Proof Let h be a nonsingular hermitian form over $(A,\sigma )$ that satisfies Property (3.1). Let $h'$ be a nonsingular hermitian form over $(A,\sigma )$ such that

(3.2) $$ \begin{align} h\perp H \simeq h'\perp H', \end{align} $$

where H and $H'$ are hyperbolic forms over $(A,\sigma )$ . Let $a\in \mathscr {P}\cap A^\times $ . By Proposition 2.16, there exist nonsingular quadratic forms $q', q_1,q_2$ over F that all have nonzero signature at P and such that $q'\otimes h'$ , $q_1\otimes H$ and $q_2 \otimes H'$ are diagonal hermitian forms of the form $\pi \otimes \langle a\rangle _\sigma \perp \nu \otimes \langle -a\rangle _\sigma $ , for some diagonal quadratic forms $\pi $ and $\nu $ with coefficients in $P^\times $ . Let $\chi :=q_h\otimes q'\otimes q_1\otimes q_2$ . It then follows from (3.2) that

$$\begin{align*}\chi \otimes h \perp \chi \otimes H \simeq \chi \otimes h' \perp \chi \otimes H'. \end{align*}$$

Observe that by taking signatures at P, the form $\chi \otimes H$ has as many entries in $\mathscr {P}\cap A^\times $ as in $-\mathscr {P}\cap A^\times $ since it is still a hyperbolic form and thus has signature zero. The same argument applies to $\chi \otimes H'$ . We now consider

$$\begin{align*}\chi \otimes h = (q_h \otimes q' \otimes q_1 \otimes q_2) \otimes h \simeq q' \otimes q_1 \otimes q_2 \otimes (q_h \otimes h).\end{align*}$$

Since the form $q_h \otimes h$ has as many entries in $\mathscr {P}\cap A^\times $ as in $-\mathscr {P}\cap A^\times $ by definition of $q_h$ (cf. Definition 3.3), the same holds for $\chi \otimes h$ . It then follows from Lemma 2.17 that the form $\chi \otimes h'$ has as many entries in $\mathscr {P}\cap A^\times $ as in $-\mathscr {P}\cap A^\times $ . Since $\operatorname {\mathrm {sign}}_P \chi \not =0$ , we conclude that $h'$ satisfies Property (3.1) with $q_{h'}=\chi $ .

Proof We have to check the following:

1. (1) $N_{\mathscr {P}}+N_{\mathscr {P}} \subseteq N_{\mathscr {P}}$ ;

2. (2) $W(F) \cdot N_{\mathscr {P}} \subseteq N_{\mathscr {P}}$ ;

3. (3) $I_{\mathscr {P}} \cdot W(A,\sigma ) \subseteq N_{\mathscr {P}}$ ;

4. (4) $N_{\mathscr {P}} \not = W(A,\sigma )$ .

We do it in order. For the verification of (1) and (2), we fix two hermitian forms $\varphi $ and $\psi $ over $(A,\sigma )$ that satisfy Property 3.1, so that $[\varphi ], [\psi ] \in N_{\mathscr {P}}$ . Therefore, there are quadratic forms $q_\varphi ,q_\psi $ over F such that

$$\begin{align*}q_\varphi \otimes \varphi \simeq \langle a_1, \ldots, a_r\rangle_\sigma \perp \langle -b_1, \ldots, -b_r\rangle_\sigma\end{align*}$$

and

$$\begin{align*}q_\psi \otimes \psi \simeq \langle c_1, \ldots, c_s\rangle_\sigma \perp \langle -d_1, \ldots, -d_s\rangle_\sigma\end{align*}$$

for some $a_1, \ldots , a_r, b_1, \ldots , b_r, c_1, \ldots , c_s, d_1, \ldots , d_s \in \mathscr {P} \cap A^\times $ , and where $\operatorname {\mathrm {sign}}_P q_\varphi \not = 0$ and $\operatorname {\mathrm {sign}}_P q_\psi \not = 0$ .

(1) We show that $[\varphi ] + [\psi ] = [\varphi \perp \psi ] \in N_{\mathscr {P}}$ by showing that $\varphi \perp \psi $ satisfies Property (3.1). We have

(3.3) $$ \begin{align} (q_\varphi \otimes q_\psi) \otimes (\varphi \perp \psi) \simeq q_\psi \otimes & (\langle a_1, \ldots, a_r\rangle_\sigma \perp \langle -b_1, \ldots, -b_r\rangle_\sigma) \perp \nonumber\\ & q_\varphi \otimes (\langle c_1, \ldots, c_s\rangle_\sigma \perp \langle -d_1, \ldots, -d_s\rangle_\sigma). \end{align} $$

Writing $q_\varphi = q_+ \perp q_-$ and $q_\psi = q^{\prime }_+ \perp q^{\prime }_-$ with $q_+, q^{\prime }_+$ positive definite at P and $q_-, q^{\prime }_-$ negative definite at P, we have that the number of entries in $\mathscr {P}\cap A^\times $ on the right-hand side of (3.3) is

$$\begin{align*}(\dim q^{\prime}_+)r + (\dim q^{\prime}_-)r + (\dim q_+)s + (\dim q_-)s = (\dim q_\psi)r + (\dim q_\varphi)s,\end{align*}$$

and that the number of entries in $-\mathscr {P}\cap A^\times $ on the right-hand side of (3.3) is

$$\begin{align*}(\dim q^{\prime}_-)r + (\dim q^{\prime}_+)r + (\dim q_-)s + (\dim q_+)s = (\dim q_\psi)r + (\dim q_\varphi)s.\end{align*}$$

Both are equal, so $\varphi \perp \psi $ satisfies Property (3.1).

(2) Since $W(F)$ is additively generated by classes of one-dimensional forms, it suffices to check that $[\langle u\rangle \otimes \varphi ] \in N_{\mathscr {P}}$ for every $u \in F^\times $ , which follows from the fact that the form $\langle u\rangle \otimes \varphi $ clearly satisfies Property (3.1).

(3) Let $\varphi $ be a nonsingular hermitian form over $(A, \sigma )$ . Since $I_{\mathscr {P}}$ is additively generated by the classes of the forms $\langle 1,-u\rangle $ for $u \in P^\times $ , it suffices to check that $\langle 1,-u\rangle \otimes \varphi $ satisfies Property (3.1) for every $u \in P^\times $ . By Proposition 2.16, there is a nonsingular quadratic form $q_\varphi $ over F such that $\operatorname {\mathrm {sign}}_P q_\varphi \not =0$ and

$$\begin{align*}q_\varphi\otimes \varphi \simeq \langle a_1,\ldots, a_r\rangle_\sigma \perp \langle -b_1,\ldots, -b_s\rangle_\sigma \end{align*}$$

for some $a_1,\ldots , a_r,-b_1,\ldots , -b_s \in \mathscr {P}\cap A^\times $ . Then

$$ \begin{align*} q_\varphi \otimes (\langle 1,-u\rangle \otimes \varphi) \simeq \langle a_1, \ldots, a_r\rangle_\sigma \perp & \langle ub_1, \ldots, ub_s\rangle_\sigma \perp\\ & \langle -ua_1, \ldots, -ua_r\rangle_\sigma \perp \langle -b_1, \ldots, -b_s\rangle_\sigma, \end{align*} $$

which shows that $\varphi $ satisfies Property (3.1).

(4) Let $a \in \mathscr {P} \cap A^\times $ . We show that $[\langle a\rangle _\sigma ] \not \in N_{\mathscr {P}}$ . Assume that it is not the case. Then, there is a nonsingular hermitian form h over $(A,\sigma )$ such that h satisfies Property (3.1) and $[h] = [\langle a\rangle _\sigma ]$ . It follows from Lemma 3.4 that $\langle a\rangle _\sigma $ also satisfies Property (3.1), and thus that

(3.4) $$ \begin{align} q_{\langle a\rangle_\sigma} \otimes \langle a\rangle_\sigma \simeq \langle a_1, \ldots, a_r\rangle_\sigma \perp \langle -b_1, \ldots, -b_r\rangle_\sigma, \end{align} $$

with $a_1, \ldots , a_r, b_1, \ldots , b_r \in \mathscr {P}\cap A^\times $ . We write

$$\begin{align*}q_{\langle a\rangle_\sigma} \simeq \langle u_1, \ldots, u_s\rangle \perp \langle -v_1, \ldots, -v_t\rangle \end{align*}$$

with $u_1, \ldots , u_s, v_1, \ldots , v_t \in P^\times $ . Since $\operatorname {\mathrm {sign}}_P q_{\langle a\rangle _\sigma } \not = 0$ we have $s \not = t$ . Equation (3.4) then becomes

$$\begin{align*}\langle u_1a, \ldots, u_sa\rangle_\sigma \perp \langle -v_1a, \ldots, -v_ta\rangle_\sigma \simeq \langle a_1, \ldots, a_r\rangle_\sigma \perp \langle -b_1, \ldots, -b_r\rangle_\sigma. \end{align*}$$

By Lemma 2.17, since the right-hand side has the same number of elements in $\mathscr {P}$ as in $-\mathscr {P}$ , we must have $s=t$ , contradiction.

Proof Let $\ell [h] = [\ell \times h] \in N_{\mathscr {P}}$ for some $\ell \in \mathbb {N}$ , where h is a nonsingular hermitian form over $(A,\sigma )$ . By Lemma 3.4, $\ell \times h$ satisfies Property (3.1). Then

$$\begin{align*}q_{\ell \times h} \otimes (\ell \times h) \simeq \langle a_1, \ldots, a_r\rangle_\sigma \perp \langle -b_1, \ldots, -b_r\rangle_\sigma\end{align*}$$

for some $a_1, \ldots , a_r, b_1, \ldots , b_r \in \mathscr {P} \cap A^\times $ , and so, clearly

$$\begin{align*}(\ell \times q_{\ell \times h}) \otimes h \simeq \langle a_1, \ldots, a_r\rangle_\sigma \perp \langle -b_1, \ldots, -b_r\rangle_\sigma,\end{align*}$$

proving that h satisfies Property (3.1) and thus $[h] \in N_{\mathscr {P}}$ .

We now prove the second statement: Assume that $[qh] \in N_{\mathscr {P}}$ for some $[q] \in W(F)$ and $[h] \in W(A,\sigma )$ . Since $I_{\mathscr {P}}$ is the kernel of $\operatorname {\mathrm {sign}}_P : W(F) \rightarrow \mathbb {Z}$ , there is $k \in \mathbb {Z}$ such that $[q] = k \bmod I_{\mathscr {P}}$ . Thus (and using that $I_{\mathscr {P}} \cdot W(A,\sigma ) \subseteq N_{\mathscr {P}}$ ), we obtain $k[h] \in N_{\mathscr {P}}$ . It follows that $k=0$ (and thus $[q]\in I_{\mathscr {P}}$ ), or that $[h] \in N_{\mathscr {P}}$ by the first part.

Proof By definition, $I_{\mathscr {P}} = \ker \operatorname {\mathrm {sign}}_P \not \ni 2$ . By [Reference Astier and Unger2, Proposition 6.5], we obtain that $N_{\mathscr {P}} = \ker \operatorname {\mathrm {sign}}^\mu _P$ . Therefore, $P\not \in \mathrm {Nil}[A,\sigma ]$ since $N_{\mathscr {P}}\not =W(A,\sigma )$ by Proposition 3.6 and the equivalence in (2.8).

Proof The hermitian form $\langle a,-b\rangle _\sigma $ trivially satisfies Property (3.1). Therefore, $[\langle a,-b\rangle _\sigma ] \in N_{\mathscr {P}} = \ker \operatorname {\mathrm {sign}}^\mu _P$ , so that $\operatorname {\mathrm {sign}}^\mu _P \langle a\rangle _\sigma = \operatorname {\mathrm {sign}}^\mu _P \langle b\rangle _\sigma $ .

Proof By [Reference Astier and Unger4, Example 3.13], it suffices to show that $\mathscr {M}^\eta _P(D,\vartheta )$ is maximal. Let $\mathscr {P}$ be a positive cone such that $\mathscr {M}^\eta _P(D,\vartheta ) \subseteq \mathscr {P}$ . By [Reference Astier and Unger4, Lemma 3.16], $\mathscr {P}$ is over P, and by Corollary 3.9, $\mathscr {P} = \mathscr {M}^\eta _P(D,\vartheta )$ .

Proof (1) Properties (P1)–(P4) are straightforward to check for $\mathscr {P}[a]$ . We show that property (P5) holds, i.e., that $\mathscr {P}[a]$ is proper. Assume $\mathscr {P}[a]$ is not proper, and let $b \in \mathscr {P}[a] \cap -\mathscr {P}[a]$ , $b \not = 0$ . Then, there exist $p_1, p_2 \in \mathscr {P}, k,r \in \mathbb {N}\cup \{0\}, u_i, v_j \in P$ , and $x_i, y_j\in D$ such that

$$\begin{align*}b = p_1 + \underbrace{\sum_{i=1}^k u_i\vartheta(x_i)ax_i}_{\alpha} = -p_2 - \underbrace{\sum_{j=1}^r v_j\vartheta(y_j)ay_j}_{\beta}.\end{align*}$$

Observe that at least one of $\alpha $ or $\beta $ is nonzero, since $\mathscr {P}$ is proper. Furthermore, $\alpha +\beta \not =0$ . (Indeed, if $\alpha +\beta =0$ , then $\alpha =-\beta \not =0$ , contradicting that $\mathscr {M}^\eta _P(D,\vartheta )$ is proper since $a\in \mathscr {M}^\eta _P(D,\vartheta )$ .) It follows that:

$$\begin{align*}p_1 + p_2 = -\sum_{i=1}^k u_i\vartheta(x_i)ax_i - \sum_{j=1}^r v_j\vartheta(y_j)ay_j =-\alpha-\beta.\end{align*}$$

The right-hand side is a nonzero sum of elements that are in $-\mathscr {M}^\eta _P(D,\vartheta )$ (since $a\in \mathscr {M}^\eta _P(D,\vartheta )$ ), and thus belongs to $-\mathscr {M}^\eta _P(D,\vartheta )$ . The left-hand side, being in $\mathscr {P}$ , does not belong to $-\mathscr {M}^\eta _P(D,\vartheta )$ by hypothesis, contradiction.

(2) Let $x = p + \underbrace {\sum _{i=1}^k u_i\vartheta (x_i)ax_i}_{\alpha } \in \mathscr {P}[a]$ . Observe that $\alpha \in \mathscr {M}^\eta _P(D,\vartheta )$ . If $\alpha =0$ , then $x = p$ and $\operatorname {\mathrm {sign}}^\eta _P \langle x\rangle _\vartheta> -m_P(D,\vartheta )$ by hypothesis. If $\alpha \not = 0$ , assume that $\operatorname {\mathrm {sign}}^\eta _P \langle x\rangle _\vartheta = -m_P(D,\vartheta )$ . We have $x-\alpha = p$ . The left-hand side is a nonzero element of $-\mathscr {M}^\eta _P(D,\vartheta )$ (the sum of two nonzero elements of a prepositive cone is nonzero by (P5)), while the right-hand side is not (by hypothesis), contradiction.

Proof Since $\mathscr {P}$ is contained in a positive cone over P, we have $P\not \in \mathrm {Nil}[D,\vartheta ]$ by Theorem 3.8. We now consider two cases.

Case 1: There is $c \in \mathscr {P}$ such that $\operatorname {\mathrm {sign}}^\eta _P \langle c\rangle _\vartheta = -m_P(D,\vartheta )$ . Then, by Lemma 3.9, $\mathscr {P} \subseteq -\mathscr {M}^\eta _P(D,\vartheta )$ .

Case 2: For every $c \in \mathscr {P}$ , $\operatorname {\mathrm {sign}}^\eta _P \langle c\rangle _\vartheta> -m_P(D,\vartheta )$ . Then, using Lemma 4.2, we can add all elements of $\mathscr {M}^\eta _P(D,\vartheta )$ to $\mathscr {P}$ and we obtain in this way a prepositive cone $\mathscr {Q}$ containing both $\mathscr {P}$ and $\mathscr {M}^\eta _P(D,\vartheta )$ . Since $\mathscr {M}^\eta _P(D,\vartheta )$ is a maximal prepositive cone (cf. Proposition 4.1), we obtain $\mathscr {Q} = \mathscr {M}^\eta _P(D,\vartheta )$ and thus $\mathscr {P} \subseteq \mathscr {M}^\eta _P(D,\vartheta )$ .

Proof Let $B \in \operatorname {\mathrm {PSD}}_\ell (\mathscr {M}_P^\eta (D,\vartheta ))$ . Then, by [Reference Astier and Unger4, Lemma 4.5], there is $G \in \operatorname {\mathrm {GL}}_\ell (D)$ such that $\vartheta (G)^tBG = \operatorname {\mathrm {diag}}(a_1, \ldots , a_\ell )$ with $a_1, \ldots , a_\ell \in \mathscr {M}^\eta _P(D,\vartheta )$ , and we may assume that

$$\begin{align*}\vartheta(G)^tBG = \operatorname{\mathrm{diag}}(a_1, \ldots, a_r,0,\ldots,0)\end{align*}$$

with $a_1, \ldots , a_r \in \mathscr {M}^\eta _P(D,\vartheta )\setminus \{0\}$ .

Since $a_i I_\ell \in \mathscr {M}^{g^{-1}(\eta )}_P(M_\ell (D),\vartheta ^t)$ , it is now easy to represent $\vartheta (G)^tBG$ , and thus B, as an element of $\mathscr {C}_P(\mathscr {M}^{g^{-1}(\eta )}_P(M_\ell (D),\vartheta ^t))$ , proving that

$$\begin{align*}\operatorname{\mathrm{PSD}}_\ell(\mathscr{M}^\eta_P(D,\vartheta)) \subseteq \mathscr{C}_P(\mathscr{M}^{g^{-1}(\eta)}_P(M_\ell(D),\vartheta^t)). \end{align*}$$

The equality follows since $\operatorname {\mathrm {PSD}}_\ell (\mathscr {M}^\eta _P(D,\vartheta ))$ is a positive cone by Theorem 4.3 and [Reference Astier and Unger4, Proposition 4.7], and $\mathscr {C}_P(\mathscr {M}^{g^{-1}(\eta )}_P(M_\ell (D),\vartheta ^t))$ is a prepositive cone by Proposition 2.14.

Proof The original proof is still valid, but the references in it to the results in [Reference Astier and Unger4] stated after [Reference Astier and Unger4, Section 5] need to be replaced by the same results obtained in the current article, as follows:

$$\begin{align*}\begin{array}[b]{ll} \underline{\text{Original proof in [4]:}}\ & \underline{\text{Corresponding statement in this article:}} \\[5pt] \text{Proposition~6.6} & \text{Theorem~3.8} \\ \text{Proposition~7.1} & \text{Theorem~4.3} \\ \text{Lemma~7.2} & \text{Lemma~4.4} \\ \text{Proposition~7.1} & \text{Theorem~4.3}\\ \text{Lemma~7.4} & \text{cf. Remark~4.5} \end{array} \end{align*}$$

Proof The first statement is clear by definition of $\mathscr {C}_P$ and since $\mathscr {M}^\mu _P(A,\sigma )$ is closed under multiplication by elements of $P\setminus \{0\}$ . The second statement follows immediately from Theorem 4.6.

Proof A positive cone $\mathscr {P}$ is a subset of $\operatorname {\mathrm {Sym}}(A,\sigma )$ , so can be identified with a map from $\operatorname {\mathrm {Sym}}(A,\sigma )$ to $\{0,1\}$ (with $\mathscr {P}(a) = 1$ iff $a \in \mathscr {P}$ ). Thus, we can view $X_{(A,\sigma )}$ as a subset of $Z := \{0,1\}^{\operatorname {\mathrm {Sym}}(A,\sigma )}$ and the topology $\mathcal {T}_\sigma $ as the topology induced by the product topology T on Z of the discrete topology on $\{0,1\}$ . Since T is compact, it suffices to show that $X_{(A,\sigma )}$ is a closed subset of Z. We slightly reformulate the prepositive cone properties (P1), (P4), and (P5) in order to make it easier to check them:

1. (P1) $0 \in \mathscr {P}$ .

2. (P4) $\forall u \in F \ u \in \mathscr {P}_F \vee -u \in \mathscr {P}_F$ . (This reformulation is equivalent to the original (P4) since $\mathscr {P}_F$ is always a preordering, so we only need to check that it is total.)

3. (P5) $\forall a \in \operatorname {\mathrm {Sym}}(A,\sigma ) \setminus \{0\} \ \neg (a \in \mathscr {P} \wedge -a \in \mathscr {P})$ .

We now show that the subset $S_i$ of Z of subsets satisfying property (Pi) is closed, for $i=1, \ldots , 5$ . The result follows since $X_{(A,\sigma )} = S_1 \cap \cdots \cap S_5$ .

$$\begin{align*}S_1 = \{\mathscr{P} \in Z \mid \mathscr{P}(0) = 1\},\end{align*}$$

which is closed in T.

$$ \begin{align*} S_2 &= \{\mathscr{P} \in Z \mid \forall a,b \in \operatorname{\mathrm{Sym}}(A,\sigma) \ a,b \in \mathscr{P} \Rightarrow a+b \in \mathscr{P}\} \\ &= \{\mathscr{P} \in Z \mid \forall a,b \in \operatorname{\mathrm{Sym}}(a,\sigma) \ \neg(a,b \in \mathscr{P}) \vee a+b \in \mathscr{P}\} \\ &= \{\mathscr{P} \in Z \mid \forall a,b \in \operatorname{\mathrm{Sym}}(a,\sigma) \ a \not \in \mathscr{P} \vee b \not \in \mathscr{P} \vee a+b \in \mathscr{P}\} \\ &= \bigcap_{a,b \in \operatorname{\mathrm{Sym}}(A,\sigma)} \{\mathscr{P} \in Z \mid \mathscr{P}(a) = 0 \vee \mathscr{P}(b) = 0 \vee \mathscr{P}(a+b) = 1\}, \end{align*} $$

which is an intersection of closed sets in T, and therefore closed.

$$ \begin{align*} S_3 &= \{\mathscr{P} \in Z \mid \forall a \in \operatorname{\mathrm{Sym}}(A,\sigma) \forall x \in A \ a \in \mathscr{P} \Rightarrow \sigma(x)ax \in \mathscr{P}\} \\ &= \{\mathscr{P} \in Z \mid \forall a \in \operatorname{\mathrm{Sym}}(A,\sigma) \forall x \in A \ a \not \in \mathscr{P} \vee \sigma(x)ax \in \mathscr{P}\} \\ &= \bigcap_{a \in \operatorname{\mathrm{Sym}}(A,\sigma),\ x \in A} \{\mathscr{P} \in Z \mid \mathscr{P}(a) = 0 \vee \mathscr{P}(\sigma(x)ax) = 1\}, \end{align*} $$

which is an intersection of closed sets in T, and therefore closed.

$$ \begin{align*} S_4 &= \{{\mathscr{P}} \in Z \mid \forall u \in F\ u \in {\mathscr{P}}_F \vee -u \in {\mathscr{P}}_F\} \\ &= \{{\mathscr{P}} \in Z \mid \forall u \in F\ (\forall a \in {\mathscr{P}} \ ua \in {\mathscr{P}}) \vee (\forall a \in {\mathscr{P}} \ -ua \in {\mathscr{P}})\} \\ &= \bigcap_{u \in F} \{{\mathscr{P}} \in Z \mid (\forall a \in {\mathscr{P}} \ ua \in {\mathscr{P}}) \vee (\forall a \in {\mathscr{P}} \ -ua \in {\mathscr{P}})\} \\ &= \bigcap_{u \in F} \{{\mathscr{P}} \in Z \mid \forall a \in {\mathscr{P}} \ ua \in {\mathscr{P}}\} \cup \{{\mathscr{P}} \in Z \mid \forall a \in {\mathscr{P}} \ -ua \in {\mathscr{P}}\} \\ &= \bigcap_{u \in F} \Big( \{{\mathscr{P}} \in Z \mid \forall a \in \operatorname{\mathrm{Sym}}(A,\sigma)\ a \not \in {\mathscr{P}} \vee ua \in {\mathscr{P}}\} \cup \\ & \qquad\qquad \{{\mathscr{P}} \in Z \mid \forall a \in \operatorname{\mathrm{Sym}}(A,\sigma)\ a \not \in {\mathscr{P}} \vee -ua \in {\mathscr{P}}\} \Big)\\ &= \bigcap_{u \in F} \Big(\bigcap_{a \in \operatorname{\mathrm{Sym}}(A,\sigma)} \{{\mathscr{P}} \in Z \mid {\mathscr{P}}(a) = 0 \vee {\mathscr{P}}(ua) = 1\}\Big) \cup \\ & \qquad \qquad \Big(\bigcap_{a \in \operatorname{\mathrm{Sym}}(A,\sigma)} \{{\mathscr{P}} \in Z \mid {\mathscr{P}}(a) = 0 \vee {\mathscr{P}}(-ua)=1\}\Big), \end{align*} $$

which is a closed set in T.

$$ \begin{align*} S_5 &= \{\mathscr{P} \in Z \mid \forall a \in \operatorname{\mathrm{Sym}}(A,\sigma) \setminus \{0\} \ \neg(a \in \mathscr{P} \wedge -a \in \mathscr{P})\} \\ &= \{\mathscr{P} \in Z \mid \forall a \in \operatorname{\mathrm{Sym}}(A,\sigma) \setminus \{0\} \ a \not \in \mathscr{P} \vee -a \not \in \mathscr{P}\} \\ &= \bigcap_{a \in \operatorname{\mathrm{Sym}}(A,\sigma) \setminus \{0\}} \{\mathscr{P} \in Z \mid \mathscr{P}(a)=0 \vee \mathscr{P}(-a)=0\}, \end{align*} $$

which is closed in T.

Proof The proof is the same as the proof of [Reference Astier and Unger4, Proposition 9.7(2)], except that we use an infinite union instead of a finite one in the final part: Let $u \in F \setminus \{0\}$ . We show that $\pi ^{-1}(H(u))$ is open. By definition,

$$\begin{align*}\pi^{-1}(H(u)) = \{\mathscr{P} \in X_{(A,\sigma)} \mid u \in \mathscr{P}_F\}.\end{align*}$$

Observe that if $c \in \mathscr {P} \setminus \{0\}$ , then $u \in \mathscr {P}_F$ if and only if $uc \in \mathscr {P}$ (indeed, $u \in \mathscr {P}_F$ or $u \in -\mathscr {P}_F$ , and only one of them occurs by (P5); the first case corresponds to $uc \in \mathscr {P}$ ). Therefore, $\mathscr {P} \in \pi ^{-1}(H(u))$ if and only if there is $c \in \operatorname {\mathrm {Sym}}(A,\sigma ) \setminus \{0\}$ such that $c \in \mathscr {P}$ and $uc \in \mathscr {P}$ . Thus

$$ \begin{align*} \pi^{-1} (H(u)) &= \bigcup_{c \in \operatorname{\mathrm{Sym}}(A,\sigma) \setminus \{0\}} (H_\sigma(c) \cap H_\sigma(uc)), \end{align*} $$

which is open in $\mathcal {T}_\sigma $ .

Proof Since $X_{(A,\sigma )}$ is compact, $X_F$ is Hausdorff, and $\pi $ is continuous, the map $\pi $ is necessarily closed.

Proof We show that $\pi (H_\sigma (a_1, \ldots , a_k))$ is open for all $k \in \mathbb {N}$ and $a_1, \ldots , a_k \in \operatorname {\mathrm {Sym}}(A,\sigma )$ . Letting $\widetilde X_F :=X_F\setminus \mathrm {Nil}[A,\sigma ]$ , we first observe that

(4.1) $$ \begin{align} X_F \setminus \pi(H_\sigma(a_1, \ldots, a_k))=\mathrm{Nil}[A,\sigma] \cup (\widetilde X_F \setminus \pi(H_\sigma(a_1, \ldots, a_k))) \end{align} $$

since $\operatorname {\mathrm {Im}}\pi \subseteq \widetilde X_F$ , and we show

(4.2) $$ \begin{align} \widetilde X_F \setminus \pi (H_\sigma(a_1, \ldots, a_k)) = \pi\big(X_{(A,\sigma)} \setminus (H_\sigma(a_1, \ldots, a_k) \cup H_\sigma(-a_1, \ldots, -a_k))\big). \end{align} $$

“ $\subseteq $ ”: Let $P \in \widetilde X_F \setminus \pi (H_\sigma (a_1, \ldots , a_k))$ and let $\mathscr {P}$ be a positive cone over P, so that $P=\pi (\mathscr {P})=\pi (-\mathscr {P})$ . We want to show that $\mathscr {P} \not \in H_\sigma (a_1, \ldots , a_k) \cup H_\sigma (-a_1, \ldots , -a_k)$ . If $\mathscr {P} \in H_\sigma (a_1, \ldots , a_k)$ , then $P \in \pi (H_\sigma (a_1, \ldots , a_k))$ , contradiction. If $\mathscr {P} \in H_\sigma (-a_1, \ldots , -a_k)$ , then $P = \pi (-\mathscr {P}) \in \pi (H_\sigma (a_1, \ldots , a_k))$ , contradiction again.

“ $\supseteq $ ”: Let $\mathscr {P} \in X_{(A,\sigma )} \setminus (H_\sigma (a_1, \ldots , a_k) \cup H_\sigma (-a_1, \ldots , -a_k))$ be over $P \in X_F$ , so that $\pi (\mathscr {P}) = P$ . If $P \in \pi (H_\sigma (a_1, \ldots , a_k))$ , then there is a positive cone $\mathscr {Q}$ over P such that $\mathscr {Q} \in H_\sigma (a_1, \ldots , a_k)$ . In particular, $\mathscr {Q} = \mathscr {P}$ or $\mathscr {Q} = -\mathscr {P}$ by Theorem 4.6 since $\mathscr {Q}$ , $\mathscr {P}$ , and $-\mathscr {P}$ are all over P. Then, $\mathscr {P} \in H_\sigma (a_1, \ldots , a_k)$ in the first case, and $\mathscr {P} \in H_\sigma (-a_1, \ldots , -a_k)$ in the second case, which are both contradictions.

The right-hand side of (4.2) is $\pi $ of a closed set, so is closed by Corollary 4.10. Therefore, the left-hand side is closed, which shows that the set $\pi (H_\sigma (a_1, \ldots , a_k))$ is open in $X_F$ by (4.1) and since $\mathrm {Nil}[A,\sigma ]$ is clopen by [Reference Astier and Unger1, Corollary 6.5].

Proof Observe that by the definition of signatures, $m_P(A,\sigma ) \leqslant n_P(A,\sigma )$ (cf. [Reference Astier and Unger3, Proposition 4.4(iii)]). For ease of notation, we assume that $(A\otimes _F F_P,\sigma \otimes \mathrm {id})= (M_{n_P}(D_P), \operatorname {\mathrm {Int}}(\Phi _P) \circ {\vartheta _P}^t)$ . Let

$$ \begin{align*} \operatorname{\mathrm{PD}}_{n_P}(D_P,\vartheta_P):=\{B \in \operatorname{\mathrm{Sym}}( M_{n_P}(D_P), {\vartheta_P}^t ) \mid \vartheta_P & (X)^t BX> 0 \text{ in }F_P,\\ & \text{for every } X \in (D_P)^{n_P}\setminus \{0\}\}. \end{align*} $$

Then, $\Phi _P\cdot \operatorname {\mathrm {PD}}_{n_P}(D_P,\vartheta _P)$ is an open subset of $ \operatorname {\mathrm {Sym}} (A\otimes _F F_P, \sigma \otimes \mathrm {id})$ by [Reference Astier and Unger5, Lemma 2.21]. Since F is dense in $F_P$ , $A\otimes 1_{F_P}$ is dense in $A\otimes _F F_P$ , and there is $a \in A$ such that $a \otimes 1_{F_P} \in \Phi _P\cdot \operatorname {\mathrm {PD}}_{n_P}(D_P,\vartheta _P)$ .

Let $\mathfrak {s}_P$ denote the hermitian Morita equivalence

$$\begin{align*}{\mathfrak{Herm}} (M_{n_P}(D_P), \operatorname{\mathrm{Int}}(\Phi_P) \circ \vartheta^t) \to {\mathfrak{Herm}} (M_{n_P}(D_P), \vartheta^t) \end{align*}$$

given by the scaling map $h\mapsto \Phi _P^{-1}h$ . Denoting the unique ordering on $F_P$ by ${\widetilde {P}}$ , we then have

$$\begin{align*}\operatorname{\mathrm{sign}}^\mu_P \langle a\rangle_\sigma = \operatorname{\mathrm{sign}}^{\mu\otimes 1}_{\widetilde{P}} \langle a\otimes 1_{F_P}\rangle_{\sigma \otimes \mathrm{id}} = \operatorname{\mathrm{sign}}^{\mathfrak{s}_P (\mu\otimes 1)}_{\widetilde{P}} \langle \Phi_P^{-1} \cdot a\otimes 1\rangle_{{\vartheta_P}^t} = \pm n_P(A,\sigma),\end{align*}$$

where the second equality follows from [Reference Astier and Unger2, Theorem 4.2], the final equality holds since we are computing the signature of a positive definite $n_P\times n_P$ matrix, and where the $\pm $ is due to the reference form $\mathfrak {s}_P (\mu \otimes 1)$ , which may induce a sign change in the result. Replacing a by $-a$ if necessary, the conclusion follows.

Proof Since F is finitely generated over $\mathbb {Q}$ , we can write $F=\mathbb {Q}(S)$ for some finite set S. By [Reference Grillet11, Chapter IV, Theorem 8.6], F has a transcendence basis $\{X_1, \ldots , X_n\}$ over $\mathbb {Q}$ which is included in S. Thus, by the primitive element theorem, $F = \mathbb {Q}(X_1, \ldots , X_n)(\alpha )$ with $\alpha $ algebraic over $\mathbb {Q}(X_1, \ldots , X_n)$ .

Let $m_{\bar X}(X)$ be the minimal polynomial of $\alpha $ over $\mathbb {Q}(\bar X)$ , where $\bar X = (X_1,\ldots ,X_n)$ . Let $P \in X_F$ and let U be a basic open set containing P. The set U is of the form

$$\begin{align*}\{Q \in X_F \mid\ g_1(\bar X, \alpha)>_Q 0, \ldots, g_r(\bar X, \alpha) >_Q 0\},\end{align*}$$

where the rational functions $g_1,\ldots , g_r$ can be chosen to be polynomials in $\mathbb {Q}[X_1,\ldots ,X_n, \alpha ]$ .

Claim There are $y_1, \ldots , y_n, \beta \in \mathbb {R}$ such that:

1. (1) $\{y_1, \ldots , y_n\}$ is algebraically independent over $\mathbb {Q}$ ;

2. (2) $g_j(y_1, \ldots , y_n, \beta )> 0$ for $j=1, \ldots , r$ (the ordering is the one from $\mathbb {R}$ );

3. (3) $\beta $ is a root of $m_{\bar y}(X)$ , where $\bar y = (y_1, \ldots , y_n)$ .

We will prove the Claim in the course of the next two lemmas, but we use it now. The map $\lambda : F = \mathbb {Q}(X_1, \ldots , X_n, \alpha ) \rightarrow \mathbb {R}$ defined by $\lambda (X_i) = y_i$ and $\lambda (\alpha ) = \beta $ is a morphism of fields, and $Q := \lambda ^{-1}(\mathbb {R}_{\geqslant 0})$ is an ordering on F such that $Q \in U$ . Writing $F' := \operatorname {\mathrm {Im}} \lambda = \mathbb {Q}(y_1, \ldots , y_m, \beta )$ , the map $\lambda $ gives an isomorphism of ordered fields $(F, Q) \cong (F', F' \cap \mathbb {R}_{\geqslant 0})$ , which is an ordered subfield of $(\mathbb {R}, \mathbb {R}_{\geqslant 0})$ and therefore archimedean.

Proof For $e \in \{1,2\}^r$ , let $g^e := g_1^{e_1} \cdots g_r^{e_r}$ and

$$\begin{align*}N_e(\bar x) := \sum_{c \in \mathbb{R}:\ m_{\bar x}(c) = 0} \operatorname{\mathrm{sgn}} g^e(\bar x, c),\end{align*}$$

where $\operatorname {\mathrm {sgn}}$ denotes the sign function. By [Reference Scheiderer19, Proposition 1.3.36], we have

$$\begin{align*}\big|\{c \in \mathbb{R} \mid m_{\bar x}(c) = 0,\ g_1(\bar x,c)>0, \ldots,\ g_r(\bar x,c) > 0\}\big| = 2^{-r} \sum_{e \in \{1,2\}^r} N_e(\bar x),\end{align*}$$

for all $\bar x\in \mathbb {R}^n$ , and thus,

$$\begin{align*}S = \Big\{\bar x \in \mathbb{R}^n \, \Big|\, \sum_{e \in \{1,2\}^r} N_e(\bar x) \geqslant 2^r\Big\}.\end{align*}$$

Using this description of S, we show that S contains an open subset. Listing all the possible values of $N_e(\bar x)$ , for $e \in \{1,2\}^r$ , whose sum gives a result greater than or equal to $2^r$ , we see that S can be expressed as a finite union of finite intersections of sets of the form

$$\begin{align*}S_{e,\ell} := \{\bar x \in \mathbb{R}^n \mid N_e(\bar x) = \ell\},\end{align*}$$

for some $e \in \{1,2\}^r$ and $\ell \in \mathbb {N} \cup \{0\}$ . More precisely, we write

$$\begin{align*}S = \bigcup_{i \in I} \bigcap_{(e,\ell) \in E_i} S_{e, \ell},\end{align*}$$

where I and each $E_i$ are finite index sets. We will show that each set $S_{e, \ell } \cap Z$ is an open subset of $\mathbb {R}^n$ (for $i \in I$ and $(e,\ell ) \in E_i$ ), where Z is an open set that will be defined below, thus proving that $S \cap Z$ is open. We will then show that $S \cap Z$ is non-empty.

We first show that $S_{e, \ell } \cap Z$ is open (this will help us decide what Z should be). For $i \in I$ and $(e, \ell ) \in E_i$ , let $(f_{e,0}, \ldots , f_{e,t_e}) \in \mathbb {Q}(x_1, \ldots , x_n)[X]$ be the Sturm sequence of m and $g^e$ , and

• • let $p_e:=(p_{e,1}, \ldots , p_{e,t_e})$ be the sequence of coefficients of the highest degree terms of $(f_{e,1}(X), \ldots , f_{e,t_e}(X))$ ;

• • let $\tilde p_e:= (\tilde p_{e,1}, \ldots , \tilde p_{e,t_e})$ be the sequence of coefficients of the highest degree terms of $(f_{e,1}(-X), \ldots , f_{e,t_e}(-X))$ .

Let $v(m_{\bar x},g^e)$ be the number of sign changes in the sequence $p_e$ and $\tilde v(m_{\bar x},g^e)$ the number of sign changes in the sequence $\tilde p_e$ . By [Reference Bochnak, Coste and Roy8, Corollary 1.2.12], $N_e(\bar x)$ is equal to $\tilde v(m_{\bar x},g^e) - v(m_{\bar x},g^e)$ , and thus

$$\begin{align*}S_{e, \ell} = \{\bar x \in \mathbb{R}^n \mid \tilde v(m_{\bar x},g^e) - v(m_{\bar x},g^e) = \ell\}.\end{align*}$$

Listing all the possible values of $\tilde v(m_{\bar x},g^e)$ and $v(m_{\bar x},g^e)$ whose difference gives $\ell $ , we see that $S_{e, \ell }$ is a finite union of sets of the form

$$\begin{align*}T_1 := \{\bar x \in \mathbb{R}^n \mid \tilde v(m_{\bar x},g^e) = k_1 \wedge v(m_{\bar x}, g^e) = k_2\},\end{align*}$$

with $k_1, k_2 \in \mathbb {N} \cup \{0\}$ . We define

$$\begin{align*}Z := \{\bar x \in \mathbb{R}^n \mid \text{all } p_{e,j} \text{ and all } \tilde p_{e,j} \text{ are} \not = 0, \text{ for all } i \in I \text{ and } (e, \ell) \in E_i\}. \end{align*}$$

The idea behind the introduction of Z is that, if $\bar x \in Z$ , then the various sequences $p_e$ and $\tilde p_e$ never contain a zero, which makes it easier to describe their sign changes.

The set Z is clearly open. We show that $T_1 \cap Z$ is open, and for this it suffices to show that if

$$\begin{align*}T_2 := \{\bar x \in \mathbb{R}^n \mid v(m_{\bar x},g^e) = k\}\end{align*}$$

with $k\in \mathbb {N}\cup \{0\}$ , then $T_2 \cap Z$ is open (since the other condition, on $\tilde v$ , can be checked in the same way).

Since, for $\bar x \in Z$ , the coefficients of $p_e$ are all non-zero, we can express that $\bar x $ is in $T_2 \cap Z$ by listing all the configurations $p_{e,i} < 0$ / $p_{e,i}>0$ that enumerate all the different ways in which k sign changes can be obtained in the sequence $(p_{e,1},\ldots , p_{e,t_e})$ . But this clearly defines an open subset of $\mathbb {R}^n$ .

We now check that $S\cap Z$ is non-empty using some basic model theory. Let $\varphi (\omega _1, \ldots , \omega _n)$ be the following first-order formula in the language of rings:

$$\begin{align*}\exists \alpha \ m_{\bar \omega}(\alpha) = 0 \wedge \bigwedge_{i=1}^r g_i(\omega_1, \ldots, \omega_n, \alpha)> 0.\end{align*}$$

By choice of $P \in X_F$ in the proof of Proposition 5.2, we have

$$\begin{align*}(F,P) \models \varphi(X_1, \ldots, X_n).\end{align*}$$

Since $\{X_1, \ldots , X_n\}$ is a transcendence basis of F over $\mathbb {Q}$ , if

$$\begin{align*}\{r_j(x_1, \ldots, x_n)\}_{j \in J}\end{align*}$$

is the finite list of all polynomials that appear in the definition of the set Z above (the polynomials $p_{e,j}$ and $\tilde p_{e,j}$ ), we have

$$\begin{align*}(F,P) \models \varphi(X_1, \ldots, X_n) \wedge \bigwedge_{j \in J} r_j(X_1, \ldots, X_n) \not = 0.\end{align*}$$

Since the formula $\varphi $ is existential, it follows that:

$$\begin{align*}(F_P, {\widetilde{P}}) \models \varphi(X_1, \ldots, X_n) \wedge \bigwedge_{j \in J} r_j(X_1, \ldots, X_n) \not = 0.\end{align*}$$

In particular,

$$\begin{align*}(F_P, {\widetilde{P}}) \models \exists x_1, \ldots, x_n \ \varphi(x_1, \ldots, x_n) \wedge \bigwedge_{j \in J} r_j(x_1, \ldots, x_n) \not = 0,\end{align*}$$

and thus, by Tarski’s transfer principle (cf. [Reference Marshall16, Corollary 11.5.4]),

$$\begin{align*}(\mathbb{R}, \mathbb{R}_{\geqslant 0}) \models \exists x_1, \ldots, x_n \ \varphi(x_1, \ldots, x_n) \wedge \bigwedge_{j \in J} r_j(x_1, \ldots, x_n) \not = 0.\end{align*}$$

The open set $S\cap Z$ is thus non-empty.

Proof By Lemma 5.3, there exist non-empty open intervals $I_1, \ldots , I_n$ of $\mathbb {R}$ such that $I_1 \times \cdots \times I_n \subseteq S$ . Let $x_1 \in I_1$ be transcendental over $\mathbb {Q}$ . Assume that we have obtained $x_1 \in I_1, \ldots , x_k \in I_k$ with $(x_1, \ldots , x_k)$ algebraically independent over $\mathbb {Q}$ , for some $k < n$ . Since $I_{k+1}$ is uncountable, there is $x_{k+1} \in I_{k+1}$ such that $(x_1, \ldots , x_{k+1})$ is algebraically independent over $\mathbb {Q}$ , and we conclude by induction.

Proof Let $\Phi $ be the collection of formulas (without parameters) in the language of ordered fields expressing that an ordered field is dense in its real closure (cf. [Reference Krapp, Kuhlmann and Lehéricy14, Remark 4.4] and the references mentioned there), with quantifiers relativized to $\underline {F}$ (the formulas will be interpreted in an L-structure such as $\mathfrak {A}$ , in which case we want them to be true if and only if they are true in the interpretation of $\underline {F}$ ).

Let $\Delta (F)$ be the complete theory (with parameters) of F in the language $L_r$ (i.e., the set of all first-order $L_r$ -formulas with parameters in F that are true in F), and where the quantifiers are relativized to $\underline {F}$ .

Fix an F-basis $\{e_1, \ldots , e_m\}$ of A. We define the structure constants $f_{ijk} \in F$ of A with respect to this basis by:

$$\begin{align*}e_i e_j = \sum_{k=1}^m f_{ijk} e_k, \text{ for } 1 \leqslant i,j \leqslant m,\end{align*}$$

and the constants $f_{\sigma ik}$ defining $\sigma $ by

$$\begin{align*}\sigma(e_i) = \sum_{k=1}^m f_{\sigma ik} e_k, \text{ for } 1\leqslant i \leqslant m.\end{align*}$$

We consider the set of L-formulas:

$$ \begin{align*} \Omega & := \Delta(F) \cup \Phi \cup \{\underline{P} \text{ ordering on } \underline{F}\} \cup \{\underline{\mathscr{P}} \text{ is a prepositive cone over } \underline{P}\} \\ &\quad \cup \{\{e_1, \ldots, e_m\} \text{ is a basis over } \underline{F}\} \cup \Big\{ e_i e_j = \sum_{r=1}^m f_{ijk} e_k \, \Big|\, 1 \leqslant i,j \leqslant m\Big\} \\ &\quad \cup \Big\{\underline{\sigma}(e_i) = \sum_{k=1}^m f_{\sigma ik} e_k \, \Big|\, 1\leqslant i \leqslant m \Big\} \\ &\quad \cup \{u \in \underline{P} \mid\ u \in P\} \cup \{a \in \underline{\mathscr{P}} \mid a \in \mathscr{P}\}. \end{align*} $$

Let S be a finite subset of $\Omega $ . Thus, S is included in

$$ \begin{align*} S' &:= \Delta(F) \cup \Phi \cup \{\underline{P} \text{ ordering on } \underline{F}\} \cup \{\underline{\mathscr{P}} \text{ is a prepositive cone over } \underline{P}\} \\ &\quad \cup \{\{e_1, \ldots, e_m\} \text{ is a basis over } \underline{F}\} \cup \Big\{ e_i e_j = \sum_{r=1}^m f_{ijk} e_k \, \Big|\, 1 \leqslant i,j \leqslant m\Big\} \\ &\quad \cup \Big\{\underline{\sigma}(e_i) = \sum_{k=1}^m f_{\sigma ik} e_k \, \Big|\, 1\leqslant i \leqslant m\Big\} \\ &\quad \cup \{u_1 \in \underline{P}, \ldots, u_k \in \underline{P}\} \cup \{a_1 \in \underline{\mathscr{P}}, \ldots, a_\ell \in \underline{\mathscr{P}}\}, \end{align*} $$

for some $u_1, \ldots , u_k \in P$ and $a_1, \ldots , a_\ell \in \mathscr {P}$ .

Consider the open set $H(u_1, \ldots , u_k)$ of $X_F$ and the open set $H_\sigma (a_1, \ldots , a_\ell )$ of $X_{(A,\sigma )}$ . Clearly, $P \in H(u_1, \ldots , u_k)$ and $\mathscr {P} \in H_\sigma (a_1, \ldots , a_\ell )$ . Recall that the map $\pi : X_{(A,\sigma )} \rightarrow X_F$ , $\pi (\mathscr {Q}) = \mathscr {Q}_F$ is open by Proposition 4.11. Therefore, $H(u_1, \ldots , u_k) \cap \pi (H_\sigma (a_1, \ldots , a_\ell ))$ is an open subset of $X_F$ containing P, and thus contains an ordering $P'$ such that $(F,P')$ is archimedean by hypothesis. Since $P' \in \pi (H_\sigma (a_1, \ldots , a_\ell ))$ , there is a positive cone $\mathscr {P}'$ on $(A,\sigma )$ over $P'$ such that $a_1, \ldots , a_\ell \in \mathscr {P}'$ .

In particular, the L-structure $(A,\sigma ,F,P',\mathscr {P}')$ is a model of $S'$ . (Recall that an archimedean ordered field is dense in its real closure. This follows directly from [Reference Prestel and Delzell17, Theorem 1.1.5].) Therefore, every finite subset of $\Omega $ has a model, so that $\Omega $ has a model $\mathcal {B} = (B, \tau , N, Q, \mathscr {S})$ by the compactness theorem.

By construction, N is an elementary extension of F, $P \subseteq Q$ , $\mathscr {P}\subseteq \mathscr {S}$ , and $(N,Q)$ is dense in its real closure.

To prove statement (2), we first check that $(B, \tau ) \cong (A \otimes _F N, \sigma \otimes \mathrm {id}_N)$ : the structure constants of B with respect to $\{e_1, \ldots , e_m\}$ are by construction the same as those of A, and therefore as those of $A \otimes _F N$ . Thus, since both B and $A \otimes _F N$ are algebras over N of the same dimension, they are isomorphic, and the isomorphism is induced by $e_i \mapsto e_i \otimes 1$ for $1\leqslant i\leqslant m$ . Similarly, the N-linear maps $\tau $ and $\sigma \otimes \mathrm {id}_N$ have the same matrix with respect to $\{e_1, \ldots , e_m\}$ and $\{e_1 \otimes 1, \ldots , e_m \otimes 1\}$ , respectively, so that the algebras with involution $(B, \tau )$ and $(A \otimes _ F N, \sigma \otimes \mathrm {id}_N)$ are isomorphic via

$$\begin{align*}\xi : B \rightarrow A \otimes_F N, \ e_i \mapsto e_i \otimes 1.\end{align*}$$

The set $\xi (\mathscr {S})$ is a prepositive cone on $(A\otimes _F N,\sigma \otimes \mathrm {id}_N)$ over Q, so is included in a positive cone $\mathscr {Q}$ on $(A \otimes _ F N, \sigma \otimes \mathrm {id}_N)$ over Q. We have $\mathscr {P} \subseteq \mathscr {S}$ , so that $\xi (\mathscr {P}) \subseteq \mathscr {Q}$ , i.e., $\mathscr {P} \otimes 1 \subseteq \mathscr {Q}$ .

Proof By Theorem 4.6, there is $\varepsilon \in \{-1,1\}$ such that $\operatorname {\mathrm {sign}}^\mu _P \langle a\rangle _\sigma = \varepsilon m_P(A,\sigma )$ for every $a \in \mathscr {P} \cap A^\times $ . Let $a \in \mathscr {P} \cap A^\times $ (cf. [Reference Astier and Unger4, Lemma 3.6]). Since $m_P(A,\sigma )\leqslant n_P(A,\sigma )$ (cf. [Reference Astier and Unger3, Proposition 4.4(iii)]) and $\operatorname {\mathrm {sign}}_P^\mu \langle -a\rangle _\sigma = - \operatorname {\mathrm {sign}}_P^\mu \langle a\rangle _\sigma $ , we prove the result by showing that $\operatorname {\mathrm {sign}}^\mu _P \langle a\rangle _\sigma = \pm n_P(A,\sigma )$ .

Fix a basis $\mathcal {B} = \{e_i\}_{i \in I}$ of A over F, and let $F_0$ be the field obtained by adding the following elements to $\mathbb {Q}$ , all determined with respect to the basis $\mathcal {B}$ : the structure constants of A, the elements of the matrix of $\sigma $ , and the coordinates of a. Let $A_0$ be the $F_0$ -algebra determined by these structure constants for a given basis $\mathcal {B}_0 = \{e^{\prime }_i\}_{i \in I}$ (i.e., we build the free $F_0$ -algebra generated by the elements $e^{\prime }_i$ and quotient out by the relations determined by the structure constants), let $\sigma _0$ be the $F_0$ -linear map on $A_0$ with the same matrix as $\sigma $ , and let $a_0 \in A_0$ be the element with the same coordinates as a (all with respect to $\mathcal {B}_0$ ). Let $\xi : A_0 \otimes _{F_0} F \rightarrow A$ , $e^{\prime }_i \otimes 1 \mapsto e_i$ . Since $A_0 \otimes _{F_0}$ , F and A are F-algebras with the same structure constants with respect to $\{e^{\prime }_i \otimes 1\}_{i \in I}$ and $\{e_i\}_{i \in I}$ , respectively, and the linear maps $\sigma _0 \otimes 1$ and $\sigma $ have the same matrices with respect to these same bases, the map $\xi $ is an isomorphism of F-algebras with involution from $(A_0 \otimes _{F_0} F, \sigma _0 \otimes \mathrm {id})$ to $(A, \sigma )$ , and $\xi (a_0 \otimes 1) = a$ . Therefore, we can assume for simplicity that $\xi $ is the identity map, so that $(A_0 \otimes _{F_0} F, \sigma _0 \otimes \mathrm {id}) = (A,\sigma )$ , $a_0 \otimes 1 = a$ , and $A_0\subseteq A$ .

Let $P_0:= F_0\cap P$ and $\mathscr {P}_0:=A_0 \cap \mathscr {P}$ . Note that $a_0 \in \mathscr {P}_0$ . By construction, $F_0$ is finitely generated over $\mathbb {Q}$ , and $(F,P)$ is an ordered extension of $(F_0,P_0)$ . By Theorem 2.9, we have, for any reference form $\mu _0$ for $(A_0, \sigma _0)$ :

$$\begin{align*}\operatorname{\mathrm{sign}}^{\mu_0}_{P_0} \langle a_0\rangle_{\sigma_0} = \operatorname{\mathrm{sign}}^{\mu_0 \otimes F}_{P} \langle a_0 \otimes 1\rangle_{\sigma_0 \otimes 1} = \operatorname{\mathrm{sign}}^{\mu_0 \otimes F}_P \langle a\rangle_\sigma = \pm \operatorname{\mathrm{sign}}^\mu_P \langle a\rangle_\sigma,\end{align*}$$

where the final equality holds since a change of reference forms at most changes the sign of the signature (cf. [Reference Astier and Unger2, Proposition 3.3(iii)]). In particular, it suffices to prove the result for $(A_0, \sigma _0)$ and $P_0$ , $\mathscr {P}_0$ , $a_0$ . Therefore, we simply use the original notation and assume that F is finitely generated over $\mathbb {Q}$ .

It follows by Proposition 5.2 that P is in the closure of the set of archimedean orderings on F. Let N, Q, and $\mathscr {Q}$ be as in Lemma 5.5. By Proposition 5.1, and using that invertible elements in a given positive cone have signature equal to $\pm m_P(A,\sigma )$ by Theorem 4.6, we know that for every $b \in \mathscr {Q}\cap (A\otimes _F N)^\times $ ,

$$\begin{align*}\operatorname{\mathrm{sign}}^{\mu \otimes N}_Q \langle b\rangle_{\sigma \otimes \mathrm{id}} = \pm n_Q(A\otimes_F N, \sigma\otimes\mathrm{id})= \pm n_P(A,\sigma), \end{align*}$$

where the final equality follows from the fact that $(N,Q)$ is an ordered extension of $(F,P)$ and Lemma 5.7 below. Since $a \otimes 1 \in (\mathscr {P} \otimes 1)\cap (A\otimes _F N)^\times \subseteq \mathscr {Q} \cap (A\otimes _F N)^\times $ , it follows that:

$$\begin{align*}\operatorname{\mathrm{sign}}^\mu_P \langle a\rangle_\sigma = \operatorname{\mathrm{sign}}^{\mu \otimes N}_Q \langle a \otimes 1\rangle_{\sigma \otimes \mathrm{id}} = \pm n_P(A,\sigma).\\[-34pt]\end{align*}$$

Proof Recall that $n_P(A,\sigma )$ is defined via the isomorphism $A\otimes _F F_P\cong M_{n_P}(D_{F_P})$ , with notation as in Sections 2.1 and 2.2. Let $(L_Q,{\widetilde {Q}})$ be a real closure of $(L,Q)$ . Observe that we may assume that $F_P\subseteq L_Q$ , and thus that $D_{F_P}\otimes _{F_P} L_Q\cong D_{L_Q}$ , because $D_{F_P} \in \{F_P, F_P(\sqrt {-1}), (-1,-1)_{F_P}\}$ . Therefore,

$$\begin{align*}A\otimes_F L_Q \cong A\otimes_F F_P \otimes_{F_P} L_Q \cong M_{n_P}(D_{F_P})\otimes_{F_P} L_Q \cong M_{n_P}(D_{L_Q}). \end{align*}$$

Hence, $n_{{\widetilde {Q}}}(A\otimes _F L_Q, \sigma \otimes \mathrm {id}) =n_P= n_P(A,\sigma )$ . The result follows since $n_Q(A\otimes _F L, \sigma \otimes \mathrm {id})= n_{{\widetilde {Q}}}(A\otimes _F L_Q, \sigma \otimes \mathrm {id})$ by definition.

Proof Up to replacing $\mathscr {P}$ by a positive cone that contains it, we may assume that $\mathscr {P}$ is a positive cone. By Theorem 4.6, we have $P \in X_F \setminus \mathrm {Nil}[A,\sigma ]$ , and so there is $a\in \operatorname {\mathrm {Sym}}(A,\sigma )$ such that $\operatorname {\mathrm {sign}}^\mu _P \langle a\rangle _\sigma \not = 0$ (cf. Remark 2.4). Therefore, by Theorem 2.9,

$$\begin{align*}\operatorname{\mathrm{sign}}^{\mu \otimes L}_Q \langle a \otimes 1\rangle_{\sigma \otimes \mathrm{id}} = \operatorname{\mathrm{sign}}^\mu_P \langle a\rangle_\sigma \not = 0,\end{align*}$$

and $Q \not \in \mathrm {Nil}[A \otimes L, \sigma \otimes \mathrm {id}]$ . In particular, there are positive cones on $(A \otimes L, \sigma \otimes \mathrm {id})$ and they are described by Theorem 4.6. By Corollary 4.7,

$$\begin{align*}\mathscr{P} = \bigcup \{D_{(A,\sigma)} \langle a_1, \ldots, a_k\rangle_\sigma \mid k \in \mathbb{N}, \ a_1, \ldots, a_k \in \mathscr{M}^\mu_P(A,\sigma)\}.\end{align*}$$

Therefore (using (a) and (b) below),

$$ \begin{align*} \mathscr{P} \otimes 1_L & \subseteq \bigcup \{D_{(A \otimes L,\sigma \otimes \mathrm{id})} \langle a_1 \otimes 1, \ldots, a_k \otimes 1\rangle_\sigma \mid k \in \mathbb{N}, a_1, \ldots, a_k \in \mathscr{M}^\mu_P(A,\sigma)\} \\& \subseteq \bigcup \{D_{(A \otimes L,\sigma \otimes \mathrm{id})} \langle b_1, \ldots, b_k\rangle_\sigma \mid k \in \mathbb{N}, b_1, \ldots, b_k \in \mathscr{M}^{\mu \otimes L}_Q(A \otimes L, \sigma \otimes \mathrm{id})\} \\& = \mathscr{C}_Q(\mathscr{M}^{\mu \otimes L}_Q(A \otimes L, \sigma \otimes \mathrm{id})), \end{align*} $$

which is a positive cone over Q by Theorem 4.6, and where:

1. (a) The second inclusion uses the fact that $a \in \mathscr {M}^\mu _P(A,\sigma )$ implies $a \otimes 1 \in \mathscr {M}^{\mu \otimes L}_Q(A \otimes L, \sigma \otimes \mathrm {id})$ which follows from the fact that $m_P(A,\sigma ) = n_P(A,\sigma ) = n_Q(A \otimes _F L, \sigma \otimes \mathrm {id}) = m_Q(A \otimes _F L, \sigma \otimes \mathrm {id})$ by Proposition 5.6 and Lemma 5.7.

2. (b) The final equality follows from Corollary 4.7.