# Are there trilinear inner products?

Is there such a thing as a "trilinear inner product"? The definition of an inner product is:

Let $$H$$ be a vector space over $$\mathbb{K}\in \{\mathbb{R,C}\}$$. An inner product is a map $$\langle \cdot|\cdot\rangle: H^2 \to \mathbb{K}$$ such that for all $$x,y,z \in H$$ and $$\lambda \in \mathbb{K}$$ the following properties hold:

• Bilinearity: $$\langle x+\lambda y | z\rangle = \langle x|z\rangle + \lambda \langle y|z\rangle$$

• Complex conjugacy: $$\overline{\langle y | x \rangle} = \langle x | y \rangle$$

• Positive definiteness: $$||x||^2:=\langle x | x \rangle$$ > 0 if $$x \neq 0$$

Can this be modified to have a trilinear map $$\langle \cdot |\cdot| \cdot \rangle : H^3 \to \mathbb{K}$$? Would it for example be possible to make $$L^3$$ into a "trilinear inner product space" like $$L^2$$ is a "bilinear inner product space"? What is so special about the number $$2$$ in this context? Of course $$2$$ is the only nubmer that is conjugate to itself in the sense that $$\frac{1}{2}+\frac{1}{2}$$, so there would be no nice identification of this trilinear inner product space with it's dual.

I guess that there is no useful notion because the complex conjugacy can't be modified to get a trilinear inner product: $$\mathbb{C}$$ is a field extension of degree $$2$$ of $$\mathbb{R}$$, but there is no field extension of degree $$3$$ of the reals this inner product could be defined over. What if the "trilinear space" is solely defined over $$\mathbb{R}$$?

• Given a one dimensional vector space over the reals, try to come up with a trilinear map with the properties you listed. – Somos Apr 17 at 22:48
• @Somos Yeah defining it as $\langle x | y | \z \rangle := xyz$ the problem is with the positive definiteness. But what about "$2n$-linear inner products"? – Jannik Pitt Apr 18 at 9:44