4
$\begingroup$

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 $1 \leq i<j \leq 4$, 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 $w \wedge x = \sum_{i<j}\left(w^i x^j - w^j x^i\right) E_i \wedge E_j$, 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<j}(w^i x^j - w^j x^i)^2 \\ &= \sum_{i<j}(w^i x^j - w^j x^i)^2 = \sum_{i\neq j} \left((w^i)^2(v^j)^2-w^i v^i w^j v^j\right) \\ &= |w|^2|v|^2-\langle w, v \rangle^2, \end{align*} 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?

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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.

I've added a correction to my online list. Thanks for pointing this out.

$\endgroup$
4
  • $\begingroup$ You're welcome, and thank you! I wonder if there's another sense in which the desired (and natural) choice of fiber metric is "unique"... $\endgroup$
    – D Ford
    Commented Mar 25, 2020 at 23:16
  • 1
    $\begingroup$ @DFord: The "real" story is the covariant analog of this inner product (i.e., on differential forms), and its relationship with the Hodge star operator. This is explained in Problem 2-18 in my book. The case of contravariant alternating tensors is much less important, but it can be handled the same way using the musical isomorphisms. $\endgroup$
    – Jack Lee
    Commented Mar 25, 2020 at 23:20
  • $\begingroup$ Dear professor @JackLee: I think the ''Stokes's Theorem" has been misspelled in your SM book. I think the correct one is ''Stokes' Theorem". Be care about COVID-19 please. $\endgroup$
    – C.F.G
    Commented Mar 27, 2020 at 14:33
  • 1
    $\begingroup$ @C.F.G: Actually, both Stokes's Theorem and Stokes' Theorem are acceptable; but most modern style guides used in academic writing (e.g., Chicago, MLA, APA) recommend the former. And I strongly prefer Stokes's Theorem, because it helps to dissuade students from writing "Stoke's Theorem," which is egregiously wrong no matter what style guide you follow. $\endgroup$
    – Jack Lee
    Commented Mar 27, 2020 at 17:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .