# 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?

• What's your doubt? Do you understand why $\partial/\partial x^\mu$ is a 1-form? – jinawee Sep 29 '13 at 18:59
• @jinawee: The components $\omega_i$ of a 1-form $\omega$ are defined by $\omega_i(p)=\omega_p\left(\frac{\partial}{\partial x^i}|_p\right)$, so I do not understand your statement. – pawel_winzig Sep 29 '13 at 19:11

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.

Let $X$ be a vector field on a semi-Riemannian manifold $(M,g)$, then in local coordinates, the vector fields $\partial_\mu = \partial/\partial x^\mu$ yield a basis for the tangent space of at every point on the manifold. This means that if $X\in TM$ is a vector field, then we can write it as a linear combination of the coordinate basis vector fields $\partial_\mu$ as follows: \begin{align} X = X^\mu\partial_\mu \end{align} The components $X^\mu$ defined by this relation are called the contravariant components of the vector field. It is then conventional to define covariant components of this vector field by "lowering" its vector index by using the components of the metric $g$; \begin{align} X_\mu = g_{\mu\nu} X^\nu \end{align} Once we have made this definition, we notice that if we also define coordinate basis vector fields $\partial^\mu$ having raised indices as follows using the components $g^{\mu\nu}$ of the inverse of the metric: \begin{align} \partial^\mu = g^{\mu\nu} \partial_\nu \end{align} then the original vector field $X$ can be written as a linear combination of its covariant components and the coordinate basis vector fields with raised indices; \begin{align} X = X_\mu \partial^\mu \end{align} Now the question becomes, what does any of this have to do with 1-forms? Well, given a semi-Riemannian manifold $(M,g)$, there is a canonical isomorphism between the tangent and cotangent spaces at a given point which we, as physicists, commonly refer to as "raising and lower indices." In coordinates this isomorphism works as follows. Using the covariant components of a vector field $X$, we define a one-form $\mathbf X$ by $$\mathbf X = X_\mu dx^\mu$$ Then the mapping $X \mapsto \mathbf X$ yields an isomorphism between $T_pM$ and $T_p^*M$ for each $p\in M$. In other words, the covariant components of $X$ can be regarded as precisely the components that its dual 1-form has under the tangent-cotangent isomorphism.

When Weinberg says that $\partial/\partial x^\mu$ is a 1-form, he's being a bit loose with terminology (as is common in physics). That object is really a vector field, but it is dual to a 1-form via the aforementioned isomorphism.

• I don't think that's what he means. The section is about manifolds in general, no metrics are involved. – MBN Sep 30 '13 at 16:23
• @MBN I never claim to capture what Weinberg "means" in the text. Weinberg seems to identify p-forms with objects that "transform in a particular way" because they have lowered indices. This is considered antiquated in modern physics, and I see no reason to propagate this idea, even if it is what he means. – joshphysics Sep 30 '13 at 18:22
• No, my comment is that your answer is based on the assumption that Weinberg identifies vectors and 1-forms, as you can always do when there is a non-degenerate metric, and I don't think that is the case. – MBN Sep 30 '13 at 20:06
• @MBN Haha look, I'm not making any assumptions about what Weinberg is doing; I don't so much care what's he's doing. My response is a description of the standard method by which one identifies vector fields with one-forms in the context of modern differential geometry on semi-Riemannian manifolds. In my view, the metric-based tangent-cotangent isomorphism is the best way to think about all of this. – joshphysics Sep 30 '13 at 20:41
• Well, but you are making assumptions (or guesses). Look at your last paragraph. Your description of is fine, but it is not what the question is about. – MBN Oct 1 '13 at 9:10

The way I understand it is that he is not referring to the tangent vectors $\frac{\partial}{\partial x^\mu}$, but to a differential operator. Notice that he doesn't use plural, he is talking about a single object. Namely the 1-form valued operator, which maps each function to its gradient. Then you can form the wedge product with this operator and a $p$-form to obtain a $(p+1)$-form, which is the exterior derivative of the original $p$-form. This is the way he defines exterior derivatives. The wedge product is introduced in the previous paragraphs. This seems like a strange choice of notations and definitions, but it may have been fairly standard at the time.

About Poincare's Lemma (since Qmecanic mentioned it): I have seen a number of places where $d^2=0$, for the exterior derivative, is called the Poincare's lemma, for example the reference, which Weinberg gives, Flanders, H. "Differential forms with Applications to the Physical Sciences".