Lie group with constant metric tensor but nonzero curvature?

For any Lie group $\mathcal{G}$, if an inner-product is defined on the tangent space at the identity element $I$, then an inner-product can be induced on the tangent space at any other point $x$ by left-invariance: $$ \big{\langle} u\ \big{|}\ v \big{\rangle}_x := \big{\langle} dL_x^{-1}(u)\ \big{|}\ dL_x^{-1}(v) \big{\rangle}_I $$

where $\langle u\ |\ v\rangle_x$ is the inner-product of any $u,v \in T_x(\mathcal{G})$ and $dL_x$ is the differential of the left group operator $L_x(y) = xy\ \ \forall x,y \in \mathcal{G}$. The resulting collection of inner-products constitute a metric tensor on $\mathcal{G}$, turning it into a Riemannian manifold. (The same mechanism can be applied using the right group operator but I think I only need to address one of the two).

If $\mathcal{G}=\text{SO(3)}$, then $dL_x^{-1}=X^\intercal$ where $X$ is an orthogonal matrix with determinant 1. Thus the left-invariant metric tensor is, \begin{align} \big{\langle} u\ \big{|}\ v \big{\rangle}_x &= \big{\langle} X^\intercal u\ \big{|}\ X^\intercal v \big{\rangle}_I\\[6pt] &= \big{\langle} u\ \big{|}\ XX^\intercal v \big{\rangle}_I\\[6pt] &= \big{\langle} u\ \big{|}\ v \big{\rangle}_I \end{align}

This leaves me with three conflicting statements:

1. The left-invariant metric tensor on SO(3) is constant.
2. Any Riemannian manifold with a constant metric tensor (in at least one coordinate system) has zero curvature.
3. With the left-invariant metric tensor, SO(3) has nonzero curvature, as shown here.

One or more of these must be false. What am I messing-up here?

I feel like #1 is a dubious assessment because surely a change of coordinates should affect that, and I don't see how coordinates even work in to the logic outlined above. Meanwhile #2 is something I've not seen proved, but only stated off-hand in numerous physics contexts. Finally, #3 seems correct but the proof is somewhat beyond my surface-level dive into the subject so far.

I appreciate the help!

Solution 1:

2. and 3. are both correct statements. More precisely, 2. says the following:

if there is a local chart $(U ,(x^i))$ of a Riemannian manifold $(M, g)$ so that $$ g = \sum_{i,j} g_{ij} dx^i dx^j$$ locally on $U$ for some constants matric $(g_{ij})$. then $g$ has zero curvature.

But 1. does not imply 2. The metric that you construct in 1. is called left-invariant and there is no local coordinates where you can write this metric as a constant matrix.

This confusion is present in every Riemannian manifold. At every point, one can find a local orthogonal frame $\{e_1, \cdots, e_n\}$: that is, $e_i$ are local vector fields on $U$ so that $g(e_i, e_j) = \delta_{ij}$. This can be found by (e.g.) gram schmidt process. This does not imply that $g$ has zero curvature.