Euclidean Space definition

[Damidami]

Hi,
I'm trying to fix in my head a very precise definition of what to mean for an euclidean space, as we use it in multivariable calculus.

The def. I had in my mind was that an ES is
(1) a real vector space
(2) of finite dimension
(3) with the "standard" (dot)
(4) inner product

I'm pretty sure of (1) and (4) (it has to be some vector space, and has to have an inner product to mesasure distances and angles). Not so sure about (2) (maybe infinite dimensional euclidean spaces are well defined?) and (3) (I've read somewhere it is not necesary that the inner product of an ES be the "dot" one, but changing the inner product doesn't change all measures (distances and angles) in the space? Does all theorems about ES still apply if I change the inner product?)

But what more bothers me is that there is no definition of a "cross product" in that definition, so I should not use it when calculating "surface integrals" in euclidean space $\mathbb{R}^3$.
Or is there a way of calculating $a \times b$ using only the inner product $a \cdot b$ ?

A little confused with those things.
Any help?
Thanks,
Damián.

 

[micromass]

Can't you just say that an Euclidean space is by definition $\mathbb{R}^n$??

 

[Damidami]

Can't you just say that an Euclidean space is by definition $\mathbb{R}^n$??

Hi micromass,
I don't understand, you mean euclidean space is simply by definition the set $\mathbb{R}^n$?
Take for example $n=2$ (we have euclidean plane $\mathbb{R}^2$), what is then the distance between $(1,1)$ and $(2,2)$?
(Remember there is no notion of distance in a "naive" set)

Thanks,
Damián.

 

[micromass]

The space $\mathbb{R}^n$ has a natural distance

$$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

 

[Damidami]

The space $\mathbb{R}^n$ has a natural distance

$$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Hi micromass,
Do you mean the distance between two vectors $(x_1, y_1)$ and $(x_2, y_2)$? So you are talking about $\mathbb{R}^2$?
Is there some other natural thing I should know about euclidean space $\mathbb{R}^2$?
I mean, that is precisely my question, what operations (structure) is defined on a euclidean space and which not. Can you naturally take the cross product of two vectors in $\mathbb{R}^2$? Or has euclidean space $\mathbb{R}^3$ some kind of special status with more operations in it?

Thanks,
Damián.

 

[Bacle2]

An additional point that you may be interested in: there is also the fact that the norm (and therefore metric) in ℝⁿ is generated by the inner-product you refer to. Thisis not true for all metric spaces (it's only so for Hilbert Spaces).

Maybe too, the issue that there is a standard topology used in ℝⁿ, i.e., the one with the basis generated by the open balls ,is/coincides with the topology generated by the metric originating from the norm, which originates from the inner-product, (which bought the cat...)

 

[micromass]

An additional point that you may be interested in: there is also the fact that the norm (and therefore metric) in ℝⁿ is generated by the inner-product you refer to. Thisis not true for all metric spaces (it's only so for Hilbert Spaces).

Maybe too, the issue that there is a standard topology used in ℝⁿ, i.e., the one with the basis generated by the open balls ,is/coincides with the topology generated by the metric originating from the norm, which originates from the inner-product, (which bought the cat...)

Indeed. Furthermore, the inner product/norm is unique! The vector space $\mathbb{R}^n$ only carries one inner product!

 

[Bacle2]

I guess you mean it is unique in the sense that it is the only inner-product that gives rise to the topology with basis the open sets? Otherwise, the norm, as a stand-alone is not unique; any two norms on finite-dimensional normed spaces are equivalent.

 

[D H]

I guess you mean it is unique in the sense that it is the only inner-product that gives rise to the topology with basis the open sets?

Putting words in micromass' mouth (or on his fingers), I pretty sure he meant to within an isomorphism. And that doesn't really count as being "different".

Otherwise, the norm, as a stand-alone is not unique; any two norms on finite-dimensional normed spaces are equivalent.

That's not true. It's true for any inner product vector spaces of the same (finite) dimensionality, but there are plenty of norms that are not inner product based and those normed (but not inner product) vector spaces are distinct from Rⁿ.

 

[Bacle2]

Putting words in micromass' mouth (or on his fingers), I pretty sure he meant to within an isomorphism. And that doesn't really count as being "different".


That's not true. It's true for any inner product vector spaces of the same (finite) dimensionality, but there are plenty of norms that are not inner product based and those normed (but not inner product) vector spaces are distinct from Rⁿ.

How am I putting words in anyone's mouth, when I say " I guess you mean"? I think this phrasing makes it clear that I'm using my own interpretation. Are you referring to inner-product space isomorphism? I prefer redundancy than conciseness that may be confusing; redundancy may be ignored, but a confusion may not be so easily-cleared.

And I guess you did not know what I meant: two norms on the same finite-dimensional space are equivalent, i.e., they both generate the same topology.

 

[D H]

I was the one putting words in micromass' mouth, not you.

And, no, I did not know what you mean. There are multiple meanings of equivalent. While the $\mathcal l_1$ norm is topologically equivalent to the $\mathcal l_2$ norm, it is not isomorphic (isometrically isomorphic) to the $\mathcal l_2$ norm.

 

[Bacle2]

O.K, no problem; this is an example of why I prefer some redundancy. By equivalent norms on the same space I meant that the topologies generated by the metric (the metric which is generated by the norm ) are the same. Take ||.||_1 and ||. ||_2 , both on the same finite-dimensional normed space V. Each gives rise to a metric d(x,y)||x-y||_i ; i=1,2 , which generates a topology, whose basis are the open balls d*x,y)<r.
The two metric topologies are equivalent , in that every set that is open in one topology is also open in the other topology.

I never bought into the full Bourbaki approach of extracting all (or, most ) content and leaving only the formal relations; I prefer to see some of the mess/background from classical math that gives some context. Anyway, just my taste.

Anyway, I'll try to be more clear next time.

 

[AntsyPants]

Can't you just say that an Euclidean space is by definition $\mathbb{R}^n$??

Feel free to correct me, but saying Euclidean space is by definition $\mathbb{R}^n$ may be a bit of an abuse.

$\mathbb{R}^n$ is a vector space generated by the span of $n$ linearly independent vectors also with dimension $n$. This space, however, needn't be Euclidean as on its own it lacks an inner product as the other user stated. In order for this space to be Euclidean, the inner product of two elements $\textbf{x}$ and $\textbf{y}$ must be given by $\textbf{x}^{T}\textbf{y}$.

On the other hand, we can in the exact same space ($\mathbb{R}^n$) define alternative inner products, such as

$<\textbf{x},\textbf{y}> = \textbf{x}^{T} \textbf{A} \textbf{y}$,

where $\textbf{A}$ is a symmetric positive-definite matrix. It can be easily shown that this defines an inner product in $\mathbb{R}^n$.

The space is exactly the same, but it isn't Euclidean, unless $\textbf{A}$ is the identity matrix.PS: I only read some of the posts diagonally because there was too much uneeded text involved in the middle.

 

[micromass]

Feel free to correct me, but saying Euclidean space is by definition $\mathbb{R}^n$ may be a bit of an abuse.

$\mathbb{R}^n$ is a vector space generated by the span of $n$ linearly independent vectors also with dimension $n$. This space, however, needn't be Euclidean as on its own it lacks an inner product as the other user stated. In order for this space to be Euclidean, the inner product of two elements $\textbf{x}$ and $\textbf{y}$ must be given by $\textbf{x}^{T}\textbf{y}$.

On the other hand, we can in the exact same space ($\mathbb{R}^n$) define alternative inner products, such as

$<\textbf{x},\textbf{y}> = \textbf{x}^{T} \textbf{A} \textbf{y}$,

where $\textbf{A}$ is a symmetric positive-definite matrix. It can be easily shown that this defines an inner product in $\mathbb{R}^n$.

The space is exactly the same, but it isn't Euclidean, unless $\textbf{A}$ is the identity matrix.


PS: I only read some of the posts diagonally because there was too much uneeded text involved in the middle.

Yes, but the space you mention is isometrically isomorphic to the Euclidean space. So have no problem calling the space with that inproduct Euclidean.

 

[Damidami]

Hi AntsyPants,
Thanks for your answer. It's the one which came closer to what I was asking.
My question is, how do you calculate the area of a paralelogram spanned by two vectors in euclidean space $\mathbb{R}^3$?

You should only use vector space operations and inner product operation, there is no cross product in euclidean space, am I right?

 

[AntsyPants]

Yes, but the space you mention is isometrically isomorphic to the Euclidean space. So have no problem calling the space with that inproduct Euclidean.

I see your point, but you still need to define a norm or else the space isn't metric, so saying it an euclidean space is by definition $\mathbb{R}^n$ still isn't correct. However, based on what you said, if you define a norm in $\mathbb{R}^n$ then the space is instantly euclidean, regardless of the norm. That I can agree with.

 

[Damidami]

Hi AntsyPants and micromass,

Yes, but the space you mention is isometrically isomorphic to the Euclidean space. So have no problem calling the space with that inproduct Euclidean.

That's interesting. Even if when we change the inner product the distance between two elements change, we can still thought it as the same space because they are isometric?
So in the definition of euclidean space it doesn't matter what the inner product is, but that some (any) inner product is defined on it?

I see your point, but you still need to define a norm or else the space isn't metric, so saying it an euclidean space is by definition $\mathbb{R}^n$ still isn't correct. However, based on what you said, if you define a norm in $\mathbb{R}^n$ then the space is instantly euclidean, regardless of the norm. That I can agree with.

Now we don't agree. An inner product has more structure that a norm (every inner product space is a normed space, but not every NS is an IPS)

And I don't think that any norm in $\mathbb{R}^n$ makes it an euclidean space, the norm has to come from an inner product. (we still need to measure angles in euclidean space)

What is starting to confuse me is that I thought there was only one norm that came from an inner product (that is $|| \cdot ||_2$), but we now have many inner products that define an eclidean space, but they don't all define the same norm, don't they?

And, I'm still asking, how do you measure the area of the paralelogram spanned by two vectors in eclidean space $\mathbb{R}^3$, because $|| a \times b ||_2$ has no sense as there is no "cross" product in an inner product space.

Thanks.

 

[Bacle2]

Antsy Pants:

If you select just any norm, and not one generated by the inner-product, you are giving up the fact that R^n is a Hilbert Space--which is , I think, an intrinsic property of R^n.

 

[Bacle2]

DamiDami wrote, in part:

"What is starting to confuse me is that I thought there was only one norm that came from an inner product (that is ||⋅||2), but we now have many inner products that define an eclidean space, but they don't all define the same norm, don't they?

And, I'm still asking, how do you measure the area of the paralelogram spanned by two vectors in eclidean space R3, because ||a×b||2 has no sense as there is no "cross" product in an inner product space.

Thanks."

I'm not sure I understand your question: the standard inner-product in ES is given by:

<(a₁,...,a_n), (b₁,...,b_n)>:=

$\sum$_i=1ⁿ (a_ib_i)

This gives rise to just one norm.

Re the parallelogram, I think this is done in general with the exterior and
wedge products.

 

[Damidami]

Re the parallelogram, I think this is done in general with the exterior and
wedge products.

But in euclidean space you don't have an exterior nor wedge product, so there has to be another way.

 

[micromass]

But in euclidean space you don't have an exterior nor wedge product, so there has to be another way.

The only thing you need is an orientation of your basis vectors. You must know what the positive and negative bases are. Once you know that, you can make the cross-product:

Indeed, the cross product of u and v is the unique vector which is
- Perpendicular to u and v
- has norm $\|u\|\|v\|\sin\theta$ with $\theta$ the angle between u and v
- the three vectors u,v and $u\times v$ have positive orientation

All these things can be done in a vector space with inner product and with orientation.

 

[Damidami]

The only thing you need is an orientation of your basis vectors. You must know what the positive and negative bases are. Once you know that, you can make the cross-product:

Indeed, the cross product of u and v is the unique vector which is
- Perpendicular to u and v
- has norm $\|u\|\|v\|\sin\theta$ with $\theta$ the angle between u and v
- the three vectors u,v and $u\times v$ have positive orientation

All these things can be done in a vector space with inner product and with orientation.

Finally, the whole point of this thread was to know what exactly is the structure that defines euclidean space.

So is it safe to say that euclidean space is a real finite dimensional vector space oriented and with an inner product?

If that is it, the only bit I still don't understand is the isometric isomorphism between all inner products in $\mathbb{R}^n$. Is it difficult to prove? The wikipedia says that rotations translations and reflections are isometries, but doesn't say all inner products spaces are isometric.

 

[micromass]

It's a consequence for Hilbert spaces. Check "Hilbert dimension". Every two Hilbert spaces with the same Hilbert dimension are isometric.

 

[D H]

If that is it, the only bit I still don't understand is the isometric isomorphism between all inner products in $\mathbb{R}^n$. Is it difficult to prove? The wikipedia says that rotations translations and reflections are isometries, but doesn't say all inner products spaces are isometric.

Any inner product $<\boldsymbol a,\boldsymbol b>$ for $\boldsymbol a,\boldsymbol b \in \mathbb{R}^n$ can be written in terms of a symmetric positive definite matrix M:
$$<\boldsymbol a,\boldsymbol b>_M = \boldsymbol a^T M\; \boldsymbol b$$
Since the matrix M is symmetric and positive definite it can be decomposed as $M=PDP^T$ where P is an orthogonal matrix and D is a diagonal matrix with positive diagonal elements. The matrix $A=PD^{-1/2}$ will transform the vectors $\boldsymbol a$ and $\boldsymbol b$ to an equivalent form in which the canonical inner product applies.