Let $Q$ be the generator of a well-behaved (not necessarily reversible) Markov process $X$ on $[n] = \{1,\dots,n\}$ and let $Q^\otimes = \sum_{m=1}^N I^{\otimes(m-1)} \otimes Q \otimes I^{\otimes(N-m)}$ be the tensor sum generating $N$ instances of $X$ (see this MO answer). Let the spectrum of $Q$ be $\sigma(Q) = \{\lambda_j\}$. It can be shown that $\sigma(Q^\otimes) = \{\sum_{m=1}^N \lambda_{j_m}\}$.

In particular, the spectral gaps of $Q$ and $Q^\otimes$ coincide.

On the other hand, the exponential decay rate of the variance of $e^{tQ^\otimes}f^\otimes$ is $N$ times the decay rate for the variance of $e^{tQ}f$ (see, e.g. here).

As I understand it in the reversible case the spectral gap and (minimal) decay rate of the variance (given by the infimum of the quotient of the Dirichlet form and the variance itself) are supposed to be the same. But in tensor product land this weird factor of $N$ seems to keep popping up. I thought fleetingly that I understood this apparent contradiction, but now realize that I still don't.

So: if the spectral gap of $Q^\otimes$ equals that of $Q$ and the decay rate for the variance of $e^{tQ^\otimes}f^\otimes$ is $N$ times that for $e^{tQ}f$, how can these various facts(?) be reconciled?

I'm probably just being obtuse, but this issue is really annoying me, and so I'll retroactively award a bounty of 100 points for the best definitive answer that I can understand and that's provided within 24 hours of this question's posting time.