On the explicit double cover of the $SO(n)$

Question

In the Wiki page for spin group we have the following:

Let $V$ be an $n$ dimensional normed space over $\mathbb R$. Let $$TV=\mathbb R\oplus V\oplus(v\otimes V)\oplus\cdots$$ be the tensor algebra of $V$. Define the Clifford algebra $Cl(V)=TV/(v\otimes v+||v||^2)$. Then $Cl(V)$ has a nayural grading $$Cl(V)=Cl^0\oplus Cl^1\oplus\cdots$$

The group $\text{Pin}(V)$ is the subgroup of $Cl(V)$ consisting of elemnets of the form $v_1\otimes\cdots\otimes v_k$.

The group $\text{Spin}(V)=\text{Pin}(V)\cap Cl^{even}$ where $Cl^{even}=Cl^0\oplus Cl^2\oplus\cdots$

Altenately, if we define a map $()^t:Cl(V)\rightarrow Cl(V)$ by $$(v_1\otimes\cdots\otimes v_k)^t=v_k\otimes\cdots\otimes v_1$$ and extend by linearity then $\text{Spin}(V)$ is the group of all $a\in \text{Pin}(V)$ such that $a\otimes a^t=1$.

With this notation, an explicit double covering is a homomorphism given by $\rho(v)a=a\otimes v\otimes a^t$ where $v\in V$.

Question : How is the above a map from $\text{Spin}(V)$ to the special orthogonal group $SO(n)$? Shouldn't the map take $a$ to a matrix in $SO(n)$? Why does $v$ come into the picture? I wonder if I'm missing something very basic here.

Edit : From the answer by lisyarus I have understood how $a\mapsto \rho(a)$ is a map from $\text{Spin}(V)$ to $O(n)$. However it is not clear to me why the determinant of $\rho(a)$ should be $1$. That is why should $\rho(a)$ land inside $SO(n)$?

Answer 1

$V$ is considered embedded into $Cl(V)$ via a natural map $V \rightarrow TV \rightarrow Cl(V)$. Now, the mapping $v \mapsto ava^{t}$ defines an automorphism of $Cl(V)$. One can show that for $v \in V$, viewed as an element of $Cl(V)$ via the aforementioned embedding, $ava^t$ is also a vector from $V$ (as opposed to some general element of $Cl(V)$).

So, the operation $\rho(a)v=ava^t$ defines an automorphism of $V$. Recall that the inner product in $V$ can be expressed through $Cl(V)$ multiplication as $\langle v,u\rangle={1\over 2}(vu+uv)$. Now, it is easy to show that $\rho(a)$ preserves inner products (remember that $aa^t=a^ta=1$)

$$\langle \rho(a)v,\rho(a)u\rangle=\langle ava^t,aua^t\rangle = \\ = {1\over 2}(ava^taua^t+aua^tava^t) = \\ = {1\over 2}(av(a^ta)ua^t+au(a^ta)va^t) = \\ = {1\over 2}(avua^t+auva^t) = \\ = a{1\over 2}(vu+uv)a^t = \\ = a\langle v,u\rangle a^t = \\ = \langle v,u\rangle aa^t= \\ = \langle v,u\rangle$$

The equality $a\langle v,u\rangle a^t = \langle v,u\rangle aa^t$ holds because $\langle v,u\rangle$ is a scalar and commutes with all elements of $Cl(V)$.

Answer 2

lisyarus already proved that $\rho(a) \in O(n)$. To see it preserves the orientation, we have $$ v-\frac{2Q(v,a)}{Q(a,a)}=\frac{vaa-(va+ava)}{Q(a,a)}=-ava$$ which is reflection in the hyperplane perpendicular to $a$. Since $\rho(a)$ is the composition of even number reflections, it preserves the orientation.