Why the canonical bundle of a complex manifold is a line bundle?

I think that I do not understand something in the definition of the line bundles. Line bundles have fibers of rank 1, that is isomorphic to $\mathbb{C}$. But I do not know how to connect this vector space with elements like $dz_i$, $\,dz_{i_1}\wedge dz_{i_{2}},$ etc. Here, $dz_i$ are locally defined one-forms for the coordinates $\{z_i\}$ of some patch of the manifold of complex dimension $m$.

So, why is the $m$-th product of forms a line bundle? What is the correspondence with $\mathbb{C}$? Can you give an example?

Also, what kind of bundle is then the collection of elements $\{dz_i\}$, $\{dz_{i_1}\wedge dz_{i_2} \}$ and so on?

• Presumably you mean the $m^{\text{th}}$ exterior product of the tangent bundle of an $m$-dimensional complex manifold forms a line bundle. This is really a linear algebra problem on the level of tangent spaces: for a vector space $V$, you define the exterior powers $\Lambda^k V$. If $V$ has dimension $n$, then $\Lambda^k V$ can readily be seen to have dimension $\binom{n}{k}$. – Dustan Levenstein Feb 4 '16 at 15:21
• Yes this is what I mean. But I do not see why the space $\Lambda^k V$ has the dimension $\frac{n!}{k!(n-k)!}$. Would you mind to expand slightly or to direct me to some specific reference? – Marion Feb 4 '16 at 15:31
• Well, the Wikipedia article on this topic does give the definitions and basis. Specifically, if $\{e_1, \ldots, e_n\}$ is a basis for $V$, then $\{e_{i_1} \land \cdots \land e_{i_k} \mid 1 \le i_1 < \cdots < i_k \le n\}$ forms a basis for $\Lambda^k V$. Since the basis is enumerated by ordered sets of $k$ distinct elements from the basis of $V$, the dimension formula follows. – Dustan Levenstein Feb 4 '16 at 15:41
• You should really read in detail some treatment of multilinear algebra and exterior powers in particular before moving any further with complex manifolds, vector bundles and what not! – Mariano Suárez-Álvarez Feb 27 '16 at 22:01
• Thanks that helps a lot LOL – Marion Feb 28 '16 at 3:17