I know nothing about computability theory, but I don't understand why determining if $\epsilon \geq 0$ gives you some kind of contradiction if, as you assert, $\min(0,\epsilon)$ is computable. What am missing?

Your action on $SU(4)$ is not an action on $SU(4)$. For example, the permutation $(12)$ changes the determinant of an element of $SU(4)$ to $-1$. Are you also, say, multiplying the last column by $(-1)^{|\sigma|}$ for a permutation $\sigma$?

@YCor: Your last message was all I needed - sorry for misunderstanding your previous claim. And thanks for your patience. (And for when I read this in the future: if $A,B\in N_{O(2n)}(\langle J_0\rangle)\setminus \langle J_0\rangle$, then it's easy to compute that $AB$ commutes with $J_0$, hence $AB\in U(n)$, hence $U(n)\subseteq N_{O(2n)}(\langle J_0\rangle)$ has index at most $2$. Since $U(n)\subseteq N_{O(2n)}(U(n))\subseteq N_{O(2n)}(\langle J_0\rangle)$, the index $2$ claim now follows.)

@YCor: It doesn't seem to be true that a J-element in $\mathrm{GL}_{2n}(\mathbf{R})$ necessarily commutes with $J_0$ (though certainly all J-elements in $\mathrm{GL}_{n}(\mathbf{C})$ do!). Consider $J_0 = \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$ and $J = \begin{bmatrix} 0 & 2\\ -1/2 & 0\end{bmatrix}.$ Anyway, I'll continue to think about this tomorrow. (If you'd like to continue via email, I'd be happy to. I'm also happy to continue in the comments, but the OP may be getting sick of all the pings.)

@YCor: If you don't mind, I'm still struggling with your proof sketch that my answer was wrong. Is the result that the normalizer of $U(n)$ is the normalizer of $\langle J\rangle$ some kind of general property of normalizers-of-centralizers, or something unique to this $U(n)$? Also, it's still not clear to me why this would force the index to be at most 2. (I do understand the result of the argument. Sorry to continue to be a bother, and thanks for your time!)

What's a reference for the statement for the case $G\in \{SU(n), Sp(n)\}$?

@Ian: They are not homogeneous unless $q=\pm 1$. This is found in Wolf's classification of space forms. I'm not sure why I wrote that above. (Also, my claim that homogeneous spaces admit Einstein metrics is wrong, as well).

I see that is misinterpreted - $d$ is the order of $\Sigma \in \theta_n$, not the order of $\theta_n$. In addition, when $k$ is odd, I think the lens spaces (in any dimension) are trivially spin, because I think for any such such space $L$, $H^\ast(L;\mathbb{Z}/2\mathbb{Z})\cong H^\ast(S^n;\mathbb{Z}/2\mathbb{Z})$. (I am also reading this a year later - so take this with a grain of salt!)

I don't think your first sentence is true. That is, not all left-invariant metrics on Lie groups are non-negatively curved, even in the compact simply connected case. Namely, the Berger metrics on $S^3$ (obtained by scaling the usual round metric in the direction of the Hopf fibers) has $2$-planes of negative curvature if you make the Hopf circles too large.

@C.F.G. I agree it's a conjecture of Hopf. Not sure why I wrote Bott there (though there is an important conjecture due to Bott regarding non-negatively curved simply connected compact manifolds)

@R.Rankin. Yes, for Lie groups, the tangent space at each point is spanned by left invariant vector fields. My answer shows that this does not occur in general for homogeeous spaces (and note that a Lie group is very special kind of homogeneous space.)

I just wanted to add that $\pi_{2k}(G)\otimes \mathbb{Q} = 0$ for any $k> 0$ and any Lie group $G$. To see this, one can reduce to connected and compact groups using the fact that a non-compact connected Lie group is topologically a product of $\mathbb{R}^n$ with a compact Lie group. Every compact Lie group is know to have the rational homotopy type of a product of spheres odd dimension (and we know exactly what spheres appear for each simple group), so we know how to compute the rational homotopy groups of any Lie group.

My board is currently covered with attempts at using Freudenthal in exactly the way you suggest (and with exactly the same predicted answer). I must just be writing down the wrong commutative diagram...