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I'm studying "A Mathematical Introduction to Conformal Field Theory" by Schottenloher and there is one point on the Osterwalder-Schrader axioms that I am a bit confused about. They are phrased in the book as follows:

We start with a countable index set $B_0$ and consider multi-indices $(i_1,\dots, i_n)\in B_0^n$. We further denote $B = \bigcup_{n=0}^\infty B_0^n$. We have functions $G_{i_1\dots i_n}:M_n\to \mathbb{C}$, where $M_n$ is the space of all $(z_1,\dots,z_n)\in \mathbb{C}^n$ with $z_i\neq z_j$. These functions are to be interpreted as Green's functions and demanded to obey

  1. Locality: The Green's functions are permutation invariant, i.e., for any $\sigma\in S_n$ the permutation group in $n$ letters we have $$G_{i_{\sigma(1)}\dots i_{\sigma(n)}}(z_{\sigma(1)},\dots, z_{\sigma(n)})=G_{i_{1}\dots i_{n}}(z_{1},\dots, z_{n}).$$

  2. Covariance: For any isometry of the plane $w(z)$ there are numbers $(h_i,\bar h_i)$ such that $$G_{i_1\dots i_n}(z_1,\bar z_1,\dots, z_n,\bar z_n)=\prod_{j=1}^n\left(\dfrac{dw}{dz}(z_j)\right)^{h_j}\left(\overline{\dfrac{d w}{dz}(z_j)}\right)^{\bar h_j}G_{i_1\dots i_n}(w_1,\bar w_1,\dots, w_n,\bar w_n).$$

  3. Reflection Positivity: There is a map $\ast : B_0\to B_0$ with $\ast^2=1$ and a continuation $\ast:B\to B$ such that $$G_{i_1,\dots,i_n}(z)=G_{(i_1,\dots,i_n)^\ast}(-z^\ast),$$ and such that $\langle \underline{f},\underline{f}\rangle \geq 0$ for all $\underline{f}\in \underline{\mathscr{S}}^+$ where $$\langle \underline{f},\underline{f}\rangle = \sum_{i,j\in B}\sum_{n,m}\int_{M_{n+m}}G_{i^\ast j}(-z_1^\ast,\dots,-z_n^\ast,w_1,\dots, w_m)f_i(z)^\ast f_j(w) d^n z d^m w.$$ Here $\underline{\mathscr{S}}^+$ is the space of all sequences $(f_i)_{i\in B}$ where $f_i\in \mathscr{S}_n^+$ when $i\in B_0^n$ with only a finite number of $f_i\neq 0$. $\mathscr{S}_n^+$ is the space of Schwarz functions on $M_n$ supported in $M_n^+$, where the later is defined as the subspace of $M_n$ in which ${\rm Re}(z_i)>0$ for all $i=1,\dots, n$.

Now, what I'm confused about is that $\ast$ map. It seems it should play the role of the adjoint, but I'm confused on why it is defined in the way it is defined. In fact, I still find it confusing for $\ast$ to act on the indices, but I feel this is because one is thinking of $(i_1,\dots, i_n)$ as associated to a particular field $\Phi_{i_1\dots i_n}$. In any case:

  1. What is the motivation to define this $\ast$ map in the way it is defined, i.e., as a map $\ast : B_0\to B_0$ with $\ast^2=1$ such that there is a continuation $\ast:B\to B$ for which $G_{i_1,\dots,i_n}(z)=G_{(i_1,\dots,i_n)^\ast}(-z^\ast)$?

  2. What is the relation of $\ast$ with the adjoint in the Hilbert space?

  3. Is this $\ast$ related to the $\ast$ map that appears in the $\ast$-algebra approach to axiomatic quantum field theory? It seems they are related, but the formalisms seem a bit distinct.

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  • $\begingroup$ Could the $*$ acting in the index set mean that you replace the fields in the Green functions by their adjoints? That makes sense as the usual definition of reflection positivity $\endgroup$
    – mike stone
    Commented Dec 27, 2023 at 17:39
  • $\begingroup$ I do think that $\Phi_{i^\ast}$ should be interpreted as the adjoint of $\Phi_i$, but I'm struggling to see that the given definition does the job. For example, in the $\ast$-algebra approach the $\ast$ operation is further demanded to obey $$(x+y)^\ast = x^\ast +y^\ast,\quad (\lambda x)^\ast = \bar \lambda x^\ast,\quad (xy)^\ast = y^\ast x^\ast,$$ for all $x,y\in \mathfrak{A}$ and $\lambda\in\mathbb{C}$. I don't see analogues of these here. I'm also a bit confused why $G_{i_1,\dots,i_n}(z)=G_{(i_1,\dots,i_n)^\ast}(-z^\ast)$ characterizes $\ast$ as the adjoint. $\endgroup$
    – Gold
    Commented Dec 27, 2023 at 18:15
  • $\begingroup$ Reflection positivity involves the inner products of states $f$ defined, in this case, on the $y$ axis. (Note that the $z\to -z^*$'s is reflection in the $y$ axis). Your last equation $(xy)^*= y^*x^*$ applies to operators not states. $\endgroup$
    – mike stone
    Commented Dec 27, 2023 at 18:20
  • $\begingroup$ I understand that the equations I wrote apply to operators. Maybe I could rephrase the question as: given that under the Hilbert space reconstruction $G_{i_1\dots i_n}(z_1,\dots, z_n) = \langle \Omega|\Phi_{i_1}(z_1)\cdots \Phi_{i_n}(z_n)|\Omega\rangle$, how does this $\ast$ relates to the Hilbert space adjoint $\dagger$? I'm trying to compute the adjoint defined by this inner product $\langle \underline{f},\underline{g}\rangle$ to see if I understand it... $\endgroup$
    – Gold
    Commented Dec 27, 2023 at 19:46

1 Answer 1

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The $*$ is not the adjoint. It is the operation of reflection in the $y$ axis.

I would change the wording so that $$ "G_{i_1,\dots,i_n}(z)=G_{(i_1,\dots,i_n)^\ast}(-z^\ast)",$$ becomes $$*:G_{i_1,\dots,i_n}(z)\mapsto G_{(i_1,\dots,i_n)^\ast}(-z^\ast).$$

O and S define the inner product of a state $$ |f\rangle \stackrel{\rm def}{=} \int_{x^0_i>0} d^dx_1 \cdots d^d x_n \Phi(x_1\ldots \Phi(x_n)|0\rangle f(x_1)f(x_2)\cdots f(x_n) $$ (where the $f$'s are supported on the positive time region) with itself as the correlator $$ \int_{x^0_i>0; \tilde x_i^0<0} dx_1 \ldots f(\tilde x_1) \cdots f(\tilde x_n) \langle\Phi^*_n(\tilde x_n) \cdots\Phi^*(\tilde x_1)\Phi(x_1) \Phi(x_n)\rangle f(x_1) \cdots f(x_n). $$ where each $\tilde x$ is located at the mirror image, due to reflection about the $x^0=0$ plane, of the corresponding $x$ point. Reflection positivity is the statement that this correlator is positive. This construction defines the vacuum state $|0\rangle$ as the object obtained this way when $n=0$. In other words the numerical values of the correlators define the Hilbert space in much the same way as the GNS construction in an abstract C$^*$ algebra defines a Hilbert space on which the algebra acts.

Under this inner product the complex variable $\Phi(x)$ that appears in a path integral becomes an operator whose adjoint $\Phi^\dagger$ is given by the $\Phi^*(x)$.

Oops: I left out the statement that in the definition of the state, the $x_i$ should be time ordered. That condition does not matter for the correlator as the correlator automatically corresponds to time-ordered fields in the opertor language.

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  • $\begingroup$ Thanks! I think I was led away by the notation and thought that $\ast$ had something to do with the adjoint. This makes more sense now. The only thing is that the book does indeed say that the Green's functions are supposed to obey $G_i(z)=G_{i^\ast}(-z^\ast)$. Isn't this usually required in the OS axioms? Maybe it is a mistake in the book? I have found another today, in the Conformal Ward Identities, so maybe this is another issue... $\endgroup$
    – Gold
    Commented Dec 27, 2023 at 23:14
  • $\begingroup$ I do not understand the $G(z)= G(-z^*)$ statement, but this might be my failure. I think there is a wikipedia article that might be clearer:en.wikipedia.org/wiki/Schwinger_function#Reflection_positivity $\endgroup$
    – mike stone
    Commented Dec 27, 2023 at 23:16

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