ScalingStacks

Using the splitting (73) of $V_{D_{2}}$ we can decompose the $\mathcal{I}^{\prime}$-cube $V_{D_{1}}$ as a direct sum of two $\mathcal{I}^{\prime}$-cubes as follows:

$$V_{D_{1}}=V^{\prime}\oplus V^{\prime\prime}$$

(78)

where

	$\displaystyle V^{\prime}(\ast 0)$	$\displaystyle=$	$\displaystyle\jmath_{a}(V_{D})$		(79)
	$\displaystyle V^{\prime}(\ast 1)$	$\displaystyle=$	$\displaystyle 0$		(80)
	$\displaystyle V^{\prime\prime}(\ast 0)$	$\displaystyle=$	$\displaystyle\iota_{a}(V_{D})$		(81)
	$\displaystyle V^{\prime\prime}(\ast 1)$	$\displaystyle=$	$\displaystyle V_{D_{1}}(\ast 1)$		(82)

Some explanation: in the formula (79) $\jmath_{a}(V_{D})$ is a subcube of $V_{D_{2}}$ and, due to (75), $\jmath_{a}(V_{D})$ sits inside $V_{D_{1}}$ as a subcube of codimension 1. Equation (80) means that $V^{\prime}(\ast 1)(\mathcal{L})=0$ for all $\mathcal{L}\subset\mathcal{I}^{\prime}.$ Thus, $V^{\prime}(\mathcal{L})=\jmath_{a}(V_{D}(\mathcal{L}))\subset V_{D_{1}}(\mathcal{L})$ for $\mathcal{L}\subset\mathcal{I}^{\prime},$ if $\mathcal{L}$ does not contain $a.$ If $\mathcal{L}$ contains $a,$ $V^{\prime}(\mathcal{L})=0.$