5
$\begingroup$

Let $X$ be a smooth, projective variety and $Z_1, Z_2$ two smooth, projective subvarieties in $X$ of the same dimension. Let $E$ be a locally free sheaf on $X$. Recall, there are natural morphims:

$$g_i:H^k_{Z_i}(E) \xrightarrow{f_i} H^k_{Z_1 \cup Z_2}(E) \to H^k(E)$$

for $i=1,2$. Let $\alpha_i \in H^k_{Z_i}(E)$ for $i=1,2$. Is it true that $g_1(\alpha_1)=g_2(\alpha_2)$ if and only if $f_1(\alpha_1)=f_2(\alpha_2)$? (If necessary assume that $Z_1 \cap Z_2$ is of codimension at least $2$ in both $Z_1$ and $Z_2$).

Any idea/reference will be most welcome.

$\endgroup$
2
$\begingroup$

I don't think this has much of a chance. The group $H^{k}(E)$ could easily be zero while the local cohomology is non-zero, so $g_1(\alpha_1)=g_2(\alpha_2)$ for any two classes, while if you have a single class $\alpha$ for which $f_1(\alpha)\neq 0$ then what you are hoping for doesn't work. More particularly, let $X=\mathbb P^n$, $E=\mathscr O_X$, and $Z_i\in\mathbb P$ different single points. Then for $k=n$ both $f_1$ and $f_2$ are injective, but their images only intersect in $0$, so $f_1(\alpha_1)=f_2(\alpha_2)$ only if $\alpha_1=\alpha_2=0$. My guess is that this fails more times than it is true. Notice that in this example $Z_1\cap Z_2=\emptyset$.

The more natural thing to look at is the Mayer-Vietoris sequence which tells you that

$$\dots \to H^{k}_{Z_1\cap Z_2}(E) \to H^k_{Z_1}(E) \oplus H^k_{Z_2}(E) \xrightarrow{\ f_1-f_2 \ } H^k_{Z_1 \cup Z_2}(E) \to H^{k+1}_{Z_1\cap Z_2}(E) \to \dots$$

is exact. Perhaps you can use that for what you needed this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.