Same complexification with different killing form signatures

Given: The two real Lie algebras $\mathfrak{s}\mathfrak{u}(2)$ and ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$, both real forms of $\mathfrak{s}={\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$, and the basis $e,f,h$ of $\mathfrak{s}$ with the Killing form $B$.

$\mathfrak{s}\mathfrak{u}(2)$ and ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ are real forms of $\mathfrak{s}={\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$; $\mathfrak{s}\mathfrak{u}(2)$ is a compact real form and ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ is a split real form (Compact and split real forms of sl two c, Compact real form of a complex semisimple Lie algebra, Split real form).

The Killing form of ${\mathfrak{s}\mathfrak{l}}_2$ satisfies $B(h,h)=8$, $B(e,f)=B(f,e)=4$, and all other pairings of the basis $e,f,h$ vanish; equivalently $B(X,Y)=4\mathrm{tr}(XY)$ (Killing form of sl_2, Killing form).

The inertia $(p,q,r)$ of a real symmetric bilinear form is a congruence invariant and classifies such forms in a fixed dimension: two forms are congruent exactly when their inertias agree (Sylvester's law of inertia: every real symmetric form is congruent to $\mathrm{diag}(I_p,-I_q,0_r)$, and $(p,q,r)$ is unique, Two real symmetric bilinear forms are congruent if and only if they have the same inertia, Positive and negative definiteness, the inertia $(p,q,r)$, rank $p+q$, and signature $p-q$ of a real symmetric bilinear or quadratic form).

Proof technique: direct computation of the two Killing forms.

1.1 The two algebras have the same complexification: by [L1] both $\mathfrak{s}\mathfrak{u}(2)$ and ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ are real forms of $\mathfrak{s}={\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$, so their complexifications are both isomorphic to $\mathfrak{s}$. [L1]

1.2 The Killing form of $\mathfrak{s}\mathfrak{u}(2)$ has inertia $(0,3,0)$: for $X\in \mathfrak{s}\mathfrak{u}(2)$ one has $X^{\ast }=-X$, and [L2] gives $B(X,X)=4\mathrm{tr}(X^2)=-4\mathrm{tr}(X{X}^{\ast })=-4{\sum }_{i,j}\mathrm{\mid }{X}_{ij}{\mathrm{\mid }}^2$, which is negative for every nonzero $X$ and zero only at $X=0$; hence $B$ is negative definite on the three-dimensional space $\mathfrak{s}\mathfrak{u}(2)$, with no positive and no null directions. [L2, algebra]

1.3 The Killing form of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ has inertia $(2,1,0)$: in the basis $(h,e,f)$ the Gram matrix of $B$ is $\left(\begin{array}{ccc}8& 0& 0\\ 0& 0& 4\\ 0& 4& 0\end{array}\right)$ by [L2], whose characteristic polynomial is $(8-\lambda )({\lambda }^2-16)$, so the eigenvalues are $8$, $4$ and $-4$; a symmetric matrix is diagonalized by an orthogonal change of basis, so the form has two positive and one negative square and is nondegenerate. [L2, algebra]

2.1 The two forms are not congruent: their inertias $(0,3,0)$ and $(2,1,0)$ differ, and by [L3] congruent forms of the same dimension have equal inertia. [step 1.2, step 1.3, L3]

3.1 No Lie-algebra isomorphism can exist between them: if $\phi \text{\hspace{0.17em}⁣}:{\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})\to \mathfrak{s}\mathfrak{u}(2)$ were an isomorphism, then ${\mathrm{ad}}_{\phi (X)}=\phi \circ {\mathrm{ad}}_X\circ {\phi }^{-1}$ would give $B_{\mathfrak{s}\mathfrak{u}(2)}(\phi X,\phi Y)=\mathrm{tr}({\mathrm{ad}}_{\phi X}{\mathrm{ad}}_{\phi Y})=\mathrm{tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)=B_{{\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})}(X,Y)$, so the two Killing forms would be congruent via the invertible matrix of $\phi$, contradicting step 2.1. [step 2.1, L2, algebra]

4.1 Consequently the complexification does not determine the inertia of the Killing form: the real forms $\mathfrak{s}\mathfrak{u}(2)$ and ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ of the same complex algebra ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$ carry Killing forms of inertia $(0,3,0)$ and $(2,1,0)$ and are not isomorphic. The compactness of $\mathfrak{s}\mathfrak{u}(2)$ corresponds exactly to the vanishing of the positive part of the inertia, while the split form has a positive-definite subspace of dimension $2$. [step 1.1, step 1.2, step 1.3, step 3.1, L1]

5.1 Endpoints and scope: both algebras are three-dimensional and nondegenerate, so the nullity is $0$ in both cases and the difference is entirely in the signature; the computation is finite, uses the explicit basis of ${\mathfrak{s}\mathfrak{l}}_2$ only, and needs no choice principle. [step 1.2, step 1.3, algebra] ∎