Suppose I have a metric $g$ and a bivector $ F $ on a four-dimensional vector space. It seems I can always decompose $ F $ into four mutually orthogonal vectors $a,b,c,d$ $$ F = a\wedge b + c\wedge d $$

Here is a sketch of a proof using tensor notation where the metric lowers indices: $$F^\mu_{\,\,\,\,\nu}\equiv F^{\mu\rho}g_{\rho\nu}.$$ We pick a unit eigenvector $u$ of $F^2$, $$ F^\mu_{\,\,\,\,\rho}F^\rho_{\,\,\,\,\nu}u^\nu = \lambda u^\mu,$$ and define the vector $v$, $$v^\mu \equiv F^\mu_{\,\,\,\,\nu}u^\nu,$$ then I can decompose $F$ as $$F^\mu_{\,\,\,\,\nu} = v^\mu u_\nu - u^\mu v_\nu + G^\mu_{\,\,\,\,\nu}.$$ Unless I'm being stupid this is the required decomposition into orthogonal simple bivectors (once we raise the index), but this is surprising to me since $g$ is arbitrary.

Can someone explain my mistake or explain why it's obvious that this is true and I shouldn't be surprised? I am familiar with the idea that any antisymmetric tensor can be put into block diagonal form using orthogonal matrices. If the metric were the Euclidean metric, this would give the required decomposition. Is there a generalization of this idea for arbitrary metric?

  • $\begingroup$ What do you mean by ${G^\mu}_\nu$? - It may be helpful not to think about tensors' matrix representations. The linear algebra still works without using those reps. Finding a block diagonal form of an antisymmetric matrix means finding a basis in which the action of the linear map on pairs of basis vectors decouples from other pairs. This is done with an orthogonal transformation. All orthogonal transformations are compositions of reflections, which are inherently metrical. There is no notion of being normal to a hyperplane otherwise. Here, you use reflections that correspond to the metric. $\endgroup$
    – Muphrid
    Commented Nov 26, 2014 at 6:55
  • $\begingroup$ $G$ is just defined as the remaining piece of $F$ after I subtract out the simple bivector in terms of $u$ and $v$. The key is that since $u$ is defined as an eigenvector, $G$ annihilates both $u$ and $v$ and thus can be expressed as an exterior product of vectors orthonormal to $u$ and $v$. $\endgroup$
    – octonion
    Commented Nov 26, 2014 at 7:26
  • $\begingroup$ @Muphrid That's exactly what I'm looking for is a explanation without components. I think I've proved that for any given metric (and so any notion of 'orthogonal' not just Euclidean) you can find an appropriate orthogonal basis where the pairs decouple. Maybe I'm not understanding jargon but when you are talking about orthogonal transformations that only corresponds to a Euclidean metric (?) $\endgroup$
    – octonion
    Commented Nov 26, 2014 at 7:33
  • $\begingroup$ A transformation can be orthogonal with respect to one metric and not orthogonal wrt another. Every metric has its own group of orthogonal transformations $\endgroup$
    – Muphrid
    Commented Nov 26, 2014 at 15:56
  • $\begingroup$ I think the question is whether $F^2$ has any eigenvectors. With the Euclidean metric we could apply the spectral theorem. The Lorentzian metric also seems fine (at least in 4D). But with metric signature $++--$ there are problems. For example, consider $$F=(e_{+1}\wedge e_{+2})+(e_{-1}\wedge e_{+2})+(e_{+1}\wedge e_{-2})-(e_{-1}\wedge e_{-2}).$$ $\endgroup$
    – mr_e_man
    Commented Feb 23, 2023 at 2:51


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