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vector space

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linear space
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Hungerford's _Algebra_ defines vector spaces as unitary modules over division rings, which means funky, noncommutative things can happen (you can have a "left" vector space which isn't a "right" vector space, for example). Knowing next to nothing about noncommutative algebra, I don't know if such beasts are useful enough to talk about. Anyone know?

My instinct is usually "no", although there are some noncommutativity partisans here who will disagree with me.

The rationale is that the noncommutative viewpoint adds so much extra complexity that it obscures the essence of the concept for those who are trying to learn it for the first time (and who obviously benefit least from the noncommutative viewpoint).

What vector space is not a metric space? Anyone can provide an example?

Thank you


None. Every vector space can be endowed with at least one metric. Here's one way of exhibiting a metric on an arbitrary vector space V.

Pick a Hamel basis $B$ for the vector space V. Then every vector v \in V can be uniquely be written as a finite linear combination of elements of B:
v = \sum_{i=1}^n c_i b_i
with $b_i \in B$. Define the norm of v as
||v|| = \sup_{i=1}^n |c_i|
It is easy to see that this meets the three criteria for norm on a vector space:

1) If ||v|| = 0, then for all i, c_i = 0, which implies that v = 0.

2) For any scalar s \ne 0, ||sv|| = \sup_{i=1}^n |s| |c_i| = |s| \sup_{i=1}^n |c_i| = |s| ||v||.

3) ||v + w|| = \sup_{i=1}^n |c_i + c'_i| \le \sup_{i=1}^n |c_i| + \sup_{i=1}^n |c'_i| = ||v|| + ||w||

.. but not all topological vector spaces are metrizable.
For example, the set of functions with compact support
has a topology which is not metrizable.

You're assuming that you have a modulus |.| in the field underlying the vector space. If you don't have a modulus, you'll have a hard time making a space into a normed space (which is what you're actually doing).
BTW: The obvious answer to the literal question is that any set is a metric space (e.g. with metric d(x,y)=1 unlesss x=y).

I agree, I did make that assumption because the usual definition of a normed vector space is based on it and the usual definition of a metric involves the real numbers.

As you say, if one takes the question literally, one can find a metric. Of course, this answer is somewhat unsatisfactory, because it has nothing to do with the underlying space being a vector space.

So we have two answers and two objections. On the one hand, the first answer prescribes a metric which is compatible with the algebraic properties of a vector space, but restricts the choice of field. On the other hand, the second method makes no assumption about the underlying field, but there is makes no connection between the metric and the algebraic properties.

The only way to resolve this dialectical tension is to look for a synthesis which combines the algebraic compatibility of the thesis with the universality of the anthithesis to the extent that this is possible. (This is the first occasion I have used Hegel's terminology which was not a discussion with Marxists. I also worry that I'm getting about as long-winded and obscure as Hegel :!( )

To do this, we should see what it means for a metric on a vector space to be compatible with the algebraic structure without assuming that the field has a modulus.

One way to start is by looking at one-dimensional subspaces. Every vector space has them. Moreover, every one dimensional vector space is isomorphic to the underlying field considered as a vector space over itself. The restriction of the metric to the one dimensional subspace will provide a metric on the vector space. Compatibility of the metric with the algebraic operations on the vector space reduces to compatibility of the metric with the algebraic operations of the field.

So, it seems that, however one chooses to define "compatibility of the metric with algebraic operations", there's no chance of a vector space having a compatible metric. If one defines compatibility of a metric with field operations to mean that the field has a modulus, then that limits the choice of field severely, although my proof shows that every vector space over that field will have a metric which is compatible with the algebraic structure. Since this assumption is quite restrictive, it would make sense to look into weaker definitions of "compatibility of the metric with the algebraic structure".

As I see it, the question now bifurcates. First, there is the question of which fields admit a metric which is compatible with the field operations under weaker definitions of compatibility. Second, there is the question of whether the fact that there exists a metric compatible with the field operations necessarily implies that there exists a metric on a vector space over that field which is compatible.

This is an interesting question. Let's keep working at it.


My professor says that some vector spaces are not metric space. He let us to give an example in an exam.

As the last few posts on this show, the answer can depend on exactly how you interpret the question. What fields are allowed as scalars? What compatibility conditions, if any, are being imposed between the metric and the algebraic sructures?

If you could you post an example of a vector space which is not a metric space according to your professor's definitions, it might help in clearing up these ambiguities.

My professor gives a such example

A set of infiniste sequences is a vector space, but not a metric space since we cannot calculate the distance between any two points.

For example (1,1,...), (2,2,...), (3,3,...)

It is possible to define a metric on the set of all infinite sequences using a Hamel basis as outlined in my first post in this thread. Therefore, this is not an example of a vector space which is not a metric space.

It is worth pointing out that the method used to show that a Hamel basis is not constructive, hence there it is not possible to compute the distance between tow elements, even though it exists. Also, the topology induced by this metric would not be a topology one would typically impose on the set of all sequences.

Could it be that you and your professor mean to say that the usual topology on this space is not a metric topology?

Naylor and Sell on p. 4 of Linear Operator Theory in Engineering and Science state that "It is possible to have metric spaces that are not linear spaces and vice-versa". My understanding is that the two-dimensional surface of a sphere in 3D is an example of a metric space that is not linear space. Similar to the original query, is there an example of a linear space that is not a metric space?

I might be a little out of it at 5am , but what if you considered a field extension of a finite field.

Say F = F_p and E = F_{p^2}. Then F/E is a vectorspace over F. Can we place a metric on F/E ?

Nevermind , I wasnt think about valuations of a field. It is possible to assign a function | | : K ----> R with the metric space properties.

I have just looked in this book and, if I understood correctly, assuming that linear spaces = vector spaces, the book cannot be correct. Every set is a metric space (it admits the discrete metric). This means that every vector space is a metric space. The book must mean something different.

One interpretation of what the book may actually mean is as follows. You may be given a topological space that is a vector space over R. Does there exist a norm on this vector space so that the topology induced by the norm is the topology given apriori? Does there exist an inner product with the same properties? Does there exist a metric with the same properties? The last question is the least relevant since metrics are not required to respect vector space operations in any way, all it asks is whether the given topology is metrizable. The answer to all of the above questions is "no, not necessarily". Indeed, it suffices to give an example of a single topological vector space so that the topology is not metrizable (since inner products induce norms induce metrics), and this was done several posts ago.

Furthermore, one direction of what this book says is correct. It is possible to have metric spaces that cannot be vector spaces. One example is Z/6, the set of integers modulo 6. It can be made into a metric space (every set can), however, it cannot be a vector space. Indeed, suppose that Z/6 is a vector space over some field k. Then it has a basis (requires AC in general). If k is infinite then Z/6 must be infinite since all k-multiples of any basis element must be distinct, which is impossible. Thus k must be finite, so the size of k is some p^i for p a prime. Let the cardinality of the basis be n. Then, via the same proof as the cardinality of finite fields being a power of a prime, the size of Z/6 must be (p^i)^n, which it is not, for any p.

As an aside, I think that all infinite sets can be made into vector spaces over Q because they are in bijection with some vector space over Q (I am not sure of this, I believe the free Q-module can be taken in most cases but maybe not) and I can always simply pull the vector space operations back via the bijection. At any rate, any set that is in bijection with a vector space can itself be made into a vector space. I believe this also requires AC but I do not know enough set theory to say for certain.

I hope this helps clear some things up,
- Eugene

While you can give any set the discrete metric (thus turning it into a topological space), if the resulting topology is not the original topology on the set, then the result is not really the same object. That's where you'll get the statement that there exist Vector spaces that are not Metric spaces - Sure, you can take your non-metrizable Vector space and give it the discrete metric, but does that space even look remotely like the space you started with?
"It's John Ashcroft's vision of hell. It's a Tom Waits song."
Simon - mhm27x5

Vector Space

Yes it is worth at least a definition. There are places in PM where
some results depend on being able to refer to a vector space over a
division ring.
To satisfy both camps (or make both unhappy) maybe a compromise
is to at least mention in your entry that in some of the literature
a vector space is defined over a division ring, using the same definition. That way those entries that need that concept can
point to yours and refer to a vector space over D , a division ring.

Is there some notion of what kind of metric spaces can be turned into vector spaces by finding suitable basis vectors? In other words, given a metric space, is there a test that can say no vector space can exist that conforms to the distance function?

I think the precise question you are trying to ask is "When is a metric space isometric to a normed vector space?"

I am not sure what the answer is, but there might not be a very nice answer; metric spaces are pretty floppy and the geometry of a vector space depends quite a bit on the base field and on finite vs infinite dimension. But if finite dimensional vector spaces over R would be enough for you, I can give a few suggestions:

First, there are some pretty clear topological restrictions coming from the fact that all metrics define the same topology on a finite dimensional real vector space. Without thinking too hard, the metric space must be complete, connected, simply connected, locally compact, and non-compact. You might be able to think of some others. Topology will give you lots of necessary conditions like this, but I doubt it will give you very many sufficient conditions; the hyperbolic plane is topologically very similar to Euclidean space.

Second, there is a basic theorem in Riemannian geometry that says that a Riemannian manifold is isometric to a Euclidean space if and only if is has constant sectional curvature K = 0. Maybe this result can be promoted to Alexandrov spaces; I'm not quite sure. Some collection of topological restrictions plus some sort of curvature condition seems the most likely to get what you want, at least for spaces where curvature makes sense.

Finally, Hilbert's 5th problem suggests a direction to try. I believe it is true that a connected locally compact topological group is a Lie group if and only if it has no small subgroups (i.e. there exists a neighborhood of the identity such that does not contain any subgroups). With the aid of the classification theorem for Abelian Lie groups, you can almost reduce the question to asking when a connected locally compact metric space is homeomorphic to a topological group.

I hope that is some help!

Edit: all NORMS define the same topology, not all METRICS.

Let me begin with a simple problem: a few days ago thee was some mention of A.Odzylko;s age.When I tried to search for this the next
day I could not find it.Mersenneforum has a search engine which
enables one to pick up the thread of any discussion & continue.Trust
my point is clear.

> How do I search for a discussion topic

You don't --- there is no search function for the forum posts.
All you can do is look through the past posts by hand.

How are called in English "shift" operators in vector spaces, that is a linear function which adds a fixed vector to the argument?

f(v) = z+v where z is fixed
Victor Porton -
* Algebraic General Topology and Math Synthesis

Usually a map like this is called translation, or, in a specific case, translation by v.

- Keenan

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