Let $G$ be a matrix group. (I am interested in $G=O_n$ in particular). Then we can define the projective version $PG$ of $G$ to be the group consisting of the tensor products $$PG=\{A\otimes A\mid A\in G\}$$ It is easy to show that this corresponds to the maybe more standard definition of the projective version as $PG=G/(\pm I)$. Indeed, take some $A\in G$ and suppose for example that $A^1_1\neq 0$ (using the "physics notation" $A^i_j$ for the entries of $A$). Then we have $[A\otimes A]^{i1}_{j1}=A^i_jA^1_1$ and in particular $[A\otimes A]^{11}_{11}=(A^1_1)^2$, so we can recover all the entries of $A$ from $A\otimes A$ up to a global sign.
Now, my question is, what if we take the antisymmetrization of the tensor product. That is, consider the set of matrices $\widetilde{PG}:=\{A\wedge A\mid A\in G\}$ with $[A\wedge A]^{ik}_{jl}=A^i_jA^k_l-A^i_lA^k_j$. Surely this is a group (a quotient of $PG$). What kind of group is this? Does it coincide with $PG$? In other words, is the map $A\otimes A\mapsto A\wedge A$ injective?
Let $V$ be a real vector space. Since $PGL(V)$ is a simple group, the kernel of $GL(V) \to GL(\wedge^2 V)$ is contained in the scalar matrices. The scalar $a$ acts by $a^2$ on both $V^{\otimes 2}$ and $\wedge^2 V$, so $a$ acts trivially on $\wedge^2 V$ if and only if it does on $V \otimes V$.
In general, each complex irreducible representation of $GL(V)$ is contained in $V_{\mathbb C}^{\otimes n} \otimes \mathrm{det}^{-m}$ for some $n, m \geq 0$, where $\mathrm{det}$ is the determinant character and $V_{\mathbb C}$ is the complexification of $V$. The kernel of $GL(V)$ on such a representation is exactly the $(n-(\dim V)m)$th roots of unity, by the same argument as above.