An equality $\ast(\delta_1 \otimes \delta_2)=(-1)^{k_1k_2}(\ast_1\delta_1)\otimes(\ast_2\delta_2)$ in huybrechts's book PAGE 37

Question

I am confused about one equatily in proposition 1.2.31 in huybrechts 's book complex geometry, $\ast(\delta_1 \otimes \delta_2)=(-1)^{k_1k_2}(\ast_1\delta_1)\otimes(\ast_2\delta_2)$.

Can anyone give me some help?Thanks a lot!

Answer 1

Context: Let $(V,\langle\,,\,\rangle)$ be an inner product space of dimension $n$, assume there is a decomposition $(V,\langle\,,\,\rangle)=(W_1,\langle\,,\,\rangle_1)\oplus(W_2,\langle\,,\,\rangle_2)$ and let $*_i$ be the Hodge star operators on $W_i$ respectively. The claimed statement basically follows from the fact that putting together orthonormal bases of $W_1$ and $W_2$ gives an orthonormal basis of $V$.

Let $\delta_i\in\bigwedge^{k_i}{W_i}^*$ for $k_i\leq\dim W_i$, we know $*(\delta_1\otimes\delta_2)$ is an $(n-k_1-k_2)$-form. Fix orthonormal bases $\{e_i\},\{f_j\}$ of $W_1$ and $W_2$ respectively, express $\delta_1=\sum_I a_I e_I^*$ and $\delta_2=\sum_J b_J f_J^*$ for multi-indices $I,J$ of size $k_1,k_2$ respectively. Then $\delta_1\otimes\delta_2=\sum_{I,J}a_Ib_J e^*_I\wedge f^*_J$. By reordering, we may assume $\{e_i,f_j\}$ is an orthonormal basis of $V$ in the correct orientation. Then $*(\delta_1\otimes\delta_2)=\sum_{I,J}a_Ib_J \epsilon\,e^*_{I'}\wedge f^*_{J'}$ where $I'$ is the multi-index given by all indices not in $I$, and $\epsilon=\operatorname{sgn}(I,J,I',J')$.

As for the right hand side, note that $*_1(\delta_1)=\sum_{I}a_I\epsilon_1 \,e^*_{I'}$ and $*_2(\delta_2)=\sum_{J}b_J\epsilon_2\, f^*_{J'}$, where $\epsilon_i$ is given by the sign of $(I,I')$ etc. Hence it suffices to prove that $\epsilon=(-1)^{k_1k_2}\epsilon_1\epsilon_2$ for any $I,J$. Since $\epsilon=\operatorname{sgn}(I,J,I',J')$ and $J$ and $I'$ have sizes $k_2$ and $k_1$ respectively, it takes $k_1k_2$ swaps to get it to $(I,I',J,J')$. At this point it is clear that $\operatorname{sgn}(I,I',J,J')=\operatorname{sgn}(I,I')\operatorname{sgn}(J,J')$ and the claim is proved.