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](https://mathoverflow.net/questions/58567/spectral-gap-of-a-product-of-markov-processes/58610#58610)). 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](https://mathoverflow.net/questions/60638/infimum-of-the-dirichlet-form-for-a-tensor-product)).


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.



Rate of decay of variance for a tensor product Markov process (100 pt bounty for good answer by 1800 EST Fri)
Working with a single copy of the Markov chain for now, you can simplify the expression for the Dirichlet form as $\mathcal D\_Q(f)=-\sum\_{j,k}p\_jQ\_{jk}f\_jf\_k$ (you obtain this from the previous expression just by noting that the row sums of the $Q$ are 0). 


Let $e^{(0)},\ldots,e^{(d-1)}$ be the eigenvectors of your $Q$. Assume reversibility of $Q$. Then $Q$ is self-adjoint with respect to the inner product $\langle f,g\rangle=\sum p\_if\_ig\_i$. This guarantees that the eigenvectors are orthogonal with respect to this inner product. Let $e^{(0)}$ be the eigenvector of 1's (with eigenvalue 0). The other eigenvectors can be chosen to have length 1 with respect to the inner product. They have negative real eigenvalues $\lambda^{(1)}>\ldots>\lambda^{(d-1)}$. 


Notice that the Dirichlet form is very simple for $f$'s that are expressed as a sum of eigenvectors: $D\_Q(\sum a\_i e^{(i)})=\sum a\_i^2D\_Q(e^{(i)})$ (using the orthogonality properties of $Q$). Further $D\_Q(e^{(i)})=-\lambda^{(i)}$ so that $D\_Q(\sum a\_ie^{(i)})=\sum a\_i^2|\lambda\_i|$. Also Var($e^{(i)}$) is 1 for $i\ne 0$ and 0 for $i=0$. 
The minimum ratio of the Dirichlet form to the variance occurs at the non-trivial eigenvector with the smallest absolute eigenvalue.


Define $Q^\otimes=\sum\_{i}c\_i I^{\otimes(i-1)}\otimes Q\otimes I^{\otimes{n-i}}$ (this is the generalization of what's in your posting with multiple $c$'s). The eigenvectors of $Q^\otimes$ are simply the tensor products of the eigenvectors of $Q$ so that $e^{(i\_1)}\otimes \cdots\otimes e^{(i\_n)}$ has eigenvalue $c\_1\lambda^{(i\_1)}+\ldots+c\_n\lambda^{(i\_n)}$. The variance of each of these eigenvectors is 1 except for the eigenvector $e^{(0)}\otimes\cdots\otimes e^{(0)}$ which has variance 0. Once again the minimum ratio of the Dirichlet form to the variance occurs at the non-trivial eigenvector with the smallest absolute value. Assuming that $c\_1\le c\_2\le c\_3\le\ldots$, this is the eigenvector $e^{(1)}\otimes e^{(0)}\otimes\cdots\otimes e^{(0)}$. In this case the ratio is given by $c\_1|\lambda\_1|$.


Hopefully this resolves the discrepancy?


