# Lecture 16

Definition. Let $q$ be a quadratic form on $\mathbb{R}^n$ or $\mathbb{C}^n$. Then the corresponding quadric is $Q = \{[v] \in \RP^n \mid q(v) =0\}$

A conic is a quadric in $\RP^2$.

How many indefinite non-degenerate quadratic forms exists in $\mathbb{R}^{n+1}$ up to change a basis?

Classification by the signature: (+++..+), (+++..+ -), …, (+ – …-), (- – – ..-), should be indefinite $\Rightarrow$ n indefinite quadrtic forms (+++..+ -), … , (+ – …-).

The number of non-empty non-degenerate quadrics in $\mathbb{R}P^n$ is $\lceil \frac{n}{2} \rceil$

Example

$n=2$: In $\RP^2$ there exist only one indefinite non-degenerate quadric/conic up to projective transformations. It has signature ( + + – ) resp. ( – – +).

$n=3$: In $\RP^3$ there exist $\lceil \tfrac32 \rceil = 2$ different quadrics up to projective transformations:

1.  (+ + + – ), ( – – – +), $Q_1 = \{[v] \in \mathbb{R}P^3 \mid v_1^2 + v_2^2 + v_3^2 – v_4^2 = 0 \}$
2.  ( + + – – ), ( – – – +),  $Q_2 = \{[v] \in \mathbb{R}P^3 \mid v_1^2 + v_2^2 – v_3^2 – v_4^2 = 0 \}$

1. is a unit sphere (or ellipse, paraboloid or 2-sheeted hyperboloid) depending on the choice of affine coordinate. To obtain a sphere as affine image of the quadric we can choose the affine coordinates  $u_1 = \frac{v_1}{v_4}$, $u_2 = \frac{v_2}{v_4}$, and $u_3 = \frac{v_3}{v_4}$.

2. is one-sheeted hyperboloid or hyperbolic paraboloid.

Let b be a degenerate bilinear form. Then the space $\{ u \in U \mid b(u, v) = 0 \forall v \in V \} = \ker(b)$ is a subspace of $V$. Consider a subspace $U_1 \subset V$ such that $V = \ker(b) \oplus U_1$, then $b\mid_{U_1}$ defines a non-degenerate quadric $Q_1$ in $P(U_1)$. The quadric $Q$ defined by $b$ is the union of lines through points in the non-degenerate quadric $Q_1$ defined by $b|_{U_1}$ and points in $P(\ker(b))$ if $Q_1 \neq \emptyset$ (see Exercise 8.1).

Example of degenerate conic.

Consider the following bilinear form in $\mathbb{R}P^2$:

$b (\begin{pmatrix} x_1\\ x_2\\x_3\end{pmatrix} , \begin{pmatrix} x_1\\ x_2\\ x_3\end{pmatrix}) = {x_1}^2 – {x_2}^2.$

Then $\ker (b) = span \{e_3\}$ and $\mathbb{R}^3 = \ker(b) \oplus \underbrace {span\{e_1, e_2\} }_{U_1}$.  Projectively, $P(\ker(b))$ is a point and the quadric defined by $b|_{U_1}$ in $P(U_1)$ consists of two points. So the degenerate/singular conic definined by $b$ in $\RP^2$ consists of two crossing lines.

Proposition. A non-degenerate non-empty quadric $Q \subset \RP^n$ determines the corresponding bilinear form up to a non-zero scalar multiple.

Proof. Let $b$  and $\tilde b$ be two linear forms defining $Q$. Consider an orthonormal basis w.r.t. $b$, i.e. $\{ e_1 .. e_p, f_1 .. f_q \}$, such that:

1. $b(e_i, e_i) = 1$ for all $i = 1, \ldots, p$;
2. $b(f_j, f_j) = -1$ for all $j = 1, \ldots, q$; and
3. $b(e_i, e_k) = b(f_j, f_l) = b(e_i, f_j) = 0$ for all
$i, k = 1, \ldots, p$ with $i \neq k$ and
$j, l = 1, \ldots, q$ with $j \neq l$.

Consider the vectors $e_i \pm f_j$. Then

\begin{align*}
& b( e_i \pm f_j, e_i \pm f_j) = \underbrace{b(e_i, e_i)}_{=1} \pm \underbrace{2b(e_i, f_j)}_{=0} + \underbrace{b( f_j, f_j)}_{-1} = 0 \\
\Rightarrow& 0 = \tilde b( e_i \pm f_j, e_i \pm f_j) =  \tilde b(e_i, e_i) \pm \tilde 2b(e_i, f_j) + \tilde b ( f_j, f_j)\\
\Leftrightarrow& \pm 2 \tilde b (e_i, f_j) = – \tilde b(e_i, e_i) – \tilde b(f_j, f_j) \\
\Rightarrow& \tilde b (e_i, f_j) = 0 \\
\Rightarrow& \tilde b(e_i, e_i) = – \tilde b (f_j, f_j).
\end{align*}

So setting $\tilde b (e_1, e_1) = \lambda \neq 0$ implies that $\tilde b(e_i, e_i) = \lambda$ for all $i = 1, \ldots, p$ and $\tilde b(f_j, f_j) = – \lambda$ for all $j = 1, \ldots, q$.

Orthogonal Transformations

Definition. Let $b$ be a non-degenerate symmetric bilinear form on $\mathbb{R}^{n+1}$. Then $F: \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1}$ is orthogonal (wrt. $b$) if $b(F(v), F(w)) = b(v,w) \forall v,w \in \mathbb{R}^{n+1}$.

The group of orthogonal transformations for a bilinear form of signature $(p, q)$ with $p+q = n+1$ is denoted by $O(p,q)$. If $q=0$ we obtain the “usual” group of orthogonal transformations $O(n+1) = O(n+1, 0)$

If $Q \subset \RP^n$ is a non-degenerate quadric defined by $b$ and  $F: \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1}$ is an orthogonal transformation, then the map $f: \mathbb{R}P^n \rightarrow \mathbb{R}P^n$ with $[x] \mapsto f([x]) = [F(x)]$ maps the quadric onto itself: $f(Q) = Q$.

Proposition. If the signature of a non-degenerate quadric $Q$ is $(p,q)$ with $p \neq q$ and $p \neq 0 \neq q$, then any projective transformation with $f(Q) = Q$ is induced by an orthogonal transformation $F$ w.r.t. $b$ that defines $Q$.

Proof. Let $b$ be the bilinear form defining $Q$ and $F$ the linear transformation defining the projective transformation $f$. Then $\tilde b(v,w):= b(F(v), F(w))$ is another bilinear form defining $Q$.

According to the previous proposition there exists $\lambda \neq 0$ such that

$b=\lambda \tilde b.$

If $\lambda > 0$ then $\frac{1}{\sqrt \lambda} F$ is an orthogonal transformation, since $b( \frac{1}{\sqrt \lambda} F(v), \frac{1}{\sqrt \lambda} F(w)) = \frac{1}{\lambda} b (F(v), F(w)) = \frac{1}{\lambda}\tilde b (v,w) = b(v,w)$.

So we need to show that $\lambda > 0$.

If $\{ e_1, \ldots, e_{n+1}\}$ is an orthogonal basis of $\mathbb{R}^{n+1}$ w.r.t. $b$ then $\{F(e_1), \ldots, F(e_{n+1}\}$ is also orthogonal with

• If $b(e_i, e_i) = 1$ then $\tilde b(e_i, e_i) = \lambda$
• $b(e_i, e_i) = -1 \Rightarrow \tilde b(e_i, e_j) = – \lambda$.

But the signature is invariant w.r.t. to change of basis (i.e. w.r.t. F). So since $p \neq q$ we have that the $\lambda$ is positive. $\square$

If $p = q$ (neutral signature), for example $p=q=2$ for a quadric  in $\RP^3$, then there exists a projective transformation preserving the quadric not induced by an orthogonal transformation. Let $b(x,x) = {x_1}^2 + {x_2}^2 – {x_3}^2 – {x_4}^2$. Then the map:

$f : \RP^3 \rightarrow \RP^3 , \sqvector{x_1\\x_2\\x_3\\x_4} \mapsto \sqvector{x_3\\x_4\\x_1\\x_2}$

yields $b(F(v), F(w)) = – b (x,x) = {x_3}^2 + {x_4}^2 – {x_1}^2 – {x_2}^2$. So $f$ preserves the quadric but it is not induced by an orthogonal transformation $F$.

# Exercise sheet 8

In the tutorial we discussed doubly ruled quadrics, i.e., quadrics of signature (+,+,-,-)  in $\mathbb{R}P^3$. The term “doubly ruled” expresses the fact that those quadrics contain two families of lines, where each family of lines generates the quadric. Two lines of those families are skew if they are contained in the same family and they intersect if they are contained in different families. Therefore, each point of the quadric can be described as the intersection point of two unique lines, which also span the tangent plane to the quadric at the intersection point.

It is an important fact that any three skew lines in $\mathbb{R}P^3$ determine a unique doubly ruled quadric that contains the three given lines. You may use this statement for the solution of Exercise 8.3.

Happy holidays,

Emanuel

# Exercise Sheet 6

In the exercise sheet 6 exercise 1 you are talking about singular conics. We have never defined these, are these the same as degenerate ones?

# Lecture 15

### Pole-Polar Relationship

• b is non-degenerate symmetic bilinear form (e.g.scalar product on $\mathbb{R}^n$)
• orthogonal complement of a vector subspace $U\subseteq V$
$U^{\perp}= \left\lbrace v\in V\mid b\left(u,v\right) = 0,\forall u\in U\right\rbrace$
• $dimU^{\perp} + dim U = dim V$, but $U^{\perp} \oplus U \neq V$.
• If $P\left( U\right)$ is a projective subspace of $P\left(V\right)$,then $P\left(U^{\perp}\right)$ yields another subspace of $P\left(V\right)$.
• in $\RP^2 = P(\mathbb{R}^3)$ we obtain:
Point $\left[ P\right] \leftrightarrow$ line $P\left( \left\lbrace p\right\rbrace ^{\perp}\right)$
Line $P\left( U\right)\leftrightarrow$ point $P\left(U^{\perp}\right)$

# Lecture 14

### Polarity: Pole – Polar Relationship

Let $b: V \times V \rightarrow \mathbb{F}$ be a symmetric non-degenerate bilinear form.

Definition. Let $U \leq V$ be a vector subspace. Then

$U^{\perp} = \{ v \in V \: | \: b(u,v) = 0, \: \forall u \in U \}.$

If $U = \{u_1,…,u_k\}$ then

$U^{\perp} = ( \text{span} \; U )^{\perp} = \{ v \in V \: | \: b(u_i,v) = 0, \: \forall i = 1,…,k \}.$

We call $U^{\perp}$ the orthogonal complement of $U$.

Remark. dim $U^{\perp} =$ dim $U^0 = n – k$, if dim $U = k$ and dim $V = n$.