$(P_\alpha)_{\alpha \in A}$ be a net of hyperfinite rank projections tending towards the identity map. We have that $P_\alpha A$ also must have hyperfinite rank, and so $P_\alpha A \dashv_\# A$ in \#-norm sense. Well, then there exists some $\varepsilon \approx 0, \varepsilon > 0$ and some $v_\alpha \in H_1$ such that $\|(A - P_\alpha A)v_\alpha\|_\# > \varepsilon$. Since these $v_\alpha$ are in $H_1$, we can apply 3 to get some subnet such that $Av_\alpha \dashv_\# v$ in \#-norm. Then we have:

$$ 
\begin{array}{l}
0 < \varepsilon < \|(A - P_\alpha A)v_\alpha\|_\# = \|v - P_\alpha v + (1 - P_\alpha)(v_\alpha - v)\|_\# \le \dots \\
\dots \le \|v - P_\alpha v\|_\# + \|(1 - P_\alpha)(Av_\alpha - v)\|_\# \le \|1 - P_\alpha\|_\# (\|v\|_\# + \|Av_\alpha - v\|_\#) \to_\# 0.
\end{array}
$$

A contradiction. Hence, $3 \Rightarrow 1$.

**Corollary 1.2.1.** A \#-compact $\Leftrightarrow A(\cdot = \mathrm{Ext}-\sum_{n \in \mathbb{N}} \alpha_n \langle \cdot, v_n \rangle_\# w_n$ where $\{v_n\}_{n \in \mathbb{N}}$, $\{w_n\}_{n \in \mathbb{N}} \subset H^\#$, and $\{\alpha_n\}_{n \in \mathbb{N}} \subset \mathbb{C}_c^\#$ and s.t. \#-lim$_{n \to \infty} \alpha_n = 0$.

The \#-convergence on the RHS is with respect to the operator \#-norm.

## §1.3 Basic Spectral Theory

Spectrum is a generalisation of eigenvalues which is crucial for understanding operator algebras. Much of it is built upon whether operators or aren't invertible.

**Definition 1.3.1.** $A \in B(H^\#)$ is said to be invertible if there exists a $B \in B(H^\#)$ such that $AB = BA = 1$. If $X$ is an algebra, we define $\mathrm{Inv}(X) = \{x \in X \mid x \text{ is invertible}\}$.

**Lemma 1.3.1.** Neumann Series is \#-convergent

Let $\|A\|_\# < 1$. Then, $1 - A$ is invertible with inverse

$$ (1 - A)^{-1} = \mathrm{Ext}-\sum_{n \in \mathbb{N}} A^n. \qquad (1.3.1) $$

Where

$$ \mathrm{Ext}-\sum_{n \in \mathbb{N}} A^n = \#\mathrm{lim}_{N \to \infty} \left( \mathrm{Ext}-\sum_{n=0}^{N} A^n \right) \qquad (1.3.2) $$

where $N \in \mathbb{N}_\infty$.

**Proof.** The first question to ask is whether the series on the right hand side even \#-converges. It does as by lemma 1.2.2 one obtains

$$ \|\mathrm{Ext}-\sum_{n \in \mathbb{N}} A^n\|_\# \le \mathrm{Ext}-\sum_{n \in \mathbb{N}} \|A^n\|_\# \le \mathrm{Ext}-\sum_{n \in \mathbb{N}} \|A\|_\#^n = (1 - \|A\|_\#)^{-1} $$

Say it \#-converges to $B$. Then, we see that because we have a telescoping sum

$$ \mathrm{Ext}-\sum_{n=0}^{N} (A^n)(1-A) = 1 - A^{N+1} = (1-A)\left(\mathrm{Ext}-\sum_{n=0}^{N} A^n\right) $$

Hence, it is sufficient to check that $1 - A^n \dashv_\# 1$ as $n \to \infty$. Fix $\varepsilon \approx 0, \varepsilon > 0$, and choose $N \in \mathbb{N}_\infty$, $N > \mathrm{Ext}-\log_{\|A\|_\#} \varepsilon$. Then we have for $n > N$

$$ \|1 - A^n - 1\|_\# = \|A^n\|_\# \le \|A\|_\#^n \le \|A\|_\#^N < \varepsilon $$

**Lemma 1.3.2.** $\mathrm{Inv}(B(H^\#))$ is \#-open in $B(H^\#)$. Furthermore, the map

$$ {}^{-1}: \mathrm{Inv}(B(H^\#)) \to \mathrm{Inv}(B(H^\#)), A \mapsto A^{-1} $$

is \#-continuous with respect to the operator \#-norm.

**Proof.** Say $A \in \mathrm{Inv}(B(H^\#))$. Then, if $\|B - A\|_\# < \|A^{-1}\|_\#^{-1}$, we have

$$ \|BA^{-1} - 1\|_\# = \|(B - A)A^{-1}\|_\# \le \|B - A\|_\# \|A^{-1}\| < 1 $$

Which by lemma 1.3.1 gives

$$ 1 - (BA^{-1} - 1) = BA^{-1} \in \mathrm{Inv}(B(H^\#)) \Rightarrow BA^{-1}A = B \in \mathrm{Inv}(B(H^\#)). $$

Then if we consider $BA^{-1}$, we can note that $1/(1 - \|1 - BA^{-1}\|) > 1$ and hence

$$ \|BA^{-1}\|_\# \le \mathrm{Ext}-\sum_{n \in \mathbb{N}} \|1 - BA^{-1}\|_\#^n = \mathrm{Ext}-\sum_{n \in \mathbb{N}} \|(A - B)A^{-1}\|_\#^n \le $$

$$ \le \mathrm{Ext}-\sum_{n \in \mathbb{N}} \|(A - B)\|_\#^n \|A^{-1}\|_\#^n = 1/(1 - \|(A - B)\|_\# \|A^{-1}\|_\#). $$
$$ \varphi_i \otimes \psi_i = \mathrm{Ext} - \sum_{kl} \alpha_{kl}^i (\eta_k \otimes \theta_l) \qquad (6.1.2) $$

and

$$ \varphi'_j \otimes \psi'_j = \mathrm{Ext} - \sum_{kl} \beta_{kl}^j (\eta_k \otimes \theta_l). \qquad (6.1.3) $$

Since the two expressions for $f$ give the same vector, $\mathrm{Ext} - \sum_i c_i \alpha_{kl}^i = \mathrm{Ext} - \sum_j d_j \beta_{kl}^j$ for each pair $(k, l)$. Thus,

$$ \begin{aligned} (A \otimes B)[\mathrm{Ext} - \sum_i c_i (\varphi_i \otimes \psi_i)] &= \mathrm{Ext} - \sum_{kl} (\mathrm{Ext} - \sum_i c_i \alpha_{kl}^i) (A\eta_k \otimes B\theta_l) = \\ \mathrm{Ext} - \sum_{kl} (\mathrm{Ext} - \sum_j d_j \beta_{kl}^j) (A\eta_k \otimes B\theta_l) &= (A \otimes B)[\mathrm{Ext} - \sum_j d_j (\varphi'_j \otimes \psi'_j)] \end{aligned} \qquad (6.1.4) $$

so $A \otimes B$ is well defined. If $g$ is any vector in $D(A^*) \otimes D(B^*)$ then $\langle (A \otimes B)f, g \rangle_{\#} = \langle f, (A^* \otimes B^*)g \rangle_{\#}$ so

$$ D(A^*) \otimes D(B^*) \subset D((A \otimes B)^*). \qquad (6.1.5) $$

If $A$ and $B$ are #-closable, $D(A^*)$ and $D(B^*)$ are #-dense. Therefore, in that case $(A \otimes B)^*$ is #-densely defined which proves that $A \otimes B$ is #-closable. Similarly, if $A$ and $B$ are #-closable then $A \otimes I + I \otimes B$ defined on $D(A) \otimes D(B)$ is #-closable.

**Definition 6.1.1.** Let $A$ and $B$ be #-closable operators on a non Archimedean Hilbert spaces $H_1^\#$ and $H_2^\#$. The tensor product of $A$ and $B$ is the #-closure of the operator $A \otimes B$ defined on $D(A) \otimes D(B)$. We will denote the #-closure by $A \otimes B$ also. Usually $A \oplus B$ will denote the #-closure of $A \otimes I + I \otimes B$ on $D(A) \otimes D(B)$.

**Proposition 6.1.1.** Let $A$ and $B$ be bounded in $\mathbb{R}_c^\#$ operators on a non Archimedean Hilbert spaces $H_1^\#$ and $H_2^\#$. Then $\|A \otimes B\|_{\#} = \|A\|_{\#} \times \|B\|_{\#}$.

**Proof** Let $\{\varphi_k\}$ and $\{\psi_l\}$ be orthonormal bases for $H_1^\#$ and $H_2^\#$ and suppose $\mathrm{Ext} - \sum_{kl} c_{kl} \varphi_k \otimes \psi_l$ is a gyperfinite sum. Then

$$ \begin{aligned} \|(A \otimes I)[\mathrm{Ext} - \sum_{kl} c_{kl} \varphi_k \otimes \psi_l]\|_{\#}^2 &= \mathrm{Ext} - \sum_l \|\mathrm{Ext} - \sum_k c_{kl} A \varphi_k\|_{\#}^2 = \\ &\le \mathrm{Ext} - \sum_l \|A\|_{\#}^2 (\mathrm{Ext} - \sum_k |c_{kl}|^2) = \|A\|_{\#}^2 \|\mathrm{Ext} - \sum_k c_{kl} \varphi_k \otimes \psi_l\|_{\#}^2. \end{aligned} \qquad (6.1.6) $$

Since the set of such gyperfinite sums is #-dense in $H_1^\# \otimes H_2^\#$, we conclude that $\|A \otimes I\|_{\#} \le \|A\|_{\#}$. Thus $\|A \otimes B\|_{\#} \le \|A \otimes I\|_{\#} \times \|B \otimes I\|_{\#} \le \|A\|_{\#} \times \|B\|_{\#}$. Conversely, given $\varepsilon \approx 0, \varepsilon > 0$, there exist unit vectors $\varphi \in H_1^\#$, $\psi \in H_2^\#$ so that

$$ \|A\varphi\|_{\#} \ge \|A\|_{\#} - \varepsilon \qquad (6.1.7) $$

and

$$ \|B\psi\|_{\#} \ge \|B\|_{\#} - \varepsilon. \qquad (6.1.8) $$

Then

$$ \|A \otimes B(\varphi \otimes \psi)\|_{\#} = \|A\varphi\|_{\#} \times \|B\psi\|_{\#} \ge \|A\|_{\#} \times \|B\|_{\#} - \varepsilon \|A\|_{\#} - \varepsilon \|B\|_{\#} + \varepsilon^2. \qquad (6.1.9) $$

Since $\varepsilon > 0$ is arbitrary $\|A \otimes B\|_{\#} \ge \|A\|_{\#} \times \|B\|_{\#}$, which concludes the proof.

**Remark 6.1.1.** We notice that both of the above propositions have natural generalizations to arbitrary hyperfinite tensor products of operators. This can be proven directly or by using the associativity of the hyperfinite tensor product of a non Archimedean Hilbert spaces.

**Remark 6.1.2.** We turn now to questions of self-adjointness and spectrum. Let $(A_k)_{k=1}^N$ be a hyperfinite family of operators, $A_k$ self-#-adjoint on $H_k^\#$. We will denote the
Therefore $\|A^{-1} - B^{-1}\|_{\#} = \|A^{-1}(AB^{-1})^{-1}(B - A)A^{-1}\|_{\#} \le \|A^{-1}\|_{\#}^2 \|B - A\|_{\#} \|(AB^{-1})^{-1}\|_{\#} \le \|A^{-1}\|_{\#}^2 \|B - A\|_{\#} [1/(1 - \|(A - B)\|_{\#} \|A^{-1}\|_{\#})]$
We can see then that as $\|A - B\|_{\#} \to 0$, $\|A^{-1} - B^{-1}\|_{\#} \to 0$ as required.

**Definition 1.3.2. (Spectrum)**

Let $A \in B(H^{\#})$. We define the spectrum of $A$, denoted $\sigma(A)$ by
$\sigma(A) := \{\lambda \in *\mathbb{C}_{c}^{\#} | (A - \lambda \cdot 1) \notin \text{Inv}(B(H^{\#}))\}$, i.e. the set of all complex numbers
such that $A - \lambda 1$ is not invertible. We denote the complement of $\sigma(A)$ by $\varphi(A)$.

**Lemma 1.3.3.** Spectrum is a generalisation of an eigenvalue ($\mathbf{eigen}(A) \subset \sigma(A)$), i.e. if $\lambda$ is an eigenvalue of $A, \lambda \in \sigma(A)$

**Proof.** Say $\lambda$ is an eigenvalue of $A$. Then $v \in H^{\#} \setminus \{0\}$ s.t. $(A - \lambda)v = 0$. However, by linearity, $(A - \lambda)0 = 0$. As $(A - \lambda)$ is not injective it cannot be invertible, hence $\lambda \in \sigma(A)$

**Corollary 1.3.1.** Let $H^{\#}$ be hyperfinitefinite dimensional. Then, if $A \in B(H^{\#})$ we have that spectrum agrees with the eigenvalues, i.e. **eigen**(A) = σ(A).

**Proof.** By lemma 1.3.3, we only need to check the other direction. Up to choosing bases, we can assume $H^{\#} = *\mathbb{C}_{c}^{\dim(H^{\#})}$. In this case, $B(H^{\#})$ is just the $\dim(H^{\#}) \times \dim(H^{\#})$ square matrices. By standard results in linear algebra, we have that $(A - \lambda) \notin \text{Inv}(B(H^{\#}))$ iff $\text{Ext-det}(A - \lambda) = 0$ iff $\lambda$ is an eigenvalue.

**Lemma 1.3.4.** If $A \in B(H^{\#})$ then $\sigma(A)$ is #-closed as a subset of the complex plane $*\mathbb{C}_{c}^{\#}$.

Moreover, it is a subset of the disc of radius $\|A\|_{\#}$ centred at the origin.

**Proof.** Say $\lambda > \|A\|_{\#}$. Then, $-\lambda^{-1}\|A\|_{\#} < 1$ so $1 - \lambda^{-1}A$ is invertible by lemma 1.3.1.

Then, $\lambda \notin \sigma(A)$. Now examine $\sigma(A)^c = \varphi(A) = \{\lambda \in *\mathbb{C}_{c}^{\#} | A - \lambda \in \text{Inv}(B(H^{\#}))\}$.

Say $\lambda \in \varphi(A)$. By lemma 1.3.2 we have that there exists some $\varepsilon \approx 0, \varepsilon > 0$ such that $\|A - \lambda - B\|_{\#} < \varepsilon \Rightarrow B \in \text{Inv}(B(H^{\#}))$.

Now, we see that if $|\lambda - \hat{\lambda}| < \varepsilon$ we have:

$$ \|A - \lambda - (A - \hat{\lambda})\|_{\#} = |\lambda - \hat{\lambda}| < \varepsilon $$

Hence, $A - \hat{\lambda} \in \text{Inv}(B(H^{\#}))$. Then $\varphi(A)$ is #-open, and so $\sigma(A)$ is #-closed.
We need the following lemmas to show that $\sigma(A) \neq \emptyset$ ever.

**Lemma 1.3.5.** Let $A \in B(H^{\#})$. Then let $\gamma: B(H^{\#}) \to *\mathbb{C}_{c}^{\#}$ be an arbitrary linear functional $(\gamma \in B(H^{\#})^{*})$. We have that the map

$$ f_{A,\gamma}: \varphi(A) \to *\mathbb{C}_{c}^{\#}, \lambda \mapsto \gamma(1/(A-\lambda)) \text{ is #-analytic on } \varphi(A), \text{ and has } \#\text{-lim}_{\lambda \to *\infty} f_{A,\gamma}(\lambda) = 0 $$

**Proof.** For $\lambda, \lambda_0 \in \varphi(A)$ we have that

$$ \frac{1}{A-\lambda} - \frac{1}{A-\lambda_0} = \frac{A-\lambda_0 - A + \lambda}{(A-\lambda)(A-\lambda_0)} = \frac{\lambda-\lambda_0}{(A-\lambda)(A-\lambda_0)}. $$

Then,

$$ \#\text{-lim}_{\lambda \to *\infty} \lambda_0 \frac{f_{A,\gamma}(\lambda) - f_{A,\gamma}(\lambda_0)}{\lambda - \lambda_0} = \gamma\left(\frac{1}{(A-\lambda)(A-\lambda_0)}\right) = \gamma\left(\#\text{-lim}_{\lambda \to *\infty} \lambda_0 \frac{\lambda - \lambda_0}{(A-\lambda)(A-\lambda_0)}\right) = \dots $$

Where we use linearity of $\gamma$ in the first equality, and #-continuity of $\gamma$ in the second.
Then, by lemma 1.3.2 we have that

$$ \dots = \gamma\left(\frac{\lambda - \lambda_0}{\#\text{-lim}_{\lambda \to *\infty} \lambda_0 (A - \lambda)(A - \lambda_0)}\right) = \gamma\left(\frac{1}{(A - \lambda_0)^2}\right). $$

Hence, $f_{A,\gamma}$ is #-analytic on $\varphi(A)$. By the estimate

$$ \| \frac{1}{A-\lambda} \|_{\#} = \frac{1}{|\lambda|} \| \frac{1}{1-\lambda^{-1}A} \|_{\#} = \frac{1}{|\lambda|} \left( \| \text{Ext-} \sum_{n \in *\mathbb{N}} (\lambda^{-1}A)^n \|_{\#} \right) \le $$
\#-closure of $I_1 \otimes \cdots \otimes A_k \otimes \cdots \otimes I_N$ on $D = \text{Ext}-\otimes_{k=1}^N D(A_k)$ by $A_k$ also. Let $P(x_1, \ldots, x_N)$ be a polynomial with $\mathbb{R}_c^\#$-valued coefficients of degree $n_k$ in $x_k$. Then, the operator $P(A_1, \ldots, A_N)$ makes sense on $\text{Ext}-\otimes_k D(A_{n_k})$ since $D(A_{n_k}) \subset D(A_l)$ for all $l \le n_k$. In fact, $P(A_1, \ldots, A_N)$ is essentially self-\#-adjoint on that domain.

**Theorem 5.1.1.** Let $A_k$ be a self-\#-adjoint operator on $H_k^\#$. Let $P(x_1, \ldots, x_N)$ be a polynomial with $\mathbb{R}_c^\#$-valued coefficients of degree $n_k$ in the $k$-th variable and suppose that $D_k^l$ is a domain of essential self-\#-adjointness for $A_k^{n_k}$. Then,
(a) $P(A_1, \ldots, A_N)$ is essentially self-adjoint on $D^l = \text{Ext}-\otimes_{k=1}^N D_k^l$.

(b) The spectrum of $\overline{\text{Ext}-P(A_1, \ldots, A_N)}$ is the \#-closure of the range of $P(A_1, \ldots, A_N)$ on the product of the spectra of the $A_k$. That is $\sigma(\overline{\text{Ext}-P(A_1, \ldots, A_N)}) = \overline{\text{Ext}}(\sigma(A_1), \ldots, \sigma(A_N))$.

**Proof** We will first prove that $P(A_1, \ldots, A_N)$ is essentially self-\#-adjoint on $D = \text{Ext}-\otimes_{k=1}^N D(A_k^{n_k})$. By the spectral theorem, there is a \#-measure space $(M_k, \mu_k^\#)$ so that $A_k$ is unitarily equivalent to multiplication by a $\mathbb{R}_c^\#$-valued \#-measurable function $f_k$ on $L_2^\#(M_k, d^\# \mu_k^\#)$. Thus we may assume that $\mu_k^\#$ is hyperfinite and that $f_k \in \bigcap_{1 \le p < *\infty} L_p^\#(M_k, d^\# \mu_k^\#)$. Furthermore $\text{Ext}-\otimes_{k=1}^N L_2^\#(M_k, d^\# \mu_k^\#)$ is naturally isomorphic to $L_2^\#(\text{Ext}-\otimes_{k=1}^N M_k, \text{Ext}-\otimes_{k=1}^N d^\# \mu_k^\#)$. Under this isomorphism $P(A_1, \ldots, A_N)$ corresponds to multiplication by $P(f_1, \ldots, f_N)$ and $D$ corresponds to the set of hyperfinite linear combinations of hyperfinite linear combinations of functions $\text{Ext}-\prod_{i=1}^N \phi_i(m_i)$ such that $f_k^{n_k} \phi_k \in L_2^\#(M_k, d^\# \mu_k^\#)$.

To prove essential self-\#-adjointness we use result from functional calculus. First, since $\mu_k^\#$ is hyperfinite and $f_k^{n_k} \phi_k \in L_2^\#(M_k, d^\# \mu_k^\#)$ we conclude that $f_k^l \in L_p^\#(M_k, d^\# \mu_k^\#)$ for $1 \le p < *\infty$. From this it follows immediately that $P(f_1, \ldots, f_N)$ is in $L_p^\#$ for all such $p$; in particular $P(f_1, \ldots, f_N) \in L_4^\#(\text{Ext}-\otimes_{k=1}^N M_k, \text{Ext}-\otimes_{k=1}^N d^\# \mu_k^\#)$. Since $f_k^{n_k}$ is self-\#-adjoint on $D_k$, $D_k$ contains the characteristic functions of \#-measurable sets in $M_k$.

Thus $D$ contains all hyperfinite linear combinations of the characteristic functions of rectangles. By the property on product \#-measures we conclude that the characteristic function of any \#-measurable set in $M_k$ is equal to such a hyperfinite linear combination except on a set of arbitrarily small $\text{Ext}-\otimes_{k=1}^N d^\# \mu_k^\#$ \#-measure. Thus the simple functions on $\text{Ext}-\otimes_{k=1}^N M_k$ can be approximated in the $L_p^\#$ sense with $1 \le p < *\infty$ by elements of $D$. In particular $D$ is \#-dense in $L_4^\#(\text{Ext}-\otimes_{k=1}^N M_k, \text{Ext}-\otimes_{k=1}^N d^\# \mu_k^\#)$. Essential self-\#-adjointness now follows from Proposition 5.1.2.

To show that $P(A_1, \ldots, A_N)$ is essentially self-adjoint on $D^l$ we need only show that $\overline{\text{Ext}-P(A_1, \ldots, A_N)} \uparrow D^l$ extends $P(A_1, \ldots, A_N) \uparrow D$. Suppose $\text{Ext}-\otimes_{k=1}^N \phi_k \in D$. Then $\phi_k \in D(A_k^{n_k})$, so since $D_k^l$ is a domain of essential self-\#-adjointness of $A_k^{n_k}$ there is a hyper infinite sequence $(\phi_k^l)_{l=1}^{*\infty}$ so that $\phi_k^l \to_* \phi_k$ and $A_k^{n_k} \phi_k^l \to_* A_k^{n_k} \phi_k$. An easy estimate shows that this implies that $A_k^m \phi_k^l \to_* A_k^m \phi_k$ for all $1 \le m \le n_k$. Therefore $\text{Ext}-\otimes_{k=1}^N \phi_k^l \to_* \text{Ext}-\otimes_{k=1}^N \phi_k$ and $P(A_1, \ldots, A_N)(\text{Ext}-\otimes_{k=1}^N \phi_k^l) \to_* P(A_1, \ldots, A_N)(\text{Ext}-\otimes_{k=1}^N \phi_k)$ The same argument works for hyperfinite linear combinations of vectors of the form $\text{Ext}-\otimes_{k=1}^N \phi_k$ so $\overline{\text{Ext}-P(A_1, \ldots, A_N)} \uparrow D^l$ extends $P(A_1, \ldots, A_N) \uparrow D$. This completes the proof of (a).

To prove (b), suppose that $\lambda \in \sigma(\overline{\text{Ext}-P(A_1, \ldots, A_N)})$. If $I$ is any \#-open interval about $\lambda$ then $P^{-1}(A_1, \ldots, A_N)(I)$ contains a product $\text{Ext}-\otimes_{k=1}^N I_k$ of open \#-intervals so that $I_k \cap \sigma(A_k) \neq \emptyset$. Since $\sigma(A_k) = \text{\#-ess range}(f_k^{n_k})$, $\mu_k^\#[(f_k^{n_k})^{-1}(I_k)] \neq 0$ so $\mu[P(f_1, \ldots, f_N)(I)] \neq 0$. That is, $\lambda \in \text{\#-ess range}(P(f_1, \ldots, f_N))$ which equals
$$ \leq \frac{1}{|\lambda|} \left( \mathrm{Ext} - \sum_{n \in \mathbb{N}} \|(\lambda^{-1}A)^n\|_{\#}^n \right) = \frac{1}{|\lambda| - \|A\|_{\#}} . $$

It is clear that $\frac{1}{A-\lambda} \xrightarrow{\lambda \to \infty} 0$ as $\lambda \to \infty$ and hence by #-continuity of $f_{A_\gamma}$ we are done.

**Theorem 1.3.1.** If $A \in B(H^#)$ then $\sigma(A) \neq \emptyset$.

**Proof.** Say $\exists A \in B(H^#)$ such that $\sigma(A) = \emptyset$. For this $A$, we have that $f_{A_\gamma}$ is:

(i) $|f_{A_\gamma}(z)|$ bounded by positive constant $K \in \mathbb{R}_{c,+}^\#$

(ii) $f_{A_\gamma}(z)$ is #-entire function, that is $f_{A_\gamma}(z)$ is a $\mathbb{C}_c^\#$-valued function #-holomorphic on the whole $\mathbb{C}_c^\#$

(iii) $f_{A_\gamma}(z)$ has $\#\text{-lim}_{\lambda \to \infty} f_{A_\gamma}(\lambda) = 0$.

The only map satisfying these three properties is the zero map. But since $\gamma$ was arbitrary, this implies that an arbitrary functional is the zero functional, which is clearly a contradiction. Hence, $\sigma(A) \neq \emptyset$

In particular this means that $\sigma(A) \neq \text{eigen}(A)$ if $\text{eigen}(A)$ is empty.

**Theorem 1.3.2.** (Generalized Gelfand Mazur theorem)

If $\mathrm{Inv}(B(H^#)) = B(H^#) \setminus \{0\}$, Then $B(H^#) \cong \mathbb{C}_c^\#$.

**Proof.** Let $A \in B(H^#)$ then let $\lambda_A \in \sigma(A)$ we have $A - \lambda_A = 0$. So $\lambda_A$ is unique. Our isomorphism is then $\psi:B(H^#) \to \mathbb{C}_c^\# A \mapsto \lambda_A$.

**Theorem 1.3.3.** (Generalized Spectral Mapping Theorem)

Let $A \in B(H^#)$, $f \in \mathbb{C}_c^\#[z]$. Then we have: $\sigma(f(A)) = f(\sigma(A))$.

**Proof.** Let $\lambda \in \sigma(A)f(z) = \mathrm{Ext}-\sum_{n=0}^N a_n z^n$. Then

$$ f(A) - f(\lambda) = \mathrm{Ext}-\sum_{n=0}^{N} a_n (A^n - \lambda^n) = (A - \lambda) \left( \mathrm{Ext}-\sum_{n=0}^{N} a_n \left( \mathrm{Ext}-\sum_{j<n} (A^j \lambda^{n-j-1}) \right) \right). $$

$$ \text{So } f(\lambda) \in \sigma(f(A)). \text{ Say } \mu \notin f(\sigma(A)) \text{ Then, we can write } f(z) - \mu = a_N \left( \text{Ext} - \prod_{n=0}^{N} (z - \lambda_n) \right). $$

Then as $\mu - f(\lambda) \neq 0 \forall \lambda \in \sigma(A)$ (the zero operator isn't invertible) we have that

$$ \lambda_n \notin \sigma(A) (n \le N). \text{ Therefore, } f(A) - \mu = a_N \left( \text{Ext} - \prod_{n=0}^{N} (A - \lambda_n) \right), \text{ must be invertible, and } \mu \notin \sigma(f(A)). $$

This theorem has many forms and generalises much more than for $f$ being a polynomial.

**Definition 1.3.3.** (Spectral Radius)

Given $A$ in $B(H^#)$ the spectral radius, denoted $r(A)$, of $A$ is defined by $r(A) = \sup_{\lambda \in \sigma(A)} |\lambda|$.

We note by lemma 1.3.4 the supremum exists and is attained. In fact, the following lemma tells us what the spectral radius of a given operator is in terms of a #-limit.

**Lemma 1.3.6.** Let $A \in B(H^#)$. Then the #-limit: $\#\text{-lim}_{n \to \infty} \|A_n\|_{\#}^{1/n}$ exists, and is equal to $r(A)$, the spectral radius of $A$.

**Proof.** By theorem 1.3.4 and lemma 1.3.4 we have that

$$ [r(A)]^n = r(A^n) \le \|A_n\|_{\#} \Rightarrow r(A) \le \|A_n\|_{\#}^{1/n}, \quad n \in \mathbb{N} \Rightarrow r(A) \le \#\text{-lim} \inf_{n \to \infty} \|A_n\|_{\#}^{1/n}. $$

For the other direction, examine again the function from lemma 1.3.5, but this time restricted to $\Omega = \{z \in \mathbb{C}_c^\# \|z| > r(A)\}$. We know that $f_{A_\gamma}$ is analytic in $\Omega \subset \varphi(A)$. So

it has laurent expansion $\mathrm{Ext}-\sum_{n \in \mathbb{Z}_-} a_n z^n$ and also that $\#\text{-lim}_{z \to \infty} f_{A_\gamma}(z) = 0$. So in fact, we have laurent expansion $\mathrm{Ext}-\sum_{n \in \mathbb{N}} \frac{a_n}{z^n}$. To determine the coefficients we

know that for $z \in \Omega$, $\|\frac{A}{z}\|_{\#} < 1$ and hence, by lemma 1.3.1 one obtains

$$ \frac{1}{z-A} = \frac{1}{z(1-z^{-1}A)} = \frac{1}{z} \left( \mathrm{Ext} - \sum_{n=0}^{\infty} \frac{A^n}{z^n} \right) = \mathrm{Ext} - \sum_{n=0}^{\infty} \frac{A^n}{z^{n+1}} = \mathrm{Ext} - \sum_{n=1}^{\infty} \frac{A^{n-1}}{z^n}. $$
$\sigma(\#\overline{P(A_1, \dots, A_N)})$.

Conversely if $\lambda \notin \#\overline{P(\sigma(A_1), \dots, \sigma(A_N))}$ then $(\lambda - P(f_1, \dots, f_N))^{-1}$ is bounded #-a.e. on $Ext \times_{k=1}^N M_k$ so $\lambda \in \rho(\#\overline{P(A_1, \dots, A_N)})$.

**Remark 6.1.3.** If $A_1, \dots, A_N, N \in *\mathbb{N}$ are bounded in $*\mathbb{R}_c^\#$, $P(A_1, \dots, A_N)$ is #-closed, but in general it is not.

**Corollary 6.1.1.** Let $A_1, \dots, A_N, N \in *\mathbb{N}$ be self-#-adjoint operators on $H_1^\#, \dots, H_N^\#$ and suppose that, for each $k, D_k$ is a domain of essential self-#-adjointness for $A_k$. Then, (a) The operators $A_\pi = \text{Ext} \otimes_{k=1}^N A_k$ and $A_\Sigma = \text{Ext} - \sum_{k=1}^N A_k$ are essentially self-#-adjoint on $D = \text{Ext} \otimes_{k=1}^N D_k$.

(b) $\sigma(A_\pi) = \#\overline{\text{Ext} - \prod_{k=1}^N \sigma(\overline{A}_k)}$ and $\sigma(A_\Sigma) = \#\overline{\text{Ext} - \sum_{k=1}^N \sigma(\overline{A}_k)}$.

**Example 5.1.1.** Suppose that $V(x)$ is a potential so that $H_1 = -\nabla_x^\# + V(x)$ is essentially self-#-adjoint on $S^\#(*\mathbb{R}_c^3)$. Then $H_2 = -\nabla_x^\# + V(x) - \nabla_y^\# + V(y)$ is essentially self-#-adjoint on the set of hyperfinite sums of products $\varphi(x)\psi(y)$, with $\varphi, \psi \in S^\#(*\mathbb{R}_c^3)$. Further $\sigma(H_2) = \#\overline{\sigma(H_1) + \sigma(H_1)}$. is obvious.

## § 6.2. Non-Archimedean Fock spaces.

Let $H^\#$ be a non-Archimedean Hilbert space and denote by $H^{\#n}, n \in *\mathbb{N}$ the $n$-fold tensor product $H^{\#n} = \text{Ext} \otimes_{k=1}^n H^\#$ and define

$$ \mathcal{F}^{\#}(H^{\#}) = \text{Ext} - \bigotimes_{n=0}^{\infty} H^{\#n} \qquad (6.2.1) $$

$\mathcal{F}^{\#}(H^{\#})$ is called a non-Archimedean Fock space over $H^{\#}$; it will be *-separable if $H^{\#}$ is. For example, if $H^{\#} = L_2^\#(*\mathbb{R}_c^3)$, then an element $\psi \in \mathcal{F}(H^{\#})$ is a hyper infinite sequence of functions

$$ \psi = \{\psi_0, \psi_1(x_1), \psi_1(x_1, x_2), \psi_1(x_1, x_2, x_3), \dots, \} \qquad (5.2.2) $$

so that

$$ |\psi_0|^2 + \text{Ext} - \sum_{n=1}^{\infty} \left( \text{Ext} - \int_{*\mathbb{R}_c^{\#n}} \psi_n(x_1, \dots, x_n) d^{\#n}x \right) < *\infty, \qquad (5.2.3) $$

where $\psi_0 \in *\mathbb{C}_c^\#$, $d^{\#n}x = \text{Ext} - \prod_{i=1}^n d^{\#n}x_i$. Actually, it is not $\mathcal{F}^{\#}(H^{\#})$ itself, but two of its subspaces which are used in quantum field theory. These two subspaces are constructed as follows: Let $\mathbf{P}_n, n \in *\mathbb{N}$ be the permutation group on $n \in *\mathbb{N}$ elements and let $\{\varphi_k\}_{k=1}^{*\infty}$ be a basis for $H^{\#}$. For each $\sigma \in \mathbf{P}_n$ we define an operator (which we also denote by a) on basis elements of $H^{\#(n)}, n \in *\mathbb{N}$ by

$$ \sigma(\text{Ext} - \bigotimes_{i=1}^n \varphi_{k_i}) = \text{Ext} - \bigotimes_{i=1}^n \varphi_{k_{\sigma(i)}} \qquad (5.2.4) $$

a extends by linearity to a bounded in $*\mathbb{R}_c^\#$ operator (of #-norm one) on $H^\#$ so we can define

$$ \mathbf{S}_n = \left( \frac{1}{n!^{\#}} \right) \text{Ext} - \sum_{\sigma \in \mathbf{P}_n} \sigma. \qquad (5.2.5) $$

It is easy to show that $\mathbf{S}_n^2 = \mathbf{S}_n$ and $\mathbf{S}_n^* = \mathbf{S}_n$, so $\mathbf{S}_n$ is an orthogonal projection. The range of $\mathbf{S}_n$ is called the $n$-fold symmetric tensor product of $H^\#$. In the case where $H^\# = L_2^\#(*\mathbb{R}_c^\#)$ and $H^{\#n} = \text{Ext} \otimes_{k=1}^n L_2^\#(*\mathbb{R}_c^\#) = L_2^\#(*\mathbb{R}_c^{\#n})$, $\mathbf{S}_n H^{\#n}$ is just the subspace of $L_2^\#(*\mathbb{R}_c^{\#n})$ of all functions left invariant under any permutation of the variables. We now define
Hence, $f_{A,\gamma}(z) = \gamma(\frac{1}{z-A}) = \gamma(\text{Ext}-\sum_{n=1}^{\infty} \frac{A^{n-1}}{z^n}) = \text{Ext}-\sum_{n=1}^{\infty} \frac{\gamma(A^{n-1})}{z^n}$.

So we have $\lim_{n \to \infty} \frac{\gamma(A^{n-1})}{z^n} = 0$, for all functionals $\gamma \in H^{\#*}$. It follows that

$\#\text{-lim}_{n \to \infty} \frac{\|A^{n-1}\|_{\#}}{|z|^n} = 0$ and so $\forall z \in \Omega, |z| > \#\text{-lim sup}_{n \to \infty} \|A_n\|_{\#}^{1/n}$

then, $\forall z \in \#\overline{\Omega} |z| \ge \#\text{-lim sup}_{n \to \infty} \|A_n\|_{\#}^{1/n}$. In particular then, $r(A) \ge \#\text{-lim sup}_{n \to \infty} \|A_n\|_{\#}^{1/n}$ and so we are done.

**Remark 1.3.1.** If A is self adjoint, $\|A^2\|_{\#} = \|A\|_{\#}^2$ so by induction $\|A^{2n}\|_{\#} = \|A\|_{\#}^{2n}$ and therefore $r(A) = \#\text{-lim}_{n \to \infty} \|A^{2n}\|_{\#}^{1/2n} = \|A\|$.

## §1.4. $l_2^\#(G)$ and $B(l_2^\#(G))$.

**Definition 1.4.1.** Let G be a discrete, $\*-countable$ group. Then define

$$l_2^\#(G) = \{f:G \to *C_c^\# \mid \text{Ext}-\sum_{g \in G} |f(g)|^2 < *\infty\} \quad (1.4.1)$$

This is a non-Archimedean Hilbert space with respect to the inner product

$$\langle f | h \rangle_{\#} = \text{Ext}-\sum_{g \in G} f(g) \bar{h}(g). \quad (1.4.2)$$

**Lemma 1.4.1.** Let $g \in G, f \in l_2^\#(G)$ then define $g * f \in l_2^\#(G)$ by $g * f(h) = f(g^{-1}h)$. This defines a group action on $l_2^\#(G)$.

**Proof.** Fix $f \in l_2^\#(G)$. We verify directly, $\forall h, g_1, g_2 \in G$:

$$(g_1 \cdot g_2) * f(h) = f((g_1 \cdot g_2)^{-1}h) = f(g_2^{-1}g_1^{-1}h) = g_2 * f(g_1^{-1}h) = g_1 * (g_2 * f(h)).$$

**Definition 1.4.1.** Let $g \in G$, we define $T_g \in B(l_2^\#(G))$ as $T_g:l_2^\#(G) \to l_2^\#(G)$, $f \mapsto g * f$. Where $g * f$ is the group action as in lemma 1.4.1.

**Lemma 1.4.2.** $T_g$ has the following properties:

$$(i) \quad T_{g_1} \cdot T_{g_2} = T_{g_1 \cdot g_2} \quad (ii) \quad T_g^* = T_{g^{-1}}.$$

**Proof.** (i) This follows clearly from lemma 1.4.1.

(ii) Let $f, h \in l_2^\#(G)$. Then,

$$\begin{align*}
\langle T_g f | h \rangle_{\#} &= \text{Ext}-\sum_{a \in G} T_g f(a) \overline{h(a)} = \text{Ext}-\sum_{a \in G} g * f(a) \overline{h(a)} = \\
&= \left[ \text{Ext}-\sum_{a \in G} f(g^{-1}(ga)) \overline{h(a)} \right] \left[ \text{Ext}-\sum_{a \in G} f(g^{-1}a) \overline{h((ga))} \right] = \\
&= \text{Ext}-\sum_{a \in G} f(a) \overline{g^{-1} * h(a)} = \text{Ext}-\sum_{a \in G} f(a) \overline{T_{g^{-1}}h(a)} = \langle f | T_{g^{-1}}h \rangle_{\#}.
\end{align*}$$

## §1.5. Topologies on $B(H^\#)$.

In order to study a non-Archimedean von Neumann algebras, one needs to look into useful topologies on $B(H^\#)$. Since all operators are bounded in $*\mathbb{R}_c^\#$ we have the operator $\#\text{-norm}$ and therefore the induced topology.

**Definition 1.5.1.** $\#\text{-Norm Topology}$.

Using this norm, we can define a metric topology, using the induced $*\mathbb{R}_c^\#$-valued metric $d:B(H^\#) \times B(H^\#) \to R, d(T_1, T_2) = \|T_1 - T_2\|_{\#}$.

This topology is useful for many reasons, but for the purposes of looking at non-Archimedean von Neumann Algebras is somehow too “fine”. We need coarser topologies to enable us to have nice examples.

**Definition 1.5.2.** Strong Operator Topology (s.o.t.)

We define the strong operator topology as the coarsest topology such that $\forall v \in H^\#$ the map $\psi_v:B(H^\#) \to *\mathbb{R}_c^\#$, $T \mapsto \|T_v\|_{\#}$ is $\#\text{-continuous}$.
$$ \mathcal{F}_s^{\#}(H^\#) = \mathrm{Ext} - \bigoplus_{n=0}^{*\infty} S_n H^{\#n} \qquad (5.2.6) $$

$\mathcal{F}_s^{\#}(H^\#)$ is called the symmetric non Archimedean Fock space over $H^\#$ or the non Archimedean Boson Fock space over $H^\#$.

## § 6.3. Second quantization of the free Hamiltonian.

Let $H^\#$ be a non Archimedean Hilbert space, $\mathcal{F}^{\#}(H^\#)$ the associated non Archimedean Fock space over $H^\#$. Suppose that $A$ is a self-#-adjoint operator on $H^\#$ with a domain of

essential self-#-adjointness $D$. Corresponding to each such $A$ we can define an operator $d\Gamma^{\#}(A)$ on $\mathcal{F}^{\#}(H^\#)$ as follows. Let

$$ A^{(n)} = A \otimes I \otimes \cdots \otimes I + I \otimes A \otimes \cdots \otimes I + I \otimes \cdots \otimes I \otimes A \qquad (5.3.1) $$

on $\mathrm{Ext}\otimes_{i=1}^n D$ as follows. Let $D_A \subset \mathcal{F}^{\#}(H^\#)$ be the set of $\{\psi_0, \psi_1, \dots\}$ such that $\psi_n = 0$ for $n$ large enough and $\psi_n \in \mathrm{Ext}\otimes_{k=1}^n D$ for each $n$. $D_A$ is #-dense in $\mathcal{F}^{\#}(H^\#)$ since $D$ is #-dense in $H^\#$. Define $A^{(0)} = 0$ and $d\Gamma^{\#}(A) = \mathrm{Ext}\sum_{n=0}^{*\infty} A^{(n)}$. $d\Gamma^{\#}(A)$ makes sense on $D_A$ and obviously to be symmetric. By Theorem 5.1.1, $A^{(n)}$ is essentially self-#-adjoint on $\mathrm{Ext}\otimes_{k=1}^n D$. Thus $A^{(n)} + \mu i$ has a #-dense range on $\mathrm{Ext}\otimes_{k=1}^n D$ whenever $\mu \in *\mathbb{R}_c^\#$ and $\mu \neq 0$. From this it follows that $d\Gamma^{\#}(A) \pm i$ has a #-dense range on $D_A$. Thus $d\Gamma^{\#}(A)$ is essentially self-#-adjoint on $D_A$. If $A$ is the quantum mechanical operator which corresponds to the free energy, $d\Gamma^{\#}(A)$ is called the second quantization of the free energy. $d\Gamma^{\#}(A)$ commutes with the projections onto the symmetric and antisymmetric non Archimedean Fock spaces and it follows that $d\Gamma^{\#}(A) \upharpoonright \mathcal{F}_s^{\#}(H^\#)$ and $d\Gamma^{\#}(A) \upharpoonright \mathcal{F}_a^{\#}(H^\#)$

are essentially self-#-adjoint on $D \cap \mathcal{F}_s^{\#}(H^\#)$ and $D \cap \mathcal{F}_a^{\#}(H^\#)$ respectively.

## § 7.1. #-Convergence of unbounded in $*\tilde{\mathbb{R}}_c^\#$$^{\#}$ operators

One of the main difficulties with unbounded in $*\tilde{\mathbb{R}}_c^\#$$^{\#}$ operators is that they are only #-densely defined. This difficulty is especially troublesome when one wants to find a notion of #-convergence for a hyper infinite sequence $A_n \to_* A$, $n \in *\mathbb{N}$ of unbounded in $*\tilde{\mathbb{R}}_c^\#$$^{\#}$ operators since the domains of the operators $A_n$ may have no vector in common. For example, if $A_n = (1 - n^{-1})x$ on $L_2(*\tilde{\mathbb{R}}_c^\#)$, it is clear that in some sense $A_n \to_* A = x$; yet we could have been given domains $D(A_n)$ and $D(A)$ of essential self-#-adjointness for these operators which have no nonzero vector in common. Of course, in this simple case the #-closures of $A_n$ and $A$ all have the same domain, but in general this will not be true, and in any case, one is often forced to deal with domains of essential self-adjointness since closures of operators are sometimes difficult to compute. It is very natural to say that self-#-adjoint operators are “close” if certain bounded in $*\tilde{\mathbb{R}}_c^\#$$^{\#}$ functions of them are “close.” Most of this section is devoted to this approach. However, we also introduce graph #-limits, a topic which will be explored further.

**Definition 7.7.1.** Let $(A_n)_{n \in *\mathbb{N}}$ and $A$ be self-#-adjoint operators. Then $A_n$ is said to #-converge to $A$ in the #-norm resolvent sense (or #-norm generalized sense) if $R_\lambda(A_n) \to_* R_\lambda(A)$ in #-norm for all $\lambda$ with $\mathrm{Im}\,\lambda \neq 0$. $A_n$ is said to #-converge to $A$ in the strong resolvent sense (or strong generalized sense) if $R_\lambda(A_n) \to_* R_\lambda(A)$
**Example 1.5.1.** For $H^# = '2(N)$, let $T_n : H^# \to H^# v \mapsto (v, en)en$ We have that $T_n \to 0$ in the s.o.t. but not in the #-norm topology.

$T_n \to 0$ in #-norm sense, since $\|Tn\| \ge \|Tn(en)\|H^# = \|en\| = 1 \forall n$.

However in the strong operator topology, we have that

$$T_n \to_\# 0 \Leftrightarrow \psi_v(T_n) \to_\# 0 \forall v \in H^#$$

In this case, $v \in '2$, so in particular the entries of $v$ always tend to zero, i.e.

$$\psi_v(T_n) \to_\# 0 \forall v \in H^#.$$

This distinguishes the strong and the #-norm topologies. Making use of the adjoint, we define a finer topology

**Definition 1.5.3.** Strong-\*Operator Topology (s\*.o.t.)

We define the strong-\* operator topology as the coarsest topology such that $\forall v \in H^#$ the two maps

$$\psi_v:B(H^#) \to *\mathbb{R}_c^#, T \mapsto \|Tv\|_{\#} \quad (1.5.1)$$

and

$$\psi_v^*:B(H^#) \to *\mathbb{R}_c^#, T \mapsto \|T^*v\|_{\#} \quad (1.5.2)$$

are both #-continuous.

And finally, making use of the inner product we define the weak operator topology:

**Definition 1.5.4.** Weak Operator Topology (w.o.t.)

We define the weak operator topology (w.o.t.) as the coarsest topology such that $\forall v, w \in H^#$ the map $\psi_{vw}:B(H^#) \to *\mathbb{R}_c^#, T \mapsto |\langle v, Tw \rangle_{\#}|$ is #-continuous.

**Lemma 1.5.1.** For the topologies as in Definitions 1.5.1, 1.5.2, 1.5.3 and 1.5.4 we have that w.o.t. < s.o.t. < s-\*.o.t < #-norm topology.

**Lemma 1.5.2.** A basis for the strong operator topology is given by

$$\beta = \{N(A, \{v_i\}_{i=1}^N, ) | A \in B(H^#) v_i \in H^# > 0\}$$

$$\text{where } N(A, \{v_i\}_{i=1}^N, ) := \{B \in B(H^#) \mid \|(A - B)v_i\|_{\#} < i = 1, 2, \dots, N\}$$

**Proof.** First we need to check it is a basis for a topology.

(i) It covers $B(H^#)$ since for example

$$N(A, \{0\}, 1) = \{B \in B(H^#) \mid \|(A - B)0\|_{\#} < 1\} = \{B \in B(H^#) \mid \|0\|_{\#} < 1\} = B(H^#).$$

(ii) It is closed under intersection since for $C \in N(A, \{v_i\}_{i=1}^N, \varepsilon) \cap N(B, \{\hat{v}_i\}_{i=1}^N, \hat{\varepsilon})$.

We have that $C \in N(C, \{v_i\}_{i=1}^N \cup \{\hat{v}_i\}_{i=1}^N, \min\{\varepsilon - \|(C - A)v_i\|_{\#}, \hat{\varepsilon}\} - \|(C - B)\hat{v}_i\|_{\#})$). The only thing we need to verify for this is that (w.l.o.g.)

$$\forall D \in N(C, \{v_i\}_{i=1}^N \cup \{\hat{v}_i\}_{i=1}^N, \min\{\varepsilon - \|(C - A)v_i\|_{\#}, \hat{\varepsilon}\} - \|(C - B)\hat{v}_i\|_{\#}), \\ D \in N(A, \{v_i\}_{i=1}^N, \varepsilon). \text{ This is clear since for all } i$$

$$\|(D - A)v_i\|_{\#} \le \|(D - C)v_i\|_{\#} + \|(C - A)v_i\|_{\#} \le \varepsilon - \|(C - A)v_i\|_{\#} + \|(C - A)v_i\|_{\#} = \varepsilon.$$

Now we need to show that for all topologies such that $\forall v \in H^#$ the map

$$\psi_v:B(H^#) \to *\mathbb{R}_c^#, T \mapsto \|Tv\|_{\#} \text{ is #-continuous, subsets of this form are #-open.}$$

$$\text{Noting that } N(A, \{v_i\}_{i=1}^N, ) = \bigcap_{i=1}^N \psi_{\hat{v}_i}^{-1}(\|Av_i\|_{\#} - \varepsilon, \|Av_i\|_{\#} + \varepsilon). \text{ This is clear.}$$

**Lemma 1.5.3.** A basis for the weak operator topology is given by:

$$\beta = \{N(A, \{v_n\}_{n \le N}, \{w_n\}_{n \le N}, \varepsilon) \mid A \in B(H^#), \{v_n\}_{n \le N}, \{w_n\}_{n \le N} \subset H^#, \varepsilon > 0\}.$$

$$\text{Where } N(A, \{v_n\}_{n \le N}, \{w_n\}_{n \le N}, \varepsilon) = \{B \in B(H^#) \mid \|(B - A)v_n, w_n\|_{\#} < \varepsilon \forall n \le N\}.$$

We omit the proof of this result. It is similar to the proof of the basis of the SOT.

**Lemma 1.5.4.** Let $f:B(H^#) \to *\mathbb{C}_c^#$ be a linear functional. The following are equivalent:

(i) $\exists \{v_n\}_{n \le N}, \{w_n\}_{n \le N} \subset H^#,$ such that $f(A) = \text{Ext-}\sum_{n \le N} \langle Av_n, w_n \rangle_{\#} \forall A \in B(H^#).$
strongly for all $\lambda$ with $\mathrm{Im}\,\lambda \neq 0$.

We have not introduced the notion of weak resolvent #-convergence since weak resolvent #-convergence implies strong resolvent #-convergence. The following theorem shows that #-norm resolvent #-convergence is the right generalization of #-norm convergence for bounded in $^*\mathbb{R}_c^\#$ self-#-adjoint operators. A similar result holds for strong resolvent #-convergence, but the analogue for weak #-convergence is not true.

**Theorem 7.1.1.** Let $(A_n)_{n=1}^\infty$ and $A$ be a family of uniformly bounded in $^*\mathbb{R}_c^\#$

self-#-adjoint operators. Then $A_n \to_A A$ as $n \to \infty$ in the #-norm resolvent sense if and only if $A_n \to_A A$ as $n \to \infty$ in #-norm.

**Proof.** Let $A_n \to_A A$ as $n \to \infty$ in #-norm. Then if $\mathrm{Im}\,\lambda \neq 0$, $(A_n - A)(A - \lambda)^{-1} \to_A 0$ in #-norm. Thus, using the equality $(A_n - \lambda)^{-1} = (A - \lambda)^{-1}[I + (A_n - A)(A - \lambda)^{-1}]^{-1}$ we obtain that $(A_n - \lambda)^{-1} \to_A (A - \lambda)^{-1}$ in #-norm as $n \to \infty$.

Conversely, suppose $A_n \to_A A$ as $n \to \infty$ in the #-norm resolvent sense. Then, since $A_n - A = (A_n - i)(A_n - i)^{-1}[(A - i)^{-1} - (A_n - i)^{-1}](A - i)$, we conclude that $\|A_n - A\|_{\#} \le (\sup_n \|A_n\|_{\#} + 1)\|(A - i)^{-1} - (A_n - i)^{-1}\|_{\#}(\|A\|_{\#} + 1) \to_A 0$, as $n \to \infty$.

The following theorem shows that to prove generalized convergence one need only show #-convergence of the resolvents at one point off the hyperreal axis $^*\tilde{\mathbb{R}}_c^\#$.

**Theorem 7.1.2.** Let $(A_n)_{n=1}^\infty$ and $A$ be self-#-adjoint operators, and let $\lambda_0 \in \mathbb{C}_c^\#$.

(a) If $\mathrm{Im}\,\lambda_0 \neq 0$ and $\|R_{\lambda_0}(A_n) - R_{\lambda_0}(A)\|_{\#} \to_A 0$, then $A_n \to_A A$ as $n \to \infty$ in the #-norm resolvent sense.

(b) If $\mathrm{Im}\,\lambda_0 \neq 0$, and if $R_{\lambda_0}(A_n)\varphi - R_{\lambda_0}(A)\varphi \to_A 0$, as $n \to \infty$ in the strong resolvent sense.

**Proof** (a) Both $R_\lambda(A)$ and $R_\lambda(A_n)$ are analytic in the half-plane of $^*\tilde{\mathbb{C}}_c^\#$ containing $\lambda_0$ and have hyper infinite power series around $\lambda_0$,

$$ 
\begin{align} 
R_\lambda(A) &= \mathrm{Ext} - \sum_{m=0}^{\infty} (\lambda_0 - \lambda)^m [R_{\lambda_0}(A)]^{m+1}, \tag{7.1.1} \\ 
R_\lambda(A_n) &= \mathrm{Ext} - \sum_{m=0}^{\infty} (\lambda_0 - \lambda)^m [R_{\lambda_0}(A_n)]^{m+1} \nonumber 
\end{align} 
$$

which #-converge in #-norm in the circle $|\lambda - \lambda_0| < |\mathrm{Im}\,\lambda_0|^{-1}$. Since $R_{\lambda_0}(A_n) \to_A R_{\lambda_0}(A)$ in #-norm, $R_\lambda(A_n) \to_A R_\lambda(A)$ in #-norm for $A$ in this circle. Therefore, by repeating this process, we get #-convergence for all $A$ in the half-plane of $^*\tilde{\mathbb{C}}_c^\#$ containing $\lambda_0$.

Furthermore, since

$$ 
\begin{align} 
\|R_{\lambda_0}(A_n) - R_{\lambda_0}(A)\|_{\#} &= \|(R_{\lambda_0}(A_n) - R_{\lambda_0}(A))^*\|_{\#} = \tag{6.1.2} \\ 
\|R_{\lambda_0}(A_n) - R_{\lambda_0}(A)\|_{\#} &\to_A 0 \quad \text{as } n \to \infty \nonumber 
\end{align} 
$$

the same argument shows that the resolvents converge in #-norm in the half-plane of $^*\tilde{\mathbb{C}}_c^\#$ containing $\lambda_0$.

(b) The proof is the same as the proof of (a) except for two things. First, we consider the vector-valued functions $R_\lambda(A_n)\varphi$ and $R_\lambda(A)\varphi$. Secondly, since the map $T \to T^*$ is not #-continuous in the strong topology, one needs a separate argument to get from one half-plane of $^*\tilde{\mathbb{C}}_c^\#$ to the other. Suppose that $\lambda_0$ is in the lower half-plane of $^*\tilde{\mathbb{C}}_c^\#$. Then, as in (a), we get #-convergence everywhere in the lower half-plane of $^*\tilde{\mathbb{C}}_c^\#$,
(ii) $f$ is #-continious in the weak sense

(iii) $f$ is #-continious in the strong sense

**Proof.** It is clear that the first implies the second, and by lemma 1.5.1 that the second implies the third. Hence all we must show is that for all $f$ is #-continious in the strong sense then we can find $\{v_n\}_{n \le N}$, $\{w_n\}_{n \le N} \subset H^#$, such that $\forall A \in B(H^#)$ :

$f(A) = \text{Ext}-\sum_{n \le N} \langle Av_n, w_n \rangle_{\#} \quad \forall A \in B(H^#)$. Suppose $f$ is #-continious in the strong sense, then the inverse image of the #-open ball in $\mathbb{C}_c^\#$ is #-open in the strong operator topology. Considering our basis elements, then there is some constant $\kappa > 0$ and $\{v_n\}_{n \le N}$ such that $|f(A)|^2 \le \kappa (\text{Ext}-\sum_n \|Tv_n\|_{\#}^2)$. Now consider the subspace of $\text{Ext}-H^# \oplus H^# \dots \oplus H^#$ given by $\{\oplus_{n \le N} Av_n|A \in B(H^#)\}$ we can define a linear functional on this set by $\oplus_{n \le N} Av_n \mapsto f(A)$. Then by the generalized Riesz representation theorem, $\exists \{w_n\}_{n \le N}$ such that $f(A) = \text{Ext}-\sum_{n \le N} (Av_n, w_n) \forall A \in B(H^#)$ as required.

#2. Non-Archemedean Banach algebras and $C_{\#}^*$-Algebras.

## §2.1. Initial Definitions and #-Continious Functional Calculus.

von Neumann Algebras are a specific type of $C_{\#}^*$ algebra, and so it is important to understand well the theory of $C_{\#}^*$ algebras before non-Archemedean von Neumann Algebras.

**Definition 2.1.1.** A non-Archemedean Banach algebra $A_{\#}$ is a complex algebra over field

$\mathbb{C}_c^*$ which is a non-Archemedean Banach space under a $\mathbb{R}_c^*$-valued #-norm which is submultiplicative:

$$ \|xy\|_{\#} \le \|x\|_{\#} \|y\|_{\#} \qquad (2.1.1) $$

for all $x, y \in A_{\#}$.

**Definition 2.1.2.** An involution on a non-Archemedean Banach algebra $A_{\#}$ is a conjugate-linear #-isometric antiautomorphism of order two, usually denoted $x \mapsto x^*$. In other words,

$$ 1.(x^*)^* = x, \|x^*\| = \|x\| $$

$$ 2.(x + y)^* = x^* + y^*, $$

$$ 3.(xy)^* = y^*x^*, $$

$$ 4.(\lambda x)^* = \bar{\lambda}x^*, \text{ for all } x, y \in A, \lambda \in \mathbb{C}_c^*. $$

**Definition 2.1.3.** Spectrum (of an element of some a non-Archemedean algebra)
Let $A_{\#}$ be some a non-Archemedean algebra and $a \in A_{\#}$ we define

$$ \sigma(a) = \{\lambda \in \mathbb{C}_c^* | a - \lambda \text{1 is not invertible}\}. $$

**Definition 2.1.4.** A Banach #-algebra is a non-Archemedean Banach algebra $A_{\#}$ with an

involution. An $C_{\#}$-algebra is a Banach #-algebra $A_{\#}$ satisfying the $C_{\#}^*$-axiom: for all $x \in A_{\#}$

$$ \|x^*x\|_{\#} = \|x\|_{\#}^2. \qquad (2.1.2) $$

**Example 2.1.1.** $B(H^#)$ is a $C_{\#}^*$ Algebra)

We see this is an immediate consequence of **lemma 1.4**

**Lemma 2.1.1.** $K \subset B(H^#)$ is a $C_{\#}^*$ Algebra iff
in particular at $\lambda = -i$. The formula

$$ (A_n - i)^{-1} - (A - i)^{-1} = \qquad (6.1.3) \\ [(A_n + i)(A_n - i)^{-1}] [(A_n + i)^{-1} - (A + i)^{-1}] [(A_n + i)(A_n - i)^{-1}] $$

which follows from elementary calculations, can then be used to prove that hyper infinite sequence $(A_n - i)^{-1}, n \in *\mathbb{N} \#\text{-converges strongly to } (A - i)^{-1}$. The above argument then shows that hyper infinite sequence $R_\lambda(A_n)$, $n \in *\mathbb{N} \#\text{-converges strongly to } R_\lambda(A)$ everywhere in the upper half-plane of $*\tilde{\mathbb{C}}_c^\#$.

For alternative ways of proving that strong #-convergence, $R_\lambda(A_n) \xrightarrow{s} \# R_\lambda(A)$ in one half-plane implies strong #-convergence in the other half-plane, see Theorem 6.1.9. We will investigate several aspects of generalized #-convergence. First, we ask how resolvent #-convergence is related to the #-convergence of other bounded functions of $A_n$ and $A$. Secondly, we investigate the relationship between the spectra of $A_n$ and the spectrum of $A$ if $A_n \to_{\#} A$ in a generalized sense. Finally, we give criteria on the operators $A_n$, $A$ themselves which are sufficient to guarantee that $A_n \to_{\#} A$ as $n \to *\infty$ in a generalized sense.

**Theorem 7.1.3.** Let $A_n$ and $A$ be self-#-adjoint operators.

(a) If $A_n \to_{\#} A$ as $n \to *\infty$ in the #-norm resolvent sense and $f$ is a #-continuous function on $*\tilde{\mathbb{R}}_c^\#$ vanishing at $*\infty$, then $\|f(A_n) - f(A)\|_{\#} \to_{\#} 0$ as $n \to *\infty$

(b) If $A_n \to_{\#} A$ in the strong resolvent sense and $f$ is a bounded in $*\tilde{\mathbb{R}}_c^\#$ #-continuous function on $*\tilde{\mathbb{R}}_c^\#$, then $f(A_n)\varphi \to_{\#} f(A)\varphi$ as $n \to *\infty$, for all $\varphi \in H^\#$.

**Proof** By the generalized Stone-Weierstrass theorem, polynomials in $(x + i)^{-1}$ and $(x - i)^{-1}$ are #-dense in $C^{*\infty}(*\tilde{\mathbb{R}}_c^\#)$, the #-continuous functions vanishing at hyper infinity. Thus, given $\varepsilon \approx 0, \varepsilon > 0$, we can find an hyperfinite polynomial $P(s, t)$ so that

$$ \|f(x) - P((x + i)^{-1}, (x - i)^{-1})\|_{s\infty} \le \frac{\varepsilon}{3}. \qquad (7.1.4) $$

Therefore,

$$ \|f(A_n) - P((A_n + i)^{-1}, (A_n - i)^{-1})\|_{s\infty} \le \frac{\varepsilon}{3} \qquad (7.1.5) $$

and

$$ \|f(A) - P((A + i)^{-1}, (A - i)^{-1})\|_{s\infty} \le \frac{\varepsilon}{3}. \qquad (7.1.6) $$

If $A_n \to_{\#} A$ as $n \to *\infty$ in the #-norm resolvent sense, then

$$ P((A_n + i)^{-1}, (A_n - i)^{-1}) \to_{\#} P((A + i)^{-1}, (A - i)^{-1}) \qquad (7.1.7) $$

in #-norm as $n \to *\infty$, and thus for hyperfinite $n$ large enough, $\|f(A_n) - f(A)\|_{\#} \le \varepsilon$. This proves (a).

To prove (b) we first note that the same proof as above shows that if $A_n \to_{\#} A$ in the strong resolvent sense and $h \in C^{*\infty}(*\tilde{\mathbb{R}}_c^\#)$, then $h(A_n)\varphi \to_{\#} h(A)\varphi$. Let $\psi \in H^\#$ and $\varepsilon \approx 0, \varepsilon > 0_{*\tilde{\mathbb{R}}_c^\#}$ be given and define $g_m(x) = \text{Ext}-\text{exp}(-x^2/m)$. Since $g_m(x) \to_{\#} 1$ pointwise,

$g_m(A)\psi \to_{\#} \psi$ by spectral theorem, so we can find an $m$ with $\|g_m(A)\psi - \psi\|_{\#} \le \varepsilon(6\|f\|_{s\infty})^{-1}$. Furthermore since $g_m \in C^{*\infty}(*\tilde{\mathbb{R}}_c^\#)$, $g_m(A_n)\psi \to_{\#} g_m(A)\psi$ by the remark above, so we can find an $N_0$, so that $n > N_0$ implies $\|g_m(A_n)\psi - g_m(A)\psi\|_{\#} \le \varepsilon(6\|f\|_{s\infty})^{-1}$. Therefore, if $n \ge N_0$,
(i) K is an algebra over $*C_c^#$

(ii) $K = K^*$

(iii) K is #-closed with respect to the #-norm topology.

**Proof.** It is clear that if $K$ is a $C_{\#}^*$ algebra it must be closed with respect to the #-norm topology and an algebra. To see the other direction, we note that the only conditions we must check are conditions of #-closure by **lemma 1.4** all of the operations work algebraically as they should. We have

1. $K$ is #-closed under taking sums, scalar multiples and products as it is an algebra.

2. $K$ is #-closed under taking adjoints by the second bullet point

3. $K$ is #-closed with respect to the #-norm topology by the third bullet point.

Therefore, $K$ is a $C_{\#}^*$ algebra.

**Example 2.2.** $K(B(H^{\#}))$ is a $C_{\#}^*$ algebra. This follows clearly from lemma 2.1.1 and theorem

**1.22.** as $K(B(H^{\#})) = \{A \in B(H^{\#}) \mid \Re(A) < *\infty\} = \mathcal{F}\Re(B(H^{\#}))$ and $\mathcal{F}\Re(B(H^{\#})) = \{A \in B(H^{\#}) \mid \Re(A) < *\infty\}$ is a *-algebra.

**Example 2.1.3.** The set $\mathcal{F}\Re(B(H^{\#}))$ is in general a *-subalgebra of $B(H^{\#})$ but is not a $C_{\#}^*$ algebra if $H^{\#}$ is hyper infinite. This can be seen by considering an orthonormal basis $\{e_i\}_{i \in \mathbb{X}}$ and

considering $p_i$ to be the orthonormal projection into the line spanned by $e_i(p_i(e_j) = \delta_{ij})$ then the hyper infinite sequence $(q_N)_{N \in \mathbb{N}}$ where $q_N = \text{Ext}-\sum_{i=1}^N p_i$ #-converges in #-norm to the identity, which would not be hyperfinite rank.

As promised, we return to spectral theory, with a more general version of theorem

1.34.

**Theorem 2.1.1.** #-Continious functional calculus

Let $K_1, K_2$ be $C_{\#}^*$ algebras and $A \in K_1$ normal, then we have:

(i) The map $\psi:C^{\#}(\sigma(A)) \to K_1f \mapsto f(A)$ is a homomorphism.

(ii) For all $f \in C^{\#}(\sigma(A))$ we have $\sigma(f(A)) = f(\sigma(A))$

If $\Psi:K_1 \to K_2$ is a $C_{\#}^*$-homomorphism, then $\Psi(f(A)) = f(\Psi(A))$

This of course raises a few questions, how for example, would one take the square root of

an operator? For the purposes of these notes we don't look too deeply into this, but one

way to define this we can take any sequence $f_n \in *C_c^#[z]$ which approximates $f$ locally uniformly well,and take $f(A) := \#\lim_{n \to *\infty} f_n(A)$.

Most of these definitions we get are intuitive, for example for $f(z) = |z|^2$, we take
$f(A) = A^*A$

§ 2.2 *C\#-valued States

**Definition 2.2.1.** If $K$ is a $*$ algebra, a state is a linear $*C_c^#$-valued functional that is positive and normalised. That is: $\omega:K \to *C_c^#$ such that:

(i) $\omega(A^*A) \geq 0 \forall A \in K$

(ii) $\omega(1) = 1$.

**Notation 2.2.1.** We denote the space of all states on $A$ by $S^#(A)$.

Throughout the rest of this subsection, $K$ will refer to a $C_{\#}^*$ algebra and we will consider states on $K$.

**Example 2.2.2.** Let $K = M_n(*C_c^#), n \in *\mathbb{N}$ the $n \times n$ matrices with complex coefficients.
$$ \|g_m(A_n)\psi - \psi\|_{\#} \le \varepsilon(3\|f\|_{*\infty})^{-1}. \qquad (7.1.8) $$

Since $fg_m$ is #-continuous and goes to zero at $*\infty$, there is an $N_1$ so that $n \ge N_1$ implies

$$ \|f(A_n)g_m(A_n)\psi - f(A)g_m(A)\psi\|_{\#} \le \frac{\varepsilon}{3}. \qquad (7.1.9) $$

Let $N = \max(N_0, N_1)$. Then for $n \ge N$,

$$ \|f(A_n)\psi - f(A)\psi\|_{\#} \le \|f(A_n)g_m(A_n)\psi - f(A)g_m(A)\psi\|_{\#} + \|A_n\|_{\#}\|g_m(A_n)\psi - \psi\|_{\#} + \|A\|_{\#}\|g_m(A)\psi - \psi\|_{\#}. \qquad (7.1.10) $$

Since $\psi$ and $\varepsilon$ were arbitrary, this proves (b).

As an example of an application of part (a) let $(A_n)_{n=1}^{*\infty}$ and $A$ be positive self-#-adjoint operators. Then, if $A_n \to_* A$ in the #-norm resolvent sense $\mathrm{Ext}\text{-exp}(-tA_n)$ #-converges in #-norm to $\mathrm{Ext}\text{-exp}(-tA)$ for each positive $t$. To see that part (a) does not extend to all of $C^#(*\tilde{\mathbb{R}}_c^#)$, notice that on $L_2^#(*\tilde{\mathbb{R}}_c^#)$ the operators $A_n = (1-n^{-1})x$ #-converge to the operator $A = x$ in the #-norm resolvent sense but $\|\mathrm{Ext}\text{-exp}(iA_n) - \mathrm{Ext}\text{-exp}(iA_n)\|_{\#} = 1$ for all $n \in *\mathbb{N}$.

An important application of part (b) is the following generalization of the classical Trotter theorem.

**Theorem 7.1.4.** Let $(A_n)_{n=1}^{*\infty}$ and $A$ be self-#-adjoint operators. Then $A_n \to_* A$ in the strong resolvent sense if and only if $\mathrm{Ext}\text{-exp}(itA_n)$ #-converges strongly to $\mathrm{Ext}\text{-exp}(itA)$ for each $t$.

**Proof** Since $\mathrm{Ext}\text{-exp}(itx)$ is a bounded #-continuous function of $x$, Theorem 7.1.3 implies that if $A_n \to_* A$ in the strong resolvent sense, then $\mathrm{Ext}\text{-exp}(itA_n) \to_* \mathrm{Ext}\text{-exp}(itA)$ as $n \to *\infty$, strongly for each $t$.

To prove the theorem in the other direction, we first derive a formula for the resolvent of a self-#-adjoint operator $A$. Suppose that $\mathrm{Im}\,\mu < 0_{*\tilde{\mathbb{R}}_c^#}$. Then, by the functional calculus

$$ \begin{aligned} \langle \psi, R_{\mu}(A)\varphi \rangle_{\#} &= \mathrm{Ext} - \int_{*\tilde{\mathbb{R}}_c^#} \left( \frac{1}{\mu - \lambda} \right) d^{\#} \langle \psi, P_{\lambda}\varphi \rangle_{\#} = \\ & \mathrm{Ext} - \int_{*\tilde{\mathbb{R}}_c^#} \left( \mathrm{Ext} - \int_{0}^{*\infty} i[\mathrm{Ext}\text{-exp}(-it\mu)][\mathrm{Ext}\text{-exp}(it\lambda)]d^{\#}t \right) = \\ & \mathrm{Ext} - \int_{0}^{*\infty} i[\mathrm{Ext}\text{-exp}(-it\mu)]^{\#} \langle \psi, \mathrm{Ext}\text{-exp}(itA)\varphi \rangle_{\#} = \\ & \left\langle \psi, \mathrm{Ext} - \int_{0}^{*\infty} i[\mathrm{Ext}\text{-exp}(-it\mu)][\mathrm{Ext}\text{-exp}(itA)]\varphi d^{\#}t \right\rangle_{\#} \end{aligned} \qquad (6.1.11) $$

Therefore,

$$ R_{\mu}(A)\varphi = \mathrm{Ext} - \int_{0}^{*\infty} i[\mathrm{Ext}\text{-exp}(-it\mu)][\mathrm{Ext}\text{-exp}(itA)]\varphi d^{\#}t \qquad (7.7.12) $$

where the #-integral is a Riemann #-integral. The third step in the computation uses
Then for all $A$ positive, $\omega(A):K \to *\mathbb{C}_c^\#; B \mapsto \frac{\mathrm{Ext}\text{-}\mathrm{Tr}(AB)}{\mathrm{Ext}\text{-}\mathrm{Tr}(A)}$.

Where $\mathrm{Ext}\text{-}\mathrm{Tr}(C)$ is the external sum of the diagonal entries of $C$ (or equivalently the external sum of the eigenvalues of $C$). Indeed, since $\mathrm{Ext}\text{-}\mathrm{Tr}(AB) = \mathrm{Ext}\text{-}\mathrm{Tr}(BA)$ and $\mathrm{Ext}\text{-}\mathrm{Tr}(A) \ge 0$ if $A$ is positive, letting $A = C^*C$ we see also

$$ \mathrm{Ext}\text{-}\mathrm{Tr}(AB^*B) = \mathrm{Ext}\text{-}\mathrm{Tr}(BAB^*) = \mathrm{Ext}\text{-}\mathrm{Tr}(BC^*CB^*) = \mathrm{Ext}\text{-}\mathrm{Tr}((CB^*)^*(CB^*)) \ge 0. $$

So $\omega(A)$ is positive, it is also normalised clearly and therefore a state.

**Definition 2.2.2.** We say that a linear $\omega^{*}\mathbb{C}_c^\#$-valued functional $\psi$ is hermitian if
$$ \forall A \in K[\psi(A^*) = \overline{\psi(A)}]. $$

We for some state $\omega$ are interested in the bilinear form $f_\omega:K \times K \to *\mathbb{C}_c^\#$,
$(A, B) \mapsto \omega(B^*A)$.

This is because it has many properties similar to an inner product. The first we show is that states are hermitian, which implies something similar to conjugate symmetry for $f_\omega$.

**Lemma 2.2.1.** Let $\omega \in S^\#(A)$ then $\omega$ is hermitian.

**Proof.** First suppose $A = A^*$ i.e. $A$ is self #-adjoint. Then let

$$ A_+ = \mathrm{Ext}\text{-}\sum_{\lambda \in \sigma(A), \lambda > 0} \lambda p_\lambda, A_- = \mathrm{Ext}\text{-}\sum_{\lambda \in \sigma(A), \lambda < 0} (-\lambda) p_\lambda $$

Noting that both of these are positive, we have that

$$ \omega(A) = \omega(A_+ - A_-) \in *\mathbb{R}_c^\# \Rightarrow \omega(A^*) = \omega(A) = \overline{\omega(A)}. $$

Then for any $A \in K$ we can write $A = A_1 + iA_2$ where $A_1, A_2$ are both self #-adjoint.

Then we have

$$ \omega(A^*) = \omega(A_1) - i\omega(A_2) = \overline{\omega(A_1) + i\omega(A_2)} = \overline{\omega(A_1) + i\omega(A_2)} = \overline{\omega(A)}. $$

**Corollary 2.2.1.** Let $f_\omega$ be the bilinear form as defined before. Then it is conjugate symmetric i.e. $f_\omega(A, B) = f_\omega(B, A)$.

**Proof.** Using that states are hermitian we see clearly

$$ f_\omega(BA) = \omega(A^*B) = \omega((A^*B)^*) = \omega(B^*A) = f_\omega(A, B). $$

Next, we show the cauchy schwarz for states.

**Lemma 2.2.2.** (Cauchy Schwarz)

Let $\omega \in S^\#(K)$, then we have, $|\omega(AB^*)|^2 \le \omega(A^*A)\omega(B^*B)$.

**Proof.** If $B = 0$ this is clear. Otherwise, by positivity we have for

$C = \omega(BB^*)A - \omega(AB^*)B :$

$$ \begin{align*} 0 \le \omega(CC^*) &= \omega((\omega(BB^*)A - \omega(AB^*)B)(\omega(BB^*)A^* - \omega(AB^*)B^*)) = \dots = \\ &= \omega(BB^*)(\omega(BB^*)\omega(AA^*) - \omega(AB^*)\omega(BA^*) - \omega(AB^*)\omega(AB^*) + \omega(AB^*)\omega(AB^*)). \end{align*} $$

Then using that states are hermitian (lemma 2.13) we can simplify

$$ \dots = \omega(BB^*)(\omega(BB^*)\omega(AA^*) - |\omega(AB^*)|^2). $$

Then by positivity, $\omega(BB^*) \ge 0$ and so $\omega(BB^*)\omega(AA^*) - |\omega(AB^*)|^2 \ge 0$ as required.

**Corollary 2.2.2.** $|f_\omega(A, B)| \le \sqrt{f_\omega(A, A)f_\omega(B, B)}$.

We see now that $f_\omega$ is very similar to an inner product, but fails on positive definiteness, as seen in the following example.

**Example 2.2.3.** In $M_2(*\mathbb{C}_c^\#)$, we can set $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$. Then as $A$ is positive we can

define the state $\omega_A$ as before: $\omega_A:M_2(*\mathbb{C}_c^\#) \to *\mathbb{C}_c^\#; B \mapsto \mathrm{Ext}\text{-}\mathrm{Tr}(AB)$.

Then for $B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \ne 0$ we have $\omega_A(B^*B) = \omega_A(B) = \mathrm{Ext}\text{-}\mathrm{Tr}(AB) = \mathrm{Ext}\text{-}\mathrm{Tr}(0) = 0$.
generalized Fubini's theorem. Applying (7.1.12) to the operators $A_n$ and $A$ we obtain

$$ \|R_{\mu}(A_n)\varphi - R_{\mu}(A)\varphi\|_{\#} \le \text{Ext-} \int_{0}^{\infty} [\text{Ext-} \exp(t \text{Im} \mu)] \|[\text{Ext-} \exp(itA_n) - \text{Ext-} \exp(itA)]\varphi\|_{\#} d^{\#}t \quad (7.1.13) $$

so if $\text{Ext-} \exp(itA_n) \to_{\#} \text{Ext-} \exp(itA)$ as $n \to \infty$ for each $t$, $\|R_{\mu}(A_n)\varphi - R_{\mu}(A)\varphi\|_{\#} \to_{\#} 0$ as $n \to \infty$ by the generalized Lebesgue dominated convergence theorem. Using a formula similar to (7.1.12) one concludes in the same way that $\|R_{\mu}(A_n)\varphi - R_{\mu}(A)\varphi\|_{\#} \to_{\#} 0$. We remark that it is possible to show that if $A_n \to_{\#} A$ in the strong resolvent sense, then $\text{Ext-} \exp(itA_n) \to_{\#} \text{Ext-} \exp(itA)$ for each $\varphi$ uniformly for $t$ in any hyperfinite interval.

**Theorem 7.1.5.** (Generalized Trotter-Kato theorem) Let $(A_n)_{n=1}^{\infty}$ be a sequence of self-#-adjoint operators. Suppose that there exist points, $\lambda_0$ in the upper half-plane of $\tilde{\mathbb{C}}_c^\#$ and $\mu_0$ in the lower half-plane of $\tilde{\mathbb{C}}_c^\#$ so that $R_{\lambda_0}(A_n)\varphi$ and $R_{\mu_0}(A_n)\varphi$ #-converge as $n \to \infty$ for each $\varphi \in H^\#$. Suppose further that one of the limiting operators, $T_{\lambda_0}$ or $T_{\mu_0}$, has a #-dense range. Then there exists a self-#-adjoint operator $A$ so that $A_n \to_{\#} A$ as $n \to \infty$ in the strong resolvent sense.

# Chapter VII. Non-Archimedean Banach spaces endroved with $\tilde{\mathbb{R}}_c^\#$-valued norm.

## 1. Definitions and examples

A non-Archimedean normed space with $\tilde{\mathbb{R}}_c^\#$-valued norm (#-norm) is a pair $(X, \|\cdot\|_\#)$ consisting of a vector space $X$ over a non-Archimedean scalar field $\tilde{\mathbb{R}}_c^\#$ or complex field $\tilde{\mathbb{C}}_c^\#$ together with a distinguished norm $\|\cdot\|_\# : X \to \tilde{\mathbb{R}}_c^\#$. Like any norms, this #-norm induces a translation invariant distance function, called the canonical or (norm) induced non-Archimedean $\tilde{\mathbb{R}}_c^\#$-valued metric for all vectors $x, y \in X$, defined by

$$ d^{\#}(x, y) = \|x - y\|_{\#} = \|y - x\|_{\#}. \quad (1.1) $$

Thus (1.1) makes $X$ into a metric space $(X, d^\#)$. A hyper infinite sequence $(x_n)_{n=1}^\infty$ is called $d^\#$-Cauchy or Cauchy in $(X, d^\#)$ or $\|\cdot\|_\#$-Cauchy if for every hyperreal $r \in \tilde{\mathbb{R}}_c^\#$, $r > 0$, there exists some $N \in \mathbb{N}^\#$ such that

$$ d^{\#}(x_n, x_m) = \|x_n - x_m\|_{\#} < r, \quad (1.2) $$

where $m$ and $n$ are greater than $N$. The canonical metric $d^\#$ is called a #-complete metric if the pair $(X, d^\#)$ is a #-complete metric space, which by definition means for every $d^\#$-Cauchy sequence $(x_n)_{n=1}^\infty$ in $(X, d^\#)$, there exists some $x \in X$ such that

$$ \#\lim_{n \to \infty} \|x_n - x\|_{\#} = 0 \quad (1.3) $$

where because $\|x_n - x\|_{\#} = d^\#(x_n, x)$, this hyper infinite sequence's #-convergence to $x$ can equivalently be expressed as: $\#\lim_{n \to \infty} x_n = x$ in $(X, d^\#)$.

**Definition 1.1.** The normed space $(X, \|\cdot\|_\#)$ is a non-Archimedean Banach space endroved with $\tilde{\mathbb{R}}_c^\#$-valued norm if the #-norm induced metric $d^\#$ is a #-complete metric, or said differently, if $(X, d^\#)$ is a #-complete metric space. The #-norm $\|\cdot\|_\#$ of a #-normed space $(X, \|\cdot\|_\#)$ is called a #-complete #-norm if $(X, \|\cdot\|_\#)$ is a
This motivates the next definition.

**Definition 2.2.3.** We define for each $\omega \in S^{\#}(K)$

$$J_{\omega} = \{A \in K \mid \omega(A^*A) = f_{\omega}(A, A) = 0\}.$$

The fact that our candidate is not positive definite is not an issue, so long as we can use some devices from abstract algebra (namely quotient objects) to “forget” about the problem areas. For this, we need to find an appropriate ideal of $K$.

**Lemma 2.2.3.** Let $J_{\omega}$ be as before. Then $J_{\omega}$ is a left ideal.

**Proof.** Say $A, B \in J_{\omega}$. Then

$$(A+B)^*(A+B) \leq (A+B)^*(A+B) + (A-B)^*(A-B) = 2A^*A + 2B^*B$$
so $0 \leq f_{\omega}(A+B, A+B) \leq 2f_{\omega}(A) + 2f_{\omega}(B) = 0$. So $J_{\omega}$ is a #-closed linear subspace.

We also see that for all $A \in J_{\omega}$ and $B \in K(BA)^*(BA) \leq \|B\|_{\#}^2 A^*A$, and so $BA \in J_{\omega}$

**Lemma 2.2.4.** If $\omega$ is a positive linear functional on $K$, then the operator #-norm of $\omega$,

$$\|\omega\|_{\#} = \sup_{A \in K \setminus \{0\}} \frac{\omega(A)}{\|\omega\|_{\#}} \text{ satisfies } \|\omega\|_{\#} = \omega(1).$$

**Proof.** We know that $\|\omega\|_{\#} \geq \omega(1)$ since $\|1\|_{\#} = 1$. Now let $A \in K \setminus \{0\}$. We have that $\|A + A^*\|_{\#} = \|1 - (A + A^*)\|_{\#} \geq 0$ and so $\omega(A + A^*) \leq \|A + A^*\|_{\#} \omega(1)$. But also, we have

$$\left|\omega\left(\frac{A+A^*}{2}\right)\right| \le \frac{|\omega(A)|+|\omega(A^*)|}{2} = |\omega(A)| = \frac{\omega(A+A^*)}{2} + \frac{\omega(A-A^*)}{2} \le \frac{\omega(A+A^*)}{2} = \\ = \left|\frac{\omega(A+A^*)}{2}\right|. \text{ And so we have equality everywhere and that } |\omega(A)| = \left|\frac{\omega(A+A^*)}{2}\right|.$$

Putting this together we have

$$|\omega(A)| = \left| \frac{\omega(A+A^*)}{2} \right| \le \frac{\|A+A^*\|_{\#} \omega(1)}{2} \le \frac{\|A\|_{\#} + \|A^*\|_{\#} \omega(1)}{2} = \|A\|_{\#} \omega(1).$$

And so $\|\omega\|_{\#} \leq \omega(1)$. In fact this relationship is equivalent.

**Lemma 2.2.5.** Let $\omega$ be a linear functional on $K$. The following statements are equivalent

1. $\omega$ is positive

2. $\|\omega\|_{\#} = \omega(1)$.

**Proof.** 1. ⇒ 2. is lemma 2.18. 2. ⇒ 1.

Let $A$ be positive, and say $\omega(A) = a + ib, a, b \in *\mathbb{R}_c^\#$. Then for all $t \in *\mathbb{R}_c^\#$ we have:

$$a^2 + (b + t\|\omega\|_{\#})^2 = \|A + it\|_{\#}^2 \le \|A + it\|_{\#}^2 \|\omega\|_{\#}^2 \le (\|A\|_{\#}^2 + t^2) \|\omega\|_{\#}^2.$$

Subtracting $2t \|\omega\|_{\#}$ from both sides we have $2bt \le \|A\|_{\#}^2$ and hence $b = 0$. Then,

$$\|A\|_{\#} \|\omega\|_{\#} - a = \omega(\|A\|_{\#} - A) = \|\omega\|_{\#} \|A\|_{\#} - A \|\omega\|_{\#} \le \|\omega\|_{\#} \|A\|_{\#} \quad \text{So } a \ge 0.$$

From the theory so far, we can relate the spectrum of some element $A \in K$ to some states on $K$.

**Lemma 2.2.6.** Let $A \in K$ then for each $\lambda \in \sigma(A)$ there exists a state $\omega_{A\lambda}: K \to *\mathbb{C}_c^\#$ such that $\omega_{A\lambda}(A) = \lambda$.

**Proof.** We define the linear functional on the subspace $*\mathbb{C}_c^\# \cdot A + *\mathbb{C}_c^\# \cdot 1$ by $\omega_0(aA + b_1) = a\lambda + b$. It is clear then that $\omega_0(aA + b_1) = a\lambda + b \in \sigma(aA + b1)$ and

$$\text{hence by lemma 1.29 } 1 = \omega_0(1) \le \|\omega_0\|_{\#} = \sup_{a,b \in *\mathbb{C}_c^\#} \left[ \frac{|a\lambda + b|}{\|aA + b1\|_{\#}} \le 1 \right].$$

Then by generalized Hahn-Banach theorem, there exists an extension of $\omega_0$ to $K, \omega_{A\lambda}$ with $\|\omega_{A\lambda}\|_{\#} = 1 = \omega_{A\lambda}(1)$ by **lemma 2.19**, $\omega_{A\lambda}$ is a state.

The next lemma shows us how even though we don't have positive definiteness, we can conclude an equivalence between $A = 0$ and $\omega(A) = 0 \forall \omega$.

**Lemma 2.21.** Let $K$ be a $\mathbb{C}_c^*$ algebra, and $A \in K$. Then we have
non-Archimedean Banach space endroved with *$\mathbb{R}_c^\#$-valued #-norm.

**Remark 1.1.** For any #-normed space $(X, \|\cdot\|_{\#})$, there exists an L-semi-inner product $\langle\cdot, \cdot\rangle_{\#}: X \times X \to *\mathbb{R}_c^{\#}$ such that $\|x\|_{\#} = \sqrt{\langle x, x\rangle_{\#}}$ for all $x \in X$; in general, there may be infinitely many L-semi-inner products that satisfy this condition. L-semi-inner products are a generalization of inner products, which are what fundamentally distinguish non-Archimedean Hilbert spaces from all other non-Archimedean Banach spaces. Characterization in terms of hyper infinite series, see ref. [1].

The vector space structure allows one to relate the behavior of hyper infinite Cauchy sequences to that of #-converging hyper infinite series of vectors.

**Remark 1.2.** A #-normed space $X$ is a non-Archimedean Banach space if and only if each absolutely #-convergent hyper infinite series $\text{Ext}-\sum_{n=1}^{\infty} v_n$ in $X$ #-converges in $X$, i.e., $\text{Ext}-\sum_{n=1}^{\infty} \|v_n\| < \infty^{\#}$ implies that $\text{Ext}-\sum_{n=1}^{\infty} v_n$ #-converges in $X$.

## 2. Linear operators, isomorphisms

If $X$ and $Y$ are #-normed spaces over the same ground field $*\tilde{\mathbb{R}}_c^\#$, the set of all #-continuous $*\tilde{\mathbb{R}}_c^\#$-linear maps $T: X \to Y$ is denoted by $B^\#(X, Y)$. In hyper infinite-dimensional spaces, not all linear maps are #-continuous. A linear mapping from a #-normed space $X$ to another normed space is #-continuous if and only if it is bounded or hyper bounded on the #-closed unit ball of $X$. Thus, the vector space $B^\#(X, Y)$ can be endroved with the operator norm

$$ \|T\| = \sup\{\|Tx\|_{\#Y} \mid x \in X, \|x\|_{\#X} \le 1\}. \quad (2.1) $$

For $Y$ a non-Archimedean Banach space, the space $B^\#(X, Y)$ is a Banach space with respect to this #-norm.

If $X$ is a non-Archimedean Banach space, the space $B^\#(X) = B^\#(X, X)$ forms a unital Banach algebra; the multiplication operation is given by the composition of linear maps.

**Definition 2.1.** If $X$ and $Y$ are #-normed spaces, they are #-isomorphic #-normed spaces

if there exists a linear bijection $T: X \to Y$ such that $T$ and its inverse $T^{-1}$ are #-continuous. If one of the two spaces $X$ or $Y$ is #-complete then so is the other space. Two #-normed spaces $X$ and $Y$ are #-isometrically isomorphic if in addition, $T$ is an #-isometry, that is, $\|T(x)\| = \|x\|$ for every $x \in X$.

**Definition 2.2.** Let $\{X, \|\cdot\|\}$ be standard Banach space. For $x \in *X$ and $\varepsilon > 0$, $\varepsilon \approx 0$ we define the open $\varepsilon$-ball about $x$ of radius $\varepsilon$ to be the set $B_\varepsilon(x) = \{y \in *X|*\|x-y\| < \varepsilon\}$.

**Definition 2.3.** Let $\{X, \|\cdot\|\}$ be standard Banach space, $Y \subset X$ thus $*Y \subset *X$ and let $x \in *X$. Then $x$ is an $*$-accumulation point of $*X$ if for every $\varepsilon > 0$, $\varepsilon \approx 0$, $Y \cap (B_\varepsilon(x)\setminus\{x\}) \neq \emptyset$.

**Definition 2.4.** Let $\{X, \|\cdot\|\}$ be a standard Banach space, let $Y \subset *X$, $Y$ is $*$-closed if every $*$-accumulation point of $Y$ is an element of $Y$.

**Definition 2.5.** Let $\{X, \|\cdot\|\}$ be standard Banach space. We shall say that internal hyper infinite sequence $\{x_n\}_{n=1}^{n=\infty}$ in $*X$ $*$-converges to $x \in *X$ as $n \to *\infty$ if for any $\varepsilon > 0$, $\varepsilon \approx 0$ there is $N \in *\mathbb{N}$ such that for any $n > N$: $*\|x_n - x\| < \varepsilon$.
$A = 0 \Leftrightarrow \omega(A) = 0 \forall \omega \in S^{\#}(K).$

**Proof.** ⇒ is clear by linearity of $\omega$.

⇔ can be seen by the string of implications

$\omega(A) = 0 \forall \omega \in S^{\#}(K) \Rightarrow \sigma(A) = \{0\} \Rightarrow A = 0.$

We see in fact there are a huge number of results in analogy to those discussed in subsection 2.1 using that $\sigma(A) \subset \{\omega(A) \mid \omega \in S^{\#}(K)\}$. For example

**Lemma 2.22.** Let $K$ be a $C_{\#}^*$ algebra and let $A \in K$.

(i) $A = A^* \Leftrightarrow \omega(A) \in *\mathbb{R}_c^{\#} \forall \omega \in S^{\#}(K)$

(ii) $A \ge 0 \Leftrightarrow \omega(A) \ge 0 \forall \omega \in S^{\#}(K)$.

**Proof.** (i)-(ii) ⇔ follows since $\sigma(A) \subset \{\omega(A) \mid \omega \in S^{\#}(K)\}$ and ⇒ since $\omega$ is hermitian.

(i)-(ii) ⇔ follows since $\sigma(A) \subset \{\omega(A) \mid \omega \in S^{\#}(K)\}$ and ⇒ since $\omega$ is positive.

## §2.3. Representations and the generalized Gelfand-Naimark-Segal Construction.

**Definition 2.3.1.** Let $K$ be a $C_{\#}^*$ algebra. A representation is a $*$-homomorphism

$\pi:K \to B(H^{\#})A \mapsto \pi[A]$

**Definition 2.3.2.** Let $K$ be a $C_{\#}^*$ algebra, represented in $B(H^{\#})$ by $\pi$. Suppose further that $H_0^{\#} \subset H^{\#}$ is a subspace such that $\{\pi[A]H_0^{\#}\}A \in K \subset H_0^{\#}$ (i.e. $\pi$ is stable in $H_0^{\#}$). Then the restriction of $\pi$ to this subspace, $\pi_0:K \to B(H_0^{\#})A \mapsto \pi[A]$ is called a subrepresentation.

**Example 2.3.1.** For all representations $\pi$ we always have the trivial subrepresentations where we restrict the domain of $\pi[A]$ to $\{0\}$ or $H^{\#}$

**Definition 2.3.3.** A representation $\pi:K \to B(H^{\#})A \mapsto \pi[A]$ is called irreducible if the only subrepresentations are the restrictions to $\{0\}$ or $H^{\#}$ there are no nontrivial subrepresentations.

**Definition 2.3.4.** (Equivalent Representations)

Say $\pi_1:K \to B(H_1^{\#})A \mapsto \pi_1[A]\pi_2:K \to B(H_2^{\#})A \mapsto \pi_2[A]$

Are two representations of the same $C_{\#}^*$ algebra such that there exists a unitary linear map $U:H_1^{\#} \to H_2^{\#}$ such that $\forall A \in K:U\pi_1(A) = \pi_2 U(A)$
Then they are called equivalent.

**Example 2.3.2.** (Direct Sum over representations)

Say $\pi_i:K \to B(H_i^{\#})$, $i \le N \in *\mathbb{N}$ are a finite or hyperfinite family of representations.

Then we can define a representation $\pi:K \to B(\text{Ext}-\bigoplus_i H_i^{\#})A \mapsto \pi[A]$ where if $v$ is uniquely decomposed into $\text{Ext}-\sum_i v_i$ (where each $v_i \in H_i^{\#}$) we have

$\pi[A](v) = \sum_i \pi_i[A](v_i)$. Then for each $i$ we have a subrepresentation equivalent to the representation in $H_i^{\#}$, given by the restriction of $\pi$ to the subspace $0 \oplus \dots \oplus 0 \oplus H_i^{\#} \oplus 0 \oplus \dots \oplus 0$. We can imagine representations like this in terms of hyperfinite matrices:

$$ \pi[A](v) = \begin{bmatrix} \pi_1[A] & 0 & \dots & 0 \\ 0 & \pi_2[A] & \dots & 0 \\ 0 & 0 & \dots & 0 \\ 0 & 0 & \dots & \pi_N[A] \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_N \end{bmatrix} $$

We explore this concept later in greater detail.

**Lemma 2.3.1.** Let $\pi:K \to B(H^{\#})$ be a representation and say $v \in H^{\#}$ has $\#$-norm 1.
**Definition 2.6.** Let {$X, \|\cdot\|$}, {$Y, \|\cdot\|$} be a standard Banach spaces. A linear internal operator $A: D(A) \subseteq X \to Y$ is $*$-closed if for every internal hyper infinite sequence {$x_n$}$_{n=1}^{n=*\infty}$ in $D(A)$ $*$-converging to $x \in *X$ such that $Ax_n \to y \in *Y$ as $n \to *\infty$ one has $x \in D(A)$ and $Ax = y$. Equivalently, $A$ is $*$-closed if its graph is $*$-closed in the direct sum $*X \oplus *Y$.

Given a linear operator $A: *X \to *Y$, not necessarily $*$-closed, if the $*$-closure of its graph in $*X \oplus *Y$ happens to be the graph of some operator, that operator is called the $*$-closure of $A$, and we say that $A$ is $*$-closable. Denote the $*$-closure of $A$ by $*\bar{A}$.
It follows that $A$ is the restriction of $*\bar{A}$ to $D(A)$.

$A$ $*$-core (or $*$-essential domain) of a $*$-$\bar{A}$ closable operator is a subset $C \subset D(A)$ such that the $*$-closure of the restriction of $A$ to $C$ is $*\bar{A}$.

**Definition 2.7.** The graph of the linear transformation $T: H \to H$ is the set of pairs $\{\langle\phi, T\phi\rangle | (\phi \in D(T))\}$.

The graph of $T$, denoted by $\Gamma(T)$, is thus a subset of $H \times H$ which is a non-Archimedean

Hilbert space with inner product $\langle\phi_1, \psi_1\rangle, \langle\phi_2, \psi_2\rangle)$.

$T$ is called a \#-closed operator if $\Gamma(T)$ is a \#-closed subset of $H \times H$.

**Definition 2.8.** Let $T_1$ and $T$ be operators on $H$. If $\Gamma(T_1) \supset \Gamma(T)$, then $T_1$ is said to be an

extension of $T$ and we write $T_1 \supseteq T$. Equivalently, $T_1 \supseteq T$ if and only if $D(T_1) \supseteq D(T)$ and $T_1\phi = T\phi$ for all $\phi \in D(T)$.

**Definition 2.9.** An operator $T$ is \#-closable if it has a \#-closed extension. Every \#-closable

operator has a smallest \#-closed extension, called its \#-closure, which we denote by \#-$\bar{T}$.

**Theorem 2.1.** If $T$ is \#-closable, then $\Gamma(\#-\bar{T}) = \#-\Gamma(\bar{T})$.

**Definition 2.10.** Let $T$ be a \#-densely defined linear operator on a non-Archimedean Hilbert space $H$. Let $D(T^*)$ be the set of $\phi \in H$ for which there is an $\xi \in H$ with $(T\psi, \phi) = (\psi, \xi)$ for all $\psi \in D(T)$.

For each $\phi \in D(T^*)$, we define $T^*\phi = \xi$. $T^*$ is called the \#-adjoint of $T$. Note that $\phi \in D(T^*)$ if and only if $|(T\psi, \phi)| \le C\|\psi\|$ for all $\psi \in D(T)$. We note that $S \subset T$ implies $T^* \subset S^*$.

**Theorem 2.2.** Let $T$ be a \#-densely defined operator on a non-Archimedean Hilbert space $H$.

Then:(i) $T^*$ is \#-closed.

(ii) $T$ is \#-closable if and only if $D(T^*)$ is \#-dense in which case $T = T^{**}$.

(iii) If $T$ is \#-closable, then $(\#-\bar{T})^* = T^*$.

**Definition 2.11.** Let $T$ be a \#-closed operator on a Hilbert space $H$. A complex number $\lambda \in *C_c^\#$ is in the resolvent set, $\rho(T)$, if $\lambda I - T$ is a bijection of $D(T)$ onto $H$ with a finitely or hyper finitely bounded inverse. If $\lambda \in \rho(T)$, $R_\lambda(T) = (\lambda I - T)^{-1}$ is called the resolvent of $T$ at $\lambda$.

The definitions of spectrum, point spectrum, and residual spectrum are the same for unbounded operators as they are for bounded operators. We will sometimes refer to the spectrum of nonclosed, but closable operators. In this case we always mean the spectrum of the closure.
Then the map $\omega_v:K \to *\mathbb{C}_c^\#A \mapsto (\pi[A]_v,v)H^\#$ defines a state on K

**Proof.** It is clear that $\omega_v$ is linear and by cauchy schwarz we have

$$|\omega_v(A)| \le \sqrt{\|\pi[A]v\|_{\#} \|v\|_{\#}} \le \sqrt{\|\pi[A]v\|_{\#} \|v\|_{\#}^2} = \sqrt{\|\pi[A]v\|_{\#}} \|v\|_{\#}.$$

So $\omega_v$ is bounded in $*\mathbb{R}_c^\#$. It is positive since $\langle\pi[A^*A]v,v\rangle = \langle\pi[A]v,\pi[A]v\rangle = \|\pi[A]v\|_{\#}^2$.

And so by **lemma 2.19** we have

$$\|\omega_v\|_{\#} = \langle\pi[1]v, v\rangle_{\#} = \langle v, v\rangle_{\#} = \|v\|_{\#}^2 = 1 \text{ as required.}$$

In fact, every state on K arises in this fashion, as shown in the GNS construction.
We break down the proof of the GNS construction into a few lemmas.

27

**Lemma 2.3.2.** The non-Archimedean Hilbert space completion of the space $K/J_\omega$ with respect to the $*\mathbb{C}_c^\#$-valued inner product

$$\langle \cdot | \cdot \rangle_{\#} : K/J_{\omega} \times K/J_{\omega} \to *\mathbb{C}_c^{\#}([A], [B]) \mapsto \omega(B^*A) = f_{\omega}(A, B)$$

is a non-Archimedean Hilbert space.

**Proof.** We have seen in lemma 2.17 that $J_\omega$ is a left ideal. Therefore this quotient object makes sense, and furthermore the inner product is well defined. It is clearly linear in the first argument, as well as positive definite by lemma 2.14 and conjugate symmetric by corollary 2.13.1. Therefore it is an inner product on the quotient space, and the Hilbert space completion defines a Hilbert space clearly.

**Definition 2.3.5.** Given a $C_\#^*$ algebra K and a state $\omega$, we define the Hilbert space completion of $K/J_\omega$ with respect to the inner product to be $L_2^\#(K, \omega)$.

**Lemma 2.3.3.** Given $K, \omega$ as before, we can define a representation $\pi:K \to B(L_2^\#(K, \omega))$

Such that $\omega(A) = (\pi[A]1_\omega | 1_\omega)$. Where $1_\omega \in L_2^\#(K, \omega)$ is the unit cyclic vector

**Proof.** For $A \in K$ we consider the map $\pi_0(A):K/J_\omega \to K/J_\omega[B] \mapsto [AB]$.

It is clear to see since $J_\omega$ is a left ideal that this is well defined and since

$$\|\pi_0[A](B)\|_{\#}^2 = \omega((AB)^*(AB)) \le \|A\|_{\#}^2 \omega(B^*B) \text{ this extends to a bounded in } *\mathbb{R}_c^\# \text{operator}$$

$\pi(A) \in B(L_2^\#(K, \omega))$ Then we have that the map $\pi:K \to B(L_2^\#(K, \omega)), A \mapsto \pi(A)$.

Is a homomorphism clearly but moreover for all $C,D \in K, ([C]\omega|\pi(A^*)([D]\omega) =$
$= \omega(D^*A^*C) = (\pi(A)[C]\omega|[D]\omega)$. And so $\pi$ is a $*$-homomorphism. Also, since
$1_\omega = [1]_\omega \in K/J_\omega$, we have $(\pi(A)1_\omega | 1_\omega) = \omega(1_\omega^*A1_\omega) = \omega(A)$.

**Theorem 2.33.** (The non-Archimedean GNS construction)

Let K be a $C_\#^*$ algebra. For every state $\omega \in S^\#(K)$ then there is a
non-Archimedean Hilbert space $L_2^\#(K, \omega)$ and a unique (up to equivalence)
representation $\pi:K \to B(L_2^\#(K, \omega))$ such that $\omega(A) = (\pi[A]1_\omega | 1_\omega), A \in K$.
Where $1_\omega \in L_2^\#(K, \omega)$ is the unit cyclic vector.

**Proof.** By **lemma 2.32** it remains to show uniqueness.

Say $\rho:A \to B(H^\#)$ is representation with $\iota \in H^\#$ cyclic and $\omega(A) = \langle\rho(A)\iota,\iota\rangle_\#$ then we
can consider the map $U_0:K/J_\omega \to H^\#$, $[A] \mapsto \rho(A)\iota$. We then would have
$(U_0(A),U_0(B)) = (\rho(A)\iota,\rho(B)\iota) = (\rho(B^*A)\iota,\iota) = \omega(B^*A) = ([A][B])$.

So $U_0$ is well defined, and an isometry. Furthermore for any $A,B \in K$ we have
$U_0(\pi(A)[B]\omega) = U_0([AB]\omega) = \rho(AB)\iota = \rho(A)U_0([B])$.

So $U_0$ extends to an isometry

$$U: L_2^\#(K, \omega) \to H^\# \text{ such that } \forall A \in K: U\pi(A) = \rho(A)U$$

Since $\iota$ is cyclic, and we have $\rho(K)\iota \subset U(L_2^\#(K, \omega))$ it follows that $U$ must be unitary.
The following corollary tells us that we can think of any $C_\#^*$ algebra as a subset of
$B(H^\#)$ for some $H^\#$.
### 3. Symmetric and self-#-adjoint operators: the basic criterion for self-#-adjointness.

**Definition 3.1.** A #-densely defined operator $T$ on a non-Archimedean Hilbert space is called symmetric (or Hermitian) if $T \subset T^*$, that is, if $D(T) \subset D(T^*)$ and $T\varphi = T^*\varphi$ for all $\varphi \in D(T)$.

Equivalently, $T$ is symmetric if and only if $(T\varphi, \psi) = (\varphi, T\psi)$ for all $\varphi, \psi \in D(T)$.

**Definition 3.2.** $T$ is called self-adjoint if $T = T^*$, that is, if and only if $T$ is symmetric and $D(T) = D(T^*)$.

A symmetric operator is always #-closable, since $D(T^*) \supset D(T)$ is #-dense in $H$. If $T$ is symmetric, $T^*$ is a closed extension of $T$ so the smallest #-closed extension $T^{**}$ of $T$ must be contained in $T^*$. Thus for symmetric operators, we have $T \subset T^{**} \subset T^*$. For #-closed symmetric operators, $T = T^{**} \subset T^*$ and, for self-adjoint operators, $T = T^{**} = T^*$.

From this one can easily see that a #-closed symmetric operator $T$ is self-adjoint if and only if $T^*$ is symmetric.

The distinction between #-closed symmetric operators and self-adjoint operators is very

important. It is only for self-adjoint operators that the spectral theorem holds and it is only self-adjoint operators that may be #-exponentiated to give the one-parameter unitary groups which give the dynamics in QFT. Chapter X is mainly devoted to studying methods for proving that operators are self-adjoint. We content ourselves here with proving the basic criterion for selfadjointness.

First, we introduce the useful notion of essential self-adjointness.

**Definition 3.3** A symmetric operator $T$ is called essentially self- #-adjoint if its #-closure $\bar{\#}-T$ is self- #-adjoint. If $T$ is #-closed, a subset $D \subset D(T)$ is called a core for $T$ if

$$ \overline{\# - T \upharpoonright D} = T. $$

If $T$ is essentially self-#-adjoint, then it has one and only one self-#-adjoint extension. The importance of essential self-#-adjointness is that one is often given a nonclosed symmetric operator $T$. If $T$ can be shown to be essentially self-#-adjoint, then there is uniquely associated to $T$ a self-adjoint operator $T = T^{**}$. Another way of saying this is that if $A$ is a self-#-adjoint operator, then to specify $A$ uniquely one need not give the exact domain of $A$ (which is often difficult), but just some #-core for $A$.

## Chapter VIII. Semigroups of operators on a non-Archimedean Banach spaces.

### §1. Semigroups on non-Archimedean Banach spaces and their generators.

A family of #-bounded operators $\{T(t)|0 < t < \infty\}$ on external hyper infinite dimensional

non-Archimedean Banach space $X$ endowed with $^*\tilde{\mathbb{R}}_{c,+}^\#$ - valued norm $\|\cdot\|_\#$ is called a strongly #-continuous semigroup if:

(a) $T(0) = I$
**Corollary 2.33.1.** Let $K$ be a $C_{\#}^*$ algebra. Then there exists a faithful representation of $K$.

**Proof.** Let $\pi$ be the direct sum over all GNS representations corresponding to states. Then by **lemma 2.21** this representation is faithful. This result is very deep, and shows that there is a one to one correspondence.

References.

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(b) $T(s)T(t) = T(s+t)$ for all $s, t \in \mathbb{R}_{c,+}^{\#}$

(c) For each $\varphi \in X, t \mapsto T(t)$ is #-continuous mapping.

We will see that strongly continuous semigroups are the “exponentials,”
$T(t) = \text{Ext-}\exp(-tA)$, of a certain class of operators. .

We begin by studying a special class of semigroups:

**Definition 1.1.** A family $\{T(t)|0 < t < \infty\}^\#$ of bounded or hyper bounded operators on external hyper infinite dimensional Banach space $X$ is called a contraction semigroup if it is a strongly #-continuous semigroup and moreover $\|T(t)\|_\# < 1$ for all $t \in [0, \infty)$.
Note that the all theorems about general strongly #-continuous semigroups are easy generalizations of the corresponding theorems for #-contraction semigroups. Thus, we study the special case first. We then briefly discuss the general theory and conclude the section by studying another special class, #-holomorphic semigroups.

**Proposition 1.1.** Let $T(t)$ be a strongly #-continuous semigroup on a non-Archimedean Banach space $X$ and set $A\varphi = \#\lim_{r \to \#} A_r\varphi$ where $D(A) = \{\varphi | \#\lim_{r \to \#} A_r\varphi \text{ exists}\}$. Then $A$ is

\#-closed and \#-densely defined. $A$ is called the infinitesimal generator of $T(t)$. We will also say that $A$ generates $T(t)$ and write $T(t) = \text{Ext-}\exp(-tA)$.

**Proof.** Let $T(t)$ be a contraction semigroup on a Banach space $X$. We obtain the generator of $T(t)$ by #-differentiation. Set $A_t = t^{-1}(I - T(t))$ and define

$$D(A) = \{\varphi | \#\lim_{t \to \#} A_t \varphi \text{ exists}\}.$$

For $\varphi \in D(A)$, we define $A\varphi = \#\lim_{t \to \#} A_t\varphi$. Our first goal is to show that $D(A)$ is \#-dense. For $\varphi \in X$, we set

$$\varphi_s = \text{Ext-} \int_0^s T(t) \varphi d^\#t. \qquad (2.1)$$

For any $r > 0$, we get

$$T(r)\varphi_s = \text{Ext-} \int_0^r T(t+r)\varphi d^\#t \qquad (2.2)$$

thus

$$\begin{aligned} A_r \varphi_s = -\frac{1}{r} \left( \text{Ext-} \int_0^s [T(t+r)\varphi - T(t)\varphi] d^\# t \right) = & \\ -\frac{1}{r} \left( \text{Ext-} \int_s^{r+s} T(t)\varphi d^\# t \right) + \frac{1}{r} \left( \text{Ext-} \int_s^r T(t)\varphi d^\# t \right) & \end{aligned} \qquad (2.3)$$

From Eq.(2.3) one obtains $\#\lim_{r \to \#} A_r \varphi_s = -T(s)\varphi + \varphi$. Therefore, for each $\varphi \in X$

and $s > 0$, $\varphi_s \in D(A)$. Since $s^{-1}\varphi_s \to \# \varphi$ as $\to \# 0$, $A$ is \#-densely defined.

Furthermore, if $\varphi \in D(A)$, then $A_r T(t)\varphi = T(t)A_r\varphi$, so $T(t) : D(A) \to D(A)$ and

$$\frac{d^\#}{d^\# t} T(t) \varphi = -A T(t) \varphi = -T(t) A \varphi \qquad (2.4)$$

$A$ is also \#-closed, for if $\varphi_n \in D(A)$, $\#\lim_{n \to \infty} \varphi_n = \varphi$, and $\#\lim_{n \to \infty} A\varphi_n = \psi$, then
[12] Foukzon, J. (2024). Model $\lambda(\varphi^{2n})_4, n \ge 2$ quantum field theory: A nonstandard approach based on nonstandard pointwise-defined quantum fields.
Jaykov Foukzon 2024 J. Phys.: Conf. Ser. 2701 012113 DOI
10.1088/1742-6596/2701/1/012113

[18] Foukzon, J. (2022). Internal Set Theory IST# Based on Hyper Infinitary Logic with Restricted Modus Ponens Rule: Nonconservative Extension of the Model Theoretical NSA. Journal of Advances in Mathematics and Computer Science, 37(7), 16–43. https://doi.org/10.9734/jamcs/2022/v37i730463

[19] E. Nelson, Internal Set Theory, an axiomatic approach to nonstandard analysis, Bull. Am. Math. Soc., 83:6 (1977) 1165-1198.

[20] Vaught, Robert L. Set theory an introduction. Springer Science & Business Media, 2001.

[21] Raymond M. Smullyan, Set Theory and the Continuum Problem, ISBN-13 978-0486474847

[22] K. Kuratowski, Set Theory, with an Introduction to Descriptive Set Theory, Hardcover Published February 26, 1976 by North-Holland ISBN 9780720404708 (ISBN10: 0720404703)

[23] G. Birkhoff, Lattice Theory, third edition (American Mathematical Society Colloquium Publications, Volume 25) Hardcover – January 1, 1967

[24] Foukzon, J., There Is No Standard Model of ZFC and ZFC2 with Henkin Semantics Advances in Pure Mathematics > Vol.9 No.9, September 2019 https://doi.org/10.4236/apm.2019.99034

[25] Foukzon, J., There is No Standard Model of ZFC and ZFC_2 with Henkin semantics. Generalized Lob's Theorem. Strong Reflection Principles and Large Cardinal Axioms. Consistency Results in Topology. arXiv:1301.5340v15 [math.GM]

[26] Foukzon, J., Relevant First-Order Logic LP# and Curry's Paradox Resolution, Pure and Applied Mathematics Journal, Volume 4, Issue 1-1, January 2015
Pages: 6-12 Published: Jan. 19, 2015
https://doi.org/10.11648/j.pamj.s.2015040101.12
arXiv:0804.4818 [math.LO]

[27] Axiomatic Theories of Truth Stanford Encyclopedia of Philosophy https://plato.stanford.edu/entries/truth-axiomatic/#TypeFreeTrut

[28] Liar Paradox Stanford Encyclopedia of Philosophy https://plato.stanford.edu/entries/liar-paradox/
$$
\begin{align*}
\#\lim_{r \to \#} 0 A_r \varphi &= \#\lim_{r \to \#} 0 \#\lim_{n \to \infty \#} \left[ -\frac{1}{r} (T(r)\varphi_n - \varphi_n) \right] = \\
&= \#\lim_{r \to \#} 0 \#\lim_{n \to \infty \#} \frac{1}{r} \left( \text{Ext-} \int_s^r T(t) A \varphi_n d^\#t \right) = \\
&= \#\lim_{r \to \#} 0 \frac{1}{r} \left( \text{Ext-} \int_s^r T(t) \psi d^\#t \right) \tag{2.5}
\end{align*}
$$

so $\varphi \in D(A)$ and $A\varphi = \psi$.

The formal Laplace transform

$$
\frac{1}{\lambda + A} = - \left( \text{Ext} - \int_{0}^{\infty} (\text{Ext}\text{-exp}(-\lambda t))(\text{Ext}\text{-exp}(-tA))d^{\#}t \right) \quad (2.6)
$$

suggests that all $\mu \in *\mathbb{C}_c^\#$ with $\text{Re}\,\mu < 0$ are in $\rho(A)$. This is in fact true and the formula (2.6) holds in the strong sense. For suppose that $\text{Re}\,\lambda > 0$. Then, since $\|\text{Ext}\text{-exp}(-tA)\| < 1$, the formula (2.7)

$$
R\varphi = \text{Ext} - \int_{0}^{\infty} (\text{Ext}\text{-exp}(-\lambda t))(\text{Ext}\text{-exp}(-tA)\varphi)d^{\#}t \quad (2.7)
$$

defines a hyper bounded linear operator of #-norm less than or equal to $(\text{Re}\,\lambda)^{-1}$.
Moreover, for $r > 0$,

$$
\begin{align}
A_r R \varphi = -\frac{1}{r} \left( \text{Ext} - \int_0^{\infty} (\text{Ext}\text{-exp}(-\lambda t))(\text{Ext}\text{-exp}(-(t+r)A) - \text{Ext}\text{-exp}(-tA))\varphi d^\#t \right) = \right. \nonumber \\
\left. \frac{1 - \text{Ext}\text{-exp}(\lambda r)}{r} \left( \text{Ext} - \int_0^{\infty} (\text{Ext}\text{-exp}(-\lambda t))(\text{Ext}\text{-exp}(-tA))\varphi d^\#t \right) + \right. \nonumber \\
\left. \frac{\text{Ext}\text{-exp}(\lambda r)}{r} \left( \text{Ext} - \int_0^r (\text{Ext}\text{-exp}(-\lambda t))(\text{Ext}\text{-exp}(-tA))\varphi d^\#t \right) \right) \tag{2.8}
\end{align}
$$

so as $r \to \# 0$, $A_r R \varphi \to \# (\varphi - \lambda R \varphi)$. Thus $R \varphi \in D(A)$ and $AR \varphi = \varphi - \lambda R \varphi$ which
implies $(\lambda + A)R \varphi = \varphi$. In addition, for $\varphi \in D(A)$ we have $AR \varphi = RA \varphi$ since

$$
\begin{align}
A \left( \text{Ext} - \int_0^\infty (\text{Ext}\text{-exp}(-\lambda t))(\text{Ext}\text{-exp}(-tA))\varphi d^\#t \right) &= \nonumber \\
&\quad \text{Ext} - \int_0^\infty (\text{Ext}\text{-exp}(-\lambda t))A(\text{Ext}\text{-exp}(-tA))\varphi d^\#t = \tag{2.9} \\
&\quad \text{Ext} - \int_0^\infty (\text{Ext}\text{-exp}(-\lambda t))(\text{Ext}\text{-exp}(-tA))A\varphi d^\#t. \nonumber
\end{align}
$$

The first equality follows by approximation with external hyperfinite Riemann
sums (see [1]) from the facts that $(\operatorname{Ext}\operatorname{-exp}(-\lambda t))(\operatorname{Ext}\operatorname{-exp}(-tA))\varphi$ and
$A(\mathrm{Ext}-\exp(-\lambda t))(\mathrm{Ext}-\exp(-tA))$ are \#-integrable, $A$ is \#-closed. Thus, for $\varphi \in D(A)$,
$R(\lambda + A)\varphi = \varphi = (\lambda + A)R\varphi$ which implies that

$$
R = (\lambda + A)^{-1}. \tag{2.10}
$$

The properties of A which we have derived are also sufficient to guarantee that A generates a contraction semigroup. In fact, we only need information about real positive A.

**Theorem 1.1.** (Generalized Hille-Yosida theorem) A necessary and sufficient condition that a \#-closed

linear operator $A$ on a Banach space $X$ generate a contraction semigroup is that

$$
(i) \ (-\infty^{\#}, 0) \subset \rho(A)
$$

$$
(ii) \ \|(\lambda + A)^{-1}\|_{\#} \text{ for all } \lambda > 0.
$$

Furthermore, if A satisfies (i) and (ii), then the entire \#-open left half-plane is
contained in $\rho(A)$ and

$$
(\lambda + A)^{-1} \varphi = - \mathrm{Ext} - \int_{0}^{\infty} (\mathrm{Ext} - \exp(-\lambda t))(\mathrm{Ext} - \exp(-tA))d^{\#}t \quad (2.11)
$$

for all $\varphi \in X$ and $\lambda$ with $\mathrm{Re} \, \lambda > 0$. Finally, if $T_1(t)$ and $T_2(t)$ are contraction semigroups generated by $A_1$ and $A_2$ respectively, then $T_2(t) \neq T_1(t)$ for some $t$ implies that
$A_1 \neq A_2$.

**Proof.** Since we showed above that conditions (i) and (ii) are necessary and that
(2.11)

holds, we need only show sufficiency. So, suppose that $A$ is a \#-closed operator on $X$
satisfying (i) and (ii). For $\lambda > 0$, define $A^{(\lambda)} = \lambda - \lambda^2(\lambda + A)^{-1}$. We will show that as
$\lambda \to \infty^{\#}$, $A^{(\lambda)} \to_{\#} A$ strongly on $D(A)$ and then construct $\mathrm{Ext}\text{-exp}(-tA)$ as the strong
\#-limit of the semigroups $\mathrm{Ext}\text{-exp}(-tA^{(\lambda)})$. For $\varphi \in D(A)$, $A^{(\lambda)}\varphi = \lambda(\lambda + A)^{-1}A\varphi$.
Moreover, by (ii),

$$
\#\lim_{\lambda \to \infty} [\lambda(\lambda + A)^{-1}\varphi - \varphi] = \#\lim_{\lambda \to \infty} [-(\lambda + A)^{-1}A\varphi] = 0. \quad (2.12)
$$

By condition (ii) the family ${\lambda(\lambda + A)^{-1}|\lambda > 0}$ is \#-uniformly hyperfinitely bounded
in \#-norm, so since $D(A)$ is \#-dense, $\#\lim_{\lambda \to \infty} [\lambda(\lambda + A)^{-1}\psi] = \psi$ for all $\psi \in X$.
Thus $\#\lim_{\lambda \to \infty} A^{(\lambda)}\varphi = A\varphi$ for all $\varphi \in D(A)$. Since $A$ is hyperfinitely bounded, the
semigroups $\mathrm{Ext}\text{-exp}(-tA^{(\lambda)})$ can be defined by hyper infinite power series. Since

$$
\begin{align}
\| \mathrm{Ext} - \exp(-tA^{(\lambda)}) \|_{\#} &= \| (\mathrm{Ext} - \exp(-\lambda t))(\mathrm{Ext} - \exp(t\lambda^2(\lambda + A)^{-1})) \|_{\#} \le \nonumber \\
&\le (\mathrm{Ext} - \exp(-\lambda t)) \left( \mathrm{Ext} - \sum_{n=0}^{\infty} \frac{t^n \lambda^{2n}}{n!} \| (\lambda + A)^{-1} \|_{\#}^{n} \right) \le 1 \tag{2.13}
\end{align}
$$

they are contraction semigroups. For all $\mu, \lambda, t > 0$, and all $\varphi \in D(A)$, we have

$$
[\mathrm{Ext} - \exp(-tA^{(\lambda)}))\varphi - [\mathrm{Ext} - \exp(-tA^{(\mu)}))\varphi = \\
\mathrm{Ext} - \int_{0}^{t} \frac{d^{\#}}{d^{\#}s} (\mathrm{Ext} - \exp(-sA^{(\lambda)}))((\mathrm{Ext} - \exp(-(t-s)A^{(\lambda)}))\varphi) d^{\#}s \quad (2.14)
$$

SO,
$$ \|[Ext-\exp(-tA^{(\lambda)})]\varphi - [Ext-\exp(-tA^{(\mu)})]\varphi\|_{\#} \le $$

$$ Ext-\int_{0}^{t} \|(Ext-\exp(-sA^{(\lambda)}))((Ext-\exp(-(t-s)A^{(\lambda)})))\|_{\#} \|A^{(\mu)}\varphi - A^{(\lambda)}\varphi\|_{\#} d^{\#}s \le \quad (2.15) $$

$$ \le t \|A^{(\mu)}\varphi - A^{(\lambda)}\varphi\|_{\#}. $$

We have used the fact that $Ext-\exp(-tA^{(\lambda)})$ and $[Ext-\exp(-(t-s)A^{(\mu)})]$ commute since $\{A^{(\lambda)}|\lambda > 0\}$ is a commuting family. Since we have proven above that $\#\lim_{\lambda \to \infty} A^{(\lambda)}\varphi = A\varphi$, $\{Ext-\exp(-tA^{(\lambda)})\}$ is Cauchy as $\lambda \to \infty$ for each $t > 0$ and $\varphi \in D(A)$. Since $D(A)$ is $\#$-dense and the $Ext-\exp(-tA^{(\lambda)})$ are uniformly hyperfinitely bounded, the same statement holds for all $\varphi \in X$. Now, define

$$ T(t)\varphi = \#\lim_{\lambda \to \infty} [Ext-\exp(-tA^{(\lambda)})\varphi]. \quad (2.16) $$

$T(t)$ is a semigroup of contraction operators since these properties are preserved under strong $\#$-limits. The above inequality shows that the $\#$-convergence in Eq. (2.16) is uniform for $t$ restricted to a hyperfinite interval, so $T(t)$ is strongly $\#$-continuous since $Ext-\exp(-tA^{(\lambda)})$ is. Thus, $T(t)$ is a contraction semigroup. It remains to show that the infinitesimal generator of $T(t)$, call it $\tilde{A}$, is equal to $A$. For all $t$ and $\varphi \in D(A)$,

$$ [Ext-\exp(-tA^{(\lambda)})\varphi] - \varphi = - \left[ Ext-\left[ \int_{0}^{t} Ext-\exp(-sA^{(\lambda)}) \right] A^{(\lambda)}\varphi d^{\#}s \right] \quad (2.17) $$

so, since $\#\lim_{\lambda \to \infty} A^{(\lambda)}\varphi = A\varphi$, we have

$$ T(t)\varphi - \varphi = - \left[ Ext-\int_{0}^{t} T(s)A\varphi d^{\#}s \right]. \quad (2.18) $$

Thus, $\tilde{A}_t\varphi \to_{\#} A\varphi$ as $t \to 0$. Therefore $D(\tilde{A}) \supset D(A)$ and $\tilde{A} \upharpoonright D(A) = A$. For $\lambda > 0$, $(\lambda + A)^{-1}$ exists by hypothesis and $(\lambda + \tilde{A})^{-1}$ exists by the necessity part of the theorem.

## §2 Hypercontractive semigroups

In the previous section we discussed $\mathcal{L}_{\#}^p$-contractive semigroups. In this section we will prove a self-adjointness theorem for operators of the form $A + V$ where $V$ is a multiplication operator and $A$ generates an $\mathcal{L}_{\#}^p$-contractive semigroup that satisfies a strong additional property.

**Definition 2.1.** Let $\langle M, \mu^{\#} \rangle$ be a $\#$-measure space with $\mu^{\#}(M) = 1$ and suppose that $A$ is a positive self-adjoint operator on $\mathcal{L}_{\#}^2(M, d^{\#}\mu^{\#})$. We say that $Ext-\exp(-tA)$ is a hypercontractive semigroup if:

(i) $Ext-\exp(-tA)$ is $\mathcal{L}_{\#}^p$-contractive;

(ii) for some $b > 2$ and some constant $C_b$, there is a $T > 0$ so that $\|Ext-\exp(-tA)\varphi\|_{\#b} \le C_b \|\varphi\|_{\#2}$ for all $\varphi \in \mathcal{L}_{\#}^2(M, d^{\#}\mu^{\#})$.

By Theorem X.55, condition (i) implies that $Ext-\exp(-tA)$ is a strongly $\#$-continuous contraction semigroup for all $p < \infty$. Holder's inequality shows that
$$ \|\cdot\|_{\#q} \le \|\cdot\|_{\#p} \qquad (1) $$

if $p \ge q$. Thus the $\mathcal{L}_{\#}^p$-Spaces are a nested family of spaces which get smaller as $p$ gets larger; this suggests that (ii) is a very strong condition. The following proposition shows

that $b$ plays no special role.

**Proposition 2.1.** Let $\mathrm{Ext}\text{-exp}(-tA)$ be a hypercontractive semigroup on $\mathcal{L}_{\#}^2(M, d^\#\mu^\#)$. Then for all $p, q \in (1, \infty^\#)$, there is a constant $C_{pq}$ and a $t_{pq} > 0$ so that if $t > t_{pq}$ then $\|\mathrm{Ext}\text{-exp}(-tA)\varphi\|_{\#p} \le C_{pq}\|\varphi\|_{\#q}$ for all $\varphi \in \mathcal{L}_{\#}^q$.

**Proof.** The case where $p < q$ follows immediately from (i) and (1). So suppose that $p > q$. Since $\mathrm{Ext}\text{-exp}(-tA) : \mathcal{L}_{\#}^2 \to \mathcal{L}_{\#}^b$ and $\mathrm{Ext}\text{-exp}(-tA) : \mathcal{L}_{\#}^{\infty^\#} \to \mathcal{L}_{\#}^{\infty^\#}$, the generalized Riesz-Thorin theorem implies that there is a constant $C$ so that for all $r \ge 2$, $\|\mathrm{Ext}\text{-exp}(-tA)\varphi\|_{\#r} \le C\|\varphi\|_{\#br/2}$. We now consider two cases. First, if $q \ge 2$ we choose $n$ large enough so that $2(b/2)^n > p$. Then $\|\mathrm{Ext}\text{-exp}(-nTA)\varphi\|_{\#2(b/2)^n} \le C\|\varphi\|_{\#2}$ so the conclusion follows if $2 < q, p > 2(b/2)^n$, by using (1), and hypothesis (i). If $1 < q < 2$, then we choose $n$ large enough so that $2(b/2)^n > p$ and $q > c$ where $c^{-1} + (2(b/2)^n)^{-1} = 1$. Since $A$ is self-adjoint and $\mathrm{Ext}\text{-exp}(-nTA)\varphi$ is a bounded or hyper bounded map from $\mathcal{L}_{\#}^2$ to $\mathcal{L}_{\#}^{2(b/2)^n}$, $(\mathrm{Ext}\text{-exp}(-nTA))^* = \mathrm{Ext}\text{-exp}(-nTA)$ is a bounded or hyper bounded map from $\mathcal{L}_{\#}^c$ to $\mathcal{L}_{\#}^2$. Thus $\mathrm{Ext}\text{-exp}(-2nTA)$ is a bounded or hyper bounded map from $\mathcal{L}_{\#}^c$ to $\mathcal{L}_{\#}^{2(b/2)^n}$. Since $c < q < p < 2(b/2)^n$, (1) implies the proposition.

**Theorem 2.1.** The operator $-\frac{1}{2}d^{\#2}/d^\#x^2 + xd^\#/d^\#x$ on $\mathcal{L}_{\#}^2(\tilde{\mathbb{R}}_{\#}^{\#}, \pi_{\#}^{-1/2}\mathrm{Ext}\text{-exp}(-x^2)d^\#x)$ is positive and essentially self-adjoint on the set of hyperfinite linear combinations of Hermite polynomials, and generates a hypercontractive semigroup. As a preparation for our main theorem, we prove the following result.

**Theorem 2.2** Let $\langle M, \mu \rangle$ be a #-measure space with $\mu(M) = 1$ and let $H_0$ be the generator of a hypercontractive semigroup on $\mathcal{L}_{\#}^2(M, d\mu)$. Let $V$ be a real-valued measurable function on $\langle M, \mu^\# \rangle$ such that $V \in \mathcal{L}_{\#}^p(M, d^\#\mu^\#)$ for all $p \in [1, \infty^\#)$ and $\mathrm{Ext}\text{-exp}(-tV) \in \mathcal{L}_{\#}^1(M, d^\#\mu^\#)$ for all $t > 0$. Then $H_0 + V$ is essentially self-#-adjoint on $C^\infty(H_0) \cap D(V)$ and is bounded below. $C^\infty(H_0) = \cap_{p \in \mathbb{N}^\#} D(H_0^p)$

# Chapter IX. Singular Perturbations of Selfadjoint Operators on a non-Archimedean Hilbert space.

## §1. Introduction

We study the sum $A + B$ of two #-selfadjoint operators on a non-Archimedean Banach spaces, and we find sufficient conditions for $C = A + B$ to be #-selfadjoint. Our technique is to approximate $B$ by a hyperinfinite sequence of bounded #-selfadjoint

operators $B_n, n \in \mathbb{N}$ and so to approximate $C$ by #-selfadjoint operators $C_n = A + B_n$. We answer three questions separately:

1. When do the operators $C_n$ have a #-lim $C$? 2. When is $C$ a #-selfadjoint operator?

3. When is $C = A + B$?

In Theorem 8 we give a set of estimates on the relative size of $A$ and $B$ which ensure a positive answer to all three questions. Hence these estimates show that $A + B = C$ is #-selfadjoint. In another paper [5], we use Theorem 8 to prove
the existence of a self-interacting, causal quantum field in 4-dimensional space-time. Formally this field theory is Lorentz covariant and has non-trivial scattering; this application was the motivation for the present work.

In order to investigate the meaning of #-lim$_{n→∞}$ C$_n$, we give a new definition for the strong #-convergence of a hyperinfinite sequence of operators. Consequences of this definition

are worked out in Section 2. In Section 3 we give estimates on operators C$_n$ which are sufficient to ensure that the #-lim$_{n→∞}$ C$_n$ = C exists and that C is maximal symmetric or #-selfadjoint. This result is given in Theorem 5 and Corollary 6.

In Section 4 we investigate whether #-lim$_{n→∞}$ C$_n$ = C is equal to A + B.

We combine this work in Theorem 8, our second main theorem, where B is a singular, but nearly positive #-selfadjoint perturbation of a positive #-selfadjoint operator A. To illustrate this theorem, let A ≥ I and let B be essentially #-selfadjoint on

$$D^{\#} = \bigcap_{n \in \mathbb{N}} D(A^n). \qquad (1.0)$$

Assume now that, for some $\beta > 0$ and some $\alpha$,

$$A^{-(1-\beta)}BA^{-(1-\beta)} \text{ and } A^{\beta}BA^{\alpha} \qquad (1.1)$$

are #-densely defined, bounded operators. Also, for some positive $a, \varepsilon \in *\tilde{\mathbb{R}}_{c+}^{\#}$ satisfying $2a + \varepsilon < 1$, suppose that there is a constant $b \in *\tilde{\mathbb{R}}_{c}^{\#}$ such that, as bilinear forms on $D \times D$,

$$0 \le aA + B + b \qquad (1.2)$$

and

$$0 \le \varepsilon A^2 + [A^{1/2}, [A^{1/2}, B]] + b. \qquad (1.3)$$

Then $A + B$ is #-selfadjoint.

We see from this example that neither the operator B nor the bilinear form B need be bounded relative to A.

While it may not appear evident, the conditions (1.1)-(1.3) are closely related to a more easily understandable estimate on $D^{\#} \times D^{\#}$,

$$A^2 + B^2 c(A + B)^2 + c. \qquad (1.4)$$

In fact, estimates (1.1)-(1.3) are chosen because they allow us not only to prove (1.4), but also the similar inequality where B is replaced by $B_n$.

Let us now see that if $A + B$ is #-selfadjoint, then (1.4) must hold for every vector in $D(A + B) = D(A) \cap D(B)$.

**Proposition 1.1.** Let A and B be #-closed operators. Then A + B is #-closed if and only if there is a constant $c \in *\tilde{\mathbb{R}}_{c}^{\#}$ such that for all $\psi \in D(A + B)$

$$\|A\psi\|_{\#} + \|B\psi\|_{\#} \le \|(A+B)\psi\|_{\#} + c\|\psi\|_{\#} \qquad (1.5)$$

and (1.5) is equivalent to (1.4) on $D(A + B) \times D(A + B)$.

**Proof:** Certainly (1.5) implies that $A + B$ is #-closed. Conversely, assume that $A + B$ is #-closed and introduce the #-norms on $D(A + B) = D(A) \cap D(B)$,

$$\|\psi\|_{\#1} \triangleq \|\psi\|_{\#} + \|A\psi\|_{\#} + \|B\psi\|_{\#} \qquad (1.6)$$

and

$$\|\psi\|_{\#2} \triangleq \|\psi\|_{\#} + \|(A+B)\psi\|_{\#} \qquad (1.7)$$
Then $D(A+B)$, $\|\cdot\|_{\#2}$ is a non-Archimedean Banach space because $A+B$ is \#-closed.
The identity map from $D(A+B)$, $\|\cdot\|_{\#2}$ to $D(A+B)$, $\|\cdot\|_{\#1}$ has a \#-closed graph because $A,B$, and $A+B$ are \#-closed. By the \#-closed graph theorem, the identity map is \#-continuous; hence

$$ \|\psi\|_{\#1} \le c \|\psi\|_{\#2}. \qquad (1.7') $$

**Proposition 1.2.** Let $A \ge I, B$ be \#-selfadjoint operators with $D^\# \subset D(B)$ and suppose (1.2) and (1.3) hold. Then (1.4) is valid on $D^\# \times D^\#$.

**Proof** The operators $A^2, B^2, AB, BA$, and $A^{1/2}BA^{1/2}$ define bilinear forms on $D^\# \times D^\#$. Using (1.2) and (1.3), we have the inequality:

$$ A^2 + B^2 = (A + B)^2 - 2A^{1/2}BA^{1/2} - [A^{1/2}, [A^{1/2}, B]] \le (A + B)^2 + (2a + \varepsilon)A^2 + 2Ab + b \quad \text{which establishes (1.4).} $$

## §2. Strong \#-Convergence of Operators

Let $\mathcal{L}(C)$ be the graph of the operator $C$. For any hyperinfinite sequence $\{C_n\}$, $n \in *\mathbb{N}$ of \#-densely defined operators we define

$$ \mathcal{L}_{*\infty}(C) = \{\phi, \chi | \phi = \#\text{-lim}_{n \to *\infty} \phi_n, \phi_n \in D(C_n), \chi = \#\text{-lim}_{n \to *\infty} C_n \phi_n\}. \quad (8) $$

In general, $\mathcal{L}_{*\infty}$ will not be the graph of an operator. If the hyperinfinite sequence $\{C_n^*\}$, $n \in *\mathbb{N}$\#-converges strongly on a \#-dense domain $D$ to an operator $C^*$, namely,

$$ C^*\psi = \#\text{-lim}_{n \to *\infty} C_n^*\psi, \psi \in D, $$

then $\mathcal{L}_{*\infty}$ is the graph of some operator $C^*$. In particular, if each $C_n$ is self \#-adjoint, and if the $C_n$ \#-converge on a \#-dense set $D$ to an operator $C$ defined on $D$, then $\mathcal{L}_{*\infty} = \mathcal{L}_{*\infty}(C_{*\infty})$ and $C_{*\infty}$ is a symmetric extension of $C$.

**Definition 2.1.** $G$ \#-CONVERGENCE. The hyperinfinite sequence of operators $C_n, n \in *\mathbb{N}$ \#-converge strongly to $C_{*\infty}$ in the sense of graphs, written

$$ C_n \to_{\#G} C_{*\infty} \qquad (8') $$

if $\mathcal{L}_{*\infty}$ is the graph of a \#-densely defined operator $C_{*\infty}$.

**Remark 2.1.** Note that for a hyperinfinite sequence of uniformly bounded operators $\{C_n^*\}_{n \in *\mathbb{N}}$ such that $C_n \to_{\#G} C_{*\infty}$, $C_{*\infty}$ is the usual strong \#-limit of the operators $C_n, n \in *\mathbb{N}$ and is everywhere defined.

**Definition 2.2.** $R$ \#-CONVERGENCE. Let the resolvents $R_n(z) = (C_n - z)^{-1}, n \in *\mathbb{N}$ exist for some $z \in *\tilde{\mathbb{C}}_c^\#$, and be uniformly bounded in $n$. The operators $C_n$ \#-converge strongly to $C_{*\infty}$ in the sense of resolvents, written

$$ C_n \to_{\#R} C_{*\infty} \qquad (8'') $$

if the resolvents $R_n(z)$ \#-converge strongly to an operator $R(z)$, which has a \#-densely defined inverse.

**Remark 2.2.** Note that in that case, the operator $C_{*\infty} = R^{-1}(z) + z$ exists for all $z \in *\tilde{\mathbb{C}}_c^\#$ for which the strong \#-limit of the $R_n(z)$ exists, and $R^{-1}(z) + z$ is independent of $z$.

**Remark 2.3.** Note that $G$ \#-convergence is weaker than $R$ \#-convergence, in the case $C_n = C_n^*$ at least, because, as we shall show, in this case $C_n \to_{\#R} C_{*\infty}$ implies $C_n \to_{\#G} C_{*\infty}$. It seems likely that $G$ \#-convergence is strictly weaker than $R$ \#-convergence; this could be established by giving an example for which
$C_n^* = C_n \to_{\#G} C_{\infty}$ with $C_{\infty}$ not maximal symmetric. The importance of $G$ #-convergence is that it is technically easier to verify-and gives less information about the #-limit-than $R$ #-convergence, while automatically selecting the correct domain in the case that $R$ #-convergence also holds. The most familiar examples of $G$ #-convergence occur where there is $C_n$ strong #-convergence on a #-dense domain. A less trivial example occurs where there is $D(C_n)$ is independent of $n$, but apparently

$$D(C) \cap D(C_n) = \{0\}.$$

We have the following connection between $G$ and $R$ #-convergence for a hyperinfinite sequence of #-selfadjoint operators.

**Proposition 3.** Let $C_n, n \in *\mathbb{N}$ be #-selfadjoint.

(a) The domain $D_{\infty}^* = \{\phi | \{\phi, \chi\} \in \mathcal{L}_{\infty}^* \text{ for some } \chi\}$ is #-dense in $H$ and only if $C_n \to_{\#G} C_{\infty}^*$, and in this case $C_{\infty}^*$ is necessarily symmetric.

(b) If $R_n(z) = (C_n - z)^{-1}$, $n \in *\mathbb{N}$ #-converges to a bounded operator $R(z)$ for an unbounded set of $z$'s with $\|zR, (z)\|_{\#}$ bounded uniformly in $z \in *C_c^{\#}$ and $n \in *\mathbb{N}$ and if $C_n \to_{\#G} C_{\infty}^*$, then each $R(z)$ is invertible.

(c) If $R_n(z)$ #-converges to an invertible $R(z)$, then $C_n \to_{\#R} C$.

(d) If $C_n \to_{\#R} C$, then $C_n \to_{\#G} C_{\infty}^*$, $\mathcal{L}_{\infty}^* = \mathcal{L}(C)$, and $C$ is maximal symmetric.

(e) Conversely, if $C_n \to_{\#G} C$, where $C$ is maximal symmetric, then $C_n \to_{\#R} C$.

In case the #-limit of the $C_n, n \in *\mathbb{N}$ is actually selfadjoint, there are further connections between $G$ and $R$ #-convergence.

**Theorem 4.**

(a) $C_n \to_{\#G} C$, and $C = C^*$.

(b) $C_n \to_{\#R} C$, and $C = C^*$.

(c) The hyper infinite sequences $\{R_n(z)\}$ and $\{[R_n(z)]^*\}$, $n \in *\mathbb{N}$ #-converge strongly and $\#\lim_{n \to \infty} R_n(z)$ is invertible for some $z$.

(d) Statement (c) holds for all non-real $z \in *\tilde{C}_c^*$.

## §3. Estimates on a $G$ #-convergent hyper infinite sequence

In this section we give estimates which are sufficient to assure that it $G$ #-convergent sequence of operators is $R$ #-convergent, and that the limit is maximal symmetric or selfadjoint. In order to measure the rate of #-convergence, we introduce a selfadjoint operator $N \ge I$ and the associated non-Archimedean Hilbert spaces $H_\lambda$ with the scalar product

$$\langle \psi, \psi \rangle_{\#\lambda} = \langle N^{\lambda/2} \psi, N^{\lambda/2} \psi \rangle_{\#}. \qquad (3.1)$$

By standard identifications we have for $\lambda \ge 0 : H_\lambda \subset H_0 \subset H_{-1}$ and $H_0 = H$. If $D : H_\alpha \to H_\beta$ is a #-densely defined, bounded operator from $H_\alpha$ to $H_\beta$, we let $\|D\|_{\#\alpha,\beta}$ denote its #-norm. Setting $\|D\|_{\#} = \|D\|_{\#0,0}$ we obtain

$$\|D\|_{\#\alpha,\beta} = \|N^{\beta/2}DN^{-\alpha/2}\|. \qquad (3.2)$$

Let $C_n, n \in *\mathbb{N}$ be a hyper infinite sequence of #-selfadjoint operators, and consider the following three conditions.

(i) Suppose that $C_n - C_m$ is a #-densely defined, bounded operator from $H_\lambda$ to $H_{-\lambda}$, for
some $\lambda$, and that as $n, m \to *\infty$

$$ \|C_n - C_m\|_{\#\lambda, -\lambda} \to \# 0. \qquad (3.3) $$

(ii) Suppose that, for some $p$ and for an unbounded set of $z = x + iy \in *\tilde{C}_c^\#$ in the sector $|x| \le \text{const} \times |y|$,

$$ \|R_n(z)\|_{\#\mu, \lambda} \le M(z), \qquad (3.4) $$

where the bound $M(z)$ is uniform in $n \in *\mathbb{N}$.

(iii) Suppose that, for the above $z$'s,

$$ \|R_n(\bar{z})\|_{\#\mu, \lambda} \le M(z). \qquad (3.5) $$

**Theorem 5.** Let $C_n, n \in *\mathbb{N}$ be a hyper infinite sequence of #-selfadjoint operators with a common domain, such that

$$ C_n \to_{\#G} C. $$

If conditions (i) and (ii) hold, then

$$ C_n \to_{\#R} C $$

and $C$ is maximal symmetric.

**Corollary 6.** If in addition to the hypothesis of Theorem 5, condition (iii) also holds, then $C$ is #-selfajoint.

**Remark 3.1.** (1) If $\mu = 0$ in (ii), then the resolvents #-converge uniformly.

(2) If the $C_n$ are uniformly semibounded from below, then we may choose the $z$ in condition (ii) to be infinite large negative numbers. In that case the conclusion of Theorem 5 is that $C_n \to_{\#R} C = C^*$.

# § 4. Estimates for singular perturbations

In this section we consider a singular perturbation $B$ of a #-selfadjoint operator $A$. We give estimates on $B$ which ensure that the sum $A + B$ is #-selfadjoint.

**Abbreviation 4.1.** We abbreviate $A^{\#-}$ instead $\#-\bar{A}$.

**Definition 4.1.** A #-core of an operator $C$ is a domain $D$ contained in $D(C)$ such that $C = (C \upharpoonright D)^{\#-}$.

**Lemma 7.** Let $A, A_n, n \in *\mathbb{N}, B, B_n, n \in *\mathbb{N}$ and $C_n = A, +B_n, n \in *\mathbb{N}$ be #-selfadjoint operators with a common #-core $D$. Assume the hypotheses of Theorem 5 and Corollary 6 for $C_n, n \in *\mathbb{N}$ and suppose also that, for $\theta \in D$,

$$ \|(A - A_n)\theta\|_{\#} + \|(B - B_n)\theta\|_{\#} \to \# 0 \text{ as } n \to *\infty \qquad (4.9) $$

and

$$ \|A_n\theta\|_{\#}^2 + \|B_n\theta\|_{\#}^2 \le \text{const.} \times \|\theta\|_{\#}^2 + \text{const.} \times \|C_n\theta\|_{\#}^2, \qquad (4.10) $$

with constants independent of $n$. Then $A + B$ is #-selfadjoint and $C_n \to_{\#R} A + B$.

**Proof** Let $\psi_n = R_n(z)\chi$. We have

$$ \|A_n \psi_n\|_{\#} \le \text{const.} \times \|\psi_n\|_{\#} + \text{const.} \times \|C_n \psi_n\|_{\#} \le \text{const.} \times \|(C_n - z)\psi_n\|_{\#} = \text{const.} \times \|\chi\|_{\#} $$

Thus,

$$ |\langle A\theta, \psi \rangle_{\#}| \le \|A\theta\|_{\#} \|\psi - \psi_n\|_{\#} + \|(A - A_n)\theta\|_{\#} \times \|\psi_n\|_{\#} + \text{const.} \|\theta\|_{\#} \times \|\psi\|_{\#} $$

and therefore

$$ |\langle A\theta, \psi \rangle_{\#}| \le \text{const.} \|\theta\|_{\#} \times \|\psi\|_{\#} $$
for $\theta \in D$ and $\psi = R(z)\chi$. It follows that $\psi \in D(A \upharpoonright D) = D(A)$ and similarly $\psi \in D(B)$. Since $D(C)$ is the range of $R(z)$, we have shown that $D(C) = D(A + B)$. Since

$$ \langle \theta, (A+B)\psi \rangle_{\#} = \langle (A+B)\theta, \psi \rangle_{\#} = \#\text{-lim}_{n \to *\infty} \langle C_n\theta, \psi_n \rangle_{\#} = \\ \#\text{-lim}_{n \to *\infty} \langle \theta, C_n\psi_n \rangle_{\#} = \langle \theta, C\psi \rangle_{\#} $$

we have $C \subset A + B$. Thus $A + B$ is a symmetric extension of the #-selfadjoint operator $C$, and so $A + B = C$.

**Remark 4.1.** As hypothesis for our next theorem, our second main result, we assume that $N \le A$ and that $N$ and $A$ commute. Let

$$ D^{*\infty}(A) = \bigcap_{n \in \mathbb{N}} A(A^n) \qquad (4.11) $$

the elements of $D^{*\infty}(A)$ are called $C^{*\infty}$ vectors for $A$. Assume that $D^{*\infty}(A)$ is a #-core for the #-selfadjoint operator $B$. Also assume that, as bilinear forms on $D^{*\infty} \times D^{*\infty}$, and for some $\alpha$ and $\varepsilon$ in the indicated ranges,

$$ 0 \le \alpha N + B + \text{const.}, 0 \le \alpha < 1/2 \qquad (4.12) $$

and

$$ 0 \le \varepsilon A^2 + \text{const} \times B + [A^{1/2}, [A^{1/2}, B]] + \text{const.}, 2\alpha + \varepsilon < 1. \qquad (4.13) $$

Let $B$ be a bounded operator from $H_v$ to $H_{-v}$ and from $H_\alpha$ to $H_\beta$ for some $\alpha, \beta$ and $v, \beta > 0$ ($H_\alpha$ is defined following Theorem 4.) If $v \ge 2$, assume that for all $\varepsilon > 0$

$$ 0 \le \varepsilon N^{\mu+2} + [N^{(\mu+1)/2}, [N^{(\mu+1)/2}, B]] + \text{const.} \qquad (4.14) $$

as bilinear forms on $D^{*\infty} \times D^{*\infty}$, for some $\mu > v - 2$.

**Theorem 8.** Under the above hypothesis, $A + B$ is #-selfadjoint.

**Proof** Let $N = \text{Ext-} \int_0^{*\infty} \lambda d^\# E_\lambda$, and let $B_n = (E_n B E_n)^{\#-}$, $A_n = A$. $E_n$ leaves $D$ invariant and so the domain of $B_n$ contains $D$. For $\theta \in D$

$$ |\langle \theta, B_n \theta \rangle_{\#}| = |\langle E_n \theta, B E_n \theta \rangle_{\#}| \le \|B\|_{\#\lambda, -\lambda} \times \|N^{v/2} E_n \theta\|_{\#}^2 \le n^v \|B\|_{\#\lambda, -\lambda} \times \|\theta\|_{\#}^2 \qquad (4.15) $$

and so $B_n$ is bounded in $\#\tilde{\mathbb{R}}_c^\#$ operator and essentially #-selfadjoint on $D$.

By Corollary 7, $C_n = A + B_n$ is essentially #-selfadjoint on $D$ and $D(C_n) = D(A)$.

Let $D_0 \subset D$ be those vectors with #-compact support relative to the spectral #-measure of $A$. In other words, if $\psi \in D_0$, there exist constants $\alpha$ and $\beta$ such that, for all $\mu \ge 0$, $\|A^\mu \psi\|_{\#}^2 \le \alpha \beta^\mu$. It is clear that the vectors $(C_n + b)D_0$ are #-dense in $H^\#$ and are #-analytic vectors for $N^{\mu/2}$ for any $\mu$. Thus by generalized Nelson's theorem [19], $N^{\mu/2}$ is essentially #-selfadjoint on $(C_n + b)D_0$, and hence $(C_n + b)D_0$ is #-dense in $H_\mu$ every $\mu$. In terms of $H^\#$

$$ N^{\mu/2}(C_n + b)D \supset N^{\mu/2}(C_n + b)D_0 \qquad (4.16) $$

is #-dense in $H^\#$ for every $\mu$. For $\lambda > v$ and for $m \le n$

$$ \|B_n - B_m\|_{\#\lambda, -\lambda} = \|N^{-\lambda/2}(B_n - B_m)N^{-\lambda/2}\|_{\#\lambda, -\lambda} \le \|N^{-(\lambda-v)/2}(B_n - B_m)^{-(\lambda-v)/2}\|_{\#\lambda, -\lambda} \qquad (4.17) \\ \le 2m^{-(\lambda-v)/2}\|B\|_{\#\lambda, -\lambda} $$

The inequality (13) is preserved under the substitution $B \to B_n$. To see this, we multiply (13) by $E_n$ on the left and right and notice that
$$ \varepsilon E_n A^2 E_n + \text{const.} \times E_n \le \varepsilon A^2 + \text{const.} \quad (4.18) $$

Similarly we see that (12) is preserved under the substitution $B \to B_n$. Thus the bounds (10) follow from Proposition 1.2 applied to the operators $A$ and $B_n$. To prove (9), let $\theta \in D$. Then,

$$ \| (B - B_n)\theta \|_{\#} \le \|(I - E_n)B\theta \|_{\#} + \|B(I - E_n)\theta \|_{\#} \le n^{-\beta/2} \|B\|_{\#\alpha\beta} \times \|N^{\alpha/2}\theta \|_{\#} + n^{-1} \|B\|_{\#\alpha\beta} \times \|N^{1+(\alpha/2)}\theta \|_{\#}. \quad (4.19) $$

We set $\mu = \nu + 2$ and use Lemma 7. Because of the uniform lower bound (12), we can find infinite large negative numbers $-c$ bounded away from the spectrum of $C_n$. If $\nu < 2$, then we set $\mu = 0, \lambda = 2$, and, because of (10), we have

$$ \|R_n(-c)\|_{\#\mu\lambda} = \|NR_n(-c)\|_{\#\mu\lambda} \le \|AR_n(-c)\|_{\#\mu\lambda} \le \text{const.} \quad (4.20) $$

Now we assume $\nu \ge 2$ and we use (4.14) to bound

$$ \|R_n(-c)\|_{\#\mu\lambda} = \|N^{(\mu+2)/2}R_n(-c)N^{-\mu/2}\|_{\#\mu\lambda} \quad (4.21) $$

Since a bounded in $\tilde{\mathbb{R}}_c^\#$ operator is determined by its action on any #-dense domain and since $N^{\mu/2}(C_n + c)$ maps onto a #-dense subset of $H^\#$, it is sufficient to show that

$$ N^{\mu+2} \le \text{const.} \times (C_n + c) N^\mu (C_n + c) \quad (4.22) $$

as a bilinear form on $D \times D$. We expand the right side as

$$ (C_n + c)N^\mu(C_n + c) = \varepsilon^2 N^{\mu+2} + P + \varepsilon(A - \varepsilon N + B_n + c)N^{\mu+1} + \varepsilon N^{\mu+1}(A - \varepsilon N + B_n + c) \quad (4.23) $$

with $0 \le P$ and so it is sufficient to show that, for some $\varepsilon > 0, \varepsilon \approx 0$

$$ 0 \le \varepsilon N^{\mu+2} + 2N^{(\mu+1)/2}(A - \varepsilon N + B_n + c/2)N^{(\mu+1)/2} + [N^{(\mu+1)/2}, [N^{(\mu+1)/2}, B_n]] \quad (4.24) $$

on the domain $D \times D$. In these inequalities, $c$ may be chosen independently of $\varepsilon$ and hence $c$ is an arbitrarily large constant. By (4.14), the sum of the first, third and fourth terms are positive and, by (12), the second term is positive. We have verified the hypotheses of Lemma 7, and the theorem follows; $A + B$ is #-selfadjoint.

**Corollary 9.** With the hypothesis of Theorem 8, $A + \gamma B \to_{\#R} A$ as $y \to_{\#} 0$

Proof: This is a special case of Corollary 10(a).

**Corollary 10.** Let $B, B_j$ be singular perturbations of $A$, each satisfying the hypothesis of Theorem 8, with constants independent of $j$.

(a) If $B_j - B_l$ is a #-densely defined bounded in $\tilde{\mathbb{R}}_c^\#$ operator from $H_\lambda$ to $H_{-\lambda}$ for a sufficiently large $\lambda$, if $\|B_j - B_l\|_{\#\lambda, -\lambda} \to_\# 0$ and $B_j \to_{\#G} B$, then

$$ A + B_j \to_{\#R} A + B. \quad (4.25) $$

(b) If (a) is true for $\lambda \le 2$ and if there is a #-core $D$ of $C$ with $D \subset \cap_j D(C_j)$, then the resolvents of $A + B_j$ #-converge in #-norm to the resolvent of $A + B$.

Proof (a) Let $Bnj$ be the sequence of operators constructed in Theorem 8

approximating $Bj$. Then, $Bnj \to Bj$ and as in (15) for $m \le n$

$$ (4.26) $$

which tends to zero as $n, j$ and $m, 1$ tend to infinity. Also (14) is valid with $B$ replaced by $B nj$, and as in the proof of Theorem 8,
(4.27)

with $M$ independent of $n$ and $j$. By Theorem 5,

(4.28)

A double limit, both exist, and thus

(4.29)

proving (a).

(b) Under our hypotheses,

(4.30)

cf., (16). Since 9 is a core for C, (C - z) 9 is dense in 2. Let

(4.31)

Then,

which proves the uniform convergence.

**Theorem 9.** Let $T: H^{\#} \to H^{\#}$ be a #-selfadjoint operator. Then there is a $\delta > 0$ such that any #-closed symmetric operator $S$ with $\hat{\delta}(S, T) < \delta$ is #-selfadjoint, where $\hat{\delta}(S, T)$ denotes the gap between $S$ and $T$.

**Corollary 7.** Let $T, T_n$ be #-closed symmetric operators and let $\{T_n\}$ #-converge to $T$ in the generalized sense (see IV-§ 2.4). If $T$ is #-selfadjoint, then $T_n$ is #-selfadjoint for sufficiently infinite large $n \in *NN$.

# Chapter IX.

## §1. Free scalar field

Let $H^{\#}$ be a #-complex Hilbert space over field $\tilde{\mathbb{C}}^{\#}$ and let $\mathcal{F}(H^{\#}) = \bigoplus_{n=0}^{\infty} H_{\#}^{(n)}$

(where $H_{\#}^{(n)} = \bigoplus_{k=1}^{n} H^{\#}$) be the Fock space over $H^{\#}$. Our goal is to

define the abstract free field on $\mathcal{F}_s(H^{\#})$, the Boson subspace of $\mathcal{F}(H^{\#})$; to do this we need to introduce several other families of operators and some terminology. Let $f \in H^{\#}$ be
fixed. For vectors in $\mathbf{H}_{\#}^{(n)}$ of the form $\eta = \psi_1 \otimes \psi_2 \otimes \cdots \otimes \psi_n$ we define a map $b^{-}(f) : \mathbf{H}_{\#}^{(n)} \to \mathbf{H}_{\#}^{(n-1)}$ by

$$b^{-}(f)\eta = (f, \psi_{1})(\psi_{2} \otimes \cdots \otimes \psi_{n}) \qquad (1)$$

$b^{-}(f)$ extends by linearity to finite linear combinations of such $\eta$, the extension is well defined, and $||b^{-}(f)\eta|| \le ||f|| \times ||\eta||$. Thus $b^{-}(f)$ extends to a bounded map (of norm $||f||$) of $\mathbf{H}_{\#}^{(n)}$ into $\mathbf{H}_{\#}^{(n-1)}$. Since this is true for each $n$ (except for $n = 0$ in which case we define $b^{-}(f) : \mathbf{H}_{\#}^{(0)} \to 0$), $b^{-}(f)$ is in a natural way a bounded operator of norm $||f||$ from $\mathcal{F}(\mathbf{H}^{\#})$ to

$\mathcal{F}(\mathbf{H}^{\#})$. It is easy to check that $b^{+}(f) = (b^{-}(f))^{*}$ takes each $\mathbf{H}_{\#}^{(n)}$ into $\mathbf{H}_{\#}^{(n+1)}$ with the action

$$b^{+}(f)\eta = f \otimes \psi_{1} \otimes \psi_{2} \otimes \cdots \otimes \psi_{n} \qquad (2)$$

on product vectors. Notice that the map $f \mapsto b^{+}(f)$ is linear, but $f \mapsto b^{-}(f)$ is antilinear.

Let $S_n$ be the symmetrization operators introduced in Section II.4. Then $S = \bigoplus_{n=0}^{\infty} S_n$ is

the projection onto the symmetric Fock space $\mathcal{F}_s(\mathbf{H}^{\#}) = \bigoplus_{n=0}^{\infty} S_n \mathbf{H}^{\#(n)}$. We will write

$S_n \mathbf{H}^{\#(n)} = \mathbf{H}_s^{\#(n)}$ and call $\mathbf{H}_s^{\#(n)}$ the $n$-particle subspace of $\mathcal{F}_s(\mathbf{H}^{\#})$. Notice that $b^{-}(f)$ takes

$\mathcal{F}_s(\mathbf{H}^{\#})$ into itself, but that $b^{+}(f)$ does not. A vector $\Psi = \{\psi^{(n)}\}_{n=1}^{\infty}$ for which $\psi^{(n)} = 0$ for all except finitely many $n$ is called a finite particle vector. We will denote the set of finite particle vectors by $F_0$. The vector $\Omega_0 = \langle 1, 0, 0, \dots \rangle$ plays a special role; it is called the vacuum.

Let $A$ be any self-adjoint operator on $\mathbf{H}^{\#}$ with domain of essential selfadjointness $D$.

Let $D_A = \{\Psi \in F_0 | \psi^{(n)} \otimes \otimes_{k=1}^n D \text{ for each } n \in \mathbb{N}^n\}$ and define $d\Gamma^{\#}(A)$ on $D_A \cap \mathbf{H}_s^{\#(n)}$ as

$$d\Gamma^{\#}(A) = A \otimes I \cdots \otimes I + I \otimes A \otimes \cdots \otimes I + \cdots + \otimes I \cdots \otimes I \otimes A. \qquad (3)$$

Note that $d\Gamma^{\#}(A)$ is essentially self-adjoint on $D_A$; $d\Gamma^{\#}(A)$ is called the second quantization of $A$. For example, let $A = I$. Then its second quantization $N = d\Gamma^{\#}(I)$ is essentially self-adjoint on $F_0$ and for $\psi \in \mathbf{H}_s^{\#(n)}$, $N\psi = n\psi$. $N$ is called the number operator. If $U$ is a unitary operator on $\mathbf{H}^{\#}$, we define $d\Gamma^{\#}(U)$ to be the unitary operator on $\mathcal{F}_s(\mathbf{H}^{\#})$ which equals $\text{Ext}\otimes \otimes_{k=1}^n U$ when restricted to $\mathbf{H}_s^{\#(n)}$ for $n > 0$, and which equals

the identity on $\mathbf{H}_s^{\#(0)}$. If $\text{Ext}\text{-exp}(itA)$ is a $\#$-continuous unitary group on $\mathbf{H}^{\#}$, then $\Gamma^{\#}(\text{Ext}\text{-exp}(itA))$ is the group generated by $d\Gamma^{\#}(A)$, i.e., $\Gamma^{\#}(\text{Ext}\text{-exp}(itA)) = \text{Ext}\text{-exp}[itd\Gamma^{\#}(A)]$.

**Definition 1.1.** We define the annihilation operator $a^{-}(f)$ on $\mathcal{F}_s(\mathbf{H}^{\#})$ with domain $F_0$ by

$$a^{-}(f) = \sqrt{N+1} b^{-}(f) \qquad (4)$$

$a^{-}(f)$ is called an annihilation operator because it takes each $(n+1)$-particle subspace into the $n$-particle subspace. For each $\psi$ and $\eta$ in $F_0$,

$$(\sqrt{N+1} b^{-}(f)\psi, \eta) = (\psi, Sb^{+}(f)\sqrt{N+1}). \qquad (5)$$

Then Eq.(5) implies that

$$(a^{-}(f))^{*} \upharpoonright F_0 = Sb^{+}(f) \sqrt{N+1} \qquad (6)$$

The operator $(a^{-}(f))^{*}$ is called a creation operator. Both $a^{-}(f)$ and $(a^{-}(f))^{*} \upharpoonright F_0$ are
\#-closable; we denote their \#-closures by $a^{-}(f)$ and $a^{-}(f)^{*}$ also.

**Example 1.1.** If $\mathbf{H}^{\#} = L_{2}^{\#}(M, d^{\#}\mu)$, then $\bigotimes_{i=1}^{n} L_{2}^{\#}(M, d^{\#}\mu) = L_{2}^{\#}(\times_{i=1}^{n} M, \bigotimes_{i=1}^{n} d^{\#}\mu)$ and that $S \bigotimes_{i=1}^{n} L_{2}^{\#}(M, d^{\#}\mu) = L_{2,s}^{\#}(\times_{i=1}^{n} M, \bigotimes_{i=1}^{n} d^{\#}\mu)$, where $L_{2,s}^{\#}$ is the set of functions in $L_{2}^{\#}$ which are invariant under permutations of the coordinates. The operators $a^{-}(f)$ and $a^{-}(f)^{*}$ are given by

$$ 
\begin{aligned}
a^{-}(f)\psi^{(n)}(m_1, \dots, m_n) &= \sqrt{n+1} \left( \mathrm{Ext} - \int_M \tilde{f}(m)\psi^{(n+1)}(m, m_1, \dots, m_n)d^{\#}\mu \right) \\
a^{-}(f)^{*}\psi^{(n)}(m_1, \dots, m_n) &= \frac{1}{\sqrt{n}} \sum_{i=1}^{n} f(m_i)\psi^{(n-1)}(m_1, \dots, \tilde{m}_i, \dots, m_n)
\end{aligned}
\quad (7) $$

where $\tilde{m}_i$ means that $m_i$ is omitted. If $A$ operates on $L_2^\#(M, d^\#\mu)$ by multiplication by the \#\mathbb{R}_c^\#\text{-valued function } \omega(m)$, then

$$ (d\Gamma^{\#}(A)\psi)^{(n)}(m_1, \dots, m_n) = \left(\sum_{i=1}^{n} \omega(m_i)\right) \psi^{(n)}(m_1, \dots, m_n) \quad (8) $$

Eq.(6) implies that the Segal field operator $\Phi_S^\#(f)$ on $F_0$ defined by

$$ \Phi_S^\#(f) = \frac{1}{\sqrt{2}} [a^{-}(f) + a^{-}(f)^{*}] \quad (9) $$

is symmetric and essentially self-\#-adjoint. The mapping from $\mathbf{H}^{\#}$ to the self-\#-adjoint operators on $\mathcal{F}_s(\mathbf{H}^{\#})$ given by

$$ f \mapsto \Phi_S^\#(f) \quad (10) $$

is called the Segal quantization over $\mathbf{H}^{\#}$. Notice that the Segal quantization is a real (but not complex) linear map since $f \mapsto a^{-}(f)$ is antilinear and $f \mapsto a^{-}(f)^{*}$ is linear. The following theorem gives the properties of the Segal quantization.

**Theorem 1.1.** Let $\mathbf{H}^{\#}$ be hyper infinite dimensional Hilbert space over field

$\tilde{\mathbb{C}}_c = \tilde{\mathbb{R}}_c^\# + i\tilde{\mathbb{R}}_c^\#$ and $\Phi_S^\#(f)$ the corresponding Segal quantization. Then:

(a) (self-adjointness) For each $f \in \mathbf{H}^{\#}$ the operator $\Phi_S^\#(f)$ is essentially self-adjoint on $F_0$, the hyperfinite particle vectors.

(b) (cyclicity of the vacuum) $\Omega_0$ is in the domain of all hyperfinite products $\prod_{i=1}^{n} \Phi_S^\#(f_i), n \in \mathbb{N}^\#$

and the set $\{\prod_{i=1}^{n} \Phi_S^\#(f_i)\Omega_0 | f_i \text{ and } n \text{ arbitrary}\}$ is \#-total in $\mathcal{F}_s(\mathbf{H}^{\#})$.

(c) (commutation relations) For each $\psi \in F_0$ and $f, g \in \mathbf{H}^{\#}$

$$ [\Phi_S^\#(f)\Phi_S^\#(g) - \Phi_S^\#(g)\Phi_S^\#(f)]\psi = i \mathrm{Im}(f,g)_{\mathbf{H}^\#}\psi. \quad (11) $$

Further, if $W(f)$ denotes the external unitary operator $\mathrm{Ext}\text{-exp}(i\Phi_S^\#(f))$ then

$$ W(f+g) = \left[ \mathrm{Ext}\text{-exp}\left( \frac{-i \mathrm{Im}(f,g)_{\mathbf{H}^{\#}}}{2} \right) \right] W(f)W(g) \quad (12) $$

(d) (\#-continuity) If $\{f_n\}_{n=1}^{\infty}$ is hyper infinite sequence such as \#-lim$_{n \to \infty}$ $f_n = f$ in $\mathbf{H}^{\#}$, then: (i) \#-lim$_{n \to \infty}$ $W(f_n)\psi$ exists for all $\psi \in \mathcal{F}_s(\mathbf{H}^{\#})$ and

$$ \#\text{-lim}_{n \to \infty} W(f_n)\psi = W(f)\psi \quad (13) $$

(ii) \#-lim$_{n \to \infty}$ $\Phi_S^\#(f_n)\psi$ exists for all $\psi \in F_0$ and

$$ \#\text{-lim}_{n \to \infty} \Phi_S^\#(f_n)\psi = \Phi_S^\#(f)\psi. \quad (14) $$

(e) For every unitary operator $U$ on $\mathbf{H}^{\#}, \Gamma^{\#}(U) : D(\overline{\Phi_S^\#(f)}) \to D(\overline{\Phi_S^\#(Uf)})$ and for $\psi \in D(\overline{\Phi_S^\#(Uf)})$
$$ \Gamma^{\#}(U) \overline{\Phi_{S}^{\#}(f)} \Gamma^{\#}(U)^{-1} \psi = \overline{\Phi_{S}^{\#}(Uf)} \psi \quad (15) $$

for all $f \in \mathbf{H}^{\#}$.

**Proof.** Let $\psi \in \mathbf{H}_{s}^{\#(n)}$. Since $\Phi_{S}^{\#}(f) : F_{0} \to F_{0}$, $\psi$ is in $C^{\infty}(\Phi_{S}^{\#}(f))$. Further, it follows from Eq.(5)-Eq.(6), and the fact that $\|b^{-}(f)\| = \|f\|$, that

$$ \|(a \star (f))^{k} \psi\|_{\#} \leq (\mathrm{Ext} - \prod_{i=1}^{k} \sqrt{p+i}) \|f\|_{\#}^{k} \|\psi\|_{\#} \quad (16) $$

where $a \star (f)$ represents either $a^{-}(f)$ or $a^{-}(f)^{*}$. Therefore,

$$ \|\Phi_S^\#(f)^k \psi\|_{\#} \le 2^{k/2} ((n+k)!)^{1/2} \|f\|_{\#}^k \|\psi\|_{\#} \quad (17) $$

Since $\mathrm{Ext}-\sum_{k=0}^{\infty} t^k 2^{k/2}((n+k)!)^{1/2} \|f\|_{\#}^k \|\psi\|_{\#} < \infty$ for all $t, \psi$ is an #-analytic vector for $\Phi_S^\#(f)$. Since $F_0$ is #-dense in $\mathcal{F}_s(\mathbf{H}^\#)$ and is left invariant by $\Phi_S^\#(f)$ is essentially self-adjoint on $F_0$ by generalized Nelson's analytic vector theorem (Theorem). The proof of (b) is obviously.

To prove (c) one first computes that if $\psi \in F_0$, then

$$ a^{-}(f)a^{-}(g)^{*} \psi - a^{-}(g)^{*}a^{-}(f)\psi = (f,g)\psi \quad (18) $$

Eq.(11) follows immediately. Although Eq.(11) and Eq.(12) are formally equivalent, Eq.(11) by itself does not imply Eq.(12) We sketch a proof of Eq.(12) which uses special properties of the vectors in $F_0$. Let $\psi \in \mathbf{H}_s^{\#(p)}$. Then

$$ \|\Phi_S^\#(f)^n \Phi_S^\#(g)^m \psi\|_{\#} \le 2^{(n+m)/2} (\mathrm{Ext} - \prod_{i=1}^{n+m} \sqrt{p+i}) \|f\|_{\#}^n \|g\|_{\#}^m \|\psi\|_{\#} \quad (19) $$

which implies that hyper infinite series $\mathrm{Ext}-\sum_{n=0,m=0}^{\infty} \|\Phi_S^\#(f)^n \Phi_S^\#(g)^m \psi\|_{\#}/n!m!$ #-converges for all $t \in \mathbb{R}_c^\#$. Since $\psi$ is an #-analytic vector for $\Phi_S^\#(g)$,

$\mathrm{Ext}-\sum_{m=0}^{\infty} ((i\Phi_S^\#(g)^m)/m!)\psi = (\mathrm{Ext}-\exp[i\Phi_S^\#(g)])\psi$. Further, for each $n \in \mathbb{N}^\#$, $(\mathrm{Ext}-\exp[i\Phi_S^\#(g)])\psi$ is in the domain of $(\overline{\Phi_S^\#(f)})^n$ since any finite and hyperfinite sum

$$ \mathrm{Ext}-\exp \sum_{m=0}^{M} \frac{(i \Phi_S^\#(g))^m}{m!} \psi $$

with $M \in \mathbb{N}^\#$ is in it and $\Phi_S^\#(f)^n (\mathrm{Ext}-\sum_{m=0}^M ((i\Phi_S^\#(g)^m)/m!)\psi)$ #-converges as $M \to \infty^\#$.

Thus the estimate $\mathrm{Ext}-\sum_{n=0,m=0}^{\infty, \infty} \|\Phi_S^\#(f)^n \Phi_S^\#(g)^m \psi\|_{\#}/n!m! t^n t^m \le \infty^\#$ shows that $(\mathrm{Ext}-\exp[i\Phi_S^\#(g)])\psi$ is an #-analytic vector for $\Phi_S^\#(f)$ and therefore can be computed by the external hyper infinite power series. Thus

$$ (\mathrm{Ext}-\exp[i\Phi_S^\#(f)])(\mathrm{Ext}-\exp[i\Phi_S^\#(g)])\psi = \mathrm{Ext}-\sum_{n=0,m=0}^{\infty, \infty} \frac{(i\Phi_S^\#(f))^n (i\Phi_S^\#(g))^m}{n!m!} \psi. \quad (20) $$

Similarly one obtains

$$ \left(\mathrm{Ext}-\exp\left[-\frac{it^2}{2} \mathrm{Im}(f,g)_{\mathbf{H}^\#}\right]\right)\left(\mathrm{Ext}-\exp\left[it\Phi_S^\#(f+g)\right]\right)\psi = \mathrm{Ext}-\sum_{n=0,m=0}^{\infty, \infty} \frac{1}{n!m!} \left[\left(-\frac{it^2}{2} \mathrm{Im}(f,g)_{\mathbf{H}^\#}\right)^m (it\Phi_S^\#(f+g))^n\right] \psi \quad (21) $$

where the hyper infinite series in RHS of Eq.(21) #-converges absolutely. Direct computations using Eq.(11) now show that Eq.(12) holds by a term-by-term comparison of the #-convergent external hyper infinite power series.

To prove (d) let $\psi \in \mathbf{H}_s^{\#(k)}$ and suppose that $\#\mathrm{lim}_{n \to \infty} f_n = f$ in $\mathbf{H}^\#$. Then

$$ \|\Phi_S^\#(f_n)\psi - \Phi_S^\#(f)\psi\| \le \sqrt{2(k+1)} \|f_n - f\|_{\#} \|\psi\|_{\#} \quad (22) $$

so $\#\mathrm{lim}_{n \to \infty} \Phi_S^\#(f_n) = \Phi_S^\#(f)$. Thus, $\Phi_S^\#(f_n)$ #-converges strongly to $\Phi_S^\#(f)$ on $F_0$.
Since $F_0$ is a core for all $\Phi_S^\#(f_n)$ and $\Phi_S^\#(f)$, \#-converge theorems imply that
$\#\lim_{n \to \infty} (\mathrm{Ext}-\exp[it\Phi_S^\#(f_n)]\psi) = \mathrm{Ext}-\exp[it\Phi_S^\#(f)]\psi$ for all $\psi \in \mathcal{F}_S(\mathbf{H}^\#)$.

To prove (e), let $\eta \in \mathbf{H}^{\#(n)}$ be of the form $\eta = \psi_1 \otimes \cdots \otimes \psi_n$. Then

$$
\begin{align*}
\Gamma^\#(U)b^-(f)\Gamma^\#(U)^{-1}\eta &= \Gamma^\#(U)b^-(f)(U^{-1}\psi_2 \otimes \cdots \otimes U^{-1}\psi_n) = \\
&\quad \Gamma^\#(U)(f, U^{-1}\psi_1)(U^{-1}\psi_2 \otimes \cdots \otimes U^{-1}\psi_n) = (Uf, \psi_1)(\psi_2 \otimes \cdots \otimes \psi_n) = b^-(Uf)\eta.
\end{align*}
$$

Since finite linear combinations of such $\eta$ are dense in $\mathbf{H}^{\#(n)}$ and $b^{-}(g)$ has norm $\|g\|$, we conclude that $\Gamma^{\#}(U)b^{-}(f)\Gamma^{\#}(U)^{-1} = b^{-}(Uf)$. But $\mathbf{N}$ and $\mathbf{S}$ commute with $\Gamma^{\#}(U)$ so this immediately implies that $\Gamma^{\#}(U)a^{-}(f)\Gamma^{\#}(U)^{-1} = a^{-}(Uf)$ on $F_0$. Taking adjoints and restricting to $F_0$ we also have $\Gamma^{\#}(U)(a^{-}(f))^{*}\Gamma^{\#}(U)^{-1} = (a^{-}(Uf))^{*}$.

Thus for $\psi \in F_0$, $\Gamma^\#(U)\Phi_S^\#(f)\Gamma^\#(U)^{-1}\psi = \Phi_S^\#(Uf)\psi$. Since the operators on both the right- and left-hand sides of this equality are essentially self-\#-adjoint on $F_0$, we conclude that $\Gamma^\#(U)\overline{\Phi_S^\#(f)}\Gamma^\#(U)^{-1} = \overline{\Phi_S^\#(Uf)}$.

**Remark 1.1.** Henceforth we use $\Phi_S^\#(f)$ to denote the \#-closure of $\Phi_S^\#(f)$.

**Definition 1.1.** For each $m > 0, m \in \mathring{\mathbb{R}}_{c,\mathrm{fin}}^\#$ let

$$
H_m^\# = \{p \in \mathring{\mathbb{R}}_c^{\#4} p \cdot \tilde{p} = m^2, p_0 > 0\}, \quad (23)
$$

where $\tilde{p} = (p^0, -p^1, -p^2, -p^3)$. The sets $H_m^\#$, which are called mass hyperboloids, are invariant under $\sigma\mathcal{L}_+^\uparrow$. Let $j_m$ be the \#-homeomorphism of $H_m^\#$ onto $\mathring{\mathbb{R}}_c^{\#3}$ (or in the case $m=0$ onto $\mathring{\mathbb{R}}_c^{\#3}\setminus\{0\}$) given by $j_m: \langle p_0, p_1, p_2, p_3 \rangle \mapsto \langle p_1, p_2, p_3 \rangle = \mathbf{p}$. Define a \#-measure $\Omega_m^\#$ on $H_m^\#$ by

$$
\Omega_m^\#(E) = \mathrm{Ext} - \int_{j_m(E)} \frac{d^{\#3} \mathbf{p}}{\sqrt{|\mathbf{p}|^2 + m^2}} \quad (24)
$$

for any measurable set $E \subset H_m^\#$. The measure $\Omega_m^\#(E)$ can easily be seen to be
$\sigma\mathcal{L}_+^\uparrow$-invariant. In fact, up to a constant multiple, $\Omega_m^\#$ is the only $\sigma\mathcal{L}_+^\uparrow$-invariant measure
on $H_m^\#$. Furthermore, every polynomially bounded $\sigma\mathcal{L}_+^\uparrow$-invariant measure on $\bar{V}_+$ is the
sum of a multiple of $\delta$ and an integral of the measures $\Omega_m^\#$. We state this fact as a
theorem.

**Theorem 1.2.** Let $\mu^\#$ be a polynomially bounded \#-measure with support in $\bar{V}_+$. If $\mu^\#$ is $\sigma\mathcal{L}_+^\uparrow$-invariant, there exists a polynomially bounded \#-measure $\rho$ on $[0, \infty^\#)$ and a constant $c$ so that for any $f \in S^\#(\mathring{\mathbb{R}}_c^{\#4})$

$$
\mathrm{Ext} - \int_{\mathring{\mathbb{R}}_c^{\#4}} f d^\# \mu^\# = c f(0) + \mathrm{Ext} - \int_0^{\infty^\#} d^\# \rho^\#(m) \left( \mathrm{Ext} - \int_{H_m^\#} f d^\# \Omega_m^\# \right). \quad (25)
$$

**Theorem 1.3.**

We can now use the Segal quantization to define the free Hermitian scalar field of mass $m$. We take $\mathbf{H}^\# = \mathcal{L}_2^\#(H_m^\#, d^\#\Omega_{m,x})$, where $H_m^\#, m > 0$, is the mass hyperboloid in $\mathring{\mathbb{R}}_c^{\#4}$ consisting of those $p \in \mathring{\mathbb{R}}_c^{\#4}$ satisfying $p \cdot \tilde{p} - m^2 = 0$ and $p_0 > 0$, and $d^\#\Omega_m^\#$ is the Lorentz invariant \#-measure.

For each $f \in S^\#(\mathring{\mathbb{R}}_c^{\#4})$ we define $Ef \in \mathbf{H}^\#$ by $Ef = 2\pi_\# \hat{f} \mid_{H_m^\#}$ where the Fourier

transform

$$
(2\pi\#)^{-2} \left( \mathrm{Ext} - \int \left( \mathrm{Exp} - \exp \left[ i \left( p \cdot \tilde{x} \right) \right] \right) f(x) d^{\#4}x \right) \quad (26)
$$

is defined in terms of the Lorentz invariant inner product $p \cdot \tilde{x}$. The reason for the
extra $\sqrt{2\pi\#}$ in our definition of $E$ and the plus sign in the definition of Fourier transform
is that if $f$ is the distribution $f(x) = g(x)\delta^\#(t)$, then $\sqrt{2\pi\#} \hat{f}$ is the ordinary
three-dimensional

Fourier transform of g. If $\Phi_S^\#(\cdot)$ is the Segal quantization over $\mathcal{L}_2^\#(H_m^\#, d^\#\Omega_{m,x}^\#)$, we define

for each $\tilde{\mathbb{R}}_c^\#$-valued $f \in S^\#(\tilde{\mathbb{R}}_c^{\#4})$

$$ \Phi_{m,x}^{\#}(f) = \Phi_{S}^{\#}(Ef). \qquad (27) $$

For $\tilde{\mathbb{C}}_c^\#$-valued function $f \in S^\#(\tilde{\mathbb{R}}_c^{\#4})$ we define

$$ \Phi_{m,x}^{\#}(f) = \Phi_{m,x}^{\#}(\mathrm{Re}f) + i\Phi_{m,x}^{\#}(\mathrm{Im}f) \qquad (28) $$

The mapping $f \mapsto \Phi_m^\#(f)$ is called the free Hermitian scalar field of mass $m$.

On $\mathcal{L}_2^\#(H_m^\#, d^\#\Omega_m)$ we define the following unitary representation of the restricted

Poincare group:

$$ (U_m(a, \Lambda)\psi)(p) = (\mathrm{Exp}-\exp[i(p \cdot \tilde{a})])\psi(\Lambda^{-1}p) \qquad (29) $$

where we are using $\Lambda$ to denote both an element of the abstract restricted Lorentz group

and the corresponding element in the standard representation on $\tilde{\mathbb{R}}_{st}^4 = \mathbb{R}^4$.

**Remark 1.3.** Recall that a #-conjugation on a Hilbert space $\mathbf{H}^\#$ is an antilinear #-isometry $\mathbf{C}^\#$ so that $\mathbf{C}^{\#2} = \mathbf{I}$.

**Definition 1.2.** Let $\mathbf{H}^\#$ be a $\tilde{\mathbb{C}}_c^\#$-complex Hilbert space, $\Phi_S^\#(\cdot)$ the associated Segal quantization. Let $\mathbf{C}^\#$ be a #-conjugation on $\mathbf{H}^\#$ and define $\mathbf{H}_{\mathbf{C}^\#}^\# = \{|\mathbf{C}^\#f = f\}$. For each $f \in \mathbf{H}_{\mathbf{C}^\#}^\#$ we define $\varphi^\#(f) = \Phi_S^\#(f)$ and $\pi^\#(f) = \Phi_S^\#(if)$. The map $f \mapsto \varphi^\#(f)$ is called the canonical free field over the doublet $\langle \mathbf{H}^\#, \mathbf{C}^\# \rangle$ and the map $f \mapsto \pi^\#(f)$ is called the canonical conjugate momentum. We often drop the $\langle \mathbf{H}^\#, \mathbf{C}^\# \rangle$ and just write $\mathbf{H}^\#$ if the intended #-conjugation is clear.

**Remark 1.4.** Note that the set of elements of $\mathbf{H}^\#$ for which the maps $f \mapsto \varphi^\#(f)$ and $f \mapsto \pi^\#(f)$ are defined depends on the #-conjugation $\mathbf{C}^\#$.

**Theorem 1.4.** Let $\mathbf{H}^\#$ be a $\tilde{\mathbb{C}}_c^\#$-complex Hilbert space with #-conjugation $\mathbf{C}^\#$. Let $\varphi^\#(\cdot)$ and $\pi^\#(\cdot)$ be the corresponding canonical fields. Then:

(i) For each $f \in \mathbf{H}_{\mathbf{C}^\#}^\#$, $\varphi^\#(f)$ is essentially self-adjoint on $F_0$.

(ii) $\{\varphi^\#(f) | f \in \mathbf{H}_{\mathbf{C}^\#}^\#\}$ is a commuting family of self-adjoint operators.

(iii) $\Omega_0$ is a #-cyclic vector for the family $\{\varphi^\#(f) | f \in \mathbf{H}_{\mathbf{C}^\#}^\#\}$.

(iv) If #-lim$_{n \to \infty}$ $f_n = f$ in $\mathbf{H}_{\mathbf{C}^\#}^\#$, then
$$ \#-\lim_{n \to \infty} \varphi^\#(f_n)\psi = \varphi^\#(f)\psi \quad \text{for all } \psi \in F_0 $$

and

$$ \#-\lim_{n \to \infty} (\mathrm{Exp}-\exp[i\varphi^\#(f_n)]\psi) = \mathrm{Exp}-\exp[i\varphi^\#(f)]\psi \quad \text{for all } \psi \in \mathcal{F}_s(\mathbf{H}^\#) $$

(v) Properties (i)-(iv) hold with $\varphi^\#(f)$ replaced by $\pi^\#(f)$.

(vi) If $f, g \in \mathbf{H}_{\mathbf{C}^\#}^\#$, then

$$ \varphi^\#(f)\pi^\#(g)\psi - \pi^\#(g)\varphi^\#(f)\psi = i(f,g)\psi \qquad (30) $$

for all $\psi \in F_0$ and

$$ (\mathrm{Exp}-\exp[i\varphi^\#(f)]) (\mathrm{Exp}-\exp[i\pi^\#(g)]) = (\mathrm{Exp}-\exp[i(f,g)]) (\mathrm{Exp}-\exp[i\pi^\#(g)]) (\mathrm{Exp}-\exp[i\varphi^\#(f)]). \qquad (31) $$

**Proof.** (i) and (iv) follow immediately from the corresponding properties of $\Phi_S^\#(\cdot)$ proven in Theorem 1.1. To see that $\{\varphi^\#(f) | f \in \mathbf{H}_{\mathbf{C}^\#}^\#\}$ is a commuting family,
notice that (12) implies

$$ (\mathrm{Exp}-\exp[\mathrm{it}\varphi^{\#}(f)]) (\mathrm{Exp}-\exp[\mathrm{is}\varphi^{\#}(g)]) = (\mathrm{Exp}-\exp[-\mathrm{its}\mathrm{Im}(f,g)]) (\mathrm{Exp}-\exp[\mathrm{is}\varphi^{\#}(g)]) (\mathrm{Exp}-\exp[\mathrm{it}\varphi^{\#}(f)]) \quad (32) $$

where we have used the fact that $\varphi^{\#}(\cdot)$ is real linear. If $f,g \in \mathbf{H}_{C^{\#}}^{\#}$, then it follows from polarization that $(f,g) = (C^{\#}f, C^{\#}g) = (g,f)$, so $\mathrm{Im}(f,g) = 0$. Thus

$$ (\mathrm{Exp}-\exp[\mathrm{it}\varphi^{\#}(f)]) (\mathrm{Exp}-\exp[\mathrm{is}\varphi^{\#}(g)]) = (\mathrm{Exp}-\exp[\mathrm{is}\varphi^{\#}(g)]) (\mathrm{Exp}-\exp[\mathrm{it}\varphi^{\#}(f)]) \quad (33) $$

for $s$ and $t$. Therefore, by Theorem VIII. 13, $\varphi^{\#}(g)$ and $\varphi^{\#}(f)$ commute. The proof of (b) is similar to the proof of (a). (X.70) and (X.71) follow immediately from (X.64), (X.65), and the fact that if $f,g \in \mathbf{H}_{C^{\#}}^{\#}$, then $\mathrm{Im}(f,ig) = \mathrm{Re}(f,g) = (f,g)$.

**Definition 1.3.** We write $f \in \mathcal{L}_{2}^{\#}(H_{m}^{\#}, d^{\#}\Omega_{m,x}^{\#})$ as $f(p_{0}, \mathbf{p})$ and define now the \#-conjugation by $(\mathbf{C}^{\#}f)(p_{0}, \mathbf{p}) = \overline{f(p_{0}, -\mathbf{p})}$.

**Remark 1.4.** Note that $\mathbf{C}^{\#}$ is well-defined on $\mathcal{L}_{2}^{\#}(H_{m}^{\#}, d^{\#}\Omega_{m,x}^{\#})$ since $\langle p_{0}, \mathbf{p} \rangle \in H_{m}^{\#}$ if and only if $\langle p_{0}, -\mathbf{p} \rangle \in H_{m}^{\#}$. $\mathbf{C}^{\#}$ is clearly a \#-conjugation.

**Definition 1.4.** We denote the canonical fields corresponding to $\mathbf{C}^{\#}$ by $\varphi^{\#}(\cdot)$ and $\pi^{\#}(\cdot)$ and define $\varphi_{m}^{\#}(f) = \varphi^{\#}(Ef)$ and $\pi_{m}^{\#}(f) = \pi^{\#}(\mu Ef)$, $\mu = \sqrt{\mathbf{p}^{2} + m^{2}}$ for $\mathbf{R}_{c}^{\#}$-valued $f \in \mathcal{L}(\mathbf{R}_{c}^{\#4})$, extending to all of $\mathcal{L}(\mathbf{R}_{c}^{\#4})$ by linearity. In terms of $a^{-}(f)$,

$$ \varphi_{m}^{\#}(f) = \{ (a^{-}(Ef))^{*} + a^{-}(\mathbf{C}^{\#}Ef) \} / \sqrt{2}, \quad (34) \\ \pi_{m}^{\#}(f) = i \{ (a^{-}(Ef))^{*} + a^{-}(\mathbf{C}^{\#}\mu Ef) \} / \sqrt{2}. $$

**Remark 1.5.** Note that the $a$'s in these last formulas differ from those most often used in discussing the free field and that the correct space-time free field is $\mathbf{\Phi}_{m}^{\#}$ and not $\varphi_{m}^{\#}$ as we will discuss below, $\varphi_{m}^{\#}$ and $\pi_{m}^{\#}$ are useful for discussing the time-zero field. The maps $f \mapsto \varphi_{m}^{\#}(f)$ and $f \mapsto \pi_{m}^{\#}(f)$ are complex linear and $\varphi_{m}^{\#}(f), \pi_{m}^{\#}(f)$ are self-adjoint if and only if $Ef \in \mathbf{H}_{C^{\#}}^{\#}$.

Because of the projection $E$ we can extend the class of functions on which $\varphi_{m}^{\#}(\cdot)$ and $\pi_{m}^{\#}(\cdot)$ are defined to include distributions of the form $\delta(t - t_{0})g(x_{1}, x_{2}, x_{3})$ where $g \in \mathbf{R}_{c}^{\#3}$. In particular, if $t_{0} = 0, g$ is $\mathbf{R}_{c}^{\#}$-valued, and $\mathrm{Ext}\-\hat{g}$ is the usual Fourier transform on $\mathbf{R}_{c}^{\#3}$, then

$$ (\mathbf{C}^{\#}E\hat{\delta g})(p_0, -\mathbf{p}) = (2\pi_{\#})^{-1/2} \overline{\hat{g}(-\mathbf{p})} = (2\pi_{\#})^{-1/2} \hat{g}(-\mathbf{p}) = E\hat{\delta g}. \quad (35) $$

Thus $E(\delta g)$ and $\mu E(\delta g)$ are in $\mathbf{H}_{C^{\#}}^{\#}$. Therefore $\varphi_{m}^{\#}(\delta g)$ and $\pi_{m}^{\#}(\delta g)$ are self-adjoint if $g \in \mathcal{L}(\mathbf{R}_{c}^{\#3})$ is real. For obvious reasons, the maps $g \mapsto \varphi_{m}^{\#}(\delta g), g \mapsto \pi_{m}^{\#}(\delta g)$ are called the time-zero fields. From now on we will only use test functions of the form $\delta g$ in $\varphi_{m}^{\#}(\cdot)$ and $\pi_{m}^{\#}(\cdot)$ and write $\varphi_{m}^{\#}(g)$ and $\pi_{m}^{\#}(g)$ if $g \in S^{\#} \mathbf{R}_{c}^{\#3}$ instead of $\varphi_{m}^{\#}(\delta g)$ and $\pi_{m}^{\#}(\delta g)$.

If $f$ and $g$ are $\mathbf{R}_{c}^{\#}$-valued functions in $\mathcal{L}(\mathbf{R}_{c}^{\#3})$, then (X.70) implies that for $\psi \in F_{0}$,

$$ [\varphi_m^{\#}(f), \pi_m^{\#}(g)]\psi = i \left( \mathrm{Ext} - \int_{H_m} \overline{\hat{f}(p)\hat{g}(p)} \mu(p)\psi d\Omega_{m,x}^{\#} \right). \quad (36) $$

For convenience and also so that our notation coincides with the standard terminology,
we now transfer the fields we have constructed from the Fock space built up from $\mathcal{L}_2^\#(H_m^\# , d\Omega_{m,x}^\#)$ to the Fock space built up from $\mathcal{L}_2^\#(*\mathbb{R}_c^{\#3})$. For notational simplicity, we define for $f \in \mathcal{L}_2^\#(H_m^\# , d\Omega_{m,x}^\#)$

$$a^{\dagger}(f) = (a^{-}(f))^{*}, a(f) = a^{-}(C^{\#}f). \qquad (37)$$

First notice that each function $f(p) \in \mathcal{L}_2^\#(H_m^\# , d\Omega_{m,x}^\#)$ is in a natural way a function $f(\mathbf{p}) = f(\mu(\mathbf{p}), \mathbf{p})$ on $*\mathbb{R}_c^{\#3}$. For each $f \in \mathcal{L}_2^\#(H_m^\# , d\Omega_{m,x}^\#)$, we define

$$(Jf)(\mathbf{p}) = f(\mu(\mathbf{p}), \mathbf{p}) / \sqrt{\mu(\mathbf{p})}. \qquad (38)$$

$J$ is a unitary map of $\mathcal{L}_2^\#(H_m^\# , d\Omega_{m,x}^\#)$ onto $\mathcal{L}_2^\#(*\mathbb{R}_c^{\#3})$, so $\Gamma^\#(J)$ is a unitary map of $\mathcal{F}_s(\mathcal{L}_2^\#(H_m^\# , d\Omega_{m,x}^\#))$ onto $\mathcal{F}_s(\mathcal{L}_2^\#(*\mathbb{R}_c^{\#3}))$. The annihilation and creation operators on $\mathcal{F}_s(\mathcal{L}_2^\#(*\mathbb{R}_c^{\#3}))$, $\tilde{a}(\cdot)$, $\tilde{a}^{\dagger}(\cdot)$, are related to $a(\cdot)$ and $a^{\dagger}(\cdot)$ by the formulas

$$\tilde{a}\left(\frac{f(\mathbf{p})}{\sqrt{\mu(\mathbf{p})}}\right) = \Gamma^{\#}(J)a(f)\Gamma^{\#}(J)^{-1} \\ \tilde{a}^{\dagger}\left(\frac{f(\mathbf{p})}{\sqrt{\mu(\mathbf{p})}}\right) = \Gamma^{\#}(J)a^{\dagger}(f)\Gamma^{\#}(J)^{-1} \qquad (39)$$

We use the unitary map $\Gamma^\#(J)$ to carry the Wightman fields over to $\mathcal{F}_s(\mathcal{L}_2^\#(*\mathbb{R}_c^{\#3}))$ by defining:(i) for $*\mathbb{R}_{c,\text{fin}}^\#$-valued $f \in \mathcal{L}_{\text{fin}}^\#(*\mathbb{R}_c^{\#4})$

$$\tilde{\Phi}_{m,x}(f) = \Gamma^{\#}(J)\Phi_{m,x}(f)\Gamma^{\#}(J)^{-1} = \\ \frac{1}{\sqrt{2}}\left\{\tilde{a}\left(\tilde{\mathbf{C}}^{\#} \frac{Ef}{\sqrt{\mu}}\right) + \tilde{a}^{\dagger}\left(\frac{Ef}{\sqrt{\mu}}\right)\right\} \qquad (40)$$

(ii) for $*\mathbb{R}_{c,\text{fin}}^\#$-valued $f \in \mathcal{L}_{\text{fin}}^\#(*\mathbb{R}_c^{\#3})$

$$\tilde{\varphi}_{m,x}(f) = \Gamma^{\#}(J)\varphi_{m,x}(f)\Gamma^{\#}(J)^{-1} = \\ \frac{1}{\sqrt{2}}\left\{\tilde{a}\left(\tilde{\mathbf{C}}^{\#} \frac{E(f\delta)}{\sqrt{\mu}}\right) + \tilde{a}^{\dagger}\left(\frac{E(f\delta)}{\sqrt{\mu}}\right)\right\} \qquad (41)$$

where $\tilde{\mathbf{C}}^{\#} = J\mathbf{C}^{\#}J^{-1}$ acts by $(\tilde{\mathbf{C}}^{\#}g)(\mathbf{p}) = \overline{g(-\mathbf{p})}$. Having established this correspondence,

we now drop the ~ and the bold face letters; from now on we will only deal with the fields

on $\mathcal{F}_s(\mathcal{L}_2^\#(*\mathbb{R}_c^{\#3}))$ and three-dimensional momenta. Further, we recall that the restriction of

the four-dimensional Fourier transform that we have been using in this section to functions of the form $\delta(x_0)g(x_1, x_2, x_3)$ the usual three-dimensional Fourier transform. Notice that

$$\tilde{f} = \text{Ext-}\tilde{h}, h = (C^{\#}\hat{f}) \qquad (42)$$

so $C^{\#}\hat{f} = \hat{f}$ if and only if $f$ is $*\mathbb{R}_c^{\#}$-valued. For $f$ and $g$ $*\mathbb{R}_c^{\#}$-valued, (36) becomes

$$[\varphi_m^\#(f), \pi_m^\#(g)] \approx i(\text{Ext-} \int f(x)g(x)) d^{\#3}x. \qquad (43)$$
(43) is the space form of the canonical commutation relations (CCR).

In the Appendix to this section we prove that for each $m > 0$, this representation of the CCR is irreducible and for different $m$, the representations are inequivalent. Thus, the time-zero fields in the free scalar field theories give rise to different representation of the

CCR.

As a final topic before turning to interacting fields we will show how the structures developed above are related to the “fields” and “annihilation and creation operators” introduced in physics texts. We let now

$$D_{S_{\text{fin}}^{\#}} = \{\psi | \psi \in F_0, \psi^{(n)} \in S_{\text{fin}}^{\#} (*\mathbb{R}_c^{\#3n}), n \in \mathbb{N}\} \quad (44)$$

and for each $p \in *\mathbb{R}_c^{\#3}$ we define an operator $a(p)$ on $\mathcal{F}_s(\mathcal{L}_2^{\#}(*\mathbb{R}_c^{\#3}))$ with domain $D_{S_{\text{fin}}^{\#}}$ by

$$(a(p)\psi)^{(n)}(k_1, \dots, k_n) = \sqrt{n+1} \psi^{(n+1)}(p, k_1, \dots, k_n). \quad (45)$$

The adjoint of the operator $a(p)$ is not a #-densely defined operator since it is given formally by

$$(a^{\dagger}(p)\psi)^{(n)}(k_1, \dots, k_n) = \frac{1}{\sqrt{n}} \sum_{i=1}^{n} \delta(p - k_i) \psi^{(n+1)}(p, k_1, \dots, k_{i-1}, k_{i+1}, \dots, k_n). \quad (46)$$

However, $a^{\dagger}(p)$ is a well-defined quadratic form on $D_{\mathcal{L}_{\text{fin}}^{\#}} \times D_{\mathcal{L}_{\text{fin}}^{\#}}$. For example, if $\psi_1 = \{0, \psi^{(1)}, 0, \dots\}$, and $\psi_2 = \{0, 0, \psi^{(2)}, 0, \dots\}$, then

$$(\psi_2, a^{\dagger}(p)\psi_1) = \frac{1}{\sqrt{2}} \{ \text{Ext-} \int \overline{[\psi^{(2)}(k_1, p)\psi^{(1)}(k_1) + \psi^{(2)}(p, k_1)\psi^{(1)}(k_1)]} d^{\#}k_1 \}. \quad (47)$$

**Remark 1.1.** Note that the formulas

$$a(g) = \text{Ext-} \int_{*\mathbb{R}_c^{\#3}} a(p)g(-p)d^{\#}p \quad (48)$$

and

$$a^{\dagger}(g) = \text{Ext-} \int_{*\mathbb{R}_c^{\#3}} a^{\dagger}(p)g(p)d^{\#3}p \quad (49)$$

hold for all $g \in S_{\text{fin}}^{\#}(*\mathbb{R}_c^{\#3})$ if the equalities are understood in the sense of quadratic forms. That is, (48) means that for $\psi_1, \psi_2 \in D_{S_{\text{fin}}^{\#}}$ we have

$$(\psi_1, a(g)\psi_2) = \text{Ext-} \int_{*\mathbb{R}_c^{\#3}} (\psi_1, a(p)\psi_2)g(-p)d^{\#3}p \quad (50)$$

and similarly for (X.76b).

Since $a(p) : D_{\mathcal{L}_{\text{fin}}^{\#}} \to D_{\mathcal{L}_{\text{fin}}^{\#}}$ the powers of $a(p)$ are well-defined operators on $D_{\mathcal{L}_{\text{fin}}^{\#}}$.

As before we can write down a formal expression for $(a^{\dagger}(p))^{n}$, but it does not make sense as operator, only as $*\mathbb{C}_c^{\#}$-valued quadratic form on $D_{\mathcal{L}_{\text{fin}}^{\#}} \times D_{\mathcal{L}_{\text{fin}}^{\#}}$.

Notice that

$$(\psi_1, (a^{\dagger}(p))^{n}\psi_2) = ((a(p))^{n}\psi_1, \psi_2) \quad (51)$$

so for each $n$, $(a^{\dagger}(p))^{n}$ and $(a(p))^{n}$ are formally adjoints in the sense of $*\mathbb{C}_c^{\#}$-valued quadratic forms. We could of course have defined the quadratic form $(a^{\dagger}(p))^{n}$ by (50) and then calculated that it arises by taking the $n$-th power of the formal object given by (45). Since $a(p_1) : D_{\mathcal{L}_{\text{fin}}^{\#}} \to D_{\mathcal{L}_{\text{fin}}^{\#}}$, $(\psi_1, a^{\dagger}(p_2)a(p_1)\psi_2)$ is a well-defined $*\mathbb{C}_c^{\#}$-valued
quadratic form for all $\langle p_1, p_2 \rangle \in *\mathbb{R}_c^{\#3} \times *\mathbb{R}_c^{\#3}$. Notice, however, that
($\psi_1, a(p_1)a^\dagger(p_2)\psi_2$) does not make sense since $a^\dagger(p_2)$ is only a quadratic form. In
general any product $\prod_{i=1}^{N_1} a(f_i)$ is a
well-defined operator from $D_{\mathcal{L}_{\text{fin}}^{\#}}$ to $D_{\mathcal{L}_{\text{fin}}^{\#}}$ and $\prod_{i=1}^{N_1} a^\dagger(f_i)$ is a well-defined quadratic
form on $D_{\mathcal{L}_{\text{fin}}^{\#}} \times D_{\mathcal{L}_{\text{fin}}^{\#}}$. Thus

$$
\left( \psi_1, \left( \prod_{i=N_1+1}^{N_2} a^\dagger(p_i) \right) \left( \prod_{i=1}^{N_1} a^\dagger(-p_i) \right) \psi_2 \right) \qquad (52)
$$

is also well-defined *C<sub>c</sub><sup>#</sup>-valued quadratic form on D<sub>L<sub>fin</sub><sup>#</sup></sub> × D<sub>L<sub>fin</sub><sup>#</sup></sub>. One can check
directly that if f ∈ L<sub>fin</sub><sup>#</sup>(*R<sub>c</sub><sup>#3</sup>) then as *C<sub>c</sub><sup>#</sup>-valued quadratic forms

$$
\[
\left( \prod_{i=N_1+1}^{N_2} a^{\dagger}(f_i) \right) \left( \prod_{i=1}^{N_1} a^{\dagger}(f_i) \right) = \\
\mathrm{Ext} - \int_{*\mathbb{R}_c^{\#3N_2}} \left( \prod_{i=N_1+1}^{N_2} a^{\dagger}(p_i) \right) \left( \prod_{i=1}^{N_1} a^{\dagger}(-p_i) \right) \left( \prod_{i=1}^{N_2} f_i(p_i) \right) d^{\#}p_1 \dots d^{\#}p_{N_2} \tag{53}
\]
$$

and

$$
N = \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#3}} a^\dagger(p) a(p) d^\# p \qquad (54)
$$

The generator of time translations in the free scalar field theory of mass *m* is given by

$$
\mathbf{H}_0 = \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#3}} \mu(p) a^\dagger(p) a(p) d^\# p \quad (54)
$$

$\mathbf{H}_0$ is called the free Hamiltonian of mass $m$. (52), (53), and (54) involve no formal manipulations, but are mathematical statements about quadratic forms.

**Theorem X.44** Let $n_1$ and $n_2$ be nonnegative integers and suppose that
$W \in \mathcal{L}_2^{\#} (*\mathbb{R}_c^{\#3}(n_1+n_2))$. Then there is a unique operator $T_W$ on $\mathcal{F}_s(\mathcal{L}_2^{\#} (*\mathbb{R}_c^{\#3}))$ so that
$D_{\mathcal{L}_{\text{fin}}^{\#}} \subset D(T_W)$ is a core for $T_W$ and

(a) as *C<sub>c</sub><sup>#</sup>-valued quadratic forms on D<sub>L<sub>fin</sub><sup>#</sup></sub> × D<sub>L<sub>fin</sub><sup>#</sup></sub>

$$
T_W = \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#3(n_1+n_2)}} W(k_1, \dots, k_{n_1}, p_1, \dots, p_{n_2}) \left( \prod_{i=1}^{n_1} a^\dagger(k_i) \right) \left( \prod_{i=1}^{n_2} a(p_i) \right) d^{\#n_1}kd^{\#n_2}p \quad (55)
$$

(b) If $m_1$ and $m_2$ are nonnegative integers so that $m_1 + m_2 = n_1 + n_2$, then
$(1+N)^{-m_1/2}T_W(1+N)^{-m_2/2}$ is a bounded operator with

$$
\|(1 + N)^{-m_1/2} T_W (1 + N)^{-m_2/2} \| < C(m_1, m_2) \|W\|_{\mathcal{L}_2^{\#}}. \quad (56)
$$

In particular, if $m_1 = n_1$ and $m_2 = n_2$, then

$$
\| (1 + N)^{-n_1/2} T_W (1 + N)^{-n_2/2} \| < C(m_1, m_2) \| W \|_{\mathcal{L}_2^{\#}}. \quad (57)
$$

(c) As *C<sub>c</sub><sup>#</sup>-valued quadratic forms on D<sub>L<sub>fin</sub><sup>#</sup></sub> × D<sub>L<sub>fin</sub><sup>#</sup></sub>

$$
T_W^* = \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#3(n_1+n_2)}} \overline{W(k_1, \dots, k_{n_1}, p_1, \dots, p_{n_2})} \left( \prod_{i=1}^{n_2} a^\dagger(k_i) \right) \left( \prod_{i=1}^{n_1} a(p_i) \right) d^{\#n_1}kd^{\#n_2}p \quad (58)
$$

(d) If $W_n \to_* W$ in $\mathcal{L}_2^\# (*\mathbb{R}_c^{\#3(n_1+n_2)})$, then $T_{W_n} \to_* T_W$ strongly on $D_{\mathcal{L}_{\text{fin}}^\#}$.

(e) $F_0$ is contained in $D(T_W)$ and $D(T_W^*)$, and on vectors in $F_0$, $T_W$ and $T_W^*$ are given
by the explicit formulas

$$ (T_W \psi)^{(l-n_2+n_1)} = K(l, n_1, n_2) \mathbf{S} \times \left[ \mathrm{Ext} - \int_{\mathbb{R}_c^{\#3n_2}} W(k_1, \dots, k_{n_1}, p_1, \dots, p_{n_2}) \psi^{(l)}(p_1, \dots, p_{n_2}, k_{n_1+1}, \dots, k_{n_1+l-n_2}) d^{\#n_2} p \right] \quad (59) $$

$$ (T_W \psi)^n = 0 \text{ if } n < n_1 - n_2 $$

$$ (T_W^* \psi)^{(l-n_1+n_2)} = K(l, n_2, n_1) \mathbf{S} \times \left[ \mathrm{Ext} - \int_{\mathbb{R}_c^{\#3(n_1)}} \frac{W(k_1, \dots, k_{n_1}, p_1, \dots, p_{n_2}) \psi^{(l)}(k_1, \dots, k_{n_1}, p_{n_2+1}, \dots, p_{n_2+l-n_1}) d^{\#n_1} k} {((l-n_2)!)^2} \right] \quad (60) $$

$$ (T_W^* \psi)^n = 0 \text{ if } n < n_2 - n_1 \text{ where } \mathbf{S} \text{ is the symmetrization operator and} $$

$$ K(l, n_1, n_2) = \left[ \frac{l!(l+n_1-n_2)!}{((l-n_2)!)^2} \right]^{1/2} \quad (61) $$

**Proof.** For vectors in $D_{\mathcal{L}_{\mathrm{fin}}^{\#}}$, we define $T_W \psi$ by the formula (X.82a). By the Schwarz inequality and the fact that $\mathbf{S}$ is a projection,

$$ \|(T_W \psi)^{(l-n_2+n_1)}\|^2 \le K(l, n_1, n_2) \|\psi^{(l)}\|^2 \|W\|^2. \quad (62) $$

If we now define an operator $T_W^* \psi$, on $D_{\mathcal{L}_{\mathrm{fin}}^{\#}}$ by using the formula in (62), then for all $\varphi$ and $\psi$ in $D_{\mathcal{L}_{\mathrm{fin}}^{\#}}$ one easily verifies that $(\varphi, T_W \psi) = (T_W^* \varphi, \psi)$. Thus, $T_W$ is #-closable and $T_W^*$ is the restriction of the adjoint of $T_W$ to $D_{\mathcal{L}_{\mathrm{fin}}^{\#}}$. From now on we will use $T_W$ to denote $\bar{T}_W$ and $T_W^*$ to denote the adjoint of $T_W$. By the definition of $T_W$, $D_{\mathcal{L}_{\mathrm{fin}}^{\#}}$ is a #-core and further, since $T_W$ is bounded on the $l$-particle vectors in $D_{\mathcal{L}_{\mathrm{fin}}^{\#}}$, we have $F_0 \subset D(T_W)$. Since the right-hand side of (59) is also bounded on the $l$-particle vectors, (X.82a) represents $T_W$ on all $l$-particle vectors. The proof of the statements in (e) about $T_W^*$ are the same. To prove (b), let $\psi \in D_{\mathcal{L}_{\mathrm{fin}}^{\#}}$. Then by the above computation

$$ \left\| \left( (1+N)^{-m_1/2} T_W (1+N)^{-m_2/2} \psi \right)^{(l-n_2+n_1)} \right\|^2 \le \left[ \frac{K(l, n_1, n_2)}{(1+l-n_2+n_1)^{m_1/2} (1+l)^{m_2/2}} \right]^2 \|\psi^{(l)}\|^2 \|W\|^2 \quad (63) $$

so that

$$ \left\| \left( (1+N)^{-m_1/2} T_W (1+N)^{-m_2/2} \psi \right)^{(l-n_2+n_1)} \right\| \le \left[ \sup_{l \in \mathbb{N}} \frac{K(l, n_1, n_2)}{(1+l-n_2+n_1)^{m_1/2} (1+l)^{m_2/2}} \right] \|\psi^{(l)}\| \|W\| \le C(m_1, m_2) \|\psi^{(l)}\| \|W\| \quad (64) $$

where

$$ C(m_1, m_2) = \sup_{l \in \mathbb{N}} \frac{K(l, n_1, n_2)}{(1+l-n_2+n_1)^{m_1/2} (1+l)^{m_2/2}} < \infty^{\#} \quad (65) $$

since $m_1 + m_2 = n_1 + n_2$. In all the sup's only $l$ so that $l - n_2 + n_1 > 0$ occur since the other terms are annihilated by the action of $T_W$. Thus, $(1+N)^{-m_1/2} T_W (1+N)^{-m_2/2}$ extends to a hyper bounded operator on $\mathcal{F}_s(\mathbf{H}^{\#})$ with norm less than or equal to $C(m_1, m_2)$. If $m_1 = n_1$ and $m_2 = n_2$, then $C(m_1, m_2) = 1$.
To prove (d) we need only note that if $\psi = (0, \dots, \psi^{(l)}, 0, \dots) \in D_{\mathcal{L}_{\text{fin}}^{\#}}$ and $W_n \to_{\#} W$ in $\mathcal{L}_2^{\#}$, then

$$ \|T_{W_n}\psi - T_W\psi\| = \|(T_{W_n-W})\psi\| \le K(l, n_1, n_2) \|W_n - W\| \|\psi\|, \quad (66) $$

where $\#\text{-lim}_{n \to \infty} K(l, n_1, n_2) \|W_n - W\| \|\psi\| = 0$.

Since $D_{\mathcal{L}_{\text{fin}}^{\#}}$ consists of finite linear combinations of such vectors, we have shown that $T_{W_n} \#\text{-converges strongly on } D_{\mathcal{L}_{\text{fin}}^{\#}}$ to $T_W$ if $W_n \to_{\#} W$ in $\mathcal{L}_2^{\#}$.

To prove (a) let $\psi_1, \psi_2 \in D_{\mathcal{L}_{\text{fin}}^{\#}}$ with $\psi_1 = (0, \dots, \psi^{(l-n_2+n_1)}, 0, \dots)$ and $\psi_1 = (0, \dots, \psi^{(l)}, 0, \dots)$.

Then, if $W = (\prod_{i=1}^{n_1} f_i(k_i)) (\prod_{i=1}^{n_2} g_i(k_i))$ the definition of the form $(\prod_{i=1}^{n_1} a^\dagger(k_i)) (\prod_{i=1}^{n_2} a_i(k_i))$ shows that

$$ \begin{aligned} (\psi_1, T_W \psi_2) &= \text{Ext-} \int_{\mathbb{R}_c^{\#3n_2}} W(k_1, \dots, k_{n_1}, p_1, \dots, p_{n_2}) \times \\ & \qquad (\psi_1, (\prod_{i=1}^{n_1} a^\dagger(k_i)) (\prod_{i=1}^{n_2} a_i(k_i)) \psi_2) d^{\#n_1} k d^{\#n_2} p \end{aligned} \quad (67) $$

Since both sides of (X.83) are linear in $W$, the relationship continues to hold for the all such $W$'s that are hyperfinite linear combinations of such products. Since

$$ (\psi_1, (\prod_{i=1}^{n_1} a^\dagger(k_i)) (\prod_{i=1}^{n_2} a_i(k_i)) \psi_2) \in \mathcal{L}_2^{\#} (*\mathbb{R}_c^{\#3(n_1+n_2)}) \quad (68) $$

and since (d) holds, both the right- and left-hand sides of (X.83) are continuous linear functionals on $(*\mathbb{R}_c^{\#3(n_1+n_2)})$. Since they agree on a $\#$-dense set, they agree everywhere. Finally, (68) extends by linearity to all of $D_{\mathcal{L}_{\text{fin}}^{\#}} \times D_{\mathcal{L}_{\text{fin}}^{\#}}$.

This proves (a); the proof of (c) is similar. |

Finally, we note that as quadratic forms on $D_{\mathcal{L}_{\text{fin}}^{\#}}$ we can express the free scalar field and the time zero fields in terms of $a^\dagger(k)$ and $a(k)$:

$$ \Phi_{m,x}(x,t) = $$

$$ \frac{1}{(2\pi\#)^{3/2}} \int_{|p| \le x} \{[\text{Ext-exp}(\mu(p)t - ipx)]a^\dagger(p) + [\text{Ext-exp}(-\mu(p)t + ipx)]a(p)\} \frac{d^{\#3}p}{\sqrt{2\mu(p)}} \quad (69) $$

$$ \varphi_{m,x}^{\#}(x) = \frac{1}{(2\pi\#)^{3/2}} \int_{|p| \le x} \{[\text{Ext-exp}(-ipx)]a^{\dagger}(p) + [\text{Ext-exp}(ipx)]a(p)\} \frac{d^{\#3}p}{\sqrt{2\mu(p)}} \quad (70) $$

$$ \pi_{m,x}^{\#}(x) = \frac{1}{(2\pi\#)^{3/2}} \int_{|p| \le x} \{[\text{Ext-exp}(-ipx)]a^{\dagger}(p) - [\text{Ext-exp}(ipx)]a(p)\} \sqrt{\frac{\mu(p)}{2}} d^{\#3}p. \quad (71) $$

## § 2. $Q^{\#}$-space representation of the Fock space structures

In this section the construction of $Q^{\#}$-space and $L_2^{\#}(Q^{\#}, d^{\#}\mu)$, another representation of the Fock space structures are presented. In analogy with the one degree of freedom case where $\mathcal{F}^{\#}(*\mathbb{R}_c^{\#})$ is isomorphic to $L_2^{\#}(*\mathbb{R}_c^{\#}, d^{\#}x)$ in such a way that $\Phi_s(1)$ becomes multiplication by $x$, we will construct a $\#$-measure space $\langle Q^{\#}, \mu^{\#} \rangle$, with $\mu(Q^{\#}) = 1$, and a unitary map $S: \mathcal{F}_s^{\#}(*\mathbb{R}_c^{\#}) \to L_2^{\#}(Q^{\#}, d^{\#}\mu)$ so that for each $f \in \mathbf{H}_c^{\#}$, $S\varphi^{\#}(f)S^{-1}$ acts on $L_2^{\#}(Q, d^{\#}\mu^{\#})$ by multiplication by a $\#$-measurable function. We can then

show that in the case of the free scalar field of mass $m$ in 4-dimensional space-time,
$V = SH_I(g)S^{-1}$ is just multiplication by a function $V(q)$ which is in $L_p^{\#}(Q, d^{\#}\mu)$ for each $p \in \mathbb{N}^{\#}$. Let $\{f_n\}_{n=1}^{\infty}$ be an orthonormal basis for $\mathbf{H}^{\#}$ so that each $f_n \in \mathbf{H}_{\mathbb{C}^{\#}}^{\#}$ and let $\{g_k\}_{k=1}^{N}, N \in \mathbb{N}^{\#}$ be a finite or hyperfinite subcollection of the $\{f_n\}_{n=1}^{\infty}$. Let $\mathbf{P}_N$ be a set of the all external hyperfinite polynomials $\mathrm{Ext}-\mathbf{P}[u_1, \dots, u_N]$ and $\mathcal{F}_N^{\#}$ be the #-closure of the set

$$ \{\mathrm{Ext}-\mathbf{P}[\varphi^{\#}(g_1), \dots, \varphi^{\#}(g_N)]\}_{P \in \mathbf{P}_N} \qquad (2.1) $$

in $\mathcal{F}_s^{\#}(\mathbf{H}^{\#})$ and define $F_0^N = \mathcal{F}_N^{\#} \cap F_0$. From Theorem X.43 (and its proof) it follows that $\varphi^{\#}(g_k)$ and $\pi^{\#}(g_l)$, for all $1 \le k, l \le N$ are essentially self-adjoint on $F_0^N$ and that

$$ (\mathrm{Ext}-\exp[it\varphi^{\#}(g_k)])(\mathrm{Ext}-\exp[is\pi^{\#}(g_l)]) = (\mathrm{Ext}-\exp[-ist\delta_{kl}])(\mathrm{Ext}-\exp[is\pi^{\#}(g_l)])(\mathrm{Ext}-\exp[it\varphi^{\#}(g_k)]). \qquad (2) $$

Thus we have a representation of the generalized Weyl relations in which the vector $\Omega_0$ satisfies $([\varphi^{\#}(g_k)]^2 + [\pi^{\#}(g_k)]^2 - 1)\Omega_0 = 0$ and is #-cyclic for the operators $\{\varphi^{\#}(g_k)\}_{k=1}^N, N \in \mathbb{N}^{\#}$. Therefore there is a unitary map $\tilde{\mathbf{S}}^{(N)} : \mathcal{F}_N^{\#} \to L_2^{\#}(\mathbb{R}_c^{\#N})$ so that

$$ \begin{aligned} \tilde{\mathbf{S}}^{(N)} \varphi^{\#}(g_k) (\tilde{\mathbf{S}}^{(N)})^{-1} &= x_k \\ \tilde{\mathbf{S}}^{(N)} \pi^{\#}(g_k) (\tilde{\mathbf{S}}^{(N)})^{-1} &= \frac{1}{i} \frac{d^{\#}}{dx_k^{\#}} \end{aligned} \qquad (3) $$

and

$$ \tilde{\mathbf{S}}^{(N)} \mathbf{Q}_0 = \pi_{\#}^{-N/4} \left\{ \mathrm{Ext} - \exp \left[ - \left( \mathrm{Ext} - \sum_{k=1}^{N} \frac{x_k^2}{2} \right) \right] \right\}. \qquad (4) $$

It is convenient to use the Hilbert space

$$ L_2^{\#} \left( *\mathbb{R}_c^{\#N}, \pi_{\#}^{-N/2} d^{\#N} x \left\{ \mathrm{Ext} - \exp \left[ - \left( \mathrm{Ext} - \sum_{k=1}^{N} \frac{x_k^2}{2} \right) \right] \right\} \right) $$

instead of $L_2^{\#}(\mathbb{R}_c^{\#N})$ so let $d^{\#}\mu_k = \pi_{\#}^{-1/2} \exp(-x_k^2/2) d^{\#}x_k$ and define

$$ (Tf)(x) = \pi_{\#}^{N/4} \left[ \mathrm{Ext} - \exp \left( \mathrm{Ext} - \sum_{k=1}^{N} \frac{x_k^2}{2} \right) \right] f(x). \qquad (5) $$

Then $T$ is a unitary map of $L_2^{\#}(\mathbb{R}_c^{\#N})$ onto $L_2^{\#}(\mathbb{R}_c^{\#N}, \mathrm{Ext} - \prod_{k=1}^N d^{\#}\mu_k^{\#})$ and if we let $\mathbf{S}^{(N)} = T\tilde{\mathbf{S}}^{(N)}$ we get

$$ \begin{aligned} \mathbf{S}^{(N)} : \mathcal{F}_N^{\#} &\to L_2^{\#} (*\mathbb{R}_c^{\#N}, \mathrm{Ext} - \prod_{k=1}^N d^{\#}\mu_k^{\#}), \\ \mathbf{S}^{(N)} \varphi^{\#}(g_k) (\mathbf{S}^{(N)})^{-1} &= x_k, \\ \mathbf{S}^{(N)} \pi^{\#}(g_k) (\mathbf{S}^{(N)})^{-1} &= -\frac{x_k}{i} + \frac{1}{i} \frac{d^{\#}}{d^{\#}x_k}, \\ \mathbf{S}^{(N)} \mathbf{Q}_0 &= 1, \end{aligned} \qquad (6) $$

where 1 is the function identically one. Note that each $\mu_k^{\#}$ has mass one, which implies that
$$
\begin{align}
& \langle \mathbf{Q}_0, (\mathrm{Ext} - \prod_{k=1}^N P_k[\varphi^\#(g_k)]) \mathbf{Q}_0 \rangle = \nonumber \\
& \int_{*\mathbb{R}_c^{\#N}} (\mathrm{Ext} - \prod_{k=1}^N P_k[x_k]) (\mathrm{Ext} - \prod_{k=1}^N d^\# \mu_k^\#) = \nonumber \\
& \mathrm{Ext} - \prod_{k=1}^N \int_{*\mathbb{R}_c^\#} P[x_k] d^\# \mu_k^\# = \mathrm{Ext} - \prod_{k=1}^N \int_{*\mathbb{R}_c^\#} \langle \mathbf{Q}_0, P_k[\varphi^\#(g_k)] \mathbf{Q}_0 \rangle, \tag{7}
\end{align}
$$

where $P_1, \dots, P_N$ are external hyperfinite polynomials. This formula (7) can also be
proven by direct computations on $\mathcal{F}_s^\#(\mathbf{H}^\#)$.

Now it is easy to see how to construct $\langle Q^{\#}, d^{\#} \mu^{\#} \rangle$. We define $Q^{\#} = \times_{k=1}^{\infty} * \mathbb{R}_c^{\#}$. Take the
$\sigma^{\#}$-algebra generated by hyper infinite products of #-measurable sets in $* \mathbb{R}_c^{\#}$ and set
$\mu^{\#} = \otimes_{k=1}^{\infty} \mu_k^{\#}$. We denote the points of $Q^{\#}$ by $q = \langle q_1, q_2, \dots \rangle$. Then $\langle Q^{\#}, d^{\#} \mu^{\#} \rangle$ is a
\#-measure space and the set of functions of the form $P(q_1, q_2, \dots)$, where $P$ is a
polynomial and $n \in \mathbb{N}^{\#}$ is arbitrary, is \#-dense in $\mathcal{L}_2^{\#}(Q^{\#}, d^{\#} \mu^{\#})$. Let $P$ be a polynomial in
$N \in \mathbb{N}^{\#}$ variables

$$
P(x_{k_1}, \dots, x_{k_N}) = \mathrm{Ext} - \sum_{l_1, \dots, l_N} c_{l_1, \dots, l_N} x_{k_1}^{l_1} \dots x_{k_N}^{l_N} \quad (8)
$$

and define

$$
S : P(\varphi^{\#}(f_{k_1}), \dots, \varphi^{\#}(f_{k_N})) \mathbf{Q}_0 \to P(q_{k_1}, \dots, q_{k_N}). \quad (9)
$$

Then

$$
\begin{equation}
\begin{split}
P(\varphi^{\#}(f_{k_1}), \dots, \varphi^{\#}(f_{k_N})) \mathbf{Q}_0 = \operatorname{Ext} - & \sum_{l, \mathbf{m}} c_l \bar{c}_{\mathbf{m}}(\mathbf{Q}_0, \varphi^{\#}(f_{k_1})^{l_1+m_1}, \dots, \varphi^{\#}(f_{k_N})^{l_N+m_N} \mathbf{Q}_0) = \\
& \sum_{l, \mathbf{m}} c_l \bar{c}_{\mathbf{m}} \int_{*\mathbb{R}_c^{\#N}} q_{k_1}^{l_1+m_1} \cdots q_{k_N}^{l_N+m_N} \left( \operatorname{Ext} - \prod_{i=1}^N d^\# \mu_{k_i}^\# \right) = \int_{\mathcal{Q}^\#} |P(x_{k_1}, \dots, x_{k_N})|^2 d^\# \mu^\#
\end{split}
\tag{10}
\end{equation}
$$

by (X.92) and the fact that each $\mu_k^\#$ has mass one. Since $\mathbf{Q}_0$ is cyclic for polynomials
in the fields (Theorem X.42), $S$ extends to a unitary map of $\mathcal{F}_s^\#(\mathbf{H}^\#)$ onto $\mathcal{L}_2^\#(Q^\#, d^\#\mu^\#)$.
Clearly

$$
S\varphi^{\#}(f_k)S^{-1} = q_k \quad \text{and} \quad SQ_0 = 1. \tag{11}
$$

**Theorem 1.** Let $\varphi_{m,x}^{\#}(f), x \in *\mathbb{R}_c^{\#1} *\mathbb{R}_{c,fin}^{\#}$ be the free scalar field of mass $m$ (in 4-dimensional space-time) at time zero. Let $g \in \mathcal{L}_1^{\#}(*\mathbb{R}_c^{\#3}) \cap \mathcal{L}_2^{\#}(*\mathbb{R}_c^{\#3})$ and define

$$
H_{l,x,\lambda}(g) = \lambda(x) \int g(x) : \varphi_{m,x}^{\#}(x)^4 : d^{\#3}x, \quad (12)
$$

where $\lambda(x) \in *\mathbb{R}_c^\#$, $\lambda(x) \approx 0$. Let $S$ denote the unitary map of $\mathcal{F}_s^\#(\mathbf{H}^\#)$ onto $\mathcal{L}_2^\#(Q^\#, d^\#\mu^\#)$ constructed above. Then $V = SH_{l,x,\lambda}(g)S^{-1}$ is multiplication by a function $V_{x,\lambda}(q)$ which satisfies:

(a) $V_{x,\lambda}(q) \in \mathcal{L}_p^\#(Q^\#, d^\#\mu^\#)$ for all $p \in \mathbb{N}^\#$.

(b) $\mathrm{Ext}\text{-exp}(-t\mathrm{V}_{x,\lambda}(q)) \in \mathcal{L}_1^\#(Q^\#, d^\#\mu^\#)$ for all $t \in [0, \infty)$.

**Proof.** We will prove (a). By Eq.() we get

$$
\varphi_{m,x}^{\#}(x) = \frac{1}{(2\pi\#)^{3/2}} \int_{|p| \le x} \{[\mathrm{Ext}\text{-exp}(-ipx)]a^{\dagger}(p) + [\mathrm{Ext}\text{-exp}(ipx)]a(p)\} \frac{d^3p}{\sqrt{2\mu(p)}}. \quad (13)
$$

Then $\varphi_{m,x}^{\#}(x)$ is a well-defined operator-valued function of $x \in *\mathbb{R}_c^{\#3}$. We define
By Theorem X.44 : $\phi_{m,x}^{\#}(x)^4$ is also a well-defined operator for each $x \in *\mathbb{R}_c^{\#3}$ and : $\phi_{m,x}^{\#}(x)^4$ takes $F_0$ into itself. Thus for each $x \in *\mathbb{R}_c^{\#3}$,

where the coefficients $d_2(x)$ and $d_0(x)$ are independent of $x$. For each $x \in *\mathbb{R}_c^{\#3}$, $S\phi_{m,x}^{\#}(x)S^{-1}$ is just the operator on #-measurable space $\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})$ which operates by multiplying by the function

$$ : \phi_{m,x}^{\#}(x)^4 := \phi_{m,x}^{\#}(x)^4 + d_2(x)\phi_{m,x}^{\#}(x)^2 + d_0(x) \quad (14) $$

so $S\phi_{m,x}^{\#}(x)^4 S^{-1}$ and $S\phi_{m,x}^{\#}(x)^2 S^{-1}$ are in $\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})$ and the $\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})$ norms are uniformly bounded in $x$. Therefore, since $g \in \mathcal{L}_1^{\#}(*\mathbb{R}_c^{\#3})$, $SH_{I,x,\lambda}(g)S^{-1}$ operates on $\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})$ by multiplication by an $\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})$ function which we denote by $V_{x,\lambda}(q)$. Consider now the expression for $H_{I,x}(g)Q_0$. This is a vector $(0,0,0,0,\psi^{(4)},0,\dots)$

$$ \mathrm{Ext}-\sum_{k=1}^{\infty} c_k(x,x)q_k \quad (15) $$

where

$$ c_k(x,x) = (2\pi\#)^{-3/2} (f_k, \mathrm{Ext}-\exp(ipx)(\mu(p)))^{-1/2}. \quad (16) $$

Furthermore,

$$ \mathrm{Ext}-\sum_{k=1}^{\infty} |c_k(x,x)|^2 = (2\pi\#)^{-3/2} \|(\mu(p))^{-1/2}\|_2^2, \quad (17) $$

so $S\phi_{m,x}^{\#}(x)^4 S^{-1}$ and $S\phi_{m,x}^{\#}(x)^2 S^{-1}$ are in $\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})$ and the $\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})$ norms are uniformly bounded in $x$. Therefore, since $g \in \mathcal{L}_1^{\#}(*\mathbb{R}_c^{\#3})$, $SH_{I,x,\lambda}(g)S^{-1}$ operates on $\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})$ by multiplication by an $\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})$ function which we denote by $V_{x,\lambda}(q)$. Consider now the expression for $H_{I,x}(g)Q_0$. This is a vector $(0,0,0,0,\psi^{(4)},0,\dots)$

$$ \begin{aligned} \psi^{(4)}(p_1,p_2,p_3,p_4) &= \mathrm{Ext}-\int_{*\mathbb{R}_c^{\#3}} \frac{\lambda g(x)[\mathrm{Ext}-\exp(-ix \sum_{i=1}^4 p_i)]d^{\#3}x}{(2\pi\#)^{3/2} \prod_{i=1}^4 (2\mu(p_i))^{1/2}} = \\ &= \frac{\lambda \hat{g}(\sum_{i=1}^4 k_i)}{(2\pi\#)^{9/2} \prod_{i=1}^4 (2\mu(p_i))^{1/2}} \end{aligned} \quad (18) $$

where $|p_i| \le x, 1 \le i \le 4$. We choose now the parameter $\lambda = \lambda(x) \approx 0$ such that $\|\psi^{(4)}\|_2 \in \mathbb{R}$, thus

$$ \|H_{I,x,\lambda(x)}(g)Q_0\|_2 \in \mathbb{R}, \quad (19) $$

since $\|H_{I,x,\lambda(x)}(g)Q_0\|_2 = \|\psi^{(4)}\|_2$. But, since $SQ_0 = 1$, we get

$$ \|H_{I,x,\lambda(x)}(g)Q_0\|_2 = \|SH_{I,x,\lambda(x)}(g)S^{-1}\|_{\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})} = \|V_{x,\lambda(x)}(q)\|_{\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})} \quad (20) $$

From (19) and Eq.(20) we get that $\|V_{x,\lambda(x)}(q)\|_{\mathcal{L}_2^{\#}(Q^{\#}, d^{\#}\mu^{\#})}$ is finite. It is easily verify that each $P(q_1,q_2,\dots,q_n), n \in \mathbb{N}^{\#}$ is in the domain of $V_{x,\lambda(x)}(q)$ and $SH_{I,x,\lambda(x)}(g)S^{-1} = V_{x,\lambda(x)}(q)$ on that domain. Since $Q_0$ is in the domain of $[H_{I,x,\lambda(x)}(g)]^p$ for all $n \in \mathbb{N}^{\#}$, 1 is in the domain of $[V_{x,\lambda(x)}(q)]^n$ for all $n \in \mathbb{N}^{\#}$. Thus, for all $n \in \mathbb{N}^{\#}$, $V_{x,\lambda(x)} \in \mathcal{L}_{2n}^{\#}(Q^{\#}, d^{\#}\mu^{\#})$. Since $\mu^{\#}(Q^{\#}) < \infty$, $V_{x,\lambda(x)} \in \mathcal{L}_p^{\#}(Q^{\#}, d^{\#}\mu^{\#})$ for all $p < \infty$.

# Chapter X. A non-Archemedean Banach algebras and $C_{\#}^*$-Algebras.

## §1. A non-Archemedean Banach algebra $B(H^{\#})$
§1.1. Basic Properties

**Definition 1.1.** An linear operator $T$ on a non-Archemedean Hilbert space $H^\#$ is a linear map $H^\# \to H^\#$. We can define a \#-norm by

$$ \|T\|_{\#} \triangleq \sup_{v \in H^{\#} \setminus \{0\}} \frac{\|Tv\|_{\#}}{\|v\|_{\#}} \qquad (1.1) $$

if supremum in RHS of (1.1) exists.

This is a \#-norm since

1. By definition of the \#-norm on $H^\#$, it is always positive.

2. We have that $T = 0 \Leftrightarrow \forall v \in H^\#$, $Tv = 0 \Leftrightarrow \forall v \in H^\# \setminus \{0\}$,

$$ \frac{\|Tv\|_{\#}}{\|v\|_{\#}} = 0 \Leftrightarrow \|T\|_{\#} = 0. $$

$$ 3. \quad \|\lambda T\|_{\#} = \sup_{v \in H^{\#} \setminus \{0\}} \frac{\|\lambda T v\|_{\#}}{\|v\|_{\#}} = |\lambda| \sup_{v \in H^{\#} \setminus \{0\}} \frac{\|T v\|_{\#}}{\|v\|_{\#}} = |\lambda| \|T\| $$

$$ 4. \quad \|T_1 + T_2\|_{\#} = \sup_{v \in H^{\#} \setminus \{0\}} \frac{\|T_1 v + T_2 v\|_{\#}}{\|v\|_{\#}} \le \sup_{v \in H^{\#} \setminus \{0\}} \frac{\|T_1 v\|_{\#} + \|T_2 v\|_{\#}}{\|v\|_{\#}} \le $$

$$ \le \sup_{v \in H^{\#} \setminus \{0\}} \frac{\|T_1 v\|_{\#}}{\|v\|_{\#}} + \sup_{v \in H^{\#} \setminus \{0\}} \frac{\|T_2 v\|_{\#}}{\|v\|_{\#}} = \|T_1\|_{\#} + \|T_2\|_{\#}. $$

**Definition 1.2.** Let $H^\#$ be a non-Archemedean Hilbert space over $\mathbb{C}_c^\#$. A linear map $A: H^\# \to H^\#$ is called bounded in $\mathbb{R}_c^\#$ operator iff $\|A\|_{\#} < *\infty$.

**Definition 1.3.** Let $H^\#$ be a non-Archemedean Hilbert space over $\mathbb{C}_c^\#$. We denote by $B(H^\#)$ the set of all bounded in $\mathbb{R}_c^\#$ operators $A: H^\# \to H^\#$.

**Definition 1.4.** Algebra $A$ is called an algebra over $\mathbb{C}_c^\#$ if it is a vector space over $\mathbb{C}_c^\#$ and a binary map $\cdot: A \times A \to A$ Satisfying:

1. Left distributivity: $\forall v, w, u \in A[(v + w) \cdot u = v \cdot u + w \cdot w]$

2. Right distributivity: $\forall v, w, u \in A[v \cdot (w + u) = v \cdot w + v \cdot u]$

3. $\forall v, w \in A, \forall \alpha, \beta \in \mathbb{C}_c^\# [\alpha\beta v \cdot w = (\alpha v) \cdot (\beta w)]$

We note that $B(H^\#)$ is an algebra over $\mathbb{C}_c^\#$ where for $A, B \in B(H^\#)$, $\lambda \in \mathbb{C}_c^\#$ we define:

$$ \lambda A: H^\# \to H^\#; v \mapsto \lambda A v $$

$$ A + B: H^\# \to H^\#; v \mapsto A v + B v $$

$$ A \cdot B: H^\# \to H^\#; v \mapsto A(B(v)) $$

In $B(H^\#)$ we have the \#-adjoint operator. This maps each $A$ to the unique $A^*$ such that for all $v, w \in H^\#$ we have $\langle Av, w \rangle_\# = \langle v, A^*w \rangle_\#$. We denote the adjoint of an operator $A$ by $A^*$ and define the adjoint of a subset $M \subset B(H^\#)$ by $M^* \triangleq \{A^* \in B(H^\#) | A \in M\}$. The adjoint has the following key properties:

**Lemma 1.4.** Adjoint Properties (Algebraic)

$\forall B, A \in B(H^\#)$ we have

1. $A^*$ always exists is unique.

2. If $A$ is bounded in $\mathbb{R}_c^\#$, then $A^*$ is also bounded in $\mathbb{R}_c^\#$.

3. $A^{**} = A$ (Involutivity)

$$ 4. \quad \|A\|_{\#} = \|A^*\|_{\#} $$

5. If $A$ is invertible, $A^*$ also is, with $(A^*)^{-1} = (A^{-1})^*$

$$ 6. (A + B)^* = A^* + B^*, (\lambda A)^* = \bar{\lambda} A^* $$

$$ 7. (AB)^* = B^*A^* $$

$$ 8. \quad \|A^*A\|_{\#} = \|A\|_{\#} $$
**Proof.** 1. Let $x \in H^{\#}$ and consider the bounded in $^*\mathbb{R}_c^{\#}$ linear functional $f: H^{\#} \to ^*\mathbb{C}_c^{\#}$,
$f(v) \mapsto \langle Av, x \rangle_{\#}$ we have $\|f\| \le \|A\|_{\#} \|x\|_{\#}$. By generalized Riesz representation
theorem there exists a unique $y \in H^{\#}$ with $f(v) = \langle v, y \rangle_{\#} \forall v \in H^{\#}$. So we set $A^*x = y$.
Then for any $y, z \in H^{\#}$ and $\forall \alpha \in ^*\mathbb{C}_c^{\#}$ we have:

$$
\begin{align*}
\langle v, A^*(\alpha y + z) \rangle_{\#} &= \langle Av, \alpha y + z \rangle_{\#} = \bar{\alpha} \langle \langle v, A^*y \rangle_{\#} + \langle v, A^*z \rangle_{\#} = \\
&= \langle v, \alpha A^*y + A^*z \rangle_{\#} \quad \forall v \in H^{\#}. \text{ In particular, if we choose } v = A^*(\alpha y + z) - \alpha A^*y + A^*z, \\
&\text{we see that } \|v\|_{\#} = 0 \Rightarrow v = 0 \Rightarrow A^* \text{ is linear.}
\end{align*}
$$

2. Following from 1. we have

$$
\|A^*x\|_{\#} = \|y\|_{\#} = \|f\|_{\#} \le \|A\|_{\#} \|y\|_{\#}.
$$

3. We can see this as

$$
\langle A^{**}v, w \rangle_{\#} = \langle v, A^{*}w \rangle_{\#} = \langle Av, w \rangle_{\#} \quad \forall v, w \in H^{*}.
$$

4. Combining the estimate from above and involutivity, we have

$$
\|A^{**}\|_{\#} \le \|A^{*}\|_{\#} \le \|A\|_{\#} = \|A^{**}\|_{\#}.
$$

So we must have equality everywhere.

5. We have $\langle v, (A^{-1})^*A^*w \rangle_\# = \langle A^{-1}v, A^*w \rangle_\# = \langle AA^{-1}v, w \rangle_\# = \langle v, w \rangle_\# \quad \forall v, w \in H^\#$.

Hence, $(A^{-1})^*A^* = 1$. The argument for $A^*(A^{-1})^* = 1$ is the same.

6. This follows clearly from conjugate linearity in the second argument of an inner product.

7. This is clear since, $\langle ABv, w \rangle_{\#} = \langle Bv, A^*w \rangle_{\#} = \langle v, B^*A^*w \rangle_{\#} \quad \forall v, w \in H^*.$

8. For this we have $\|T\|_{\#}^2 = \sup_{\|x\|_{\#}=1} \|Tx\|_{\#}^2 = \sup_{\|x\|_{\#}=1} |\langle Tx, Tx \rangle_{\#}| =$
$\sup_{\|x\|_{\#}=1} |\langle T^*Tx, x \rangle_{\#}| \le \sup_{\|x\|_{\#}=1} \|T^*Tx\|_{\#} \|x\|_{\#} = \|T^*T\|_{\#}.$ But also,
$\|T^*T\|_{\#} \le \|T^*\|_{\#} \|T\|_{\#} = \|T\|_{\#}^2$, and so there is equality everywhere.

§1.2 Types of Operators

**Definition 1.2.1.** A is called *normal* if $A^*A = AA^*$.

**Definition 1.2.2.** A is called positive if $A = B^*B$ for some $B \in B(H^\#)$

**Definition 1.2.3.** A is called *self #-adjoint* if $A^* = A$.

**Lemma 1.2.1.** Let $A \in B(H^{\#})$. Then $A = A_1 + iA_2$ where $A_1$ and $A_2$ are both self \#-adjoint.

**Proof.** Let $A_1 = \frac{A + A^*}{2}$, $A_2 = \frac{iA^* - iA}{2}$.

It is then clear from basic algebra.

**Definition 1.2.4.** *U* is called *unitary* if *U*\* *U* = *U* *U*\* = 1

**Example 1.2.1.** If $U$ is unitary, we have $\forall h, k \in H^{\#}, \langle h, k \rangle_{\#} = \langle Uh, Uk \rangle_{\#}$. This is
because $\langle Uh, Uk \rangle_{\#} = \langle h, 1k \rangle_{\#} = \langle h, k \rangle_{\#}$.

**Definition 1.2.5.** A is called *isometric* if $A^*A = 1$.

We also have a relaxed definition, a partial isometry.

**Definition 1.2.6.** A is called a partial isometry if it is an isometry on the orthogonal complement of its kernel, i.e. $A^*A v = v, \forall v \in \text{Ker}(A)^\perp =$
$= \{v \in H^\# | \langle v, w \rangle_\# = 0, \forall w \in \text{Ker}(A)\}$.

**Definition 1.2.7.** $p \in B(H^\#)$ is called a *projection* if $p = p^* = p^2$.

**Example 1.2.2.** Consider $H^\# = I_2^*(\mathbb{N})$ the set of all square summable $^*\mathbb{C}_c^\#$-valued series. An example of a projection would be:

$$
p_n:H^{\#} \rightarrow H^{\#}, (a_1, a_2, \dots, a_n, a_{n+1}, a_{n+2}, \dots) \mapsto (a_1, a_2, \dots, a_n, 0, 0, \dots).
$$

We see this is self #-adjoint as $\langle p_n a, b \rangle_\# = \text{Ext}-\sum_{k=1}^n a_k \bar{b}_k = \langle a, p_n b \rangle_\#$ and idempotent
as $p_n^2 = p_n$.
**Lemma 1.2.2.** Multiplication and #-norm property

$$ \forall A, B \in B(H^{\#}), \|A \cdot B\|_{\#} \le \|A\|_{\#} \|B\|_{\#} $$

**Proof.** For all $h \in H^{\#}$, we always have the estimate $\|Ah\|_{\#} \le \|A\|_{\#} \|h\|_{\#}$.

Using this we have

$$ \|AB\|_{\#} = \sup_{h \in H \setminus \{0\}} \|(AB)h\|_{\#} / \|h\|_{\#} = \sup_{h \in H \setminus \{0\}} \|A(Bh)\|_{\#} / \|h\|_{\#} $$

$$ \le \sup_{h \in H \setminus \{0\}} \|A\|_{\#} \|Bh\|_{\#} / \|h\|_{\#} = \|A\|_{\#} \|B\|_{\#} $$

**Lemma 1.2.3.** $(B(H^{\#}), \|\cdot\|_{\#})$ is complete, i.e. if $(A_n)_{n \in \mathbb{N}} \subset B(H^{\#})$ is cauchy with respect to the operator #-norm $\|\cdot\|_{\#}$, it #-converges in #-norm to some element $A \in B(H^{\#})$.

**Proof.** Let $(A_n)_{n \in \mathbb{N}}$ be cauchy with respect to the operator #-norm. This means that

$$ \forall \varepsilon(\varepsilon \approx 0, \varepsilon > 0) \exists N \in \mathbb{N}_{\infty} \ [n, m > N \Rightarrow \|A_n - A_m\|_{\#} < \varepsilon]. $$

In particular then $\|A_n\|$ is bounded above, say by $K \in \mathbb{R}_{c,+}^{\#}$. Now fix $v \in H^{\#}$ and let $N, m, n \in \mathbb{N}_{\infty}$ be as before. We have that

$$ \|A_n v - A_m v\|_{\#} \le \|A_n - A_m\|_{\#} \|v\|_{\#} \le \|v\|_{\#}. $$

Hence, $(A_n v)_{n \in \mathbb{N}}$ is cauchy in $H^{\#}$. By completeness of $H^{\#}$, we have a #-limit and can define $A: H^{\#} \to H^{\#}, v \mapsto \lim_{n \to \infty} A_n v$, this is our candidate for our #-limit.

A is linear since (by algebra of #-limits)

$$ A(\alpha v + w) = \#\lim_{n \to \infty} A_n(\alpha v + w) = \alpha (\#\lim_{n \to \infty} A_n v) + \#\lim_{n \to \infty} A_n w = \alpha Av + Aw $$
and bounded in $\mathbb{R}_{c,+}^{\#}$ because

$$ \|Av\|_{\#} = \#\lim_{n \to \infty} \|A_n v\|_{\#} = \#\lim_{n \to \infty} \|A_n v\|_{\#} \le \#\lim_{n \to \infty} \|A_n\|_{\#} \|v\|_{\#} \le K \|v\|_{\#}. $$

Hence, $A \in B(H^{\#})$. Finally we show convergence in #-norm. Fix $\varepsilon \approx 0, \varepsilon > 0$ and let $N \in \mathbb{N}_{\infty}$ be as in the definition of cauchy. If $n > N$ we have:

$$ \|A - A_n\|_{\#} = \sup_{\|v\|_{\#}=1} \|\#\lim_{m \to \infty} (A_m - A_n)v\|_{\#} \le $$

$$ \sup_{\|v\|_{\#}=1} \#\lim_{m \to \infty} \|(A_m - A_n)v\|_{\#} \|v\|_{\#} \le \#\lim_{m \to \infty} = \varepsilon. $$

**Definition 1.2.8.** $A \in B(H^{\#})$ is #-compact if for all bounded subsets $\beta$ of $H^{\#}$ the image of $A$ restricted to $\beta$ has #-compact #-closure:

$$ A \in B(H^{\#}) \text{ #-compact} \Leftrightarrow \forall \beta \text{ bounded, } \#A\beta \text{ is #-compact.} $$

We denote by $K^{\#}(B(H^{\#}))$ the set of all #-compact operators in $B(H^{\#})$.

**Lemma 1.2.4.** A is #-compact iff

$$ \forall (v_\alpha)_{\alpha \in X} \text{ bounded} \Rightarrow (A v_\alpha)_{\alpha \in X} \text{ has a #-convergent subsequence.} $$

**Definition 1.2.9.** For $A \in B(H^{\#})$ the rank $\Re(A)$ of $A$ is the dimension of the range of $A$

**Lemma 1.2.5.** If $A$ has rank $N \in \mathbb{N}$, then we can write

$$ A(\cdot) = \mathrm{Ext} - \sum_{n \le N} \alpha_n \langle \cdot, v_n \rangle_{\#} w_n \quad \text{where} \quad \{v_n\}_{n \le N}, \{w_n\}_{n \le N} \subset H^{\#}, \{\alpha_n\}_{n \le N} \subset \mathbb{C}_c^{\#} $$

**Proof.** This follows immediately from the generalized Riesz representation theorem noting that if $\{w_n\}_{n \le N}$ forms a basis of $A(H^{\#})$, then $\langle A \cdot, w_n \rangle_{\#}$ is a linear functional $H^{\#} \to \mathbb{C}_c^{\#}$. So for some $v_n$, $\langle A \cdot, w_n \rangle_{\#} = \langle \cdot, v_n \rangle_{\#}$ so setting $\alpha_n = 1/\|w_n\|_{\#}$ we have

$$ A \cdot = \mathrm{Ext} - \sum_n \alpha_n \langle A \cdot, w_n \rangle_{\#} w_n = \mathrm{Ext} - \sum_{n \le N} \alpha_n \langle \cdot, v_n \rangle_{\#} w_n. $$

We denote by $\mathcal{F}^{\#}\Re(B(H^{\#}))$ the set of hyperfinite rank operators in $B(H^{\#})$.

**Lemma 1.2.6.** Any operator with hyperfinite rank is #-compact

**Proof.** Say $A$ has hyperfinite rank, then if $(v_\alpha)_{\alpha \in X}$ is bounded, then $(A v_\alpha)_{\alpha \in X}$ is bounded, and lies in a hyperfinite dimensional Hilbert space. By generalized Bolzano Weirstrass theorem [1], we have that it omits a #-convergence subsequence so by lemma 1.2.5, $A$ is #-compact.

**Theorem 1.2.1.** Let $H_1 = \{h \in H^{\#} \mid \|h\|_{\#} < 1\}$ The following are equivalent:
1. $A \in \overline{\{C \mid \Re(C) < *\infty\}} = \overline{\mathcal{F}^{*}\Re(B(H^{*}))}$ where the #-closure is with respect to the
   #-norm topology

2. $A \in K^\#(B(H^\#))$

3. $A(H_1)$ has #-compact #-closure

**Proof.** $1 \Rightarrow 2.$

If $A \in \overline{\{C \mid \Re(C) < *\infty\}}$ then the #-compactness $A$ is clear, since hyperfinite rank
operators are in $K^\#(B(H^\#))$ and $K^\#(B(H^\#))$ must be #-closed with respect to the #-norm
topology.

2 $\Rightarrow$ 3.

This is clear by definition since $H_1$ is bounded subset of $H^\#$.

3 $\Rightarrow$ 1.

Say this did not hold, i.e. we had some $A$ that has property 3 but not 1. Then, let
$(P_\alpha)_{\alpha \in A}$ be a net of hyperfinite rank projections tending towards the identity map. We
have that $P_\alpha A$ also must have hyperfinite rank, and so $P_\alpha A \dashv\# A$ in #-norm sense.
Well, then there exists some $\varepsilon \approx 0, \varepsilon > 0$ and some $v_\alpha \in H_1$ such that
$\|(A - P_\alpha A)v_\alpha\|_\# > \varepsilon$. Since these $v_\alpha$ are in $H_1$, we can apply 3 to get some subnet such
that $Av_\alpha \to_\# v$ in #-norm. Then we have:

$$
0 < \varepsilon < \|(A - P_{\alpha}A)v_{\alpha}\|_{\#} = \|v - P_{\alpha}v + (1 - P_{\alpha})(v_{\alpha} - v)\|_{\#} \le \dots
$$

$$
\dots \le \|v - P_{\alpha}v\|_{\#} + \|(1 - P_{\alpha})(Av_{\alpha} - v)\|_{\#} \le \|1 - P_{\alpha}\|_{\#} (\|v\|_{\#} + \|Av_{\alpha} - v\|_{\#}) \to_{\#} 0.
$$

A contradiction. Hence, 3 $\Rightarrow$ 1.

**Corollary 1.2.1.** A #-compact $\Leftrightarrow A(\cdot) = \text{Ext} - \sum_{n \in *\mathbb{N}} a_n \langle \cdot, v_n \rangle_{\#} w_n$ where $\{v_n\}_{n \in *\mathbb{N}}$,
$\{w_n\}_{n \in *\mathbb{N}} \subset H^\#$, and $\{a_n\}_{n \in *\mathbb{N}} \subset *\mathbb{C}^\#$ and s.t. #-lim$_{n \to *\infty} a_n = 0$.

The #-convergence on the RHS is with respect to the operator #-norm.

§1.3 Basic Spectral Theory

Spectrum is a generalisation of eigenvalues which is crucial for understanding
operator

algebras. Much of it is built upon whether operators or aren't invertible.

**Definition 1.3.1.** $A \in B(H^\#)$ is said to be invertible if there exists a $B \in B(H^\#)$ such that $AB = BA = 1$. If $X$ is an algebra, we define $\text{Inv}(X) = \{x \in X \mid x \text{ is invertible}\}$.

**Lemma 1.3.1.** Neumann Series is #-convergent

Let $\|A\|_{\#} < 1$. Then, $1 - A$ is invertible with inverse

$$
(1 - A)^{-1} = \mathrm{Ext} - \sum_{n \in *\mathbb{N}} A^n. \qquad (1.3.1)
$$

Where

$$
\mathrm{Ext} - \sum_{n \in *\mathbb{N}} A^n = \#\mathrm{lim}_{N \to *\infty} \left( \mathrm{Ext} - \sum_{n=0}^{N} A^n \right) \quad (1.3.2)
$$

where $N \in *\mathbb{N}_{\infty}$.

**Proof.** The first question to ask is whether the series on the right hand side even
\#-converges. It does as by lemma 1.2.2 one obtains

$$
\| \mathrm{Ext} - \sum_{n \in *\mathbb{N}} A^n \|_{\#} \le \mathrm{Ext} - \sum_{n \in *\mathbb{N}} \|A^n\|_{\#} \le \mathrm{Ext} - \sum_{n \in *\mathbb{N}} \|A\|_{\#}^n = (1 - \|A\|_{\#})^{-1}
$$

Say it #-converges to $B$. Then, we see that because we have a telescoping sum

$$
\mathrm{Ext} - \sum_{n=0}^{N} (A^n)(1-A) = 1 - A^{N+1} = (1-A) \left( \mathrm{Ext} - \sum_{n=0}^{N} A^n \right)
$$

Hence, it is sufficient to check that **1** − *A*<sup>*n*</sup> →<sub>**#**</sub> **1** as *n* → *∞*. Fix ε ≈ 0, ε > 0, and choose *N* ∈ *eries, *N* > Ext-log<sub>||A||<sub>||#||</sub></sub> ε Then we have for *n* > *N*
$$ \|\mathbf{1} - A^n - \mathbf{1}\|_{\#} = \|A^n\|_{\#} \le \|A\|_{\#}^{n} \le \|A\|_{\#}^{N} < \varepsilon $$

**Lemma 1.3.2.** $\mathrm{Inv}(B(H^{\#}))$ is #-open in $B(H^{\#})$. Furthermore, the map $^{-1}: \mathrm{Inv}(B(H^{\#})) \to \mathrm{Inv}(B(H^{\#})), A \mapsto A^{-1}$

is #-continuous with respect to the operator #-norm.

**Proof.** Say $A \in \mathrm{Inv}(B(H^{\#}))$. Then, if $\|B - A\|_{\#} < \|A^{-1}\|_{\#}^{-1}$, we have
$$ \|BA^{-1} - \mathbf{1}\|_{\#} = \|(B - A)A^{-1}\|_{\#} \le \|B - A\|_{\#}\|A^{-1}\|_{\#} < 1 $$

Which by lemma 1.3.1 gives

$$ 1 - (BA^{-1} - 1) = BA^{-1} \in \mathrm{Inv}(B(H^{\#})) \Rightarrow BA^{-1}A = B \in \mathrm{Inv}(B(H^{\#})). $$

Then if we consider $BA^{-1}$, we can note that $1/(1 - \|\mathbf{1} - BA^{-1}\|) > 1$ and hence

$$ \begin{align*} \|BA^{-1}\|_{\#} &\le \mathrm{Ext} - \sum_{n \in \mathbb{N}} \|1 - BA^{-1}\|_{\#}^{n} = \mathrm{Ext} - \sum_{n \in \mathbb{N}} \|(A - B)A^{-1}\|_{\#}^{n} \le \\ &\le \mathrm{Ext} - \sum_{n \in \mathbb{N}} \|(A - B)\|_{\#}^{n} \|A^{-1}\|_{\#}^{n} = 1/(1 - \|(A - B)\|_{\#} \|A^{-1}\|_{\#}). \end{align*} $$

$$ \text{Therefore } \|A^{-1} - B^{-1}\|_{\#} = \|A^{-1}(AB^{-1})^{-1}(B - A)A^{-1}\|_{\#} \le $$

$$ \|A^{-1}\|_{\#}^{2} \|B - A\|_{\#} \|(AB^{-1})^{-1}\|_{\#} \le \|A^{-1}\|_{\#}^{2} \|B - A\|_{\#} [1/(1 - \|(A - B)\|_{\#} \|A^{-1}\|_{\#})] $$

We can see then that as $\|A - B\|_{\#} \to 0$, $\|A^{-1} - B^{-1}\|_{\#} \to 0$ as required.

**Definition 1.3.2.** (Spectrum)

Let $A \in B(H^{\#})$. We define the spectrum of $A$, denoted $\sigma(A)$ by
$$ \sigma(A) := \{\lambda \in \mathbb{C}^{\#} \mid (A - \lambda \cdot \mathbf{1}) \notin \mathrm{Inv}(B(H^{\#}))\}, \text{i.e. the set of all complex numbers} $$
such that $A - \lambda \mathbf{1}$ is not invertible. We denote the complement of $\sigma(A)$ by $\varphi(A)$.

**Lemma 1.3.3.** Spectrum is a generalisation of an eigenvalue ($\mathrm{eigen}(A) \subset \sigma(A)$), i.e. if $\lambda$ is an eigenvalue of $A, \lambda \in \sigma(A)$

**Proof.** Say $\lambda$ is an eigenvalue of $A$. Then $\nu \in H^{\#} \setminus \{0\}$ s.t. $(A - \lambda)\nu = 0$. However, by linearity, $(A - \lambda)0 = 0$. As $(A - \lambda)$ is not injective it cannot be invertible, hence $\lambda \in \sigma(A)$

**Corollary 1.3.1.** Let $H^{\#}$ be hyperfinitefinite dimensional. Then, if $A \in B(H^{\#})$ we have that spectrum agrees with the eigenvalues, i.e. $\mathrm{eigen}(A) = \sigma(A)$.

**Proof.** By lemma 1.3.3, we only need to check the other direction. Up to choosing bases, we can assume $H^{\#} = *\mathbb{C}_{c}^{\dim(H^{\#})}$. In this case, $B(H^{\#})$ is just the $\dim(H^{\#}) \times \dim(H^{\#})$ square matrices. By standard results in linear algebra, we have that $(A - \lambda) \in \mathrm{Inv}(B(H^{\#})) \text{ iff } \mathrm{Ext}\text{-det}(A - \lambda) = 0 \text{ iff } \lambda \text{ is an eigenvalue.}$

**Lemma 1.3.4.** If $A \in B(H^{\#})$ then $\sigma(A)$ is #-closed as a subset of the complex plane $*\mathbb{C}_{c}^{\#}$.

Moreover, it is a subset of the disc of radius $\|A\|_{\#}$ centred at the origin.

**Proof.** Say $\lambda > \|A\|_{\#}$. Then, $-\lambda^{-1}\|A\|_{\#} < 1$ so $1 - \lambda^{-1}A$ is invertible by lemma 1.3.1. Then, $\lambda \notin \sigma(A)$. Now examine $\sigma(A)^c = \varphi(A) = \{\lambda \in *\mathbb{C}_{c}^{\#} \mid A - \lambda \in \mathrm{Inv}(B(H^{\#}))\}$.

Say $\lambda \in \varphi(A)$. By lemma 1.3.2 we have that there exists some $\varepsilon \approx 0, \varepsilon > 0$ such that $\|A - \lambda - B\|_{\#} < \varepsilon \Rightarrow B \in \mathrm{Inv}(B(H^{\#}))$.

Now, we see that if $|\lambda - \hat{\lambda}| < \varepsilon$ we have:
$$ \|A - \lambda - (A - \hat{\lambda})\|_{\#} = |\lambda - \hat{\lambda}| < \varepsilon $$

Hence, $A - \hat{\lambda} \in \mathrm{Inv}(B(H^{\#}))$. Then $\varphi(A)$ is #-open, and so $\sigma(A)$ is #-closed. We need the following lemmas to show that $\sigma(A) \ne \emptyset$ ever.

**Lemma 1.3.5.** Let $A \in B(H^{\#})$. Then let $\gamma:B(H^{\#}) \to *\mathbb{C}_{c}^{\#}$ be an arbitrary linear functional $(\gamma \in B(H^{\#})^*)$. We have that the map

$f_{A_{\gamma}}: \varphi(A) \to *\mathbb{C}_{c}^{\#}, \lambda \mapsto \gamma(1/(A - \lambda))$ is #-analytic on $\varphi(A)$, and has $\#\text{-lim}_{\lambda \to \infty} f_{A_{\gamma}}(\lambda) = 0$
**Proof.** For $\lambda, \lambda_0 \in \varphi(A)$ we have that
$$ \frac{1}{A - \lambda} - \frac{1}{A - \lambda_0} = \frac{A - \lambda_0 - A + \lambda}{(A - \lambda)(A - \lambda_0)} = \frac{\lambda - \lambda_0}{(A - \lambda)(A - \lambda_0)}. $$

Then,

$$ \#\lim_{\lambda \to \#} \lambda_0 \frac{f_{A_\lambda}(\lambda) - f_{A_\lambda}(\lambda_0)}{\lambda - \lambda_0} = \gamma\left(\frac{1}{(A - \lambda)(A - \lambda_0)}\right) = \gamma\left(\#\lim_{\lambda \to \#} \lambda_0 \frac{\lambda - \lambda_0}{(A - \lambda)(A - \lambda_0)}\right) = \dots $$

Where we use linearity of $\gamma$ in the first equality, and #-continuity of $\gamma$ in the second.
Then, by lemma 1.3.2 we have that

$$ \dots = \gamma\left(\frac{\lambda - \lambda_0}{\#\lim_{\lambda \to \#} \lambda_0 (A - \lambda)(A - \lambda_0)}\right) = \gamma\left(\frac{1}{(A - \lambda_0)^2}\right). $$

Hence, $f_{A_\lambda}$ is #-analytic on $\varphi(A)$. By the estimate

$$ \begin{aligned} \left\| \frac{1}{A - \lambda} \right\|_{\#} &= \frac{1}{|\lambda|} \left\| \frac{1}{1 - \lambda^{-1}A} \right\|_{\#} = \frac{1}{|\lambda|} \left( \left\| \mathrm{Ext} - \sum_{n \in \mathbb{N}} (\lambda^{-1}A)^n \right\|_{\#} \right) \le \\ &\le \frac{1}{|\lambda|} \left( \mathrm{Ext} - \sum_{n \in \mathbb{N}} \|(\lambda^{-1}A)^n\|_{\#}^n \right) = \frac{1}{|\lambda| - \|A\|_{\#}} \dots \end{aligned} $$

It is clear that $\frac{1}{A - \lambda} \xrightarrow{\lambda \to \#} 0$ as $\lambda \to \infty$ and hence by #-continuity of $f_{A_\lambda}$ we are done.

**Theorem 1.3.1.** If $A \in B(H^#)$ then $\sigma(A) \neq \emptyset$.

**Proof.** Say $\exists A \in B(H^#)$ such that $\sigma(A) = \emptyset$. For this $A$, we have that $f_{A_\lambda}$ is:

(i) $|f_{A_\lambda}(z)|$ bounded by positive constant $K \in \mathbb{R}_{c,+}^\#$

(ii) $f_{A_\lambda}(z)$ is #-entire function, that is $f_{A_\lambda}(z)$ is a $\mathbb{C}_c^\#$-valued function #-holomorphic on the whole $\mathbb{C}_c^\#$

(iii) $f_{A_\lambda}(z)$ has $\#\lim_{\lambda \to \#} f_{A_\lambda}(\lambda) = 0$.

The only map satisfying these three properties is the zero map. But since $\gamma$ was arbitrary, this implies that an arbitrary functional is the zero functional, which is clearly a contradiction. Hence, $\sigma(A) \neq \emptyset$

In particular this means that $\sigma(A) \neq \text{eigen}(A)$ if $\text{eigen}(A)$ is empty.

**Theorem 1.3.2.** (Generalized Gelfand Mazur theorem)

If $\mathrm{Inv}(B(H^#)) = B(H^#) \setminus \{0\}$, Then $B(H^#) \cong \mathbb{C}_c^\#$.

**Proof.** Let $A \in B(H^#)$ then let $\lambda_A \in \sigma(A)$ we have $A - \lambda_A = 0$. So $\lambda_A$ is unique. Our isomorphism is then $\psi:B(H^#) \to \mathbb{C}_c^\# A \mapsto \lambda_A$.

**Theorem 1.3.3.** (Generalized Spectral Mapping Theorem)

Let $A \in B(H^#)$, $f \in \mathbb{C}_c^\#[z]$. Then we have: $\sigma(f(A)) = f(\sigma(A))$.

**Proof.** Let $\lambda \in \sigma(A)f(z) = \mathrm{Ext} - \sum_{n=0}^N a_n z^n$. Then

$$ f(A) - f(\lambda) = \mathrm{Ext} - \sum_{n=0}^{N} a_n (A^n - \lambda^n) = (A - \lambda) \left( \mathrm{Ext} - \sum_{n=0}^{N} a_n \left( \mathrm{Ext} - \sum_{j<n} (A^j \lambda^{n-j-1}) \right) \right). $$

So $f(\lambda) \in \sigma(f(A))$. Say $\mu \notin f(\sigma(A))$. Then, we can write $f(z) - \mu = a_N \left( \mathrm{Ext} - \prod_{n=0}^N (z - \lambda_n) \right)$.

Then as $\mu - f(\lambda) \neq 0 \forall \lambda \in \sigma(A)$ (the zero operator isn't invertible) we have that

$\lambda_n \notin \sigma(A)(n \le N)$. Therefore, $f(A) - \mu = a_N \left( \mathrm{Ext} - \prod_{n=0}^N (A - \lambda_n) \right)$, must be invertible, and $\mu \notin \sigma(f(A))$.

This theorem has many forms and generalises much more than for $f$ being a polynomial.

**Definition 1.3.3.** (Spectral Radius)

Given $A$ in $B(H^#)$ the spectral radius, denoted $r(A)$, of $A$ is defined by $r(A) = \sup_{\lambda \in \sigma(A)} |\lambda|$.

We note by lemma 1.3.4 the supremum exists and is attained. In fact, the following lemma tells us what the spectral radius of a given operator is in terms of a #-limit.

**Lemma 1.3.6.** Let $A \in B(H^#)$. Then the #-limit: $\#\lim_{n \to \#} \|A_n\|_\#^{1/n}$ exists, and is equal to