Lemma 18.19.1.AFE454C19DE7 Let $N\geq 1$, let $X_ j$, $1\leq j\leq N$, be connected compact complex manifolds with smooth real $(1,1)$-forms $\theta _ j$, and let $u_ j,\varphi _ j\in \mathrm{PSH}(X_ j,\theta _ j)$. On $X=\prod _ jX_ j$, with projections $\operatorname {pr}_ j$, put
\[ \vartheta =\sum _ j\operatorname {pr}_ j^*\theta _ j,\qquad u=\sum _ j u_ j\circ \operatorname {pr}_ j,\qquad \varphi =\sum _ j\varphi _ j\circ \operatorname {pr}_ j. \]
Then
\[ u\in \mathcal E_{\mathrm{env}}(X,\vartheta ,\varphi ) \quad \Longleftrightarrow \quad u_ j\in \mathcal E_{\mathrm{env}}(X_ j,\theta _ j,\varphi _ j) \quad \text{for every }j. \]
Proof.
The ordinary order between the sums is equivalent to the ordinary order in each coordinate. Indeed, the latter follows from the former by fixing the other coordinates at points where their potentials are finite and restricting the inequality, and the converse follows by adding the coordinate inequalities.
Suppose that each $u_ j$ belongs to its relative class. For $b>1$, choose stretching competitors $h_ j$ as in definition 16.28.61. The sum $h=\sum _ jh_ j\circ \operatorname {pr}_ j$ is $\vartheta $-plurisubharmonic and satisfies
\[ b^{-1}h+(1-b^{-1})\varphi \leq u. \]
This convex inequality is meaningful at poles. For $b=1$, use $h=u$. Thus $u$ belongs to the relative class on the product.
Conversely, fix $b>1$ and a stretching competitor $h$ on the product. Choose a point $x=(x_1,\ldots ,x_ N)$ at which $h$ and all the coordinate potentials are finite; such points have full smooth volume. For each $j$, let $h^{(j)}$ be the restriction of $h$ to the slice through $x$ in the $j$th factor. It is $\theta _ j$-plurisubharmonic and not identically $-\infty $, since it is finite at $x_ j$. Put
\[ c_ j=\sum _{\ell \ne j}u_\ell (x_\ell ),\qquad d_ j=\sum _{\ell \ne j}\varphi _\ell (x_\ell ). \]
Restricting the product inequality to the slice gives
\[ b^{-1}\bigl (h^{(j)}-bc_ j+(b-1)d_ j\bigr ) +(1-b^{-1})\varphi _ j\leq u_ j. \]
This gives a stretching competitor in the $j$th coordinate for every $b>1$. Together with the ordinary order and the competitor $u_ j$ for $b=1$, it proves the converse.
$\square$
Pome
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