Sphere

Definition

As a subset of Euclidean space

The unit $n$-sphere $S^n$ is defined as the subset of Euclidean space ${\mathbb{R}}^{n+1}$ comprising those points whose distance from the origin is $1$.

$S^n=\{(x_0,x_1,\dots ,x_n)\mid x_0^2+x_1^2+\dots +x_n^2=1\}$

Particular cases

$n$	sphere $S^n$
0	$S^0$ -- discrete two-point space
1	circle
2	2-sphere
3	3-sphere

Equivalent spaces

Space	How strongly is it equivalent to the circle?
boundary of the $(n+1)$-hypercube	homeomorphic; not diffeomorphic because of sharp edges
boundary of the $(n+1)$-simplex	homeomorphic; not diffeomorphic because of sharp edges
ellipsoid in ${\mathbb{R}}^{n+1}$	equivalent via affine transformation
one-point compactification of ${\mathbb{R}}^n$	homeomorphic via stereographic projection
universal cover of real projective space $\mathbb{R}{\mathbb{P}}^n$, which is the space of lines in ${\mathbb{R}}^{n+1}$	homeomorphic, diffeomorphic, also isometric if we choose the natural metric.

Algebraic topology

Homology groups

With coefficients in $\mathbb{Z}$, the $n$-sphere $S^n$ has $H_0(S^n)\cong H_n(S^n)\cong \mathbb{Z}$ and $H_k(S^n)=0$ for $k\notin \{0,n\}$. In particular, the $n$-sphere is $(n-1)$-connected.

Interpretations in terms of various homology theories:

Fill this in later

With coefficients in any $R$-module $M$ for a ring $R$, the $n$-sphere $S^n$ has $H_n(S^n)=M$ and $H_k(S^n)=0$ for all $k\ne n$.

Cohomology groups and cohomology ring

With coefficients in $\mathbb{Z}$, the $n$-sphere $S^n$ has $H^0(S^n)\cong H^n(S^n)\cong \mathbb{Z}$ and $H^k(S^n)=0$ for $k\notin \{0,n\}$. In particular, the $n$-sphere is $(n-1)$-connected.

With coefficients in any $R$-module $M$ for a ring $R$, the $n$-sphere $S^n$ has $H^n(S^n)=M$ and $H^k(S^n)=0$ for all $k\ne n$.

The cohomology ring is isomorphic to $\mathbb{Z}[x]/(x^2)$, where $x$ is a generator of the <amth>n^{th}</math> cohomology.

Homotopy groups

Further information: n-sphere is (n-1)-connected

For $k<n$, the homotopy group ${\pi }_k(S^n)$ is the trivial group. ${\pi }_n(S^n)\cong \mathbb{Z}$, with the identity map $S^n\to S^n$ being a generator.

The cases $k>n$ are discussed below:

Case	What can we say?
$n=1$	${\pi }_k(S^n)$ is trivial for all $k>1$
$k=2n-1$	${\pi }_k(S^n)\cong \mathbb{Z}$
$k>n,k\ne 2n-1$	${\pi }_k(S^n)$ is a finite abelian group