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Typically, in a normed vector space, norms are regarded to represent lengths, somehow or other. In particular, in an inner product space $(V,\langle\cdot,\cdot\rangle)$, the Euclidean norm $\lVert\cdot\rVert$ on $V$ is defined in terms of the inner product $\langle\cdot,\cdot\rangle$ by

$$\forall \mathbf{v} \in V, \enspace \lVert\mathbf{v}\rVert =\sqrt{\langle\mathbf{v},\mathbf{v}\rangle}$$

My question is the following:

Does there exist an absolute notion of length on a vector space (or a way to define such a notion) that always coincides with the pure geometric notion of the length between two points, WITHOUT depending on any choice of basis, norm or inner product whatsoever?

I ask this question because of the following:

If we consider any two points $A$ and $B$ in the Euclidean space $\mathbb{R}^3$ and work in a direct orthonormal frame $(O,\vec{i},\vec{j},\vec{k})$ with the standard dot product as our choice for the inner product, then the set $\{\vec{i},\vec{j},\vec{k}\}$ forms an orthonormal basis of $\mathbb{R}^3$. In this setting, the Euclidean norm coincides with the pure geometric notion of the length $AB$ (i.e. the length of the line segment $[AB]$ between the points $A$ and $B$) since, by the Pythagorean theorem, one has $\lVert\overrightarrow{AB}\rVert=AB$, where $\overrightarrow{AB}$ is the vector representing the directed line segment from $A$ to $B$ in terms of the basis we are working in which, in this case, is the basis $\{\vec{i},\vec{j},\vec{k}\}$.

If I am not mistaken, however, the length $AB$ depends only on the location of $A$ and $B$ and so, unlike $\lVert\overrightarrow{AB}\rVert$, does not depend on any choice of basis, norm or inner product. For instance, if we change the basis to $\{2\vec{i},2\vec{j},2\vec{k}\}$ while keeping everything else unchanged, then one would have $\lVert\overrightarrow{AB}\rVert= 2(AB) ≠ AB$. In other words, whether or not the equality $\lVert\overrightarrow{AB}\rVert=AB$ holds true seems to depend on the choice of basis, norm and inner product. Thus, the notion of norm does not always coincide with the pure geometric notion of length.

This concludes my concerns. The same question goes analogously for the notion of oriented angles between line segments. Unfortunately, I was unable to find a definitive answer anywhere. Please do kindly correct me if I am wrong and request further clarification if necessary.

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  • $\begingroup$ I have the impression that when you say "length $AB$", you are already implicitly fixing a norm (or metric) on that space. For example, in the case of $\Bbb R^3$ it is the norm $\|(x,y,z)\|^2=x^2+y^2+z^2$. So I would say your notion of length is not independent of a norm to begin with. How would you properly define "geometric notion of length" at all? $\endgroup$ – M. Rumpy Apr 9 at 21:37
  • $\begingroup$ By "length $AB$", I mean the actual distance between the two points $A$ and $B$ located in space, which is fixed independently of any choice of coordinates. But you might be right. A priori, I am currently unsure of a way to define this notion formally. Essentially, in my question, I would like to know if there does exist such way to define it. $\endgroup$ – JJD1120 Apr 9 at 21:43
  • $\begingroup$ The standard definitions of inner product and norm in Euclidean space do not depend on the choice of basis. What is Euclidean space? The set of fixed length tuples of real numbers. When you change the basis and redefine the inner product in terms of the new coordinate vectors, you've changed the definition of inner product. That's you changing the definition of the inner product, not that the original defn depends on the choice of basis. There are multiple inner products that you can define on the Euclidean space. That doesn't mean one depends on the choice of basis. $\endgroup$ – Myath Apr 9 at 22:02
  • $\begingroup$ In a vector space every metric comes with a natural norm and every norm induces a natural metric so they are inherently entangled. Proving this is a good exercise. Use your intuition about the relationship between length and distance to structure the proof. $\endgroup$ – CyclotomicField Apr 9 at 22:31
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The answer to your question is No. One always needs to make some choice.

Let me demonstarte you that not even in real physical space the notion of "actual distance between the two points $A$ and $B$" is well-defined without making any choice. You do not directly fix a norm or basis, but assigning a number to every pair of points is not intrinsic to the physical space either.

Ask first: what is distance? It could be some number. But in the real world, you are not 1 apart, but 1 meter apart. That is, you need units. To define a unit of length you need to find two points which are now apart exactly 1 by definition $-$ a ruler so to say $-$ as well as a way to move this "ruler" around to places where you actually want to know the distance.

In mathematical language, this might be modelled as follows:

For any vector space $V$ you can fix a point $p\in V$ as well as a subgroup $G\subseteq\mathrm{Lin}(V)$ of the linear functions. The point $p$ is "defined to be 1 away from the origin" (the ruler), and the group $G$ is the set of ways to move your ruler without changing its length (by definition). That is, the points of distance 1 from the origin are exactly the point $Tp$ for all $T\in G$ by definition. This is exactly how it is done in the real world. In the real world, the point $p$ would correspond to the protoype meter. Luckily we do not have to think too much about $G$ in the real world, as all motions we can do easily to a sufficiently rigid body preserve length.

There are some conditions on $p$ and $G$ to make them a good system of measurement. For example, $p\not=0$. Furthermore, you would want that $\alpha I\not\in G$ for all $\alpha\in\Bbb R\setminus\{-1,1\}$ (where $I$ denotes the identity transform). This is because you would like to define the point $\alpha p$ to be at distance $|\alpha|$ from the origin. But if $\alpha I\in G$, then the point $\alpha p=(\alpha I)p$ would be at distance 1 by definition. Further, you would want that for any $q\in V$ there is an $\alpha\in\Bbb R$ and a $T\in G$ so that $\alpha q=Tp$. This means that every point has a distance from the origin.

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  • $\begingroup$ I see. I was unable to fully grasp the last paragraph, but the beginning sufficed well enough. Thank you for your clarification. :) $\endgroup$ – JJD1120 Apr 9 at 22:37
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To specify a point in 2-D space, you could define 2 by 2 matrix A where both column vectors in A are linearly independent. Thus any linear combination of A column vector can span any point in 2-D space. You would express any point as Ax where x is 2 by 1 vector. Length of vector formed from origin to point is |Ax|. Any 2 column vector can be picked as long as they are linearly independent. With different column vectors, x (the weight applied to columns) would be different. But |Ax| would produce same result. This idea can be extended to any space

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