Self-dual Lie Algebra representation

Question

If $V$ is a Lie algebra $L$ module, then I want to know whether if $V$ and $V^*$ are isomorphic as $L$ modules then there is a basis for $V$ in which the matrices representing the action of $L$ are all skew-symmetric. If there is a basis $(e_i)$ for which the isomorphism takes $e_i$ to $e_i^*$ then I can do the problem. But is there such a basis?

Answer 1

Suppose that $L$ is a real semi-simple algebra. Let $K$ its killing form, it is not degenerated, there exists an isomorphism $L\rightarrow L^*$ defined by $L(x)=K(x,.)$, but if $L$ is for example $sl(2,\mathbb{R})$, the matrices of the adjoint representation are not all skew-symmetric for a given basis since you have an element $h$ such that $ad_h$ has $2$ as an eigenvalue.

Answer 2

While Tsemo Aristide's answer shows via striking examples that the answer is "no", I would like to quickly explain what I think is the mistake in the question. Full disclosure: This is partly due to me making the same mistake at first when thinking about the related question Under what conditions can we find a basis of a Lie algebra such that the adjoint representation acts by skew-symmetric matrices? today.

An isomorphism of $\mathfrak g$-modules $V \simeq V^*$ indeed is equivalent to the existence of a non-degenerate $\mathfrak g$-invariant bilinear form $\langle \cdot , \cdot \rangle$ on $V$ (NB, $\mathfrak g$-invariant for Lie algebra representations i.e. $\mathfrak g$-modules means $\langle x v ,w\rangle = -\langle v, x w\rangle$). (Compare also https://math.stackexchange.com/a/3781591/96384 .)

Further, in the case of a simple $\mathfrak g$-module, and $\mathfrak g$ being absolutely simple, this form is even unique up to a scalar (as follows from Schur's Lemma.)

Now in general this form does not need to be symmetric, it can be alternating. E.g. for the fundamental representation of $\mathfrak {sl}_2(\mathbb C)$ on $V:=\mathbb C^2$, check that up to a scalar, that non-degenerate bilinear form is the alternating $\langle \pmatrix{x\\y}, \pmatrix{w\\z} \rangle := \det(\pmatrix{x &w \\y &z})$. However e.g. for the adjoint representation, one can show that it is always symmetric. Indeed, up to a scalar it is the Killing form.

So why is the assertion still not (generally) true in that case, as Tsemo Aristide's answer and this here show? Because to get to skew-symmetric matrices from a symmetric bilinear form with the method shown by Malkoun in their answer there, you need an orthonormal basis, and for that you need the form to be much more than non-degenerate, you need it to have an orthonormal basis i.e. being equivalent up to scaling to the form given by the symmetric matrix $S= Id$ (where in general, you represent a symmetric BLF $b(\cdot, \cdot)$ by any symmetric matrix $S$ via $b(x,y) := x^{tr}Sy$). Now over an algebraically closed field like $\mathbb C$, every symmetric BLF is equivalent to that standard one. But e.g. over $\mathbb R$, where due to Sylvester's Law of Inertia symmetric BLFs are classified by their index, this is equivalent to the form being anisotropic i.e. (positive or negative) definite. Which over the real numbers it is only in the case of semisimple Lie algebras which come from compact Lie groups. Indeed, since the Killing form is indefinite as soon as our Lie algebra contains nonzero nilpotent elements, this shows that over any subfield of $\mathbb R$ there is no hope to represent all $ad(x)$ for $x$ in a given semisimple Lie algbera with skew-symmetric matrices, unless the Lie algebra contains only semisimple elements i.e. is anisotropic ("compact").