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# 数学代写|线性代数代写Linear algebra代考|Witt’s Cancellation Theorem

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## 数学代写|线性代数代写Linear algebra代考|Witt’s Cancellation Theorem

We now come to one of the major results of orthogonal geometry, due to Witt. To wit:

Theorem 11.27 (Witt’s cancellation theorem) Let $\mathrm{V}$ be a nonsingular orthogonal geometry over a field $F$, with $\operatorname{char}(\mathrm{F}) \neq 2$. Suppose that
$$\mathrm{V}=\mathrm{S} \oplus \mathrm{S}^{\perp}=\mathrm{T} \oplus \mathrm{T}^{\perp}$$
where $\mathrm{S}$ and $\mathrm{T}$ are nonsingular. Then
$$\mathrm{S} \approx \mathrm{T} \Rightarrow \mathrm{S}^{\perp} \approx \mathrm{T}^{\perp}$$
Proof. Let $\tau: \mathrm{S} \rightarrow \mathrm{T}$ be an isometry. We proceed by induction on $\operatorname{dim}(\mathrm{S})$. Suppose first that $\operatorname{dim}(\mathrm{S})=1$, and that $\mathrm{S}=\operatorname{span}{\mathbf{s}}$. Then $\mathrm{T}=\operatorname{span}{\tau(\mathbf{s})} \quad$ and $\quad\langle\tau(\mathbf{s}), \tau(\mathbf{s})\rangle=\langle\mathbf{s}, \mathbf{s}\rangle . \quad$ According to Theorem 11.25, there is a symmetry $\sigma$ for which $\sigma(\mathbf{s})=\epsilon \tau(\mathbf{s})$, where $\epsilon= \pm 1$. Hence, $\sigma$ is an isometry of $\mathrm{V}$ for which $\sigma(\mathrm{S})=\mathrm{T}$. It follows that
$$\mathbf{x} \in \mathrm{S}^{\perp} \Leftrightarrow\langle\mathbf{x}, \mathbf{s}\rangle=0 \Leftrightarrow\langle\sigma(\mathbf{x}), \sigma(\mathbf{s})\rangle=0 \Leftrightarrow\langle\sigma(\mathbf{x}), \tau(\mathbf{s})\rangle=0 \Leftrightarrow \sigma(\mathbf{x}) \in \mathrm{T}^{\perp}$$
and so the restriction $\left.\sigma\right|_{\mathrm{S}^{\perp}}$ is an isometry from $\mathrm{S}^{\perp}$ to $\mathrm{T}^{\perp}$, which shows that $\mathrm{S}^{\perp} \approx \mathrm{T}^{\perp}$.

Now suppose the theorem is true for $\operatorname{dim}(S)<\mathrm{k}$, and let $\operatorname{dim}(\mathrm{S})=\mathrm{k}$. Let $\tau: \mathrm{S} \rightarrow \mathrm{T}$ be an isometry. Since $\mathrm{S}$ is nonsingular, we can choose a nonnull vector $s \in S$, and write
$$\mathrm{S}=\operatorname{span}{\mathbf{s}} \oplus$$
where $U$ is nonsingular. Moreover,
$$\mathrm{T}=\operatorname{span}{\tau(\mathrm{s})}(1) \tau(\mathrm{U})$$
Thus,
$$\mathbf{V}=\operatorname{span}{\mathbf{s}}(1) \mathrm{U}(1) \mathrm{S}^{\perp}$$
and
$$\mathrm{V}=\operatorname{span}{\tau(\mathbf{s})} \text { (1) } \tau(\mathrm{U}) \oplus \mathrm{T}^{\perp}$$

## 数学代写|线性代数代写Linear algebra代考|Maximal Hyperbolic Subspaces

Since a hyperbolic space is completely determined (up to isometry) by its dimension, it is of interest to know something about maximal hyperbolic subspaces of a nonsingular orthogonal geometry. (In the symplectic case, if $\mathrm{V}$ is nonsingular, then $\mathrm{V}$ is hyperbolic.) We will denote a hyperbolic space by $\mathfrak{K}^6$, and a hyperbolic space of dimension $2 \mathrm{k}$ by $36_{2 \mathrm{k}}$, thus
$$\mathfrak{H}{2 \mathrm{k}}=\mathrm{H}_1 \oplus \cdots \oplus \mathrm{H}{\mathrm{k}}$$
where each $\mathrm{H}_{\mathrm{i}}$ is a hyperbolic plane.
Note that a two-dimensional space is a hyperbolic plane if and only if it is nonsingular and contains a null vector. (We assume that $\operatorname{char}(\mathrm{F}) \neq 2$.)

Suppose that $\mathrm{V}$ is isotropic, that is, $\mathrm{V}$ contains a null vector. If $\mathrm{U}{\mathrm{k}}$ is a nonempty null subspace of $\mathrm{V}$ of dimension $\mathrm{k}$, then $\operatorname{Rad}\left(\mathrm{U}{\mathrm{k}}\right)=\mathrm{U}_{\mathrm{k}}$, and so we may apply Theorem 11.28 , to deduce that

$$\mathrm{U}{\mathrm{k}} \subset \mathcal{F}{2 \mathrm{k}}=\mathrm{H}1 \oplus \cdots \oplus \mathrm{H}{\mathrm{k}}$$
where $H_i$ is generated by a hyperbolic pair $\left(\mathbf{x}{\mathrm{i}}, \mathbf{y}{\mathrm{i}}\right)$. Thus, any null subspace $\mathrm{U}{\mathbf{k}}$ is contained in a hyperbolic space $3 \mathscr{6}{2 \mathbf{k}}$ with $\operatorname{dim}\left(\mathfrak{F}{2 \mathbf{k}}\right)=2 \operatorname{dim}\left(\mathrm{U}{\mathbf{k}}\right)$. This implies that the Witt index of $\mathrm{V}$ is at most $\operatorname{dim}(\mathrm{V}) / 2$.
On the other hand, suppose that
$$\mathcal{F}{2 \mathrm{k}}=\mathrm{H}_1 \oplus \cdots \oplus \mathrm{H}{\mathbf{k}}$$
is a hyperbolic space in $\mathrm{V}$, and that $\mathrm{H}{\mathrm{i}}$ is generated by the hyperbolic pair $\left(\mathbf{x}{\mathbf{i}}, \mathbf{y}{\mathrm{i}}\right)$. Then the set $\mathscr{B}=\left{\mathbf{x}_1, \ldots, \mathbf{x}{\mathbf{k}}\right}$ is independent, for if
$$\mathrm{r}1 \mathbf{x}_1+\cdots+\mathrm{r}{\mathrm{k}} \mathbf{x}{\mathbf{k}}=0$$ it follows that $$0=\left\langle r_1 \mathbf{x}_1+\cdots+r_k \mathbf{x}{\mathbf{k}}, \mathbf{y}{\mathbf{j}}\right\rangle=\mathrm{r}{\mathbf{j}}\left\langle\mathbf{x}{\mathbf{j}}, \mathbf{y}{\mathbf{j}}\right\rangle=\mathrm{r}{\mathbf{j}}$$ for all $\mathrm{j}$. Moreover, since $\left\langle\mathbf{x}{\mathrm{j}}, \mathbf{x}{\mathrm{j}}\right\rangle=0$ for all $\mathrm{i}, \mathrm{j}$, the subspace $\mathrm{U}{\mathrm{k}}=$ $\operatorname{span}{\mathscr{B}}$ is a k-dimensional null space. Thus, any hyperbolic space ${ }^3{ }{2 k}$ in $\mathrm{V}$ contains a null space $\mathrm{U}{\mathbf{k}}$. This implies that if $\mathfrak{H}{2 \mathrm{~m}}$ is a maximal hyperbolic subspace of $\mathrm{V}$, then $\mathrm{m} \leq w(\mathrm{~V})$. Furthermore, since $\mathrm{V}$ must contain a null space $\mathrm{U}{w(\mathrm{~V})}$, it must also contain a hyperbolic subspace of dimension $2 w(\mathrm{~V})$. In other words, the maximum dimension of a hyperbolic subspace of $\mathrm{V}$ is $2 w(\mathrm{~V})$.

## 数学代写|线性代数代写Linear algebra代考|Witt’s Cancellation Theorem

$$\mathrm{V}=\mathrm{S} \oplus \mathrm{S}^{\perp}=\mathrm{T} \oplus \mathrm{T}^{\perp}$$

$$\mathrm{S} \approx \mathrm{T} \Rightarrow \mathrm{S}^{\perp} \approx \mathrm{T}^{\perp}$$

$$\mathbf{x} \in \mathrm{S}^{\perp} \Leftrightarrow\langle\mathbf{x}, \mathbf{s}\rangle=0 \Leftrightarrow\langle\sigma(\mathbf{x}), \sigma(\mathbf{s})\rangle=0 \Leftrightarrow\langle\sigma(\mathbf{x}), \tau(\mathbf{s})\rangle=0 \Leftrightarrow \sigma(\mathbf{x}) \in \mathrm{T}^{\perp}$$

$$\mathrm{S}=\operatorname{span} \mathbf{s} \oplus$$

$$\mathrm{T}=\operatorname{span} \tau(\mathrm{s})(1) \tau(\mathrm{U})$$

$$\mathbf{V}=\operatorname{span} s(1) \mathrm{U}(1) \mathrm{S}^{\perp}$$

$$\mathrm{V}=\operatorname{span} \tau(\mathbf{s})(1) \tau(\mathrm{U}) \oplus \mathrm{T}^{\perp}$$

## 数学代写|线性代数代写Linear algebra代考|Maximal Hyperbolic Subspaces

$$\mathfrak{H} 2 \mathrm{k}=\mathrm{H}1 \oplus \cdots \oplus \mathrm{Hk}$$ 毎个 $\mathrm{H}{\mathrm{i}}$ 是一个双曲平面。

$$\mathrm{Uk} \subset \mathcal{F} 2 \mathrm{k}=\mathrm{H} 1 \oplus \cdots \oplus \mathrm{Hk}$$

$$\mathcal{F} 2 \mathrm{k}=\mathrm{H}_1 \oplus \cdots \oplus \mathrm{Hk}$$

$$\mathrm{r1} \mathbf{x}_1+\cdots+\mathrm{rkxk}=0$$

$$0=\left\langle r_1 \mathbf{x}_1+\cdots+r_k \mathbf{x} \mathbf{k}, \mathbf{y} \mathbf{j}\right\rangle=\mathbf{r} \mathbf{j}\langle\mathbf{x} \mathbf{j}, \mathbf{y} \mathbf{j}\rangle=\mathrm{r} \mathbf{j}$$

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