Generalizing Rigid Motions Group w/ Metric

[topsquark]

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

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) [math]||x|| = \sqrt{ \sum_i x_i^2 }[/math].

 

[I like Serena]

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

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) [math]||x|| = \sqrt{ \sum_i x_i^2 }[/math].

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".