Let $f:T^2\to Y$ be a resolution of singularities where $Y$ is a torus with two "pinched" points (or, if you prefer, two copies of $\mathbb{P}^1$ meeting at two points). I'm interested in using the Leray spectral sequence to calculate the cohomology of the constant sheaf on $Y$ . My goal is to better understand spectral sequences and this looks like a nice example to me.

To begin with, $Rf_*\mathbb{Q}_X$ has cohomology sheaves $R^0f_*\mathbb{Q}_X = \mathbb{Q}_Y$ and $R^1f_*\mathbb{Q}_X = \mathcal{S}_{\{a,b\}}$, the skyscraper sheaf with stalk $\mathbb{Q}$ at the pinched points $a,b$.

The $E_2$ page is \begin{align*} \begin{matrix} H^2(Y; \mathbb{Q}_Y) & 0 & 0\\ H^1(Y; \mathbb{Q}_Y) & 0& 0\\ H^0(Y; \mathbb{Q}_Y) & H^0(Y;\mathcal{S}_{\{a,b\}})=\mathbb{Q}^{\oplus 2} & 0 \end{matrix} \end{align*} and the $E_3 (=E_\infty)$ page is \begin{align*} \begin{matrix} H^2(Y; \mathbb{Q}_Y)/ \text{im}(d_2^{0,1}) & 0 &0 \\ H^1(Y; \mathbb{Q}_Y) & 0& 0\\ H^0(Y; \mathbb{Q}_Y) & \text{ker}(d_2^{0,1}) & 0 \end{matrix} \end{align*}

Using the fact that the cohomology $H^n(T^2)$ is filtered by these objects, we find $H^0(T^2) = E_\infty^{0,0} = E_2^{0,0} = H^0(Y)$ and which obviously makes sense, as well as filtrations $$E_3^{0,1}\hookrightarrow H^1(X)$$ where $H^1(X) / \text{ker}(d^{0,1}_2) = E^{1,0}_3 = H^1(Y)$ and $$E_3^{0,2}\hookrightarrow (?) \hookrightarrow H^2(X)$$ where $H^2(X) / (?) = E^{2,0}_3 = H^2(Y)/\text{im}(d^{0,1}_2)$ and $(?)/E^{0,2}_3 = E^{1,1}_3$. Since $E^{1,1}_3 = 0 = E^{0,2}_3$, this implies $H^2(Y)/\text{im}(d^{0,1}_2) = H^2(X)$.

How can I get my hands on the differential $d_2^{0,1}$ in order to finish this calculation?


1 Answer 1


I think the most direct way to figure out the mystery differential is using the edge map $$ \mathbf Z^2 \cong H^1(X,\mathbf Z) \to H^0(Y,R^1 f_\ast \mathbf Z) \cong \mathbf Z^2.$$ Let's first think about how this map is defined: a class in $H^q(X)$ restricts to a class in $H^q(F_x)$ for each fiber $F_x$, and $H^q(F_x)$ is the stalk of $R^q f_\ast \mathbf Z$ at $x$. These classes are in fact compatible, i.e. come from a global section, which means that we get an element of $H^0(Y, R^q f_\ast \mathbf Z)$ from an element of $H^q(X)$. Now $H^1(X)$ has rank $2$, spanned by a meridian and a longitude. If I model $Y$ as the result of collapsing two disjoint meridian circles, then the longitude class restricts to zero in each fiber and the meridian circle restricts to the same generator in both special fibers. In particular the cokernel of the edge map is isomorphic to $\mathbf Z$, and the rank of the mystery differential is $1$.

(Also, I think you wrote your edge maps the wrong way around, possibly because your spectral sequences look transposed from how they're usually written.)

As a sanity check one can also argue geometrically that $Y \simeq S^2 \vee S^2 \vee S^1$, which gives the same cohomology groups. Indeed we can construct a CW decomposition of $Y$, taking the 1-skeleton to be a circle, and gluing on two $2$-cells. But both attaching maps are nonsurjective and therefore nullhomotopic.

  • $\begingroup$ Oh yes I think you're right about how I've written them. A classic case of getting lost in the indices. But your answer definitely clears up the nontrivial part, thanks!! $\endgroup$
    – EJAS
    Commented Jan 5, 2021 at 15:07

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