# Definition of sphere without using a metric

In differential geometry the $$n$$-sphere $$S^n$$ seems to be always defined as the set of points in $$\mathbb{R}^{n+1}$$ with distance $$1$$ from the origin. I am interested in a more topological definition, which does not depend on the presence of a metric in the ambient space.

Let us take a $$2$$-dimensional real vector space $$V$$ and consider $$V \setminus \{0\}$$. Define a relation $$\sim$$ between vectors in $$V \setminus \{0\}$$ such that $$v \sim w$$ if and only if $$v = \lambda w$$ with $$\lambda > 0$$. This is not the projective space $$\mathbb{RP}^1$$, as $$\lambda$$ is required to be positive. Now let us put the quotient topology on this set.

This looks like a definition of a $$1$$-sphere in $$V$$ independent of any inner product on $$V$$. Is this a commonly used definition? Can we take it as a definition of $$1$$-sphere?

This is not so commonly used, The definitions I mostly see are either the one point compactification of the $$n$$-dimensional vector space, respectively the CW-complex consisting of one $$n$$ cell and one $$0$$ cell, glued together in the obvious way.

Where, if you know a little topology, you can see that both are actually equivalent. Especially the second is used quite often, since it is immediately a CW-complex, a structure used very often in topology.

• You are right, I was considering the most common definition in differential geometry. I edited the question. Yes I know the definitions you are mentioning. – Gibbs Oct 30 '18 at 10:12
• yes you can, however you should be careful since there are occasionally issues arising, especially in differential geometry regarding a puncture of a point or a whole disk (seeing this is one advantage of the mapping class group). Purely topologically it works out of course. However, for calculations I would not recommend it, since the restriction of $\lambda > 0$ is quite unhandy, and tough to generalize to $\mathbb{C}$. – Enkidu Oct 30 '18 at 10:17

Yes, this definition works. It is sometimes used to define the sphere bundle associated to a real vector bundle without having to choose a metric on the bundle.

I do not think it is commonly used, but it obviously is a valid approach. Let us first consider $$V = \mathbb{R}^n$$. Then $$r : \mathbb{R}^n \setminus \{ 0 \} \to S^{n-1}, r(x) = x/\lVert x \rVert ,$$ is a strong deformation retraction. It is easily verified that $$r$$ is a quotient map. Defining $$x \sim y$$ iff $$r(x) = r(y)$$, we see that $$r$$ induces a homeomorphism $$\hat{r} : (\mathbb{R}^n \setminus \{ 0 \})/ \sim \phantom{} \to S^{n-1} .$$ But $$x \sim y$$ iff $$x = \lambda y$$ for some $$\lambda > 0$$ which shows that your definition yields a space canonically isomorphic to $$S^{n-1}$$.

Note, however, that it is not sufficient to consider an $$n$$-dimensional real vector space $$V$$. What we need additionally is a topology on $$V$$. It is well-known that there exists a unique topology making $$V$$ a Hausdorff topological vector space - this topology is the one that has to be used. It is induced by any norm on $$V$$.

• Right, I did not say that, but I am already considering a topology on $V$ (otherwise I could not define the quotient topology). – Gibbs Nov 2 '18 at 22:25