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Yang–Mills existence and mass gap problem is officially stated by Clay Mathematics Institute:

Yang–Mills Existence and Mass Gap.'' Prove that for any compact simple gauge group G, a non-trivial quantum Yang-Mills theory exists on $\mathbb{R}^4$ and has a mass gap Δ > 0. Existence includes establishing axiomatic properties at least as strong as those cited in Streater-Wightman 1964, Osterwalder-Schrader 1973) and (Osterwalder-Schrader 1975).

However, if we want to define a mass gap or an energy gap for the quantum theory, it is usually helpful to have a Hamiltonian description of the theory with a Hilbert space or Fock space, instead of only the Lagrangian or path-integral description.

To have a Hamiltonian description with a Hilbert space for this Yang-Mills theory, it is usually more physical to have the space-time to be the Lorentizian or Minkowski $$ \mathbb{R}^{3,1}, \text{ with metric signature (+,+,+,-)} $$ instead of the Euclidean $$ \mathbb{R}^{4}, \text{ with metric signature (+,+,+,+)}.$$

My Question: Is there a reason why the official statement is on the less natural $\mathbb{R}^{4}$? instead of the $\mathbb{R}^{3,1}$ that may be more natural to have a Hamiltonian or Schrodinger equation description of Yang-Mills theory with a Hilbert space?

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    $\begingroup$ Did you look at the Osterwalder-Schrader papers cited right there in your quotation? Also, Fock spaces are known not to work for interacting quantum field theories, so that idea is known to be ruled out. $\endgroup$
    – Robert Furber
    Oct 5, 2021 at 16:23
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    $\begingroup$ Looking at the description of the problem by Jaffe and Witten on the CMI website, which your quote is excerpted from, one observes that they do not use the notation $\mathbb R^{3,1}$ to denote Minkowski space but rather say "$\mathbb R^4$ with a Minkowski signature" as they do on p. 5, the page before that quote. So I don't think you can conclude that they mean to exclude the Minkowski case. $\endgroup$
    – Will Sawin
    Oct 5, 2021 at 16:36
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    $\begingroup$ As Robert said, Fock spaces are irrelevant to the question. As Will said, I am sure Jaffe and Witten would be happy with a solution to the problem regardless of whether it uses the Euclidean or the Minkowski formulation. Now another comment is, in the Euclidean approach, one has access to the Hilbert space and the Hamiltonian. For Markovian Euclidean QFTs, the Hilbert space is the $L^2$ space of the marginal probability distribution of the time zero restriction of the field. For more general OS positive models one takes observables in a half space with the inner product given by OS... $\endgroup$
    – Abdelmalek Abdesselam
    Oct 5, 2021 at 17:25
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    $\begingroup$ ...positivity. One needs completion and a quotient operation, then one gets the HIlbert space. The Hamiltonian then is obtained as the infinitesimal generator of time translations. All this is explained in the book by Glimm-Jaffe Chapter 6. $\endgroup$
    – Abdelmalek Abdesselam
    Oct 5, 2021 at 17:27
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    $\begingroup$ Forgot to mention, time translations in the Euclidean setting give a contraction semigroup $e^{-tH}$ rather than a unitary one-parameter group $e^{-itH}$. So you get $H$ by Hille-Yoshida instead of Stone. $\endgroup$
    – Abdelmalek Abdesselam
    Oct 5, 2021 at 17:41

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