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I am reading a textbook on rigorous quantum mechanics and quantum field theory in which there often appears statements such as

Let $A(\phi)$ be a polynomial function defined on $\mathcal{S}'(\mathbb{R}^d).$

where $\mathcal{S}'(\mathbb{R}^d)$ is the set of all tempered distributions.

Since it is well known that products of distributions are not always well-defined I am having trouble interpreting something like $A(\phi)$. How are polynomials of tempered distributions defined?

Abdelmalek Abdesselam
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CBBAM
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  • Maybe for each $\psi$, they mean something like $A(\phi)(\psi) = \int_{\mathbb{R}^d} p(\phi)(x) \psi(x) \ dx$, where $p$ is some polynomial which we ``plug'' $\phi$ into? Which textbook is this from? – Mousedorff Apr 26 '24 at 18:23
  • @Mousedorff This is from Glimm & Jaffe's Quantum Physics: A Functional Integral Point of View, Theorem 6.3.1 on page 106. – CBBAM Apr 26 '24 at 19:14
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    Is it possible that the meaning is that $A(\phi) = P(\phi(f_1), \dots, \phi(f_n))$ for some multivariate polynomial $P$ and some test functions $f_i$? This would be well-defined and objects of this type form a common useful class of functionals (that I more typically see being called cylindrical functions) – Rhys Steele Apr 27 '24 at 21:25
  • @RhysSteele I think you are right, this definition would make the most sense in the context of the textbook. Thank you very much! – CBBAM Apr 29 '24 at 02:21

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This is probably an oversight, but it indeed seems that Glimm and Jaffe did not define what they meant by a polynomial function $\phi\mapsto A(\phi)$, on the the space of temperate distributions $\mathscr{S}'(\mathbb{R}^d)$, before they invoked this notion in Theorem 6.3.1. From the general context in the book, a reasonable guess is that they meant a finite linear combination of products of functions of the form $\phi\mapsto \phi(f)$ where $f$ is a test function in Schwartz space $\mathscr{S}(\mathbb{R}^d)$. This is what was already mentioned in the comment by Rhys. However, this is an ad hoc definition, and it might be worthwhile to recall the more canonical definition of a polynomial function on $\mathscr{S}'(\mathbb{R}^d)$, by first revisiting this notion in the simpler case of a finite dimensional vector space $V$, using a little bit of modern multilinear algebra.

The space of degree one homogeneous polynomial functions, is the dual space $V'$. The space of degree $n$ homogeneous polynomial functions on $V$ is the symmetric power ${\rm Sym}^n(V')\subset (V')^{\otimes n}$. Finally, the space of polynomial functions is the direct sum $\bigoplus_{n\ge 0}{\rm Sym}^n(V')$.

Now, lets redo this with $V=\mathscr{S}'(\mathbb{R}^d)$. The key remark is that $V'=(\mathscr{S}'(\mathbb{R}^d))'=\mathscr{S}(\mathbb{R}^d)$, because Schwartz space is reflexive. For this one needs to always use the canonical choice of topology, namely the strong dual topology, when taking duals of topological vector spaces, instead of the ad hoc but more popular weak-$\ast$ topology. Then taking topological (completed) tensor products, the analogue of $(V')^{\otimes n}$ is $\mathscr{S}(\mathbb{R}^{nd})$. Finally, the correct notion of polynomial function is: functions of the form $$ \phi\longmapsto \sum_{n=0}^N \langle \phi^{\otimes n}, f_n\rangle $$ where $(x_1,\ldots,x_n)\mapsto f_n(x_1,\ldots,x_n)$ is a Schwartz function on $(\mathbb{R}^d)^n$, i.e., $f_n\in \mathscr{S}(\mathbb{R}^{nd})$. Here $\phi^{\otimes n}\in \mathscr{S}'(\mathbb{R}^{nd})$ is built using the notion of tensor product of Schwartz distributions.

Note that one can choose to require the $f_n$ to be symmetric or not, but it does not matter since they are paired with a distribution which is symmetric.

Also note that in the finite dimension case the two types of definitions are the same because tensors are finite sums of decomposable tensors. For $V=\mathscr{S}'(\mathbb{R}^d)$, this is not true, but the polynomial functions in my answer can be approximated by functions as in Glimm-Jaffe and the comment by Rhys.

Abdelmalek Abdesselam
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  • Thank you very much! Why must we use the strong dual topology instead of the weak-* topology? – CBBAM May 01 '24 at 04:11
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    The short answer is for the same reason one uses the standard Euclidean topology on $\mathbb{R}^n$ instead of the examples of alternate topologies one can find for instance in the book by Munkres, mainly used as a source of counterexamples. The strong dual topology satisfies a number a good properties, see https://math.stackexchange.com/questions/2623515/schwartz-kernel-theorem-and-dual-topologies/2647815#2647815 Moreover, since $S'$ is isomorphic to a space of temperate sequences one can compare the strong and weak-star topologies with their explicit seminorms side-by-each... – Abdelmalek Abdesselam May 02 '24 at 13:53
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    ...in the sequence setting and it is then clear one topology is natural and the other one is a bit strange and not really convenient. See https://mathoverflow.net/a/362536/7410 – Abdelmalek Abdesselam May 02 '24 at 13:54
  • I will take a look at those links, thanks again for all the help! – CBBAM May 02 '24 at 16:19
  • Could you suggest any reference explaining exactly how the polynomial functions as in GJ definition approximate yours? Any help appreciated! – Keith Jun 25 '24 at 11:10