# How to geometrically think of a covector field

Hello Physicsforums!
It is quite often that I find the geometrical visualization of a covector field as a field where the covector for each point bends along the manifold, M. That is to say it is "contained" in the surface - in contrast to the tangent field in which the vectors streches out from M ... as just tangents. Is there any good site for the justification of this visualization? :/

I am now reading 'Introduction to Smooth Manifolds' by John M. Lee where he writes that every nonzero functional can be determined by its kernel and the hyperplane for which w_p(X) = 1. Futhermore he writes that the kernel is of "codimension-1" - what does this mean?

Let us take the functional [1, 0, 0, 0] for the cotangent space for a point in R^4. If I understood everything this will have a kernel of dimension 3 in the tangentspace spanned by [0, 1, 0, 0]^T, [0, 0, 1, 0]^T and [0, 0, 0, 1]^T. It is not of dimension 1.. Have I missunderstood something fundamental here?

Thanks soooo much!

// Daniel

Yes, you're missing something fundamental. Luckily it's only a definition--your understanding of the mathematics is good. The codimension of a subspace is its dimension subtracted from the dimension of the ambient space. So in your case, codimension 1 is the same as dimension 3, as you computed.

Maybe you were thinking "co"-dimension of the kernel should just mean its dimension in the "co"-tangent space, but that's not the definition.

Aha! That makes much more sense. :)
Lee has however a totally new way of picturing the covariant field than the way I described.. Still I haven't found any mathematical arguments why the one that I described is to prefer or even correct.

It is quite often that I find the geometrical visualization of a covector field as a field where the covector for each point bends along the manifold, M. That is to say it is "contained" in the surface - in contrast to the tangent field in which the vectors streches out from M ... as just tangents. Is there any good site for the justification of this visualization? :/

It's not very clear what you mean.

It sounds like a foliation, which is problematic because foliations don't always exist, even locally. The big theorem that tells you whether or not there is a foliation is the Frobenius theorem.

One way I would visualize a covector field is just as a plane field of codimension 1, consisting of the kernel of the covector at each point.

I am now reading 'Introduction to Smooth Manifolds' by John M. Lee where he writes that every nonzero functional can be determined by its kernel and the hyperplane for which w_p(X) = 1.

You don't need the whole hyperplane, just one vector in it. It's determined by its kernel and its effect on some vector that is not in the kernel. Just basic linear algebra. Take a basis for the kernel, extend it to a basis of the whole tangent space. You know where that basis goes, so you know where everything goes. I guess he mentions the hyperplane because that helps to visualize how far apart the level surfaces are.

One last question that popped up when I was reviewing though..

Smith writes that a pushforward fulfils: :

$(F_* X)(f) = X (f \circ F).$

Where:
$F: M \rightarrow N$,
$X$ is in $T_pM$,
$(F_* X)$ is in $T_{F(p)}N$ and
$f: N \rightarrow$ R

However this means that the righthand side is a derivative that belongs to a tangent space for M acting on a function defined on N..... I cant make it add up. It feels like there is a typo in this equation.

It's not very clear what you mean.

It sounds like a foliation, which is problematic because foliations don't always exist, even locally. The big theorem that tells you whether or not there is a foliation is the Frobenius theorem.

I am refering to the following sentence (not found in Smiths book):

When acting on scalar functions, the basis vectors $e_\mu \rightarrow \partial_\mu$ are tangential to the vector space; the 1-forms $\omega^\mu \rightarrow dx^\mu$ lie along it.

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Smith writes that a pushforward fulfils: :

(F∗X)(f)=X(f∘F).

Where:
F:M→N,
X is in TpM,
(F∗X) is in TF(p)N and
f:N→ R

However this means that the righthand side is a derivative that belongs to a tangent space for M acting on a function defined on N..... I cant make it add up. It feels like there is a typo in this equation.

The composition is a function from M to R.

When acting on scalar functions, the basis vectors eμ→∂μ are tangential to the vector space; the 1-forms ωμ→dxμ lie along it.

I don't know what he's talking about.

The composition is a function from M to R.
yeah thnx, I honestly dont know what I was thinking..

I don't know what he's talking about.

Do you mean it´s sounds suspicous or that you want more context? :)

I don't know what he's talking about.

Do you mean it´s sounds suspicous or that you want more context? :)

Maybe both. What vector space is he talking about?

Hmm, the sentence is taken from a chapter in field theory - so maybe it should be assumed that we are talking about the more relevant spaces in physics, e.g. minkowskispace. But the metric matrix is not choosen and the equations are in their general forms.
There is not much to say.. really. After the above quotation he introduces affine connections, $\Gamma_{\mu \nu}^{\;\;\;\; \lambda}$. No example is given until the end of the chapter we're euclidean spaces and minkowski space is introduced/reviewed.

The physical applications aren't going to change anything. I was just wondering which vector space he is referring to. Just the tangent space?

I wouldn't take the statement seriously. It's not really important. Just think of a covector as a plane field.

Gah, this whole covector field chapter has made me confused..
Lee writes: $df_p(X_p) = X_pf$.
Now, since $X_p$ is a derivation I assume that I can write the RHS as: $X_p f = v^i \frac{\partial f}{\partial x^i}(p)$, this is obviously vector for which the components are not of infinitesimal order, since we are dividing differentials of order 1 with differentials of order 1.. just like the elementary calculus derivative we could maybe get, lets say: $\frac{\partial f}{\partial x^2}(p) = 3$.
The LHS however looks like differentials! If $df_p =\frac{\partial f}{\partial x^i}(p) dx^i|_p$ are the components of $df_p(X_p)$, cant we somehow just apply:

$[\frac{\partial f}{\partial x^1}(p) dx^1|_p, …, \frac{\partial f}{\partial x^n}(p) dx^n|_p] \cdot [v^1 \frac{\partial }{\partial x^1}(p), …, v^n \frac{\partial f}{\partial x^n}(p)]^T$ ?

Haha, I think you can see how confused I am here.. The field $df$ is in someway the gradientfield in covector form.. right? I cant get the both sides to add up. The LHS seems to be a vector of infinitesimal order, while the RHS is not.

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To summarize the previous post:
It seems like we are setting a vector with infinitesimal components equal to one that is not infinitesimal.