# Double Hodge star without coordinates.

The proofs I've found of the fact that $**\alpha=(-1)^{k(n-k)}\alpha$, or, equivalently, the fact that the Hodge star is an isometry, all use an orthonormal basis. Is there a basis-free proof of either of these facts directly from the definition of the Hodge star by

$$\alpha\wedge*\beta=\langle\alpha,\beta\rangle\mathrm{vol}?$$

• I don't agree with the "equivalently." The definition I prefer to use for $\star$ uses an orthonormal basis, so I'm not sure what definition you mean. Commented Jun 7, 2018 at 1:01
• @TedShifrin I think that he is right claiming that the facts are equivalent. Please see my partial answer below. Commented Jun 7, 2018 at 7:25
• @YuriVyatkin, thanks for adding those details; that's exactly the question. I'm not sure which of the two facts ($*$ is an isometry; $**=(-1)^{k(n-k)}$) would be easier to prove without coordinates, but either fact is enough. Commented Jun 9, 2018 at 17:21
• Related question? math.stackexchange.com/questions/3967980/… @TedShifrin Yasha Berchenko-Kogan
– BCLC
Commented Dec 31, 2020 at 13:56
• The statement is obviously incorrect if the metric on the exterior algebra are not induced from the metric on the original vector space. For we can choose two totally unrelated metrics on $k$ forms and $n-k$ forms separately. So I think we need a coordinate independent definition of induced metrics on forms first. Commented Mar 14, 2021 at 8:51

Let me make just a partial contribution as an answer, as it is too long for a comment.

Assuming we know that $\star$ is an isometry, and taking the identity in the question as a definition,

\begin{align} \langle \alpha, \beta \rangle \mathrm{vol} & = \langle \star \alpha, \star \beta \rangle \mathrm{vol} = \star \alpha \wedge \star \star \beta = (- 1)^{k(n-k)} \star \star \beta \wedge \star \alpha \\ & = (- 1)^{k(n-k)} \langle \star \star \beta , \alpha \rangle \mathrm{vol} = (- 1)^{k(n-k)} \langle \alpha , \star \star \beta \rangle \mathrm{vol} \end{align}

so, we get the desired fact from the non-degeneracy of the metric.

Conversely, if we know, that $**\alpha=(-1)^{k(n-k)}\alpha$ holds, then, considering that

$$\langle \star \alpha, \star \alpha \rangle \mathrm{vol} = \star \alpha \wedge \star \star \alpha = (-1)^{k(n-k)} \star \alpha \wedge \alpha = \alpha \wedge \star \alpha = \langle \alpha, \alpha \rangle \mathrm{vol}$$

which implies that $\star$ is an isometry.

Now, we probably want to prove in a coordinate-free manner, that $\star$ is an isometry. However, it looks like a delicate matter, as we can see from the discussion around this question, so I feel that I need to take a timeout :)

• Related question? math.stackexchange.com/questions/3967980/…
– BCLC
Commented Dec 31, 2020 at 13:56
• Some insights could be sourced from this answer math.stackexchange.com/a/4483178/2002 (I still don't have much time for that, sadly). Commented Jul 2, 2022 at 11:35
• If the determinant of $g_{ij}$ in an orthonormal basis (and hence in each orthonormal basis) is $-1$, then the Hodge operator is not an isometry, is it? Commented Aug 6, 2022 at 16:36
• At least thats what these notes claim, page 32, third line. Commented Aug 6, 2022 at 16:39
• @Filippo You are right in regard that if we wanted to extend this reasoning to the pseudo-Riemannian signature (called Lorentzian in those notes), then we would need to care about the sign. However, the point of my (partial) answer was to find a way to avoid coordinates and frames, and I did not succeed :) Commented Aug 6, 2022 at 23:39

As remarked above by Yuri Vyatkin, the Hodge * being an isometry is more or less the same as the double Hodge being $$(-1)^{k(n-k)}$$ times the identity.

This is my attempt to prove it is an isometry (well,"isometry like": the sign of the innerproduct/nondegenerate symmetric bilinear form is allowed to flip sign) without using a basis. Only for simple (fully decomposable) $$k$$-vectors however. By (multi)linearity it should work on all of $$\Lambda^k(V)$$. I guess this is how many properties on $$\Lambda(V)$$ are defined.

Consider the Grassmann complement. Given a special $$n$$-covector $$\omega$$ and also a privileged $$n$$-vector $$\Omega$$, the Grassmann complement is a map $$*:\Lambda^k(V)\to\Lambda^{n-k}(V^*)$$ defined by $$i_{a_1\wedge\dots\wedge a_k}\omega(u_1,\dots,u_{n-k})=\omega(a_1,\dots,a_k,u_1,\dots,u_{n-k}).$$ Similarly $$*:\Lambda^k(V^*)\to\Lambda^{n-k}(V)$$ - with $$\Omega$$ instead of $$\omega$$. (Both directions being called * as well as the Hodge dual denoted * is confusing. Sorry.) The action of a simple $$k$$-covector on a simple $$k$$-vector is given by the determinant $$\langle a_1\wedge\dots\wedge a_k,\alpha_1\wedge\dots\wedge \alpha_k\rangle =\left|\begin{array}{ccc} \langle a_{1},\alpha_{1}\rangle & \cdots & \langle a_{1},\alpha_{k}\rangle\\ \vdots & \ddots\\ \langle a_{k},\alpha_{1}\rangle & & \langle a_{k},\alpha_{k}\rangle \end{array}\right|.$$ To find $$\langle *(\alpha_{1}\wedge\dots\wedge\alpha_{k}), *(a_{1}\wedge \dots\wedge a_k)\rangle$$, we can use that the adjoint of contraction with $$\alpha\in V^*$$ is wedging with $$\alpha$$. $$\langle *(\alpha_{1}\wedge\dots\wedge\alpha_{k}), *(a_{1}\wedge \dots\wedge a_k)\rangle=\langle i_{\alpha_{1}\wedge\dots\wedge\alpha_{k}}\Omega, i_{a_{1}\wedge \dots\wedge a_k} \omega\rangle\\ =\langle \Omega,\alpha_{1}\wedge\dots\wedge\alpha_{k}\wedge i_{a_{1}\wedge \dots\wedge a_k} \omega\rangle$$ We also need $$i_a (\sigma \wedge \tau)=(i_a \sigma)\wedge \tau + (-1)^{\deg \sigma}\sigma\wedge i_a \tau$$ ("$$i_a$$ is an anti-derivation"). For $$k=1$$, we simply get $$\alpha\wedge i_a\omega = -i_a(\alpha\wedge\omega)+\langle a,\alpha\rangle \omega=\langle a,\alpha\rangle \omega.$$ Next, $$k>1$$. By bringing the $$i_{a_k}$$ to the front in the LHS below, using the anti-derivational character of $$i_a$$ you get an expansion of the determinant above. (When working with orthonormal/dual bases this step is particularly easy). We get $$\alpha_{1}\wedge\dots\wedge\alpha_{k}\wedge i_{a_{k}}\dots i_{a_{1}}\omega=\langle a_1\wedge\dots\wedge a_k,\alpha_1\wedge\dots\wedge \alpha_k\rangle\omega$$ So $$\langle a_1\wedge\dots\wedge a_k,\alpha_1\wedge\dots\wedge \alpha_k\rangle\langle \Omega,\omega\rangle =\langle *(\alpha_{1}\wedge\dots\wedge\alpha_{k}), *(a_{1}\wedge \dots\wedge a_k)\rangle.$$ In the case of the Hodge * dual we have the volume form $$\omega=\sqrt{|g|}dx^1\wedge\dots\wedge dx^n$$ and also a special $$\Omega=\varphi(\omega)$$, where $$\varphi$$ is the Riesz isomorphism (is it called that on Minkowski space too?). We should have $$\langle\Omega,\omega\rangle=\sqrt{|g|}\cdot\sqrt{|g|}/\det{g}=\text{sign}(\text{metric})$$

(The Clifford answer by Nicholas Todoroff looks pretty neat to me btw. I don't fully understand it.) First post, can't comment.

$$\newcommand\Ext{{\bigwedge}} \newcommand\form[1]{\langle#1\rangle} \newcommand\Cl{\mathrm{Cl}} \newcommand\rev\widetilde$$We consider the Hodge star on the exterior algebra $$\Ext V$$ arising from $$I \in \mathop{\Ext^{\!n}}V$$ and the non-degenerate symmetric bilinear form $$\form{\cdot,\cdot}$$ on $$V$$ which extends to $$\Ext V$$ where $$\form{I, I} = \pm1$$. We assume $$V$$ is a finite-dimensional vector space over a field of characteristic $$\not=2$$. The bilinear form gives rise to a Clifford algebra $$\Cl(V)$$ which is canonically isomorphic to $$\Ext V$$, so we may consider $$I$$ as an element of $$\Cl(V)$$ and $$\form{\cdot,\cdot}$$ as a bilinear form on $$\Cl(V)$$. It is a fact that $$\form{A, B} = \form{\rev AB}_0,\quad A, B \in \Cl(V)$$ where $$\rev A$$ is the reverse of $$A$$ (the main anti-involution which fixes vectors and reverses all products), $$\rev AB$$ is the Clifford product, and $$\form{\cdot}_0$$ is the scalar part (i.e. grade 0 projection). It follows from the definition of the Hodge star and well-known properties of the Clifford product that when $$A$$ and $$B$$ are $$k$$-vectors we have $$A\wedge\star B = \form{\rev AB}_0I = \form{A\rev B}_0I = A\wedge(\rev BI).$$ Since this is true for every $$A$$, we see $$\star B = \rev BI,$$ and this equation is true for all $$B \in \Cl(V) \cong \Ext V$$ since it is true for each grade and is linear in $$B$$.

If $$\form{\cdot,\cdot}$$ is Euclidean then $$\rev II = \form{\rev II}_0 = \form{I, I} = 1$$ and also $$I^2 = -1$$. The fact that the Hodge star is an isometry is now trivial: $$\form{\star A, \star B} = \form{\rev{\rev AI}\rev BI}_0 = \form{\rev IA\rev BI}_0.$$ At this point, it is well-known that $$X \mapsto \rev IXI$$ is an orthogonal transformation (this being one of the main features of Clifford algebras; $$I$$ is an element of the Pin group, more specifically the Spin group). We could also argue algebraically to get $$\form{\rev IA\rev BI}_0 = \form{A\rev BI\rev I}_0 = \form{A\rev B}_0 = \form{A, B}.$$ The double Hodge star formula is also trivial; if $$B$$ is a $$k$$-vector $$\star\star B = \rev{\rev BI}I = \rev IBI = (-1)^{n(n-k)}\rev IIB = (-1)^{n(n-k)}B.$$ I will explain this it bit further. Let e.g. $$[B] = \{x \in V \;:\; x\wedge B = 0\}$$ be the subspace represented by $$B$$. $$I$$ factors as $$B^\perp B$$ where $$[B^\perp] = [B]^\perp$$, meaning that $$I = B^\perp B = B^\perp\wedge B$$. (Clearly, $$B^\perp$$ is proportional to $$\star B$$.) The grade of $$B$$ is $$k$$ and that of $$B^\perp$$ is $$n-k$$ so $$BI = BB^\perp B = (B\wedge B^\perp)B = (-1)^{n(n-k)}(B^\perp\wedge B)B = (-1)^{n(n-k)}IB.$$ Note that the double Hodge star formula relies on $$\form{\cdot,\cdot}$$ being Euclidean, and changes for other bilinear forms.