Lemma 9

Let $\boldsymbol{a}= (a_1, . . . , a_n)\in A^n$. Let $p\in \mathbf{N}$ and $\boldsymbol{a}'$ = $(a_{n+1}, . . . , a_{n+p})\in A^p$. Then we have $\Theta_{(\boldsymbol{a},\boldsymbol{a}')}\circ \pi_{n,n+p}^*= \Theta_{\boldsymbol{a}}$. In particular, we have $\Theta_{\boldsymbol{a}}(1) = 1$.

Since one has $\pi_{n,n+p}=\pi_{n,n+1}\circ \cdots \circ \pi_{n+p-1,n+p}$, it is enough to prove the first assertion when $p= 1$, which we henceforth suppose. We write simply $\pi =\pi_{n,n+1}$. It is then enough to prove that, for every open neighbourhood U of Sp$^n(\boldsymbol{a})$, and every function $f\in \mathscr{O}(U; A)$, one has $\Theta_{(\boldsymbol{a},a_{n+1})}(f\circ \pi ) = \Theta_{\boldsymbol{a}}(f)$. Let $g=f\circ \pi$. Let $h$ (resp. $h'$) be a mapping of $\mathbf{C}^n$ into $\mathbf{C}$ (resp. of $\mathbf{C}$ into $\mathbf{C}$), indefinitely differentiable, equal to 1 in a neighbourhood of Sp$^n(\boldsymbol{a})$ (resp. of Sp($\boldsymbol{a}'$)), with compact support contained in U (resp. in $\mathbf{C}$). There exist mappings $(u_1, . . . , u_n)$ of $\mathbf{C}^n$ into A, indefinitely differentiable, such that the sequence $(h, u_1, . . . , u_n)$ is adapted to $\boldsymbol{a}$ (lemma 3 of I, p. 54), and an indefinitely differentiable mapping $u_{n+1}$ of $\mathbf{C}$ into A such that the pair $(h', u_{n+1})$ is adapted to $a_{n+1}($loc. cit.)

For $\boldsymbol{z}\in \mathbf{C}^n$ and $z_{n+1}\in \mathbf{C}$, write $h''(\boldsymbol{z}, z_{n+1}) =h(\boldsymbol{z})h'(z_{n+1})$ and $u''_{n+1}(\boldsymbol{z}, z_{n+1}) =h(\boldsymbol{z})u_{n+1}(z_{n+1})$. The functions $h''$ and $u''_{n+1}$ are indefinitely differentiable in $\mathbf{C}^{n+1}$. The function $h''$ is equal to 1 in a neighbourhood of Sp$^{n+1}(\boldsymbol{a}, a_{n+1})$, and has compact support contained in $U\times \mathbf{C}$. For every $\boldsymbol{w}= (\boldsymbol{z}, z_{n+1})\in \mathbf{C}^{n+1}$, one has

which proves that the sequence $(h'', u_1\circ \pi , . . . , u_n\circ \pi , u''_{n+1})$ is adapted to $(\boldsymbol{a}, a_{n+1})$. Let $\omega$ be the associated differential form.

The differential form $du_1\wedge dz_1\wedge \cdots \wedge du_n\wedge dz_n\wedge dh$ on $\mathbf{C}^n$ is of degree $2n+ 1$, hence is zero. Therefore

Since $g=f\circ \pi$, formula (5) and lemma 8 imply

$\Theta_{\boldsymbol{a},\boldsymbol{a}'}(g) =$ (2$(ni\pi +$ 1)!$)^{n+1}\int_{U\times\mathbf{C}}g\omega =$ (2$(ni\pi +$ 1)!$)^{n+1}(\int_{\mathbf{C}}du_{n+1}\wedge dz_{n+1})$

On the one hand, one has

by corollary 5. On the other hand, part c) of lemma 4 of I, p. 54 and the fact that the integral of a closed form is zero (VAR, R2, p. 48, 11.2.4) imply

$\int_Uf du_1\wedge dz_1\wedge \cdots \wedge du_n\wedge dz_n=(2i\pi n$!$)^n\Theta_{\boldsymbol{a}}(f)$.

Thus one obtains

$\Theta_{\boldsymbol{a},\boldsymbol{a}'}(g) =$ (2$(ni\pi +$ 1)!$)^{n+1}\times ((2ni\pi +$ 1)!$)^n\Theta_{\boldsymbol{a}}(f)\times 2i\pi = \Theta_{\boldsymbol{a}}(f)$.

Finally, the formula $\Theta_{\boldsymbol{a}}(1) = 1$ results from the foregoing and from the corollary of Prop. 1.