0
$\begingroup$

I have already addressed this problem on my previous question but I still have trouble understanding Brydges' RG maps on his lecture notes, so I'll try to elaborate my question a little better.

Let $\Lambda \subset \mathbb{Z}^{d}$ be finite. We have a positive-definite $|\Lambda|\times |\Lambda|$ matrix $C$ which can be decomposed as a sum $C=C_{1}+\cdots+C_{N}$ where, for each $j=1,...,N$, $C_{j}$ is again positive-definite. Let $\varphi:\Omega \to \mathbb{R}^{|\Lambda|}$ be a random vector $\varphi = (\varphi(x))_{x\in \Lambda}$ with distribution $\mu_{C}$ being a Gaussian measure on $\mathbb{R}^{|\Lambda|}$ associated to $C$. Moreover, for each $j=1,...,N$, $\xi_{j}:\Omega \to \mathbb{R}^{|\Lambda|}$ is a random vector $\xi = (\xi(x))_{x\in \Lambda}$ whose distribution $\mu_{j}$ is a Gaussian measure associated to $C_{j}$. In addition, to each $j=0,1,...,N$ we define new random vectors as partial sums $\varphi_{j} :=\sum_{k>j}^{N}\xi_{k}$, where $\varphi_{N}:=0$ and $\varphi_{0}:=\varphi$.

Given $X\subset \Lambda$, let $\mathcal{N}_{j}(X)$ be the algebra of functions measurable with respect to the $\sigma$-algebra generated by $\{\varphi_{j}(x), x\in X\}$ and $\tilde{\mathcal{N}}_{j}(X)$ is the algebra of functions measurable with respect to the $\sigma$-algebra generated by $\{\xi_{j+1}(x),\varphi_{j+1}(x), x\in X\}$. If $X=\Lambda$, we write $\mathcal{N}_{j}(\Lambda) \equiv \mathcal{N}_{j}$ and similarly for $\tilde{\mathcal{N}}_{j}$.

Now, take $F^{\Lambda}$ to be a function which is $\tilde{\mathcal{N}}_{j}$-measurable (equation (2.20) in Brydges notes). Then, Brydges evaluates the expectation: \begin{eqnarray} \mathbb{E}_{j+1}F^{\Lambda} \tag{1}\label{1} \end{eqnarray} which should be just the expectation with respect to the measure $\mu_{j+1}$. This is where I got in trouble: the measure $\mu_{j+1}$ is on $\mathbb{R}^{|\Lambda|}$, but as far as I'm concerned, $F^{\Lambda}$ is a random variable defined on some underlying probability space, whose $\sigma$-algebra is generated by $\{\xi_{j+1}(x),\varphi_{j+1}(x),x\in\Lambda\}$. So I don't understand the meaning of (\ref{1}).

My guess is that Brydges is using some sort f isomorphic version of the actual $F^{\Lambda}$. In fact, because $F^{\Lambda}$ is $\tilde{\mathcal{N}}_{j}$-measurable, there exists a function $f: \mathbb{R}^{2|\Lambda|}\to \mathbb{R}$ such that $F^{\Lambda}(\cdot) = f(\xi_{j+1},\varphi_{j+1})(\cdot)$ and this $f$ can be integrated with respect to $\mu_{j+1}$ as in (\ref{1}). But I don't know if my guess is correct and, if it is, I don't know how to justify why can $F^{\Lambda}$ be "replaced" by $f$ in (\ref{1}).

$\endgroup$
2
$\begingroup$

I think Brydges is (tacitly) assuming that $\Omega=\mathbb{R}^{|\Lambda|}$. This, of course, is bound to create confusion between elements of $\mathbb{R}^{|\Lambda|}$ and random elements of $\mathbb{R}^{|\Lambda|}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It was so obvious! Thank you so much! It all makes sense now! $\endgroup$ – IamWill Feb 6 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.