Tangential and Anisotropic Bases in Quadratic Spaces

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Tangential and Anisotropic Bases in Quadratic Spaces

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Alexander Leones,Pedro Hurtado,

John Moreno,

Adolfo Pimienta^*

Alexander Leones,Pedro Hurtado,

John Moreno,

Adolfo Pimienta^*

This version is not peer-reviewed

Submitted:

12 July 2024

Posted:

15 July 2024

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Abstract

This paper presents an explicit method for constructing tangential and anisotropic bases in quadratic spaces. We will also provide three theorems that allow us to explicitly calculate vectors that are orthogonal or tangent to given vectors or that provide necessary and sufficient conditions for the existence of such vectors. A classical theorem on quadratic space states that every finite-dimensional quadratic space has an orthogonal basis. A result that seems to be unanswered is under what conditions a finite-dimensional quadratic space has a basis consisting of isotropic vectors. A second problem is whether or not every finite-dimensional quadratic space has a basis consisting of mutually tangent vectors, where two vectors v,w are tangent if (u·v)2=(u·u)(v·v). In this paper, we attack these two problems.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

MSC: 11E88; 11E04; 15A63

1. Introduction

A detailed and basic study to understand quadratic forms appears in[4,5].The anisotropic part of a quadratic form over a field has been studied by P. Koprowski and B. Rothkegel in [3]. On the other hand, the algorithms that allow us to understand the anisotropic part of a quadratic form are systematically presented in the works of P. Koprowski and A. Czogała (see [1,2]). If V is a quadratic space, that is, a vector space with inner product “·”, the equation

u \cdot v = 0

, with

u, v \in V

, is usually interpreted as the orthogonality of the vectors u and v there are numerous geometric examples that justify this point of view. However, little mention is made of the condition

{(u \cdot v)}^{2} = (u \cdot u) (v \cdot v)

. In the present work, we propose to convince the reader, with several examples, that this ĺast condition expresses the fact that the vectors u and v are tangents, prior identification of vectors with everyday geometric entities. We will also provide three theorems that allow us to explicitly calculate vectors that are orthogonal or tangent to given vectors or that provide necessary and sufficient conditions for the existence of such vectors.

2. Preliminary Facts

For the sake of simplicity and convenience to the reader, we will begin this section with some basic and necessary concepts for understanding this work.

Definition 1.

Let V an vector space over an abstract field

F

, an inner product over V is a map

φ : V \times V \to F

which satisfies the following properties:

(i): $φ (u_{1} + u_{2}, v) = φ (u_{1}, v) + φ (u_{2}, v)$ , $u_{1}, u_{2}, v \in V$ ;
(ii): $φ (λ u, v) = λ φ (u, v)$ , $λ \in F$ , $u, v \in V$ ;
(iii): $φ (u, v) = φ (v, u)$ , $u, v \in V$ .

Usually write

u \cdot v

instead of

φ (u, v)

. Two vectors

u, v \in V

are called orthogonal if

u \cdot v = 0

and tangent if

{(u \cdot v)}^{2} = (u \cdot u) (v \cdot v)

An quadratic space V is a vector space over a field

F

endowed with an inner product “·”.

We say that V is regular, if

u \cdot v = 0

for every

u \in V

implies that

v = 0

Definition 2.

Let

u \neq 0

be a vector in the quadratic space V, we call u isotropic if

u \cdot u = 0

and anisotropic if

u \cdot u \neq 0

A subespace S of V is totally isotropic if for every pair

u, v \in S

, we have

u \cdot v = 0

Definition 3.

Let V be a quadratic space of finite dimension n. The index of V (ind V) is the maximum dimension of a totally isotropic subspace of V. So, ind V=0 if and only if V is anisotropic.

The Grammian of a basis B

= {v_{1}, v_{2}, \dots, v_{n}}

of V is the determinant

| v_{i} \cdot v_{j} |

. We shall write GramB

= | v_{i} \cdot v_{j} |

We will have the opportunity to use, the following results, the tests can be consulted in detail in each reference.

Theorem 1

(See [5], 42.1). Every finite-dimensional quadratic space has an orthogonal basis.

Theorem 2

(See [5], 42.2-42.4). The following assertions are equivalent in a finite-dimensional quadratic space V:

V es regular;
V has an orthogonal anisotropic basis;
For every basis B of V, GramB $\neq 0$ .
There exists a basis B of V such that GramB $\neq 0$ .

For each subset

A \subset V

, we define:

A^{⊥} = {u \in V : a \cdot u = 0 \forall a \in A} .

(1)

Note that

A^{⊥}

is a subspace of V.

Theorem 3

(See [5], 42.6). If U is an arbitrary subspace of V, we have:

dim U + dim U^{⊥} = dim V + dim U \cap V^{⊥} .

Corollary 1.

If V is regular, for every subspace of V, we have:

dim U + dim U^{⊥} = dim V and U = U^{⊥ ⊥} .

3. Tangency and Orthogonality

In this section, we will study vectors tangent or orthogonal to a fixed family of vectors.

Let V be a vector space over a field

F

. A vectorial determinant is an expresión of the type:

v = |\begin{matrix} v_{1} & v_{2} & \dots & v_{n} \\ a_{11} & a_{12} & \dots & a_{1 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n - 1, 1} & a_{n - 1, 2} & \dots & a_{n - 1, n} \end{matrix}|

(2)

where

v_{1}, v_{2}, \dots, v_{n} \in V

and each

a_{i j} \in F

, for

i \in {1, 2, \dots, n - 1}

and

j \in {1, 2, \dots, n}

. The development of this determinant is always on the first row and therefore

v = λ_{1} v_{1} + λ_{2} v_{2} + \dots + λ_{n} v_{n}

where

λ_{i}

is the cofactor of

a_{0 i}

in the matrix:

v = (\begin{matrix} a_{01} & a_{02} & \dots & a_{0 n} \\ a_{11} & a_{12} & \dots & a_{1 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n - 1, 1} & a_{n - 1, 2} & \dots & a_{n - 1, n} \end{matrix})

where

a_{o i} = 1

for each

i = 1, \dots, n

Theorem 4.

Let V a regular quadratic space on finite dimension n over a field

F

; let B

= {w_{1}, w_{2} \dots, w_{n}}

be a basis of V and let

{v_{1}, v_{2} \dots, v_{n - 1}}

be a collection of linearly independent vectors of V. Define

v \in V

by a means of the vectorial determinant:

v = |\begin{matrix} w_{1} & w_{2} & \dots & w_{n} \\ w_{1} \cdot v_{1} & w_{2} \cdot v_{1} & \dots & w_{n} \cdot v_{1} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ w_{1} \cdot v_{n - 1} & w_{2} \cdot v_{n - 1} & \dots & w_{n} \cdot v_{n - 1} \end{matrix}|

Then

v \neq \bar{0}

v \cdot v_{i} = 0

for each

i = 1, \dots, n - 1

and

v \cdot v = det (w_{i} \cdot w_{j}) det (v_{i} \cdot v_{j})

An immediate corollary of this theorem states that the vectors

{v_{1}, v_{2} \dots, v_{n - 1}}

are orthogonal to an anisotropic vector if and only if

det (v_{i} \cdot v_{j}) \neq 0

Theorem 5

(See [4,5]). Let V be a regular quadratic space on finite dimension n over a field

F

. Let B

= {v_{1}, v_{2} \dots, v_{n}}

be a basis of V and let

α_{1}, α_{2} \dots, α_{n - 1} \in F

be arbitrary scalars. Consider the vector:

v = |\begin{matrix} v_{1} & v_{2} & \dots & v_{n} & 0 \\ v_{1} \cdot v_{1} & v_{1} \cdot v_{2} & \dots & v_{1} \cdot v_{n} & α_{1} \\ ⋮ & ⋮ & ⋱ & ⋮ & ⋮ \\ v_{n - 1} \cdot v_{1} & v_{n - 1} \cdot v_{2} & \dots & v_{n - 1} \cdot v_{n} & α_{n - 1} \\ v_{n} \cdot v_{1} & v_{n} \cdot v_{2} & \dots & v_{n} \cdot v_{n} & μ \end{matrix}|

where μ is a root of the quadratic polynomial.

p (X) = |\begin{matrix} α_{1} & α_{2} & \dots & α_{n - 1} & X & 1 \\ v_{1} \cdot v_{1} & v_{1} \cdot v_{2} & \dots & v_{1} \cdot v_{n - 1} & v_{1} \cdot v_{n} & α_{1} \\ ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ v_{n - 1} \cdot v_{1} & v_{n - 1} \cdot v_{2} & \dots & v_{n - 1} \cdot v_{n - 1} & v_{n - 1} \cdot v_{n} & α_{n - 1} \\ v_{n} \cdot v_{1} & v_{n} \cdot v_{2} & \dots & v_{n} \cdot v_{n - 1} & v_{n} \cdot v_{n} & X \end{matrix}| .

Let

Δ = det (v_{i} \cdot v_{j}) =

GramB. Then

v \cdot v_{i} = {(- 1)}^{n - 1} α_{i} Δ

for each

i = 1, \dots, n - 1

v \cdot v_{n} = {(- 1)}^{n - 1} μ Δ

and

v \cdot v = Δ^{2}

Proof.

\begin{matrix} v \cdot v_{i} & = |\begin{matrix} v_{1} \cdot v_{i} & \dots & v_{n} \cdot v_{i} & 0 \\ v_{1} \cdot v_{1} & \dots & v_{1} \cdot v_{n} & α_{1} \\ ⋮ & ⋱ & ⋮ & ⋮ \\ v_{n - 1} \cdot v_{1} & \dots & v_{n - 1} \cdot v_{n} & α_{n - 1} \\ v_{n} \cdot v_{1} & \dots & v_{n} \cdot v_{n} & μ \end{matrix}| = |\begin{matrix} 0 & \dots & 0 & - α_{i} \\ v_{1} \cdot v_{1} & \dots & v_{1} \cdot v_{n} & α_{1} \\ ⋮ & ⋱ & ⋮ & ⋮ \\ v_{n - 1} \cdot v_{1} & \dots & v_{n - 1} \cdot v_{n} & α_{n - 1} \\ v_{n} \cdot v_{1} & \dots & v_{n} \cdot v_{n} & μ \end{matrix}| \\ = {(- 1)}^{n - 1} α_{i} det (v_{i} \cdot v_{j}) = {(- 1)}^{n - 1} α_{i} Δ . \end{matrix}

In a similar way, we prove that

v \cdot v_{n} = {(- 1)}^{n - 1} μ Δ

. On the other hand,

\begin{matrix} v \cdot v & = |\begin{matrix} v \cdot v_{1} & \dots & v \cdot v_{n} & 0 \\ v_{1} \cdot v_{1} & \dots & v_{1} \cdot v_{n} & α_{1} \\ ⋮ & ⋱ & ⋮ & ⋮ \\ v_{n - 1} \cdot v_{1} & \dots & v_{n - 1} \cdot v_{n} & α_{n - 1} \\ v_{n} \cdot v_{1} & \dots & v_{n} \cdot v_{n} & μ \end{matrix}| \\ = {(- 1)}^{n - 1} Δ |\begin{matrix} α_{1} & \dots & α_{n - 1} & μ & 0 \\ v_{1} \cdot v_{1} & \dots & v_{1} \cdot v_{n - 1} & v_{1} \cdot v_{n} & α_{1} \\ ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ v_{n - 1} \cdot v_{1} & \dots & v_{n - 1} \cdot v_{n - 1} & v_{n - 1} \cdot v_{n} & α_{n - 1} \\ v_{n} \cdot v_{1} & \dots & v_{n} \cdot v_{n - 1} & v_{n} \cdot v_{n} & μ \end{matrix}| \\ = {(- 1)}^{n - 1} Δ [p (μ) + {(- 1)}^{n - 1} Δ] = {(- 1)}^{n - 1} Δ [0 + {(- 1)}^{n - 1} Δ] = Δ^{2} . \end{matrix}

□

From the previous theorem, we can obtain the following two corollaries.

Corollary 2.

In case

v_{n} \cdot v_{i} = 0

for each

1, 2, \dots, n - 1

and if D is the non-singular matrix:

D = (\begin{matrix} v_{1} \cdot v_{1} & \dots & v_{1} \cdot v_{n - 1} \\ ⋮ & ⋱ & ⋮ \\ v_{n - 1} \cdot v_{1} & \dots & v_{n - 1} \cdot v_{n - 1} \end{matrix})

(3)

and

B = (α_{1}, \dots, α_{n - 1})

, then the roots of

p (X)

are the numbers:

μ = \pm \sqrt{(1 - B D^{- 1} B^{T}) (v_{n} \cdot v_{n})}

(These numbers are in

F

if and only if there exists a scalar

λ \in F

such that

\sqrt{(1 - B D^{- 1} B^{T}) (v_{n} \cdot v_{n})} = λ^{2}

Proof.

If we develop the determinant giving

p (X)

through the last row, we obtain:

p (X) = {(- 1)}^{n - 1} X^{2} | D | - {(- 1)}^{n - 1} (v_{n} \cdot v_{n}) S

where:

S = |\begin{matrix} α_{1} & α_{2} & \dots & α_{n - 1} & 1 \\ v_{1} \cdot v_{1} & v_{1} \cdot v_{2} & \dots & v_{1} \cdot v_{n - 1} & α_{1} \\ ⋮ & ⋮ & ⋱ & ⋮ & ⋮ \\ v_{n - 1} \cdot v_{1} & v_{n - 1} \cdot v_{2} & \dots & v_{n - 1} \cdot v_{n - 1} & α_{n - 1} \end{matrix}| .

Since the matrix D is invertible, we can find a vector

Λ = (α_{1}, α_{2}, \dots, α_{n - 1})

such that

Λ D = B

. Subtract from the first row os S the linear combination

α_{1} S_{2} + \dots + α_{n - 1} S_{n}

where

S_{2}, \dots, S_{n}

denote the second, third, …, n-th row of S. We obtain then:

S = |\begin{matrix} 0 & \dots & 0 & 1 - Λ B^{T} \\ v_{1} \cdot v_{1} & \dots & v_{1} \cdot v_{n - 1} & α_{1} \\ ⋮ & ⋱ & ⋮ & ⋮ \\ v_{n - 1} \cdot v_{1} & \dots & v_{n - 1} \cdot v_{n - 1} & α_{n - 1} \end{matrix}| = {(- 1)}^{n - 1} | D | (1 - Λ B^{T}) .

Replacing

Λ

B D^{- 1}

, we finally obtain:

S = {(- 1)}^{n - 1} | D | (1 - B D^{- 1} B^{T}) .

Therefore:

p (X) = {(- 1)}^{n - 1} | D | [X^{2} - (1 - B D^{- 1} B^{T}) (v_{n} \cdot v_{n})]

and the roots of

p (X)

are the numbers

μ = \pm \sqrt{(1 - B D^{- 1} B^{T}) (v_{n} \cdot v_{n})}

. In this case, the vector v in Theorem 5 is:

v = {(- 1)}^{n - 1} | D | μ v_{n} - (v_{n} \cdot v_{n}) |\begin{matrix} v_{1} & \dots & v_{n - 1} & 0 \\ v_{1} \cdot v_{1} & \dots & v_{1} \cdot v_{n - 1} & α_{1} \\ ⋮ & ⋱ & ⋮ & ⋮ \\ v_{n - 1} \cdot v_{1} & \dots & v_{n - 1} \cdot v_{n - 1} & α_{n - 1} \end{matrix}| .

□

Corollary 3.

α_{i}^{2} = v_{i} \cdot v_{i}

for each

i = 1, 2, \dots, n - 1

, there exist

2^{n - 1}

possibilities for the sequence

α_{1}, \dots, α_{n - 1}

. If we additionally suppose that

v_{i} \cdot v_{i} = 1

for

i = 1, \dots, n - 1

we conclude that there exist at most

2^{n - 1}

vectors that are simultaneously tangent to

v_{1}, v_{2}, \dots, v_{n - 1}

. If

K = R

and we fix

B = (α_{1}, \dots, α_{n - 1})

where

α_{i}^{2} = v_{i} \cdot v_{i}

for each

i = 1, \dots, n - 1

, then there exist two, one, or none linearly independent vectors tangent to

v_{1}, \dots, v_{n - 1}

according to

(1 - B D^{- 1} B^{T}) > 0

;

B D^{- 1} B^{T} = 1

(1 - B D^{- 1} B^{T}) < 0

. These vectors may be written in the form:

v = \pm | D | \sqrt{\frac{1 - B D^{- 1} B^{T}}{v_{n} \cdot v_{n}}} v_{n} |\begin{matrix} v_{1} & \dots & v_{n - 1} & 0 \\ v_{1} \cdot v_{1} & \dots & v_{1} \cdot v_{n - 1} & α_{1} \\ ⋮ & ⋱ & ⋮ & ⋮ \\ v_{n - 1} \cdot v_{1} & \dots & v_{n - 1} \cdot v_{n - 1} & α_{n - 1} \end{matrix}| .

4. Tangential and Anisotropic Bases

Let us remember that in a quadratic space V of dimension n a base B

= {v_{1}, v_{2}, \dots, v_{n}}

is unitary (resp. antiunitary) if

v_{i} \cdot v_{j} = 1

(resp.

v_{i} \cdot v_{j} = - 1

) for each

i \in {1, 2, \dots, n}

Theorem 6.

K = R

, V is regular and

dim V = 2 ind V + 1

, then V has a tangential basis which is unitary or antiunitary.

Proof.

We can write V as following:

V = 〈 v_{1}, v_{1}^{'} 〉 ⊥ 〈 v_{2}, v_{2}^{'} 〉 ⊥ \dots ⊥ 〈 v_{s}, v_{s}^{'} 〉 ⊥ 〈 v 〉

, where

ind V = s

v_{i} \cdot v_{i} = v_{i}^{'} \cdot v_{i}^{'} = 1

for each

i = 1, 2, \dots, s

and

v \cdot v \neq 0

. Since

K = R

, we can suppose, without loss of generality, that

v \cdot v = 1

v \cdot v = - 1

We, define

w_{i} = v_{i} + v

w_{i}^{'} = 2 v^{'} - \frac{v}{v \cdot v}

w_{0} = v

. We have then

w_{i} \cdot w_{i} = v \cdot v = w_{i}^{'} \cdot w_{i}^{'}

. Also,

w_{i} \cdot w_{i}^{'} = (v_{i} + v) \cdot (2 v_{i}^{'} - \frac{v}{v \cdot v}) = 2 - 1 = 1

for each

i = 1, 2, \dots, s

and

w_{i} \cdot w_{j}^{'} = w_{i}^{'} \cdot w_{j} = - 1

for

i \neq j

w_{i} \cdot v = (v_{i} + v) \cdot v = v \cdot v

w_{i}^{'} \cdot v = (2 v_{i}^{'} - \frac{v}{v \cdot v}) \cdot v = - 1

. Then,

{w_{1}, w_{1}^{'}, \dots, w_{s}, w_{s}^{'}, v}

is tangential basis of V which is unitary or antiunitary depending if

v \cdot v = 1

v \cdot v = - 1

. □

Before generalizing this theorem, we need the next lemma.

Lemma 1.

Let

b \in R

b > 1

and let

V = 〈 u_{1} 〉 ⊥ 〈 u 〉 ⊥ \dots ⊥ 〈 v_{n} 〉

be an anisotropic quadratic space over

R

. We suppose that

u_{i} \cdot u_{i} = 1

for each

i \in {1, \dots, n}

(positive case) or

u_{i} \cdot u_{i} = - 1

for each

i \in {1, \dots, n}

(negative case). Let

u_{1}^{'}, u_{2}^{'}, \dots, u_{n - 1}^{'}

be linearly independent vectors in V such that

u_{i}^{'} . \cdot u_{j}^{'} = b

for

i \neq j

and each

u_{i}^{'} \cdot u_{i}^{'} = b^{2}

, in the positive case, or

u_{i}^{'} \cdot u_{j}^{'} = - b

for

i \neq j

and

u_{i}^{'} \cdot u_{i}^{'} = - b^{2}

, in the negative case. Then there exists a vector

u^{'} \in V

such that

u_{1}^{'}, \dots, u_{n - 1}^{'}, u^{'}

are linearly independent and

u^{'} \cdot u_{i}^{'} = b

(

u^{'} \cdot u^{'} = - b^{2}

) in the positive (negative) case.

Proof.

Let

w \in V

such that

w \cdot u_{i}^{'} = 0

for each

i \in {1, 2, \dots, n - 1}

and

w \cdot w = \pm 1

, (

+ 1

in the positive case and

- 1

in the negative case). Let:

u_{0}^{'} = λ \sum_{i = 1}^{n - 1} u_{i}^{'}, where λ = \frac{1}{n - 2 + b} .

We see that:

(n - 1) λ = \frac{n - 1}{(n - 1) + (b - 1)} < 1, and then {(n - 1)}^{2} λ^{2} < 1 .

Hence:

\begin{matrix} u_{0}^{'} \cdot u_{0}^{'} & = & λ^{2} (\sum_{i = 1}^{n - 1} u_{i}^{'} \cdot u_{i}^{'} + 2 \sum_{i < j} u_{i}^{'} \cdot u_{j}^{'}) \\ = & λ^{2} ((n - 1) b^{2} + (n - 1) (n - 2) b) \\ = & λ^{2} (n - 1) b (b + n - 2) \\ = & λ (n - 1) b . \end{matrix}

Since

λ (n - 1) = \frac{n - 1}{(b - 1) + (n - 1)} < 1

, we conclude that

u_{0}^{'} \cdot u_{0}^{'} < b < b^{2}

. Then, if

μ = \sqrt{b^{2} - u_{0}^{'} \cdot u_{0}^{'}}, = u_{0}^{'} + μ w

is the vector we were looking for. In the negative case we have

u_{0}^{'} \cdot u_{i}^{'} = - b

for each

i \in {1, 2, \dots, n - 1}

and:

u_{0}^{'} \cdot u_{0}^{'} = λ^{2} [- (n - 1) b^{2} - (n - 1) (n - 2) b] = - λ^{2} (n - 1) b λ^{- 1} = - λ (n - 1) b .

Clearly

- λ (n - 1) b > - b^{2}

since

λ (n - 1) < 1 < b

. Therefore,

u_{0}^{'} \cdot u_{0}^{'} > - b^{2}

. Defining

μ^{'} = \sqrt{b^{2} + u_{0}^{'} \cdot u_{0}^{'}}

, the vector

u^{'} = u_{0}^{'} + μ^{'} w

satisfies

u^{'} \cdot u^{'} = - b^{2}

and

u^{'} \cdot u_{i}^{'} = - b

for each

i \in {1, 2, \dots, n - 1}

. □

The next theorem is the most important result of this paper:

Theorem 7.

Let V be a regular quadratic spaces over

R

such that

dim V = n > 2 ind V > 0

. Then V has unitary tangential basis or an antiunitary tangential basis.

Proof.

Let

s = ind V

, we can find isotropic vectors

v_{1}, v_{1}^{'}, \dots, v_{s} v_{s}^{'}

on V with

v_{i} \cdot v_{i}^{'} = 1

for each

i = 1, 2, \dots, s

and anisotropic mutually orthogonal vectors

u_{1}, \dots, u_{t}

such that

u_{j} \cdot u_{j} = 1

for every

j \in {1, 2, \dots, t}

(positive case) or such that

u_{j} \cdot u_{j} = - 1

for every

j \in {1, 2, \dots, t}

(negative case) and such that:

V = 〈 v_{1}, v_{1}^{'} 〉 ⊥ \dots ⊥ 〈 v_{s}, v_{s}^{'} 〉 ⊥ 〈 u_{1} 〉 ⊥ \dots 〈 u_{t} 〉 .

Using theorem 6, we can find a tangential basis

B = {w_{1}, w_{1}^{'}, w_{2}, w_{2}^{'}, \dots, w_{s}, w_{s}^{'}, u_{1}}

of the subspace

V_{0} = 〈 v_{1}, v_{1}^{'} 〉 ⊥ \dots ⊥ 〈 v_{s}, v_{s}^{'} 〉 ⊥ 〈 u_{1} 〉

which is unitary in the positive case or antiunitary in the negative case.

By Lemma 1 we can find linearly independent vectors

u_{2}^{'}, u_{3}^{'}, \dots, u_{t}^{'}

〈 u_{2} 〉 ⊥ \dots ⊥ 〈 u_{t} 〉

such that

u_{j}^{'} \cdot u_{j}^{'} = 4

and

u_{i}^{'} \cdot u_{j}^{'} = 2

for

i \neq j

in the positive case and such that

u_{j}^{'} \cdot u_{j}^{'} = - 4

and

u_{i}^{'} \cdot u_{j}^{'} = - 2

for

i \neq j

in the negative case.

In the positive case we define

u = - v_{s} + 2 v_{s}^{'} - u_{1} \in V_{0}

. If

i < s

w_{i} \cdot u = (v_{i} + u_{1}) \cdot (- v_{s} + 2 v_{s}^{'} - u_{1}) = - u_{1} \cdot u_{1} = - 1

and

w_{i}^{'} \cdot u = (2 v_{i}^{'} - u_{1}) \cdot (- v_{s} + 2 v_{s}^{'} - u_{1}) = u_{1} \cdot u_{1} = 1,

where

w_{i}

and

w_{i}^{'}

are defined as in theorem 6. The next step is to prove that

w_{1}, w_{1}^{'}, \dots, w_{s}, w_{s}^{'}, u_{1}, u + u_{2}^{'}, u + u_{3}^{'}, \dots, u + u_{t}^{'}

is a tangential basis for V. Take

1 < j \leq t

. If

i < s

\begin{matrix} w_{i} \cdot (u + u_{j}^{'}) & = & w_{i} \cdot u = - 1, \\ w_{i}^{'} \cdot (u + u_{j}^{'}) & = & w_{i}^{'} \cdot u = 1, \end{matrix}

and if

i = s

\begin{matrix} w_{s} \cdot (u + u_{j}^{'}) & = & w_{s} \cdot u = (v_{s} + u_{1}) \cdot (- v_{s} + 2 v_{s}^{'} - u_{1}) = 1, \\ w_{s}^{'} \cdot (u + u_{j}^{'}) & = & w_{s}^{'} \cdot u = (2 v_{s}^{'} - u_{1}) \cdot (- v_{s} + 2 v_{s}^{'} - u_{1}) = - 1 . \end{matrix}

Continuing in the positive case, let us calaculate

(u + u_{i}^{'}) \cdot (u + u_{j}^{'})

. If

i \neq j

(u + u_{i}^{'}) \cdot (u + u_{j}^{'}) = u \cdot u + u_{i}^{'} \cdot u_{j}^{'} = - 3 + 2 = - 1

and if

i = j

(u + u_{i}^{'}) \cdot (u + u_{i}^{'}) = u \cdot u + u_{i}^{'} \cdot u_{i}^{'} = - 3 + 4 = 1

. We conclude then that

w_{1}, w_{1}^{'}, \dots, w_{s}, w_{s}^{'}, u_{1}, u + u_{2}^{'} + u + u_{3}^{'}, \dots, u + u_{t}^{'}

is a tangential unitary basis for V.

In the negative case, we apply again lemma 1 and find linearly independet vectors

u_{2}^{'}, u_{3}^{'}, \dots, u_{t}^{'}

〈 u_{2} 〉 ⊥ \dots ⊥ 〈 u_{t} 〉

such that

u_{t}^{'} \cdot u_{j}^{'} = - 2

for

i \neq j

and

u_{i}^{'} \cdot u_{i}^{'} = - 4

for each

i \in {2, 3, \dots, t}

. The proposed antiunitary tangential basis is

w_{1}, w_{1}^{'}, \dots, w_{s}, w_{s}^{'}, u_{1}, u + u_{2}^{'}, \dots, u + u_{t}^{'}

where

u = - v_{s} - 2 v_{s}^{'} - u_{1}

and

w_{i}, w_{i}^{'}

are as in the positive case. We have now, if

i < s

\begin{matrix} w_{i} \cdot u & = & (v_{i} + u_{1}) \cdot (- v_{s} - 2 v_{s}^{'} - u_{1}) = - u_{1} \cdot u_{1} = 1 . \\ w_{i}^{'} \cdot u & = & (2 v_{i}^{'} - u_{1}) \cdot (- v_{s} - 2 v_{s}^{'} - u_{1}) = - u_{1} \cdot u_{1} = 1 . \end{matrix}

and if

i = s

\begin{matrix} w_{s} \cdot u & = & (v_{s} + u_{1}) \cdot (- v_{s} - 2 v_{s}^{'} - u_{1}) = - 2 (v_{s} \cdot v_{s}^{'}) = - u_{1} \cdot u_{1} = - 2 + 1 = - 1 . \\ w_{s}^{'} \cdot u & = & (2 v_{s}^{'} + u_{1}) \cdot (- v_{s} - 2 v_{s}^{'} - u_{1}) = - 2 (v_{s} \cdot v_{s}^{'}) = - u_{1} \cdot u_{1} = - 2 + 1 = - 1 . \end{matrix}

1 < j \leq t

and

i < s

, we have:

\begin{matrix} w_{i} \cdot (u + u_{j}^{'}) & = & w_{i} \cdot u = 1; \\ w_{i}^{'} \cdot (u + u_{j}^{'}) & = & w_{i}^{'} \cdot u = 1, \end{matrix}

and for

i = s

\begin{matrix} w_{s} \cdot (u + u_{j}^{'}) & = & w_{s} \cdot u = - 1; \\ w_{s}^{'} \cdot (u + u_{j}^{'}) & = & w_{s}^{'} \cdot u = - 1 . \end{matrix}

Therefore,

{[w_{i} \cdot (u + u_{j}^{'})]}^{2} = (w_{i} \cdot w_{i}) [(u + u_{j}^{'}) \cdot (u + u_{j}^{'})] = 1

, since

(u + u_{j}^{'}) \cdot (u + u_{j}^{'}) = u \cdot u + u_{j}^{'} \cdot u_{j}^{'} = 3 - 4 = - 1

Finally, if

1 < i

j \leq t

\begin{matrix} (u + u_{i}^{'}) \cdot (u + u_{j}^{'}) & = & u \cdot u + u_{i}^{'} \cdot u_{j}^{'} = 3 - 2 = 1 for i \neq j \\ (u + u_{i}^{'}) \cdot (u + u_{i}^{'}) & = & u \cdot u + u_{i}^{'} \cdot u_{i}^{'} = 3 - 4 = - 1 . \end{matrix}

Therefore,

w_{1}, w_{1}^{'}, \dots, w_{s}, w_{s}^{'}, u_{1}, u + u_{2}^{'}, \dots, u + u_{t}^{'}

is a tangential antiunitary basis for V. □

The following theorem gives us the necessary and sufficient conditions for a quadratic space to have a tangential base.

Theorem 8.

Let

V = U \oplus V^{⊥}

, where U is a regular subspace of V. Then V has a tangential basis if and only if U has a tangential basis.

Proof.

Suppose

dim U = n

and

dim V^{⊥} = m

. Suppose V has a tangential basis

{v_{1}, v_{2}, \dots, v_{n + m}}

, every

v_{i}

can be written in the form

v_{i} = u_{i} + {\bar{u}}_{i}

u_{i} \in U

and

{\bar{u}}_{i} \in V^{⊥}

. We claim the vectors

{u_{1}, u_{2}, \dots, u_{n + m}

generate U. Indeed, if

w \in U

is arbitrary, we have scalars

λ_{1}, λ_{2}, \dots, λ_{n + m}

such that:

w = λ_{1} v_{1} + λ_{2} v_{2} + \dots + λ_{n + m} v_{n + m} .

Therefore,

w - (λ_{1} u_{1} + λ_{2} u_{2} + \dots + λ_{n + m} u_{n + m}) = λ_{1} {\bar{u}}_{1} + λ_{2} {\bar{u}}_{2} + \dots + λ_{n + m} {\bar{u}}_{n + m}

. Since

w - (λ_{1} u_{1} + λ_{2} u_{2} + \dots + λ_{n + m} u_{n + m}) \in U \cap V^{⊥}

, we have

w = λ_{1} u_{1} + λ_{2} u_{2} + \dots + λ_{n + m} u_{n + m}

as was to be proved. Therefore, a subfamily of

{u_{1}, u_{2}, \dots, u_{n + m}

with n elements is a basis for U, say

U = 〈 u_{1}, u_{2}, \dots, u_{n} 〉

. Observe now:

{(u_{i} \cdot u_{j})}^{2} = {((u_{i} + {\bar{u}}_{i}) \cdot (u_{j} + {\bar{u}}_{j}))}^{2} = {(v_{i} \cdot v_{j})}^{2} = (v_{i} \cdot v_{i}) (v_{j} \cdot v_{j}) = (u_{i} \cdot u_{i}) (u_{j} \cdot u_{j}) .

Therefore,

{u_{1}, \dots, u_{n}}

is a tangential basis U.

Suppose now that U has a tangential basis

{u_{1}, \dots, u_{n}}

. Let

{{\bar{u}}_{1}, \dots, {\bar{u}}_{m}}

be a basis for

V^{⊥}

. Denote

w_{i} = u_{1} + {\bar{u}}_{i}

, for each

i \in {1, \dots, m}

. We calculate

{(u_{i} \cdot w_{j})}^{2}

and

{(w_{i} \cdot w_{j})}^{2}

\begin{matrix} {(w_{i} \cdot w_{j})}^{2} & = & {(u_{i} \cdot (u_{1} + {\bar{u}}_{j}))}^{2} = {(u_{i} \cdot u_{1})}^{2} = (u_{i} \cdot u_{i}) (u_{1} \cdot u_{1}) = (u_{i} \cdot u_{i}) (w_{j} \cdot w_{j}), \\ {(w_{i} \cdot w_{j})}^{2} & = & {((u_{1} + {\bar{u}}_{i}) \cdot (u_{1} + {\bar{u}}_{j}))}^{2} = (u_{1} \cdot u_{1}) (u_{1} \cdot u_{1}) = (w_{i} \cdot w_{i}) (w_{j} \cdot w_{j}) . \end{matrix}

Since

V = U \oplus V^{⊥}

{u_{1}, u_{2}, \dots, u_{n}, w_{1}, \dots, w_{m}}

is a tangential basis for V. □

We will finish this work with the definition of isometries.

Definition 4.

Two quadratic spaces

V_{1}, V_{2}

over the same field

F

areisometricif there exists an isomorphism

ψ : V_{1} \to V_{2}

such that

v \cdot v^{'} = ψ (v) \cdot ψ (v^{'})

for each pair of vectors

v, v^{'} \in V_{1}

. Such mapping ψ is called anisometrybetween

V_{1}

and

V_{2}

The following theorem will be very useful for our purposes, its proof can be consulted in detail in ([5], Theorem 42:16).

Theorem 9.

Let U, V be isometric regular subspaces of a quadratic space W. Then

U^{⊥}

and

V^{⊥}

are isometric too.

The problem about the existence of isotropic basis of a regular space V over the field of real numbers is solved with the following theorem:

Theorem 10.

F = R

, V is regular and if

s = ind V > 0

. Then V has an isotropic basis.

Proof.

If V is a hyperbolic space and we keep the notation of theorem 6, it is clear that

{v_{1}, v_{1}^{'}, \dots, v_{s}, v_{s}^{'}}

is the desired basis. Suppose now that

t = dim V - 2 ind V > 0

. Let us consider the extension

V^{*} = V ⊥ 〈 u_{t + 1} 〉

where

u_{t + 1} \cdot u_{t + 1} = u_{i} \cdot u_{i}

for each

i \in {1, 2, \dots, t}

. By theorem 7,

V^{*}

has a tangential basis

{v_{1}, v_{2}, \dots, v_{2 s + t + 1}}

where

v_{i} \cdot v_{i} = 1

v_{i} \cdot v_{i} = - 1

for each

i \in {1, 2, \dots, 2 s + t + 1}

. We define the vectors

w_{i} = v_{i} \pm v_{2 s + t + 1}

, where

i \in {1, 2, \dots, 2 s + t}

and the sing + is taken if

v_{i} \cdot v_{2 s + t + 1} \neq v_{2 s + t + 1} \cdot v_{2 s + t + 1}

and the sing − is taken if

v_{i} \cdot v_{2 s + t + 1} = v_{2 s + t + 1} \cdot v_{2 s + t + 1}

. It is clear that every

w_{i}

is isotropic and

{〈 v_{2 s + t + 1} 〉}^{⊥} = 〈 w_{1}, w_{2}, \dots, w_{2 s + t} 〉

Since the subspaces

〈 v_{2 s + t + 1} 〉

and

〈 v_{t + 1} 〉

V^{*}

are isometric, theorem 9 implies that their orthogonal complements are isometric too. Therefore,

V = {〈 u_{t + 1} 〉}^{⊥}

is isometric to

{〈 u_{2 s + t + 1} 〉}^{⊥}

and V has an isotropic basis. □

Corollary 4.

V = U \oplus V^{⊥}

, U is regular and

ind U > 0

, then V has an isotropic basis.

Proof.

By theorem 10, U has an isotropic basis B. If

\bar{B}

any basis of

V^{⊥}

, then B

\cup \bar{B}

is an isotropic basis of V. □

We will finish the paper with some remarks and two conjectures.

Using induction and theorem 5, it may be proved easily that every quadratic space V over the field of complex numbers has a tangential basis provided that V is regular and

dim V \geq 3

We also observe that if V is a hyperbolic space of index

s > 0

, say

V = 〈 v_{1}, v_{1}^{'} 〉 ⊥ \dots ⊥ 〈 v_{s}, v_{s}^{'} 〉

where

v_{i} \cdot v_{i} = 0 = v_{i}^{'} \cdot v_{i}

and

v_{i} \cdot v_{i}^{'} = 1

for each

i = 1, 2, \dots, s

, then there exists a linear automorphism

ϕ : V \to V

such that

ϕ (v_{i}) = v_{i}^{'}

and

ϕ (v_{i}^{'}) = - v_{i}

for each

i = 1, 2, \dots, s

ϕ

satisfies the following property:

$u \cdot w = - ϕ (u) \cdot ϕ (w)$ for every pair $u, w \in V$ .

Hence, if the hyperbolic space V has a unitary tangential basis and also an antiunitary tangential basis.

We say a quadratic space V is bitangential if V has a unitary tangential basis and also an antiunitary tangential basis.

Therefore, if a hyperbolic space V has a unitary tangential basis , then V is bitangential.

The previous comments motivate the following conjectures:

Problem 1.

If V is a regular quadratic spaces over

R

of any finite dimension, then V cannot be tangential.

A weaker conjecture is:

Problem 2.

F = R

, no hyperbolic space over

R

can be have a tangential basis.

If this last conjecture is true, we can finally state: If

F = R

, the only regular quadratic spaces over

R

that have a tangential basis are those of positive index and which are not hyperbolic spaces.

5. Conclusions

The results presented in this article allow us to obtain an explicit method to construct tangential and anisotropic bases in quadratic spaces. We will also provide three theorems that allow us to explicitly calculate vectors that are orthogonal or tangent to given vectors or that provide necessary and sufficient conditions for the existence of such vectors.

Author Contributions

All authors, A.L., P.H., J.M., and A.P., contributed equally to writing this article. .

Funding

The last author was supported by Vicerrectoría de Investigación e Innovación de la Universidad Simón Bolivar (sede de Barranquilla).

Data Availability Statement

Not applicable.

Acknowledgments

The last author thanks Dr. Adalberto García-Máynez (R.I.P.) for several fruitful conversations about the problem’s solution in the title.

Conflicts of Interest

The authors declare no conflicts of interest.

References

P. Koprowski, Algorithms for quadratic forms, J. Symb. Comput, 43 (2008) 140-152. [CrossRef]
P. Koprowski and A. Czogała, Computing with quadratic forms over number fields, J. Symb. Comput, 89 (2018) 129-145.
P. Koprowski and B. Rothkegel, The anisotropic part of a quadratic form over a number field, J. Symb. Comput, 115 (2023) 32-52. [CrossRef]
T. Y. Lam, Introduction to quadratic forms over fields, Graduate Studies in Mathematics, vol. 67. American Mathematical Society, Providence, RI, 2005.
O.T. O’Meara, Introduction to Quadratic Forms, Springer, Berlin, (1973).

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Tangential and Anisotropic Bases in Quadratic Spaces

Abstract

1. Introduction

2. Preliminary Facts

3. Tangency and Orthogonality

4. Tangential and Anisotropic Bases

5. Conclusions

Author Contributions

Funding

Data Availability Statement

Acknowledgments

Conflicts of Interest

References

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