Theorems of elementary geometry that can be generalized to inner product space over $\mathbb{R}$

Question

In inner product spaces on $\mathbb{R}$, length and angles can be defined. Hence, something similar to elementary geometry can be done there.

For example, the parallelogram law can be generalized to inner product spaces.

Are there other examples?

Answer 1

In the following sense, all theorems in elementary geometry hold in inner product spaces.

Let $V$ be an inner product space over $\mathbb{R}$, and $W$ be a 2- or 3-dimensional subspace. $W$ is isomorphic to the 2- or 3-dimensional Euclidean space, so all theorems of plane or spatial geometry hold in $W$.

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According to a comment, a first order property of an inner product space that holds in all finite dimensional spaces holds in all inner product spaces. arxiv.org/abs/0904.3482

Answer 2

Not necessarily. Scalar multiples of a single vector yield points on a straight line while satisfying the axioms of an inner product space. You need at least two linearly independent vectors to get Euclidean Plane Geometry.

One can formulate the axioms of Euclidean Geometry in terms of some inner product spaces. Afterwards, what can be proven from those follows.

Axiom 1: A line can be drawn between any two points: If we have positive vector $\vec{p_1}$ and positive vector $\vec{p_2}$, we have $-\vec{p_1}$ because an inner product space is a vector space. For similar reasons we have $\vec{p_2}-\vec{p_1}$, a vector point from $\vec{p_1}$ to $\vec{p_2}$.

Axiom 2: A line segment can be extended in any direction direction an arbitrary distance: Suppose we have $\vec{l}=\vec{p_2}-\vec{p_1}$. Then $\vec{r}=\vec{p_1}+c\vec{l}$ is an extension of $\vec{l}$.

Axiom 3: Given a point as a center C and another point R, a circle can be drawn through R centered at C. Let $\vec{c}$ be the center of a circle. Are points such that $(\vec{c}-\vec{R})^2=r^2$ in the space?

It suffices to find vectors about $\vec{0}$ that are a linear combination of $\vec{c}$ and $\vec{d}$ yielding a vector of squared norm $r^2$. Then add $\vec{c}$ to get a new point on the circle.

$(m\vec{c}+n\vec{d})^2=r^2=m^2c^2+2mn\vec{c}\cdot \vec{d}+n^2d^2=r^2$

$m=\frac{-2n\vec{c}\cdot \vec{d}\pm \sqrt{4n^2(\vec{c}\cdot \vec{d})^2-4c^2(n^2d^2-r^2)}}{2c^2}=\frac{-n\vec{c}\cdot \vec{d} \pm \sqrt{c^2r^2-n^2[c^2d^2-(\vec{c}\cdot \vec{d})^2]}}{c^2}$

A solution exists for $r^2\ge n^2d^2$.

Axiom 4: All right angles are congruent: Angles are congruent if they retain their size under rotations or translations. Any point on the unit circle can be represented as as the linear combination of two linearly independent vectors and are thus in hour IPS. Let $\vec{v}$ be on the unit circle. Using arguments similar to those used for Axiom 3, we can find $\vec{w}$ on the unit circle so that $ \vec{w}\cdot \vec{v}=0$.

$\vec{v}=p\vec{c}+q\vec{d}$

$\vec{w}=m\vec{c}+n\vec{d}$

$v^2=1=p^2c^2+q^2d^2+2pq \vec{c}\cdot \vec{d}$

$\vec{v}\cdot \vec{w}=0= mpc^2+(pn+mq)\vec{c}\cdot \vec{d}+nqd^2$ implies $m$ and $n$ have a linear relationship depending on $p$, $q$ and the basis vectors perpendicular to $\vec{v}$.

$w^2=1=m^2c^2+n^2d^2+2mn\vec{c}\cdot \vec{d}$ Determines points on that line that intersect the unit circle producing a right angle to any radius of the unit circle. Translate the corresponding vectors to produce any angle elsewhere in the plane.

Axiom 5: Given a line and a point not on that line, there is one and only one line passing through that point parallel to the line.

A point and a slope determine a line. A common point with a different slope is a different line. Lines are parallel iff they have the same slope. Points-slope form has an isometric representation in an inner product space.

So the theorems should follow.