# What is the difference between a 1-form and a co-vector?

My understanding has been that co-vectors are defined on a (pseudo-) Riemannian manifold as dual vectors with respect to the given metric. However, 1-forms do not need the notion of a metric and have nothing to do with vectors except that one feeds on the other.

However, while reading this answer on the local representation of a tautological form (which is defined on a general manifold irrespective of the presence of a metric structure), I cannot seem to comprehend the difference between a 1-form and a co-vector, which difference seems important to the answer.

Please let me know what I am missing. If possible, please also answer the linked question bearing in mind my lack of comprehension of this difference. Thank you.

• @PVAL-inactive The professor who taught me differential geometry made a clear distinction between a 1-form field (defined as a section of the cotangent bundle) and a 1-form (which I believe you call a co-vector). I have never quite come across the vector/co-vector terminology, which is why when I read "co-vector is the dual of the vector", I assumed it to be with respect to the isomorphism induced by the metric. I resent your saying that my education is inadequate. Perhaps you should read my answer to the linked question. – Nanashi No Gombe Jan 5 '17 at 13:02

A vector at $p$ in a manifold $M$ 'lives' in the tangent space to $p$ of $M$ denoted $T_pM$. A co-vector lives in the dual space to this tangent space $(T_pM)^*$. Usually either the tangent or cotangent space is constructed in a way so as to be a finite dimensional vector space and then the dual space is given in the usual way as linear maps to $\Bbb{R}$.

For smooth manifolds you can see the tangent vectors as directional derivatives of smooth functions real valued functions on the manifold, and usually the basis for the tangent space is written as $\left\{\frac{\partial}{\partial x^i}\big|_p\right\}_{i=1}^{\mathrm{dim}(M)}$. The $x^i$ are the co-ordinates of some chosen chart that $p$ lies in. Then to agree with older notation (and until the exterior derivative is introduced) the dual basis is notated $\left\{dx^i\big|_p\right\}_{i=1}^{\mathrm{dim}(M)}$.

A tangent vector is of the form $$\sum_iv^i \frac{\partial}{\partial x^i}\Big|_p,$$

with the $v^i$ real numbers, whilst a cotangent vector or co-vector is of the form

$$\sum_i\omega_i dx^i\Big|_p,$$

again with the $\omega_i$ real numbers.

A differential form (sometimes called a co-vector field) on the other hand, rather than being just at a point is defined on an open set for example, or on possibly the whole manifold.

$$\omega=\sum_i \omega_i(p)dx^i\Big|_p$$

Now the $\omega_i$ are (smooth) functions, noticing the point dependence, and for comparison, a vector field in local co-ordinates would look like

$$V=\sum_i v^i(p)\frac{\partial}{\partial x^i}\Big|_p.$$

As PVAL's comments mentioned, you can think of differential forms and vector fields as sections of the Co-Tangent and Tangent Bundles, and I prefer this because it makes it clear the point dependence, even without local co-ordinates.

I also recommend Tu's book 'An Introduction to Manifolds' as I think it's very readable.

Evidently (see comment from Nanashi No Gombe under question) this was only a matter of mismatched naming conventions, but here is a TL;DR answer for future readers, which (in my experience) matches the way most everyone uses these terms:

A co-vector is to a $$1$$-form as a vector is to a vector field.

That is, if you have a $$1$$-form $$\alpha$$ on some manifold $$M$$ and you pick any $$x \in M$$, the value of $$\alpha$$ at $$x$$ is a co-vector.

If you think of a co-vector $$\alpha|_x$$ as a map $$T_x M \rightarrow \mathbb{R}$$, then you can extend this to thinking of a $$1$$-form as a map $$\alpha: \Gamma (TM) \rightarrow \mathcal{C}^{\infty}(X)$$.