Tom Scrubytom-scruby

Ahh, I see. Thanks. A quick follow-up question then. In the next subsection the cup product is defined in terms of a product of R-valued functions on arrays, but in example 2.1 it seems to act directly on the arrays themselves. How should I understand the functions and the ring in this case? Is there an easy way to see why the product you give is the only non-trivial option (or, alternatively, is there a good place I can read more about this)?

A very minor question where I'm probably missing something basic, but it seems like the coboundary operator at the top of page 8 is mapping a function on a length p+1 array to a linear combination of functions on length p arrays. Wouldn't this make it a boundary operator rather than a coboundary one?

Congratulations to the authors on a very nice result!

I'll also use this as an opportunity to note that, following discussions with the authors of this work, my coauthors and I have updated our related work (arxiv.org/abs/2408.13130) and modified our claims r.e. achieving $\gamma \rightarrow 0$ with our codes. The new version of our paper (released today) is now consistent with the discussion of remark 1.2 of this work, and we'd like to thank the authors for bringing these details to our attention.

Apologies for what I'm sure is a very naive question, but what exactly is meant by the claim of "unconditional quantum advantage" in cases such as this? As someone who is fairly ignorant of complexity theory I would have thought that an unconditional proof that a certain problem can be efficiently solved by a quantum computer but not by a classical one would be equivalent to a proof that $BQP \neq BPP$, but this is presumably not what is being claimed here. Does "unconditional" actually mean something like "unconditional as long as $P \neq NP$", or is there something else that I'm misunderstanding? Thanks in advance.