Generalizing Rigid Motions Group w/ Metric

[topsquark]

Define:
\(\displaystyle Euc(n) = \{ T \in End( \mathbb{R}^n )| ~ ||Tx - Ty|| = ||x - y||~\forall x,y \in \mathbb{R}^n \}\)

This is defined as the Euclidean group of rigid motions.

Can we generalize this group to be defined with any metric (well actually inner product, I suppose)? Obviously it won't be Euclidean any more. Would that represent a group of "rigid motions" as defined by that metric?

-Dan

Edit: I should mention that in the definition of Euc(n) \(\displaystyle ||x|| = \sqrt{ \sum_i x_i^2 }\).

 

[I like Serena]

Define:
\(\displaystyle Euc(n) = \{ T \in End( \mathbb{R}^n )| ~ ||Tx - Ty|| = ||x - y||~\forall x,y \in \mathbb{R}^n \}\)

This is defined as the Euclidean group of rigid motions.

Can we generalize this group to be defined with any metric (well actually inner product, I suppose)? Obviously it won't be Euclidean any more. Would that represent a group of "rigid motions" as defined by that metric?

-Dan

Edit: I should mention that in the definition of Euc(n) \(\displaystyle ||x|| = \sqrt{ \sum_i x_i^2 }\).

Yes.
It's called an isometry.
Literally translated: equal metric.
You only need a metric to define it. Neither an inner product, nor a norm are required.

 

[topsquark]

Yes.
It's called an isometry.
Literally translated: equal metric.
You only need a metric to define it. Neither an inner product, nor a norm are required.

Ah! Yes, I have heard of isometries. (I must actually be starting to understand how some of this stuff crosses course boundaries.) Thanks for the info!

-Dan

 

[Deveno]

Here is how I think of it:

A METRIC is a "spatial thing", it tells us "how separated" (we use "how far apart" as measured by the metric) two points are.

A NORM is a "vector thing"-there is this requirement of "homogeneity" (also called absolute scalability)which tells us that the norm is "consistent" with scalar multiplication:

$\|\alpha v\| = |\alpha|\|v\|$

Given a normed vector space, one can define a metric $d$ by:

$d(u,v) = \|u - v\|$.

Now one of the properties of a norm $\|\cdot\|$ is that:

$\|v\| = 0 \iff v = 0$.

This implies $d(u,v) = 0 \iff u = v$.

A norm must also satisfy the TRIANGLE INEQUALITY:

$\|u + v\| \leq \|u\| + \|v\|$

From this we have:

$0 = \|v + -v\| \leq \|v\| + \|-v\| = \|v\| + \|(-1)v\| = \|v\| + |-1|\|v\| = 2\|v\|$

so a norm is positive-definite, and so $d: V \times V \to \Bbb R_0^+$.

The triangle property of a norm tells us:

$d(u,w) = \|u - w\| = \|u - v + v - w\| \leq \|u - v\| + \|v - w\| = d(u,v) + d(v,w)$, so we have a bona-fide metric.

However, not ALL metrics on a vector space are "compatible" with the vector space structure:

We can define, for example, the "discrete metric":

$d(v,v) = 0$, for all $v \in V$.

$d(u,v) = 1$, if $u \neq v$, which is not absolutely scalable, so does not correspond to a norm.

It is possible to give examples of two metrics which induce the same topology on a vector space $V$, but one is a norm, and one is not, but I will not do so here.

Even MORE restrictive, is the case of an inner product space. An inner product $\langle \cdot,\cdot\rangle$ induces a norm by:

$\|v\| = \sqrt{\langle v,v \rangle}$

But not all norms arise in this way, an example is the $p$-norm for $p \neq 2$:

$\displaystyle \|v\|_p = \left(\sum_i^n |v_i|^p\right)^{1/p}$ which is NOT an inner product.

$\Bbb R^n$ is rather special: It's an inner product space with a topology induced by the metric which is induced by the norm induced by its inner product. Moreover, the vector space addition and scalar multiplication are continuous maps.

Note that $\text{End}(\Bbb R^n)$ has a natural ring structure (the ring of Endomorphisms of the $\Bbb R$-module $\Bbb R^n$), which is isomorphic to the ring of $n \times n$ matrices over $\Bbb R$ (this isomorphism is not "canonical" but depends on a choice of basis, so speaking in terms of "matrices" depends on picking a "coordinate system", while speaking of $\text{End}(\Bbb R^n)$ is "basis-free")).

The group of units of this ring, is the general linear group, which can also be written $\text{Aut}(\Bbb R^n)$. Note that LINEAR isometries (under any metric for for a finite-dimensional vector space) are automatically bijective (isometries are always injective, via positivity of the metric, and in a finite-dimensional vector space setting, $T \in \text{End}(V)$ is injective if and only if it is surjective, by the rank-nullity theorem).

In the "physical world", invertible linear maps correspond to "reversible" linear transformations of a space: in other words, "no loss of information". In more abstract structures, just as with ordinary integers, 0 continues to play "the bad guy".

 

[blue_raver22]

I can confirm that the group Euc(n) is indeed the Euclidean group of rigid motions, where n represents the number of dimensions in the vector space. This group is defined as the set of endomorphisms (linear transformations from a vector space to itself) that preserve the Euclidean metric, meaning that the distance between any two points remains the same after the transformation is applied.

It is possible to generalize this group to be defined with any metric, as long as the metric satisfies certain properties. Specifically, the metric must be invariant under rigid motions, meaning that the distance between any two points remains the same after a rigid motion is applied. This would represent a group of "rigid motions" as defined by that particular metric.

However, it is important to note that this group would no longer be Euclidean, as the Euclidean metric is specific to Euclidean space. In other words, the metric would change depending on the geometry of the space. Nonetheless, this generalized group would still represent rigid motions in that particular space.