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 LeviCivita connections. Set $\nabla_t:=(1t)\nabla_0+t\nabla_1$. Then $\nabla_t$ is a torsion free connection. Does there exist a continuous family of metrics $g_t$ such that $\nabla_t$ is the LeviCivita connection of $g_t$?

$\begingroup$ what about $g_t(v,w)=t g_0(v,w)+(1t)g_1(v,w)$ ? $\endgroup$– GiulioNov 21 '16 at 18:40

4$\begingroup$ @Giulio, please note that the LeviCivita connection depends nonlinearly on the metrics tensor. $\endgroup$– Maxim BravermanNov 21 '16 at 20:00

3$\begingroup$ No. This does not hold, even in dimension 2. Just compute a few examples, and you'll see. $\endgroup$– Robert BryantNov 22 '16 at 1:24
Here's a more specific approach that explains why you shouldn't expect this: For simplicity, I'll work in the 2dimensional 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 LeviCivita 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 torsionfree 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 $$ \theta = (1t)\,\theta_0 + t\,\theta_1\,, $$ which is the connection $1$form on $S$ associated to the connection $\nabla = (1t)\,\nabla_0 + t\,\nabla_1$. 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 LeviCivita connection of any Riemannian metric.
The answer of Professor Bryant shows, that the space of LeviCivita connections is not convex. On the other hand, there are natural convex subspaces, at least in dimension 2, namely the space of LeviCivita 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 complexvalued complex linear 1forms 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 1forms 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 $$t\tilde\nabla+(1t)\nabla=\nabla^02(t\partial\tilde\lambda+(1t)\partial\lambda)$$ which is the LeviCivita connection for $e^{2(t\tilde\lambda+(1t)\lambda)}g_0.$

1$\begingroup$ Your argument can be generalized. Suppose in some frame the matrices which represent $g_1$ and $g_0$ commute. Let $G$ and $G_0$ be these matrices. Then we can write $G_1= e^{A_1}$, $G_0= e^{A_0}$ where $A_1$ and $A_0$ a symmetric matrices which commute which each other. Then basically repeating your argument we obtain that $\nabla_t$ is the LeviCivita connection of $G_t:= e^{(1t)A_0+tA_1}$. $\endgroup$ Nov 24 '16 at 13:08

1$\begingroup$ @MaximBraverman: Your argument cannot be right because any two metrics can be simultaneously diagonalized in some frame, and hence their matrices will commute. $\endgroup$ Nov 24 '16 at 13:24

$\begingroup$ @RobertBryant I was not accurate. What one needs is that the matrices with respect to some coordinate system commute. Then setting $A(t)= (1t) A_0+tA_1$ and $G(t)=e^{A(t)}$ we get $\partial G/\partial x^j= e^{A}\partial A/\partial x^j$ and the Christoffel symbols are linear in $A(t)$. $\endgroup$ Nov 24 '16 at 14:19