relative sign in Hodge star of tensor product

Question

Let $V$ be a vector space of arbitrary (finite) dimension and let $(V, \langle \ ,\ \rangle, I) = (W_1, \langle\ ,\ \rangle_1, I_1) \oplus (W_2, \langle\ ,\ \rangle_2, I_2)$ be a direct sum decomposition, with respect to scalar products and complex structures, where $W_{1,2}$ are even-dimensional.

On $\bigwedge^\bullet V^* = \bigwedge^\bullet W_1^* \otimes \bigwedge^\bullet W_2^*$, for $\delta_i \in \bigwedge^{k_i}W_i^*$, $i=1,2$, the Hodge $\star$-operator of $\delta_1 \otimes \delta_2$ is given by $$ \star(\delta_1 \otimes \delta_2)=(-1)^{k_1k_2}(\star_1\delta_1) \otimes(\star_2 \delta_2)$$

We can limit ourselvers here to the vector spaces and their exterior structures, forgetting about continuous dependence on the point in some manifold, etc. (i.e., this should be a linear algebra question); my QUESTION is: where does the $(-1)^{k_1k_2}$sign come from?

(This comes from the proof of Proposition 1.2.31 in the book Compelx Geometry:an Introduction, by Huybrechts)

I have an IDEA how to do it, using the fact that (with appropriate notation) $$ (\alpha_1 \otimes \alpha_2)\wedge \star(\beta_1 \otimes \beta_2)= (\alpha_1 \wedge \star_1 \beta_1)\otimes (\alpha_2 \wedge \star_2\beta_2)$$ and then plugging in the desired sign when I swap $\star_1\beta_1$ and $\alpha_2$, but I don't understand why, roughly speaking, the tensor behaves like the wedge, i.e. (e.g. for one-forms) $$ (e_1 \otimes f_1) \wedge (e_2 \otimes f_2)= - (e_1 \wedge e_2)\otimes (f_1 \wedge f_2)$$

Answer 1

I studied that part in Huybrechts' book some time ago. Looking at my notes, I think that a decent way to prove this is to use the standard property of the Hodge star, i.e. $ \star \alpha \wedge \beta = <\alpha, \beta> \eta$ where $\eta$ is the volume form. With that in mind the trick seems to be to once use this property on the combined space $W_1 \oplus W_2$ and once on each of the $W_i$ each.

$$ \star (\alpha_1 \otimes \alpha_2) \wedge (\beta_1 \otimes \beta_2) = <\alpha_1, \beta_1>_1 <\alpha_2, \beta_2>_2 \eta_1 \eta_2 = (-1)^{k_1 k_2} \star_1 \alpha_1 \wedge \star_2 \alpha_2 \wedge \beta_1 \wedge \beta_2 $$

I.e. the overall sign comes from commuting the differential forms in the last step.

Answer 2

$ \newcommand\Cl{\mathrm{Cl}} \newcommand\tensor\otimes \newcommand\gtensor{\mathbin{\hat\otimes}} \newcommand\Ext{{\bigwedge}} \newcommand\rev\widetilde $

Let $W_1, W_2$ have dimensions $n_1, n_2$.

Since we have a bilinear form, it doesn't matter whether we consider $V$ or $V^*$, so I will consider $V$ for notational simplicity. We consider the associated Clifford algebras, which follow the same sort of decomposition: $$ \Cl(V) \cong \Cl(W_1)\gtensor\Cl(W_2) $$ where $\gtensor$ is the graded tensor product. (Of course, each Clifford algebra uses the associated bilinear form.) There is a canonical linear isomorphism $\Ext V \cong \Cl(V)$, and we may think of a Clifford algebra as an exterior algebra together with a Clifford product; we will use the notation $\Cl^k$ to refer to $k$-vectors in the Clifford algebra. Now choose $\omega_1 \in \Cl^{n_1}(W_1)$ and $\omega_2 \in \Cl^{n_1}(W_2)$ such that $|\omega_1^2| = |\omega_2^2| = 1$; in other words, we make a choice of unital volume forms. Then there is canonical unital volume form $\omega \in \Cl^{n_1+n_2}(V)$ given by $$ \omega = \omega_1\omega_2 = \omega_2\omega_1. $$ These commute since they have even grade and $W_1, W_2$ are orthogonal subspaces of $V$. The Hodge stars on $\star, \star_1, \star_2$ are realized in the Clifford algebra as $$ \star X = \rev X\omega,\quad \star_1 X = \rev X\omega_1,\quad \star_2 X = \rev X\omega_2 $$ where $\rev X$ is the reversal. It follows that for any $X \in \Cl^{k_1}(W_1)$ and $Y \in \Cl^{k_2}(W_2)$ $$ \star(XY) = \rev{XY}\omega = \rev Y\rev X\omega_1\omega_2 \overset{(1)}= (-1)^{k_1k_2}\rev X\omega_1\rev Y\omega_2 = (-1)^{k_1k_2}(\star_1X)(\star_2Y). $$ (1) follows since $\rev Y$ has grade $k_2$ and $\rev X$ has grade $k_1$, so they commute with factor $(-1)^{k_1k_2}$. $\rev Y$ and $\omega_1$ then commute for the same reason.

We can, however, write a more general equation: for any $X \in \Cl(W_1)$ and $Y \in \Cl(W_2)$, $$ \star(XY) = \rev Y\rev X\omega_1\omega_2 = \rev Y\omega_2\rev X\omega_1 = (\star_2Y)(\star_1X), $$ which follows since $\omega_2$ is even and so commutes with $\rev X\omega_1$ which lives in an orthogonal space.