Proof of $\mathsf{SU}(3)/T^2$ is not a symmetric space 
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*How to see that $\mathsf{SU}(3)/T^2$ is not a symmetric space? This is not obvious to me.


*How to see that it admits a metric of positive curvature? The only clue that I know is the submersion $\mathsf{SU}(3)\to \mathsf{SU}(3)/T^2$ if I am not mistaken. Then O'Neill's formula guarantees its curvature is positive?
No matter which definition of symmetric spaces is assumed. e.g.  $\nabla R=0$.
 A: *

*There are (at least) two interpretations of this.  First, you could be asking why $SU(3)/T^2$ does not admit any Riemannian metric which makes it a symmetric space.  Second, you could be asking whether the canonical metric on $SU(3)$ descends to a symmetric metric on $SU(3)/T^2$.

I'll answer the first version since it has the second as a special case.
First note that $SU(3)/T^2$ is simply connected.  Simply connected symmetric spaces are completely classified by Cartan.  The short version is that each is isometric to a product of so-called irreducible pieces and all the possible pieces have been classified up to isometry.  The pieces come in two varieties:  they are either isometric to a compact simple simply connected Lie group $G$ with bi-invariant metric, or given as a quotient $G/H$ where $G$ is simple and simply connected, and $H$ is the identity component fixed point set of non-trivial involution $\sigma:G\rightarrow G$.  From the table under "classification" here, note that for every $G/H$, $H$ has an at most one-dimensional circle factor.  In particular, $G/H$ has second Betti number at most $1$.  Since $SU(3)/T^2$ has second Betti number $=2$ (from the long exact sequence in homotopy groups together with the Hurewicz thoerem), $SU(3)/T^2$ cannot appear on either of these two lists.
So now it's just a matter of showing the $SU(3)/T^2$ is not diffeomorphic to a product of things as in the previous paragraph.  To that end, note that $SU(3)/T^2$ is a simply connected $6$-manifold with second Betti number $2$ with Euler characteristic $6$.  Thus, it cannot be any irreducible piece, so it must be a product.  But, if $SU(3)/T^2$ were diffeomorphic to a product, it would have to be homeomorphic to $S^2\times \mathbb{C}P^2$ by Freedman's classification of simply connected $4$-manifolds.  However, $\pi_4(S^2)\neq 0$, while $\pi_4(SU(3)/T^2) = 0$.  The latter calculation follows since $\pi_4(SU(3))  = 0$, which follows from Bott Periodicity.


*This is tricky because a bi-invariant metric on $SU(3)$ does not descend to a positively curved metric on $SU(3)/T^2$.  Rather, one have to pick a particular left invariant metric which is still $T^2$-invariant on the right.  (In fact, it's $U(2)$ invariant on the right.)

The technique here is called a "Cheeger deformation".  Namely, you consider a subgroup of $K$ in $G$ and look at product metrics on $G\times K$.  There is a natural isometric $K$ action on $G\times K$ given by $k\ast(g,k') = (gk^{-1}, kk')$, and it's not too hard to see that $(G\times K)/K$ is diffeomorphic to $G$.  Since the $K$ action is isometric, this induces a new metric on $G$ which turns out to be left $G$-invariant and right $K$-invariant.  Since it's a submersion metric, O'Neill's formulas imply that the new metric on $G$ is non-negatively curved, but it actually has fewer zero-curvature planes than a bi-invariant metric on $G$, at least if $K$ is non-abelian.
This is the metric you use.  When you try to this on $U(2)\subseteq SU(3)$, what ends up happening is that for orthonormal $X,Y\in T_e G$, the sectional curvature of of the plane spanned by $X$ and $Y$ is non-negative, and is $0$ iff both $[X,Y]=0$ and $[X_\mathfrak{k},Y_{\mathfrak{k}}] = 0$.
For $SU(3)/T^2$ one shows that if there orthornomal $X,Y\in T_{eT^2} SU(3)/T^2$ spanning a zero-curvature plane, then their horizontal lifts to $T_e G$ satisfy the previous two equations.   Of course "horizontal" adds another equation, and one can then show the three equations are incompatible.  (I guess it's really six equations since three more are hiding in "orthonormal".)
