# Is $SO(n)$ a topological space?

I am reading some articles about covering space in Wikipedia. It says that $\operatorname{Spin}(n)$ is the universal cover of $SO(n)$ for $n>2$.

I cannot understand how people view groups as topological spaces. Does it mean that $SO(n)$ is a group with topological structure (topological group), or a fundamental group of some topological spaces?

Thanks.

• $SO(n)$ is a subset of $M_{n\times n}$, the space of $n\times n$ matrices. If it's over $\Bbb R$ (or $\Bbb C$), that means there is a topology on that matrix space as $\Bbb R^{n^2}$. Subspaces inherit a topology. That's my initial intuition, at least. – Arthur Sep 3 '15 at 7:18
• Everything may be a topological space as there is always a trivial topology. – Rubi Shnol Sep 3 '15 at 12:24
• @ValerySaharov Oh! my statement in the title isn't clear. My question is how people treat such groups in algebraic topology. Therefore not trivial topology nor discrete topology is the answer. – autodavid Sep 3 '15 at 13:43

$SO(n)$ is a subset of the space $M(n)$ of $n \times n$ matrices. This is a real vector space of dimension $n^2$ and is given the standard topological structure on a real vector space, induced by whatever your favorite norm is.
Any subset of a topological space is a topological space with the subspace topology. $SO(n)$ is not just a topological space and a group - it's both at the same time (a topological group). This is a topological space whose multiplication map $\mu: G \times G \to G$ and inversion map $\iota: G \to G$ are both continuous. This is true for $SO(n)$, because matrix multiplication is continuous (it's polynomial in the entries of the matrix; and/or because you can prove by hand that in the operator norm, $\|AB\| \leq \|A\|\|B\|$), and inversion is continuous (it's a polynomial function in the entries of the matrix by Cramer's rule, divided by the determinant, which never vanishes ofr the matrices we're considering and is also continuous). So $SO(n)$ is a topological group in this topology.
Virtually every matrix group you can think of ($GL_n, SL_n, U(n), SU(n), Sp(n),\ldots$) is a topological group, in fact a Lie group, which is a group with a "smooth structure" in which you can take derivatives.
Now pick a covering map $p: \tilde G \to G$. Then any element $\tilde e \in \tilde G$ with $p(\tilde e) = e$ determines a unique topological group structure on $\tilde G$ such that $p: \tilde G \to G$ is a continuous group homomorphism. The groups you get for all the different choices of $\tilde e$ are isomorphic. Thus we obtain the universal covering group $\text{Spin}(n) \to SO(n)$ for $n>3$. (For $n = 2$ it's just the double cover, and in this case we end up having $\text{Spin}(2) \cong SO(2)$.)