Vogan and Satake diagrams give equivalent real form classifications

We first justify choice independence for Satake diagrams. Use $K=(\mathrm{Aut}({\mathfrak{g}}_0)\cap O(B_{\theta }){)}^0$. This is compact and connected by [L7], has Lie algebra $\mathrm{ad}{\mathfrak{k}}_0$, and commutes with $\theta$: preservation of both $B$ and $B_{\theta }$ is equivalent to that commutation. Differentiation, centerlessness and exponentiation give the Lie algebra assertion. Its elements are products of $\exp (\mathrm{ad}X)$ with $X\in {\mathfrak{k}}_0$, since an exponential identity neighborhood generates a connected group. Changes of involution can be absorbed by real automorphisms by [L2]. For fixed $\theta$, the split part ${\mathfrak{a}}_0$ of a maximally split Cartan is maximal abelian in ${\mathfrak{p}}_0$: [L6] gives $Z_{{\mathfrak{g}}_0}({\mathfrak{a}}_0)=\mathfrak{m}\oplus {\mathfrak{a}}_0$ with $\mathfrak{m}\subseteq {\mathfrak{k}}_0$. Thus [L2] supplies $k\in K$ aligning the split parts of two choices. After this alignment their compact parts are maximal abelian in $\mathfrak{m}=Z_{{\mathfrak{k}}_0}({\mathfrak{a}}_0)$, since any further commuting element would centralize the Cartan and hence belong to it by self-normalization. Maximal-torus conjugacy in $Z_K({\mathfrak{a}}_0{)}^0$ then aligns those compact parts without moving ${\mathfrak{a}}_0$. Hence the whole maximally split Cartans are $K$-conjugate. It remains to compare compatible positive systems on one Cartan. [A1, L2, L6, L7, Cartan subalgebra, algebra]

The decorated diagram determines the full root involution $t$. Its action on black roots is the identity. For a white simple root $\alpha$, write $t\alpha =-ϵ\alpha +v$, with $v\in {\mathrm{span}}_{\mathbb{R}}{\mathrm{\Delta }}_0$ by [L3]. For every black $\delta$, orthogonality and $t\delta =\delta$ give $(v,\delta )=(\alpha +ϵ\alpha ,\delta )$. The Gram matrix on the independent black roots is positive definite, so these equations determine $v$ uniquely. The root metric can be chosen canonically from the Killing form on each simple component: the trace formula $B(H,H^{\mathrm{\prime }})={\sum }_{\gamma \in \mathrm{\Phi }}\gamma (H)\gamma (H^{\mathrm{\prime }})$ shows that a based root isomorphism preserves this normalization, including permuted isomorphic components. Thus a decorated diagram isomorphism intertwines the two root involutions.

Each reflection of a restricted-root hyperplane is realized in $K$. Choose $0\mathrm{\ne }E\in {\mathfrak{g}}_0^{\lambda }$. By [L6], $[E,\theta E]$ lies in ${\mathfrak{a}}_0$, and pairing it with $H\in {\mathfrak{a}}_0$ gives $[E,\theta E]=B(E,\theta E)H_{\lambda }$, where $B(H_{\lambda },H)=\lambda (H)$ and $B(E,\theta E)<0$. Rescale $E$ positively so this coefficient is $-2\mathrm{/}\mathrm{\mid }\lambda {\mathrm{\mid }}^2$, and put $h=2H_{\lambda }\mathrm{/}\mathrm{\mid }\lambda {\mathrm{\mid }}^2$. Then $(E,-\theta E,h)$ is an ${\mathfrak{s}\mathfrak{l}}_2$ triple. The automorphism $k=\exp ((\pi \mathrm{/}2)\mathrm{ad}(E+\theta E))$ belongs to $K$, fixes $\ker \lambda$ in ${\mathfrak{a}}_0$, and sends $h$ to $-h$: on $h$ and $E-\theta E$ the adjoint generator gives a rotation through $\pi$. Thus it induces the reflection and permutes restricted roots. These reflections act transitively on restricted chambers: a generic polygonal path between two regular points crosses one distinct root hyperplane at a time, and reflection in the crossing wall interchanges its two adjacent chambers while preserving the entire finite arrangement. Induction on the crossings proves transitivity. Coincident hyperplanes from $\lambda$ and $2\lambda$ are a single wall, so nonreduced systems cause no problem.

Suppose two real forms have isomorphic Satake diagrams. In each choose normalized simple triples with compact conjugation $\kappa =\theta \sigma$: the argument in [L1] uses only theta-stability of the Cartan, not maximal compactness. Explicitly $\mathfrak{u}={\mathfrak{k}}_0\oplus i{\mathfrak{p}}_0$ is compact, $-B(Z,\kappa Z)>0$, and after positive rescaling $f_i=-\kappa e_i$ satisfies $[e_i,f_i]=h_i$. Match their based diagrams and use [L5] to identify all $e_i,f_i,h_i$. This identifies the compact conjugations, since their actions are $e_i\mapsto -f_i$, $f_i\mapsto -e_i$, $h_i\mapsto -h_i$, antilinearly. Hence work in one complex algebra with one $\kappa$ and two compact-preserving involutions $\theta ,{\theta }^{\mathrm{\prime }}$ having the same root action $t$, by step 1.2.

Choose a product of these reflections to align the restricted positive systems. Its representative $k\in N_K({\mathfrak{a}}_0)$ might move ${\mathfrak{t}}_0$. Both $k{\mathfrak{t}}_0$ and ${\mathfrak{t}}_0$ are maximal abelian in $\mathfrak{m}$: any further commuting compact vector centralizes the Cartan and hence belongs to it. The compact group $M=Z_K({\mathfrak{a}}_0)$ has Lie algebra $\mathrm{ad}\mathfrak{m}$. Applying maximal-torus conjugacy in $M^0$, using the maximal-abelian/torus correspondence of [L2], gives $m\in M^0$ aligning these compact parts. Thus $mk$ normalizes the whole Cartan while still aligning restricted positives. After transport, the two compatible positive systems differ only on imaginary roots. Those roots form the reduced root subsystem $\mathrm{\Phi }\cap {\mathrm{span}}_{\mathbb{R}}{\mathrm{\Delta }}_0$: a positive root restricting to zero has no white coefficient, since each nonzero simple restriction is positive on a common regular split vector. Apply [L7] to this subsystem to align its positive systems. Each of its reflections fixes ${\mathfrak{a}}_0$ and commutes with $t$, so preserves all nonzero root restrictions. Consequently it aligns the full positive systems and preserves the black vertices and pair relation. Their Satake diagrams are therefore isomorphic. Empty restricted or imaginary systems require no reflections. Real isomorphisms transport all these data, proving the Satake assignment is well defined.

The automorphism $A={\theta }^{\mathrm{\prime }}{\theta }^{-1}$ fixes the Cartan pointwise and multiplies each root line by a scalar $c(\alpha )$. These scalars extend to a character $c:Q\to U(1)$. Indeed the simple scalars define a character on the free lattice; the corresponding diagonal assignment $e_i\mapsto c_i{e}_i$, $f_i\mapsto c_i^{-1}{f}_i$, $h_i\mapsto h_i$ preserves all Serre relations and agrees with $A$ on the generators, so agrees everywhere. Compact preservation gives $\mathrm{\mid }{c}_i\mathrm{\mid }=1$ by pairing $e_i$ with $-\kappa e_i$. From ${\theta }^{\mathrm{\prime }2}=1$ follows $c(t\lambda )=c(\lambda {)}^{-1}$. On every black simple root both involutions act as $+1$, because the root is compact imaginary, so $c$ is trivial on $Q_0$.

The character $c$ is trivial on $Q^t=\{\lambda \in Q:t\lambda =\lambda \}$. To prove it, reduce the simple-root coefficient vector of such a $\lambda$ modulo $Q_0$. By [L3], $t$ acts on white classes by minus the permutation $ϵ$. A fixed vector has zero coefficient at every unpaired white vertex and opposite coefficients at the two vertices of each pair. For a chosen member $\alpha$ of each pair, the invariant vector $\alpha +t\alpha$ has white coefficient vector $[\alpha ]-[ϵ\alpha ]$. Subtract the appropriate integral multiples of these invariant vectors from $\lambda$; the remainder lies in $Q_0$. But $c(\alpha +t\alpha )=c(\alpha )c(t\alpha )=1$ by step 3.2 and $c(Q_0)=1$, proving the claim. This uses the invariant lattice, not just its real span.

Define a character on $(t-1)Q$ by $b((t-1)\lambda )=c(t\lambda )$. It is well defined: two preimages differ by a member of $Q^t$, on which $c$ is one. It extends to a character $b:Q\to U(1)$. In fact Smith normal form in [L5] gives a basis $q_1,\dots ,q_r$ of $Q$ in which $(t-1)Q$ is generated by $d_1{q}_1,\dots ,d_s{q}_s$, with positive integers $d_j$. Choose a $d_j$-th root in $U(1)$ of each prescribed value to define $b(q_j)$, and set the remaining basis values to one. Only finitely many roots are selected. Let $D$ be the diagonal Serre automorphism with root scalar $b(\alpha )$. Unit modulus implies $D\kappa =\kappa D$ on all generators. On the $\alpha$ root line, $D\theta D^{-1}$ differs from $\theta$ by $b(t\alpha )\mathrm{/}b(\alpha )=c(t\alpha )$, exactly the factor in ${\theta }^{\mathrm{\prime }}=A\theta$. The maps also agree on the Cartan, hence $D\theta D^{-1}={\theta }^{\mathrm{\prime }}$. Consequently $D$ intertwines $\sigma =\theta \kappa$ with ${\sigma }^{\mathrm{\prime }}={\theta }^{\mathrm{\prime }}\kappa$ and restricts to an isomorphism of real forms. Satake classes are therefore injective on real isomorphism classes.

By [L1] Vogan classes classify the real forms. By step 3.1 and step 5.1 Satake classes do also, with surjectivity onto the realized classes by their definition. Composing these two classifications gives the asserted equivalences and bijection. The zero algebra has one empty diagram; all-black diagrams have $Q_0=Q$, while all-white diagrams have $Q_0=0$, and the invariant-lattice proof treats both. Rank-one lattices and permuted components are covered by the same finite-basis argument. The stated difference between the decorations is their defining choice of Cartan and root labels, not an assertion that their drawings coincide.