TITLE: Orthogonal representations of graphs
QUESTION [8 upvotes]: A faithful orthonormal representation of a graph $G=(V,E)$ on $n$ vertices $\{1,2,\dotsc,n\}$ is an assignment of unit vectors $v_1,v_2,...,v_n \in \mathbb{R}^d$ to the vertices of $G$ such that $\langle v_i,v_j \rangle =0 \Leftrightarrow ij \in E(G)$, and in addition $|\langle v_i , v_j \rangle| \neq 1$ if $i \neq j$, i.e., distinct vertices are assigned non-parallel vectors. Note that this definition of orthonormal representation is slightly rarer in the literature and differs (by graph complementation) from the definition in [1] where $ij \in E(\bar{G}) \Rightarrow \langle v_i , v_j \rangle = 0$.  
The question is: Given a graph $G$ that has a faithful orthonormal representation in dimension $d$, form a new graph $G'$ by deleting an edge $uv$ from $G$. Does $G'$ then also have a faithful orthonormal representation in the same dimension $d$?
[1] L. Lovász, On the Shannon Capacity of a Graph, IEEE Trans. Inf. Theory, 25 (1):1-7 (1979).

REPLY [4 votes]: The following would be a counterexample if we require only $v_i \ne v_j$ but don't forbid different vertices to become antipodes on the sphere. Let us call this a weakly faithful orthogonal representation.
The graph of the octahedron has a weakly faithful representation in (real) dimension 2, given by $\pm e_i$. If you remove one edge, there will be no faithful representation in dimension 2: the remaining edges force the two non-connected vertices to lie at distance $\pi/2$ on the sphere.
So, for faithful orthogonal representations the question is: can one force the distance (in the standard spherical metric) between two points in the projective space to be $\pi/2$ by imposing distances $\pi/2$ between some pairs of points?

REPLY [3 votes]: Thanks to a commenter for pointing out that the following was hasty: the fallacy is of course in the statement that the edgeless graph had no faithful orthogonal representation. I'll leave it at that, but add a warning. 

No. Fallacious proof by contradiction; the red is wrong.  if this were true, then, since you required the graph to be finite, iterating the (hypothetical) statement $\lvert E\rvert$-many times would lead us to the conclusion that the edgeless graph on $n$ vertices has a faithful orthogonal representation in dimension $d$, $\color{red}{\text{which is absurd}}$. 
This proves that the answer cannot possibly be yes. 

Useful addition: the relevant technical term is monotone graph property. And the above gives a reason why 

for any fixed $n$ and $d$, the property(=isomorphism-invariant class of graphs) of all those graphs which admit a faithful orthogonal representation in dimension $d$ is not a monotone decreasing graph property.