In physics literature, the covariance of a Gaussian measure $\mu$ on a function space is denoted as $C(x,y)$. Moreover, they say that if the covariance is translation-invariant, then actually $C(x,y)=\widetilde{C}(x-y)$.
I am extremely confused about exact meaning of such statement. For example, let us consider the following Hilbert space
\begin{equation}
V:=\{ f \in L^2(S^1) \mid f(x)=-f(-x) \text{ and } f \text{ is real-valued}\}
\end{equation}
where $S^1$ is the circle.
Then, by Minlos theorem, there exists a centered Gaussian measure $d\mu$ on $V$ such that
\begin{equation}
\exp\Bigl(-\frac{1}{2}\bigl\langle f, (-\Delta)^{-1} f \bigr\rangle \Bigr)=\int_{V} e^{i\langle f, g\rangle}d\mu(g)
\end{equation}
for all $f \in V$. Here $\langle, \rangle$ is the $L^2$ inner product.
$(-\Delta)^{-1} : V \to V$ clearly commutes with the translation operation. Also, the $2-$point moment of $d\mu$ is defined as the bilinear map $C : V \times V \to \mathbb{R}$
\begin{equation}
C(f,g):=\int_{V} \langle f, h \rangle \langle g,h\rangle d\mu(h).
\end{equation}
I suspect that what physics literature mean by "covariance" is $C(f,g)$ here. But in what sense is $C(f,g)$ translation-invariant?
Such things are so frustatingly confusing…Could anyone please clarify?
Best Answer
Here's my guess at what is meant by translation invariance in your question. Let $\mathbb{S}^1 = \mathbb{R} \ / \ \mathbb{Z}$ be the circle. Define $T_x \colon L^2 ( \mathbb{S}^1 ) \to L^2 ( \mathbb{S}^1 )$ by $$ ( T_x f ) ( y ) = f ( x + y ). $$
Since $$ \langle T_x f, h \rangle = \int_{\mathbb{S}^1} f ( x + y ) \, h ( y ) \, d y = \langle f, T_{-x} h \rangle, $$ we have $$ C ( T_x f, T_x g ) = \int_V \langle f, T_{-x} h \rangle \, \langle g, T_{-x} h \rangle \, d\mu ( h ) = \int_V \langle f, h \rangle \, \langle g, h \rangle \, d (T_{-x})_* \mu ( h ), $$ where $(T_{-x})_* \mu$ is the pushforward of $\mu$ under $T_{-x}$. Since $(-\Delta)^{-1}$ commutes with $T_x$, $$ \langle T_x f, ( -\Delta)^{-1} T_x f \rangle = \langle T_x f, T_x ( -\Delta)^{-1} f \rangle = \langle f, ( -\Delta)^{-1} f \rangle, $$ we have that $(T_{-x})_* \mu = \mu$, which gives $C ( T_x f, T_x g ) = C ( f, g )$.