Is the homology of the based loop space of a compact globally symmetric space a polynomial ring?

Question

Let $X$ be a space. Then the homology group $H_*(\Omega X;\mathbb{Q})$ of the based loop space of $X$ is a $\mathbb{Q}$-algebra with the Pontryagin product given by loop concatenation.

When $X=G$ is a compact simply-connected Lie group, we know that $H_*(\Omega G;\mathbb{Q})$ is a polynomial algebra concentrated in even degrees. (In fact, I don't have any reference for this fact. I guess it is proved by showing that the loop concatenation and the group structure of $G$ give rise to a Hopf algebra structure, and then using the classification. Please correct me if I am wrong.)

Now my question is how about when $X=G/H$ is a compact globally symmetric space: Is $H_*(\Omega X;\mathbb{Q})$ also a polynomial algebra?

Remark: I ask this question because it seems to me that when talking about loop spaces, compact Lie groups and symmetric spaces (where the first is a special case of the second) should be treated in a unified way. For example, if we are only interested in $H_*(\Omega X;\mathbb{Q})$ as vector space, then by the work of R. Bott and later W. Ziller, we know a basis such that the degree of each of the vectors is completely determined by the (restricted) root datum associated to the symmetric space $X$.

Answer 1

It is not. For an explicit counterexample we can take $X = \mathbb{CP}^2$. The main technical result we'll use to do the computation is the following, which is Theorem 16.13 in Felix, Halperin, and Thomas' Rational Homotopy Theory.

Theorem: Let $X$ be simply connected such that $H_{\bullet}(X, \mathbb{Q})$ has finite type (finite-dimensional in each degree). Then $H_{\bullet}(\Omega X, \mathbb{Q})$ is the universal enveloping algebra (in the graded sense) of the (graded) Lie algebra $\pi_{\bullet}(\Omega X, \mathbb{Q})$ given by the rational homotopy groups of $X$ (shifted by one degree) with the Whitehead bracket.

The rational homotopy groups of $\mathbb{CP}^2$ can be determined as follows. We know that it's simply connected, and the Hurewicz theorem gives $\pi_2(\mathbb{CP}^2) \cong \mathbb{Z}$. The long exact sequence in homotopy applied to the fiber sequence $S^1 \to S^5 \to \mathbb{CP}^2$ gives that its higher homotopy groups are the same as those of $S^5$; rationally this is a $\mathbb{Q}$ in degree $5$ and nothing else, so overall we get that the rational homotopy of $\mathbb{CP}^2$ is a $\mathbb{Q}$ in degrees $2$ and $5$, and nothing else. Taking loop spaces we get that the rational homotopy of $\Omega \mathbb{CP}^2$ is a $\mathbb{Q}$ in degrees $1$ and $4$, and nothing else; in particular all Whitehead brackets vanish.

It follows that $H_{\bullet}(\Omega \mathbb{CP}^2, \mathbb{Q})$ is the free graded-commutative algebra on generators $\alpha, \beta$ in degrees $1$ and $4$. This algebra is not a polynomial algebra, because graded-commutativity (equivalently, the vanishing of the Whitehead bracket $[\alpha, \alpha] = 2 \alpha^2$) forces $\alpha^2 = 0$, so this algebra has nilpotents.