Show that the canonical divisor of \(\PP^n\) satisfies \[ K_{\PP^n}\sim -(n+1)H, \] where \(H\subseteq\PP^n\) is a hyperplane.
Solution
1. The Euler sequence on \(\PP^n\) is\[ 0 \longrightarrow \OO_{\PP^n} \longrightarrow \OO_{\PP^n}(1)^{\oplus(n+1)} \longrightarrow T_{\PP^n} \longrightarrow 0. \]
This is the Euler sequence for projective space T-MODEULER.
2. Taking determinants in step 1 gives\[ \boxed{\det T_{\PP^n}\cong\OO_{\PP^n}(n+1)}. \]
For a short exact sequence of locally free sheaves \[ 0\to\mce'\to\mce\to\mce''\to0, \] one has \[ \det\mce\cong\det\mce'\tensor\det\mce''. \] Applying this to the Euler sequence yields \[ \det\bigl(\OO(1)^{\oplus(n+1)}\bigr) \cong \det(\OO)\tensor\det(T_{\PP^n}). \] The left side is \[ \OO(1)^{\tensor(n+1)} \cong \OO(n+1), \] and \(\det(\OO)\cong\OO\). Hence \[ \det T_{\PP^n}\cong\OO(n+1). \]
3. The canonical line bundle is\[ \boxed{\omega_{\PP^n}\cong\OO_{\PP^n}(-n-1)}. \]
Because \(\PP^n\) is smooth of dimension \(n\), \[ \omega_{\PP^n} = \det\Omega^1_{\PP^n} = \dualof{(\det T_{\PP^n})}. \] Step 2 therefore gives \[ \omega_{\PP^n} \cong \dualof{\OO(n+1)} \cong \OO(-n-1). \]
4. If \(H\) is a hyperplane, then\[ \boxed{K_{\PP^n}\sim -(n+1)H}. \]
The hyperplane divisor \(H\) corresponds to the line bundle \[ \OO_{\PP^n}(H)\cong\OO_{\PP^n}(1). \] Thus \[ \OO_{\PP^n}\bigl(-(n+1)H\bigr) \cong \OO_{\PP^n}(-n-1). \] By step 3 this is the canonical line bundle. Therefore the canonical divisor class is \[ K_{\PP^n}\sim -(n+1)H. \]
5. Q.E.D.
Step 4 is the asserted linear equivalence, and step 3 is its equivalent canonical-bundle form.