In the comments above, I've been trying to figure out what the vector bundle of antiholomorphic forms is. José Figueroa-O'Farrill writes

I think that holomorphic and antiholomorphic refer, respectively, to (1,0) and (0,1) forms.

I write

It ... seems that "antiholomorphic forms" is the dual vector bundle to "holomorphic forms", that is to say, it is the holomorphic tangent bundle.

The point of this answer is to explain that we are both right.

First of all, I am used to distinguishing between two notions. For me, a "$(1,0)$-form" is a $\mathbb{C}$-valued $1$-form which locally looks like $\sum f_i(z) d z_i$, for some $C^{\infty}$ functions $f_i$. In a "holomorphic $1$-form" we also require the $f_i$ to be holomorphic functions. Apparently, not everyone makes this distinction. That's fine, but in this post I am going to use my language because it appears to be more precise.

Let $X$ be a complex manifold. The sheaf of $(1,0)$-forms on $X$ is a sheaf of modules for the sheaf of $C^{\infty}$ functions. Similarly, the sheaf of holomorphic $1$-forms is a sheaf of modules for the sheaf of holmorphic functions. Using the appropriate version of the Serre-Swan theorem in each case; we get a complex vector bundle over $X$, in the $C^{\infty}$ and holomorphic categories respectively. Taking the forgetful functor from holomorphic vector bundles to smooth vector bundles, we get naturally isomorphic vector bundles in either case, and this bundle is the cotangent bundle $T^*(X)$. So, no matter how you think about it, the "bundle of holomorphic $1$-forms" is the cotangent bundle.

That's what happens with $(1,0)$-forms. What happens with $(0,1)$-forms? In this case, I think it is best to generalize the $C^{\infty}$ approach above. That is to say, we consider the sheaf of all $1$-forms that locally look like $\sum f_i(z) d \overline{z}_i$, for some $C^{\infty}$ functions $f_i$. Again, Serre-Swan gives us a $C^{\infty}$ complex vector bundle which I'll call $A$. From this perspective, $A$ does not have an obvious holomorphic structure.

However, I claim that $A$ is noncanonically isomorphic to the tangent bundle to $X$. Here is the isomorphism. Choose a positive definite Hermitian structure on tangent bundle of $X$. This is always possible by a partition of unity argument. (My convention is that Hermitian forms are linear in the second argument and conjugate linear in the first.) For any $(1,0)$-vector field $v$, the $1$-form $\langle \ , v \rangle$ is a $(0,1)$-form.

The tangent bundle to $X$, of course, does have a natural holomorphic structure; the holomorphic sections are of the form $\sum f_i(z) \partial/\partial z_i$ where the $f_i$ are holomorphic. But this structure is very hidden in the presentation of $A$ as the bundle of $(0,1)$-forms, and that was the point that was confusing me.