general relativity - Covariant and contravariant vectors

Reading Weinberg's Gravitation and Cosmology, I came across the sentence (p.115, above equation (4.11.8))

The partial derivative operator $\partial/\partial x^\mu$ is a covariant vector, or in other words a 1-form, [...]

Now, a section of $T^*M$ is called a covector field or a 1-form. Elements of $T^*_pM$ are called tangent covectors. In books about differential geometry I read that tangent COvectors are called covariant vectors. How do I have to understand the quote from Weinberg's book?

Answer

OP is formally correct that $$\frac{\partial}{\partial x^{\mu}}~\in~ \Gamma(TM|_{U}) $$ is a vector field (defined in a local coordinate neighborhood $U$), and not a one-form. What Weinberg simply means by casually saying that

The partial derivative operator $\partial/\partial x^\mu$ is a covariant vector, or in other words a 1-form,[...]

is just that

The local basis of vector fields
$$\tag{1} \frac{\partial}{\partial x^{\mu}} ~=~\frac{\partial y^{\nu}}{\partial x^{\mu}} \frac{\partial}{\partial y^{\nu}}$$ transforms in the same way as the components $$\tag{2} \eta^{(x)}_{\mu}~=~\frac{\partial y^{\nu}}{\partial x^{\mu}}\eta^{(y)}_{\nu} $$ of a 1-form/covector [or a covariant (0,1) tensor] $$\eta~=~\eta^{(x)}_{\mu} dx^{\mu} ~=~\eta^{(y)}_{\nu} dy^{\nu}$$
under a local coordinate transformation $x^{\mu} ~\to~ y^{\nu}=y^{\nu}(x)$.

The point is that the (traditional) physicist often thinks of a $(r,s)$ tensor $$T~=~ \frac{\partial}{\partial x^{\mu_1}}\otimes \cdots \otimes \frac{\partial}{\partial x^{\mu_r}} ~T^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s} ~ dx^{\nu_1}\otimes \cdots \otimes dx^{\nu_s}$$ merely in terms of its components $T^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s}$, and in particular, the transformation property thereof under local coordinate transformations. Local basis elements, such as, e.g., $\frac{\partial}{\partial x^{\mu}}$ and $dx^{\nu}$ are often viewed as merely bookkeeping devices.

In conclusion, Ref. 1 is probably not the best textbook to learn differential geometry from. For instance, already in eq. (4.11.12) on the very next page 116 Weinberg claims that the fact that the exterior derivative squares to zero, $d^2=0$,

is known as Poincare's Lemma.

This is definitely not correct, cf. Wikipedia: The identity $d^2=0$ means that exact forms are closed, while Poincare's Lemma states that the opposite holds locally: closed forms are locally exact (except for zero-forms).

References:

1. S. Weinberg, Gravitation and Cosmology, 1972.