owned this note
owned this note
Published
Linked with GitHub
# Example
2019-09-20 11:47:20 Stefan's talk
Let
$$X = \{x \in M_2(C): \det x = 0\}\subset C^4$$
and
$$X\times P^1 \supset \tilde X = \{(x,\ell): x\in X, \ell \subset\ker x\}\xrightarrow{p} X$$
Note $p$ is proper, $U = X\setminus \{0\}$ is smooth, and $p:p^{-1}(U)\xrightarrow{\sim} U$.
We want to compute $p_* C_{\tilde X}$ aka fill in the table
| $\deg$ | $U$ | $\{0\}$ |
|---|---|---|
|2 | | |
|1 | | |
|0 | | |
In the $U$ column, by base change, we have
$$
\iota^* p_* C_{\tilde X} \cong (p\big|_{U}) _* \tilde \iota_U^* C_{\tilde X}
$$
where $\tilde\iota_U:p^{-1}(U)\to \tilde X$ and $\iota : U\to X$. Thus $p_*C_{\tilde X}|_U\simeq C_U$, and this has cohomology $C$ concentrated in degree $0$.
In the $\{0\}$ column we use the fact that $p$ is proper. Because of this, the stalk $(p_* C_{\tilde X})_{\{0\}}$ is
$$
H^\bullet (p^{-1}(\{0\});C)=H^\bullet(P^1;C)
$$
Therefore
| $\deg$ | $U$ | $\{0\}$ |
|---|---|---|
|2 | 0 | C |
|1 | 0 | 0 |
|0 | C | C |
Now let $j: U\to X$. We'll compute $j_* C_U$ aka fill in
| $\deg$ | $U$ | $\{0\}$ |
|---|---|---|
|$\vdots$| | |
|2 | | |
|1 | | |
|0 | | |
On $U$, we use that $j^*j_*C_U\simeq C_U$.
At $\{0\}$, we'll use $G_m$ localization.
So we need to compute $\iota^*j_* C_U = p_*j_* C_U = H^\bullet(U;C)$. Our setup is
$$
U\xrightarrow{\iota} C^4 \setminus\{0\} \simeq S^7 \xleftarrow{j} GL(2,C)
$$
For "completely formal reasons," we have the following triangle
$$
\iota_*\iota^! C_{C^4\setminus\{0\}}\to C_{C^4\setminus \{0\}}\to j_*C_{GL(2)}
$$
This triangle is the same as the triangle
$$
\iota_* C_U [-2]\to C_{C^4 \setminus\{0\}}\to j_* C_{GL(2)}
$$
As we see after we compute
$$
\begin{aligned}
\omega_{C^4\setminus \{0\}} &= C_{C^4\setminus\{0\}} [8] \\
\iota^! C_{C^4\setminus \{0\}} [8] &= C_U [6]\\
\iota^! C_{C^4\setminus \{0\}} &= C_U [-2]
\end{aligned}
$$
and
$$
\begin{aligned}
H^\bullet (GL(2);C) &\cong H^\bullet(U(2);C) \\
&\cong H^\bullet(S^1\times SU(2);C) \\
&\cong H^\bullet(S^1\times S^3;C) \\
&\cong \begin{cases}
C & \bullet = 0,1,3,4 \\
0 & \text{else}
\end{cases}
\end{aligned}
$$
The long exact sequence in cohomology from this triangle is therefore
$$
H^{-2}(U;C)\to H^0(S^7) \xrightarrow{\sim} H^0(GL(2)) \to H^{-1}(U;C)\to \cdots
$$
Therefore
| $\deg$ | $U$ | $\{0\}$ |
|---|---|---|
|5| 0 | C |
|4| 0 | 0 |
|3| 0 | C |
|2 | 0 | C |
|1 | 0 | 0 |
|0 | C | C |
Lastly we will compute the dualizing complex aka the Verdier dual of the constant sheaf on $X$, $\omega_X = D(C_X)$.
On $U$, $j^* = j^!$ because $j$ is an open embedding. So $j^!\omega_X = \omega_U = C_U [6]$ so out table will have the constant sheaf $C$ in degree 6 and 0 everywhere else.
To calculate at $\{0\}$, we we truncate $j_*C_U$ we recover the constant sheaf on $X$ i.e. $\tau^{\le 0} j_* C_U = C_X$ (look at the table of local systems).
The $t$-structure axioms now provide us a distinguished triangle
$$
\tau^{\le 0} j_* C_U \to j_* C_U \to \tau^{\ge 1} j_* C_U
$$
and
$$
C_X \to j_* C_U\to \tau^{\ge 1} j_* C_U
$$
We apply $D$ to get
$$
(DC_X)_0\leftarrow (j _! DC_U)_0 \leftarrow (D \tau^{\ge 1} j _* C_U)_0
$$
Note $D C_U = C_U[6]$. The middle term is 0 since $0\not\in U$ and $j_!$ is given on stalks by extension by zero. Thus
$$
(D\tau^{\ge 1}j_* C_U)_0 \cong (\omega_X[-1])_0
$$
The lefthand side is the same as $D((\tau^{\ge 1} j_* C_U)_0)$ so we end up with nonzero cohomology in degrees $-5,-3,-2$: The truncation $\tau^{\geq 1}j_*{C}_U$ is (see the table above) a complex with cohomology in degrees 5,3, and 2. Taking stalks gives a complex of vector spaces with cohomology in the same degrees, and ${D}$ just dualizes this complex (note that, in order to give a cochain complex, degrees must be multiplied by $-1$).
Therefore $(\omega_X[-1])_0$ has cohomology in degrees $-5,-3$ and $-2$. Unshifting we have
| $\deg$ | $U$ | $\{0\}$ |
|---|---|---|
|-3| 0 | C |
|-4| 0 | C |
|-5| 0 | 0 |
|-6| C | C |
## Another example
Let $U = A^1\setminus\{0\}\xrightarrow{j} A^1 \xleftarrow{i} \{0\}$. We want to see that $j_! C_U [1]$ and $C^{\{0\}}$ are perverse.
Note $j_* C_U [1]$ is perverse iff $j_! C_U [1]$ is.
Consider the following triangle.
$$
j_!C_U\to C_X\to C^{\{0\}}\to$$
Rotating, we get
$$
\underbrace{C^{\{0\}} = IC(Z,X)}_ {\text{socle}} \to j_! C_U [1] \to \underbrace{C_X [1] = IC(U,X)}_{\text{ head }}
$$