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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?

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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.

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  • $\begingroup$ 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. $\endgroup$ – Gibbs Oct 30 '18 at 10:12
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    $\begingroup$ 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 $\mathbb{C}$. $\endgroup$ – Enkidu Oct 30 '18 at 10:17
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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.

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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 $V$.

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

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