I wanted to show that having such a bilnear form is the same as being reductive but this tends to be false.

Yes you are right, It works for semi simple algebras because the non degenerate Killing form induces on an ideal, the Killing form of the ideal.

If you have an invariant, symmetric and non-degenerate bilinear form as stated, and, if $\mathfrak i$ is an ideal in $\mathfrak g$. Then, the orthogonal $\mathfrak i ^{\perp}$ of $\mathfrak i$ is also an ideal and your algebra $\mathfrak g$ splits as $\mathfrak i + \mathfrak i ^\perp$. So, for a non reductive algebra, you don't even have one such a bilinear form.

It does not split as a Lie algebra. In general you found yourself with a semi direct product.

Yes, sorry but it's not clear in your question. The phrase "Compute the Lehmer-Permutation πk from k on n numbers" is really confusing.

So you mean you first take a word consisting of numbers, say $w$. Then, from $w$ you have the Lehmer encoding that gives you a permutation $\sigma$. Thirdly, you apply $\sigma$ to the ordering of your letter in your intial word $w$ ?

For $f(x) = x^2$, and $f(y) = y^2$ the Jacobian is $4xy$ vanishing at 0

Taking such an elliptic familly on $Y$ parametrized by $X$. For any $x$ in $X$, you have a legit elliptic pseudo-differential operator on $Y$. This way you can reduce $KK(Y, X)$ to the $K$ homology of $Y$. This is induced on $KK$ level by the map $ev_x : C(X) \to \mathbb C$

And by such not of research level. See MathSE for questions of this type.

Then take $x = (1, 0, \cdots, 0)$. The point is that your question is just on comparaison of classical norms in a finite dimensional real vector space.

oops, indeed, thanks for pointing this out

The special unitary group has a center of dimension $n-1$ no ? Not finite in any case but $n=1$

@DenisNardin You mean by taking action on the whole $\mathbb C$ ?