Let $X=(B^6\times T^2)/\mathbb{Z}_k\subset (\mathbb{C}^3\times T^2)/\mathbb{Z}_k$ where $k=2,3,4,6$, where the generator of $\mathbb{Z}_k$ acts on $\mathbb{C}^3$ by the multiplication by a primitive $k$-th root of unity, $B^6$ is the ball of unit radius in $\mathbb{C}^3$, and the $T^2$ is given a flat metric such that it has an nontrivial isometry of order $k$. Then the orbifold singularities in $X$ coming from the $\mathbb{Z}_k$ action are as follows:
- $k=2$: four of the form $\mathbb{C}^4/\mathbb{Z}_2$
- $k=3$: three of the form $\mathbb{C}^4/\mathbb{Z}_3$
- $k=4$: two of the form $\mathbb{C}^4/\mathbb{Z}_4$ and one of the form $\mathbb{C}^4/\mathbb{Z}_2$
- $k=6$: one of the form $\mathbb{C}^4/\mathbb{Z}_6$, one of the form $\mathbb{C}^4/\mathbb{Z}_3$, and another of the form $\mathbb{C}^4/\mathbb{Z}_2$.
Let $X'$ be the space obtained by removing from $X$ small balls $/\mathbb{Z}_l$ around each of the orbifold singularities. The "outer" boundary $Y_o$ of $X'$ is a $T^2$ bundle over $S^5/\mathbb{Z}_k$, whereas the "inner" boundary $Y_i$ of $X'$ is of $\bigsqcup S^7/\mathbb{Z}_l$ around the orbifold singularities.
Now I'm interested in the relation of $H^4(Y_o,\mathbb{Z})$ and $H^4(Y_i,\mathbb{Z})$ obtained by restricting $H^4(X',\mathbb{Z})$ to $Y_{o,i}$.
$H^4(Y_i,\mathbb{Z})$ are easy to compute: they are just
- $k=2$: $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus \mathbb{Z}_2$
- $k=3$: $\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus\mathbb{Z}_3$
- $k=4$: $\mathbb{Z}_4\oplus\mathbb{Z}_4\oplus\mathbb{Z}_3$
- $k=6$: $\mathbb{Z}_6\oplus\mathbb{Z}_3\oplus\mathbb{Z}_2$
whereas the Leray-Serre spectral sequence tells me that $H^4(Y_o,\mathbb{Z})$ has a standard filtration $F^2\supset F^3 \supset F^4$ with the quotient $F^2/F^3$, $F^3/F^4$, $F^4$ given as follows:
- $k=2$: $\mathbb{Z}_2$, $\mathbb{Z}_2\oplus \mathbb{Z}_2$, $\mathbb{Z}_2$
- $k=3$: $\mathbb{Z}_3$, $\mathbb{Z}_3$, $\mathbb{Z}_3$
- $k=4$: $\mathbb{Z}_4$, $\mathbb{Z}_2$, $\mathbb{Z}_4$
- $k=6$: $\mathbb{Z}_6$, $\mathbb{Z}_1$, $\mathbb{Z}_6$
So I see that $H^4(Y_i)$ and $H^4(Y_o)$ are "almost the same". My question is the precise relationship.
My guess is, from various other stupid stringy computations, that
$F^2/F^3\simeq \mathbb{Z}_k$ of $H^4(Y_o)$ is given by the "sum" of the summands of $H^4(Y_i)$, where we use a natural map $\mathbb{Z}_3\to \mathbb{Z}_6$ that sends 1 mod 3 to 2 mod 6 to take the sum. So $1\oplus 1\oplus 1 \in \mathbb{Z}_6\oplus\mathbb{Z}_3\oplus\mathbb{Z}_2$ is sent to 0, while
the map from $F^4 \simeq\mathbb{Z}_k$ of $H^4(Y_o)$ to $H^4(Y_i)$ is given by a natural restriction to each components, e.g. $1\in \mathbb{Z}_6$ is sent to $1\oplus 1\oplus 1\in\mathbb{Z}_6\oplus\mathbb{Z}_3\oplus\mathbb{Z}_2$.
Is there an (easy) way to check this?
Disclaimers: I know I should be able to do the computation by constructing e.g. an explicit cell decomposition of $X'$, but I haven't done my exercises. I also know that this might not quite be a "research level" question on the math side, since any math graduate student should be able to compute this, but this is related to my string theory research ...