensional vector space over an alg. closed field of characteristic $\neq 2$, and $B(\cdot, \cdot)$ a non-degenerate bilinear form on $V$, then there is a basis $e_1, ..., e_n$ of $V$ with $B(e_i, e_j) = \delta_{ij}$. – Torsten Schoeneberg Feb 16 '23 at 06:59
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    @TorstenSchoeneberg$:$ If we start with a basis of $\mathfrak g$ containing $x$ and if $y$ is another element of the basis, then in order to find a vector orthogonal to $x$ we need to subtract from $y$ the projection of $y$ onto the line through $x.$ The vector I am referring to here is $y_1 = y - \frac {K(x,y)} {K(x,x)} x.$ My question is $:$ Can we conclude that $K(y_1, y_1) \neq 0\ $? We have $K(y_1, y_1) = K(y,y) - \frac {{K(x,y)}^2} {K(x,x)}.$ So $K(y_1, y_1) = 0$ if and only if $K(x,y)^2 = K(x,x) K(y,y).$ – Anacardium Feb 16 '23 at 07:21
  • I see the issue. Maybe Gram-Schmidt is not the best way to go. Search this site for "Hermite reduction", Will Jagy has posted some answers on this, it should bring any symmetric bilinear form into diagonal form. Non-degenerate means all diagonal entries are non-zero, and in an algebraically closed field you can scale them to $1$. – Torsten Schoeneberg Feb 17 '23 at 06:05
  • See https://math.stackexchange.com/q/1388421/96384, https://math.stackexchange.com/a/1371679/96384, https://math.stackexchange.com/a/1382315/96384, https://math.stackexchange.com/a/1324099/96384, and more links there. – Torsten Schoeneberg Feb 17 '23 at 06:30

1 Answers1

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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

Anacardium
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Anil Bagchi.
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