Is the space of Levi-Civita connections convex More precisely, suppose we a given two metrics $g_0$ and $g_1$ on a manifold $M$. Let $\nabla_0$ and $\nabla_1$ be the corresponding Levi-Civita connections. Set $\nabla_t:=(1-t)\nabla_0+t\nabla_1$. Then $\nabla_t$ is a torsion free connection. Does there exist a continuous family of metrics $g_t$ such that $\nabla_t$ is the Levi-Civita connection of $g_t$? 
 A: The answer of Professor Bryant shows, that the space of Levi-Civita connections is not convex. On the other hand, there are natural convex subspaces, at least in dimension 2, namely the space of Levi-Civita connections for metrics of a fixed conformal class. To prove this, let us assume we are in the oriented case, so we are dealing with Riemann surfaces. Let $K\to\Sigma$ denote the canonical bundle, i.e., its sections are the complex-valued complex linear 1-forms on $\Sigma$. A conformal metric on $\Sigma$ in its conformal class defines a unique connection $\nabla$ on $K$ whose $(0,1)$-part is just the exterior derivative, which is the natural holomorphic structure on $K$. If $g=e^{2\lambda}g_0$ their connection 1-forms  differ by  $$\nabla-\nabla^0=-2\partial\lambda.$$ For $\tilde g=e^{2\tilde\lambda}g_0$ we obtain $$\tilde\nabla-\nabla^0=-2\partial\tilde\lambda,$$ and $$t\tilde\nabla+(1-t)\nabla=\nabla^0-2(t\partial\tilde\lambda+(1-t)\partial\lambda)$$
which is the Levi-Civita connection for $e^{2(t\tilde\lambda+(1-t)\lambda)}g_0.$
A: Here's a more specific approach that explains why you shouldn't expect this:  For simplicity, I'll work in the 2-dimensional case, where it's probably the clearest.  Let
$$
\omega = \begin{pmatrix}\omega^1\\\omega^2\end{pmatrix}
$$
be a coframing on a surface $S$.  A metric $g$ is then defined by a $2$-by-$2$ matrix $G$ of functions on $S$, i.e., $g = {}^t\omega G\omega$, where $G$ is symmetric and positive definite.  The Levi-Civita connection of $g$ is then represented by a $2$-by-$2$ matrix of $1$-forms $\theta$ satisfying the equations
$$
\mathrm{d}\omega = -\theta\wedge\omega
\qquad\text{and}\qquad \mathrm{d}G = G\,\theta + {}^t\theta\,G,
$$
the first equation being the torsion-free condition and the second equation being metric compatibility.  The curvature matrix $\Theta = \mathrm{d}\theta + \theta\wedge\theta$ then satisfies
$$
0 =  G\,\Theta + {}^t\Theta\,G,
$$
i.e., $G\,\Theta$ is skewsymmetric, so it can be written in the form
$$
G\,\Theta = \det(G)\begin{pmatrix}0 & K\,\omega^1\wedge\omega^2\\
                           -K\,\omega^1\wedge\omega^2 & 0\end{pmatrix}
$$
for some unique function $K$ on $S$ (i.e., the Gauss curvature).  Thus, in particular,
$$
\Theta = K\,\det(G)\,G^{-1}\begin{pmatrix}0 & 1\\
                           -1\, & 0\end{pmatrix}\,\omega^1\wedge\omega^2
= H\,\omega^1\wedge\omega^2,
$$
where $\det(H) = K^2\det(G)\ge 0$, while $\mathrm{tr}(H)=0$.
Now suppose that we have two metrics $g_0$ and $g_1$ on $S$ with corresponding matrices $G_0$ and $G_1$ and connection forms $\theta_0$ and $\theta_1$.  Let $t$ be a real number satisfying $0<t<1$ and consider the $1$-form 
$$
\theta = (1-t)\,\theta_0 + t\,\theta_1\,,
$$
which is the connection $1$-form on $S$ associated to the connection $\nabla = (1-t)\,\nabla_0 + t\,\nabla_1$. Its curvature satisfies
$$
\begin{aligned}
\Theta &= \mathrm{d}\theta + \theta\wedge\theta
= (1{-}t)\,\Theta_0 + t\,\Theta_1 
   - t(1{-}t)\,(\theta_1{-}\theta_0)\wedge(\theta_1{-}\theta_0)\\
&= \bigl((1{-}t)H_0 + t\,H_1 - t(1{-}t)\,Q\bigr)\,\omega^1\wedge\omega^2.
\end{aligned}
$$
Now, at any given point of $S$, it is easy to construct metrics $g_0$ and $g_1$ so that $G_0$, $G_1$, $K_0$, $K_1$ and $\theta_1-\theta_0$ are arbitrary, subject only to the conditions that the $G_i$ are positive definite, and $(\theta_1-\theta_0)\wedge\omega = 0$.  Using this, one can arrange that, at the given point, the matrix 
$$
H = (1{-}t)H_0 + t\,H_1 - t(1{-}t)\,Q\,,
$$
which is traceless, satisfies $\det(H) <0$.  But, as we have seen above, this is not possible if $\theta$ satisifes $\mathrm{d}G = G\,\theta + {}^t\theta\,G$ for some positive definite $G$.  Thus, when $\det(H)$ is negative at some point, $\theta$ cannot be the Levi-Civita connection of any Riemannian metric. 
