Let $V$ be a finite-dimensional real inner product space. I am fairly sure that the following operation cannot exist, and would like a nice conceptual explanaition for its impossibility.
Question: Can there exist a product $(x,y) \mapsto xy : V \times V \to V$ satisfying
- Bilinearity
- Commutativity
- Orthogonality meaning $xy$ is perpindicular to $x$ (and $y$ too, by commutativity) for all $x,y \in V$?
Added: As was pointed out in the comments, I want to exclude the zero product.
My thoughts: I didn't think of anything too useful, just that the orthogonality gives us that for fixed $x$, the map $y \mapsto xy$ is an antisymmetric transformation. So we have the identity $(xy,z) = -(y,xz)$. This lets us play around a bit, e.g.
$$(xy,z) = (yx,z) = -(x,yz) = -(x,zy) = \ldots$$
but I didn't get anywhere this way.
I claim that the only such product is the trivial product.
Observe that the datum of a bilinear form $m : V \times V \to V$ is equivalent to the datum of a tensor $\omega \in (V^\ast)^{\otimes 3}$ via $$ \forall x, y, z \in V, \quad \omega(x,y,z) = \langle x,m(y,z) \rangle. $$ Then $m$ is commutative if and only if $\omega \in V^\ast \otimes (S^2 V^\ast)$ and satisfies the orthogonality condition if and only if $\omega \in (\wedge^2 V^\ast) \otimes V^\ast$, so that $m$ is an operation of the desired form if and only if $$ \omega \in V^\ast \otimes (S^2 V^\ast) \cap (\wedge^2 V^\ast) \otimes V^\ast. $$ But now, suppose we have such a tensor $\omega$. Then for any $x$, $y$, $z \in V$, $$ \omega(x,y,z) = -\omega(y,x,z) = -\omega(y,z,x) = \omega(z,y,x) = \omega(z,x,y) = -\omega(x,z,y) = -\omega(x,y,z), $$ so that $\omega(x,y,z) = 0$.
If you like, the underlying reason why any product of the desired form must be zero is that the subspaces $V^\ast \otimes (S^2 V^\ast)$ (corresponding to symmetry) and $(\wedge^2 V^\ast) \otimes V^\ast$ (corresponding to the orthogonality condition) of $(V^\ast)^{\otimes 3}$ have trivial intersection, as a result of the behaviour of the canonical representation of the permutation group $\mathfrak{S}_3$ on $(V^\ast)^{\otimes 3}$ by permutation of indices. Explicitly, since $V^\ast \otimes (S^2 V^\ast)$ is the $+1$ eigenspace of the permutation $(2\,3) \in \mathfrak{S}_3$ and $(\wedge^2 V^\ast) \otimes V^\ast$ is the $-1$ eigenspace of the permutation $(1\,2) \in \mathfrak{S}_3$, it follows that $V^\ast \otimes (S^2 V^\ast) \cap (\wedge^2 V^\ast) \otimes V^\ast$ is in the $-1$ eigenspace of $$ (2\,3)(1\,2)(2\,3)(1\,2)(2\,3)(1\,2) = e \in \mathfrak{S}_3, $$ which, of course, is trivial.