The manifold of (symmetric) rank-$k$-matrices is given by
$$M_n(k) = \left\{ X \in \mathbb R^{n\times n} : \mbox{rank} (X) = k \right\}$$
Why is the manifold of rank-$k$-matrices locally curved everywhere?
I am not familiar with formalizations of curvature, I would already be happy with some intuition.
Note: The tangent space at $X\in M_n(k)$ with singular value decomposition $X = U D U^\top$, where $U \in \mathbb R^{n\times k}$ and $D$ is diagonal, is given by
$$ T_{M_n(k)}X = \left\{ Y U^\top + U Y^\top : Y \in \mathbb R^{n \times k} \right\}$$
Figured out some intuition by myself:
For each rank-k-matrix $X$ there are other rank-$k$-matrices infinitesimally close with a slightly different tangent spaces since the row- respective columnspace has been perturbed (keep the form of the tangent space in mind). Consider for example the following rank-2-matrices for small $\varepsilon>0$ $$\begin{pmatrix}0 & 0 & 1\\ 0 & 0 & 1\\ 1 & 1 & 1\end{pmatrix}, \quad \begin{pmatrix}0 & 0 & 1\\ 0 & 0 & 1+\varepsilon\\ 1 & 1+\varepsilon & 1\end{pmatrix}.$$ Since the tangent spaces changes locally there must be nonzero curvature.