Homotopy groups of quotient of SU(n)

Question

Let $X$ be the quotient topological space obtained by identifying the matrices $A$ and $\overline{A}$ in the topological group $\mathrm{SU}(n)$ (here $\overline{A}$ denotes entry-wise complex conjugation). Are there any techniques for finding the second homotopy group of $X$? For example, is the quotient map $p:\mathrm{SU}(n) \rightarrow X$ a fibration?

Answer 1

Here is an answer to a related question, which might or might not be useful. There is a map $q\colon SU(n)\to S^{2n-1}$, sending a matrix $A$ to its last column. This induces a map $X\to Y$, where $Y$ is the conjugation-quotient of $S^{2n-1}$, so it is potentially interesting to understand the homotopy type of $Y$. We can identify $\mathbb{C}^n$ with $U\oplus V$, where $U=\mathbb{R}^n$ and $V=i\mathbb{R}^n$, so the conjugation action is $(u,v)\mapsto(u,-v)$. The sphere $S^{2n-1}=S(U\oplus V)$ can be identified with the join $S(U)*S(V)=S^{n-1}*S^{n-1}$, and this gives a homeomorphism $Y=S^{n-1}*\mathbb{R}P^{n-1}$, and thus a homotopy pushout square $\require{AMScd}$ \begin{CD} S^{n-1}\times\mathbb{R}P^{n-1} @>>> \mathbb{R}P^{n-1} \\ @VVV @VVV\\ S^{n-1} @>>> Y. \end{CD} In general, suppose we have spaces $A$ and $B$ with nondegenerate basepoints $a_0$ and $b_0$. Then in $A*B$ we have contractible subsets $A*\{b_0\}$ and $\{a_0\}*B$ whose intersection is the interval $\{a_0\}*\{b_0\}$, so the union of these two subsets is still contractible, so we can collapse it out without changing the homotopy type. This gives $A*B\simeq\Sigma(A\wedge B)$. By applying this to the case at hand, we get a homotopy equivalence $Y\simeq\Sigma^n\mathbb{R}P^{n-1}$. If $n=1$ we see that $Y$ is contractible (which is easy to see directly). If $n=2$ then $\mathbb{R}P^{n-1}=S^1$ and $Y\simeq S^3$. If $n>2$ then the first nontrivial homotopy group of $Y$ is $\pi_{n+1}(Y)\simeq H_{n+1}(Y)\simeq \mathbb{Z}/2$.

If we let $X'$ be the conjugation-quotient for $SU(n-1)$, we have an evident sequence $X'\xrightarrow{j}X\xrightarrow{q}Y$. However, one can check that $q^{-1}\{u\}$ is homeomorphic to $X'$ if $u$ is real and to $SU(n-1)$ if $u$ is not real. This means that $q$ is not a fibre bundle projection, so it is not so easy to extract information about $X$ from information about $Y$. It may be better to compare $X/SO(n)$ with $Y/S^{n-1}$ and then work back to $X$. (This is related to Moishe Kohan's suggestion in the comments.)