ANNALES MATHÉMATIQUES BLAISE PASCAL

BERTIN DIARRA

*p*-adic Clifford algebras

Annales mathématiques Blaise Pascal, tome 1, n° 1 (1994), p. 85-103

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# P-ADIC CLIFFORD ALGEBRAS

Bertin DIARRA

In a previous paper [2], we gave the index of the standard quadratic form of rank $n$ over the field of p-adic numbers. Here, we recover, as a consequence, the structure of the associated Clifford algebra.

The classification of all (equivalence classes of) quadratic forms over a p-adic field is well known (cf. [5]), with this classification, one is able to classify all p-adic Clifford algebras.

## I - INTRODUCTION

Let $K$ be a field of characteristic $\neq 2$ and $E$ a vector space over $K$ of finite dimension $n$. A mapping $q: E \to K$ is a quadratic form over $E$ if there exists a bilinear symmetric form $f: E \times E \to K$ such that

$$q(x) = f(x, x) \quad \text{and} \quad f(x, y) = \frac{1}{2}[q(x + y) - q(x) - q(y)]$$

We assume that $q$ is regular, that is $f$ is non-degenerated.

An element $x \in E$ is isotropic if $q(x) = 0$. Let $V$ be a subspace of $E$; the orthogonal subspace of $V$ is the set $V^{\perp} = \{y \in E/f(x, y) = 0 \text{ for all } x \in V\}$. The subspace $V$ is called totally isotropic if $V \subset V^{\perp}$. It is well known (cf. for example [1]) that any totally isotropic subspace is contained in a maximal totally isotropic subspace. The maximal totally isotropic subspaces have the same dimension $\nu$, called the index of $q$ and $2\nu \leq n$. If $2\nu = n$, then $(E, q)$ is called a hyperbolic space and for the case $n = 2$, one says hyperbolic plane. The index $\nu = 0$ iff $q(x) \neq 0$ for $x \neq 0$ i.e. $(E, q)$ is anisotropic.
Let $E = K^n$ and $B = (e_1, \dots, e_n)$ be the canonical basis of $E$; the standard quadratic form $q_0$ is the quadratic form associated to the bilinear form

$$ \langle x, y \rangle = \sum_{j=1}^{n} x_j y_j \quad ; \quad \text{where } x = \sum_{j=1}^{n} x_j e_j \text{ and } y = \sum_{j=1}^{n} y_j e_j \text{;} \\ \text{hence } q_0(x) = \langle x, x \rangle = \sum_{j=1}^{n} x_j^2. $$

Let $(E, q)$ be a quadratic space, possibly non regular; an algebra $C = C(E, q)$ over $K$, with unit 1, is said to be a *Clifford algebra* for $(E, q)$ if

(i) There exists a one-to-one linear mapping $\rho : E \to C$ such that $\rho(x)^2 = q(x) \cdot 1$.

(ii) For every algebra $A$ with unit 1 and linear mapping $\phi : E \to A$ satisfying $\phi(x)^2 = q(x) \cdot 1$, there exists an algebra homomorphism $\tilde{\phi} : C \to A$ such that $\tilde{\phi} \circ \rho = \phi$.

Clifford algebra exists and is unique up algebra isomorphism (cf. for instance [1] or [3]). For example, let $K < X_1, \dots, X_n >$ be the free algebra with free system of generators $X_1, \dots, X_n$ and $I$ be the two-sided ideal of $K < X_1, \dots, X_n >$ generated by $X_i X_j + X_j X_i - 2f(e_i, e_j) \cdot 1, 1 \le i, j \le n$, where $(e_1, \dots, e_n)$ is an orthogonal basis of $(E, q)$; then $C(E, q) = K < X_1, \dots, X_n > / I$.

## II - THE P-ADIC STANDARD QUADRATIC FORM $q_0$

### II - 1. The index of $q_0$

Let $p$ be a prime number and $Q_p$ be the p-adic field i.e. the completion of the field of rational numbers $Q$ for the p-adic absolute value.

We denote by $[\alpha]$ the integral part of the real number $\alpha$.

**Proposition 1** [2]

The standard quadratic form $q_0(x) = \sum_{j=1}^{n} x_j^2$ over $E = \mathbb{Q}_p^n$ has index

(i) $\nu = \lfloor \frac{n}{2} \rfloor$ if $p \equiv 1 \pmod{4}$

(ii) $\nu = \lfloor \frac{n}{2} \rfloor$ if $p \equiv 3 \pmod{4}$ and $n \neq 2 \pmod{4}$

(iii) $\nu = \lfloor \frac{n}{2} \rfloor - 1$ if $p \equiv 3 \pmod{4}$ and $n \equiv 2 \pmod{4}$
**Proof:**

1°) If $p \equiv 1 \pmod{4}$, it is well known that $i = \sqrt{-1} \in \mathbb{Q}_p$. Let $\nu = \lfloor \frac{n}{2} \rfloor$ and $\epsilon_j = i \epsilon_{2j-1} + \epsilon_{2j}$, $1 \le j \le \nu$, then $V = \bigoplus_{j=1}^{\nu} \mathbb{Q}_p \epsilon_j$ is a maximal totally isotropic subspace of $E = \mathbb{Q}_p^n$.

2°) $p \equiv 3 \pmod{4}$

Therefore $i \notin \mathbb{Q}_p$ and if $n=2$ the index of $q_0$ is 0.

If $n=3$, applying Chevalley's theorem and Newton's method to $q_0(x) = x_1^2+x_2^2+x_3^2$ we find $a, b \in \mathbb{Q}_p, a \ne 0, b \ne 0$, such that $a^2+b^2+1=0$. Therefore $\epsilon_1 = a \epsilon_1 + b \epsilon_2 + \epsilon_3$ is isotropic in $\mathbb{Q}_p^3$ and $\nu = \lfloor \frac{3}{2} \rfloor = 1$.

(a) For $n = 4m$, put $\epsilon_{2j-1} = a \epsilon_{4j-3} + b \epsilon_{4j-2} + \epsilon_{4j-1}$ and $\epsilon_{2j} = -b \epsilon_{4j-3} + a \epsilon_{4j-2} + \epsilon_{4j}$, $1 \le j \le m$. It is clear that $q_0(\epsilon_{2j-1}) = q_0(\epsilon_{2j}) = a^2+b^2+1=0$ and $\langle \epsilon_{2j-1}, \epsilon_{2j} \rangle = -ab+ab=0$. Therefore $V = \bigoplus_{j=1}^m (\mathbb{Q}_p \epsilon_{2j-1} \oplus \mathbb{Q}_p \epsilon_{2j})$ is a totally isotropic subspace of $\mathbb{Q}_p^n$ and $\nu = 2m = \lfloor \frac{n}{2} \rfloor$.

If $n = 4m + 1$, with the same notations as above the subspace $V$ is totally isotropic in $\mathbb{Q}_p^n$ and $\nu = 2m = \lfloor \frac{n}{2} \rfloor$.

On the other hand if $n = 4m + 3$ the subspaces $V = \bigoplus_{j=1}^m (\mathbb{Q}_p \epsilon_{2j-1} \oplus \mathbb{Q}_p \epsilon_{2j})$ and $\mathbb{Q}_p \epsilon_{2m+1}$ where $\epsilon_{2m+1} = a \epsilon_{4m+1} + b \epsilon_{4m+2} + \epsilon_{4m+3}$, are totally isotropic and orthogonal. Therefore $V_o = V \oplus \mathbb{Q}_p \epsilon_{2m+1}$ is totally isotropic and $\nu = 2m + 1 = \lfloor \frac{n}{2} \rfloor$.

(b) If $n = 4m + 2$, let $V = \bigoplus_{j=1}^m (\mathbb{Q}_p \epsilon_{2j-1} \oplus \mathbb{Q}_p \epsilon_{2j})$ be as above. It is easy to verify that if $x \in \mathbb{Q}_p^n$ is isotropic and $x$ is orthogonal to $V$ then $x \in V$. Therefore $V$ is a maximal totally isotropic subspace of $\mathbb{Q}_p^n$ and $\nu = 2m = \lfloor \frac{n}{2} \rfloor - 1$.
**Proposition 2 :** Let $p = 2$.

Let $n = 8m + s$, $0 \le s \le 7$.

The standard quadratic form $q_0(x) = \sum_{j=1}^{n} x_j^2$ over $E = \mathbb{Q}_2^n$ has index

(i) $v = 4m$ if $0 \le s \le 4$

(ii) $v = 4m + t$ if $s = 4 + t$, $1 \le t \le 3$

**Proof :**

1°) If $1 \le n \le 4$, then the index of $q_0$ is 0.

Indeed, this is clear when $n = 1$.

If $n = 2$, let $x = x_1e_1 + x_2e_2 \in \mathbb{Q}_2^2$ be isotropic and different from 0 i.e. $q_0(x) = x_1^2 + x_2^2 = 0$ and say $x_2 \ne 0$. Therefore $1 + a^2 = 0$ with $a = x_1x_2^{-1}$ and $v_2(a) = 0$ i.e. $a = 1 + 2^{2\mu}a_0$, $\mu \ge 1$, $v_2(a_0) = 0$.

Then $1 + a^2 = 2 + 2^{2\mu+1}a_0 + 2^{2\mu}a_0^2 = 0$ or $1 + 2^{2\mu}a_0 + 2^{2\mu-1}a_0 = 0$; in other words $1 \equiv 0 \pmod{2}$; a contradiction.

In the same way, one shows that if $n = 3$ or $4$, the index of $q_0$ is 0.

$$2^{\circ)} \qquad n = 5$$

Let $x_0 = 2e_1 + e_2 + e_3 + e_4 + e_5 \in \mathbb{Q}_2^5$, then $q_0(x_0) = 8$ and $\frac{\partial q_0}{\partial x_j}(x_0) = 2 \ne 0 \pmod{4}$,
$2 \le j \le 5$

By Newton's method there exists

$x = \sum_{j=1}^{5} x_j e_j \in \mathbb{Q}_2^5$ such that $q_0(x) = 0$ with $x_1 \equiv 2 \pmod{8}$, $x_j \equiv 1 \pmod{8}$, $2 \le j \le 5$.

Put $a = x_1x_5^{-1}$, $b = x_2x_5^{-1}$, $c = x_3x_5^{-1}$, $d = x_4x_5^{-1}$, then $a^2 + b^2 + c^2 + d^2 + 1 = 0$.

The two following elements of $\mathbb{Q}_2^5$

$$\epsilon_1 = a e_1 + b e_2 + c e_3 + d e_4 + e_5$$

$$\epsilon'_1 = -a e_1 - b e_2 - c e_3 - d e_4 + e_5$$

are isotropic with $\langle \epsilon_1, \epsilon'_1 \rangle = 2$. Hence $H = \mathbb{Q}_2\epsilon_1 \oplus \mathbb{Q}_2\epsilon'_1$ is a hyperbolic plane in $\mathbb{Q}_2^5$. Let $U = H^{\perp}$ be the orthogonal subspace of $H$ in $\mathbb{Q}_2^5$. The following three elements of $\mathbb{Q}_2^5$:

$$u_1 = b e_1 - a e_2 + d e_3 - c e_4$$

$$u_2 = e_1 - \frac{ac + bd}{c^2 + d^2} e_3 + \frac{bc - ad}{c^2 + d^2} e_4$$
$$u_3 = e_2 + \frac{ad - bc}{c^2 + d^2} e_3 - \frac{ac + bd}{c^2 + d^2} e_4$$

are elements of $U$, with

$$q_0(u_1) = -1, \quad q_0(u_2) = -\frac{1}{c^2+d^2} = q_0(u_3)$$

Furthermore $<u_i, u_j> = 0$ if $1 \le i \ne j \le 3$, and $(u_1, u_2, u_3)$ is a basis of $U$.

For every $u = y_1u_1+y_2u_2+y_3u_3 \in U$ we have $q_0(u) = y_1^2q_0(u_1)+y_2^2q_0(u_2)+y_3^2q_0(u_3) = -\frac{c^2y_1^2+d^2y_1^2+y_2^2+y_3^2}{c^2+d^2}$ and $q_0(u) = 0$ iff $u = 0$ because the standard quadratic form of rank 4 is anisotropic. In other words $(U, q_0)$ is anisotropic and $\mathbb{Q}_2^5 = H \perp U$ is a Witt decomposition of $(\mathbb{Q}_2^5, q_0)$. Hence the index of $q_0$ is 1.

$$3^{\circ)} \qquad \underline{n = 8m+s}, \quad 0 \le s \le 4. \\ \text{Put, for } 0 \le j \le m-1$$

$$ (1) \qquad \left\{ \begin{array}{lclclcl} \epsilon_{j,1} & = & a e_{8j+1} & + & b e_{8j+2} & + & c e_{8j+3} & + & d e_{8j+4} & + & e_{8j+5} \\ \epsilon_{j,2} & = & -b e_{8j+1} & + & a e_{8j+2} & + & d e_{8j+3} & - & c e_{8j+4} & + & e_{8j+6} \\ \epsilon_{j,3} & = & -d e_{8j+1} & + & c e_{8j+2} & - & b e_{8j+3} & + & a e_{8j+4} & + & e_{8j+7} \\ \epsilon_{j,4} & = & c e_{8j+1} & + & d e_{8j+2} & - & a e_{8j+3} & - & b e_{8j+4} & + & e_{8j+8} \end{array} \right. $$

and

$$ (2) \qquad \left\{ \begin{array}{lclclcl} \epsilon'_{j,1} & = & -a e_{8j+1} & - & b e_{8j+2} & - & c e_{8j+3} & - & d e_{8j+4} & + & e_{8j+5} \\ \epsilon'_{j,2} & = & b e_{8j+1} & - & a e_{8j+2} & - & d e_{8j+3} & + & c e_{8j+4} & + & e_{8j+6} \\ \epsilon'_{j,3} & = & d e_{8j+1} & - & c e_{8j+2} & + & b e_{8j+3} & - & a e_{8j+4} & + & e_{8j+7} \\ \epsilon'_{j,4} & = & -c e_{8j+1} & - & d e_{8j+2} & + & a e_{8j+3} & + & b e_{8j+4} & + & e_{8j+8} \end{array} \right. $$

A straightforward computation shows that $<\epsilon_{i,k}, \epsilon_{j,l}> = 0 = <\epsilon'_{i,k}, \epsilon'_{j,l}>$, $0 \le i, j \le m-1$; $1 \le k, l \le 4$ and $<\epsilon_{j,l}, \epsilon'_{j,l}> = 2$; $0 \le j \le m-1$; $1 \le l \le 4$. Furthermore $<\epsilon_{i,k}, \epsilon'_{j,l}> = 0$ if $(i, k) \ne (j, l)$.

Hence the subspaces $V = \bigoplus_{\substack{j=0 \\ 1 \le l \le 4}}^{m-1} \mathbb{Q}_2\epsilon_{j,l}$ and $W = \bigoplus_{\substack{j=0 \\ 1 \le l \le 4}}^{m-1} \mathbb{Q}_2\epsilon'_{j,l}$ are isotropic with $V \cap W = (0)$

Therefore $H = V \oplus W$ is a hyperbolic subspace of $E = \mathbb{Q}_2^{8m+s}$, with $\dim V = \dim W = 4m$.
But $E = E_m \perp E_s$ (orthogonal sum) where $E_m = \bigoplus_{j=1}^{8m} Q_2 e_j$ and $E_s = \bigoplus_{k=1}^{s} Q_2 e_{8m+k} \simeq Q_2^s$.

If $s = 0$, we have $E = E_m = V \oplus W = H$ and $(E, q_0)$ is a hyperbolic space with index $4m$.

If $1 \le s \le 4$; $E = E_m \perp E_s$ with $E_m = V \oplus W = H$. Since $1 \le \dim E_s = s \le 4$, the standard quadratic space $(E_s, q_0)$ is anisotropic. Consequently $E = (V \oplus W) \perp E_s$ is a Witt decomposition of $E$ and the index of $q_0$ is $4m$.

$$4^\circ) \quad \underline{n = 8m + 4 + t}, \quad 1 \le t \le 3.$$

$$a) \quad \underline{n = 8m + 5}$$

With the same notations as above, we have $E = E_m \perp E_5$ where $E_5 = \bigoplus_{k=1}^{5} Q_2 e_{8m+k} \simeq Q_2^5$.

Let us write, as for $n = 5$,

$$(3) \quad \left\{ \begin{array}{l} \epsilon_{4m+1} = a e_{8m+1} + b e_{8m+2} + c e_{8m+3} + d e_{8m+4} + e_{8m+5} \\ \epsilon'_{4m+1} = -a e_{8m+1} - b e_{8m+2} - c e_{8m+3} - d e_{8m+4} + e_{8m+5} \end{array} \right.$$

and

$$(4) \quad \left\{ \begin{array}{l} u_{m+1} = b e_{8m+1} - a e_{8m+2} + d e_{8m+3} - c e_{8m+4} \\ u_{m+2} = e_{8m+1} - \frac{ac+bd}{c^2+d^2} e_{8m+3} + \frac{bc-ad}{c^2+d^2} e_{8m+4} \\ u_{m+3} = e_{8m+2} + \frac{ad-bc}{c^2+d^2} e_{8m+3} - \frac{ac+bd}{c^2+d^2} e_{8m+4} \end{array} \right.$$

The subspace $U_5 = \bigoplus_{h=1}^3 Q_2 u_{m+h}$ of $E_5$ is anisotropic. On the other hand, $q_0(\epsilon_{4m+1}) = 0 = q_0(\epsilon'_{4m+1})$; $\langle \epsilon_{4m+1}, \epsilon'_{4m+1} \rangle = 2$ and $\epsilon_{4m+1}, \epsilon'_{4m+1}$ are orthogonal to $U_5$. Therefore $V_0 = V \oplus Q_2 \epsilon_{4m+1}$ and $W_0 = W \oplus Q_2 \epsilon'_{4m+1}$ are isotropic subspaces of $E$ and $E = (V_0 \oplus W_0) \perp U_5$ is a Witt decomposition of $E$. Hence the index of $q_0$ is $\dim V_0 = \dim W_0 = 4m+1$.

$$(b) \quad \underline{n = 8m+6}.$$

As before, we have $E = E_m \perp E_6$ where $E_6 = \bigoplus_{k=1}^{6} Q_2 e_{8m+k} \supset E_5$; hence $\epsilon_{4m+1}$ and $\epsilon'_{4m+1} \in E_6$.
Let us put

$$
(5) \quad \left\{
\begin{array}{l}
\epsilon_{4m+2} = -b e_{8m+1} + a e_{8m+2} + d e_{8m+3} - c e_{8m+4} + e_{8m+6} \\
\epsilon'_{4m+2} = b e_{8m+1} - a e_{8m+2} - d e_{8m+3} + c e_{8m+4} + e_{8m+6}
\end{array}
\right.
$$

and

$$
(6) \quad \left\{
\begin{aligned}
\omega_{m+1} &= e_{8m+1} + \frac{bd-ac}{c^2+d^2} e_{8m+3} - \frac{ad+bc}{c^2+d^2} e_{8m+4} \\
\omega_{m+2} &= e_{8m+2} - \frac{bc+ad}{c^2+d^2} e_{8m+3} + \frac{ac-bd}{c^2+d^2} e_{8m+4}
\end{aligned}
\right.
$$

The subspace $U_6 = Q_2\omega_{m+1} \oplus Q_2\omega_{m+2}$ of $E_6$ is anisotropic. Moreover, $q_0(\epsilon_{4m+2}) = 0 = q_0(\epsilon'_{4m+2})$; $\langle \epsilon_{4m+2}, \epsilon'_{4m+2} \rangle = 2$ and $\epsilon_{4m+2}, \epsilon'_{4m+2}$ are orthogonal to $U_6$. Therefore $V_1 = V_0 \oplus Q_2\epsilon_{4m+2}$ and $W_1 = W_0 \oplus Q_2\epsilon'_{4m+2}$ are isotropic subspaces of $E$ and $E = (V_1 \oplus W_1) \perp U_6$ is a Witt decomposition of $E$. Hence the index of $q_0$ is $\dim V_1 = \dim W_1 = 4m+2$.

$$
(c) \quad \underline{n=8m+7}.
$$

We have $E = E_m \perp E_7$, where $E_7 = \bigoplus_{k=1}^{7} Q_2 e_{8m+k} \supset E_6$.

Let us write

$$
(7) \quad \left\{
\begin{array}{l}
\epsilon_{4m+3} = -d e_{8m+1} + c e_{8m+2} - b e_{8m+3} + a e_{8m+4} + e_{8m+7} \\
\epsilon'_{4m+3} = d e_{8m+1} - c e_{8m+2} + b e_{8m+3} - a e_{8m+4} + e_{8m+7}
\end{array}
\right.
$$

and

$$
(8) \quad u_m = c e_{8m+1} + d e_{8m+2} - a e_{8m+3} - b e_{8m+4}
$$

The subspace $U_7 = Q_2 u_m$ of $E_7$ is anisotropic. Furthermore $q_0(\epsilon_{4m+3}) = 0 = q_0(\epsilon'_{4m+3})$; $\langle \epsilon_{4m+3}, \epsilon'_{4m+3} \rangle = 2$ and $\epsilon_{4m+3}, \epsilon'_{4m+3}$ are orthogonal to $U_7$. Therefore $V_2 = V_1 \oplus Q_2 \epsilon_{4m+2}$ and $W_2 = W_1 \oplus Q_2 \epsilon_{4m+2}$ are isotropic subspaces of $E$ and $E = (V_2 \oplus W_2) \perp U_7$ is a Witt decomposition of $E$. Hence the index of $q_0$ is $\dim V_2 = \dim W_2 = 4m+3$.

Remark

Let $K$ be a non formally real field. The level of $K$ is the least integer $s$ such that
$$ -1 = \sum_{j=1}^{s} a_j^2 \text{ where } a_j \in K, a_j \neq 0. \text{ It is well known that } s = 2^r, r \ge 0 \text{ (c f. [3] or [4])}. $$
The level of a p-adic field is 1 if $p \equiv 1 \pmod 4$; 2 if $p \equiv 3 \pmod 4$ and 4 if $p = 2$.

If the level of a field $K$ is $1$ (resp. $2$, resp.4) then the index of the standard quadratic form over $K^n$ is given by Proposition 1 - (i) [resp. Prop.1 - (ii) - (iii), resp. Prop.2].

More generally let $K$ be a field of level $s = 2^r, r \ge 0$. If we write for any integer $n, n = m2^{r+1} + a$ where $0 \le a \le 2^{r+1} - 1$; then the index of the standard quadratic form over $K^n$ is

$$
(i) \quad \nu = m2^r \quad \text{if} \quad 0 \le a \le 2^r
$$

$$
(ii) \quad \nu = m2^r + t \quad \text{if} \quad a = 2^r + t, \ 1 \le t \le 2^r - 1.
$$

II - 2 The Clifford algebra C(Qₚⁿ, q₀)

The following results can be deduced from a general setting (cf. [3] p. 128-129). Here
we establish them by using the computation of the index of q₀ made in II-1.

Let us recall that if *E* is a vector space over a field *K* then the exterior algebra Λ(*E*)
is the Clifford algebra associated to the null quadratic from over *E*.

On the other hand, let (E, q) be a regular quadratic space over K. If E = V ⊕ W is a hyperbolic space (V and W being maximal totally isotropic subspaces ), it is well known that the Clifford algebra C(E, q) is isomorphic to End(∧(V)), the space of linear endomorphisms of the vector space ∧(V). Furthermore the subalgebra of the even elements of C(E, q), say C+(E, q) is isomorphic to End (∧+(V)) × End (∧-(V)) where ∧+(V) (resp. ∧-(V)) is the subspace of the even ( resp. odd ) elementts of ∧(V).

Generally, if $E = (V \oplus W) \perp U$ is a Witt decomposition of $E$, then $C(E,q) \simeq \text{End}(\wedge(V)) \otimes_2 C(U,q)$, the tensor product of $\mathbb{Z}/2\mathbb{Z}$ - graded algebras (cf. for example [1]).

If $dimE = n$, then $dimC(E, q) = 2^n = dim \wedge (E)$.

If $a, b \in K^*$, we denote by $\left(\frac{a, b}{K}\right)$ the associated quaternion algebra : i.e. the algebra
generated by $i, j$ with $i^2 = a$ ; $j^2 = b$ ; $ij = -ji$. Also $\left(\frac{a, b}{K}\right)$ is the Clifford algebra
of the rank 2 quadratic form $q(x) = ax_1^2 + bx_2^2$.

Let us write $M(n, K)$ the algebra of the $n \times n$ matrices with coefficients in $K$.
**Theorem 1**: $p \equiv 1 \pmod{4}$

(i) If $n = 2m$, then $C(Q_p^n, q_0) \simeq M(2^m, Q_p)$

(ii) If $n = 2m + 1$, then $C(Q_p^n, q_0) \simeq M(2^m, Q_p) \oplus M(2^m, Q_p)$

**Proof**

Indeed, if $n = 2m$, then $(Q_p^n, q_0)$ is a hyperbolic space.

It follows that $C(Q_p^n, q_0) \simeq \text{End}(\wedge(Q_p^m))$.

And, if $n = 2m + 1$, we have a Witt decomposition $Q_p^n = (V \oplus W) \perp U$ where $U = Q_p u$

It follows that $C(U, q_0) \simeq Q_p \oplus Q_p$ which gives (ii)

**Theorem 2**: $p \equiv 3 \pmod{4}$

(i) If $n = 4m$, then $C(Q_p^n, q_0) \simeq M(2^{2m}, Q_p)$

(ii) If $n = 4m + 1$, then $C(Q_p^n, q_0) \simeq M(2^{2m}, Q_p) \oplus M(2^{2m}, Q_p)$

(iii) If $n = 4m + 2$, then $C(Q_p^n, q_0) \simeq M(2^{2m+1}, Q_p)$

(iv) If $n = 4m + 3$, then $C(Q_p^n, q_0) \simeq M(2^{2m+1}, Q_p[i])$

with $i = \sqrt{-1}$.

**Proof**:

The case (i) is evident, since $Q_p^{2m}$ is a hyperbolic space.

If $n = 4m + 1$, we have a Witt decomposition $Q_p^n = (V \oplus W) \perp U$ where $U = Q_p u$ with $u = a e_{4m-3} + b e_{4m-2} + e_{4m-1} - e_{4m+1}$ and $q_0(u) = a^2 + b^2 + 1 + 1 = 1$. It follows that $C(U, q_0) \simeq Q_p \oplus Q_p$, which gives (ii).

If $n = 4m + 2$, we have a Witt decomposition $Q_p^n = (V \oplus W) \perp U$ where $U = Q_p u_1 \oplus Q_p u_2$ and $u_1 = a e_{4m-3} + b e_{4m-2} + e_{4m-1} + a e_{4m+1} + b e_{4m+2}$
$u_2 = -b e_{4m-3} + a e_{4m-2} + e_{4m} - b e_{4m+1} + a e_{4m+2}$

Furthermore $\langle u_1, u_2 \rangle = 0$, $q_0(u_1) = -1 = q_0(u_2)$ and $C(U, q_0) \simeq (\frac{-1, -1}{Q_p})$. This quaternion algebra contains an element $z$ with $N(z) = a^2 + b^2 + 1 = 0$. Hence $(\frac{-1, -1}{Q_p}) \simeq M(2, Q_p)$ and finally we have $C(Q_p^n, q_0) \simeq M(2^{2m}, Q_p) \otimes_2 M(2, Q_p) \simeq M(2^{2m+1}, Q_p)$.

If $n = 4m + 3$, we have a Witt decomposition $Q_p^n = (V \oplus W) \perp U$ where $U = Q_p u$, with $u = -b e_{4m+1} + a e_{4m+2}$ and $q_0(u) = b^2 + a^2 = -1$. Hence $C(U, q_0) \simeq Q_p[i]$, because $u^2 = q_0(u) = -1$.
We conclude that $C(Q_p^n, q_0) \simeq M(2^{2m+1}, \mathbb{Q}_p[i])$.

In the proof of the foregoing theorem, one needs the following lemma

**Lemma:**

Let $K$ be a field (char. $\neq 2$), $c, d \in K^*$ such that $c^2 + d^2 \neq 0$.

$$ \text{If } \sigma = \frac{1}{c^2+d^2}, \text{ then } \left(\frac{-\sigma, -\sigma}{K}\right) \simeq \left(\frac{-1, -1}{K}\right). $$

If the two-rank quadratic forms $q_1(x) = -\sigma x_1^2 - \sigma x_2^2$ and $q_2(x) = -x_1^2 - x_2^2$ are equivalent, then their Clifford algebras are isomorphic. But, putting $x_1 = cx_1' + dx_2'$ and $x_2 = dx_1' - cx_2'$, we have $q_1(u(x')) = -\sigma(dx_1' + dx_2')^2 - \sigma(dx_1' - cx_2')^2 = -\sigma(c^2 + d^2)(x_1'^2 + x_2'^2) = q_2(x')$. Hence $q_1$ and $q_2$ are equivalent and the lemma is proved.

**Remark**

The quaternion algebra $\left(\frac{-1, -1}{\mathbb{Q}_2}\right) = \mathbb{H}_2$ is a skew field.

Indeed, for any $z \in \mathbb{H}_2 = \left(\frac{-1, -1}{\mathbb{Q}_2}\right)$, $z \neq 0$, the norm of $z$ is $N(z) = x_0^2+x_1^2+x_2^2+x_3^2 \neq 0$ (the standard quadratic form of rank 4 over $\mathbb{Q}_2$ is anisotropic).

**Theorem 3 :** $p=2$

The Clifford algebra $C(Q_2^n, q_0)$ is isomorphic to :

(0) $\mathrm{End}(\wedge(Q_2^{4m})) \simeq \mathrm{M}(2^{4m}, \mathbb{Q}_2)$, if $n=8m$

(1) $\mathrm{M}(2^{4m}, \mathbb{Q}_2) \oplus \mathrm{M}(2^{4m}, \mathbb{Q}_2)$, if $n = 8m + 1$

(2) $\mathrm{M}(2^{4m+1}, \mathbb{Q}_2)$, if $n = 8m + 2$

(3) $\mathrm{M}(2^{4m+1}, \mathbb{Q}_2[i])$, with $i = \sqrt{-1}$, if $n = 8m + 3$

(4) $\mathrm{M}(2^{4m+1}, \mathbb{H}_2)$, if $n = 8m + 4$

(5) $\mathrm{M}(2^{4m+1}, \mathbb{H}_2) \oplus \mathrm{M}(2^{4m+1}, \mathbb{H}_2)$, if $n = 8m + 5$

(6) $\mathrm{M}(2^{4m+2}, \mathbb{H}_2)$, if $n = 8m + 6$

(7) $\mathrm{M}(2^{4m+3}, \mathbb{Q}_2[i])$, if $n = 8m + 7$

**Proof**

According to the proof of Proposition 2, if $n = 8m + s, 0 \le s \le 7$, then $Q_2^n = (V \oplus W) \perp E_s$ where $V$ and $W$ are totally isotropic subspaces of dimension $4m$, and $(E_s, q_0) \simeq$
($\mathbb{Q}_2^s, q_0$). It follows that $C(\mathbb{Q}_2^n, q_0) \simeq \text{End}(\wedge(\mathbb{Q}_p^{4m})) \otimes_2 C(\mathbb{Q}_2^s, q_0)$. It is easy to see that
$C(\mathbb{Q}_2, q_0) \simeq \mathbb{Q}_2 \oplus \mathbb{Q}_2$ ; $C(\mathbb{Q}_2^s, q_0) \simeq (\frac{1,1}{\mathbb{Q}_2}) \simeq M(2, \mathbb{Q}_2)$ and $C(\mathbb{Q}_2^3, q_0) \simeq M(2, \mathbb{Q}_2[i])$.

If $s = 4$, the subalgebra, generated by $e_1e_2, e_2e_4$ and $e_1e_4$, is isomorphic to $\left(\frac{-1, -1}{\mathbb{Q}_2}\right) =$
$\mathbb{H}_2$. Hence $C(\mathbb{Q}_2^4, q_0) \simeq M(2, \mathbb{H}_2)$.

If $s = 5$, then $\mathbb{Q}_2^5 = F \perp U$, where $F$ is a hyperbolic plane and $U$ a three-dimensional anisotropic subspace, with orthogonal basis $(u_1, u_2, u_3)$ satisfying $q_0(u_1) = -1, q_0(u_2) = -\sigma = q_0(u_3)$. ($\sigma = \frac{1}{c^2+d^2}$ and $a, b, c, d \in \mathbb{Q}_2$ such that $a^2 + b^2 + c^2 + d^2 + 1 = 0$).

Therefore $C_+(U, q_0) \simeq \left(\frac{-\sigma_1 - \sigma_2}{\mathbb{Q}_2}\right) \simeq \mathbb{H}_2$ ; $C_+$ stands for the even subalgebra. But in
$C(U, q_0), (u_1 u_2 u_3)^2 = \sigma^2$ is a square in $\mathbb{Q}_2$ ; therefore $C(U, q_0) \simeq \mathbb{H}_2 \oplus \mathbb{H}_2$. Furthermore
$C(\mathbb{Q}_2^5, q_0) \simeq C(F, q_0) \otimes_2 C(U, q_0) \simeq M(2, \mathbb{H}_2) \oplus M(2, \mathbb{H}_2)$, because $C(F, q_0) \simeq M(2, \mathbb{Q}_2)$.

If $s = 6$, then $\mathbb{Q}_2^6 = F \perp U$, where $F$ is a hyperbolic space of dimension 4 and $U$ a two-dimensional anisotropic subspace with an orthogonal basis $(u_1, u_2)$ satisfying $q(u_1) = -\sigma = q(u_2)$. Therefore $C(U, q_0) \simeq \left(\frac{-\sigma_1 - \sigma_2}{\mathbb{Q}_2}\right) \simeq \mathbb{H}_2$. And consequently $C(\mathbb{Q}_2^6, q_0) \simeq C(F, q_0) \otimes_2 C(U, q_0) \simeq M(2^2, \mathbb{H}_2)$.

If $s = 7$, then $\mathbb{Q}_2^7 = F \perp U$, where $F$ is a hyperbolic space of dimension 6 and $U = \mathbb{Q}_2 u$, with $q_0(u) = -1$. Hence $C(U, q_0) \simeq \mathbb{Q}_2[i]$ and $C(\mathbb{Q}_2^7, q_0) \simeq M(2^3, \mathbb{Q}_2[i])$.

One deduces the isomorphisms of the theorem from $C(\mathbb{Q}_2^n, q_0) \simeq M(2^{4m}, \mathbb{Q}_2) \otimes_2 C(\mathbb{Q}_2^s, q_0)$.

N.B : A classical way to prove the above theorems is based on the isomorphisms

$$
\begin{array}{l}
C(K^{n+2}, q_0) \simeq C(K^n, -q_0) \otimes C(K^2, q_0) \\
\text{and } C(K^{n+2}, -q_0) \simeq C(K^n, q_0) \otimes C(K^2, -q_0) \\
\text{which give first 8-periodicity, etc ...} \\
(-q_0 \text{ is the opposite of the standard quadratic form } q_0)
\end{array}
$$
# III - THE FAMILIES OF P-ADIC CLIFFORD ALGEBRAS

## III-1. Equivalent classes of the p-adic quadratic forms

Let $a, b \in \mathbb{Q}_p^* = \mathbb{Q}_p \setminus \{0\}$. The Hilbert symbol $(a, b)$ is defined by $(a, b) = 1$ if the quadratic form of rank 3, $q'(x) = x_0^2 - ax_1^2 - bx_2^2$ is isotropic $(a, b) = -1$ otherwise.

$$ \text{N.B.} \quad (a, b) = 1 \quad \text{iff} \quad \left(\frac{a, b}{\mathbb{Q}_p}\right) \simeq \mathcal{M}(2, \mathbb{Q}_p). $$

Let $E$ be a vector space over $\mathbb{Q}_p$ of dimension $n$. Let us consider a regular quadratic form $q$ over $E$. If $(e_j)_{1 \le j \le n}$ is an orthogonal basis of $E$ and $a_j = q(e_j)$; then the discriminant $d(q)$ of $q$ is equal to $a_1 \dots a_n$ in the group $M_p = \mathbb{Q}_p^*/\mathbb{Q}_p^{\cdot 2}$. Let $\epsilon(q) = \prod_{1 \le i < j \le n} (a_i, a_j)$.

### Theorem A

(i) The p-adic regular quadratic forms $q$ and $q'$ of rank $n$ are equivalent iff $d(q) = d(q')$ and $\epsilon(q) = \epsilon(q')$.

(ii) Let $d \in M_p$ and $\epsilon = \pm 1$. There exists a p-adic regular quadratic form $q$ such that $d(q) = d$ and $\epsilon(q) = \epsilon$ iff

$$ (a) \quad n = 1 \quad \text{and} \quad \epsilon = 1 $$

$$ (b) \quad n = 2 \quad \text{and} \quad (d, \epsilon) \neq (-1, -1) $$

$$ (c) \quad n \ge 3 $$

### Proof: cf. [5]

According to that proof of Theorem A, one can give, explicitly, representatives of the equivalence classes of p-adic regular quadratic forms.

Let us recall that $M_2 = \{\pm 1, \pm 2, \pm 5, \pm 10\}$ and $M_p = \{1, p, \omega, \omega p\}$ if $p \neq 2$, where $\omega$ is a unit such that $\begin{pmatrix} \omega \\ p \end{pmatrix} = -1$; $\begin{pmatrix} - \\ p \end{pmatrix} = \text{the Legendre symbol.}$ Furthermore $-1 = 1$ in $M_p$ if $p \equiv 1 \pmod 4$ and $M_p = \{1, p, -1, -p\}$ if $p \equiv 3 \pmod 4$.

We are content ourself here, with the primes $p$ different from 2. Then a complete set of representatives of the equivalent classes of regular p-adic quadratic forms is obtained as follows.
(a) $n=1$

Then $q^a(x) = ax^2, a \in M_p$; and the Clifford algebras $C(Q_p, q^a)$ are isomorphic respectively to $Q_p \oplus Q_p$, $Q_p[\sqrt{p}], Q_p[\sqrt{\omega}]$ and $Q_p[\sqrt{\omega p}]$.

(b) $n=2$

Then we have over $\mathbb{Q}_p^2$ (with $\omega = -1$ if $p \equiv 3 \pmod{4}$)

$$
\begin{array}{ll}
q_0(x) = x_1^2 + x_2^2 & q_4(x) = p x_1^2 + \omega p x_2^2 \quad \text{if } p \equiv 1 \pmod{4} \\
q_1(x) = x_1^2 + p x_2^2 & q_4(x) = p x_1^2 + p x_2^2 \quad \text{if } p \equiv 3 \pmod{4} \\
q_2(x) = \omega x_1^2 + \omega p x_2^2 & q_5(x) = x_1^2 + \omega p x_2^2 \\
q_3(x) = x_1^2 + \omega x_2^2 & q_6(x) = p x_1^2 + \omega x_2^2
\end{array}
\quad \text{(resp.)}
$$

Furthermore $\epsilon(q_\ell) = 1$ if $\ell = 0, 1, 3, 5$ and $\epsilon(q_\ell) = -1$ if $\ell = 2, 4, 6$.

N.B : If $p=2$, then for $n=2$, one has

8 regular quadratic forms $q$ such that $\epsilon(q) = 1$

and 7 regular quadratic forms $q$ such that $\epsilon(q) = -1$.

(c) $n=3$

If $(e_1, e_2, e_3)$ is the canonical basis of $\mathbb{Q}_p^3$, then

$$
\bullet \quad q'_{\ell}(x) = q_{\ell}(x_1 e_1 + x_2 e_2) + x_3^2, \quad 0 \le \ell \le 6
$$
and

$$
\bullet \quad q'_{7}(x) = p x_{1}^{2} + \omega x_{2}^{2} + \omega p x_{3}^{2} = q_{6}(x_{1}e_{1} + x_{2}e_{2}) + \omega p x_{3}^{2} \quad \text{if } p \equiv 1 \pmod{4} \\
\phantom{\bullet \quad q'_{7}(x) = p x_{1}^{2} + \omega x_{2}^{2} + \omega p x_{3}^{2} = } \text{resp.}
$$

$$
\bullet \quad q'_{7}(x) = p x_{1}^{2} - x_{2}^{2} + p x_{3}^{2} = q_{6}(x_{1}e_{1} + x_{2}e_{2}) + p x_{3}^{2} \quad \text{if } p \equiv 3 \pmod{4}
$$
Furthermore $d(q'_{\ell}) = d(q_{\ell}), \epsilon(q'_{\ell}) = \epsilon(q_{\ell}), 0 \le \ell \le 6 \text{ and } d(q'_{7}) = -1, \epsilon(q'_{7}) = -1$.

(d) $n \ge 4$

Let $(e_j)_{1 \le j \le n}$ be the canonical basis of $\mathbb{Q}_p^n$, then

$$
\bullet \quad q_{\ell}''(x) = q_{\ell}(x_1 e_1 + x_2 e_2) + \sum_{j=3}^{n} x_j^2, \quad 0 \le \ell \le 6.
$$

In other words $q_{\ell}''(x) = q_{\ell}(x_1 e_1 + x_2 e_2) + q_0(\sum_{j=3}^{n} x_j e_j)$

$$
\text{i.e.}(\mathbb{Q}_p^n, q_\ell'') \cong (\mathbb{Q}_p^2, q_\ell) \perp (\mathbb{Q}_p^{n-2}, q_0), \quad 0 \le \ell \le 6
$$
and
$$
\bullet \quad q_7''(x) = q_7'\left(\sum_{j=1}^{3} x_j e_j\right) + \sum_{j=4}^{n} x_j^2 = q_7'\left(\sum_{j=1}^{3} x_j e_j\right) + q_0\left(\sum_{j=4}^{n} x_j e_j\right)
$$
i.e. $(\mathbb{Q}_p^n, q_7'') \simeq (\mathbb{Q}_p^3, q_7') \perp (\mathbb{Q}_p^{n-3}, q_0)$

N.B : $p=2$

If $n=3$, then the classes of regular quadratic forms have 15 representative forms $q'$ with $\epsilon(q') = 1$,resp.$\epsilon(q') = -1$ and $d(q') \neq -1$, obtained from corresponding representative quadratic forms of ranks 2 by adding the rank 1 form $x_3^2$. The other representative form is $q'_{15}(x) = -x_1^2 - x_2^2 - x_3^2$ with $\epsilon(q'_{15}) = -1$ and $d(q'_{15}) = -1$.

And if $n \ge 4$, one proceeds as above.

III - 2 The p-adic Clifford algebras

With the above notations, we have the following concrete propositions

**Proposition 3** : $p \neq 2$

(i) $C(\mathbb{Q}_p^2, q_\ell) \simeq M(2, \mathbb{Q}_p)$ if $\ell = 0, 1, 3, 5$.

(ii) $C(\mathbb{Q}_p^2, q_\ell) \simeq (\frac{p, \omega}{\mathbb{Q}_p}) = \mathbf{H}_p = \text{the } p\text{-adic quaternion field, if } \ell = 2, 4, 6.$

Proof

(i) Indeed, if $\ell = 0, 1, 3, 5$; then $\epsilon(q_\ell) = 1$. Therefore $C(\mathbb{Q}_p^2, q_\ell) \simeq M(2, \mathbb{Q}_p)$.

(ii) If $\ell = 2, 4, 6$ then the Clifford algebras $C(\mathbb{Q}_p^2, q_\ell)$ are isomorphic to the quaternion algebras with norm respectively,

$$
N_2(z) = x_0^2 - \omega x_1^2 - \omega p x_2^2 + \omega^2 p x_3^2 ;
$$

$$
N_4(z) = x_0^2 - p x_1^2 - \omega p x_2^2 + \omega p^2 x_3^2 \text{ if } p \equiv 1 \pmod{4};
$$

(resp. $N_4(z) = x_0^2 - p x_1^2 - p x_2^2 + p^2 x_3^2$ if $p \equiv 3 \pmod{4}$)

and $N_6(z) = x_0^2 - p x_1^2 - \omega x_2^2 + \omega p x_3^2$.

It is easily seen that these quadratic forms are anisotropic and equivalent. Therefore
$C(\mathbb{Q}_p^2, q_2) \simeq C(\mathbb{Q}_p^2, q_4) \simeq C(\mathbb{Q}_p^2, q_6) \simeq (\frac{p, \omega}{\mathbb{Q}_p}) = \mathbf{H}_p$ is a skew fied. Hence $\mathbf{H}_p$ is the unique
quaternion field over $\mathbb{Q}_p$ ( according isomorphism). This result obtained directly here is a
general result for local fields (cf. [3]).
**Proposition 4 :** $p \equiv 1 \pmod{4}$

The Clifford algebra $C(Q_p^3, q_l')$ is isomorphic to

(i) $M(2, Q_p) \oplus M(2, Q_p)$ if $\ell = 0$

(ii) $M(2, Q_p[\sqrt{p}])$ if $\ell = 1, 2$

(iii) $M(2, Q_p[\sqrt{\omega}])$ if $\ell = 3, 4$

(iv) $M(2, Q_p[\sqrt{\omega p}])$ if $\ell = 5, 6$

(v) $H_p \oplus H_p$ if $\ell = 7$

Similarly we have

**Proposition 4' :** $p \equiv 3 \pmod{4}$

The Clifford algebra $C(Q_p^3, q_l')$ is isomorphic to

(i) $M(2, Q_p[i])$ if $\ell = 0, 4$

(ii) $M(2, Q_p[\sqrt{-p}])$ if $\ell = 1, 2$

(iii) $M(2, Q_p) \oplus M(2, Q_p)$ if $\ell = 3$

(iv) $M(2, Q_p[\sqrt{p}])$ if $\ell = 5, 6$

(v) $H_p \oplus H_p$ if $\ell = 7$

**Proof of Propositions 4 and 4'**

Let us recall that if $(E, q)$ is a regular quadratic space over a field $K$ with $n = \dim E$ odd, then $C(E, q) \simeq Z \otimes C_+(E, q)$, where $Z$ is the centre of $C(E, q)$ and $C_+(E, q)$ the subalgebra of even elements. Furthermore, if $(e_1, \dots, e_n)$ is an orthogonal basis of $(E, q)$ then $u = e_1 \dots e_n$ is such that $u^2 = (-1)^{[\frac{n}{2}]}d(q)$ and $Z = K[u]$.

In particular for $n = 3$ and $q(x) = \alpha x_1^2 + \beta x_2^2 + \gamma x_3^2$, we have $e_1^2 = \alpha, e_2^2 = \beta, e_3^2 = \gamma$; $u^2 = -\alpha\beta\gamma = \delta \neq 0$ and $C_+(E, q) = \langle 1, e_1e_2, e_1e_3, e_2e_3 \rangle$ subspace generated by $1, \dots, e_2e_3$. Put $E_1 = e_1e_2, E_2 = e_1e_3, E_3 = -\alpha e_2e_3$, hence $C_+(E, q) = \langle 1, E_1, E_2, E_3 \rangle$ with $E_1^2 = -\alpha\beta, E_2^2 = -\alpha\gamma, E_1E_2 = E_3 = -E_2E_1$. Therefore $C_+(E, q) \simeq (\frac{-\alpha\beta, -\alpha\gamma}{K})$. Consequently (1) if $\delta \in K^{*2}$, then $Z \simeq K \oplus K$ and $C(E, q) \simeq (\frac{-\alpha\beta, -\alpha\gamma}{K}) \oplus (\frac{-\alpha\beta, -\alpha\gamma}{K})$ (2) if $\delta \notin K^{*2}$, then $Z = K[u]$ is a field and $C(E, q) \simeq (\frac{-\alpha\beta, -\alpha\gamma}{K[u]})$.

Applying these remarks to Propositions 4 and 4', one finds the desired isomorphisms. For example if $p \equiv 1 \pmod{4}$ and $\ell = 2$, then $\delta = -\omega^2p = (i\omega)^2p$ and $Z = Q_p[\sqrt{p}]$, hence
$C(\mathbb{Q}_p^3, q_2') \simeq \left(\frac{-\omega^2 p, -\omega}{\mathbb{Q}_p[\sqrt{p}]\sqrt{p}}\right) = \left(\frac{p, \omega}{\mathbb{Q}_p[\sqrt{p}]\sqrt{p}}\right) \simeq M(2, \mathbb{Q}_p[\sqrt{p}]) : \tilde{q}(v) = p x^2 + \omega y^2$ represents 1 over $\mathbb{Q}_p[\sqrt{p}]$. Also if $\ell = 7$, then $\delta = -p^2\omega^2 = (i\omega p)^2$, hence $Z \simeq \mathbb{Q}_p \oplus \mathbb{Q}_p$ and since $\left(\frac{-p\omega, -\omega p^2}{\mathbb{Q}_p}\right) \simeq \left(\frac{p\omega, \omega}{\mathbb{Q}_p}\right) \simeq \mathbb{H}_p$ we have $C(\mathbb{Q}_p^3, q_7') \simeq \mathbb{H}_p \oplus \mathbb{H}_p$.

In the case $p \equiv 3 \pmod{4}$, for example if $\ell = 0(\text{resp. } \ell = 3)$ we have $\delta = -1(\text{resp. } \ell = 1)$ and $Z \simeq \mathbb{Q}_p[i]$, $(\text{resp. } Z \simeq \mathbb{Q}_p \oplus \mathbb{Q}_p)$. Hence $C(\mathbb{Q}_p^3, q_0') \simeq \left(\frac{-1, -1}{\mathbb{Q}_p[\sqrt{-1}]\sqrt{-1}}\right) \simeq M(2, \mathbb{Q}_p[i])$, $(\text{resp. } C(\mathbb{Q}_p^3, q_3') \simeq \left(\frac{-1, -1}{\mathbb{Q}_p}\right) \oplus \left(\frac{-1, 1}{\mathbb{Q}_p}\right) \simeq M(2, \mathbb{Q}_p) \oplus M(2, \mathbb{Q}_p))$.

The other verifications are left to the reader.

**Lemma 2 :** $p \neq 2$
$C(\mathbb{Q}_p^4, q_7'') \simeq M(2, \mathbb{H}_p)$. $\square$

Indeed, since $q_7'' = p x_1^2 + \omega x_2^2 + \omega' p x_3^2 + x_4^2$ where $\omega' = \omega$ if $p \equiv 1 \pmod 4$ and $\omega = -1, \omega' = 1$ if $p \equiv 3 \pmod 4$; we have $C(\mathbb{Q}_p^4, q_7'') \simeq \left(\frac{p, \omega}{\mathbb{Q}_p}\right) \otimes_2 \left(\frac{\omega'p, 1}{\mathbb{Q}_p}\right) \simeq \mathbb{H}_p \otimes_2 M(2, \mathbb{Q}_p) \simeq M(2, \mathbb{H}_p)$.

**Theorem 4 :** $p \equiv 1 \pmod 4$ ; $n \ge 4$

1°) If $n = 2m$, then the Clifford algebra $C(\mathbb{Q}_p^n, q_\ell'')$ is isomorphic to

(i) $M(2^m, \mathbb{Q}_p)$ if $\ell = 0, 1, 3, 5$

(ii) $M(2^{m-1}, \mathbb{H}_p)$ if $\ell = 2, 4, 6, 7$

2°) If $n = 2m + 1$, then the Clifford algebra $C(\mathbb{Q}_p^n, q_\ell'')$ is isomorphic to

(i) $M(2^m, \mathbb{Q}_p) \oplus M(2^m, \mathbb{Q}_p)$ if $\ell = 0$

(ii) $M(2^m, \mathbb{Q}_p[\sqrt{\tau}])$ if $\ell = 1, 2, 3, 4, 5, 6$

with $\tau = p(\text{resp. } \omega, \text{resp. } \omega p)$ for $\ell = 1, 2$ $(\text{resp. } \ell = 3, 4$; resp. 5, 6).

(iii) $M(2^{m-1}, \mathbb{H}_p) \oplus M(2^{m-1}, \mathbb{H}_p)$ if $\ell = 7$

**Proof :**

1°) $n = 2m$

Notice that $C(\mathbb{Q}_p^n, q_\ell'') \simeq C(\mathbb{Q}_p^{n-2}, q_0) \otimes_2 C(\mathbb{Q}_p^2, q_\ell)$, $0 \le \ell \le 6$. But by Proposition 3, we have $C(\mathbb{Q}_p^2, q_\ell) \simeq M(2, \mathbb{Q}_p)$ if $\ell = 0, 1, 3, 5$ and $C(\mathbb{Q}_p^2, q_\ell) \simeq \mathbb{H}_p$ if $\ell =$
2, 4, 6. Since $C(Q_p^{n-2}, q_0) \simeq M(2^{m-1}, Q_p)$ by Theorem 1 - (i) -, we have $C(Q_p^n, q_\ell'') \simeq M(2^{m-1}, Q_p) \otimes_2 M(2, Q_p) \simeq M(2^m, Q_p)$ if $\ell = 0, 1, 3, 5$ and $C(Q_p^n, q_\ell'') \simeq M(2^{m-1}, Q_p) \otimes_2 H_p \simeq M(2^{m-1}, H_p)$ if $\ell = 2, 4, 6$.

For $\ell = 7$, applying Lemma 2 and Theorem 1 - (i) - we obtain $C(Q_p^n, q_\ell'') \simeq C(Q_p^{n-4}, q_0) \otimes_2 C(Q_p^4, q_\ell'') \simeq M(2^{m-2}, Q_p) \otimes_2 M(2, H_p) \simeq M(2^{m-1}, H_p)$.

$$2^{\circ)} \qquad \underline{n = 2m+1}$$

If $1 \le \ell \le 6$, then we have $C(Q_p^n, q_\ell'') \simeq C(Q_p^{n-3}, q_0) \otimes_2 C(Q_p^3, q_\ell') \simeq M(2^{m-1}, Q_p) \otimes_2 C(Q_p^3, q_\ell')$.

Applying Proposition 4, we obtain the isomorphism $C(Q_p^n, q_\ell'') \simeq M(2^m, Q_p[\sqrt{\tau}])$ as claimed.

The case $\ell = 0$ is Theorem 1 - (ii) -

If $\ell = 7$, then $C(Q_p^3, q_7') \simeq H_p \oplus H_p$ and $C(Q_p^n, q_7'') \simeq M(2^{m-1}, Q_p) \otimes_2 (H_p \oplus H_p) \simeq M(2^{m-1}, H_p) \oplus M(2^{m-1}, H_p)$.

**Theorem 5 :** $p \equiv 3(mod.4); n \ge 4$

The Clifford algebra $C(Q_p^n, q_\ell'')$ is isomorphic to the following matrix algebra or direct sum of two matrix algebras.

$$1^{\circ)} \qquad \underline{n = 4m}$$

(i) $M(2^{2m}, Q_p)$ if $\ell = 0, 1, 3, 5$

(ii) $M(2^{2m-1}, H_p)$ if $\ell = 2, 4, 6, 7$

$$2^{\circ)} \qquad \underline{n = 4m+1}$$

(i) $M(2^{2m}, Q_p) \oplus M(2^{2m}, Q_p)$ if $\ell = 0, 4$

(ii) $M(2^{2m}, Q_p[\sqrt{\tau}])$ if $\ell = 1, 2, 3, 5, 6, 7$

with $\tau = p$ (resp. $-1$, res. $-p$) for $\ell = 1, 2$ (resp. $\ell = 3, 7$, resp. $\ell = 5, 6$).

$$3^{\circ)} \qquad \underline{n = 4m+2}$$

(i) $M(2^{2m+1}, Q_p)$ if $\ell = 0, 1, 3, 5$

(ii) $M(2^{2m}, H_p)$ if $\ell = 2, 4, 6, 7$
$$4^{\circ)} \quad \underline{n = 4m+3}$$

$$(i) \quad M(2^{2m+1}, \mathbf{Q}_p) \oplus M(2^{2m+1}, \mathbf{Q}_p) \quad \text{if} \quad \ell = 3$$

$$(ii) \quad M(2^{2m+1}, \mathbf{Q}_p[\sqrt{\tau}]) \quad \text{if} \quad \ell = 0, 1, 2, 4, 5, 6, \\ \text{with } \tau = -1 \text{ (resp. } -p, \text{res. } p) \text{ for } \ell = 0, 4 \text{ (resp. } \ell = 1, 2, \text{resp. } \ell = 5, 6).$$

$$(iii) \quad M(2^{2m}, \mathbf{H}_p) \oplus M(2^{2m}, \mathbf{H}_p) \quad \text{if} \quad \ell = 7.$$

**Proof:**

$$1^{\circ)} \quad \underline{n = 4m}$$

As in Lemma 2, it is readily seen that $C(Q_p^n, q_\ell'') \simeq M(2^2, \mathbf{Q}_p)$ if $\ell = 0, 1, 3, 5$ and $C(Q_p^n, q_\ell'') \simeq M(2, \mathbf{H}_p)$ if $\ell = 2, 4, 6, 7$.

If $n = 4m, m \ge 2$, we have $C(Q_p^n, q_\ell'') \simeq C(Q_p^{n-4}, q_0) \otimes_2 C(Q_p^n, q_\ell'')$. But Theorem 3 - (i) - gives $C(Q_p^{n-4}, q_0) \simeq M(2^{2m-2}, \mathbf{Q}_p)$. Therefore $C(Q_p^n, q_\ell'') \simeq M(2^{2m-2}, \mathbf{Q}_p) \otimes_2 M(2^2, \mathbf{Q}_p) \simeq M(2^{2m}, \mathbf{Q}_p)$ if $\ell = 0, 1, 3, 5$ and $C(Q_p^n, q_\ell'') \simeq M(2^{2m-2}, \mathbf{Q}_p) \otimes_2 M(2^2, \mathbf{H}_p) \simeq M(2^{2m-1}, \mathbf{H}_p)$ if $\ell = 2, 4, 6, 7$.

$$2^{\circ)} \quad \underline{n = 4m+1}$$

With notations used in the proof of Propositions 4 and 4' we have $C(Q_p^n, q_\ell'') \simeq Z \otimes C_+(Q_p^n, q_\ell'')$ and $Z = \mathbf{Q}_p[u]$ where $u^2 = d(q_\ell'')$. Hence $Z$ is isomorphic to $\mathbf{Q}_p \oplus \mathbf{Q}_p$ if $\ell = 0, 4$; resp. $\mathbf{Q}_p[\sqrt{p}]$ if $\ell = 1, 2$; resp. $\mathbf{Q}_p[\sqrt{-1}]$ if $\ell = 3, 7$; resp. $\mathbf{Q}_p[\sqrt{-p}]$ if $\ell = 5, 6$. On the other hand $C_+(Q_p^n, q_\ell'') \simeq C_+(\mathbf{Q}_p \cdot x_n, x_n^2) \otimes C(Q_p^{n-1}, -q_\ell'') \simeq C(Q_p^{n-1}, -q_\ell'') \simeq M(2^{2m}, \mathbf{Q}_p)$. Hence $C(Q_p^n, q_\ell'') \simeq Z \otimes M(2^{2m}, \mathbf{Q}_p)$ which proves the isomorphisms.

$$3^{\circ)} \quad \underline{n = 4m+2}$$

Since $n - 2 = 4m$, we obtain $C(Q_p^n, q_\ell'') \simeq C(Q_p^{4m}, q_0) \otimes_2 C(Q_p^2, q_\ell)$.

By Theorem 2 - (i) - one has $C(Q_p^{4m}, q_0) \simeq M(2^{2m}, \mathbf{Q}_p)$ and by Proposition 3, $C(Q_p^2, q_\ell) \simeq M(2, \mathbf{Q}_p)$ if $\ell = 0, 1, 3, 5$ and $C(Q_p^2, q_\ell) \simeq \mathbf{H}_p$ if $\ell = 2, 4, 6$. It follows that $C(Q_p^n, q_\ell'') \simeq M(2^{2m+1}, \mathbf{Q}_p)$ if $\ell = 0, 1, 3, 5$ and $C(Q_p^n, q_\ell'') \simeq M(2^{2m}, \mathbf{H}_p)$ if $\ell = 2, 4, 6$.

For the case $\ell = 7$, since $n - 4 = 4(m - 1) + 2$ we have $C(Q_p^n, q_7'') \simeq C(Q_p^{n-4}, q_0) \otimes_2 C(Q_p^4, q_\ell'')$. By theorem 2 - (iii) -, $C(Q_p^{n-4}, q_0) \simeq M(2^{2m+1}, \mathbf{Q}_p)$ and by Lemma 2, $C(Q_p^4, q_7'') \simeq M(2, \mathbf{H}_p)$. Hence $C(Q_p^n, q_7'') \simeq M(2^{2m}, \mathbf{H}_p)$.

Notice that in $1^{\circ)}$ and $3^{\circ)}$ the exponent of 2 is $\frac{n}{2}$.
$$4^{\circ}) \qquad \underline{n = 4m+3}$$

Here, $n-3 = 4m$ and $C(Q_p^n, q_\ell'') \simeq C(Q_p^{4m}, q_0) \otimes_2 C(Q_p^3, q_\ell')$. But $C(Q_p^{4m}, q_0) \simeq M(2^{2m}, Q_p)$ and by Proposition 4', $C(Q_p^3, q_\ell')$ is isomorphic to $M(2, Q_p) \oplus M(2, Q_p)$ if $\ell = 3$, resp. $H_p \oplus H_p$ if $\ell = 7$, resp. $M(2, Q_p[\sqrt{\tau}])$ if $\ell = 0, 1, 2, 4, 5, 6$ with $\tau = -1$ for $\ell = 0, 4$; $\tau = -p$ for $1, 2$ and $\tau = p$ for $\ell = 5, 6$.

Taking tensor product we obtain the desired isomorphisms.

**Remark :**

As for $C(Q_p^n, q_0)$, for the other Clifford algebras $C(Q_p^n, q_\ell'')$ we have 2-periodicity when $p \equiv 1 \pmod 4$ and 4-periodicity when $p \equiv 3 \pmod 4$.

**N.B.** When $p=2$, in the same way one can give as obove the table of the 2-adic Clifford algebras.

## REFERENCES

[1] R. DEHEUVELS, *Formes quadratiques et groupes classiques*, P.U.F. - Paris - 1981.

[2] B. DIARRA, *Remarques sur les matrices orthogonales (resp. symétriques) à coefficients p-adiques*, Ann. Sci. "Blaise Pascal", Clermont II, sér. Math., Fasc. 26è - 1990 - p. 31-50.

[3] T.Y. LAM, *The algebraic theory of quadratic forms*, W.C. Benjamin, Inc. - Reading, Massachusetts - 1973.

[4] P. RIBENBOIM, *L'arithmétique des corps*, Hermann - Paris - 1972.

[5] J.P. SERRE, *Cours d'arithmétique*, P.U.F. - Paris - 1970.

Université Blaise Pascal, Complexe Scientifique des Cézeaux, Laboratoire de Mathématiques Pures, 63177 AUBIERE Cedex.