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 $g\_t$? 



Is the space of Levi-Civita connections convex
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. 


