Let $\alpha = \sum_{i < j} a_{ij}\; dx_i\wedge dx_j$ be a two form on $\mathbb{R}^4$ and consider the mapping $f_\alpha \colon \beta \mapsto \beta \wedge \alpha.$ The matrix of $\alpha$ is $$ [\alpha] = \frac{1}{2}\left(\begin{matrix} 0 & a_{12} & a_{13} & a_{14}\\ -a_{12} & 0 & a_{23} & a_{24}\\ -a_{13} & -a_{23} & 0 & a_{34}\\ -a_{14} & -a_{24} & -a_{34} & 0 \end{matrix}\right ) $$ and the matrix for $f_\alpha$ with respect to the bases $$ \begin{gather*} B_1 = \{ dx_1, dx_2, dx_3, dx_4 \} \\ B_2 = \{ dx_1\wedge dx_2 \wedge dx_3, dx_1\wedge dx_2 \wedge dx_4, dx_1\wedge dx_3 \wedge dx_4, dx_2\wedge dx_3 \wedge dx_4 \} \end{gather*} $$ is $$ [f_\alpha] = \left(\begin{matrix} a_{23} & -a_{13} & a_{12} & 0 \\ a_{24} & -a_{14} & 0 & a_{12} \\ a_{34} & 0 & -a_{14} & a_{13} \\ 0 & a_{34} & -a_{24} & a_{23}\end{matrix}\right). $$ The latter matrix is almost symmetric along antidiagonal and moreover $$ \det([\alpha]) = \det([f_\alpha]) = (a_{14} a_{23} - a_{13} a_{24} + a_{12} a_{34})^2. $$

How can one see this without coordinates? Is there a generalization?


It was suggested in the comments that it is better to pick bases adapted to Hodge star with respect to the standard Euclidean product. If I didn't make another silly mistake, then with respect to this "Hodge basis" $$ B_H = \{ dx_2\wedge dx_3 \wedge dx_4 , -dx_1\wedge dx_3 \wedge dx_4, dx_1\wedge dx_2 \wedge dx_4, -dx_1\wedge dx_2 \wedge dx_3, \} $$ the matrix of $f_\alpha$ is $$ \left(\begin{matrix} 0 & a_{34} & -a_{24} & a_{23}\\ -a_{34} & 0 & a_{14} & -a_{13}\\ a_{24} & -a_{14} & 0 & a_{12}\\ -a_{23} & a_{13} & -a_{12} & 0 \end{matrix}\right ). $$

And indeed it is exactly, as Igor writes, the matrix of $*\alpha.$ Since $\alpha \wedge \alpha = \frac{1}{2}\mathrm{Pf}([\alpha])$ (See Wikipedia entry for Pfaffian.) and Hodge star is isometry (and so $\alpha\wedge \alpha = \langle \alpha | \alpha \rangle = \langle *\alpha | *\alpha \rangle = *\alpha \wedge *\alpha$), we have that $\mathrm{Pf}([\alpha]) = \mathrm{Pf}([*\alpha]) = \mathrm{Pf}([f_\alpha]).$

  • $\begingroup$ Yes, the determinant of even sized skew-symmetric matrices always looks very nice! See wikipedia for Pfaffian. $\endgroup$ May 29, 2020 at 12:17
  • $\begingroup$ @DietrichBurde Thank you for the remark. I am aware of the Pfaffian. But why does it appear in the determinant for $[f_\alpha]$? $\endgroup$ May 29, 2020 at 12:20
  • $\begingroup$ Because $[\alpha]$ is skew-symmetric? You wrote $\det ([\alpha])$. So it is a Pfaffian. (You also have $\phi_{\alpha}$ and $f_{\alpha}$, but perhaps there is a typo?) $\endgroup$ May 29, 2020 at 12:25
  • $\begingroup$ @DietrichBurde Yeah, typo. Thanks. Determinant of $\alpha$ is square of Pfaffian. That is clear. But why is it the same as the determinant of the linear mapping $\beta \mapsto \beta \wedge \alpha$? And why does this linear mapping have a matrix symmetric along antidiagonal? $\endgroup$ May 29, 2020 at 12:35
  • $\begingroup$ Well... almost symmetric. Two signs are messing up with the symmetry. $\endgroup$ May 29, 2020 at 13:23

1 Answer 1


Use the Euclidean metric to Hodge-dualize and raise indices. Then $f_\alpha\colon \beta \mapsto *(\iota_{\beta^\sharp} (*\alpha))$, up to signs. The Hodge dual in your basis is just a permutation matrix, with some signs maybe. If you choose the basis a bit more carefully, you should see an antisymmetric matrix representation of $f_\alpha$, with the same coefficients as $*\alpha$. And I believe the Pfaffians $\alpha\wedge \alpha$ and $(*\alpha) \wedge (*\alpha)$ should also be proportional by a sign, upon factoring out the Euclidean volume form.

  • 1
    $\begingroup$ One can see how to construct a new basis $B_2'$ of $\bigwedge^3 (\Bbb R^4)^*$ better adapted to the problem, too: If we declare $B_1$ to be orthonormal and pick either orientation, we get an oriented metric $g$ on $\Bbb R^4$. Then, we can define $B_2' := \{\ast_g b : b \in B_1\}$. By construction, with respect to these bases (1) the matrix representation of the Hodge star is the identity matrix, and (2) the matrix representation of $f_\alpha$ is skew-symmetric and, under the identifications made in the question statement, coincides with the matrix representation of $\ast \alpha$. $\endgroup$ May 29, 2020 at 22:57
  • $\begingroup$ Thank you guys! $\endgroup$ May 30, 2020 at 16:19

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