I want to know why $Spin(V)$ is connected. I am watching $Spin(V)$, where $V$ is a real vector space with a positive defined metric, as the products of elements of the Clifford algebra $Cl(V)$ of the form $v_{1}...v_{r}$ where $v_{i} \in Cl(V)$ has norm $1$ and $r$ an even integer.
Why is $Spin(V)$ connected? (At the moment, I amnot supposed to know that is the covering space of any space).
So $r=2n$. Let $\gamma_i:[0,1]\to V$ be a path connecting $v_i$ to a fixed vector $v_0\in V$ with $|v_0|=1$, and such that $|\gamma_i(t)|=1$ for all $t$ (this exists because the unit sphere in $V$ is path connected). Then $\Gamma:[0,1]\to Spin(V)$ defined by $\Gamma(t)=\prod_i \gamma_i(t)$ connects $\prod_i v_i$ to $(v_0)^{2n}={-1}^n \in Spin(V)$.
Now take a path $\gamma_0$ connecting $v_0$ to $-v_0$ in the unit sphere of $V$. Then $v_0\gamma_0(t)$ connects $-1$ to $1$ in $Spin(V)$.
All together, any element of $Spin(V)$ is connected by a path to either $1$ or $-1$, and those two are connected by a path. So the whole $Spin(V)$ is (path) connected.