For any Riemann Curvature $K$, does there exists an open set $G$ such that every point $p$ in $G$ has Curvature $K$? For $2$-Dimensional manifolds, for any curvature $K$, we can create an open ball $G$ such that every point has the same curvature $K$, giving rise to spaces that are spherical, flat, or hyperbolic.  But in higher dimensions, the idea of Riemann curvature gets more complicated, where each point’s curvature is dependent on $\frac{n^2(n^2-1)}{12}$ variables. I was wondering if, for any possible Riemann curvature $K$, there exists an open set $G$ where each point $p$ in $G$ has Riemann curvature $K$? Basically where every point is “indistinguishable” from every other point and has curvature $K$.
 A: First, let me try to formulate a meaningful mathematical question. If $M$ is a connected (differentiable) manifold then for a (differentiable) function $f$ on $M$ to be "the same at every point" means that $f$ is constant, equivalently, $df=0$. Now, the curvature $R$ is not a function on $M$  but a tensor. The   meaningful interpretation of the statement "$R$ is the same at every point of $M$" is that 
For the Levi-Civita connection $\nabla$ of the Riemannian metric $g$, for which $R$ is the curvature tensor, $\nabla R=0$, i.e. $R$ is parallel. 
Equivalently, if $\Pi$ denotes the parallel transport of $\nabla$, then $\nabla R=0$ means that $\Pi_c$ preserves $R$ for every smooth path $c$ in $M$. 
Riemannian metrics with parallel curvature tensor are very special: They are locally symmetric. A detailed discussion and classification of such metrics can be found for instance in Helgason's book on symmetric spaces. 
The definition of the covariant derivative of a tensor can be found in most books on differential geometry. (My personal favorite is do Carmo's "Riemannian Geometry"; the definition will be in chapter 4.) 
With this in mind, your question becomes: 
Suppose that $K$ is an "algebraic curvature tensor" on $T_pM$ for some $p\in M$. (In other words, $K$ can be interpreted as a symmetric bilinear form on $\Lambda^2 T_pM$. Equivalently, $K$ has all the symmetries of a curvature tensor.) Is there a Riemannian metric $g$ on a neighborhood $U$ of $p$ in $M$, such that $g$ is locally symmetric and its curvature tensor at $p$ equals $K$?     
This question has negative answer in all dimensions $\ge 3$. (And, of course, positive in dimension 2.) The key is that there are not that many locally symmetric metrics. 
For instance, in dimension 3, up to isometry, there is (locally) just a 1-parameter family of locally symmetric metrics; they all (locally) have the form $X_k\times {\mathbb R}$ where $X_k$ is a surface of constant curvature $k$. In contrast, $\Lambda^2 {\mathbb R}^3\cong {\mathbb R}^3$ and the space of symmetric quadratic forms (up to an isomorphism) on  ${\mathbb R}^3$ is 3-dimensional (they are classified by the three eigenvalues.) Hence, a "generic" algebraic curvature tensor $K$ on $T_pM$ cannot extend to the curvature tensor of a locally symmetric metric on a neighborhood of $p$ in $M$.  
