Extreme confusion with the exact meaning of Gaussian measure with "translation-invariant" covariance

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 $$$$V:=\{ f \in L^2(S^1) \mid f(x)=-f(-x) \text{ and } f \text{ is real-valued}\}$$$$ where $$S^1$$ is the circle.

Then, by Minlos theorem, there exists a centered Gaussian measure $$d\mu$$ on $$V$$ such that $$$$\exp\Bigl(-\frac{1}{2}\bigl\langle f, (-\Delta)^{-1} f \bigr\rangle \Bigr)=\int_{V} e^{i\langle f, g\rangle}d\mu(g)$$$$ for all $$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 $$C : V \times V \to \mathbb{R}$$ $$$$C(f,g):=\int_{V} \langle f, h \rangle \langle g,h\rangle d\mu(h).$$$$

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?

• Gaussian measure is uniquely determined by its mean and covariance. If mean is zero and covariance is translation invariant, then the measure itself is translation invariant. But there are no good ($\sigma$-finite) translation invariant measures on infinite dimensional Hilbert spaces. Could you provide some context or a reference to the literature where this terminology for this measure is used? Commented Nov 4, 2023 at 9:10
• @tsnao by translation invariance, I mean translation invariance in the argument of the functions. Commented Nov 4, 2023 at 10:04
• It is about $f(x+a)$, not $f + g$. Commented Nov 4, 2023 at 10:05
• the process associated to this measure is indexed by functions, not by points. And your covariance depends on $f$ and $g$, not on their values at some points. Do you mean that $C ( T_x f, T_x g ) = C ( f, g )$, where $T_x \colon V \to V \colon (T f)(y) = f(x+y)$? Commented Nov 4, 2023 at 10:24
• @tsnao yes, that seems like what I want. So confusing....... Commented Nov 4, 2023 at 10:37

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 $$C ( T_x f, T_x g ) = C ( f, g )$$.