# Bertschinger:Vectors versus one-forms with a metric present: "We should not think of these as being distinct 'components' of the same entity at all."

In Edmund Bertschinger's lecture notes Introduction to Tensor Calculus for General Relativity, following equation (14) he tells us:

The component $$P_\mu$$ of the one-form $$\overset{\sim}{P}$$ is often called the covariant component to distinguish it from the contravariant component $$P^\mu$$ of the vector $$\vec{P}$$ . In fact, because we have consistently treated vectors and one-forms as distinct, we should not think of these as being distinct ”components” of the same entity at all.

I really don't understand some of the distinctions Dr. Bertschinger is making. The above statement, in particular seems contrary to how I have learned tensor analysis (from many sources, including books in his bibliography). Am I correct in understanding that there is no mathematical distinction between vectors and one-forms on a (pseudo-)Riemannian manifold endowed with a metric? That is, Bertschinger is suggesting certain geometric objects are natively vectors, while others are natively one-forms, but each can be easily converted to the other.

Edit to add protest: To paraphrase Evar Nering's Linear Algebra and Matrix Theory: A homomorphism or isomorphism defined uniquely by intrinsic properties, independent of the choice of basis, is said to be natural or canonical.

• Vector fields and one-forms are defined on a smooth manifold without the help of any metric. Choosing a metric defines a one-to-one correspondence between vectors and one-forms, but if we change the metric, we change the correspondence, so the correspondence is a property of the metric, not a property of vector fields and one-forms themselves. No metric, no unique one-to-one correspondence. Bertschinger is probably alluding to this. Sep 1, 2020 at 0:36
• At every point on a Riemannian manifold with a metric there is always a locally orthonormal basis available. So there will be a bijection between the tangent and cotangent spaces. A non-singular coordinate transformation will not change that correspondence. After reading a bit further, it appears he was really saying we can't simply move the index up or down to go from vector to one-form components. We need to use the metric to toggle an index. Sep 1, 2020 at 1:30
• Exactly. A Riemannian manifold comes with a specific metric, by definition. On a manifold without any metric at all (just a smooth manifold, not Riemannian), we can still define vector fields and one-forms. But to toggle the index up and down, we need a metric. We can even think of this as the definition of a metric -- it's not the standard definition, but it's equivalent to the standard definition. Changing the correspondence is equivalent to changing the metric. (I don't mean a coordinate transformation, which only changes how the metric is described. I mean changing the metric itself.) Sep 1, 2020 at 2:54
• ... In other words, we could define a Riemannian manifold to be a smooth manifold equipped with the right type of one-to-one correspondence between vector fields and one-forms, because if the correspondence is compatible with any metric at all (that's what I mean by "right type"), then the metric is uniquely specified by the correspondence. Sep 1, 2020 at 3:02
• @ChiralAnomaly So are you agreeing or disagreeing with me? Because that is exactly my argument. Sep 1, 2020 at 23:55

Am I correct in understanding that there is no mathematical distinction between vectors and one-forms on a (pseudo-)Riemannian manifold endowed with a metric?

No. Vectors and one-forms are distinct mathematical objects with distinct transformation behavior under changes of coordinate chart. There are also subtleties when defining pushforwards and pullbacks which distinguish the two; for a generic smooth map $$\Phi:M\rightarrow N$$, one can push vectors forward from $$M$$ to $$N$$ and pull one-forms back from $$N$$ to $$M$$, but the reverse operations are not always well-defined. The fact that the space of vectors and the space of one-forms are isomorphic does not mean that the elements of those spaces are the same.

For a generic smooth manifold, there is no canonical isomorphism that naturally maps the space of vectors to the space of one-forms. Such isomorphisms exist, but there's no particular reason why we need to choose one over the other - any nondegenerate bilinear form is sufficient to do the job.

A metric is such a form, so on metric manifolds we typically choose the metric itself to define the partnership between a vector and its one-form dual. Even in GR, though, this is not always the case; when working with linearized gravity, it is conventional to use the Minkowski background metric, rather than the full metric, to map vectors to their one-form partners. On symplectic manifolds, where there is generally no metric at all, one uses the native symplectic form.

Vectors and one-forms are intimately related, insofar as they can be put into one-to-one correspondence. However, the modern perspective treats vectors and covectors as distinct objects, and to view them that way is conceptually far cleaner than the "different components of the same object" idea that is a bit more old school.

• It appears that every counter example you provided involves something other than a Riemannian manifold endowed with a metric. Sep 1, 2020 at 8:07
• @StevenThomasHatton You misunderstand - I am not providing counterexample, but rather emphasizing the lack of canonical isomorphism to “convert” vectors to one forms. Sure, if you have a metric then you can use it, but that choice is arbitrary (as illustrated by the linearized gravity example I mentioned). If the relationship between a vector and its one-form partner is arbitrary and changes depending on context, that’s a pretty good indicator that we shouldn’t pretend that they are one and the same object. Sep 1, 2020 at 13:15
• Linearized gravity seems to be a "slipshod" treatment of a Riemannian manifold. In a sense, carrying the tangent plane beyond its definition in the context of a proper Riemannian manifold. It will take a careful review of Bertschinger's notes to verify, but it appears he is developing the "canonical" isomorphism between the contragredient tangent vector spaces at points of a smooth manifold endowed with a metric. And that is what my question is about. Sep 2, 2020 at 0:08
• @StevenThomasHatton There is nothing slipshod about it. For conventional reasons, one chooses a different object than the metric to map vectors to one-forms, that's all. You still seem to be missing my point, which is that (i) a metric on a Riemannian manifold is a suitable object to define an isomorphism between vectors and one-forms, but (ii) one-forms and vectors are mathematically distinct objects, and any such choice of isomorphism is ultimately an arbitrary choice of convention. Sep 2, 2020 at 0:13
• This is a longer discussion than the comment section will allow for. One problem is that there are at least as many definitions of "differential form" as there are textbooks discussing them. I can start with rudimentary set theory (avoiding Russel and Goedel) , and the Dedekind/Peano axioms, then, without reference to any physical object, hand you a Riemann manifold, which at every point has a local Pythagorean metric and a tangent space endowed with a local Cartesian coordinate system and contragredient vector spaces. Sep 2, 2020 at 1:22

There are (at least) two occasions when it is useful to distinguish contravariant and covariant tensors from each other.

• Most objects are either naturally covariant or contravariant, and if metric variations are considered, which is pretty often, it may happen that the covariant or contravariant forms of objects do not get varied or get varied differently.
• When null hypersurfaces or hypersurfaces with null points are considered, the induced metric on the hypersurface is degenerate. In this case, contravariant and covariant tensors will behave extremely differently on the hypersurface and therefore the distinction has to be made in the bulk spacetime as well, even if the metric is nondegenerate there. See this paper by Mars and Senovilla for an example.