MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hi everybody,

We know the definition of a cone in a Real Banach Space. I want to know if there is any definition for a cone in an abstract metric space. Have you ever seen such definition anywhere?

Thanks in advance.

share|cite|improve this question
One might imagine something in terms of geodesic arcs. It would be helpful if you provided some context and indicated what features of a cone in a Banach space you consider to be basic and wish to generalize: otherwise it's just an exercise in name recognition that is probably more of a job for google and MathSciNet than a human being. For instance, suppose you have a Riemannian manifold. What do you want a cone to be in that case? – Pete L. Clark Jan 24 '10 at 7:33
Actually the feature of a cone that I want to use is that a cone induces a partial ordering relation. – Axiom Jan 24 '10 at 7:48
The question is a bit vague. If you speak of partial order, maybe you should check for "Lorentzian metrics". You have a cone there and a partial order. But the metric is not Riemanninan. In the Rieammanian case things like this could also be related to optimal control. Also there is such a structure on the universal cover of the grassmanian of Lagrangian planes -- this is related to Maslov index. – Dmitri Jan 24 '10 at 10:08
I need to define a partial order relation in a metric space via a cone like the real Banach space's case: If P is a cone in a real Banach space, then we have: x <= y iff x-y belongs to P I'm so sorry if my question was not clear. – Axiom Jan 24 '10 at 20:20
Try to explain what do you need it for. – Anton Petrunin Jan 24 '10 at 22:22

There is a natural definition of cone in the context of pointed metric spaces:

A pointed metric space $(X,p)$ is a cone if $(\lambda X,p)$ is isometric to $(X,p)$ for all $\lambda>0$.

Here, $(\lambda X,p)$ denotes the metric space obtained from $(X,p)$ by multiplying the distance function by $\lambda$. Also isometric should be understood in the pointed category, i.e., $(X,x_0)$ and $(Y,y_0)$ are isometric if there exists a distance preserving map $f\colon X\to Y$ such that $f(x_0)=y_0$. This is, for example, the definition you can find in the book by Burago, Burago & Ivanov, Def 8.2.1. It also coincides with the definition in a Banach space setting.

Obviously, you can then say that a subspace $(X,p)$ of your metric space $(\bar X,p)$ is a cone if it is a cone as an abstract metric space...

share|cite|improve this answer

First of all, the notion of $cone$ is a purely algebraic stuff, and not a metrical one. The $cone$ is naturally defined in the framework of $linear$ spaces, and not of Banach spaces. One can introduce, e.g., various "natural'' cones in a Hilbert space, without using its norm.

However, if $\left(X,\, d\right)$is a complete metric space, and $\psi:X\rightarrow\left[\,0,\,\infty\right)$is a lower semicontinuous function, then the partial ordering on $X$ defined by $x\preccurlyeq y$ iff $d\left(x,y\right)\leq\psi\left(y\right)-\psi\left(x\right)$is very useful in proving the Caristi-Kirk Fixed Point Theorem.

Another metrical variant would be to use the Ralph DeMarr' cone, combined with an Arens-Eells embedding.

share|cite|improve this answer

Maybe this works: given a point $p$ and a subset $A$ of your metric space $(X,d)$ define the cone on $A$ from $p$ to be all points that lie between $p$ and $A$, that is all points $x$ with $d(p,a)=d(p,x)+d(x,a)$ for some $a\in A$.

share|cite|improve this answer

You can isometrically embed any metric space into a Banach space via the Arens-Eells theorem (original: simpler proof by E. Michael: ). This embedding is, in some sense, canonical. Convex conse are well-defined in Banach spaces, so you could say that a point x is in the convex cone generated by x_1, ... x_n in the original metric space if f(x) is in the cone of f(x_1), ..., f(x_n) -- where f is the Arens-Eells embedding.

share|cite|improve this answer
In what sense is this embedding canonical? – Pete L. Clark Jan 24 '10 at 12:45
@Pete: I think it might be a left adjoint, IIRC; I remember seeing this mentioned in a talk of Godefroy some years ago. That is, every nonexpansive map from a (pointed?) metric space X into a Banach space E, extends canonically to a contractive linear map from F(X) into E, where F(X) is this Arens-Eells widget – Yemon Choi Jan 24 '10 at 21:59
Embedding a metric space (X,d) isometrically into a Banach space can be done in a simpler way, as in Lang's Undergraduate Analysis, exercise 3 in Chapter VI, Section 2. For each x in X, let f_x : X --> R by f_x(y) = d(x,y). This is a continuous function of y (Lang's exercise doesn't mention the continuity), although f_x need not be bounded. Fix a in X and define g_x = f_x - f_a (i.e., g_x(y) = d(x,y) - d(a,y)). Each g_x : X --> R is continuous and is bounded by the triangle inequality. The space B(X) of continuous bounded functions X --> R is a Banach space using the sup norm [continued] – KConrad May 16 '11 at 11:42
and ||g_x - g_y|| = d(x,y). Thus (X,d) embeds isometrically into B(X) by sending each x in X to the function g_x. This is not "canonical" since it depends on a choice of a. If the metric d is bounded on X then f_x is already in B(X) so we can drop the whole business with a: sending x to f_x is an isom. embedding of X into B(X). – KConrad May 16 '11 at 11:42
The construction given by KConrad can be considered as the enriched Yoneda embedding of a metric space, when seen as a category enriched in the symmetric monoidal closed structure on the preorder of real numbers (with order opposite to the standard order) equipped with the symmetric monoidal product given by addition. This is the beginning observation of Lawvere's famous paper on metric spaces and enriched category theory. – Todd Trimble May 16 '11 at 14:39

EDIT: As pointed out by Pete below, it seems I misunderstood the question, so what I write below is not relevant. Apologies!

This is not the general answer, but in riemannian geometry there is a notion of cone. If $(M,g)$ is a riemannian manifold, then its metric cone is $\mathbb{R}^+ \times M$, with $\mathbb{R}^+$ the positive real half-line parametrised by $r>0$, with metric $$dr^2 + r^2 g$$ The best example is of course $(M,g)$ the unit sphere in $\mathbb{R}^n$ and its cone is then $\mathbb{R}^n \setminus \lbrace 0\rbrace$. In this case (and in this case only) the metric extends smoothly to the origin, but in general the apex of the cone is singular.

This is used as a local model for conical (!) singularities and there is a nice interplay between the geometry of $M$ and that of its cone. The most dramatic use of the cone I know is that it turns the problem of determining which riemannian spin manifolds admit real Killing spinors into a holonomy problem, namely the determination of which metric cones admit parallel spinors.

Some of this generalises to the pseudo-riemannian setting; although this is perhaps not as useful as in the riaemannian setting as the holonomy classification in indefinite signatures (except for lorentzian) is still lacking.

share|cite|improve this answer
@Jose: I found this usage by a google search (I am not a Riemannian geometer...) but thought that it did not answer the question because the cone does not live in the original space. – Pete L. Clark Jan 24 '10 at 12:34
@Pete: Indeed... I guess I misunderstood the question :( – José Figueroa-O'Farrill Jan 24 '10 at 14:11
@Jose: Well,'s certainly not written so as to be easy to understand. – Pete L. Clark Jan 24 '10 at 14:23
I need to define a partial order relation in a metric space via a cone like the real Banach space's case: If P is a cone in a real Banach space, then we have: x <=y iff x-y belongs to P I'm so sorry if my question was not clear. – Axiom Jan 24 '10 at 20:18

link textcone is a special type of set which shoud stisfy some properties. let P is cone then it shoud be closed non empty having atleast 2 elements all linera combination by positive constant shoud also lies in P if x lies in p then its assitive inverse may not lies in p

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.