For real Lie algebras, is any invariant bilinear form a scalar multiple of the Killing form? I know that for a complex Lie algebra, Schur's lemma can be used to show that any invariant bilinear form on a simple Lie algebra is a scalar multiple of the killing form, but Schur's lemma does not hold for $\mathbb{R}$, so I guess this isn't true for real simple Lie algebras. Does someone know a example? I can't seem to find one.
 A: Actually for a Lie algebra $\mathfrak{g}$ over a field $K$ and a field extension $K\subset L$, we have (by a straightforward linear algebra argument), with hopefully self-explanatory notation $$\dim_K \mathrm{InvBil}_K(\mathfrak{g})=\dim_L \mathrm{InvBil}_L(\mathfrak{g}\otimes_KL).$$
Consequence: for $\mathfrak{g}$ semisimple and $K$ of characteristic zero, $\dim_K \mathrm{InvBil}_K(\mathfrak{g})$ is equal to the number of simple factors of the "complexification" $\mathfrak{g}\otimes_K\bar{K}$.
In particular, $\dim_K \mathrm{InvBil}_K(\mathfrak{g})=1$ if and only if $\mathfrak{g}$ is absolutely simple.
Hence, to get counterexamples, it is enough to exhibit simple Lie algebras that are not absolutely simple. If $L$ is a finite extension of $K$ of degree $\ge 2$ and $n\ge 2$, $\mathfrak{sl}_n(L)$, viewed as Lie algebra over $K$, is such a Lie algebra.
A: Let $\mathfrak{g}\subset\mathfrak{gl}_n(\mathbb{C})$ be any complex Lie subalgebra such that $\mathfrak{g}_\mathbb{R}$, the “realification” of $\mathfrak{g}$ (just $\mathfrak{g}$ viewed as a real Lie algebra) has nondegenerate Killing form. In particular, $\mathfrak{g}\neq 0$. One case in which this property holds is whenever $\mathfrak{g}_\mathbb{R}$ is simple: see this post. (Also we have that $\mathfrak{g}_\mathbb{R}$ is a real simple Lie algebra iff $\mathfrak{g}$ is a simple complex Lie algebra, as they explain here.) A concrete example of this case may be the simple complex Lie algebra $\mathfrak{g}=\mathfrak{sl}_2(\mathbb{C})$.
Under the hypothesis “$\kappa_{\mathfrak{g}_\mathbb{R}}$ is nondegenerate” it happens that $\kappa_{\mathfrak{g}_\mathbb{R}}(\cdot,\cdot)$ and $\kappa_{\mathfrak{g}_\mathbb{R}}(i\cdot,\cdot)$ will be a counterexample (this idea was suggested in a comment by Torsten Schoeneberg). Specifically, we claim that these two are real-valued, real-bilinear, symmetric and invariant forms. On this case, then there is no real $\lambda$ such that $\lambda\kappa_{\mathfrak{g}_\mathbb{R}}(x,y)=\kappa_{\mathfrak{g}_\mathbb{R}}(ix,y)$ for all for $x,y\in\mathfrak{g}_\mathbb{R}$. Indeed, if such a $\lambda$ existed, then $0=\kappa_{\mathfrak{g}_\mathbb{R}}((i-\lambda)x,y)$, and by non-degeneracy of $\kappa_{\mathfrak{g}_\mathbb{R}}$ (that's our key hypothesis), it would follow that $(i-\lambda)x=0$, and since $\mathfrak{g}\neq 0$, that $\lambda=i$. But we were assuming $\lambda$ was real.
It is clear that $\kappa_{\mathfrak{g}_\mathbb{R}}(\cdot,\cdot)$ satisfies the required properties. It is also clear that $\kappa_{\mathfrak{g}_\mathbb{R}}(i\cdot,\cdot)$ is real-valued and real-bilinear. The proof of symmetry and invariance of $\kappa_{\mathfrak{g}_\mathbb{R}}(i\cdot,\cdot)$ is the typical given for a Killing form. Plus one observation: for $x\in\mathfrak{g}_\mathbb{R}$, the operators $\operatorname{ad}_{\mathfrak{g}_\mathbb{R}}(x)$ and “multiplication by $i$” (which is a real-linear endomorphism on the realification of any complex vector space) commute, due to the fact that the Lie bracket on $\mathfrak{g}_\mathbb{R}$ is just the commutator of matrices in $\mathfrak{gl}_n(\mathbb{C})$.
