I am reading Huybrechts's Complex Geometry, p.92~p.92 and stuck at showing that $K_{\mathbb{P}^n} \cong \mathcal{O}(-n-1)$ from the Euler sequence. Here, $K_{\mathbb{P}^n}$ is the canonical bundle of projective space (his book p.71) and for $k \in \mathbb{Z}$, $\mathcal{O}(k)$ is defined by
- Case 1) $k>0$ : $\mathcal{O}(k) := \mathcal{O}(1)^{\otimes k}$ ( $\mathcal{O}(1)$ is the tautological line bundle ; i.e., $\mathcal{O}(1):= \mathcal{O}(-1)^{*}$. C.f. his book p.68~69 ).
- Case 2) $k<0$ : $\mathcal{O}(k) := \mathcal{O}(-k)^{*}$.
And the Euler sequence is given by (His book p.93, Proposition 2.4.4 )
$$ 0 \to \mathcal{O} \to \oplus_{j=0}^{n}\mathcal{O}(1) \to \mathcal{T}_{\mathbb{P}^n} \to 0.$$
Let's take the dual for the Euler sequence. Then we get
$$ 0 \to \Omega_{\mathbb{P}^n} \to (\oplus_{j=0}^{n}\mathcal{O}(1))^{*} \to (\mathcal{O})^*\to 0.$$
Then we have ( c.f. his book viii) of Example 2.2.4. )
$$ \det((\oplus_{j=0}^{n}\mathcal{O}(1))^{*}) \cong \det(\Omega_{\mathbb{P}^n}) \otimes \det((\mathcal{O})^*) := K_{\mathbb{P}^n} \otimes \det((\mathcal{O})^*)$$
Here the $\det$ means the highest exterior power. Then I don't know how can we extract the desired formula $K_{\mathbb{P}^n} \cong \mathcal{O}(-n-1)$ from this relation. I think that this derivation involves calculation of determinant line bundle of direct sum and dual etc.. and it seems that there is no remark in his book that contains associated computation. I don't know what relations we may use until now. What is key point? Can anyone helps?
EDIT : I tried to make furthur progress. My first attempt is as follows.
First, I guess next statements.
Q.1. If $L$ is a line (holomorphic) bundle on $X$, then $\det(L) \cong L$ ?
Q.2. $(\mathcal{O})^* \cong \mathcal{O}$?
Second, if these questions are true, then for the right-hand side of the above isomorphic relation, $$K_{\mathbb{P}^n} \otimes \det((\mathcal{O})^*) = K_{\mathbb{P}^n} \otimes \mathcal{O} \cong K_{\mathbb{P}^n}.$$
And for the left-hand side, by using https://www.math.stonybrook.edu/~azinger/mat566-spr18/vectorbundles.pdf (final sentence),
$$ \det((\oplus_{j=0}^{n}\mathcal{O}(1))^{*}) = \det ( \oplus_{j=0}^{n} (\mathcal{O}(-1)^*)^*) = \det ( \oplus_{j=0}^n ( \mathcal{O}(-1))) \stackrel{!}{=} \det (\mathcal{O}(-1)) \otimes \cdots \otimes \det (\mathcal{O}(-1)) = (\mathcal{O}(-1)^*)^* \otimes \cdots \otimes (\mathcal{O}(-1)^*)^* \stackrel{?}{\cong} (\mathcal{O}(-1)^* \otimes \cdots \otimes \mathcal{O}(-1)^*)^* =: (\mathcal{O}(1) \otimes \cdots \otimes \mathcal{O}(1))^* =:(\mathcal{O}(n+1))^* =: \mathcal{O}(-n-1) $$
( The dualization of holomorphic vector bundle commutes with the tensor product? )
Is this argument correct? The above questions are true?
Q1: Yes, this is the definition of the determinant of a one-dimensional vector space.
Q2: Yes, the dual of the trivial line bundle is the trivial line bundle (for instance, use that a line bundle is trivial iff it has a non-vanishing global section).
Your proof is correct. (In my opinion the hard part is the part where you go from the sequence to the fact about the determinant.)