Let $G$ be a simply connected Lie group with Lie algebra $\mathfrak{g}$ and $K$ a connected closed Lie subgroup of $G$ with Lie algebra $\mathfrak{s}$. Then $G/K$ is a homogeneous space. Equip $G$ with a left-invariant Riemannian metric $Q_G$ on $G$ and let $Q_{G/K}$ be the unique Riemannian metric on $G/K$ so that $\pi: G \to G/K$ is a Riemannian submersion (that is, so that $G/K$ is a Riemannian homogeneous space). Let $\mathfrak{h}$ be the orthogonal complement of $\mathfrak{s}$ with respect to $Q_G$, so that $\mathfrak{g} = \mathfrak{s} \oplus \mathfrak{h}$.
- If $Q_G$ happens to be bi-invariant (so that $Q_{G/K}$ is $G$-invariant), then is $G/K$ necessarily symmetric?
- Conversely, if $G/K$ happens to be symmetric, then is $Q_G$ necessarily bi-invariant?
1) I believe this can be proved by showing that Cartan's decomposition holds: $$[\mathfrak{h}, \mathfrak{h}] \subset \mathfrak{s}, \ [\mathfrak{h}, \mathfrak{s}] \subset \mathfrak{h}, \ [\mathfrak{s}, \mathfrak{s}] \subset \mathfrak{s}.$$
Observe that $[\mathfrak{s}, \mathfrak{s}] \subset \mathfrak{s}$ is trivially true since $K$ is a Lie subgroup of $G$. For the other two, I thought to argue as follows:
Let $\xi \in \mathfrak{h}$, $\eta, \sigma \in \mathfrak{s}$. Because the metric is bi-invariant (and hence $\text{Ad}$-invariant),
$$0 = \left< \xi, \sigma \right> = \left< \text{Ad}_{\text{Exp}(t \eta)} \xi, \text{Ad}_{\text{Exp}(t\eta)} \sigma\right>.$$
Taking the derivative at $t = 0$ yields
$$0 = \left< [\eta, \xi], \sigma \right> + \left< \xi, [\eta, \sigma]\right> = \left< [\eta, \xi], \sigma \right>$$
since $[\eta, \sigma] \in \mathfrak{s}$. Hence $[\eta, \xi] \in \mathfrak{h}$, and so $[\mathfrak{h}, \mathfrak{s}] \subset \mathfrak{h}$. Repeating this argument with $\xi, \eta \in \in \mathfrak{h}, \sigma \in \mathfrak{s}$ similarly yields $[\mathfrak{h}, \mathfrak{h}] \subset \mathfrak{s}$. It can be seen that the map $L = \text{Id}_{\mathfrak{s}} – \text{Id}_{\mathfrak{h}}$ is an involutive automorphism of $\mathfrak{g}$ and since $G$ is simply connected, there exists an involutive automorphism $\sigma: G \to G$ with $K = G_{\sigma}$ (the fixed point set of $\sigma$). It follows that $G/K$ is symmetric.
2) I believe this is false. We can take $K = \{e\}$, from which we get $G/K \cong G$, and this post seems to suggest that there exist Lie groups which do not admit bi-invariant metrics but nevertheless symmetric spaces. Though the answer avoids taking a firm stand one way or another, and just proposes a strategy that would yield a counter-example if one existed.
Best Answer
The answer to both questions is no.
Let $G = SU(3)$, the group of $3\times 3$ unitary complex matrices with determinant $1$. In symbols, $G = \{A\in M_3(\mathbb{C}): A\overline{A}^t = I, \det(A) = 1\}$.
Let $K\subseteq G$ be defined as $K = \{A\in SU(3):Ae_3 = e_3\}$ where $\{e_1,e_2,e_3\}$ is the standard $\mathbb{C}$-basis of $\mathbb{C}^3$.
The natural action of $G$ on $\mathbb{C}^3$ is transitive on the unit sphere $S^5\subseteq \mathbb{C}^3$. The stabilizer is $K$, so we have a diffeomorphism $G/K\cong S^5$.
Proposition: Suppose we equip $G$ with bi-invariant metric. Then the induced metric on $S^5$ is not symmetric. In particular, the answer to question 1 is "No".
Proof: Let's show Cartan's decomposition does not hold.
To that end, first note that $G$ is simple, so the bi-invariant metric is unique up to scaling. And the bi-invariant metric is easy to write down. On the Lie algebra $\mathfrak{g} = \{X\in M_3(\mathbb{C}): X + \overline{X}^t = 0, tr(X) = 0\}$ of $G$, we have $\langle X,Y\rangle = -\operatorname{Re}(Tr(XY))$.
Now, we can identify $\mathfrak{s}$ with the set of matrices of the form $\begin{bmatrix} \alpha i & b & 0\\ -\overline{b} & -\alpha i& 0 \\ 0& 0 & 0 \end{bmatrix}$, where $\alpha\in\mathbb{R}$ and $b\in\mathbb{C}$.
A computation then reveals that $\mathfrak{h}$ consists of all matrices of the form $$\begin{bmatrix} \beta i & 0 & c\\ 0 & \beta i & d\\-\overline{c} & -\overline{d} & -2\beta i\end{bmatrix}$$ where $\beta \in \mathbb{R}$ and $c,d\in \mathbb{C}$.
Now, consider $X,Y\in \mathfrak{h}$ with $X = \begin{bmatrix} i & 0 & 0\\ 0 &i & 0\\0 & 0 & -2i\end{bmatrix}$ and $Y =\begin{bmatrix} 0 & 0& 0\\ 0 & 0 & 1\\ 0 & -1 & 0\end{bmatrix}.$
Then $[X,Y] = XY-YX = \begin{bmatrix}0 & 0 & 0\\0 &0 &3i\\0 &3i &0 \end{bmatrix}$, which is not an element of $\mathfrak{s}$. $\square$
By the way, your proof that $[\mathfrak{h},\mathfrak{s}]\subseteq \mathfrak{h}$ is fine, but it doesn't generalize to showing $[\mathfrak{h},\mathfrak{h}]\subseteq \mathfrak{s}$ as the example above indicates.
Proposition: There is a left invariant metric $Q_G$ on $G$ for which the induced metric on $G/K\cong S^5$ is symmetric. In particular, the answer to question $2$ is also "No".
Proof: If we equip $S^5$ with the symmetric (round) metric $Q_{S^5}$, then the natural action by $G$ is isometric.
Of course, $Q_{S^5}$ restrics to an inner product on $T_{I K} G/K$, which is then naturally identified with $\mathfrak{h}$. Hence, we obtain an inner product $Q_\mathfrak{h}$ on $\mathfrak{h}$.
Let $Q_{\mathfrak{s}}$ denote the inner product corresponding to a bi-invariant metric on $K$. And define $Q_\mathfrak{g} = Q_\mathfrak{s} + Q_\mathfrak{h}$. Finally, let $Q_G$ be the left-invariant metric on $G$ whose value at the identity is $Q_\mathfrak{g}$.
Then, by construction, $Q_G$ is is left invariant and the induced metric on $S^5$ is $Q_{S^5}$, so is symmetric.$\square$