In the introduction of some article, there is the following statement:
Let $\mathbf{G}$ be a semisimple linear connected algebraic group defined over $\mathbb{Q}$. Let $G = \mathbf{G}(\mathbb{R})$ be the real locus of G, which is a Lie group with finitely many connected components. Let $K \subset G$ be a maximal compact subgroup of G.
Then $X = G/K$ is a symmetric space of noncompact type
Why is $X$ a symmetric space and why of the noncompact type?
In order to make $G/K$ a symmetric space, one needs an involution $\sigma \colon G \to G$ such that $(G^\sigma)^0 \subseteq K \subseteq G^\sigma$. Where does this come from?$^{(1)}$
Also in order to be of the noncompact type, the Lie algebra $\mathfrak{g}$ of $G$ needs to be non-compact and if $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ is the decomposition into the eigenspaces of $\Theta = d_e\sigma$ and $B$ the Killing form of $\mathfrak{g}$, then $$ B_\Theta(X,Y) := -B(X,\Theta(Y)) $$ has to be positiv definite. I don't see why this is true.
$(1)$ Please note that $Z(G)$ is not assumed to be finite.