# Is the Eastin-Knill Theorem incorrect?

I am reading through this paper (the Eastin-Knill Theorem) and there is a step in the proof of the main theorem that I do not understand.

Let $$Q$$ be a composite quantum system supporting a quantum code $$C$$. For a quantum system of finite dimension $$d$$, let $$U(d)$$ denote the set of unitaries on that quantum system. For any $$d$$, $$U(d)$$ is a compact connected Lie group. Let $$\mathcal{T}$$ be the set of unitary product operators on $$Q$$. Being a direct product of compact and connected Lie groups, $$\mathcal{T}$$ is also compact and connected. Let $$\mathcal{G}$$ be the set of unitary product operators that are also logical operators with respect to the code $$C$$ on $$Q$$. It is shown in the paper that $$\mathcal{G}$$ is a Lie subgroup of $$\mathcal{T}$$. As a Lie group, $$\mathcal{G}$$ can be partitioned into cosets of the connected component of the identity, $$\mathcal{C}$$. The set of cosets is $$R = \mathcal{G}/ \mathcal{C}$$ which constitutes a topologically discrete group.

Let $$\mathcal{F}$$ be a set consisting of one representative from each coset of $$\mathcal{C}$$ in $$\mathcal{G}$$, that is, one element from each set in $$R$$. Since $$R$$ is discrete, it is obvious that $$\mathcal{F}$$ is discrete as well. The authors then claim that $$\mathcal{F}$$ is finite since it is a discrete subset of a compact set, namely $$\mathcal{T}$$. But this alone is not enough to guarantee finiteness (consider the discrete subset $$\{\frac{1}{n}\}_n$$ of $$[0,1]$$). If it were also (topologically) closed, that would be sufficient. But I cannot figure out a reason why this set must be closed, or any other reason why $$\mathcal{F}$$ might be finite.

Would greatly appreciate an explanation if anyone knows why.

• does this help math.stackexchange.com/questions/2257962/… Apr 17, 2022 at 14:35
• Not really, the author (of the book mentioned in the post) says that what he meant to write in the statement of the lemma was that the set is closed and discrete, but in my case, the authors never claim the set is closed (or even mention closure at all). Apr 18, 2022 at 15:55
• sorry, I was more referring to the answer of the question I linked rather than the question itself. In that answer, notice that under an alternative definition of discrete the result you are asking about is true. I suspect that something like that is happening in the paper as the Easton-Knill theorem is certainly not "incorrect". It's an unfortunate truth but definitions do vary amongst mathematical fields, especially when mixing math and physics. Apr 18, 2022 at 17:00
• The notion of discreteness being used here comes from the fact that $\mathcal{G}/\mathcal{C}$ is a discrete group when $\mathcal{G}$ is a Lie group and $\mathcal{C}$ is the identity component. This fact can be found on the wikipedia page for identity component: en.wikipedia.org/wiki/Identity_component. Following the link for discrete group, we find the usual definition of discrete, not the alternate definition proposed in the question you linked. Apr 19, 2022 at 14:34
• glad it worked out! Apr 19, 2022 at 14:38

$$\mathcal{G}$$ is a closed subgroup of $$\mathcal{T}$$, which is a closed subset of a unitary group. Unitary groups are compact. Closed subsets of compact spaces are compact. Thus $$\mathcal{G}$$ is compact.
Compact sets have finitely many connected components. $$\mathcal{C}$$ is the identity component of $$\mathcal{G}$$. So the cosets of $$\mathcal{C}$$ are the connected components of $$\mathcal{G}$$. Thus there are finitely many cosets and $$\mathcal{F}$$ is finite.