TITLE: $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m) \simeq \mathcal S(\mathbb R^{n+m})$?
QUESTION [7 upvotes]: If $S(\mathbb R^n)$ is the Scwartz space of smooth rapidly decaying functions equipped with the topology generated by the family of semi-norms
$$\mathcal N_p (\varphi)= \sum_{|\alpha|, |\beta| \leq p} \sup_{x\in \mathbb R^n} |x^\alpha \partial^\beta \varphi (x) |\, ,$$
Is it true that $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m)\simeq \mathcal S(\mathbb R^{n+m})$, where $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m)$ is the completion of $\mathcal S(\mathbb R^n) \otimes \mathcal S(\mathbb R^m)$ for the topology $\pi$ defined by the family of semi-norms
$$\mathcal N_{p,q} (\varphi\otimes \psi)=\mathcal N_p(\varphi)\mathcal N_q (\psi)\, .$$

REPLY [8 votes]: Yes, if you understand the tensor product topology in the right way.
Since the spaces are nuclear, inductive and projective tensor products coincide.
The result is theorem 51.6 of Treves: Topological vector spaces, distributions, and kernels.
Added later:
Attention: The description of  seminorms on the tensor-product given in the question is not sufficient to specify a locally convex topology on the tensor product. There are many satisfying this description; all between the the projective one and the the inductive one and even more. See the source I have given, or many other books.


TITLE: Argument for exponential decay of correlation function
QUESTION [6 upvotes]: Suppose we have some lattice classical spin system. The correlation function is defined as
$$\Gamma_{ij}=\langle S_iS_j\rangle -\langle S_i\rangle \langle S_j\rangle $$
It is often said that
$$ \Gamma_{ij}\sim e^{-|i-j|/\xi}$$
where $\xi$ is the correlation length (this is, if I understand correctly, the definition of correlation length). Is there a way to, if not prove, at least argument that this must be the case, while being agnostic to any particular choice of Hamiltonian, beyond assuming that it is local in some way? I understand that intuitively the correlation must decay, but why exponentially?
Also, can I interpret this definition of $\xi$ literally? Suppose the system is translationally invariant and $\gamma=\Gamma_{ii}$ is the same for every $i$, can I say
$$\Gamma_{ij}=\gamma e^{-|i-j|/\xi} $$ and hence
$$\xi= \frac{|i-j|}{\log\gamma-\log\Gamma_{ij}}$$
for every $i,j$? I find it hard to believe that this quantity doesn't ever depend on $i,j$. If this is indeed true, there must be some locality requirements on the Hamiltonian. What are they?

REPLY [3 votes]: As mentioned in the comments, I don't think that in every system the correlation function behaves like this. In fact, this is not the case when analyzing a point of phase transitions. This is one of the things that characterize phase transitions! There also other classes of systems that do not show this behavior. However, we can say that this is the general case, and in fact a definition for the correlation length.
To see that, let's start from the following assumption - there is a correlation length in the system. That is, a length scale that when examining chunks of the system smaller than that scale it is correlated, and above it is not correlated. Then the correlation function cannot have a power-law form $\Gamma_{i,j} \propto |r_i-r_j|^{-\alpha}$ simply because this is scale-free! This is the case in conformal symmetric systems (which characterize some phase transitions) and in some special 1d systems.
So looking for a decaying function that shows a characteristic scale we of course have $\Gamma_{i,j} \propto  \exp[-|r_i-r_j|/\xi]$. We can add other stuff on top of it, but this will be the dominant behavior. This is usually how we define the correlation length, and then we can see how it behaves when we change system parameters, and usually it will blow up to infinity at the point of phase transition, which exactly means that the other terms (the scale-free) are the dominant ones. I think that one of the definitions of a phase-transition is "the point in which there is no correlation length in the system".
Now that $\propto$ sign plays an important role here. In general the correlation function will be quite complicated and calculating it exactly in a correlated system will be an intractable problem. However, we can approximate its behavior using some methods, and we will always be interested in the dominant behavior, which far from the phase transition will be encoded in this exponential form.