TITLE: graphs with independence number = Shannon capacity
QUESTION [14 upvotes]: For $G$ a graph, let $\alpha(G)$ be its independence number and $\Theta(G)=\lim_n \sqrt[n]{\alpha(G^{\boxtimes})}$ its Shannon capacity, where $\boxtimes$ denotes strong product.
Consider graphs $G$ and $H$ satisfying $\alpha(G)=\Theta(G)$ and $\alpha(H)=\Theta(H)$. For example, $G$ and $H$ could be perfect, but the more interesting situations arise when neither of them is perfect.
Question: Does this assumption imply
(1) $\alpha(G\boxtimes H) = \alpha(G)\alpha(H)$ ?
(2) $\Theta(G\boxtimes H) = \Theta(G)\Theta(H)$ ?
(3) $\Theta(G + H) = \Theta(G) + \Theta(H)$ ?
Here, $G+H$ stands for the disjoint union of $G$ and $H$.
If my reasoning is correct, then (1) and (2) are equivalent and imply (3).
As far as I can see, neither the work of Haemers nor the results of Alon have anything directly to say about these questions. But then again, I am not an expert on this, so I might have missed something obvious.
Edit (see Will Traves' answer): Actually, I am specifically interested in those $G$ and $H$ which are well-covered.
Edit: The paper is here.

REPLY [5 votes]: By now, we have been able to resolve this question, and our revised paper contains a proof showing that the answer to all three questions is negative in a very strong sense. Let me provide a brief summary here and refer to the paper for more details.
What we show is this: there exist graphs $G$ and $H$ with $\alpha(G)=\Theta(G)$ and $\alpha(H)=\vartheta(H)$ which violate all three desired inequalities. Note that the property $\alpha(H)=\vartheta(H)$ is even stronger than the required property $\alpha(H)=\Theta(H)$.
The construction crucially relies on the definitions and results on hypergraphs discussed in our paper and on results of Haemers. We have partially translated it into pure graph-theoretic terms, but it does not seem possible to do so completely.
The counterexample turns out to be the following:

$G$ is a $108$-regular graph on $220$ vertices. The vertices correspond to the $3$-element subsets of $\{1,\ldots,12\}$ and two such vertices are adjacent whenever the subsets intersect in exactly one element. This graph was considered by Haemers, who showed that $\alpha(G) = \Theta(G) < \vartheta(G)$, and this property is the most important ingredient for us.
$H$ is a graph constructed from the complement of $G$ which turns out to have $1131460$ vertices.

I suspect that a similar construction can be carried out starting with any graph $G$ having the property that $\alpha(G) = \Theta(G) < \vartheta(G)$, but we haven't checked all the details of this.