# Counterexample: Uniqueness of fiber metric on alternating $2$-vectors

I'm working on the following problem (Lee's "Riemannian Manifolds", Problem 8-33(a)).

Suppose $$(M,g)$$ is a Riemannian manifold. Let $$\Lambda^2(TM)$$ be the bundle of $$2$$-tensors on $$M$$. Show that there is a unique fiber metric on $$\Lambda^2(TM)$$ whose associated norm satisfies $$|w \wedge x|^2 = |w|^2|x|^2-\langle w, x\rangle^2$$ for all tangent vectors $$w, x$$ at every point $$q \in M$$.

My question: Are we guaranteed uniqueness?

Existence is straightforward by taking a local orthonormal frame $$\{E_1,\ldots, E_n\}$$ of $$M$$ and declaring $$\{E_i \wedge E_j : i < j\}$$ to be an orthonormal frame. One can further show using the algebra of alternating bivectors that given any local orthonormal frame $$\{\tilde E_1, \ldots, \tilde E_n\}$$, the corresponding set $$\{\tilde E_i \wedge \tilde E_j : i < j\}$$ of contravariant $$2$$-tensor fields is orthonormal in this inner product, so this fiber bundle is smooth and well-defined on all of $$M$$.

However, I'm not sure we have uniqueness. Consider $$(M,g) = (\mathbb{R}^4, \overline g)$$, where $$\overline g$$ is the Euclidean metric, and let $$\{E_1, E_2, E_3, E_4\}$$ be the standard orthonormal coordinate frame. Define the metric $$\langle \cdot, \cdot \rangle$$ on $$\Lambda^2(T\mathbb R^4)$$ by declaring $$|E_i \wedge E_j| = 1$$ for $$1 \leq i, along with the relations $$\langle E_1 \wedge E_2, E_3 \wedge E_4 \rangle = \langle E_1 \wedge E_4, E_2 \wedge E_3 \rangle = -\langle E_1 \wedge E_3, E_2 \wedge E_4 \rangle = 1,$$ and all products of the form $$\langle E_i \wedge E_j, E_i \wedge E_k \rangle = 0$$ for $$j \neq k$$. Noting $$w \wedge x = \sum_{i, one can show by direct computation that in this metric, we have: \begin{align*} |w \wedge x|^2 &= 2\bigg((w^1 x^2 - w^2 x^1)(w^3x^4-w^4x^3) - (w^1x^3-w^3x^1)(w^2x^4-w^4x^2) + (w^1x^4-w^4x^1)(w^2x^3-w^3x^2)\bigg) \\ &\quad+ \sum_{i because the parenthetical term to the right of the $$2$$ in the first equation above simplifies to $$0$$. This is obviously a different metric from the one generally constructed in the proof of existence, so is there a reason this metric fails the conditions of the problem, or is uniqueness indeed too much to ask for?

Oh no! You're absolutely right. The equation for $$|w\wedge x|^2$$ determines the norm on all decomposable $$2$$-forms, but that's not the same as determining the norm on all $$2$$-forms.